text:
Finding one of the two unknown potentials with a given voltage of a net reaction
A half-cell consisting of palladium rod dipping into a $1\ M$ $\ce{Pd(NO_3)_2}$ solution is connected with a standard hydrogen half-cell. The cell voltage is $0.99\ volt$ and the platinum electrode in the hydrogen half-cell is the anode. Determine $E^ο$ for the reaction $\ce{Pd\longrightarrow Pd^{2+} + 2e}$.
Now the question is: can I find a numerical value of the needed $E^ο$, if I can't use half-cell potential of $\ce{Pt^{2+}\!−\!Pt}$?

You have the overall cell voltage, so you should be able to leave the E0 as unknown and solve for it.

Careful, in the standard hydrogen electrode the redox reaction is not $Pt^{2+} + 2e^- -> Pt$.


text:
How does partial reduction of esters to aldehydes occur?
Wikipedia says that esters can be reduced to aldehydes using DIBAl-H. Can anyone please explain how this reaction occurs?

DIBAL exists as a bridged dimer, and it becomes a reducing agent only after it has formed a Lewis acid–base complex, so it reduces electron-rich carbonyl groups most rapidly.

DIBAL will reduce esters even at –70 °C, and at this temperature the tetrahedral intermediate that forms during the reaction may be stable. Only in the aqueous acidic work-up does it collapse to the aldehyde (by expelling EtOH in the example reaction below) when excess DIBAL has been destroyed so that no further reduction to the alcohol is possible.

A stable tetrahedral intermediate is more likely in the reduction of lactones, and DIBAL is most reliable in the reduction of lactones to lactols.
Edit:
As requested by the OP here are some more details about the collapse of the tetrahedral intermediate: During the aqueous acidic work-up the alcohol group is turned into a good leaving group via protonation. Thus, it is easily expelled from the molecule, when a water molecule removes the $\ce{Al^{$i$}Bu_2}$ group and the $\ce{C=O}$ double bond is reformed.

A different route, where the former carbonyl-oxygen is protonated twice - thus being turned into a good leaving group - and afterwards expelled from the molecule to give an oxonium ion is possible too. But such an oxonium ion is rather unstable and highly reactive, so that the reverse reaction is very fast and the oxonium ion is not produced in any appreciable amount.

Why is that tetrahedral intermediate so stable?  In other words, why is (O-AliBu2)- a bad leaving group?  If LiAlH4 was used as the reducing agent, then (O-AlH4)2- would come off.  So what is the difference here?

@JoshuaMeyers I would guess that steric hindrance is the reason. The $\ce{{}^{$i$}Bu}$ groups make it much harder for, say, another ester molecule to attack the $\ce{Al}$ in the tetrahedral intermediate and kick out the original (now reduced) ester leading to its collapse. This is not the case for the small hydride groups.

Why is the aldehyde not further reduced to alcohol? On many sites, it's told that DIBAL can reduce aldehydes and ketones. Please reply.

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sort: votes


text:
Vacuum Distilled Medicine
If I were to buy/build a large still, that was air tight, and could hold a vacuum, and filled the reservoir with vegetable glycerin (which has a boiling point higher than the majority or desired organic compounds) and I made the atmospheric pressure inside less than normal atmospheric pressure, could I distill chemicals out of organic matter without altering them? Such as distilling willow bark to get an aspirin containing solution, or distill mint leaves without changing the chemical composition of the medicinal compounds in the plant/organic matter?
This could be a cool project. It would enable us to explore further into natural medicine. But would this procedure work?

I'm not sure I agree with your statement that "this would enable us to explore further into natural medicine" since distillation, even under vacuum, has been around for over a hundred years.

I don't know the exact proportions, but I think you'd need a truckload of bark to extract the aspirin.  There's a reason that people began synthesizing these substances artificially! :)

In general the answer would be no.
While there are many distillable compounds you could extract this way (not sure how many medicinal ones, though) there are many more that would decompose or simply not be easily extracted.
This is why there are many other ways chemists commonly extract things from mixture or from plants. For example, if you wanted to extract caffeine from coffee or from coffee beans, you could use solvent-based extraction (methylene chloride works well, for example, and has been used in industrial processes to create decaffeinated coffee). There is no single easy solution, though, and often much trial and error will be required to find the right solvent to differentially extract what you want.

Look up the industrial process.  Organic essences are usually steam distilled or pot-to-pot durped, or liquid CO2 or liquidn propane/butane extracted.  We're not making hash oil, are we?
http://www.fao.org/inpho_archive/content/documents/vlibrary/ad420e/img/AD420e26.gif
http://www.seanmichaelragan.com/img/radial_alembic_diagram.png
There is no aspirin in willow bark, there is salicylic acid.  Since you can buy everything at high purity for less than small scale process costs, unless you own a planted acre, why bother?
Two differences between "natural" and "synthetic." 1) If the molecule is chiral, the synthetic product will be racemic. R-(–)-carvone is spearmint, its mirror image S-(+)-carvone is caraway.  Big difference. 2) "Natural" contains radioactive carbon-14: 540 Bq/kg carbon.  Anything made from petroleum or coal is long since decayed cold.
Ignorance is not a form of knowing things, nor is faith.  The only empiricially valid solutions are engineering solutions.
http://www.erowid.org/archive/rhodium/chemistry/equipment/pictures/nomograph.gif

Not really. There are too many different types of compounds for only one procedure to work. If it was that easy there wouldn't be so many different ways of preparing extracts.
Try to take one plant/compound to work with at a time.

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Why do molecules have to first transform to gas before doing a reaction?
I have to calculate the enthalpy of the reaction
$$\ce{Cs(l) + 1/2I2(s) -> CsI(s)}$$
and the answer is
$$\Delta H_f = \Delta H_{vap}^{\ce{Cs}} + \frac{1}{2}\Delta H_{subl}^{\ce{I2}} + \frac{1}{2}E_{diss}^{\ce{I2}} + I_1^{\ce{Cs}} + E_A^{\ce{I}} + E_N.$$
I can see that the molecules are first put to gas state before creating an ionic bond.
Why does it have to do it ?
In which case do the molecules have to first go in gas state before reacting ?
Is it only in case of ionic bond ?
Notes:

$E_{diss}$ is the energy of dissociation between the to atoms $\ce{I}$.
$I_1$ is the energy of first ionisation.
$E_A$ is the energy of electron affinity.
$E_N$ is the energy of neutralisation between $\ce{Cs+}$ ans $\ce{I-}$.

Not all reactions require vaporization. This process looks like the Born-Haber cycle for cesium iodide, an exercise in Hess's Law usually used to fill in one of the missing values in the cycle that might be hard to measure (usually lattice energy, but not always). The Born-Haber cycle can also be used to drive general chemistry students crazy.
The $E_N$ "energy of neutralization" looks like the lattice energy, which is defined as the enthalpy/energy change accompanying the formation of a solid crystal lattice from the gas phase ions. In your case, the lattice energy of $\ce{CsI}$ is defined as the energy change accompanying the following reaction:
$$\ce{Cs+ (g) + I^- (g) -> CsI(s)}$$
Because the lattice energy is defined for gas phase ions, you need the vaporization enthalpies for this process. Not all reactions require this vaporization step.

Ok so, it is necessary to vaporize when I want to build a ionic bond right ?
Isn't it weird that there is not the deposition enthalpy of the CsI ?


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Shift of the null half-reaction potential
If we change the null half-potential, would it change half-potentials of other half-cell reactions by the same amount, so that voltages of net reactions stay the same?

Yes, you should be right: changing the standard electrode changes the half-cell potentials but the galvanic cell potentials (resulting from a pair of electrodes) stay the same.


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Why does the reaction of chlorosulfonic acid (colorless) and 1-dodecanol (colorless) produces a deep dark-red-colored solution?
I was wondering if anyone had any insight as to why the reaction of chlorosulfonic acid (colorless) and 1-dodecanol (colorless) produces a deep dark-red-colored solution? The reaction should give hydrogen lauryl sulfate, which according to this website, is a white solid.
I have also tried adding $\ce{NaCl}$ to the chlorosulfonic first, and then adding the 1-dodecanol, which surprisingly gives the sodium dodecyl sulfate soap as expected.
Is this color from over sulfonating the hydrocarbon chain on the hydrogen lauryl sulfate?  If so, what would this side product be called, and why isn't it seen much in industry if it is created along with sodium dodecyl sulfate?

Google told me that chlorosulfonic acid reacts with tellurium to form a red solution.  But that's probably not your problem unless you have very unusual contaminants in your system somewhere.   My guess is that the acid (or sulfur trioxide contaminants in it) are oxidizing the dodecanol alcohol to an aldehyde or dehydrating it to an olefin, and that aldehydes/olefins are polymerizing to give you red gunk.  When salt is added, is that anhydrous salt or a solution?  Water in the salt could help prevent dehydration.

You need to provide a lot more detail otherwise any answer you get will be a guess. What is the experimental procedure in detail for both experiments (quantities, temperature, time, mode of additions etc), literature references, what stage did you get in the 1st experiment, did you do any work-up/purification, did you isolate any product(s), do you have any data for this product or products (spectroscopic etc) and so on. For example I have done chlorosulfonic acid reactions that gave a dark red color during reaction but after work-up yielded a pure white product.


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How to determine an oxidation number of an atom in the compound, that has at least two elements, which are not hydrogen or oxygen?
Wiki says:

[Oxidation state] is defined as the charge an atom might be imagined to have when electrons are counted according to an agreed-upon set of rules:
The oxidation state of a free element (uncombined element) is zero
    for a simple (monoatomic) ion, the oxidation state is equal to the net charge on the ion.
Hydrogen has an oxidation state of 1 and oxygen has an oxidation state of −2 when they are present in most compounds. (Exceptions to this are that hydrogen has an oxidation state of −1 in hydrides of active metals, e.g. LiH, and oxygen has an oxidation state of −1 in peroxides, e.g. H2O2
          the algebraic sum of oxidation states of all atoms in a neutral molecule must be zero, while in ions the algebraic sum of the oxidation states of the constituent atoms must be equal to the charge on the ion.

The same is written in my textbook. But how am I supposed to find the ox. number of an atom, which is in compound like $\ce{K2UO4}$?

Do not abbreviate Wikipedia.org as Wiki. 2) Do not use backticks unless there is some code somewhere.

In this case, the potassium ions are +1 each, and the oxygens are -2 as normal.  That leaves uranium in a highly oxidized +6 state.  The piece of information I think is missing is that Group 1 ions are almost always in the +1 state, because they so readily lose their outer valence electron.  A molecule that is not an ion, like your example, must have an overall charge of zero.

Group 1 ions are alkali metals? What about, say, $Mn_2U_2O_7$? How do you learn what oxidation number would some atom in such or more complex compound would have? How do you know which atom's ox. number is to be learned and which - to be determined by the learned one(s) and, optionally, oxygen and/or hydrogen?

Good question! With rare exceptions, Group 1 & 2 elements always form ionic bonds and so have the +1 or +2 state, respectively.  Same for halogens (Group 17) nearly always in the -1 state. Manganese and uranium can have many oxidation states, but transition metals like that tend to be unique in most inorganic molecules; i.e. I'm not certain your example exists! In most molecules, there is only one element that can have multiple oxidation states, and all the other elements will have consistent oxidation states like oxygen. There must be rare exceptions, but I can't think of one.

Yeah, I confused it for $Mg$. Does this apply to group 16?

Not as much as you might think, because as you go down the periodic table, electronegativity decreases and you can have more oxidation states. This happen already at sulfur, which can have lots of oxidation states. If you look at a good period table, it will show the possible oxidation states. So does Wikipedia, which also shows the most common oxidation states. You start to learn what's possible and what's not.

Technically (which may not be the answer you want) you compute the oxidation numbers in a molecule by first drawing its Lewis (dot) diagram showing all of the valence electrons.  Then you assign the electron pair in each bond to the more electronegative atom.  Then refigure the "charges" on each atom.  The resulting "charge" will be the oxidation number.
Over the years folks have worked out shortcuts to this.  For example, alkali metals easily lose an electron in almost all situations.  Thus we have for example that $\ce{Na}$ has an oxidation number of +1.

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Is there a kind of table about configurations similar to the example I gave below?
If MALDI is used as an ionization method, then TOF as an analyzer is suitable since the MALDI creates singly charged ions which have high m/z ratio and only TOF can analyze that ions, and multichannel plate should be used because the TOF is very quick and MCP's duty time is enough to capture all ions.
Is there a kind of table about configurations similar to the example I gave above? A table with reasons would be great but not certainly necessary.

I always like to see mass spectrometry questions but am confused about what you are asking for.  MALDI ionization can be used with a variety of mass spectrometry detectors, including TOFs, linear ion traps, Orbitraps, and probably even quadrupoles.

Are you asking for a table with ionisation techniques on one axis and analysation techniques on the other? AFAIK, all ionisation techniques are compatible with all analysation techniques, but some have no realistic use cases.

The question is quite old. I was studying MS and was looking at my lecture notes then I observed that my lecturer had something like a table in his mind. From what I've learned there are non-offical "X ionisation technique + Y analyzer + Z detector is the best combination" kind of statements. Seeing all the combination was supposed to light a bulb on designing a theoretical setup and reasoning each component.


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Analytical solution to first-order rate laws
Does anyone know the analytical solutions to the following rate laws:
1) $\frac{\mathrm{d}\rho_\text{W}}{\mathrm{d}t} = -\rho_\text{W} \cdot (K_1+K_2+K_3)$
2) $\frac{\mathrm{d}\rho_\text{T}}{\mathrm{d}t} = \rho_\text{W} \cdot K_2 - (K_4+K_5) \cdot \rho_\text{T}$
3) $\frac{\mathrm{d}\rho_\text{C}}{\mathrm{d}t} = \rho_\text{W} \cdot K_3 + \rho_\text{T} \cdot K_5$
As an example, I have determined the following analytical solution for this rate law:
$\frac{\mathrm{d}\rho_\text{W}}{\mathrm{d}t} = -\rho_\text{W} \cdot (K_1+K_2+K_3) \qquad \Rightarrow \qquad \rho_\text{W} = \rho_{\text{W}0} \cdot \operatorname{e}^{-(K_1+K_2+K_3) \cdot t}$
where $\rho_{\text{W}0} = $ initial density, $K = $ reaction rate constant, $t =$ time
The scheme for the reactions are:

$\ce{wood} \xrightarrow{K_{1}}{} \ce{gas}$
$\ce{wood} \xrightarrow{K_{2}}{} \ce{tar}$
$\ce{wood} \xrightarrow{K_{3}}{} \ce{char}$
$\ce{tar} \xrightarrow{K_{4}}{} \ce{gas}$
$\ce{tar} \xrightarrow{K_{5}}{} \ce{char}$

Any help with the three rate laws above would be greatly appreciated.

This is the solution I get:
$$\begin{align}\rho _T(t) &= -\frac{K_2 \rho _{w0} \left(e^{\left(-K_1-K_2-K_3\right) t}-e^{t (-K4-K5)}\right)}{K_1+K_2+K_3-K4-K5}\\\rho _W(t) &= e^{\left(-K_1-K_2-K_3\right) t} \rho _{w0}\\\rho_C(t) &=\tiny{-\frac{\rho _{w0} e^{-t \left(K_1+K_2+K_3+K4+K5\right)} \left(K_2^2 K5 e^{\left(K_1+K_2+K_3\right) t} \left(e^{t (K4+K5)}-1\right)+K_2 \left(K5 (K4+K5) \left(e^{t (K4+K5)}-e^{t \left(K_1+K_2+K_3+K4+K5\right)}\right)+K_1 K5 e^{\left(K_1+K_2+K_3\right) t} \left(e^{t (K4+K5)}-1\right)-K_3 \left(-(K4+2 K5) e^{t \left(K_1+K_2+K_3+K4+K5\right)}+K5 e^{\left(K_1+K_2+K_3\right) t}+(K4+K5) e^{t (K4+K5)}\right)\right)+K_3 (K4+K5) \left(-K_1-K_3+K4+K5\right) \left(e^{t (K4+K5)}-e^{t \left(K_1+K_2+K_3+K4+K5\right)}\right)\right)}{\left(K_1+K_2+K_3\right) (K4+K5) \left(-K_1-K_2-K_3+K4+K5\right)}}\end{align}$$
Solving this by hand isn't that hard, as the coupling is unidrectional. So, we can solve for $\rho_W$ first, and substitute it in the equation for $\frac{d\rho_T}{dt}$.
Now, we have an equation of the form $\frac{dy}{dt}=Ay+Be^{kt}$, which has a solution of the form $y=Ce^{At}+De^{kt}$, where C,D can be found by back-substitution. Now, we subtitute the values for $\rho_T, \rho_W$ in the last equation, and we get an equation of the form $\frac{d\rho_C}{dt}=f(t)$ ($f(t)$ here is some combination of exponentials), which can easily be solved by integrating $f(t)$

A post for ants?

For what it's worth, problems like this can be solved very nicely with Mathematica and I find that it provides access to some really cool chemistry even when the math is a bit overwhelming for some students.  One can obtain the same solution as @ManishEarth with the following Mathematica code:
soln = DSolve[
  {
   wood'[t] == -wood[t] (k1 + k2 + k3),
   tar'[t] == wood[t]*k2 - (k4 + k5)*tar[t],
   char'[t] == wood[t]*k3 + tar[t]*k5,
   wood[0] == wood0,
   tar[0] == 0,
   char[0] == 0},
  {wood[t], tar[t], char[t]}, t]

The fun part is how easy it is to visualize the results:

And there are ways to dynamically manipulate the plot such that one can explore the effect of the rate constants on the time-dependent concentrations.

The mathematical rigor of my solution is naturally a bit weak, but oftentimes my emphasis is on teaching the science.

Wow, thanks for pointing this out in Mathematica. I use Matlab for everything at work. Do you know if you can dynamically adjust parameters in Matlab like you show in Mathematica?

I haven't used Matlab for a number of years now.  There is this code which provides some type of manipulative features, but I don't think Matlab is as advanced in this type of visualization as Mathematica is (at the moment, at least).


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Finding electron configuration of lanthanide ions
When I have a gadolinium ion ($\ce{Gd^3+}$), how can I calculate its electron configuration?
$$\ce{Gd} : \mathrm{[Xe] (4f)^7 (5d)^1 (6s)^2}$$
Do I need to first subtract the 2 electrons in the $\mathrm{s}$-orbital and then the $\mathrm{d}$-/$\mathrm{f}$-orbital?

When determining electron configurations of cations, one takes the electron configuration of the neutral atom, and then starts removing electrons from the orbital with highest n value.  If there is a tie, you then choose the orbital with highest l value.

I agree with bobthechemist's comment.
When you are determining the electronic configuration of anions/cations, you will first need to determine electronic configuration of neutral atom. Like in the case above:
$$\ce{Gd} : \mathrm{[Xe] (6s)^2 (4f)^7 (5d)^1}$$
Now arrange this configuration according to ascending value of $n$:
$$\mathrm{[Xe] (4f)^7 (5d)^1 (6s)^2}$$
Now, according to your need start subtracting electrons from highest value of $n$, like for $\ce{Gd^3+}$, the electronic configuration will be:
$$\ce{Gd^3+} : \mathrm{[Xe] (4f)^7}$$
See this to know more on calculating electronic configuration of cations and anions.


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Special electron configurations in p block on periodic table?
Gold and Chromium have weird electron configurations, because electrons are differently divided : eg an electron jumps from a s-orbital to a d-orbital to have a half-filled s orbital and a fully/half filled d-orbital, which is much more stable.
But does this also occur with the p-orbitals? eg why does Fluorine not transfer an electron from its s orbital to its p orbital to have a  half-filled s orbital and a fully filled p-orbital, that would be more stable?

The simplest answer to this problem is related to relative energy differences between the Nd and (N+1)s orbitals in transition metals and Ns and Np orbitals. The Nd orbitals in transition metals are at a slightly lower energy than the (N+1)s orbitals, so for certain atoms there are some unusual changes in the electron configurations, e.g. Cr and Cu. However, the energy differences between the Ns and Np orbitals are quite large, with Np being higher in energy. In these cases, there is a substantial energetic penalty in moving an electron from the Ns to Np orbital.
The electron configuration problem for transition metals is complex, but this J Chem Ed article has a nice description of the theory, although it is somewhat technical.


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Regarding logic behind Organic Chemistry
Was there any logic or thinking behind the experiments performed by Cannizzaro, Beckmann, Gattermann, Koch etc.?
Did they thoughtfully perform experiments or just add any two reactants and luckily discovered the reactions which are known by their name today?
I meant to ask that in Physics all problems are treated mathematically and logically; But in case of Organic Chemistry, there are lot of exceptions to all statements which are made by Cannizzaro etc. It is said that if a single observation is against a scientific law; the law is abandoned and new perfect law takes its place. But it seems that Organic Chemistry is unaware of this and "perfect" statements are not part of it.

Sometimes it was a lucky accident, sometimes there was a lot of math and logic going in the designing experiments.  The organic chemistry of that era, though, was a lot more applied and empirical than the physics of today.  They usually weren't setting out to find the absolute laws-of-nature governing chemistry, they were just trying to make better dyes, medicines, rubber, etc, and stumbled upon new phenomena along the way.
Further Explanation
Almost all these guys worked in Germany in the late 1800s, when chemical industry was the hot new thing (think of biotechnology in today's world).  There was a lot of money to be made in making better dyes, medicines, rubber, etc, so people would hack/engineer the established methods for making that stuff, and see if the resulting products were any better.
For example:
Chemical A cures headaches really well, but causes nausea.  Chemical B gives a little relief from headaches, but doesn't have any side effects.  I bet I could make chemical C (which no one had ever made before), which has a structure that's sort of a combination of chemicals A and B, and maybe that will cure headaches with no side effects.
==> {made the chemical and did some tests to find out the structure}.
What the heck!  This has a completely different structure than what I was expecting, it must have undergone a second reaction to rearrange itself.
==> {publish paper},{get famous},{people 100 years later question the logic of your experiments} :)

The problem with organic chemical reactions is that they are hard.  Organic molecules have many covalent and polar covalent bonds, which require much energy to break.  Specifying which bond you want to break may require very specific chemical conditions.  It usually requires various catalysts to speed up the reactions.  So to determine exactly what organic reaction you want can be rendered "logical", but it usually requires a lot of experimentation, which may well be sped up by "luck".  But once the experimental results have been replicated, you have a "logical" procedure for the future.  It may have not involved mathematics, but it was proven by experiment, which is the ultimate criterion in science, even in physics!


text:
Solvation of uranyl formate/acetate
Uranyl formate has a structure something like:
$\ce{UO2-(OCHO)2}$
So the structural formulae I've seen have two double-bonded oxygens directly attached to the U and each formate group is attached by single bonds to the oxygens. Like this.
The empirical formula for the dry powder is written as a monohydrate.
Can someone explain how this thing gets solvated in water and what the solvated ions are, exactly? I just realized I was assuming it dissociates into a 'uranyl' ion (whatever that looks like) and two formate ions$\ldots$but then I'm confused about where the formate's oxygen goes.

This is a uranyl ion, one of the most common forms of dissolved uranium:

Because the two U=O bonds can form a single delocalised region, there's a strong tendency towards linearity (c.f. carbon dioxide). It tends to dissolve in water by gaining equatorial water ligands (4 or 5), some of which it may then hydrolyse, depending on the pH:

The axial oxygen atoms are also valid targets for hydrogen bonding, typically gaining one hydrogen bond each.

The difficulty in dissolving uranyl formate in particular suggests that the formate ion is quite well-bound, and it seems that in this case you get a combination of water and formate ligands in the complex around the uranyl. I was expecting the formate to be a bi-dentate ligand, but according to the results of a combined computational and experimental study 1, it's uni-dentate (which incidentally allows it to form polymeric chains in solution).
Here's a picture from that study showing a computationally obtained structure of a mixed water/formate-ligated uranyl structure.

2013, Dalton Trans., Christian Lucks, André Rossberg, Satoru Tsushima, Harald Foerstendorf, Karim Fahmya, Gert Bernharda, DOI:10.1039/C3DT51711J


text:
Which are the differences between resonance Raman and fluorescence?
If the fluorescence is the re-emitting of a photon with a larger wavelenght due to the transition from a higher energy state to a lower energy state the case of resonance Raman (where there aren't any virtual states involved) seems be equal to the fluorescence. Which differences are there?

okay, old question, but:
recall that the virtual state can always be decomposed as a sum over all real states, and the contribution of every real state to the superposition is determined by the matrix element and the difference in energy from the real state to the photon energy; usual perturbation-theory sort of equation where you divide by energy; by the way this is how you get to the usual polarizability equations for non-res raman from the virtual state sort of equation;
in resonance raman, if you tune sufficiently close to a discrete absorption band, then that one state will dominate the sum, and you no longer consider things to be from a virtual state
if you've done that, resonance raman is equivalent to resonance fluorescence
recall resonant fluorescence is basically when you've made your gas dilute enough that collisional cooling is slow compared to the radiative lifetime, so you emit from whatever level you excited to, instead of cooling to v=0 upstairs before emitting
so there are two key things needed to make resonance raman the same as fluorescence: you have to tune so close to a resonance that a only a single excited state is contributing, and you have to be comparing to the fluorescence you get in the limit of no excited state relaxation occurring before the emission

The main difference between Raman scattering and fluorescence is the excited state lifetime. Fluorescence excited states are longer-lived than the 'virtual' states associated with Raman scattering.
In fluorescence, absorption of light excites an electron to a higher energy state. The lifetimes of these excited states are long enough where the geometry of the molecule can relax to accommodate this new electron configuration. However, this new geometry is not equivalent to the lowest energy structure, so when the molecule returns to the ground state, the energy of the system will be slightly higher than where it started. This means that the fluorescence energy is always lower than the excitation energy (this is true for molecules, not atoms).
For Raman scattering, light absorption promotes the molecule to a very short-lived 'virtual' excited state. When this state relaxes, it can return to a different vibrational or rotational sublevel of the ground state than where the molecule started. This leads to slightly different energies than the elastically scattered light, which is at the same energy as the incoming radiation source.
Resonance Raman is a special case where the excitation energy is tuned to a particular electronic transition of a molecule. This has the advantage that vibrational transitions associated with the electronic change are substantially enhanced compared to other Raman transitions, which simplifies the Raman spectrum. Because you are exciting a specific electronic transition, though, it is possible to get a substantial fluorescence background as well.
This J Chem Ed article has a nice description of Raman versus fluorescence. ChemWiki has a page devoted to the differences between Raman and Resonace Raman.

Thanks Ryan, but my question is specific to differences between RESONACE Raman and fluorescence and I don't see any answer to it!In resonace Raman seems that virtual energy states are not involeved. Thanks however.

@GM, yes both resonance Raman and fluorescence use excitation light that is tuned to a particular electronic transition, but the relaxation modes are different. (As that J Chem Ed article in my initial reply will explain.) Fluroescence can cause a significant background problem in RR, and vice versa. However, not all electronic transitions lead to fluorescence. It really depends on the dominant relaxation modes for a particular system.

I have no acess to the article. So the differences between fluorescence and Raman Resonace are the different relaxation modes? In particular what difference are there between theese relaxation modes? Thanks a lot.

The Raman effect (resonance or otherwise) does not involve absorption of a photon, but rather scattering  of the photon by the molecule's electrons resulting in loss or gain of vibrational & rotational energy.  This results  in a shift in the wavelength of the scattered light by these quanta, usually a few hundred wavenumbers. There is no 'virtual' state, ( just a  excited state nearby in energy) this is just a convenience to explain the effect. If absorption can also occurs, as in Resonance Raman,  this is a separate process to the Raman effect itself.


text:
Difference between 18 electron rule and effective atomic number rule
Is there any difference between 18-electron rule and effective atomic number rule? Both rules are based on attainment of a noble gas configuraion as a criteria for stability of complexes. Besides, both rules were proposed by Sidgwick. I was wondering if the mere fact that 18-electron rule applies mainly to organometallic complexes is a good enough difference.

Both rules seem to be equivalent. I found a lot of references mentioning this by a simple google search for the effective atomic number rule:
http://www.britannica.com/EBchecked/topic/179961/effective-atomic-number-EAN#ref6518
http://books.google.de/books?id=Fs4zQ-hNTz8C&pg=PA726&lpg=PA726&dq=18-electron+rule+Sidgwick&source=bl&ots=wMQ4_5FCh5&sig=hLQI_m8yNmxs3yhH7XP6ODFTBjE&hl=de&sa=X&ei=FnYPUtSqMc_xtQbh84CQBw&ved=0CC4Q6AEwADgK#v=onepage&q=18-electron%20rule%20Sidgwick&f=false
http://chemww2b.rutgers.edu/sites/chemww2b.rutgers.edu/files/coursefiles/Courses_f09/571/MainGroup-Lecture8.pdf

Both of these rules are equivalent and are related to transition metals attaining a noble gas configuration of electrons in coordination complexes. The difference between them is the number of electrons given to the central metal.
The effective atomic number uses all of the metals electrons. For example, in an $Fe^{2+}$ complex, the central iron atom has 24 electrons, and you would want the ligands coordinated to the metal to contribute another 12 electrons to reach 36 electrons, the next noble gas configuration.
The 18-electron rule is a simplification of the EAN. There is always a noble gas configuration of electrons for the metals core orbitals (e.g. 1s2s2p3s3p for $Fe^{2+}$), so you can ignore these electrons in the counting scheme. For transition metals, this means that the only important orbitals are the valence nd(n+1)s(n+1)p orbitals, which can hold a total of 18 electrons. For $Fe^{2+}$, there are 6 electrons in these orbitals. Again, the ligands would need to donate 12 electrons to reach the 18 electron limit.


text:
Flory Huggins Theory
For an ideal chain, Flory Huggins theory gives us a distribution which is like a Gaussian.
$$P(l, r) \sim \left(\frac{3}{2 \pi l^2}\right)^{3/2} e^{\frac{-3 r^2}{2 l^2}}$$
where, $l$ is the monomer to monomer distance(also known as Kuhn length) and $R$ is the size of the monomer
This is great but what are the approximation away from an ideal chain polymer? I want to understand how one can go beyond Flory Huggins Theory.

The assumptions made in the Flory-Huggins theory are

Quasi-solid lattice in the liquid
Inter-changeability of segments (not necessarily the same as the polymer structure
units) of polymer and solvent molecules in the lattice
Independence of lattice constants on composition (artificial)
Polymer molecules are of same size
Average concentration of polymer segments in cells adjacent to cells unoccupied
by the polymeric solute is taken to be equal to the overall average concentration
The expected number of available positions for each successive segment is
overestimated in the formulation as the formula includes double counting of
segments, separated by 2 or more segments in the same chain, that will fall on the
same position twice

In particular the assumption that polymer molecules are of the same size is a very unrealistic one, which was later removed in the Flory-Krigbaum theory


text:
Why do salts such as NaCl dissolve?
If we look at solubility of salts in water, we are told that they disassociate, because the positive $\ce{Na}$ ion is attracted to the partially negative oxygen in water and the negative  $\ce{Cl}$ is attracted to the partially positive hydrogen. But why does this happen? I mean, the couloumb force should be much stronger between the $+1$ and $-1$ charges on the $\ce{Na}$ and $\ce{Cl}$ than the partial charges on the polar water molecule. Why does $\ce{NaCl}$, then, disassociate? It would make much more sense for the two ions to stick together.
Also, when you cut $\ce{NaCl}$, shouldn't it stick to its pieces? Look at salt crystals: they don't have any force of attraction between them. But, if you have positive sodium ions and negative chloride ions, then they should stick together.

As it happens, the enthalpy of solution of $\ce{NaCl}$ in water (that is, the energy change associated with the dissolution of sodium chloride crystals in water) at standard conditions is very slightly positive, i.e., it is an endothermic process. At a constant temperature and pressure, these kinds of thermodynamic processes are dictated by the change in Gibbs free energy, described by the equation
$$
\Delta G = \Delta H - T\Delta S
$$
Where $\Delta G < 0$ is a necessary criterion for a spontaneous process. Given that $\Delta H$ is positive, $\Delta S$ must be positive as well, otherwise the process wouldn't occur spontaneously (which is to say, not without input of work from the surroundings, or coupling to some other strongly favorable reaction, neither of which is the case for the dissolution of salt). In other words, this is a process that is driven by the increase in entropy, which is entirely to be expected when moving from a highly ordered state (i.e., a crystalline solid) to a less ordered liquid solution. Indeed, in ideal solutions, where intermolecular forces of attraction are taken to be equal between all components, enthalpy change is necessarily zero, and entropy is always positive, so that the process of mixing in ideal solutions is always spontaneous. Of course, in real solutions, this is not the case.
Your intuition that the Coulombic forces between ions should be stronger is correct in this particular instance, as indicated by the positive enthalpy (meaning, the breaking of ionic bonds in the crystal lattice, as well as intermolecular bonds between solvent molecules, requires more energy than is released in the forming of ion-dipole bonds). The input of energy required for this process comes in the form of heat, drawn from the solvent. However, there are numerous examples of salts for which the enthalpy of solution is negative under a wide range of conditions.
The statement "[l]ook at salt crystals: they don't have any force of attraction between them" is inaccurate, though. In fact, there is strong ionic bonding. The very fact that the crystals remain solid except at very high temperatures is sufficient evidence of this, and the strength of the bonding in the crystal lattice can be quantified by calculating the lattice energy. It's a mistake to consider any substance in isolation. When you talk about "cutting" $\ce{NaCl}$, what you're talking about is applying physical force to the crystals, exposing them to the complex mixtures of gasses and mositure in the air, and bringing them in contact with, say, the metal in the knife. If left in relative isolation, the salt remains undisturbed, packed in a crystal lattice. A gas, however, which has weak forces of attraction between individual particles, will expand to fill a container without any work being done to it (at least as far as atmospheric pressure allows). A liquid will experience capillary forces, either climbing the walls of a narrow container, resulting in a concavity at its surface, or pooling away from the walls of the container, resulting in a convexity at its surface (which of the two happens depends mainly on the comparative strengths of the forces of attraction between the molecules of the liquid and those of the substance comprising the container).

So, if you touch two pieces of sodium in a vacuum (so that oxidation does not occur), will they immediately become one seamless piece?

@Gerard http://physics.stackexchange.com/questions/87107/why-dont-metals-bond-when-touched-together is a similar question, not quite the same. But I would presume they would become one seamless piece.

I read this over and over again but I still can't understand how the weaker intermolecular forces pull apart the stronger ionic bonds. Is it not possible to answer this question without mentioning entropy?

@SwedishArchitect I'd think cold welding should work with the two pieces of sodium of Gerard's comment, just as with any other metal.

When a crystal is placed into a solvent, the molecules/ions in the crystal become dissociated and are solvated by the solvent. The process stops when there is equilibrium between crystal and solution of molecules/ions plus solvent. At this point the solution is saturated but only if some solid is still present in contact with the solution. At this point the free energy is at a minimum. Some molecules/ions will always dissociate from the crystal due to the entropy gain by diluting or mixing, with the solvent. The solubility then depends on whether these molecules/ions are more stable in the solvent than in the crystal. (These process do not indicate how quickly a crystal will dissolve. That depends on the activation energy barrier to leave the crystal and enter solution, not on the relative energies of the crystal and solvated molecules/ions).
In an ionic crystal the ions are held together by Coulombic forces, which are of long range and have a $1/r$ distance dependence. To determine the accurate binding energy in a crystal all interactions with other ions have to be accounted for, not just nearest neighbours. This is a standard calculation the most difficult part of which involves finding the Madelung Constant for the particular crystal.
At the surface of the crystal the interactions are far smaller than the bulk as these molecules/ions are in contact with only half of the crystal's ions and they also interact with the solvent. To estimate the solubility, the Coulomb energy of separating two ions to a distance r in a solvent of dielectric constant (relative permittivity) $\epsilon$ is calculated. This energy is $$w =\frac{z_1z_2e^2}{4\pi\epsilon_0\epsilon r}$$
where $\epsilon _0$ is the permittivity of free space, e the electronic charge and $z_i$ the charge on ion i and the energy at $r=\infty$ is taken to be zero. The Coulomb law is not really valid at small separations, because of the molecularity of the medium makes the continuum assumption, in terms of the dielectric constant, to break down. Nevertheless it does give an insight, albeit, approximate into solvation.
The free energy for separating two ions with radii $r_+$ and $r_-$ is thus $$\Delta\mu \approx \frac{+z_1z_2e^2}{4\pi\epsilon_0\epsilon (r_+ + r_-)}$$
which is positive since the coulomb interaction will always be negative for combining opposite charges.
The Boltzmann distribution for the concentration $X_1$ and $X_2$ in two co-existing phases is
$$ X_1=X_2exp(-\frac{(\mu_1-\mu_2)}{k_BT})=X_2exp(-\frac{\Delta\mu}{k_BT})$$
If we associate the concentration $X$ as a mole fraction then this can be a measure of the solubilty, $X_s$, thus
$$ X_s \approx exp(-\frac{(\Delta\mu}{k_BT})=exp[-\frac{e^2}{4\pi\epsilon_0\epsilon k_BT}\frac{1}{(r_1+r_2)}]$$
In water the mole fraction solubility of NaCl is $0.11$ which is a larger than the calculated $X_s =0.08 $, but not by much considering the simplicity of the model.
The last equation has the general form $$X_s = exp(-C/\epsilon)$$
where $C$ is a constant depending on the ion pairs and temperature. Thus a plot of logarithm solubility (as mole fraction) vs. reciprocal of dielectric constant for the same ion pairs in different solvents should give  a straight line.  The plot below shows data for NaCl in different solvents, the approx relationship is borne out. Because of the reciprocal nature of the ionic radii in the equation, larger ions should generally be more soluble than smaller ones.

The solvents($\epsilon$) from points left to right are methylformamide(182.4), formamide(109.5), water(78.5), ethylene glycol(40.7), ethanolamine(37.72), methanol(32.6), ethanol(24.3), propanol(20.2),butanol(17.8), pentanol(13.9). ( Data adapted from J. Israelachvilli, Intermolecular and Surface Forces).
The huge range of solubility is quite dramatic, and shows that the main effect on solubility is due to the solvent dielectric constant $\epsilon$. You can appreciate this from the Coulomb law. The electric field around an ion is effectively 'quenched' in a high dielectric meaning that the interaction energy decreases rapidly with distance. Thus an ion effectively becomes isolated from all others and is surrounded only by solvent dipoles. In a low dielectric the electric field of the ion and spreads out to a far greater distance (in the ratio of dielectric constants which could be 20 times larger), thus one ion can 'feel' the effect of others nearby when they diffuse close to one another and if they are attracted then they can coalesce to form an ionic molecule, and thus are no longer soluble.
In the figure all these solvents are rather similar and hydrogen bonding solvents to a greater of lesser extent. Non hydrogen bonding solvents are less effective it seems for solvating ionic compounds, for example, the solubility of NaCl in acetone ($\epsilon = 20.7$) is $X_s=10^{-7} $ far smaller than for propanol with a similar $\epsilon$

Why salt dissolves?
The interion attraction between the sodium and chlorine is weaker than the  attraction of water to ions. Why? The answer is because the distance between the sodium and the chlorine is bigger than distance between water and ions. Besides many molecules of water take part in the dissolution they come up with greater charge than just +1 or -1. Total force of attraction to water is greater.
Why split crystals do not attach to each other?
The answer is because the distance between the atoms can not be restored to initial distance as it was in crystal lattice. As you might know there are repelling and attracting forces acting on the atoms of solid matter. The attraction forces are very rapidly decreasing with greater distance +2-3 radii of molecule, ion etc and they are gone. Two pieces of materials can not be put back together simply by pressing because also there are molecules of air that fill in the space between the two pieces. Those molecules are decreasing the effective area of the contact since they are simply in between. The effective area of a contact are number of ions approached each other for the intermolecular/ion attraction.
During any kind of deformation the molecules and atoms reorganise themselves slightly so if you put the pieces as it seems to you exactly to the same location where they were before the atoms are not any more at the same place, they have changed their position. That is another reason why the pieces do not stick to the same original piece.


text:
How many electrons can an orbital of type f hold?
I was taking a chemistry test and I encountered the following question:

How many electrons can an orbital of type f hold?
A. 6
  B. 10
  C. 2
  D. 14
  E. 1

Since there can be [-ℓ, ℓ] orientations and since the orbital type f has ℓ = 3, we should have 7 possible orientations with 2 spins, so ${7 \times 2 = 14}$, so I thought the correct answer was D (14).
However, I got it wrong and the correct answer is marked as C (2). Is it an error in the test, or am I missing something?

You're correct that there are seven possible spatial orientations for an f-type orbital, and hence seven possible orbitals in one f-type sub-shell. However, the question specifically asks for the maximum number of electrons in one such orbital, and any single atomic orbital, regardless of the sub-shell type specified by $l$, can only hold two electrons. This is by virtue of the Pauli exclusion principle. Fourteen would be the maximum number of electrons across an entire f-type sub-shell, but the question only asks about one orbital.

The question specifically ask that no.of electron an orbital of f subshell can hold.... As we know that f subshell contain 7 orbital and each orbital can hold maximum 2 electons so correct answer would be 2..we can simply understand this by taking the real life example imagine that there is house named f which consists of 7 rooms so similarly in this case house is a subshell and each room is an orbital... Thank you,

What new info this answer brings to be posted in the first place ?


text:
Why can't pure water conduct electricity since it can be reduced at cathode and oxidised at anode?
We all know that pure water can't conduct electricity. But during electrolysis, if add a small amount of $\ce{HCl}$ acid inside, water could be decomposed to hydrogen gas and oxygen gas, as it is reduced at the cathode and oxidised at the anode. So why can't pure water just conduct electricity by electrolysis?

Indeed, water can conduct electricity by electrolysis. But this is hampered by the fact that water has very high resistance. From Electrolysis of water:

Electrolysis of pure water requires excess energy in the form of
  overpotential to overcome various activation barriers. Without the
  excess energy the electrolysis of pure water occurs very slowly or not
  at all. This is in part due to the limited self-ionization of water.
  Pure water has an electrical conductivity about one millionth that of
  seawater.

And from Self-ionization of water we can see how small it is:

Water molecules dissociate into equal amounts of $\ce{H3O+}$ and $\ce{OH-}$, so their concentrations
  are equal to 1.00×10−7 mol∙dm-3 at 25 °C

Pure water has a very high resistivity, on the order of 18 MOhm/cm.  This effectively means that any applied potential is going to be converted into thermal energy.
A side note, oxidation of an $\ce{HCl}$ (or any salt with chloride ions) solution will produce $\ce{Cl_2}$ gas, so electrolysis of these solutions should be performed in a well ventilated area.

Pure water cannot conduct electricity fundamentally because of lack of rapid ion transport inside bulk solution.
In fact, pure water electrolysis has been achieved, by using nanogap electrochemical cells, where the distance between two electrodes less than Debye-length of water (less than 100nm!). But the fundamental mechanism has been changed significantly.

===============

Lack of rapid ion transport in traditional electrolyzer.

We take cathode and H3O+ ions as an example. Initially near the cathode surface water molecules can be dissociated into H3O+ and OH- ions. H3O+ ions obtain electrons from cathode leading to hydrogen evolution; while newly-generated OH- ions can only transport very slowly through the bulk solution by slow diffusion or hopping process. These lead to local OH- ions accumulation (so that the solution near cathode turns alkaline) especially at the cathode surface, reducing the reaction rate of hydrogen evolution and thus water splitting. In other words, the reaction becomes very slow or even self-limited, showing a large equivalent resistance between the cathode and the anode. That is why pure water in macrosystem cannot be split efficiently.

Pure water electrolysis in Nanogap cells

In pure water, when the counter-electrode is placed within the Debye-length, double layer regions of the cathode and the anode are overlapping with each other so that high electric field exists in the entire gap. Still at cathode, newly-generated OH- ions can be migrated rapidly from cathode towards anode due to large electric field in the entire gap. When the gap distance is small enough, initially the transport rate can be even higher than the electron-transfer rate. Once OH- ions are generated, they are immediately drawn from cathode to anode, leading to such OH- ions waiting for electron-transfer at the anode, rather than accumulated at the cathode. In this way, the whole reactions would continue even in pure water, but now are limited by electron-transfer.
https://pubs.acs.org/doi/abs/10.1021/acsnano.7b04038

A solution conducts current when positive and negative ions are randomly dissolved in it. Pure water doesn't have enough ions but when you add $\ce{HCl}$ you add a strong electrolyte so, after the addition, solution can conduct electricity very well. However you can perform an electrolysis even without $\ce{HCl}$ but this is a very slow process because even if there are positive and negatives ions (and this could induce you to think that the pure water will conduct current!) the positive ions are near the negative electrode (the cathode) to be reduce on the other hand the negative ions are near the positive electrode (the anode) to be oxidise. So in fact you mustn't imagine a solution of positive and negative ions like in a solution of a true strong electrolyte but two bunch of ions, positive in the anode and negative near the cathode, so the current can't flow very well. This is why pure water can't conduct electricity even during the electrolysis when there are many ions in the solution.

IN pure water there is no chance for electricity conduction as it has no free ions.It contains only $\ce{H2O}$ molecules. For electricity conductance, there should be ions present. In the case you are talking about some drops of $\ce{H2SO4}$ is added for the conduction of electricity.

Pure Water can't conduct electricity because it has no free ions. But when water is contaminated, it increases the ionic level of it(then its no longer a pure water)-that's why you see waters conducting electric current and you got electrocuted by it.

Pure water does not conduct electricity because there are no suspended particles of metals such as Iron and Copper. Pure water also has lack of zinc. Ions are even not present in pure water. We can make water conductive by adding drops of Sulphuric Acid.

Welcome to Chemistry.SE! Take the [tour] to get familiar with this site. Mathematical expressions and equations can be formatted using $\LaTeX$ syntax. For more information in general have a look at the [help]. At the moment this reads more like a comment than an actual answer - could you elaborate a little more. With a bit more rep, you will be able to post comments on any question/answer.

It is quite uncommon for water to contain suspended solid metallic particles. Metallic elements in a sample of water are almost always present in the form of dissolved ions.

authors: עידן יעקובוב


text:
EZ-Water - Fraud or breakthrough?
Recently the quack medicine folks online have been promoting the research of a certain Dr. Gerald Pollack who claims to have discovered a "forth phase of water", and who has recently published a book on the said topic.
Although it's clear that his research is being deliberately misinterpreted, it's not clear to me whether or not his own claims are valid in the first place. Although some papers on the topic were published in peer reviewed journals, it doesn't seem like anyone else in the scientific community has acknowledged or replicated his results.
So, is "EZ-water" a real breakthrough or just the brainchild of a deluded scientist?

Just found this on reddit, which rather strengthens my suspicions. Just wondering if any real chemists here could confirm them.

Just came across this website called Phperfect Live Water (http://www.phperfect.com.au/) that seems to be trying to sell the same concept and I was wondering whether there has been new research to support its claims or not.

I am not a chemist. I saw his tedx talk. He claims little is known about water, elaborates on this and then tries to answer the points.

EZ-water is not a breakthrough. It is not new, nor is it valid. This appears to be one of the many claims about the healthful benefits of drinking "ionized water".
The original "article" linked, contains a number of misleading statements.

Water molecules make up 99% of your body. This is true, but misleading. According to various sources, the human body is between 57% and 75% water by mass. Since water has a much lower molar mass than most biomolecules, 99% of the molecules in your body may well be water.
The journal listed in which the research is published is Water, which is not a high impact journal, despite its name. Water had an impact factor of 0.973 in 2012. That means that articles published in 2011 in Water were cited less than 1 time each by anyone in 2012. Compare with Cell, which had in impact factor of 31.957 in the same period - each article in Cell is cited an average of nearly 32 times.
Water has a "fourth phase" and no single boiling point or melting point. This is true! Any student of chemistry should be able to tell you this. Take a look at the phase diagram of water below. Water apparently has at least 18 phases (15 solid phases, plus liquid, vapor, and supercritical fluid). You can also see that the solid-liquid and liquid-vapor phase boundaries are not single points. They are curves varying in temperature and pressure. Those curves represent the conditions in which the two phases have the same vapor pressure. This is not a new discovery or a startling secret. This behavior has been known for many decades (if not centuries).

Later on the "article" tells us that EZ-water is not $\ce{H2O}$ (note that they cannot be bothered to format their chemical formulas) but $\ce{H3O2}$, which is alkaline with a negative charge. Hmmm... If it has a different formula and different properties, it must be a different compound. Note that this "formula" is essentially $\ce{H3O2 = H2O + OH}$, and hints at what happens inside these water ionizers.
EZ-water has a negative charge. It is possible for objects to build up charge, but it is not possible for them to remain permanently charged. Overall, the universe is balanced in charge. Consider lightning as a consequence of attempting to maintain separation of charge. This "negative charge" comes from a misunderstanding of pH. The water definitely has a higher pH, which means more $\ce{OH-}$ ions than $\ce{H+}$ ions, but something (probably $\ce{Na+}$) needs to replace the missing $\ce{H+}$.
I can go on, but you get the point.

A water ionizer electrolyzes water, which requires electricity (not light as suggested by Mercola). I had the good fortune to be involved in the testing and dismantling of one of these ionizers recently. The device claimed to take ordinary tap water and convert it to a "super-cleaning solution that contained only negatively charged oxygen particles" or something like that. From experiments:

The ionizer increases the pH of the water.
The ionizer does not work with deionized or distilled water!
The ionizer does not produce measurable oxygen gas (a smoldering splint will not relight in the vicinity)
The device contains an electrolysis cell and an ion exchange resin

Here is what happens:

Water is electrolyzed.
Doing so requires ions in the water already to conduct the electricity, which is why the device must use tap water (and not purified water). At the cathode, water is reduced to hydrogen and hydroxide. At the anode, it is oxidized to oxygen and protons.
\begin{align}
\ce{2H2O +2e- &-> H2 + 2OH-}\\
\ce{2H2O &-> O2 +4H+ + 4e-}
\end{align}

Protons are exchanged for sodium ions
To prevent the recombination of hydroxide and protons to form water and neutralize the solution, there is an ion exchange membrane between the anode and the rest of the solution.
$$\ce{H+ + OH- -> H2O}$$
The ion exchange resin collects the protons and releases sodium ions into solution to maintain charge neutrality, which must happen or you would have a lightning gun and not a mystic device to produce super water.

Ultimately, drinking this stuff would probably have no effect on your health. But we are no doctors, so don’t listen to us either. It might help or it might hurt, depending on your condition, it also might not matter.
First, your stomach secretes acid, so any extra base would be neutralized immediately. Let's say an ionizer gets the water's pH to 8 (slightly basic, and more basic than your blood). This means that the concentration of protons is $\ce{[H+]}=10^{-\text{M}pH}=10^{-8} \text{ M}$ (remember that pH is a logarithmic scale. The concentration of hydroxide is $10^{-14} / 10^{-8} = 10^{-6} \text{ M}$. The pH in your stomach ranges from 1.5 - 3.5, which represent proton concentrations between $10^{-3.5}= 3.16\times 10^{-4}$ and $10^{-1.5}=3.16\times 10^{-2}$. If we use the higher pH stomach acid to neutralize an equal volume of EZ-water, the final pH would be:
\begin{align}
[\ce{H+}]_f &= \frac{3.16\times 10^{-4} - 1\times 10^{-6}}{2}=\frac{3.15\times 10^{-4}}{2}=1.07_5 \times 10^{-4}\\
\text{pH} &=-\log[\ce{H+}] =-\log(1.07_5 \times 10^{-4})=3.97
\end{align}
Since your stomach acid would likely have a greater volume than the ionized water, the change in pH would be even smaller. This calculation assumes your stomach acid is not a buffer, which it is.
Second, the acid-base homeostasis of your blood prevents the pH of your body from changing. Yes, your blood pH is 7.4 or so (which is slightly basic), but that does not mean your blood is negatively charged. Outside of range of 7.38 - 7.42 all of your blood proteins (like hemoglobin) start to denature, which will kill you.
Well, to be more precise, you will suffocate before that happens as hemoglobin/myoglobin oxygen binding capacity is pH sensitive. And if anything, we can tolerate acidic blood more robustly than basic blood pH (down to 6.8 or 7, but only up to about 7.6) before death. You might want to look up acidosis.

If you want to raise the pH of your stomach fluid, which can have health benefits, save money and buy antacids

Thank you for the very detailed answer! Do you have any information about Pollack's research, that is, without Mercola's dubious commentary?

Great answer, except the part where you mention the impact factor of the journal Water. An impact factor of 1 is low, I agree, but many very decent pieces of research (although typically not very innovative) are published in low impact factor journals. A low impact factor is not at all a bad quality stamp, neither is a high impact factor a good quality stamp. To be fair, there is some connection between impact factor and quality, but it is far from 1 to 1.

Excellent answer! However, I agree with Michiel that a lot of good research is published in journals with an impact factor << 10. To give an example, J. Phys. Chem. A and J. Photochem. Photobiol. A have impact factors as "low" as 2.77 and 2.69, resp.

According to the website of Dr Pollack himself, the health effect of EZ water is a proposed research project currently applying for funding.  He has made no claim of any proven health effect, and he even claimed explicitly that he does not know if EZ water has health benefits.  He also explicitly stated that the mechanism of formation of EZ water is still under investigation.
His book is not a scientific work.  He presents his proposed yet untested ideas and theories regarding phenomena that may be attributed to EZ water; it however does not claim anything is proven at all.  It is a good thing to have ample imagination; however at least at this moment he is somewhat behind the schedule on proving these ideas.
In fact, he is currently carrying out research on many of these subjects.  According to what I saw on his website, his research seem to be sound, but at this moment no conclusive results are produced yet about any of his big hypotheses.
Actually it is fairly common to describe a proposed research to have the potential to have a revolutionary impact on the life of every man on earth.  Very few write a book like him though.
I would not call him a deluded scientist because he did not really claim anything nonconventional being discovered.  However he does put himself in a position where immense commercial profits are generated solely through exploiting and distorting results of his research.

The question was first about the existence of this so called "EZ water", way before we get to any health benefits. I think Ben has explained rather well why there is no real reason to beilive such water exists.

Ok, so they did make something, but it is NOT water.  These are highly energetic microscopic structures generated from water in very special conditions. These kind of things are not rare in surface or cluster chemistry--you bombard anything with high energy particles, you get a bunch of weird stuff. They are observed as rather short lived structures, difficult to prepare and hard to stabilize, so there is not even any possibility of drinking it.

He literally asked for money to commercialise health benefits of EZ water  https://www.indiegogo.com/projects/can-water-cure-disease#/.

Thanks for this objective analysis. Bought the book, I put it to test, considering: hydrophilic surface creates the stack of EZ layers. made a rack of microscope cover glasses 0.13mm thick separated by de-phased ones to make 0.13mm gaps, (ideally EZ-water forming between sheets and flow should "filter/purify" water). But didn't observe ANY exclusion zone at all. Random particles would just fall to the outer glass. Food dye didn't react at all. This glass is hydrophilic AF BTW, water droplets extend to max over it... So I'm confused, slowly inclined to call scam. hints/thoughts on home made EZ?


text:
What makes Gold so different from Platinum?
Looking at Wikipedia, Gold & Platinum are in adjacent seats.
    Platinum                                Gold
[Xe] 4f14 5d9 6s1                     [Xe] 4f14 5d10 6s1
Atomic number of 78                   Atomic number of 79
Atomic pass of 195.084u               Atomic mass of 196.96657 ± 0.000004 u

How does that single electron make such a huge difference?

From the electronic configuration, (and therefore the spatial organization of the AO's), it has to be known that "filling" or "non-filling" electrons play the role inside the Binding energy of the atoms, for an exquisite example the fact that Why is mercury liquid at room T°? depends only on the position of the electrons, and so the accesible quantum states for the binding.

The difference is of a single electron if you consider only one atom. 1g of gold has about 3 followed by 21 zeros more electrons than Platinum, and if that's not much of a difference for you...

Let's not forget that the number of electrons is not the only difference in the subatomic particles of Pt and Au.

I'm not quite sure what the asker is referring to when they mention the "huge difference". Arguably, gold and platinum are actually strikingly similar, even in their oddities (such as relativistic effects); if you want a huge difference, why not compare neighbours such as C and N, or among the metals, Cu and Zn? There are plenty more dissimilar pairs without invoking a fundamental change in electron distribution (like going on to the next subshell). So what exactly is the huge difference in question?

That one electron is making sure that Gold has a complete d-orbital, and therefor is much more stable. Elements want to achieve noble gas configurations, and if they can't, then they want to have a half-filled orbitals. In the case of gold the d-orbital is filled completely and the s-orbital is half filled. Whereas Platinum doesn't have that complete d-orbital filled.
Chemical properties (such as reactivity) are influenced by electrons. The amount of neutrons will determine if it can be radioactive or not(,...) look at Tritium : that is a hydrogen atom with 2 neutrons in it and it is radioactive. Whilst protium (the most common isotope of Hydrogen with 1 proton and 0 neutrons) is not radioactive.


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What is the right quantum number of copper?
Copper has $[Ar] 3d^{10} 4s^1$ electron configuration. And zinc has $[Ar] 3d^{10} 4s^2$ electron configuration.
My question is, what are the right quantum numbers of the last electron of copper? Is it $$a. n=3,l=2,m=+1,s=-1/2$$ or $$b. n=3,l=2,m=+2,s=-1/2$$
If I use (a), the electron configuration of copper is $[Ar] 3d^{9} 4s^2$ and which isn't stable. But if I use quantum numbers in (b), it is the same with the last electron of zinc.

The correct quantum numbers for last electron of Copper ($\ce{Cu}$) would be $B. n=3,l=2,m=+2,s=-1/2$.

if I use B quantum numbers, it is the same with the last electron of zinc.

Yes, last electrons of Both can have same quantum numbers because Pauli Exclusion law says that no two electrons in an atom can have all four quantum numbers same but it doesn't say that two electrons of different atoms can't have all four quantum number same.
Refer this for more.

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


text:
What non-toxic non-water substances have a freezing point very close to water's?
I was linked to a Kickstarter for a cooling ball to be placed in drinks which claims with constant marketing hyperbole how much it relies on "Phase Change" ! which is a neat concept and seems not unreasonable, but in thinking about it I'm failing to imagine (as I'm no chemist, and of ill education) what refrigerant could be inside that meets the named requirements of non-toxic food safe, as well as passing from solid to liquid (or liquid to gas) such that a typical freezer would freeze it (−2 °C to −5 °C typically?), and it would liquefy around a temperature you'd want to keep your drink (say 1–3 °C).
Obviously water meets these requirements, however due to the expansion pressures water commits I have a very hard time imagining a small stainless steel ball containing that expansion through many freeze-thaw cycles without causing eventual deformities which would be continually weakened.
So the question is in the title, and to note the container is a sealed solid stainless steel ball of unknown thickness 2" in diameter.
Let me know if this question is off-topic, more than anything I'm just curious so I realize sometimes questions not based on a problem are poor fits for SE.
Do note: I am not interested in actually trying to do something with this, it is simply a curiosity, so don't be concerned if the things that come to mind aren't FDA tested, I'll settle for answers that simply show there are such things that are likely non-toxic regardless of thorough safety testing. Also the phase change from liquid to gas is a possible one too if you know of any safe chemicals here that wouldn't cause uncontrollable internal pressure and would have that phase change at the temperature range specified.
Just trying to come up with an idea for a compound that could meet their marketing hyperbole and finding I come up short.

Just out of interest, do you have a link to the Kickstarter?

@Snipergirl the kickstarter in question is Spirit Steels - The ultimate beverage cooling device

I tried to use the curated dataset from Mathematica to answer this question.  You can ignore the following if you have no interest in Mathematica and just jump down to the end for my ideas.
The following code gets all of chemicals currently in the curated dataset and grabs the melting points and NFPA health rating.  The second line finds the chemicals with melting points between -7 and 0 degrees and the last line looks for chemicals with a health rating of 0, which is low risk.
out = {#,  ChemicalData[#,"Name"], ChemicalData[#,"MeltingPoint"],
   ChemicalData[#,"NFPAHealthRating"])& ChemicalData[];
cull1 = Cases[out, x_ /; -7 < x[[3]] < 0];
cull2 = Cases[cull1, x_ /; x[[4]] == 0];

We go from about 43000 chemicals in the first list to 220 in the second and finally 6 in the last.  I'm not sure I agree with the health ratings of these compounds but here you go (melting points in parentheses):

2-methyl-2-propanethiol (-0.5)
prehnitene (-6)
trans-1,2-dichlorocyclohexane (-6)
hexadecene (-6)
trans-1,2-dibromocyclohexane (-6)
olive oil (-6)

The only chemical I would deem "food safe" would be the last one.  Perhaps some type of food-grade oil is the mystery substance.

Thanks!! This is exactly the sort of answer I was looking for. Olive oil would probably make a spectacular choice here with a -6C freezing point that should freeze in your average freezer and definitely keep your drink cold. No telling if this is what is used but it sure could be. Altogether bravo on the approach!

You could even partially hydrogenate it to tune the melting point, if you wanted to. Mmmm, margarine cubes.

Surprised that 2-methyl-2-propanethiol is down as non-hazardous...I would imagine it's pretty poisonous, although presumably less so than butanethiol.

@J. LS https://en.wikipedia.org/wiki/Tert-Butylthiol - it stinks still terribly ;)

Jimmy,
There are most likely many chemicals that have a melting point range in that vicinity, but I can't comment on their toxicity. However, I can comment on some of the science here.
First of all, water is one of very few molecules that expands when it freezes. That has to do with the hydrogen bond network formed between water molecules in the solid state. This network is slightly less dense than low-temperature liquid water, so water will expand when it is frozen. (This is also why solid water floats on liquid water.) Most compounds form frozen solids that are more dense than their liquid state, so they will contract on freezing. This is why it is fine to freeze these metal balls if they are filled with something besides water. Although they could have water inside, but just not be completely full, I suppose.
If you were going to find a water substitute, then you would want to find a chemical that has a high heat capacity and enthalpy of fusion. Heat capacity is the amount of energy required to heat a given substance by a certain temperature. If a substance has a high heat capacity, then it take more energy to change the temperature of that material. (It gets this energy by removing heat from your beverage.) The enthalpy of fusion is the amount of energy it takes to change a substance from a solid to a liquid. Again, you would want something with a high enthalpy of fusion for the same reasons as for heat capacity. Unfortunately, water is really great on both of these fronts, which is why ice cubes are so great for cooling a drink.
I'm not sure I really answered your question, but I hope that helps a little.

http://en.wikipedia.org/wiki/Glycerol has a melting point of 20°c (in comparison to water's melting point +20°c) and it is used as ingredient in 'Bounty' sweets (E number  : E422)
I also suggest that you have a look at the E-number list. There are a lot of allowed products for human consumption (however not all are safe in high doses) you can find the list here : http://en.wikipedia.org/wiki/E_number
An exact answer to the question I don't have :(


text:
Does freezing a solution with water always cause the water to separate and form the ice lattice?
I'm curious, I was trying to look into the affect of freezing a solution with water even when the solution is completely miscible. I came across something that detailed this regarding salt water and basically said when it gets low enough to freeze, the water forms it's tell-tale lattice basically pushing the salts out so the ice itself is pure water and less dense (considerably less dense because the salt solution allowed the density to increase all the way to the lowered freezing point).
I am curious, is this a general fact of the behaviour of any solution with water regardless of miscibility? Are there water solutions which break up the lattice while still allowing freezing, or does the affect of blocking the formation of that lattice completely block the solution from freezing altogether no matter what? Are there water solutions which will freeze solid even with the lattice without just separating the non-water compounds?
Note: I recognize amorphous ice can be created with wizardry, but my question is more in regards to the typical behaviour on earth.

related: "impurities in ice"

It's worth noting that from a non-chemical perspective, if you freeze the water fast enough (with, e.g. liquid nitrogen), the impurities just get included in the glassy amorphous ice no matter what they are.

@Aesin right, I've heard if water is frozen quickly enough it can freeze before the hydrogen bonds can form so the lattice isn't created

@JimmyHoffa that is what the amorphous ice is, according to the wikipedia article :-) Also interestingly, although most ice on earth is in the lattice form, apparently in space most of it is amorphous. So likely the most common form of solid H2O is actually not the one we have in abundance

The purification that occurs when an aqueous solution freezes is called Fractional Freezing and is based on the idea that impurities of a solvent (and this can be any solvent, not just water) have a much lower solubility in the solid phase of the solvent relative to the liquid phase.   Several techniques for performing fractional freezing for purification purposes can be found here.   As the solid is formed, the impurities are expelled into the liquid that hasn't yet solidified.  The purified crystals can then be removed by physical methods (e.g. filtration).  In the case of water, where ice is less dense than the liquid phase, the purified crystals float to the top of the impurity-laden liquid.  This is probably the reason why some people have observed a layer of impurities deep within a block of ice.  As mentioned above, the process is not limited to just water, and freeze crystallization is use to eliminate the purities of silicon wafers and has been known for many years as a way to make hard liquor.  This latter use of fractional freezing is close to the OPs question, which inquires about miscible substances (in this case, ethanol and water).  Despite their miscibility, ethanol and water can be separated by fractional freezing, just like they can be separated by fractional distillation.  Certain substances cannot be separated completely using distillation, and these substances are called azeotropes.  Water and ethanol make an azeotropic mixture that is approximately 95% water.  (So the moonshine you were thinking of making using the reference above will be limited to 190 proof.)  What I find interesting is the following figure,

which shows the freezing point of water/ethanol mixtures as a function of ethanol content.  At a point around 95% there is a significant dip in the freezing point.  At this point, called the eutectic point, the solution behaves as if it is a single component (much like an azeotrope) and will not likely be separable through 'simple' fractional freezing means.  Note I use the term simple because azeotropes can be separated using special distillation techniques.
So in summary: fractional freezing will work with any aqueous solution, regardless of the miscibility of the components, unless the component ratio is close the eutectic point.  We don't need to limit ourselves to aqueous solutions, as many species can be purified in this manner.  The purification mechanism is not so much a factor of the crystal lattice, but the relative solute solubilities in the solid and liquid phases of the solvent.  Finally, many water solutions will freeze solid even with impurities.  This and other types of separations work on equilibrium processes, and multiple iterations are necessary to "completely" remove impurities.  (or more correct, remove the impurities to a level that is acceptable for your application.)

Thanks for the information, though perhaps it wasn't clear in my question, I wasn't curious about purification, my curiosity was entirely regarding freezing water without causing the lattice which I understand (maybe inaccurately?) to be the reason it becomes less dense when frozen; that is I am curious if you can freeze water in such a way as to not cause it to become less dense by way of a  solution as opposed to by way of rapid-freezing.

@JimmyHoffa If that is the case, then we need to watch the terminology used.  When you use solution in your title and miscible in your question, that suggests water with multiple components.  In my answer, I'm calling all solutes "impurities" that get removed upon freezing due to the process described.  One does not need to invoke the lattice of solid water to explain this phenomenon.

@JimmyHoffa As for freezing water in such a way to make it less dense, this can't be done without additional pressure.  See the links in this reference, one of which has some useful figures.

Thanks! That's exactly what I had figured and just wanted verification! I am way out of my depth on this topic as a whole, so my apologies for the confusion. My chemistry knowledge starts and ends with the the fact that there is a periodic table heh. Didn't know I was asking about a eutectic system, but sounds like even with a eutectic system the solid is analogous to the solid forms of the system's members alone.

The situation you are describing is known as a eutectic system. Whether a system is at the eutectic point depends upon both the materials composing the solution as well as the relative amounts of each. A solution of $\bf{A}$ and $\bf{B}$ that are not at eutectic compositions will freeze out one component until a eutectic composition is achieved. While a eutectic system will simultaneously solidify both components, many systems will freeze solid $\bf{A}$ and solid $\bf{B}$ (the two being interdeispersed) and not solid $\bf{AB}$, though some systems do solidify as mixed (metal alloys, for example).
Water and sodium chloride form a eutectic at about 23% $\ce{NaCl}$ by weight. You can see a phase diagram for water and $\ce{NaCl}$ at this University of Calgary page. Note that an eutectic can consist of more than two components, as mentioned by the University of Calgary page. A typical phase diagram of $\ce{NaCl/H2O}$ solution is as depicted below:

Note: An eutectic composition is around 27% $\ce{NaCl}$ (by $w/v$) dissolved in the water (keep in mind that saturated $\ce{NaCl}$ solution is around 30% at room temperature).

If you have a eutectic system and you freeze it, does it become consistently dispered solid A and solid B as opposed to arbitrarily dispersed; and the lattice with water will still be formed? My curiosity is regarding the lattice specifically and the ability to freeze a solution with water that freezes consistently and without the lattice. Does a eutectic freeze solid AB with an altogether different solid form than the typical solid form than the independent components?

Ah that link appears to explain it with the phase diagram, sufficient amount of NaCl and when frozen its no longer frozen H2O, but frozen 2H2O + NaCl which may logically have a different solid form than frozen H2O. Is this a correct understanding?

@JimmyHoffa Some eutectic systems freeze as solid solutions, though I'm not aware of any off the top of my head that involve water. If I come across any, I'll be sure to post it and let you know. The dot formula of NaCl and water refers the the hydrate. Read more about it here: http://en.wikipedia.org/wiki/Hydrate

So the dot formula of NaCl and water are a 'hydrate' which may have a different solid structure than the solid of water, correct? Or is it only refered to as a 'hydrate' terminologically, though the actual thing being referred to is still water structurally?

@buckminst:  Since University of Calgary page cannot be found anymore, I have included one from another site. If you like you may keep it or otherwise delete it. It's up to you.

Certain solutes, especially under pressure, actually become incorporated into the water-ice structure to form clathrate hydrates. Of these, methane hydrate is the best known, occurring naturally in oceans, where organisms generate the methane under sufficient pressure to create the clathrate phase.  These illustrations from the source linked above shows the conditions of ocean depth and temperature where methane clathrate hydrates form, and the resulting clathrate structure.

In the clathrate phase, the water molecules are packed differently from pure water ice.  While the ordinary ice structure has a lot of open space, even more space is needed to incorporate the methane or other small gas molecules, and so the water molecules are packed less densely to leave the openings for the foreign material.


text:
Can substituted acetone undergo Cannizzaro reaction?
I know the Cannizzaro reaction takes place in carboxyl compounds without $\alpha$-H with the regular examples being formaldehyde and benzaldehyde.
But can this reaction take place with say diphenylmethanone or di ter-butyl ketone (2,2,4,4-tetramethylpentan-3-one) or 2,2-dimethylpropanal ?
I couldn't find any examples of these compounds showing the cannizzaro reaction.

Possible typo : "carboxyl compounds without $\ce{\alpha-H}$..." . It should have been "carbonyl compounds without..."

As far as I'm aware, the Cannizzaro reaction can't occur with ketones, only aldehydes. The reaction mechanism involves the transfer of hydride from the former carbonyl carbon (of the tetrahedral intermediate resulting from hydroxide's nucleophilic addition to the carbonyl carbon), during which step the carbonyl reforms, yielding a carboxylic acid. Since a ketone, by definition, has no hydrogens bonded to the carbonyl carbon, it obviously can't undergo such a reaction. I would expect ketones with no $\alpha$-hydrogens to be essentially unreactive with hydroxide, since aldol reactions and keto-enol tautomerism are impossible with no $\alpha$-hydrogens, and most hydrates are unstable.

Means cross cannizzaro can be possible between any aldehyde + ketone mixture and we will get only single product as aldehyde get oxidised and ketone get reduced? Is it correct?

As Greg E. wrote in his answer, Cannizzaro reaction does not usually occur with ketones. However there are some instances where ketones give Cannizzarro products. From the Wikipedia page of Cannizarro reaction:

Certain ketones can undergo a Cannizzaro-type reaction, transferring one of their two carbon groups rather than the hydride that would be present on an aldehyde.

The paper [1] gives an instance of such reaction. A reaction from the paper is:
Here, the 3 H-atoms of the methyl group of acetophenone(1-Phenylethan-1-one) are replaced by F-atoms. Therefore, it is a ketone without any $\ce{\alpha-H}$. After nucleophilic attack by $\ce{OH^-}$ on the carbonyl carbon, the intermediate expels a trifluoromethyl carbanion, and forms a carboxyllic acid.
In the next step, the carbanion attacks another ketone molecule to form alcohol.
However, according to the paper these products are formed only when solid NaOH with DMF as solvent, is used. Hydrated form of the ketone (gem-diol) also exists in a significant amount in the product mixture.
References:

Jiang, X.; Matsukawa, S.; Kakuda, K.; Fukuzaki, Y.; Zhao, W.; Li, L.; Shen, H.; Kojima, S.; Yamamoto, Y. Efficient synthesis of tetradecafluoro-4-phenylheptan-4-ol by a Cannizzaro-type reaction and application of the alcohol as a bulky Martin ligand variant for a new anti-apicophilic phosphorane. Dalton Trans. 2010, 39 (41), 9823. DOI: 10.1039/C0DT00539H.

Means cross cannizzaro can be possible between any aldehyde + ketone mixture and we will get only single product as aldehyde get oxidised and ketone get reduced? Is it correct?

@Mohit To be fair, I don't know the answer to the question. But if I had to guess, I would say that yes, the aldehyde should get oxidised and the ketone reduced. I will check some books and try to come up with an answer.

@Mohit, Also note that cannizaro type reaction that I have wrote in the answer does not happen with any ketone, only specific ketones that can expel relatively stable carbanions.


text:
How do we actually know the chemical structure of a compound?
If a compound is placed before you and you are asked to determine what it is and what it's structure is, how do you actually do that?
This might be a baby question for most of you but I am not from chemistry background and so this amazes me!

This is a very broad question and it would help to focus it a bit.  Are you assuming that the compound in front of you is pure?  Are you assuming it is a solid or can it be a liquid or gas?  Are we assuming we know nothing about the compound?

yes , let's say we know nothing about the compound.

A full answer to this question could fill several volumes.

There are textbooks devoted to this topic, even when we assume things like compound is 1) pure, 2) organic, and 3) not a polymer.

For a true unknown a typical lab would use a combination of instrumental techniques to identify it. GC/MS, IR and NMR for instance for an organic compound. If it works, GC/MS is particularly powerful since it purifies the sample and identifies it at the same time. // An inorganic compound or mixture would use XRF & XRD. // There are massive databases of compounds to augment identification using each of these techniques as well.

The easiest way is to inquire where the compound originated, observe the compound with generic senses (color, texture, phase-gas/liquid/solid) and then make educated guesses based on that! If there remains any doubt as to the chemical or if that info is not available then a systematic process of ruling out the most likely possibilities is conducted. The single largest category in chemistry would be organic or carbon containing compounds. They may be analyzed using NMR to determine their structure with good accuracy. Any decent sized chemistry department will have an NMR available. For inorganic compounds, some simple acid-base tests could narrow the possibilities.
With the most coveted and expensive tools, it is easy to tell what elements compose a sample. I once visited a forensic laboratory where they had a scanning electron microscope that could display individual grains of gunshot residue (invisible to the naked eye)for which police swab the hands of suspects. The particular SEM I saw in action had an x-ray source connected to it, which allowed any point including an individual microscopic dot of suspected gunshot residue to be analyzed for its elemental composition. The guy there clicked on any part of the SEM image to display what element it was made of... completely automated!
One of the longest laboratory experiments I did in school was determining the percentage of iron in a mixture of iron and clay. It took 12 hrs and used only wet chemistry from 100 yrs ago.
There are many techniques that could be used but it really comes down to which one is most promising or available.

Actually, it's quite simple for most substances. Use a gas chromatograph. If the substance can be vaporized at reasonable temperatures, this will give you enough information to pin down the chemical compounds, or at least rule out a wide range of compounds. We can tell because every compound has a unique retention time based on its mass and attraction to the medium coating the column. With enough reference data and good precision, it's easy to figure out firstly how many separate compounds are in the mixture and secondly what their chemical formulae are.
Failing that, there are other kinds of chromatography, not to mention mass spectrometry.

Gas chromatography is a separation technique, not a characterization technique.  Identification using GC requires that the known compound has already been isolated.  It is not likely true that "every compound has a unique retention time".  I agree that you can estimate the number of compounds in a mixture but you cannot determine their chemical formula with GC alone.

For example, GC does a very poor job with polymers, ionic compounds, minerals, etc.

The easiest and most precise method: grow single crystals and do x-ray diffraction experiment. as long the question is not which lanthanide I have, some chemical "common sense" + x-ray structure give u full answer. Of course, you still need to do other characterisations as NMR and IR spectroscopy.

X-ray is great for crystalline materials. If you have a cross-linked polymer or a low molecular weight liquid or gas, you need something else.

Use a sample of the material and perform proton NMR, IR Spectrography, Gas Chromatography/Mass Spectrometry, then cross reference the information using COSY, DEPT and HETCOR analysis.


text:
Which computational method/basis set should I use for small molecular ions?
I apologize for a perhaps newbie question.
I would like to compute the potential energy surface for several small molecular ions, like $\ce{H2+}$ and $\ce{H3+}$, using quantum chemistry software. What are the proper method and basis set to do that and why? I am thinking about ROHF/STO-3G or UHF.

$\ce{H2+}$ or $\ce{H3+}$ have unpaired electrons because they are ions. Due to this, you should be using an unrestricted shell theoretical model which doesn't restrict electrons to paired (closed shell) orbitals. You should start with unrestricted Hartree-Fock (UHF) and can increase the level of theory to include electron correlation with something like 2nd order Møller Plesset (MP2) or 4th order (MP4).
The electron correlation models acknowledge the fact that electron locations are not totally random, repulsion between electrons is a real effect and can give more realistic results.
With regard to basis set, STO-3G would be considered a starting point. If time and resources allow, you could build up to something like a 6-31G(d,p) which adds p-orbitals to the hydrogen atoms allowing more optimization of molecular orbitals. Increasing the basis set quality approximates the radial components of the atomic orbitals with more Gaussian curves, thereby increasing the possible accuracy of the calculation.
Exploring Chemistry with Electronic Structure Methods - Foresman and Frisch

For the people who are not familiar with the terminology, could you add references?

Thanks for the suggestion, I've added some additional detail and a reference. From the OP's terminology it appears that Gaussian might be the software used so I stuck with their nomenclature.

$\ce{H3+}$ does not have unpaired electrons.

if u accurately want to make ur computations a bigger basis set and high level are necessery (CI(SDQ) or CAS), however so bigger basis set so less human readable are results. If the only aim is to see nice orbitals, minimal basis and DFT or even semiemprical methods should be enough. As stated before unrestricted methods are necessary unless u want to play with half-occupations.


text:
Thermoplatics into Thermosetting plastics
I must confess, I am not a chemistry student but have always been a fan of it, so I am not sure if this is a blunder.
I have often wondered if Thermoplastics can ever be converted to or made to behave like a Thermosetting plastic. In other words, can a Thermoplastics be made more heat resistant?

Depending on the plastic, it is possible to convert a thermoplastic to a thermoset. The way to achieve this, is by linking the polymer chains in a thermoplastic to one-another and is called 'cross-linking'. Then, the chains cannot 'slide' along-side each other and will therefor become a thermoset. This means the plastic will not melt and start to degrade when the temperature becomes to high for the molecular bonds to hold.
Note that cross linking the polymer chains is a chemical reaction which needs certain functional molecules. When the thermoplastic chains have functional groups left to attach a 'cross linker' to, a bridge between two chains can be made. When the thermoplastic has no functional groups left however, it might never be possible to convert the thermoplastic into a thermoset.
Lastly, I would like to subscribe to the idea that there are no stupid questions (or blunders). Just people stupid enough not to ask them. Good job asking this one :)
Hope this helped!

Thanks for your response. I was almost giving up. I don't totally understand your answer but I get that it is possible only if there are "functional groups" in the thermoplastic. Now giving nylon has an example, does it have the functional groups you mentioned?

@burntblark If you could post what parts you don't understand, I would gladly explain more. As for Nylon, most Nylon is Nylon-6 (wikipedia, nylon-6), which has only functional groups to connect the building blocks (or monomers) together. Then, there are no groups left to 'cross-link'. Though Nylon-6 is most common, almost all Nylon has no functional groups, other than the groups used to make the connections between the monomers. Hope this helped :)

X-ray or beta radiation will probably work with nearly any polymeric material. ;-)


text:
What is the net ionic equation for rection of liquid bromide with sodium iodide crystals?
I have this problem,
Liquid bromine is added to a container of sodium iodide crystals
I interpreted it with this molecular formula $\ce{Br_{2(l)} + 2NaI_{(s)} \rightarrow 2NaBr_{(s)} + I_{2(g)}}$.
From my understanding, since none of the products nor any of the reactants is aqueous, there shouldn't be a net ionic equation. However, my friend has proceeded with the problem as follows:
$\ce{Br_2 + 2Na^+ + 2I^- \rightarrow 2Na^+ + 2Br^- + I_2}$
$\ce{Br_2 + 2I^- \rightarrow 2Br^- + I_2}$.
Could somebody explain which one of us is correct and why?

First, iodine probably shouldn't be a gas, because it exists as a solid under standard conditions. As for how the net ionic equation should be written, I think it's a matter of convention. When I took general chemistry I, we were instructed to separate soluble ionic compounds and cancel spectator ions ($\ce{Na+}$) irrespective of whether the reaction was explicitly taking place in aqueous solution. This is a redox reaction, and sodium is not undergoing change in oxidation state, so that seems right to me. Hopefully somebody weighs in with a conclusive answer as to what the convention is.

Technically you are correct. How can there be any ions when the reactants are not in aqueous medium ? But here it is a matter of convention and maybe an illustration of how to write the ionic equations from molecular ones. Likewise, if you have done balancing of redox reactions , you might notice half of the reactions given to balance are completely impossible according to the laws of chemical combination. You'll always be at a loss when you overthink in chemistry, even if your argument is right.  Taking it as just a convention would be a better idea .
Hope this helped !