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Why is platinum used as a catalyst in fuel cell electrodes (and elsewhere)?
I have rather basic knowledge of chemistry (at the level of a first year college class). But maybe somebody can still clarify with some simple words: What makes platinum a good catalyst for the electrodes of fuel cells? And why can't other materials be used?

Imagine that a fuel cell is like electrolysis - the other way around.
In the electrolysis of water, you'd apply a certain potential (a voltage) to the electrodes and observe hydrogen and oxygen formed at the electrodes.
In theory, you need a particular potential to run the reaction. In reality - and this depends on phenomena on the surface of the electrode material - you often need more.
With platinum, however, this overpotential for hydrogen is zero.
This means that you have no losses, neither in electrolysis, nor in the fuel cell.
This might be oversimplified but I hope it's not completely wrong ;)

As pointed out in the other answer, platinum makes the oxidation of hydrogen gas to protons occur faster at any given potential, which in general allows you to get more energy out of the system at a given power level.
As for why this happens, it has to do with how platinum metal binds to hydrogen atoms. Hydrogen gas (H2) adsorbs ("sticks") to platinum metal, which causes the H-H bond to break as two H-Pt bonds form. This makes it easier to remove an electron from each H atom, generating two H+ from H2. The thing that makes platinum such a good catalyst for this process is that it binds hydrogen gas strongly enough, but not too strongly. Metals that have weaker interactions with hydrogen gas will not be as effective because the first step won't happen often enough (adsorption of hydrogen gas) and metals that bind hydrogen gas too strongly will not release the products, also resulting in lower rate.


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is there any mnemonic for the AXE method?

Is there any mnemonic that helps remembering the names of the different geometries (Shapes) that we obtain from applying the AXE method ?

Do you mean remember $AX_2E_0$ refers to "linear" $AX_3E_0$ to "Trigonal planar" and so on?

@GM yup , that's it .

I think you shouldn't abuse with mnemonic. In this case you should start with this rule of VSPR:

each atom in a molecule will be positioned so that there is minimal
repulsion between the valence electrons of that atom.

Now you should begin to train your three dimensional thinking!
This is the method I've adopted to learn AXE, of course everyone has his own but maybe it could help to you to find yours:
Learn to draw the $AX_nE_0$ series:
This for me was the first step: learn to draw the basic geometry 0 lone pair. This is quite intuitive knowing the previous law. Begin with the $A$ then add the $X$s. Note that the linear module could help you to remember it.

Add 1,2,3 lone pairs series
This is the next step. I've brutally modified the wikipedia table to show how you can easily do this step from the Basic geometry 0 lone pair. Simply begin to substitute $X$ with $E$: one $E$ for 1 lone pair, two for 2 lone pair...
Nomenclature
The last step is to add the nomenclature. First of all you should take in account that the lone pair ($E$) don't appear as a part of the final shape from which the nomenclature is derived. If not considering the $E$ group you obtain a planar molecule you have to add the word planar after the name, if you obtain a pyramidal shape you should use pyramidal after the name of the planar pyramid's base.
So imagine that $E$ groups are invisible and try use this method if you  can't find it only with pyramids try to figure out which of the following figures is more close to the actual shape of the molecule.

The first two are platonic solids  (Octhaedron, and Tetrahedron) then there is a T for a T-shaped, and a bent figure. In the original wikipedia table respective molecule configuration is already rotated to the right position, I think you can do it on your own, with your 3D thinking.
For the last, seesaw shaped, there is a powerful "mem":

Here is an old photo of two boys in a $SF_4$ molecule.Good ol' times!

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text:
reaction between Iron(ii) and Vanadium(v)
When we reduce Vanadium (V) using iron(II), why are we adding 85% Phosphoric acid?
After the above reduction, if we are going to titrate the mixture with Potassium permanganate, why are we adding dipotassiumpersulfate to the mixture before the titratioin?

Here's the procedure:
  To 5 ml of vanadium (V) solution, add 2 ml of 6 M sulfuric acid, 1 ml of phosphoric acid and 0.20 g of hydrated ferrous sulfate to reduce vanadium (V). Then swirl the solution and allow it to stand for 3 minutes. Then add 0.23 g of dipotassiumpersulfate and let to stand for 5 minutes. Stir the contents vigorously and titrate with 0.02 standard potassium permanganate.

Could you provide either 1) a link to your procedure or 2) more detail about it? These questions are hard to answer out of context.

Procedure : to 5 ml of Vanadium (v) solution, add 2 ml of 6M sulfuric acid, 1ml of Phosphoric acid and 0.20g of hydrated Ferrous sulfate to reduce Vanadium (v). then, swirl the solution and allow it to stand for 3 minutes. then add 0.23 g of dipotassiumpersulfate and let to stand for 5 minutes. stir the contents vigorously and titrate with 0.02 standard potassium permanganate.

I think I realize (a part of) the answer, concerning the conditions of a redox titration of vanadium(V), for which a metavanadate, such as $\ce{NH4VO3}$ is typically used as starting material. Under acidic conditions, i.e. in the presence of 85% $\ce{H3PO4}$, the colourless $\ce{VO2+}$ is formed.
$\ce{VO3- + 2 H+ -> VO2+ + H2O} $
Again under acidic conditions, the latter species is reduced by Fe(II) to the blue $\ce{VO^{2+}}$.
$\ce{VO2+ + Fe^{2+} + 2 H+ -> VO^{2+} + Fe^{3+} + H2O} $
Is is conceivable that the peroxodisulfate is used to oxidize the excess of Fe(II), which otherwise would be oxidized by the permanganate?

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text:
Voltage vs current density - why are both parameters important to control an electrochemical reaction?
In many electrochemistry scholarly articles/studies they talk about certain specific reactions occurring at certain voltage potentials, while in other articles they talk about current density as the determiner for which reactions happen.
Why is this? From what I know about electrical engineering you can either set the voltage, and the current will adjust accordingly, or you can set the current and the voltage will adjust to meet the current. You cannot set both of them at the same time. That would defy the laws of physics.
So what do many of these articles mean when they talk about, for instance:

Reported reactions of this type often occur at potentials smaller than -3 V

How does this relate to the current density that is set by a power supply for instance?

Reaction rate in electrochemistry is limited by diffusion of reactants from solution towards electrode. Once all reactants in near proximity for given type of reactions are exhausted, and if voltage is still applied, other reactions (if voltage is strong enough) may occur. For example, consider positive electrode in $\ce{KCl}$ solution (leaving aside paired negative electrode). It is true, that $\ce{Cl^-}$ anion is oxidized easier then water. However, once all chloride near the electrode is oxidised, the next possible substrate to be oxidised is $\ce{OH^-}$ that is always present is water via reaction $\ce{2H_2O <=> H_3O^+ + OH^-}$. Thus, depending on the current density on the electrode you will have different major products.
In short: resistance of the electrochemical cell is a complex function of temperature, voltage, current and solution composition.
The current density in electrochemistry may be adjusted not only by adjusting voltage applied, but also by adjusting surface area of the electrodes used. Also, it is critical to maintain conditions for effective transportation of reactants towards electrodes.


text:
Why does hydrogen so easily take the place of oxygen in organic molecules during reduction reactions?
It goes without saying that every reduction agent has different reducing properties and strengths in terms of what sort of bond it can reduce/hydrogenate. But, why does it generally follow that hydrogen takes the place of an oxygen molecule during a hydrogenation/reduction reaction (in the presence of the necessary hydrogen or $\ce{H}$-donor of course)?
What is it about the properties of oxygen that makes it so ready to disassociate from the molecule when it gains an electron, allowing hydrogen to take its place? I'm sure there is something simple that I'm missing, but any explanation is appreciated.

Welcome to Chemistry.SE. Could you add an example of what you are referring to as "hydrogen taking the place of oxygen"? Most organic reductions (involving oxygen-containing functional groups) are of the type where a carbon-oxygen double bond is converted into a carbon-oxygen single bond: $\ce{R2C=O -> R2CH-OH}$.

Yes that is a relatively good example. There is also the instance of catalytic transfer hydrogenolysis where R2CH-OH is converted into R2CH3 by cleavage of the oxygen. Or other instances with a strong enough hydrogenating agent were R-COOH can be converted to RCH3. My question is what is it about the intrincis properties of oxygen and hydrogen that make hydrogen so readily want to take its place in a highly reductive environment?

But, why does it generally follow that hydrogen takes the place of an oxygen molecule during a hydrogenation/reduction reaction [...]

Ok, if at all, it's an oxygen atom ;)
I don't see that general trend towards complete deoxygenation, neither in catalytic hydrogenations, nor in hydride transfer reactions.
It usually takes

rather drastical conditions
quite some special reagents
a particular type of derivatisation

or a combination of any of these to convert oxygen-containing compounds to alkanes.
Let's have a look at some examples:
$\ce{R1-CO-R2 -> R1-CH2-R2}$

Ketones and aldehydes may be directly reduced to alkanes in the Clemmensen reduction. The "mildest" variants use anhydrous $\ce{HCl}$ and amalgamated $\ce{Zn}$ in organic solvents.
The clasical Wolff-Kishner reduction, using hydrazine and sodium in diglyme, involves the in situ functionalization of the carbonyl and isn't a prototype for a mild reaction either. Admittedly, milder variants exist, using N-tert-butyldimethylsilylhydrazone and catalytical amounts of a Lewis acid, e.g. $\ce{Sc(OTf)3}$
Tosylhydrazones may be converted to alkanes using $\ce{NaCNBH3}$, $\ce{Na(CH3COO)3BH}$, or boranes.
Catalytic hydrogenation over Raney nickel is possible after converting the carbonyl to a dithiane; so it's actually a desulfurization.

$\ce{R1-CH2-OH -> R1-CH3}$
I'm not aware of any transformation that does not involve functionalization.

Alkyl tosylates can be deoxygenated using $\ce{LiEt3BH}$.
Acylation of alkohols with thiocarbonyl chlorides furnish thiocarbonates, which may be deoxygenated under using either tin hydrides ($\ce{R3SnH}$, Barton deoxygenation) or $\ce{B(CH3)3}$ in water/benzene.
Alkanes may also be obtained from alkohols under Mitsunobu conditions ($\ce{Ph3P}$, $\ce{DEAD}$, o-nitrobenzenesulfonylhydrazine).

$\ce{R1-CH2-COOH -> R1-CH3}$
Direct transformations of carboxylic acids to alkanes typically involve decarboxylation, i.e. the loss of $\ce{CO2}$. This may be achieved either by the Barton decarboxylation, or via photoinduced electron transfer reactions with electronically excited (electron) acceptors. In the course of the latter reactions, a carboxylate is oxidized to an acyloxy radical, which then undergoes decarboxylation.
To sum it up: It's not that easy ;)

In the special case of reduction to an alcohol in acid (or other moiety that becomes a leaving group under reaction conditions), the resulting (stabilized) carbocation can be further reduced to the hydrocarbon (e.g., catalytic reforming using noble metal catalyst on acid zeolite under hydrogen).  Benzylic positions often reduce to (unwanted) hydrocarbon.  You can push this by using Lewis acid $\ce{AlH3}$ rather than $\ce{LiAlH4}$,
$$\ce{3 LiAlH4 + AlCl3 → 4 AlH3 + 3 LiCl}.$$
Another entry into deoxygenation reduction is to use a potent oxophile like titanium, e.g., McMurray reaction, that is astounding effective for coupling hindered carbonyls into olefins by plucking out the oxygens.  Its cousin the Tebbe methylenation does $\ce{R2C=O}$ to $\ce{R2C=CH2}$, convert a diaryl benzil's carbonyls into methylenes. Follow with olefin metathesis and ethylene extrusion (ADMET polymerization) ton give fully functionalized polyacetlyenes as living polymers, allowing block co-polymer builds, before quench.  What fun!  Here is an example with a p-phenylene spacer,


text:
why are Ca(OH)₂ and Al(OH)₃ less soluble than Na(OH) in water
Even though it has more $\ce{OH}$ groups and still it is very weak in terms of dissolving?
please clarify to me in terms of polar and non polar solubility, how to determine whether a compound is soluble or not in water.

$\ce{OH}$ is small anion and bonds effectively with small cations. Thus, in general, the bigger (and less specific charge) of the cation, the higher is solubility of the corresponding base in polar solvents, meaning that $\ce{Ca(OH)_2}$ and $\ce{Al(OH)_3}$ having much smaller and more charged cations are almost covalent in nature and thus, do not dissociate.
In general, solubility of salt in water is determined by energy balance between breaking of the ionic bonds and hydratation of anions formed. Thus, the most soluble salts are usually formed by small and big ions, that cannot bond effectively, but at least one of them can be hydrated effectively.
Please, note: solubility for molecular compounds, like sugars and some non-dissociating compounds that looks like salts (like $\ce{Cu(CH_3COO)_2*H_2O}$) is an entirely different matter and determined by relative strength of intermolecular interactions (read: the molecule have to be able to form hydrogen bonds)


text:
Configuration Interaction matrix composition from electronic configurations
The six electronic configurations below could all be found in the matrix elements of the CI matrix at CISD level. How do I construct the 6 x 6 CI matrix composed of these six configurations and which of the elements will be zero?
How different would it be if I used MP4(SDTQ) method?

Given those six configurations, the CI matrix with be a six-by-six matrix with rows and columns defined by the respective configuration. The matrix elements themselves will be the Hamiltonian expressed in those configurations.
For example, say column one corresponds to $\psi_0$, or $\left|0\right>$ in bra-ket notation. Similarly, row one corresponds to $\psi_0$, or $\left<0\right|$ in bra-ket notation. The CI matrix element where these coincide, namely $\left<0|H|0\right>$, where H is your Hamiltonian, is simply the ground state energy (the Hartree-Fock energy, if the configurations are the result of HF orbitals). The other matrix elements, like $\left<1|H|0\right>$ or $\left<0|H|2\right>$ have rules for determining them. The best reference is the book Modern Quantum Chemistry, by Szabo and Ostlund. Dr. David Sherrill has a good reference as well, which you can find here: http://vergil.chemistry.gatech.edu/notes/ci/ci.html
Okay, to answer the second part of your question. There are a few rules for determining CI matrix elements. One of the first ones is that singly excited configurations do not mix with the ground state. This is Brillouin's theorem. I don't see any singly excited configurations in the six you gave, so we won't have any of those. The other rule is that configurations that differ by more than two orbital occupations are zero. Since you truncated to doubles, you also won't have this case. The other thing you need to check is spin. If spin is not conserved, you'll have a zero matrix element. I think you'll have this case with the first configuration and the last one, which have different multiplicities.
As to your last question, I'm not real sure what you are asking. MP4(SDQ) is a perturbative method, which is different from CI, which is a variational method. In other words you do a CI by diagonalization, whereas MP4 is not. So I think CISDQ would be a better question...anyway, the point is that you are increasing your configuration space from singles and doubles (SD) to singles, doubles and quadruples (SDQ). The point I made in the preceding paragraph still holds, and expressing your Hamiltonian between ground state and quadruply excited configurations will be zero because their occupation differs by more than two. Same for singly excited configurations and quadruples.
EDIT: Just had the thought, you should also search for "Slater's rules" when trying to construct CI matrices. These are the rules I mentioned earlier.

Your comment was very helpful. I can summarize the rules for the CI matrix elements to be zero as this:

Brillouin's theorem i.e. matrix elements between HF ref and singly excited configurations is zero
If irreducible representation of two configurations are different, Hij is zero (the H operator belongs to the totally symmetric representation)
H does not include spin so if total spin of two determinants is different, matrix element for the will be zero.
H includes 1e and 2e operators so if two determinants are different by more than 2 MO, Hij =0

However, I am not sure how to apply these rules (particularly number 2) for the specific problem in the question. And the second part of the question is that since MPn is based on HF, can I assume that these rules apply for MPn matrix too or are they differently constructed?

Number 2 is based off the symmetry of the molecule. Basically interacting MOs and interacting excitations out of the ground state MOs need to have the same symmetry. Electronic structure calculations save a lot of time by doing smaller calculations on irreducible representations (irreps) instead of one larger calculation on the whole molecule. MPn is a little different, but I think it is the same. In its most general construction with HF canonical orbitals, you do have matrix elements between determinants/configurations, so the rules should apply.

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Is there a hydrophobic gas?
I am looking into the subject of water repelling and have been led to the hydrophobic subject area. I have two questions regarding this subject.

Is there a hydrophobic gas ?
If a hydrophobic liquid is heated to a gas does it loose its water repellent qualities.

Thanks for the help guys

The trouble with the question is that hydrophobicity is not a meaningful concept in a gas.
The concept is mostly useful when characterising how surfaces behave. Typical hydrophobic compounds don't interact strongly with water (e.g. non polar hydrocarbon waxes) and tend to disrupt water's hydrogen bond network. So water doesn't "wet" the surface and tends to form droplets that seem to be trying to minimise the contact area with the surface. The contact angle that a droplet makes with the surface is used as a measure of static hydrophobicity (if the droplet spreads out because the surface likes to interact with water, the angle is low, if the surface reels water the angle is high).
It is less clear that describing liquid phases as hydrophobic is meaningful, though it is not uncommon in casual usage. Compounds that would be hydrophobic as solids will probably exist as liquids that don't mix with water or can form micelles where they cluster together in very small droplets in the water phase. The bulk effect that matters is that hydrophobic compounds don't mix with water and form separate phases. The hydrophobicity of the molecules describes why the separation happens, but the effect is immiscibility.
None of this matters in the gas phase. The point of gases (compared to solids or liquids) is that the bulk properties are not determined by molecular interactions. The molecules of a gas don't interact much and are, to a good approximation, interchangeable whatever the individual molecules are made of. Gas equations describing gas behaviour can effectively ignore the molecular interactions.
What this means is that if you take a compound that would be hydrophobic as a solid and immiscible with water as a liquid, it will mix quite happily with water vapour. So a hydrophobic compound can be a gas, but a gas can't be hydrophobic.

Nice answer! Just out of curiosity: if you look at droplets in the Cassie state, suspended on top of pillars with air in between, they get a higher contact angle than when they are purely in contact with the solid. This is (at least in an averaging sense) explained by the contact angle of water with the air between the pillars which is taken to be 180 degrees. Wouldn't that warrant the use of the term hydrophobic gas? Or would you argue that any gas is hydrophobic in this sense, thus making the term hydrophobic redundant?

@Michiel The structure of the surface (that is, its physical not chemical makeup) makes a lot of difference (so the Lotus leaf is more hydrophobic than the chemicals it is made from). So i still wouldn't use the term hydrophobic gas but stick to descriptions based on chemistry and structure of the surface.

Sure many gases are hydrophobic! A hydrophobic molecule is in fact mainly a non polar molecule. So every non polar gas without $H$ bond site is a hydrophobic molecule. The alkanes for example are all hydrophobic molecules.

So in fact I will not say that if a hydrophobic liquid is heated to a gas  it loses its water repellent qualities, because in fact it is still a non polar molecule even if he has greater kinetic energy however in gas state molecules have a really low interaction between each other so it doesn't really have a strong influence in what happen if the polarity is different. For this reason I don't know if is really useful to speak of hydrophobic gas (even if someone use it see e.g. The interface between water and a hydrophobic gas) because you can't measure contact angle or use other tests usually perform to characterize hydrophobic materials.

a hydrophobic compound can be a gas, but is the gas therefore hydrophobic? Is the concept even meaningful in a gas?

@matt_black I fully agree, I think the same but indeed the term hydrophobic gas is accepted so why don't write what gases can be called so?

is the term hydrophobic gas accepted? I'd like to see an example and an explanation of what the term could possibly mean.

@matt_black In fact is not so common however see the link I've posted or this book!

A careful reading of the paper shows that the term "hydrophobic gas" only occurs in the context of a solid electrode coating on an electrode used in a system where a gas is one of the reactants. In context, the term refers to a solid hydrophobic layer on the surface of an electrode.

You are right for an example see the link in my answer, I will edit the question I thought was more common..

Consider some zero dipole moment, "obviously" hydrophobic gases:  $\ce{CH4}$, $\ce{Xe}$, $\ce{SF6}$.  All three readily form water clathrate hydrates,
http://en.wikipedia.org/wiki/Clathrate_hydrate


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Chemical Spill in Charleston, West Virginia
I live in Charleston, WV.  On 1/9/2014 we were told not to use our tap water for anything other than flushing the toilet after a storage facility leaked over 7,400 gallons of 4-methylcyclohexane methanol into the Elk river, less than two miles up-river from the water company's main in-take.
The CDC said that concentrations under 1ppm were safe.  After a week of bathing in my sink and drinking bottled water, my area was determined to be safe, however my water still has a strong odor.
There is not much information about this chemical's health effects.  While the chemical itself is low-toxicity, I wonder if it might react with items in the water supply to create other compounds which may be more toxic.
Specifically, would this chemical react with chlorine in the water, or cast iron, PVC, and copper in the pipes?  Does anyone have any experience with this compound?

The name MCHM or "4-methycycloheaxe methanol" (as newspapers reported it) is not entirely a systematic name and is a little ambiguous. The actual compound is better called  4-methylcyclohexylmethanol and the structure is shown below:
Specific details about its toxicity are not easy to find, though it has been patented as an air freshener ingredient. It has also been around for a long time and is used in the coal industry as a frothing agent in significant quantities so if it had some particularly nasty property we would probably know by now. Some safety information is available (pdf) on the related 1,4-cyclohexanedimethanol and that isn't particularly toxic. So there doesn't seem to be any strong reason to worry about it, though it will taint the flavour of water. There are no obvious paths for it to known nasties in low concentrations in water, so the biggest worry is probably the taint.
Having said all this, the company should not have let it get into the water supply.


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Can TBE electrophoresis buffer be stored in a PETE container?
I'm doing a science project on agarose gel electrophoresis and would like to know if it is safe to store Tris/Borate/EDTA buffer solution inside a PETE container since I will be making about 10L of the buffer solution and do not have access to a spigoted carboy or any such container.

Are you planning to make 10l of 1x TBE? If so, make the solution 10x concentrated (take the ingredrients for 10l and dissolve them in 1l), than you can easily store your buffer stock in a 1l bottle. Dilute portions of it 1:10 to make the working solution. But you can also store 10l of working solution in a 10l container, we have such a reservoir in our lab.

Do you have glass containers? We've stored TBE in glass bottles for years with no problems. If you don't have a large glass bottle on hand, get some juice or something from the store, drink it, and clean that bottle out real good with distilled water.


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How do I determine crystal field splitting (for example, for [Ni(CO)4])?
How to know how many electrons are there in $T_{2g}$ and $E_g$ orbitals?
I think rather than a broad answer, an example would help. So, consider the compound $\ce{Ni(CO)4}$ and how to calculate CFSE for it?

t2g6 eg2 for strong field ligand(co) CFSE : (-4×x) +(6×y) =(-4×6) +(6×2) =-12Dq=-1.2∆o

Firstly, what geometry does the complex have?
Once you know the geometry, find the spin-state of the $d$-electrons, as well as the number of $d$-electrons applicable.
Find the crystal field diagram relevant to your geometry. Fill in the $d$-electrons according to the Aufbau principle and to Pauli's exclusion principle.
For your example, tetracarbonylnickel has a tetrahedral geometry and is a $d$10 compound. Carbon monoxide is a strong field ligand causing a high spin complex (note that since $d$ = 10, all of the $t_{2}$ and $e$ orbitals are full and hence spin-state is irrelevant).

As per the Aufbau principle, you start building up from the bottom, placing two electrons in the bottom $t_2$ orbitals and three in the top three $e$ orbitals, giving:
$e$: ↑ ↑ ↑
$t_2$: ↑ ↑
The remaining five electrons go in each half-filled orbital to give 2 filled $t_2$ orbitals and 3 filled $e$ orbitals:
$e$: ↑↓ ↑↓ ↑↓
$t_2$: ↑↓ ↑↓
To find the CSFE, the $t_2$ orbitals each stabilise by $\frac{3}{5}$ and the $e$ destabilize by $\frac{2}{5}$. Since $2 * \frac{3}{5} - 3 * \frac{2}{5}$ = 0, the CFSE = 0.
For octahedral complexes, the $e_g$ and $t_{2g}$ orbital populations can be used analogously to find the stabilisation energy.


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Why is the pressure value altered in this gas law numerical?
When I work through the following problem it is unclear to me why 0.00120 mmHg is converted to 0.00332 mmHg.

What is the temperature of 0.00332 g of mercury in the gas phase with pressure of 0.00120 mmHg and a volume of 435 L?

I converted the grams of mercury to moles and then used the ideal gas law constant of 62.36 to solve for T. Is there any reason that the pressure of 0.00120 mmHg should be converted to 0.00332 mmHg?

Your tutor is a master of confusion :-D
Would the exercise be easier if the pressure was given in stones per square acre? ;-)
Write down the equation for the ideal gas law again. Pay attention to the units!
Don't let the mercury fool you ;-)

Uhh... Square acres?  Better to give lengths in knot-seonds...


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Calculating the molarity of DNA in a cell

In the following questions use a value of 3 for $\pi$, $6 \times 10^{23}$ for Avogadro’s number and $660$ for the molecular weight of $\pu{1 bp}$ of DNA. The volume of a sphere of radius $r$ is $4/3\,πr^3$. A bacterium has a single copy of a $\pu{4 \times 10^6 bp}$ circular genomic DNA.
If the diameter of this spherical cell is 1 micrometer, what would be the molar concentration of DNA in this cell?

The volume comes to be $\pu{2 \times 10^{-5} L}$.
Amount of substance is $\pu{6.7 \times 10^-18 mol}$.
Dividing this by the volume, my answer comes $\pu{3.3 \times 10^{-13} M}$, but the answer given in my book is $\pu{3.3 \times 10^{-9} M}$.

Surely as a polymer of indeterminate length, the molar concentration of DNA is an entirely meaningless phrase?

If the stated bacterium's cell has a diameter of $\pu{1 \mu m}$, the volume can be derived in terms of liters remembering the linear relation between cubic meters:
$$
  V_\text{cell}
    =\frac{4}{3}\pi \left(\pu{0.5\times10^-6 m}\right)^3
    =\frac{4}{3}\pi \left(\pu{0.5\times10^-5 dm}\right)^3
    =\pu{5\times 10^-16 L}
$$
Inside this volume, the organism contains a certain number of base pairs; so the total amount of substance has to be known.
From Avogadro's Number, a mole is defined to be that portion (number of particles) of every substance in a defined physical phase; since this is an aqueous solution, the volume, as well as the molecular weight is meaningless: if one mole is defined by a certain number of particles, a different number of particles defines a different amount of substance:
$$
  n_{\text{tot bp}}
  =\frac{N_{\text{bp}}}{N_{\text{A}}}
  =\frac{\pu{4\times 10^6 molecules}}{\pu{6\times 10^23 molecules mol-1}}
  =\pu{6.7\times10^-18 mol}
$$
Then, by the definition of molarity, dividing the amount of substance contained inside the cell by its volume gives a decent number, for a cell:
$$
  M_\text{DNA}
  =\frac{n_\text{tot bp}}{V_\text{cell}}
  =\frac{\pu{6.7 \times 10^-18 mol}}{\pu{5 \times 10^-16 L}}
  =\pu{1.34 \times 10^-2 M}
$$
I think that the wrong result is due to the volume, because to me it seems rather surprising that one sphere of $\pu{1 \mu m}$ in diameter has:
$$
  \pu{2\times 10^-5 L}
  =\pu{2\times 10^-2 mL}
  =\pu{20 \mu L}
  =\pu{20 mm3}
  \neq \pu{5 \times 10^-8 mm3}
$$
of occupied volume. I suspect that something went wrong with the conversions, because I don't see (for now) any errors in my derivation.

I found the mistake in my volume of 2X10^-5 L.

Although I cannot find any mistake in your calculation, your answer does not match the one given in my book !

@biogirl My suggestion is to tell your professor about this mistake, show the calculation proposed and let him give you the correct advice. Maybe you should take the problem backwards with the given result: since the volume calculation is correct, there should be something missed with the moles. I'll try by myself and edit my answer if any breakthrough comes up.

There is one particle of dsDNA in the cell. Divide by $N_\mathrm{A}$ to get the amount of substance, and divide by the volume to get the concentration matching the concentration given in the answer.
\begin{align}
  V_{\mathrm{cell}}
    &= 4 \cdot {(\pu{0.5 μm})}^{3}\\[0.5ex]
    &= \pu{0.5 μm3}\\[0.5ex]
    &= \pu{5E-16 L}\\[3.5ex]
  N_{\mathrm{A}}
    &= \pu{6E23 1//mol}\\[3.5ex]
  n_{\mathrm{DNA}}
    &= \frac{1}{N_{\mathrm{A}}}\\[0.5ex]
    &= \pu{1.7E-24 mol}\\[3.5ex]
  c_{\mathrm{DNA}}
    &= \frac{n_{\mathrm{DNA}}}{V_{\mathrm{cell}}}\\[1.5ex]
    &= \frac{\pu{1.7E-24 mol}}{\pu{5E-16 L}}\\[1.5ex]
    &= \pu{3E-9 mol//L}
\end{align}

@Martin Nice vertical alignment, but what about the vertical space? OK, I looked it up: https://tex.stackexchange.com/a/228609

Keep in mind that MathJax only borrows a small subset of commands from (La)TeX, the complete overview is in the documentation: http://docs.mathjax.org/en/latest/input/tex/macros/. Manual spacing is not always a good idea, but in this specific situation it's perfectly fine, because MathJax tends to look different depending on the locally installed fonts. Maybe also interesting (loads of MathJax = slow): Hidden points of editing you probably didn't know

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


text:
Spices are absorbed or adsorbed in food?
When we cook food, we add several spices and condiments which enhance color, taste and smell. Are these spices and condiments just adsorbed on the food particles? Or are they absorbed? For example using pepper while cooking a piece of chicken, would it just stay on the top of the leg (adsorbed)? Or get absorbed?

The answer varies with the type of food and the type of condiment. Both are very complicated mixtures, so a same combination of food and condiment can display varying levels of adsorption/absorption depending on the region/compound being analyzed, but in general, I expect that while all of chemical/physical adsorption and chemical/physical absorption happen to some extent (cooking is quite a messy collection of physical and chemical interactions!), either physical absorption is the most prevalent, or none of them happens to the largest part of the spices.
First of all, adsorption is a surface phenomenon, which fundamentally limits how much matter can be considered adsorbed onto an object. Adsorbed films are usually very thin (probably not much more than a few nanometers thick, often a single molecule thick). Regular macroscopic objects have very little surface area relative to their volume, so it takes a comparatively tiny amount of substance to completely cover the object in a thin adsorbed layer, compared to say, activated charcoal. Thus we can likely exclude adsorption as the main factor.
It's also important to consider that most food is cooked with some amount of solvent (water or oil/fat), and in fact solvents come out of the food while it's being cooked. This means that a lot of the spices can very well end up as part of a liquid (albeit often quite concentrated) solution, which remains stuck on the surface simply due to its high viscosity and significant friction with the uneven surface of the food. I bet this is what goes on in most cases.

Of course, many foods consist of more than one phase. Meat, for example, contains both fat and water. some spices are fat soluble, others water soluble, so it isn't uncommon for the valour to depend on which phase the spice flavour ingredients are dissolved in.


text:
Calculating mass of krypton from mole fraction

There are two gases in a container: krypton and carbon dioxide. If the mass of the gases is 35 grams, and total pressure of the container is 0.708 atm, and the pressure of krypton is 0.250 atm. What is the mass of the krypton?

I've found the mole fraction of krypton is 0.3531. First, I think if I multiply mole fraction of krypton and mass of the gases, I'd get the mass of krypton. But my teacher said if the answer is false. So what is the correct answer?

I think your method is correct.

Let $m_{\ce{CO2}}$ be the mass of carbon dioxide and $m_{\ce{Kr}}$ be the mass of krypton in the vessel. These two are related by:
$$m_{\ce{CO2}}+m_{\ce{Kr}}=35\ \text{g}\ \ \ \ \ (1)$$
If we can get a second equation in both variables, you can solved the system for each variable.
You were able to determine the mole fractions $\chi$:
$$\chi_{\ce{Kr}}=\frac{n_{\ce{Kr}}}{n_T}=\frac{P_{\ce{Kr}}}{P_T}=\frac{0.250\ \text{atm}}{0.708 \ \text{atm}}= 0.353\ \ \ \ \ (2)$$
$$\chi_{\ce{CO2}}=1-\chi_{\ce{Kr}}=0.647 \ \ \ \ \ (3)$$
Let $n_{\ce{CO2}}$ be the number of moles of carbon dioxide and $n_{\ce{Kr}}$ be the the number of moles of krypton in the vessel (its a little easier to work in moles). We can generate two relationships between the numbers of moles using 1) the masses and 2) the mole fractions.

From the masses, since the mass of a sample of substance is equal to the number of moles times the molar mass (or formula weight).

$$m_x=n_x\cdot FW_x\ \ \ \ \ (4)$$
$$m_\ce{Kr}=n_\ce{Kr}\cdot FW_\ce{Kr}=n_\ce{Kr}\cdot (83.798\ \text{g/mol})\ \ \ \ \ (5)$$
$$m_{\ce{CO2}}=n_{\ce{CO2}}\cdot FW_{\ce{CO2}}=n_{\ce{CO2}}\cdot (44.01\ \text{g/mol})\ \ \ \ \ (6)$$
$$n_{\ce{CO2}}\cdot (44.01\ \text{g/mol})+n_\ce{Kr}\cdot (83.798\ \text{g/mol})=35\ \text{g}\ \ \ \ \ (7)$$

From mole fractions

$$\dfrac{\chi_{\ce{CO2}}}{\chi_{\ce{Kr}}}=\dfrac{\dfrac{n_{\ce{CO2}}}{n_T}}{\dfrac{n_{\ce{Kr}}}{n_T}}=\dfrac{n_{\ce{CO2}}}{n_{\ce{Kr}}}=\dfrac{0.647}{0.353}=1.832\ \ \ \ \ (8)$$
$$n_{\ce{CO2}}=1.832\cdot n_{\ce{Kr}} \ \ \ \ \ (9)$$

Substitute equation (9) into equation (7) and solve for $n_{\ce{Kr}}$ and convert to $m_{\ce{Kr}}$.


text:
What does it mean by degree of aromaticity of a molecule?
This is a fundamental question, though I didn't really get the meaning/idea what does it mean by aromaticity or degree of aromaticity. Does it mean the ability of a molecule to be aromatic?

Welcome to Chemistry.SE! Could you provide a little more context? I suspect that this term refers to the amount of aromatic resonance stability compared to benzene. For example, furan (despite meeting the Huckel requirements) is less aromatic, evidenced by its ability to act as a diene in a Diels-Alder reaction

Aromaticity might be viewed as resonance stabilzation energy, NMR shift, or altered reactivity.  Here is a way of measuring its change among related systems,
Chem. Commun.  48, 8144 (2012)
J. Am. Chem. Soc. 131, 189 (2009)
J. Am. Chem. Soc. 127, 16251 (2005)
J. Org. Chem. 67, 8798 (2002)
J. Am. Chem. Soc., 124, 13495 (2002)
Chem. Rev. 101, 1301 (2001)
Tetrahedron 57, 3689 (2001)
http://nptel.ac.in/courses/IIT-MADRAS/Engineering_Chemistry_III/pdf/1_Aromaticity.pdf
A more practical approach derives from MO theory and Hund's rule of maximum multiplicity.
http://www.youtube.com/watch?v=U6FX5RfLjd0
http://www.youtube.com/watch?v=H0so_NJ2aVw
Frost circle

I'm currently studying this and...
Degree of aromaticity can be assessed by many methods, both thermodynamic and spectroscopic but one of the favoured these days is actually computational. One of the characteristics of an aromatic system is that it gives rise to ring currents which then cause significant deshielding of the aromatic protons. This can be exploited by a technique called “Nuclear Independent Chemical Shift calculations”, or NICS. In this computational method, the shielding parameter (the CS bit) for an arbitrary dummy nucleus (the NI bit) in the centre of the ring is calculated. If the shielding parameter is negative, then the ring is aromatic, if the shielding parameter is positive, it is anti-aromatic and if zero, it’s just not a delocalised system.


text:
Ionic Product of Water
Why is the ionic product for water ($K_\text{w}$) temperature dependent? The concentration of $\ce{H+}$ and $\ce{OH-}$ ions in pure water is $[\ce{H+}] = [\ce{OH-}] = 1.0\times 10^{-7}\ \mathrm{mol/l}$ at $25\ \mathrm{^\circ C}$, and therefore the ionic product of water at $25\ \mathrm{^\circ C}$ is $1.0\times 10^{-14}$. Why it is so? What is the effect of temperature on this value?

Equilibrium constants are related to free energy and free energy is strongly dependent on temperature.

Wikipedia has some useful graphs regarding the dependence of the ionic product of water on temperature, pressure and amount of dissolved ions.

I think this link should really help you out in understanding how the equilibrium constant depends on temperature.
Now, considering the case of water dissociation:
$$\ce{H2O <=> H+ + HO-}$$
We can calculate the free energy of the reaction ($\Delta_\text{r}G$), using the Gibbs energy for the reactant and the products (in aqueous solution):
$$\Delta G(\ce{H2O}) = -237.13\ \mathrm{kJ/mol}$$
$$\Delta G(\ce{H+}) = 0\ \mathrm{kJ/mol}$$
$$\Delta G(\ce{HO-}) = -157.24\ \mathrm{kJ/mol}$$
$$\Delta_\text{r}G = \Delta G(\ce{H+}) + \Delta G(\ce{HO-}) - \Delta G(\ce{H2O}) = -157.24\ \mathrm{kJ/mol} + 237.13\ \mathrm{kJ/mol} = 79.89\ \mathrm{kJ/mol}$$
The values of the Gibbs energies were taken from the 8th edition of Atkins' Physical Chemistry.
Now, applying the formula below (which can be obtained from $\Delta_\text{r}G = -RT\ln K_\text{eq}$) for $T = 298\ \mathrm{K}$ (or $25\ \mathrm{^\circ C}$):
$$K_\text{eq} = \operatorname{e}^{-\frac{\Delta_\text{r}G}{RT}}$$
we get something like this :
$$K_\text{eq} = 2.71^{-\frac{79890\ \mathrm{J\ mol^{-1}}}{8.3145\ \mathrm{J\ mol^{-1}\ K^{-1}} \times 298\ \mathrm{K}}} = 9.93 \times 10^{-15}$$
This value can be approximated to ${10^{-14}}$.

An easier approach would be that

concentration of hydromium and hydroxyl ions will increase with
  temperature

as tendency of

water to dissociate will increase

Hence, more quantity of products means greater Ionic Product.

This is much oversimplified.  The Wikipedia graphs referred to in Nicolau's comment shows a maximum at constant pressure, beyond which the ion product drops.  Decreasing density, which kills the hydrogen bonding, eventually overrides the temperature effect we would normally expect.


text:
What is the nature and character of fire?
What is fire? Is it a material or energy or something else? Please explain it to me. My friend asked me this. I am stuck with this problem.

See here for an excellent overview of the chemistry of fire from user Ben Norris and here for some insights from the great Feynman.

I think you should read the question linked by Nicolau if this don't fulfill your thirst of knowledge try to be more specific... ;-)

@Neto why not to post this comments as an answer? So the question will be no longer appear as an unanswered question.


text:
Is tea powder better dissolved when put in hot water?
I usually make tea by adding two spoons of tea powder to water at room temp. and start to heat it. My friend said it will be better to add the powder once the water starts boiling. She said that it will dissolve tea more easily because of the sudden temperature difference experienced by tea powder.
Does it make a difference? Why?

I would say that your method is actually quicker (if that's what you mean by 'better'). My reasoning is as follows: the solubility of the powder in water increases with increasing temperature, as does the dissolution rate but the dissolution rate remains finite all the time.
1) If you add the powder and start heating at $t=0$ the concentration of dissolved powder will start to increase from $c=0$ to $c>0$ immediately. Once the water reaches the boiling point at $t=t_b$ already a certain amount (maybe even all) of the powder has dissolved i.e. some concentration of dissolved powder is present $c=c_1$.
2) If you boil the water first and then add the powder you start with $c=0$ at $t=t_b$, which is a lower concentration than in the powder-first case.
Because there is already more powder dissolved at the boiling point ($c=c_1$ vs. $c=0$ at $t=t_b$), the moment at which all powder is dissolved comes sooner for the powder-first case (situation 1).
Note that I approach this question purely from the physical point of view, I have no clue whether the two methods might result in differently tasting tea.
P.S. I ignored any possible changes in the boiling point due to the already dissolved powder. For low concentrations, as in the case of tea, this is no problem. If the concentrations become much higher then the change in boiling point might become dominant and the story changes

Essentially if put your tea in the room temperature water, the total time to dissolve will be longer, but the time to dissolve from the beginning of boiling would be shorter if you add your tea to room temperature water.

Your friend is right. As the temprature increases, the solubility of tea increases. Hence your tea will dissolve more easily when the water is heated before adding the tea powder, rather than your method.

This doesn't explain why heating first then adding powder would be better then adding powder then heating. The solubility/temperature effect is there in both situations. Could you clarify why the heating then powder would be better?!


text:
Is hydrogen peroxide safe for cleaning porcelain?
From my understanding hydrogen peroxide only reacts with bacteria in a way that it should burn them more or less.
If that is true, can I then safely use hydrogen peroxide to clean my bathroom sink which are made of porcelain?

While hydrogen peroxide is not normally used for cleaning surfaces, there is nothing in normal glazed porcelain that it would damage. Porcelain is about as unreactive to most things as glass, which is unreactive enough to be used for the bottles that store hydrogen peroxide.
But you should probably be careful with it as it can be dangerous if the solution is a strong one. Normally available hydrogen peroxide should not be as strong as the solutions used in the lab, though, which is often a 30% solution in water and should only be handled if you know what you are doing (see the safety data sheet). Consumer hydrogen peroxide is usually a 3-6% solution and isn't any more dangerous that hypochlorite bleach (and is a safer way to bleach hair).
Standard household (hypochlorite) bleach is usually better for cleaning, though.

I find the smell of household (hypochlorite) bleach to be pretty disagreeable (the ones with extra smell more so) so the ideas has some appeal.  The 3 - 9% (10, 20 and 30 volume) solutions usually available are pretty safe for surfaces.  They may bleach some dyes or materials if left for a long time much like bleach would.  Glass and ceramic should be amongst the least reactive surfaces as mentioned above.  A bonus is that it (pretty much) leaves no residue on evaporation.  It may react with some metals but of little consequence unless left for a long time, just rinse or wipe any puddles away.


text:
Is it possible to use diatomic carbon to utilize a new type of fuel that takes away the waste product of carbon dioxide being released into the air?
These days, we as a community what to polish and make our world a greener environment. I found a way to accomplish this task, however I don't know if it is applicable. I am wondering if the use of diatonic carbon would do the trick to produce a process that could regenerate the fuel in our vehicles and eliminate the releasing of carbon dioxide into the air. The purpose of this reaction would be to remove the Carbon Dioxide waste that is created by gasoline. Gasoline waste contains one less carbon than was originally in the reaction
(UNDER CONSTRUCTION)

Welcome to ChemistrySE! I think in the body of your question you should explain your question  title (not very clear in my opinion :-) not global warming!

Would you mind to elaborate the outcome of a reaction between C2 and carbon dioxide?

Wow - on several levels. No. First Law of Thermodynamics: You cannot win. Second Law of Thermodynamics: You can only break even on a very cold day. Third Law of Thermodynamics: It never gets that cold.

The answer rests on what diatomic carbon is. I have no idea what it is. But, if such a thing could exist, the issue would be: does it require more energy to create than any other way of absorbing or reacting carbon dioxide? Unless diatomic carbon were a free resource, it seems unlikely there would be any net environmental gain.

No doubt, diatomic carbon does exist - in blue flames and in comets.
The characteristic uv absorption pattern (Swan bands) has already been described almost 100 years ago.
At low temperature (77 K) it reacts with hydrocarbons by H-abstraction to form $\ce{HC#CH}$, and the addition to $\ce{C=C}$ double bonds has been reported too.
But I can't figure out any reasonable way for a reaction with carbon dioxide.


text:
Lemon juice and carbonic acid reaction
I was doing a chemistry project to see if the pH of a carbonated solution effects the time taken for it to settle after being disrupted, and I am trying to map out the reaction when the components are combined.
I was able to figure out a few, but I got stuck on carbonic acid and lemon juice.  For some reason, I was getting ethane as a by product of the reaction, and I'm sure that's not right.  Could someone explain what I'm doing wrong?
the formula is $\ce{H2CO3 + C6H8O7}$

I'm assuming $\ce{C6H8O7}$ is citric acid. Acid + acid does not necessarily do anything productive.

if you're getting ethane there is something wrong here.  are you sure it is lemon juice?

You should show what have you done...In my opinion there are no reactions between these two acid. When you make a lemonade you mix lemon juice that has a bit of citric acid with sparkling water that has carbonic acid  and you don't smell ethane.

Would you mind to explain how you determined that ethane was produced?

These two weak acids have similar disassociation constants in water.  A low Ka indicates a relatively small number of acid molecules will form an anion stabilized by water.  Two weak acids with similar disassociation constants will coexist in a solution without interacting.  If two weak acids have very different Ka then there can be a reaction.  In this case nothing would happen.   For more generalized information search for mixture of weak acids.

Carbonic Acid

$$\ce{H2CO3 + H2O <=> HCO3- + H3O+}$$
$$K_a = 4.3 \times 10^{-7} = \frac{[\ce{H3O+}][\ce{HCO3^-}]}{[\ce{H2CO3}]}$$

Citric Acid

$$\ce{H3C6H5O7 + H2O <=> H2C6H5O7- + H3O+}$$
$$K_a = 1.8 \times 10^{-6}  = \frac{[\ce{H3O+}][\ce{H2C6H5O7^-}]}{[\ce{H3C6H5O7}]}$$

By Le Chatelier's principle ,a new equilibrium will be formed for the H3O+ concentration , since H2CO3 has the lower dissociation constant.H3O+ concentration and therefore the pH will lie somewhere between the two acids


text:
Why can we smell copper?
If I can smell an object, it means that molecules of it are getting separated from it, so they can reach my nose. As far as I know, metals don't sublimate, especially not in room temperature. However, copper has a very strong and characteristic smell. Does it mean that copper will degrade pretty quickly, or are just we so sensitive to it that a few molecules are sufficient? I assume it has to do something with oxidization, but it doesn't oxidize that much naturally as other metals, for example, iron.

Related: Why does my blood taste like rust?

I looked for an answer since ever (but especially with iron). Meantime I have studied chemistry. And I am still looking for an answer (I assumed dust and body mucose play a rolel. Now I start to scroll down with hope :)

This is a nice question, as it confronts a very replicable and common experience with a well established yet seemingly contradictory fact. As you expected, the smell of metal has nothing to do with the metal actually getting into your nose, as most metals have far too low of a vapor pressure at ordinary temperatures to allow direct detection. The characteristic smell of metal, in fact, is caused by organic substances!
There has been the focus on the specific case of the smell of iron (free-access article!). There are at least two ways in which iron produces a metallic smell. Firstly, acidic substances are capable of corroding iron and steel, releasing phosphorus and carbon atoms present in the metal or alloy. These can react to form volatile organophosphorus compounds such as methylphosphine ($\ce{H3CPH2}$ which have a garlic/metallic odor at small concentrations. From the article:

The “garlic” metallic odor (see Supporting Information) of the gas product from the acidic dissolution of cast iron is dominated by these organophosphines. We measured an extremely low odor threshold for two key odorants, methylphosphine and dimethylphosphine (6 and 3 ng P/m³, respectively, garlic-metallic odor), which belong therefore to the most potent odorants known. Phosphine ($\ce{PH3}$) is not important for this odor because we found it has a much higher odor detection threshold (>10⁶ ng/m³). A “calcium carbide” (or “burned lime”/“cement”) attribute of the general “garlic” odor is probably caused by unsaturated hydrocarbons (alkynes, alkadienes) that are linked to a high carbon content of iron (Table 1, see Supporting Information).

Also, it turns out that $\ce{Fe^{2+}}$ ions (but not $\ce{Fe^{3+}}$) are capable of oxidizing substances present in oils produced by the skin, namely lipid peroxides. A small amount of $\ce{Fe^{2+}}$ ions are produced when iron comes into contact with acids in sweat. These then decompose the oils releasing a mixture of ketones and aldehydes with carbon chains between 6 and 10 atoms long. In particular, most of the smell of metal comes from the unsaturated ketone 1-octen-3-one, which has a fungal/metallic odour even in concentrations as low as $1\ \mu g\ m^{-3}$ . In short:

Sweaty skin corrodes iron metal to form reactive $\ce{Fe^{2+}}$ ions that are oxidized within seconds to $\ce{Fe^{3+}}$ ions while simultaneously reducing and decomposing existing skin lipid peroxides to odorous carbonyl hydrocarbons that are perceived as a metallic odor.

In the supporting information for the article (also free-access), the authors describe experiments performed with other metals, including copper:

Comparison of iron metal with other metals (copper, brass, zinc, etc.): When solid copper metal or brass (copper-zinc alloy) was contacted with the skin instead of iron, a similar metallic odor and GC-peak pattern of carbonyl hydrocarbons was produced and up to one μmole/dm² of monovalent cuprous ion [$\ce{Cu+}$] was detected as a corrosion product (Supporting Figs. S3 to S6). Zinc, a metal that forms $\ce{Zn^{2+}}$ but no stable $\ce{Zn+}$, was hesitant to form metallic odor, except on very strong rubbing of the metal versus skin (that could produce metastable monovalent $\ce{Zn+}$). The use of common color-tests to demonstrate directly on human palm skin the presence of low-valence ions (ferrous and cuprous) from the corrosion of iron, copper and brass alloys is shown in Supporting Figure S6. Alumina powder rubbed on skin did not produce significant odorants. These results provide additional evidence that it is not metal evaporation, but skin lipid peroxide reduction and decomposition by low valence metal ions that produces the odorants.

The last paragraphs of the article summarize the findings:

In conclusion: 1) The typical “musty” metallic odor of iron metal touching skin (epidermis) is caused by volatile carbonyl compounds (aldehydes, ketones) produced through
the reaction of skin peroxides with ferrous ions ($\ce{Fe^{2+}}$) that are formed in the sweat-mediated corrosion of iron. $\ce{Fe^{2+}}$ ion containing metal surfaces, rust, drinking water, blood etc., but also copper and brass, give rise to a similar odor on contact with the skin. The human ability to detect this odor is probably a result of the evolutionarily developed but largely dormant ability to smell blood (“blood scent”).

The “garlic-carbide” metallic odor of phosphorus- and carbon-rich cast iron and steel under attack by acid, is dominated by volatile organophosphines. Corroding cast iron
is an environmental source of C–P compounds that may lead to confusion in the verification and monitoring of the Chemical Weapons Convention (see also ref. [15])

As an aside, this may be why sometimes people recommend getting strong smells off your hands by rubbing them against a metal object. While it probably doesn't work for some metals and for some smelly compounds, it's possible that the metal catalyzes the decomposition of the malodorous substances into less strongly smelling ones.
You can read a little more in this press article on the study.

But blood tastes just like pure metallic iron 'tastes' - so how is that then explained?

@LievenB Another good question, which I thought might pop up eventually. I don't really know. Maybe the iron cations in blood can also catalyze the formation of organic compounds with a metallic smell from lipids or other substances present in the blood and mouth. Flavour comes mostly from smell, not taste (oddly enough).

I've found this very interesting link : http://phys.org/news80405535.html and I'm guessing that it's the same too with other metals such as copper,...

@LievenB The paper cited in that phys.org news article is the very same as the one cited by my Nature news article!

@LievenB It turns out the Angewandte Chemie paper had touched on the smell of blood. Also, I don't recommend trusting the phys.org link too much. For example, it mentions that iron ions have a negative charge, strangely.

Smell is associative: we smell "sweet" food not from the sugar, but from the volatile compounds that occur when sugar breaks down under heat. Only because we experienced the smell paired with tasting the sugar (there is still plenty of un-broken-down sugar left) do we think it smells sweet.


text:
Lifetime of a (reversible) hydrogen fuel cell system
I am trying to determine how long a system that constantly converts water into H and O and then back will last, as in roughly how many cycles. Now this system can either be made of say a reversible fuel cell or a non-reversible fuel cell combined with a dedicated electrolysis machine (if there is a difference in lifetime). As far as I can tell, platinum is used (almost) everywhere as catalyst and as far I can determine it seems to corrode somewhat throughout use, especially in terms of splitting the oxygen in the fuel cell. I am no professional though and I can only find limited information so I would appreciate it if someone could please tell me roughly how long a system like this will last and how.

Fuel cell manufacturers quote a range of different lifetimes which depend on many factors. Backup cells tend to have lower lifetimes as the start-stop process degrades them compared to continuous operation. On the other hand, use as engines in vehicles requires a reasonably long life.
Impure fuel will damage the platinum (CO for example will poison the catalytic surface). But platinum is recyclable so either poisoning or other factors won't necessarily be a fundamental barrier to use (some companies take used cells back at the end of life and reuse the platinum). Membranes and electrodes also degrade mechanically depending on duty cycle and any mechanical stabilisation in the design.
Nedstack quotes lifetimes ranging from 2,000 hours for backup cells (which presumably means years of actual operation as they only get used when other power is cut which isn't a routine happening) and 8,000-16,000 hours for continuous operation (which is two years). Tests of fuel cells in vehicles have reported lifetimes of 7,500 hours (which presumably relates to several years of use in practice), though they seem to need more maintenance than diesel engines.

Poisoning of the catalyst, e.g. by $\ce{CO}$, is only one aspect that reduces the lifetime of PEM fuel cells.
The Nafion polymer membrane can suffer due to mechanical stress as a result of rapid temperature changes or dehydration.

Fuel cells don't convert water into H and O.  The generation of hydorgen gas is done at some point prior to being introduced into the cell (as in the case of hydrogen combustion engines).  The first stage of the fuel cell uses a catalyst, like platinum, to split hydrogen gas (H2) into atomic hydrogen.  Since the catalyst is a conductor and is sandwiched next to a proton-permeable membrane (proton exchange membrane), the electrons from the hydrogen are also separated and conducted to the final stage of the cell.
If the fuel cell itself was generating hydrogen from H2O, the system as a whole would generate no net energy--all of the energy obtained from the hydrogen reaction at the end would need to go back into separating the H2O in the first place.

Welcome to chemistry.se! While these are all interesting and informative (and true) facts, they unfortunately do not provide an answer to the question.

I have actually purchased and run three fuel cell stacks. One was dead on arrival and never converted any hydrogen to electricity. If you ever buy one, you will notice no warranty, no return policy and no expected useful life. This led us to actually invent a way to probe PEMFC health without putting hydrogen in, and we got a patent on that. The second fuel cell ran for about 40 hours, but the performance degraded quickly over that time. The third one also degraded and died within 200 hours. These were commercial stacks used with extreme care in a research lab facility by expert engineers. Given that demonstration cars and buses also don't last very long, I think we know why fuel cells are not going anywhere. The nature of the degradation is such that there isn't much that can be done to make the PEMFC more robust. See publication by S. Page in Journal of IEEE.


text:
Suggest methods and basis sets for a variety of systems
Please help me with any/all of the cases below. In the following cases, the named method and basis set are not suitable for the chemical systems. Why aren't they? Could you suggest a suitable method/basis set for each?

1. MP2/6-31G(d) for the calculation of reaction energy of the reaction below (H2C: should be calculated in the triplet state):

MY ATTEMPTED ANSWER

The carbene H2C: has two singlet configurations that contribute to the ground state energy, so we need a multireference method to recover the single excitations. MP2 is not good for this open-shell systems since they have large spin contamination and thus large error. The basis set also needs to include diffuse functions for electron correlation methods.
2. CISD/cc-pVTZ for the calculation of the reaction energy of:

MY ATTEMPTED ANSWER

We have bond elongation in the reactant to the point of breakage. CISD is not size-consistent. It recovers less electron correlation energy for the larger reactant than for the products since the reactant is larger, so CISD cannot be used. The basis set however is fine. I don't know what method can work best for this system but would go arbitrarily with CCSD(T).
3. BLYP/6-31G(d) for geometry optimization and calculation of binding energy of the neon dimer, $\ce{Ne\bond{...}Ne}$.

MY ATTEMPTED ANSWER

The most important interactions in the Ne dimer are London forces, or van der Waals (vdW) interactions. BLYP is a hybrid DFT method that includes exact exchange term in exchange-correlation functional. However, the method is local (in LDA, the external potential is the uniform electron gas, and in a GGA, its gradient is included), and vdW interaction requires fully non-local functionals. The basis set is also too small and should include diffuse functions. I don't know what method to recommend.
4. MP2/6-311++G(2df,p) for the singlet-triplet energy splitting of para-benzyne. Note that for both singlet and triplet states unrestricted MP2 is needed.

MY ATTEMPTED ANSWER

The para-related electrons have low correlation, meaning they have small singlet-triplet separations of the frontier orbitals. MPn energy terms are related to the inverse of this separation and thus for a very small s-t separation, this term is too large for MP2 to handle. The basis set is probably fine. I think coupled cluster methods or DFT would work here.
5. CCSD/3-21G for geometry optimization of the DNA base adenine.

MY ATTEMPTED ANSWER

The basis set is too small for this molecule and a correlated method (no diffuse or polarization functions are used). Furthermore, CCSD is very costly and not good for this big molecule. Maybe handling the conjugated pi bonds is important. Suggestions?
6. CISD/6-311G(d,p) to compare the CC bond lengths of oligoenes ($2<n<20$).

MY ATTEMPTED ANSWER

CISD is not size-extensive, so as $n$ increases, the calculations become prohibitive.
7. MP3/6-311G(d,p) for the electron affinity of F2 i.e. the process:

MY ATTEMPTED ANSWER

MP3 is not good when we have an open-shell species or a system with electron congestion. Convergence will be very slow and not guaranteed at all. We assume in MPn that the zeroth-order wavefunction is a reasonable approximation, and the electron correlation contribution can be added as a small perturbation. But if HF describes this molecule very poorly, then MPn won't improve upon the zeroth-order wavefunction dramatically.
8. LSDA/6-31G(d) for the relative energies of the various conformations of n-butane obtained through rotation about the central C-C bond.

MY ATTEMPTED ANSWER

LSDA is equal to LDA for closed shell systems. Thus the external potential is the uniformly distributed positive charge and can be used only if the "constant" electron density is different for each point in space. However, here the electron density due to bond twisting is not spatially uniform. I think a GGA or B3LYP should be used. Furthermore, I think the basis set is too small and should include extra polarization and diffuse functions (I don't know why, just saying).

What method/basis set would you suggest for the following:
9. The activation energy of the Diels-Alder reaction between cyclopentadiene and acrylonitrile (reaction does not need static electron correlation treatment).

MY ATTEMPTED ANSWER

Pure LDAs or GGAs overestimate bond length, so I suggest the hybrid B3LYP with a small basis set.
10. The interaction energy between two Ar atoms in the $\ce{Ar\bond{...}Ar}$ van der Waals complex.

MY ATTEMPTED ANSWER

Use CCSD with a medium basis set, since vdW is repulsion of electron clouds at large distances as they interpenetrate and CCSD has traces of higher excitations. A small basis set can cause basis set superposition error (BSSE). Inclusion of higher order virtual orbitals as in CCSD improves the dispersion description.

Is this a homework question?  Post your attempts at answering each point and we can try to help you from there.

@AliRaf: BLYP is not the same as B3LYP -- B3LYP is a hybrid functional, while BLYP is a GGA. For (v) I'd just use a hybrid functional of some sort: B3LYP should do a perfectly adequate job. For (iii) I'd want something with excitations but I'd want to use something different for the initial geometry optimisation and then something expensive for the energy calculation, maybe at a few different points to verify the geometry.

For Problem 9, just because a method doesn't calculate the geometry well doesn't meant the energetics will be inaccurate. There is often not a huge correlation between accurate geometries and accurate energies, unfortunately. B3LYP is commonly used for reaction energetics of organics. The M06 suite of functionals is parameterized for thermochemistry, so that could be a good (more modern) choice.

This is a great question and I love it.  Thank you for taking the time to not only post the question but each attempted response.  I will try to address the parts of this question that I can at the moment.  I will list, in order, your attempted answer along with my response.

Carbene H2C: has two singlet configurations that contribute to the ground state energy we need a multireference method to recover the single excitations. MP2 is not good for this open-shell systems since they have large spin contamination and thus large error. The basis set also needs to include diffuse functions in the electron correlation methods.

Yes, a multireference method will need to be used and a complete active space will need to be defined.  Therefore, MP2 will NOT work.  The 6-31(d) basis set is atrocious and should never be used for production level work given the growth of computing power in the last decade.  This double-$\zeta$ quality basis set is much too small and at minimum a triple-$\zeta$ basis set should be used.  Diffuse functions are critical to properly describe the spatial electronic region of a weakly bound electron far away from the nucleus, but in this case, I would only include diffuse functions on non-hydrogen atoms given that the unpaired electrons are on carbon and not hydrogen (as per your picture). Lastly, polarization functions should be included on every atom, and not just the non-hydrogen atoms.  Therefore I would use a basis set such as 6-311+G(d,p), 6-311+G(2df,2pd), or even heavy-aug-cc-pVTZ.  Quadruple-$\zeta$ basis sets should be used if you can afford it but may not be feasible in this case.

We have bond elongation in the Reactant to the point of breackage. CISD is not size-consistent. It recovers less electron correlation energy for the larger reactant as for the products since reactant is larger and thus cannot be used. The basis set however is fine. I don't know what method can work best for this system but would go arbitrarily with CCSD(T).

CISD does introduce size-consistency problems but there are 'workarounds' you can use (example). Obviously we want to avoid using more and more approximations so it would be prudent to avoid CISD in this case.  The basis set could be dramatically improved by adding diffuse functions to the non-hydrogen atoms (e.g. heavy-aug-cc-pVTZ).

Most important interactions in Ne dimer are London forces of van der Waals. BLYP is a hybrid DFT method that includes exact exchange term in exchange-correlation functional. However, the method is local (in LDA, external potential is the uniform electron gas and in GGA its gradients included) but vdW interaction requires fully non-local functionals. The basis set is as well too small and should included diffuse functions. I don't know what method to recommend.

The neon dimer is a very interesting system.  How does two atoms with full valency (8 electrons) bind?  This is where electron correlation comes into play.  It is due to the correlated motions of the electrons that allow these two atoms to form a bond, albeit a very, very weak bond.  This is known as non-covalent interactions or dispersion interactions.  For these types of systems, very expensive correlated methods are absolutely required to get it right.  DFT methods fail at properly describing dispersion forces though a lot of advancements have been made to build in a dispersion correction to the canonical functionals.  These corrections are usually denoted as -DX, where X is the generation of that correction (currently we are on -D3, or the 3rd generation of these types of corrections).  BLYP has no dispersion corrections built in and therefore will likely tell you that the neon dimer is UNBOUND (i.e. the binding energy is positive at all interatomic separations).  Unsurprising.  Therefore you will have to use a correlated method such as MP2.  But even MP2 can do a poor job of describing dispersion interactions (example). In this example, MP2 overbinds the PCCP dimer by as much as 5 kcal mol$^{-1}$ with respect to CCSD(T). CCSD(T) is what we generally consider to be the correct answer.  Given that CCSD(T) binds the PCCP dimer by only 1-2 kcal mol$^{-1}$, you can see the problem here. The error of the MP2 method is about 5 times as large as the actual binding energy, and that is a major problem.  The basis set is also too small and you MUST include diffuse functions on both Neon atoms.  Without these diffuse functions, even correlated methods will have a hard time determining if this thing actually binds at all.

The basis set is too small for this molecule and a correlated method (no diffuse or polarization functions are used). Furthermore, CCSD is very costly and not good for this big molecule. Maybe also conjugated pi bonds is of essence. Suggestion?

The question deliberately states 'optimization'.  Therefore, nearly any commonly used density functional could be used (e.g. B3LYP).  If you want a slightly better result, go for some newer functionals such as the Minnesota functionals from Truhlar's group (e.g. M06-2X, M08-2X, etc.) or try some dispersion-corrected functionals.  If the question asked what you would use to determine the energetics, THEN move on to something like MP2 or CCSD(T).  I wouldn't trust the DFT energetics unless they have been extensively calibrated in previous work.  (Protip: never rely on DFT methods which haven't be calibrated for your type of system... in fact, just don't trust DFT at all :P).  The basis set is much too small.  A DFT optimization will go quick enough that a triple-$\zeta$ basis set should be used.  It goes without saying that you should include polarization functions on ALL atoms and diffuse functions on the non-hydrogen atoms (e.g. heavy-aug-cc-pVTZ, 6-311+G(2df,2pd), etc.).

CISD is not size-extensive so as n increases the calculations become prohibitive.

To be honest, I have a hard time coming up with an answer to this as the "question" is not very clear as to what the goal is.

MP3 is not good when we have an open-shell species or a system with electron congestion. Convergence will be very slow and not guaranteed at all. We assume in MPn that the zeroeth order wavefunction is reasonable approximation and the add the small perturbation contribution. But if HF describes this molecule very poorly then MPn won't improve it dramatically.

Yes, MP3 is not guaranteed to converge at all which is pretty much the only reason (in my opinion) that you should need.  By the time you figure out the convergence/divergence/oscillation of the MPn series, you would have been better off sticking to something that is more expensive but guaranteed to converge.  Stick to CC methods.

LSDA is equal to LDA for closed shell systems. Thus eternal potential is the uniformly distributed positive charge and can be used only if the "constant" electron density is different for each point in space. But here electron density due to bond twisting is not spatially uniform. I think GGA or B3LYP should be used. Furthermore I think the basis set is too small and should include polarization and diffusion (I don't know why, just saying).

Given the fact that we've come a long way since LSDA, any use of LSDA for any type of production level work should just downright be avoided.  Here I would use GGAs or meta-GGAs.  They are much better.  Also, a larger basis set should be used. (polarization functions for all atoms, diffuse on non-hydrogen atoms, blah blah blah).

pure LDA or GGA overestimate bond length so I suggest hybrid B3LYP with small basis set

This question deliberately asks how you would go about determining the activation ENERGY of this reaction. Therefore, do NOT rely on any DFT method when it comes to energetics UNLESS extensive calibration has been done.  Here's what I would do for this if I were going to turn this into a research paper.  First and foremost, optimize these guys with MP2/heavy-aug-cc-pVTZ.  They are small enough that this could be done.  Now, if I wanted to use DFT, I would pick a slew of DFT methods of various flavors including hybrids, dispersion-corrected, etc. and optimize each geometry as well.  Now, for each MP2 optimized geometry, I would do some high level single point energy computations using explicitly correlated methods such as MP2-F12 AND CCSD(T)-F12.  (The explicitly correlated methods accelerate the convergence of the correlation to the CBS limit over canonical implementations.  Essentially, MP2-F12/haTZ energetics can be on par with MP2/ha5Z energetics.  Same goes for CCSD(T).)  Given the size of these systems, I would (at the very least) bust out Dunning's correlation-consistent series of basis sets and push them as far as I could go.  From eyeballing the system, I think I could get away with doing heavy-aug-cc-pVDZ and heavy-aug-cc-pVTZ right off the bat.  I'm pretty sure I could do heavy-aug-cc-pVQZ as well.  Once I've done this, I have a darn good idea how close to the CBS limit my energetics are.  Now I can compare the DFT binding energies to the 'right answer' and boom, I've got DFT results I can rely on.  Of course, if the MP2 and CCSD(T) energetics show significant deviation, then we would have to revisit our optimization procedures but I don't think it would be an issue in this case.

CCSD with medium basis set. Since vdW is repulsion of electron clouds at large distances as they interpenetrate and CCSD has traces of higher excitations. A small basis set can cause basis set superposition error (BSSE). Inclusion of higher order virtual orbitals as in CCSD improves dispersion description.

I would not recommend CCSD.  You might as well do an MP2 calculation to be honest but for this type of system with a large dispersion component, you will want to do CCSD(T) energetics.  System is small enough to use a quadruple-$\zeta$ quality basis set.  Add diffuse and polarization functions.  Also, the frozen-core approximation should be tested.  You may want to correlate ALL the electrons to determine core-core and core-valence electron effects (use something like aug-cc-pCVQZ).

Brilliant (although I would not have the resources to follow your advice). One question regarding (i): There are two (admittedly quite old) publications, proving a triplet ground state for methylene. F. A. L. Anet , R. F. W. Bader , Anne-Marie Van der Auwera, J. Am. Chem. Soc., 1960, 82 (12), pp 3217–3218 and Charles F. Bender , Henry F. Schaefer III, J. Am. Chem. Soc., 1970, 92 (16), pp 4984–4985. Is there anything newer, how would that change your advice?


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I want to make electricity of aluminiun cans i.e. reverse the process of refinning aluminium. Any ideas?
Recycling Al cans to produce electricity,i.e. to reverse the process of electrolysis and produce electricity

Since you can make good money just recycling the aluminium (because you thereby avoid the electricity required to make it) this doesn't sound very economically viable.

There is a company whose claim to fame is this exact process(Phinergy). Make aluminum where energy is available, e.g. Alcoa plant near Niagara Falls, and oxidize it elsewhere when necessary.  http://triblive.com/business/headlines/7431480-74/battery-aluminum-alcoa , https://en.wikipedia.org/wiki/Aluminium%E2%80%93air_battery

You could "burn" the aluminum in the presence of oxygen to generate aluminum oxide.  For every two moles of aluminum consumed, a bit less than 1700 kJ of energy would be released.  You could use the energy to drive a turbine and generate electricity.

There are issues with using aluminum from cans, as they are coated with a thin layer of plastic to prevent them from reacting.  That will tend to interfere with any chemistry you might want to do to the aluminum inside.  If that's not an issue, it's pretty straightforward to do, if not economical in the slightest.
There's nothing to prevent you from setting up a galvanic cell with aluminum at the anode and a metal salt with a reduction potential greater than -1.6V at the cathode.  I've made cells using aluminum foil and copper chloride in the past, for example, though I wouldn't recommend them for electricity production.
That's not to say that there isn't promise for aluminum's use in batteries.  Researchers at Stanford U published a paper in Nature in early 2016 about an aluminum-ion battery with good voltage that could be recharged repeatedly and that charged in about 1 minute.


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If "Gas" belongs to "State of Matter" then what does "Metal" belong to?
My question is, if "gas" belongs to "State of Matter" then what does "Metal" belong to.
Gas can be substituted with any state of matter(in the analogy); metal can be substituted with metalloid or nonmetal(in the analogy). I've researched this and I can't seem to find an answer to this. So, this could read as "if 'solid' belongs to 'State of Matter', then what does 'nonmetal' belong to'"?
I'm not trying to assert that metal is a state of matter.
As most of y'all (hopefully all) know, the elements on the periodic table can be a metal, metalloid, or nonmetal. Would "Metal" belong to "Metal Status"? I'm asking because I'm writing a program to represent the PeriodicTable and I need an appropriate name for my variable. I may just opt for "Metal Status".

Possibly relevant: http://en.wikipedia.org/wiki/Periodic_table_(metals_and_non-metals)

Maybe something like "a variety of solid, which is a state of matter".

@SatwikPasani Calling metals a variety of solids runs into problems. Does gold stop being a metal if you heat it up past its melting point? And then there's mercury which is already liquid at standard temperature and pressure.

@DavidH I would prefer to stay pragmatic. Similar situation applies to liquid nitrogen, which is gas, but liquid. Words are meant to make communication easier, not the opposite way.

"Metal" is a classification used by chemists and usually means something along the lines of "this electrically conductive, usually solid element that is deformable".
The other way around: Chemists use "gas" to refer to a state of matter, not some defined gas (for example the atmosphere).
Putting these two together: "Metal" belongs to the periodic table of elements, while "solid", "liquid", "gas" and "plasma" are merely words to describe what state the element or compound is in. It is possible to have liquid, solid or even gaseous metals.
Edit: If you want a name for a variable that is either in the state Metal, Half-Metal or Metalloid and Non-Metal, I'm not sure such a name even exists. Feel free to make one up that suits your needs, comment your code sufficiently and it will not be a problem. "Metallicness" or something along those lines would be my choice.

Thanks for answer my question; I'm not sure what VINAY was answering. I was asking this question mainly because I was trying to come up with a suitable name for a variable for my variable and I didn't know if there was an official classification name that covered metal, nonmetal, and metalloid.

If you want some official statement, it's always good to check the IUPAC Gold Book where I've found a reference to metals being characterized by their conductivity.

I have edited in an answer that I think is what you're looking for.

Thanks for the edit. I might stick with metalStatus. I do like metallicness, and it's probably a better name if you consider technicalities, but it seems more like a gradient-like name.
I hear metallicness and I hear is this 0% metal, 50% metal, 75% metal, or 100% metal. It's not a big deal, but I think I'll keep 'metalStatus'. Thank you for your edit, though.

What is Metal?
A metal (from Greek "μέταλλον" – métallon, "mine, quarry, metal") is a solid material (an element, compound, or alloy) that is typically hard, opaque, shiny, and features good electrical and thermal conductivity.
What is State of matter?
In physics, a state of matter is one of the distinct forms that different phases of matter take on. Four states of matter are observable in everyday life: solid, liquid, gas, and plasma.

From the above definitions, it is clear that gas is a state of matter and metal is also the state of matter with certain specific properties. The distinction is that, all solids are not metals but all metals are solids.
An element or compound or alloy is said to be metal if it is

solid
hard
opaque
shiny
good electrical and thermal conductor.

Gold has all the above properties at standard conditions for temperature and pressure, it can be regarded as metal only at that temperature and pressure.
Most of the elements, compounds, or alloys are regarded as metal, at standard conditions for temperature and pressure.
But, mercury is regarded as metal only below the temperature of $-38.83$ degree celsius and is not regarded as metal at standard conitions for temperature and pressure. Thus, we say mercury as the only metal which exist as liquid at standard conditions for temperature and pressure. Remember it is wrong to say mercury as the only liquid metal.

What you're basically saying is that "metal" is also a state of matter? I don't find that category on this comprehensive Wikipedia article...

Any element or compound or alloy you make think, it will be one of the state of matter. Metal is also the solid state of matter with certain specific properties said above. I hope that cleared your doubt.

So you would argue that liquid gold (or mercury at standard conditions for that matter) in fact is not a metal? Although it still is opaque and a good electrical and thermal conductor? A Google Scholar search yields 282'000 results, so I suggest you rethink your position.

Yes I argue that liquid gold (or mercury at standard conditions) is not a metal. Would you like to prove the definition of wikipedia wrong?

I think I'll have to start a discussion on the wikipedia site concerning "Metal", because I have found the following "A heavy, silvery d-block element, mercury is the only metal that is liquid at standard conditions for temperature and pressure" on the wikipedia site "Mercury (element)". Clearly there are some inconsistencies regarding this subject.

You can read my last $3$ lines to understand those lines without misconception.

let us continue this discussion in chat

I know the difference between them. My point was that solid, liquid, gas, and plasma are all under the description of "State of Matter" when they are being taught.
My question was asking if there is an overall classification, like "State of Matter", that applies to Metal, Nonmetal, and Metalloid.

"Would you like to prove the definition of wikipedia wrong?"  I cringed hard at this.

I find this definition highly controversial. As nearly all of our everyday life definition of metal, are properties derived from it having "a delocalized sharing of electrons throughout the metal lattice"; which also sounds more reasonable, when one considers that a simple physical property (state of matter under room temperature) of a molecule can change the status of an element despite its similar pattern shared by other adjacent (on the periodic table) metals.
I find my comment here useless as it was only just then did I realize this post was over a year ago.


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Converting 1,4-benzenedimethanol to benzene and methanol?
I have the molecule 1,4-benzenedimethanol:

How could I produce two methanol molecules and one benzene molecule? (Is there a catalytic reaction method available for this?)
I'm guessing that hydrogen ions could be used to split off the methanol molecules on both sides?

Oxidize to the dialdehyde (Jones reagent), decarbonylate with rhodium (do it at the bench before building the pilot plant), reduce the CO to methanol.  Why would you want to do that?

I will not ask why you would want to do that.
You're right that the benzylic alkohol can be deprotonated. But that won't help you for the intended reaction.
Regarding the possible mechanisms, I suggest to do it on paper.
Draw the starting material and split the bonds. Draw the fragments that result from splitting the $\ce{Ph\bond{-}CH2OH}$ bond. Do this first for the two heterolytical pathways possible in theory. Inspect the fragments and ask yourself, how stable these might be in general, and in the presence of other molecules of the starting material.
Are they likely? ;)
Compare with the following experiment on paper: Which reactions are possible with your starting material in the presence of strong acids? (Assuming that their corresponding bases are lousy nucleophiles.) Yes, it has nothing to do with the reaction you'd like to perform. But it works ;)
Now let's inspect the homolytical case.
To cut the "lecture" short, it has indeed been observed that benzylic alcohol furnishes benzene upon sonication (PDF). The yields however are only in the ppm range and I'm not aware of a comparable study on the title compound.

Edit:
Concerning the question in your last comment, have a look at the prices for methanol and benzene at the chemical supplier of your choice:
Methanol (z.S.)         2,5 l   14.00 €
Benzene  (reinst)       1,0 l   32.90 €
1,4-Benzenedimethanol   50 g  294.30 €
Let's just say that the production of methanol and benzene the way you suggested is not a promising concept.

thanks :) (My organic chemistry knowledge is not very good) but would it be possible to create other organic molecules, derived from this molecule that then can be eventually reacted to form methanol and another compound?


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To which state of matter does the flame belong to?
I had this question from the day (9 years old, now 16) that I learned about states of matter. I have asked many of my teachers, some of them told me it's a gas some that it's a plasma. Can anyone answer my question?
Recently I've learned that the plasma state is obtained when all the electrons from the atom are removed. Obviously it's present in the sun due to high temperatures.

This is cross posted on Physics.SE http://physics.stackexchange.com/questions/94609/to-which-state-of-matter-does-the-flame-belong-to

Yes I did. Because this topic belongs to both physics and chem. And what's wrong with that.

You could at least point it out in the question, because then the answerer may look at the other answers to see if the question has already been sufficiently answered.

@user37419 Also, a cross posted question should be written to specific suit the audience on the sites it is posted on.  Also, you might want to read the faq about cross posting http://meta.stackexchange.com/questions/64068/is-cross-posting-a-question-on-multiple-stack-exchange-sites-permitted-if-the-qu

A candle flame is the product of the oxidation of wax to $\ce{CO2}$ and other small organic compounds. Since these small compounds are energetically lower, energy is released during the reaction, in the form of light and heat. So really, we start out with gas.
However, flames can also conduct electricity, and the conductivity varies with the location in the flame. The linked science fair project summary states that the lowest resistance (highest conductivity) was found at the edge of the bright yellow part of the flame the conductivity was the largest, meaning that the concentration of ions was the highest. The lowest resistance value found was $80~\Omega$, which is still very large in comparison to the resistance of plasma, which is generally considered to be $0$.
So, summarizing my answer: A flame is a gas with lightly ionized portions, but does not qualify as a real plasma.

Ok, can you tell me how did you get to know about science fair and all. Is it from web or from newspaper. And also in plasma state the electrons are removed from the atoms right, then where will the go in sun?

I googled "flame conductivity" and skimmed for something useful. The fouth entry did it for me. In a plasma, not all electrons are removed from the atoms. As to solar reactions, that would make a good additional question: "Where do the electrons from the solar plasma go to?" Feel free to open a new question about that ;)

Why don't you answer it here itself?

Because the sun has almost nothing to do with a candle flame, and because the question you asked really only referred to a flame. Also: One question per question ;-) It keeps things clearer.

While a candle flame is not a plasma, it is a fairly low energy, low temperature flame. In really hot flames, would you classify the state as more like a plasma than a gas?

A gaseous flame is hot gas seeded with free radicals and ionized species, a dilute plasma.  Note that solid flames exist.  Mix finely divided nickel and aluminum, tantalum and graphite, etc., compress into a compacted body, and heat an edge very hot until things start.  A white hot interface propagates through the body as the components react.  I would hesitate to call that a condensed phase plasma.


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Would trans hydride ligands on a metal complex be beneficial for a catalyst?
Would trans hydride ligands on a metal complex be beneficial for a catalyst?
Such as alkene metathesis or hydrodefluorination

In alkene metathesis you need large ligands that direct the stereoselectivity of the reaction, and a hydride ligand is very small.
Also for hydrodefluorination I don't see an advantage of a trans hydride ligand on the catalyst. cis hydride ligands could be beneficial, as they then take the place of the fluorine atom on the substrate.


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Where do the electrons that are released in the Sun go?
Where do the electrons released from the clutches of nuclei in the Sun's plasma go? Do they go to the product nuclei of the nuclear fusion reaction in the Sun (i.e. He)? Or do they remain on the surface of the Sun?

This almost begs the question - If the deuterium and tridium nuclear reactions each produce a positron, each of those presumably 'react' with an electron, annihilating each other, but the protons of the three hydrogen atoms starting the reaction are still around. Unless there is a source of electrons, wouldn't the positive charge in the sun accumulate from these deuterium, tritium and helium ions that are accumulating? If electrons bind with those helium nuclei to form stable atoms, where did those electrons come from?

This isn't an answer, and should probably be converted to a comment. Regardless, if there are free protons in the solar plasma, the corollary is that there are also free electrons. Sum them up along with all other charged species and you'll find the Sun is electrically neutral to a very high precision. Note that in the diagram, six free electrons are implicit in the left side. Two electrons are annihilated by positrons, leaving four free electrons on the right side, which is exactly enough to neutralise the positive charges.

Some of the electrons will react with positrons formed during the fusion processes; they annihilate each other and give off high-energy radiation (also called gamma rays):

Source
But this process  happens at the core, and there is no way that those electrons get out to the surface. So my best guess is that the electrons which don't get annihilated just stay there in the core plasma. I do not think that they combine with the $\ce{^{4}_2 He^{2+}}$ particles formed by this chain, because the temperature is still too high. So you start out with 6 protons and 6 electrons (for electrical neutrality reasons), two electrons get annihilated in the process and you end up with a doubly positively charged helium particle and two protons and four electrons. So the electrical neutrality is not violated.
There is not only the core of the sun but also the outer layers and the corona, from where we observe the solar wind, which is a stream of particles, mainly electrons and protons. So there you have some electrons that escape the sun. But these are not the electrons from the core, they come from the surface of the sun.

By which process the beryllium nucleus splits up into He and 2 protons.

It's the process called radioactive decay: $\ce{^6 Be}$ has a half-life of 5.0 zeptoseconds, so it decays very very quickly into two protons and the helium particle.

I know that. Which type of radioactive decay is that? Can you name that decay? all I know is alpha, beta +, beta- and gamma decays.

Do you even look at the pages I link to? This isotope decays under proton emission (doubly so). Here is a comprehensive list of decay modes. And if you like my answer it would be nice to accept it :)


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Difference between solubility and dissolution?
How is solubility different from dissolution? And how are they both different from solvation?

Solvation [...] is the process of attraction and association of molecules of a solvent with molecules or ions of a solute. As ions dissolve in a solvent they spread out and become surrounded by solvent molecules.
  [...]
Dissolution is a kinetic process, and is quantified by its rate. Solubility quantifies the dynamic equilibrium state achieved when the rate of dissolution equals the rate of precipitation.

Quote from Wikipedia.

Solubility is a chemical property referring to the ability of a given substance, the solute, to dissolve in a solvent. Dissolution, or solvation, is the process by which a solute interacts with and becomes dissolved in a solvent due to an increase in stability.

This is in principle the same as Tschoppi already stated.

Solubility is maximum amount of solute that will dissolve in a given solvent at a given temperature to form saturated solution. Dissociation is a more kinetic process which depend on lattice and hydration energy.

• Dissolution is the process where a solute dissolves in a solvent to form a solution, whereas solubility is the outcome of dissolution.
• Solubility is a thermodynamic entity whereas dissolution is kinetic.
• Solubility is measured in mol/kg and dissolution is measured in mol/s

solubility:

the capacity of solute to be dissolved in a given amount of solvent (how many moles of solute are dissolved)
solutes solubility is always questioned
it is a thermodynamic process
its unit is mol/kg

dissolution:

rate of solute dissolving in a solution (how fast the solute dissolves)
solute is always soluble
it is a kinetic process
its unit is mol/sec


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What is an intuitive explanation for Electron Configurations
I'm sort of self studying Chemistry. I'm now getting stuck wrapping my head around a concept (and I realize it was the same concept I had difficulty grasping years ago) and that is Electron Configurations.
I know that perhaps due to the nature of chemistry it is non-intuitive. However, I want an explanation that is more than "This is a bunch of numbers and letters that follow a pattern. Live with it" - which seems to just be the explanations I'm getting when I went to school.
So, my question is, what are Electron Configurations, and how would you explain them intuitively?

Electron configurations are a bunch of numbers and letters, and they also follow a pattern. But there is also a logic behind the whole thing.
Basically, chemists use electron configurations to see in which atomic orbitals the electrons are found. From this several things can be derived, most of all the reactivity of said electrons (which is where the chemistry happens).
The intuitive answer is this: Electron configurations are a way of telling in which atomic orbitals the electrons are put in, according to the Aufbau principle and the Pauli exclusion principle. Decoding it is pretty easy, as it always follows this pattern:
$n$$(s|p|d|f)^x$ where $n$ is the main quantum number (or shell number), $(s|p|d|f)$ denotes the atomic orbitals with the quantum number and $x$ is the number of electrons found in the orbitals.
The orbital "groups" ($p$, $d$, $f$) contain several "orbitals" (side note: an orbital is strictly seen a one-electron function) where two electrons with opposite spin fit in. $p$ has 3 ($p_x,\; p_y,\; p_z$), $d$ has 5 ($d_{xy},\; d_{xz},\; d_{yz},\; d_{x^2-y^2},\; d_{z^2}$) and $f$ even more. This means that the maximal number of electrons in the $p$ group is $6$, and in the $d$ group $10$. So when you fill in the electrons according to the aufbau principle, you have to go to the next orbital group when you reach this number.
Bottom line: Every electron in an atom is defined by the four quantum numbers $n$ (main or principal quantum number), $l$ (azimuthal), $m_l$ (magnetic) and $m_s$ (spin). Using the electron configuration we can derive the first three of these four numbers.