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Can atoms delocalize in molecules?
Electrons can delocalize over multiple equally probable bonding structures. Can atoms delocalize over multiple equally probably bonding structures in certain cases?
It seems to me that any such effect would be most pronounced with very small atoms such as hydrogen.
Furthermore, I would expect such effects to be mostly to occur with hydrogen bonds and cation-pi interactions.
One possibility I am imagining sticking the ends of a bunch of linear molecules that end in OH groups or fluorine atoms together.

I'm not exactly sure what you want. You can only talk about probability of bonding if you have sufficiently long timescale - there are quick reactions creating equilibria.

Tautomerism in organic compounds?

Can atoms delocalize over multiple equally probably bonding structures in certain cases?

If it happens at all, it exceedingly rare. Through my 10-years work in chemistry, I didn't once found a case where it may be relevant. However, what does and indeed affect chemistry is reversible rearrangements.
The simplest case is rotation of molecules around single bonds. Another one is nitrogen inversion. More complex cases also exist. Such things sometimes make interpretation of NMR quite difficult. NMR gives a peak for each non-equal nucleus in the molecule. When internal rearrangement is involved, it sometimes makes two atoms 'equal', i.e. exchanging places too fast for NMR to catch the difference and producing a single peak instead. This, however, is 'quasi-delocalisation', since there is no need in resorting to delocalisation to explain said phenomena.
Another quantum phenomena often observed is tunneling. It affects slightly effective reaction rates/barriers and vibrational spectra.
Anyway, de Broglie wave length for proton is thousand times smaller than for electrons, so quantum effects in nucleus movement usually overshadowed by quantum effects in electrons movement.

The ring-whizzer mechanism in my book (Inorganic Chemistry, Miessler et al.) is a pretty cool example of a metal bonded monohapto to a cyclopentadienyl ligand, moving around it. (https://en.wikipedia.org/wiki/Fluxional_molecule) I wonder if that counts?

Delocalization is a wave property. The less mass objects have, the more they behave like waves and the less like particles. Electrons are, by mass, 1837 times lighter than protons, so the wave character of an atomic nucleus is significantly less pronounced and usually of no consequence.
Take a look at three-center two-electron bonds (for example, in diborane) and three-center four-electron bonds (for example, in the biflouride anion $[F-H-F]^-$): This seems to relate to what you are talking about concerning the OH groups.
The wave character of nuclei and even bigger objects has been proven, though: see, for example, the experiments of Arndt et al. with $C_{60}$: http://www.nature.com/cgi-taf/DynaPage.taf?file=/nature/journal/v401/n6754/full/401680a0_fs.html&content_filetype=PDF

If what you suggest were commonplace then it would be difficult to decide what a molecule is as its structure may consist of innumerable forms. Thus x-ray crystallography would not produce a structure and nmr spectra would most likely be uninterpretable. As this is not generally the case there cannot be many (possibly any?) examples of these that do not involve H (or D) atoms.
One normally considers the ammonia inversion as H atoms tunnelling, but what about the rapid inversion ($10^{8}$ s$^{-1}$) of, for example the bowl shaped molecule corrannulene(C$_{20}$H$_{10}$ see below and web molecule-viewer.com for rotatable picture of 3D structure). Presumably corrannulene does not tunnel from one bowl shaped form to another as its mass is so great, but crosses the inversion barrier top due to thermal energy excitation. Similarly, rotation of alkyl groups in their rotational potential is thermally activated.
The important point is that there is always a competition between tunnelling and barrier crossing. All atoms can tunnel, light ones more easily than heavy ones but usually heavy atom tunnelling cannot compete with thermal barrier crossing and is not observed.

Other tunnelling examples are tautomers shown on the Wikipedia page and involve hydrogen. If tautomerism occurs in the gas phase then we should probably assume that its due to tunnelling of the H atom, but in solution it may be a different H atom that leaves and returns. Another example is the H bond, such as in the structure of DNA where the H atom can move between two other atoms (O..H..N and N..H..N) or a suggested by @Zubo .


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Clarification about Frost diagram y-axis conventions?
I was looking at the Frost diagram of Manganese, and on the UC Davis ChemWiki, I see this one:

However, the one straight from my textbook, "Descriptive Inorganic Chemistry" (Overton), looks like this:

In the first image, the y-axis is $\mathbf{nE^o}$ and the x-axis goes from 0 to +7 from left to right, but in the second image, the y-axis is $\mathbf{-nE^o}$ and the x-axis goes from +7 to 0 from left to right.
I'm wondering if reversing the oxidation state ordering changes the y-axis definition between $nE^o$ and $-nE^o$.
It seems like a silly question to me, but I still want to be sure.

yep ! Write out pairs of coordinates!

I can't see any reason why reversing the x-axis direction should affect the y-axis in any way. I'd say one of your diagrams has it wrong.

The UC Davis version is incorrect. The free energy of the system ($\Delta G$) is equal to -nFE, so y-axis is reported as $\Delta G / F$ or -nE, according to the original report by Arthur Frost (J. Am. Chem. Soc., 1951, 73 (6), 2680-2682).
Note: In the original report free energy was denoted by $\Delta F$, rather than $\Delta G$. The x-axis designation is insignificant.

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How to detect dog feces?
I am working on building a robot to detect and pickup dog poop.  I have looked at the concentrations of the various components and simple chemical/electrical ways to detect them.  The best component to detect seems to be methanethiol but I could not find a simple sensor to detect it.  I found this sensor http://www.aliexpress.com/item/Free-shipping-4CH3SH-10-sensor-methanethiol-sensor-electrochemical-gas-sensor/32381228085.html and I was wondering if it could work for this purpose, I could not find much information on the sensor.  I was also thinking something like an alcohol sensor could work.

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Would Schweizer's reagent dissolve nitrocellulose?
Schweizer's reagent, or tetraamminediaquacopper dihydroxide, is used in the production of rayon. It does this by dissolving the cellulose which can be acidified to form strands of cellulose. I have had some success with this.
So, seeing as cellulose and nitrocellulose are rather similar, would it be able to dissolve it too? If not, why? The nitrocellulose would have to be able to survive the addition of acid, which should not be a problem since it is synthesized using sulfuric and nitric acids. So, as long as I use sulfuric acid it should work fine.
Something else that might affect it is how nitrated the cellulose is? Cellulose dinitrate and cellulose hexanitrate might act differently.
And now, assuming that it can dissolve it in some way, it could be extracted from that from celluloid plastic, like that found in ping pong (table tennis) balls?
So, by adding some ping pong ball to Schweizer's reagent, then letting it dissolve and adding it to a solution of sulfuric acid, the camphor in the celluloid would be left behind and you should end up with some sort of "nitro-rayon". This is a bit of a stretch, but to me seems feasible.
I will try this when I can, but for now I hope this is enough to prompt speculation.

Schweizer's reagent dissolves cellulose because it forms a complex with hydroxyl groups of glucose moieties. Therefore, a completely nitrated cellulose will not dissolve in Schweizer's reagent. However, the reagent is quite basic and it will cause a hydrolysis of the nitrocelulose, so it may dissolve eventually, but after precipitation it will be normal cellulose again. But there is also a good information for you: nitrocellulose easily dissolves in many organic solvents e.g., acetone, ethyl acetate, diethyl ether and so on. So once you have a nitrocellulose, you do not need to play with Schweizer's reagent.

Alright. Welp, from what I've heard most organic solvents turn ping pong balls into a paste rather then separate it. Thanks for the help.

Note that celluloid in ping pong balls does not contain fully nitrated cellulose. And separating celluloid into components would be infeasible.


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Is the rate of desorption dependent on pressure?
While I was learning about Langmuir's adsorption isotherm in my chemistry class, my teacher talked about the situation when the rate of adsorption and desorption will become equal. However, he mentioned that the rate of adsorption is dependent on pressure but the rate of desorption is not and when he gave an explanation I wasn't satisfied with it.
I would imagine that greater the external pressure on the adsorbent lesser will be the tendency of the adsorbate particles to leave the surface. But that isn't what my teacher said. Please explain.

Without knowing exactly what you were told its hard to be exact, but here is a description of the situation.
The Langmuir model makes assumptions (a) adsorption is complete when the surface is filled with one gas molecule  per site, (b) all adsorption sites are equivalent (i.e. the same) and (c) adsorption and desorption are separate processes, i.e. one does not depend on the other, i.e if one site is occupied it does not affect adjacent sites an any way. Note that this is a model and this determines how we look at the problem.
The model is then $$M(g) + S(surface) = GM(surface) $$ where G is gas and S surface and rate constant for adsorption is $k_a$ and for desorption $k_d$.
We let $\theta$ be the extent of adsorption which varies from 0 to 1, i.e. how full the surface is, then the rate of change of $\theta$ depends on the rate constant for adsorption , pressure and fraction of vacant sites or $N(1-\theta)$ for N vacant sites. Thus for adsorption $$ \frac{d\theta}{dt} = k_aPN(1-\theta) $$
and for desorption $$ \frac{d\theta}{dt} = -k_dN\theta $$
At equilibrium we make the rate of change equal to zero, i.e. rate of adsorption equals rate of desorption and with a bit of algebra find that $$\theta=\frac{KP}{KP+1}$$  where K is the equilibrium constant or $K=k_a/k_d$ This equation has the fancy name as the Langmuir Isotherm.
You can see that the rate going onto the surface must be equal to the pressure (no gas, no adsorption) and the area free to adsorb onto $(1-\theta)$. (We assume in this model than no molecule double adsorbs; this is dealt with in more advanced treatments). The rate leaving has to be proportional to the amount already there , i.e. proportional to N the total number of surface sites multiplied by the fraction covered.
Ask more questions in the comments if this is not clear.

I understood Langmuir's isotherm. I didn't understand why rate of adsorption is directly proportional to pressure but rate of desorption is not.

Why is the inverse process concentration-independent?

@Marcel it has to be proportional to the number on the surface , which is $\theta N$ so is fraction of concentration. No molecules on surface  off rate zero, full surface; maximum off rate :)

@porphyrin why  is desorption not proportional to , say, the decrease of pressure?

@sajja islam because we consider adsorption and desorption as separate processes: if there are no molecules on the surface the rate of desorption must be zero and it has to be at a maximum when the surface is completely covered.

For adsorption, its forward reaction is precursor concentration dependent and the precursor concentration is represented by pressure. For desorption, the forward reaction is also precursor concentration dependent and the concentration here is theta (or one - you can treat the precursor as solid surface so you have concentration of one).

No matter in adsorption or desorption, the believed dominant driving force is the capability for forming or breaking the chemical bond between the adsorbate-surface. Other forces from the environment, such as the van der Waals or Brownian motion are considered negligible during either adsorption or desorption because the activation energy for such processes to occur is low (<20 kJ/mol) but the chemical bonding between adsorbate-surface is usually higher than 200 kJ/mol.
In another word, the chemisorption is the rate-limiting step if one want to calculate the overall rate.  Your intuitive feeling that "greater the external pressure on the adsorbent lesser will be the tendency of the adsorbate particles to leave the surface" is not making sense under Langmuir model because the tendency you mentioned can be only translated into the Brownian motion, which is a less important force.
Under this prerequisite, the overall rate for one given type of molecule/atom to be adsorbed or desorbed on/from the surface can be then calculated and should be highly dependent with the initial concentration of the molecule/atom in either process(e.g. high concentration leads to high chance to be adsorbed). The initial concentration for adsorption is related with the gas pressure while for desorption is related with number of the adsorbates on the surface already.
Therefore, the gas pressure should not affect the desorption process because the rate-limiting step for desorption is only related with how well and fast the chemical bonds between adsorbates-surface would break. The environmental pressure can do nothing about it in theory.

The scientist's name is van der Waals, not "van der Vaal". Also, logically splitting a substantial amount of text into paragraphs significantly increases readability.

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Basic strength of compunds
Why is Ammonia stronger than Phosphine in terms of basic strength?
I am confused as Nitrogen is more electronegative than Phosphorous.

The strength of a Lewis base corresponds to the strength of its conjugate acid. As usual, a weak acid will give a stronger base, therefore parent acids with higher pKa's deprotonate to stronger Lewis bases. ammonium, $\ce{NH4+}$, has pKa=9.25, while phosphonium, $\ce{PH4+}$, is a much stronger acid with pKa=~-12. Basicity decreases down the group because proton affinity decreases as electronegativity decreases. The N-H bond is stronger than the P-H bond (hence $\ce{NH4+}$ is a weaker acid than $\ce{PH4+}$).
Lewis bases donate electrons from their HOMO, and for ammonia's MO, the HOMO, which corresponds to the lone pair (LP), is very similar in energy to the nitrogen AO, therefore the LP electron density can be considered to be more localized to the nitrogen. Presumably in a MO for an analogous species with a group 15 central atom lower on the periodic table (with lower electronegativity), the AO of the central atom would be higher in energy, so the HOMO electron density would be less concentrated on that atom.
Also, ammonia is more hybridized than phosphine. The s-p energy gap is greater for phosphorous, so the energy penalty of hybridization is higher, therefore the LP will have more pure s character than sp3 character, making it less directed.
Since the ammonia LP electron density is more concentrated on the central atom, and more directed, it makes it more nucleophilic.


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Achiral universe - asymmetric synthesis
Would it be somehow possible to conduct asymmetric synthesis if all chiral molecules would exist only as racemates in nature?

Why, chirality pretty well may emerge spontaneously, upon crystallization or otherwise.

I guess that if a random "odd" number of molecules is taken from a sample, the collection will have an enantiomeric excess, but I don't see how decent enantiomeric excess can be obtained.

Suppose a compound racemizes easily in the solution, but not in the solid state. Suppose it crystallizes in an enantiopure form rather than as a racemate. Prepare a (super)saturated solution and wait until it starts to crystallize. Given the right conditions, the entire bulk may be converted to one enantiomer (which one in particular is defined by the first seed crystal, i.e. by blind luck).

Why would it crystallize so easier in the enantiopure form?

Why not? Crystal structure depends on many things and is not easy to predict; there is no universal law that a racemic crystal should be more energetically favorable than the enantiopure one. Sometimes it would, and sometimes it would not.

@Marko Any single crystal usually contains only one enantiomer. There may possibly be substances which form a special unit cell with pairs of enantiomers, but I am not aware of any specific example.

So precipitation is one way. Are there any others?

You could use polarized light to selectively/preferentially destroy one enantiomer in the the racemate.

How does that work? Please someone explain both procedures in the answer and I may accept it.

The interaction between one enantiomer in the racemate and any chiral probe (e.g. your hand, polarized light, etc.) will be diastereomeric (e.g. different energy) to the interaction of the other enantiomer and the same chiral probe.

There wouldn't be any chiral probe in the universe.

As sunlight passes through the atmosphere it becomes partially polarized; it is a chiral probe.

Looking at it from a more philosophical point of view... if there was no chirality in the world, there would be no life and therefore no chemists to worry about the racemates ;)

Haha, true, but this was more a question for contemplation.

see here for example, or google something like "preferential destruction of one enantiomer by polarized light"

@NotNicolaou perhaps it would be better to say, if there was no chirality in the world, there would be no life as we know it.

@ron You are confusing linear and circular polarization. Scattered light from the sky is partially polarized all right, but still not chiral.

@IvanNeretin Thanks for pointing that out. I'll fall back to the chirality found in some mineral and clay surfaces.

Yes! There is a simple possibility. Take one racemate and do what Pasteur did:

be lucky, in that both enantiomers crystallise separately;
be lucky again in that they form macroscopically distinguishable crystals (in this case probably: mirror images);
collect the different enantiomers. Voilà, you have enantiopure substances and can use them in synthesis.

(Thankfully, using tartaric acid immediately gives access to the Sharpless asymmetric epoxidation.)
Once you have any single enantiopure compound, you are able to generate further ones. For example, reacting a racemate with an enantiopure compound in a way that preserves the respective stereocentres will result in diastereomeric products. Diastereomers can theoretically always be separated (it may be harder for some and some may interconvert under ambient conditions but the theory stands). You could also generate diastereomeric salts, by taking an enantiopure anion (say (R)), and a racemic cation. You will receive the diastereomers (R,S) and (R,R) which can then be separated and so on and so forth.

Great combination of Pasteur experiment and Sharpless epoxidation. It could also be used for resolution either by crystallization as salts or ester derivatizaion or for kinetic resolution through Sharpless epoxidation.


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Excitation of an electron of H-atom from n=2 to any $n_2 $
I got taught that if an electron de-excites from any $n_2 $(where $n_2 >3$) to $n=2$, the  energy change can be represented by $$\Delta E=13.6\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)\ \mathrm{eV}$$
What if an electron excites from $n=2$ to $n_2 $(where $n_2 >3$)?
Do we have to reverse the signs in the brackets?
Also, if it excites like I said, will it be absorbing visible light?(Because in de-excitation, it emits visible light)

The principal quantum numbers for the energy levels go as n = 1, 2, 3, 4 .. ∞. The lowest level is $n$=1. If one goes from a higher level to a lower one a photon is emitted (as fluorescence) and the atom has lost this amount of energy, say $n$=2 to $n$=1. Absorption is the opposite $n $=1 to $n$=2 . As ΔE is positive for absorption it must be negative for emission. Draw out a diagram of energy levels and it becomes clearer. The equation for ΔE is the difference in energy of the two levels, starting at the lower level for absorption and upper for emission.

@porphyrin thank you. Put it as an answer, ill accept it.

The excitation of electrons are caused by photons while their de-excitation is causes the emission of photons of the same energy as the energy differemce between two states. The general expression for energy of n-th excited state of H-atom is given by
$$E_n=-\frac{13.6 eV}{n^2}$$
Hence by the conservation of energy, the energy released/absorbed is given by the formula you have written.
$$h\nu=\Delta E=E_{n_2}-E_{n_1}=13.6\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right) eV$$
Hence your initial state is always $n_1$ and final state is always $n_2$. The energy will be positive if photon is absorbed and negative if it is released.
So, for specific values of $n_1$ and $n_2$ ($n_1<n_2$), if the photon released while going from $n_2$ to $n_1$ is in visible range, the photon absorbed for corresponding excitation will have same frequency, and hence it has to be in visible range.
Please note that there are selection rules for the transitions, determined by the
transition moment integral. Please go through the book in refernece for details.
References:

D.J. Griffiths, Introduction to Quantum Mechanics , Pearson, 2 ed,  2004. ISBN: 9332542899

Hi @Mitradip. It is better to make units upright; as for the present context, you could , instead of using $eV,$  use $\rm eV;.$

Good answer. I know you told us to look in Griffiths, but briefly, what are the selection rules that you speak of? Because as far as I know there are no restrictions on either n_1 or n_2. (Except that they are both positive integers, but that is not a selection rule.)

@orthocresol, for hydrogen atom there is no such restriction, but for other system models (like harmonic oscillator) there are.


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Oxidizing and reducing agent

What i know about the above question is that a compound acts as an oxidising as well as reducing agent when the oxidation of central metal is somewhere between it's maximum and minimum oxidation state which is in the case for $NaNO_2,NaNO$.I don't know how to choose between the two.Any more information about the behaviour of other two compounds will be helpful.Thank you

(I guess NaNO is in fact $\ce{Na2N2O2}$.) Well, it's impossible to chose, as both compounds would act as oxidizing as well as reducing agents. This, BTW, has nothing to do with the oxidation state of metal atom, which is the same in all four compounds.

I agree with Ivan Neretin in that the sodium oxidation state is equal in all compounds, and it will not act as oxidizing or reducing agent.
My thoughts:
Na$_2$O will not act as a strong oxidizing or reducing agent, as both ions are in their preferred state.
NaO$_3$, or sodium ozonide, will likely act as an oxidative agent. However, I believe it disproportionates in water: 4 O$_3$ $^-$ + 2 H$_2$O $\rightarrow$ 5 O$_2$ + 4 OH$^-$, so that's kind of a candidate: it goes from oxidation state -1/3 to 0 and -II.
NaNO also seems strange to me - if it's Na$_2$N$_2$O$_2$, I see here that it acts as a reducing agent.
NaNO$_2$ can certainly act as both reducing and oxidizing agent, as the nitrogen atom is in oxidation state +III.
In conclusion, (4) seems to be the safest bet, considering only one answer is expected.


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Why does the composition of Ocean Water list sodium and chloride separately?
This is the site I am using.
It is also found here, but this uses mg/L instead of percents.
It lists Sodium at 30.8%, and Chloride at 55.3%. How can this be? Shouldn't the two be listed as Sodium Chloride or salt? Because as far as I know, there is no way to obtain chloride on its own.

It's not like they just float there in the form of $\ce{NaCl}$, either.

@IvanNeretin What do you mean

Forget it. Let's put it this way: yes, these two could be listed together as $\ce{NaCl}$, but the authors chose otherwise. This is entirely legitimate and meaningful, even though you can't isolate $\ce{Cl-}$ on its own.

Giving a NaCl content is nonsense. If you had put together the numbers, you would've found they don't match.  55.3 g / 35.4 (g/mol) = 1.5 chloride, 30.8 / 23 (g/mol) = 1.3 sodium. There are a lot of other ions in the sea, with magnesium and sulfate making up most of the difference. https://en.wikipedia.org/wiki/Seawater#Origin

First of all, of course you could provide NaCl content as a weight or molar percentage, and I can demonstrate if necessary.
However, this doesn't really reflect the reality: ocean water is a solution of many salts. Chloride does not necessarily only come from NaCl, but can also come from KCl, MgCl$_2$, and so on. All the ions are solved individually, and if you start evaporating the water, you will get a mix of salts.
So there's really no point in showing NaCl percentage.
Of course, you're right in that it is impossible to obtain pure chloride ions. But the chlorides you will obtain will not necessarily be those of sodium.

Thanks! Is there anyway I can find somewhere where it lists it as NaCl and MgCl2, etc

@SuperNinja741DoesGaming Do you mean like a brand of water, or are you interested in the calculation? I'm not sure what you are asking about.

still for ocean water

@SuperNinja741DoesGaming sorry for the late answer, I was away... here's the thing - if you want the composition listed in compounds rather than ions (i.e. NaCl - x g/L instead of Na+ - y g/L, Cl- - z g/L), then that might depend on the way the sea water is evaporated, and will very much depend on what sea water we are talking about (concentrations differ a lot). I've found this paper http://jsedres.sepmonline.org/content/57/5/928.full.pdf+html , but can't access it at the moment. It may contain what you are interested in.

@SuperNinja741DoesGaming Also, there's this http://www.hydrochemistry.eu/exmpls/sea_evap.html  , but I'm not sure if that helps. Sorry for not having a better answer. I can, of course, show how to calculate the THEORETICAL MAXIMUM of NaCl in a certain sea water, but I'm not sure if that would actually precipitate in reality. Let me know if you are still interested.


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Why does Raoult's law work?
I am trying to look for the entropy explanation but I do not understand anything I can find. Please keep it simple but talk about it from a entropy point of view. Use water and a volatile solute as an example if you need one. I do not understand how the addition of a volatile solute to water increases the vapour pressure of the solution.

Adding a volatile solute does not increase the vapor pressure unless you form an azeotrope, which is driven by enthalpy rather than entropy.

An explanation from a similar question: http://chemistry.stackexchange.com/questions/27997/raoults-law-explained-with-thermodynamics-free-energy/41843#41843

Raoult's law is an approximation for ideal solutions.
Those links have more detail, but the basic idea is that when you assume that the interactions between solute and solvent molecules don't depend on the identity of the molecules, but the vapor pressure does, then it makes sense that the vapor pressure of the mixture would be a linear combination of the vapor pressure of the molecules in the mixture.
Think about it like a bunch of slightly sticky spheres bouncing around in a bucket. A certain percentage will randomly bounce out of the bucket - that's the vapor pressure. If you have a different type of sphere with less mass, more will bounce out in the same amount of time - the vapor pressure will be higher. If you mix them, the total vapor pressure is given by the total number of both types that bounce out. The more volatile ones leave more easily, so the total increases.
Since ideal gases don't interact with each other and since ideal mixtures have negligible enthalpy of solvation, all that is left is entropy terms. If you are looking for a connection to entropy, that is it.
In cases where the ideal approximation is not realistic (water and a volatile solute, for example), Raoult's law doesn't really hold, but it is still a reasonable approximation when the concentration is low simply because most functions look approximately linear for small deviations. It also tends to work reasonably well because most things that will form a solution tend to have similar interaction energies - otherwise they are less likely to mix in the first place.
Of course, there are many, many practical cases where Raoult's law doesn't work, and so we have corrections for that: fugacity and activity.


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What is the mathematical reason behind the gradual increase in osmotic pressure?
Question:

Arrange the following in increasing order of osmotic pressure, $0.1~\mathrm{M}$ cane sugar, $0.1~\mathrm{M}$ $\ce{NaCl}$ solution, $0.1~\mathrm{M}$ $\ce{H2SO4}$.

Solution: As the number of particles/effective concentration of  the solutions increase in the following order:
Cane sugar $(0.1~\mathrm{M})$ > $\ce{NaCl}$ $(0.2~\mathrm{M})$ > $\ce{H2SO4}$ $(0.3~\mathrm{M})$
So the osmotic pressure which is related to concentration by the relation $P=cRT$ varies as:
Cane sugar > $\ce{NaCl}$ > $\ce{H2SO4}$
Is the following reasoning correct?

That formula for the osmotic pressure is known as the van 't Hoff law. It is described in some detail at the Wikipedia page for osmotic pressure, which also includes a brief (albeit incomplete and rather sloppy) derivation. To reach the van't Hoff law
$$\Pi = cRT$$
you have to approximate $\ln x_v$ by a Taylor series $(\ln x_v = \ln (1 - x_\mathrm{solute}) \approx -x_\mathrm{solute}$ which holds for small values of $x_\mathrm{solute})$ to get
$$\Pi = \frac{x_\mathrm{solute}RT}{V_\mathrm{m}}$$
The Wikipedia page uses $V$ to refer to the molar volume of the solvent (which should really be represented by $V_\mathrm{m}$).[1] In any case, $x_\mathrm{solute} = n_\mathrm{solute}/n_\mathrm{tot}$ and $V_\mathrm{m} \approx  V/n_\mathrm{tot}$ such that
$$\frac{x_\mathrm{solute}}{V_\mathrm{m}} \approx \frac{n_\mathrm{solute}}{V} = c_\mathrm{solute}$$
and if the solute is a 1:1 electrolyte, such as $\ce{NaCl}$, then you have to multiply by two to account for the total number of solute particles.
I am not sure about the term "effective concentration" (I personally have never seen it, and if I had to, I would probably write something like the "total concentration of solute particles") but the logic used is correct as long as you are in the regime where the van 't Hoff law is applicable i.e. very dilute solvent.
Lastly you should probably use $<$ instead of $>$ in your ordering to be clear.

[1] In the actual derivation this should strictly be the partial molar volume of the solvent
$$\overline{V}_{\!\!\mathrm{m,solvent}} = \left(\frac{\partial V}{\partial n_\mathrm{solvent}}\right)_{\!n_\mathrm{solute}}$$
but this can to a good extent be approximated by the molar volume of the pure solvent.


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How to measure ppm of copper
What is the best way to measure ppm of copper in colloidal solution.  My TDS meter only shows at 2ppm which cannot be correct.

provide qualitative solution composition please.

The absolute approach:
Note: colloidal copper is small flecks of copper mixed with purified water. Typically colloidal copper is sold in dilute quanitities (like 100 ppm).
Measure and record the density of the colloidal copper solution (in grams per liter). Measure and record the weight of an empty flask. Put a sample of the colloidal copper solution in the flask. Measure and record the total weight of the sample (and flask). Subtract the weight of the flask from the total weight of the sample and record the total weight of the sample (only). Boil off the aqueous portion of the sample. Weigh and record the total weight the container of dried copper. Subtract the weight of the flask from the total weight of the container of dried copper and record the weight the dried copper.
Use the following conversions to determine the ppm of copper: 1 ppm = 1 mg per liter; and sample weight/density = liters of sample.
Example:

sample density = 1000g/L
(total) sample weight = 250g
(dry) copper weight = 25mg

250g/(1000g/L) = 0.25L
25mg/0.25L = 100 ppm.


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Water contamination of Polyol F0350
I just got asked this question on an interview:
I transported Polyol F0350 in 3 tanks. Upon arrival one tank was determined to have an unacceptable water content. It may have been contaminated with seawater, industrial water or distilled water. What tests do I perform to determine the water source and what type of mitigation do I recommend to be most cost productive - off blending the product or offering it for sale at a reduced price?

What was your answer? How would you think about the problem? What are the differences in composition of the three water sources?

As far as the first part of your question is concerned, think in the conductivity of destilled water vs seawater. The second part of your question more related to quality management and the policies for nonconformity.

Tricky question, but it probably has a trivially simple answer if you have experience with this problem, such as the 'look' of the mixture in the tanker. Not quantitative of course but possibly accurate enough, thus it could be worth asking others with experience for their opinion.
Assuming that water is intimate with the organic, i.e. dissolved or as an emulsion, some water needs to be separated out then there is the Distillation method which uses an immiscible organic with a boiling point higher than water, say xylene. The water is quantitatively carried over with solvent and can then be analysed.
Conductivity would be one way but not conclusive, as both industrial water and sea water may conduct or industrial water may not conduct and so be similar to distilled water. However, the distilled water may also conduct as its not de-ionised water. A simple test unlikely to be definitive definitive, but if its sea water, then a simple flame test should show sodium, as may industrial water, but distilled water far less so. Industrial water may contain soluble organics as well as various inorganic ions so may be coloured, acidic or alkaline and have a distinct smell. The sea water and distilled water have known pH.
Ultimately you will want to do a proper analysis, say GCMS.
Finally, after separation, water could still be present in small amounts in the organic then there is the Turbidity method or the Karl-Fisher titration to determine the amount of water left in the organic layer. However, I suspect you are asked to to a test at the back of the tanker or at least with very simple apparatus.
Mitigation is decided by company policy and this probably depends on what the cost is to correct the problem.
I hope that someone with real experience of this can help, that is when they have stopped laughing at my amateurish ideas.


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How can alkali metals form covalent bonds?
Alkali metals have spherical outer s orbitals unlike elements like carbon which have protruding p orbitals that can "attach" onto other orbitals to form covalent compounds. Even in cases like methane elements like carbon can form hybrid orbitals that still have a protruding p like character.
So how can covalent alkali metals form covalent bonds given that their orbitals are spherical and not protruding?
Does the s orbital distort to be more p like in covalent compounds?
Can the s orbital hybridize and steal a more protruding like character from internal orbitals or from the orbitals of another atom?

hydrogen atom also only has s orbital. They can form covalent bond in H2.

You should really read about bonding and antibonding orbitals.

A covalent bond can be described as a bond between two atoms, where the bonding electrons are "shared" more-or-less equally between both atoms.
Therefore, for an alkali metal to form a covalent bond, it will need to be bonding to an element of similar electronegativity.
If this were to happen, then the s-orbital of the alkali metal will hybridize with an orbital from the bonded atom to form 2 molecular orbitals: one that is bonding (has significant electron density between both atoms) and one that is anti-bonding.
As mentioned above, the molecular orbitals of $\ce{H2}$ are a good example of how s-orbitals can form bonding molecular orbitals.

For the most part, they don't.
However, lithium sometimes can. This is due to their valence shells being relatively close to the nucleus, so they have some electronegativity and can form somewhat covalent bonds.
For more info see this.


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Barium sulfide vs Ba2S2
I was predicting a very simple chemical equation
$$\ce{Ba + S ->}$$
And I had solved it as
$$\ce{Ba + S -> BaS}$$
The answer should make sense as barium has a charge of +2, and sulfide has a charge of −2; however, the book had said that the answer was
$$\ce{Ba + S -> Ba2S2}$$
I do indeed know how and why they got the answer, but I just want to know why my answer isn't correct, or if it is and I'm just acting silly.

Don't know what your book is saying, no such compound exists, whereas BaS is well-known https://en.wikipedia.org/wiki/Barium_sulfide

see this:-http://chemistry.stackexchange.com/questions/8438/reaction-between-zinc-and-sulfur

As orthocresol pointed out, $\ce{Ba_2S_2}$ does not exist as it's own substance. Perhaps you've misread the question and it instead asks for something else? Otherwise it is wrong, or at least not simplified. Your answer is in effect the correct one, although I'm not sure if sulfur and barium would react like that exactly.
It might in addition generate multiple polysulfides, such as $\ce{BaS_2}$, $\ce{BaS_3}$, following the rule of $\ce{BaS_{$x$}}$. However, I believe your answer is correct in that the initial reaction between barium and sulfur would produce $\ce{BaS}$, while further reactions might add more sulfurs onto it.

BaS2 does exist, BaS3 and BaS4 too (also @ortho). You should not answer like that, I'm tempted to downvote you.

@Mithoron Alright. I can see what you mean, however, I do not know what exactly it would yield. My answer does not add any information on what the actual product would be, so I will likely remove it soon if I cannot come up with a better one.

@Mithoron It seems it might generate multiple polysulfides, as you mentioned. I will try and explain this well enough, but since my understanding is rudimentary, my answer might not be the best. If it is still not up to par, downvote and I'll remove it.

Welp, it has now been changed. I have received a downvote before the edit, so I will wait for one more.

Wait, there was Ba2S2... Oh my, sorry I didn't down vote you and question is bad. Oops!

@Mithoron Thanks for clearing it up. I'm relatively new to chemistry, so I am trying to make sure my answers are correct or do not exist.

Obligatory poke towards using \ce{...}

"Exist" is not quite the right word. Why don't we write $\ce{BaS}$ as $\ce{Ba2S2}$, really? Sure, it does not contain isolated particles $\ce{Ba2S2}$, but it does not contain $\ce{BaS}$ particles either. It is not that $\ce{Ba2S2}$ is "not true"; it is that writing it like that has no point.

I am just curious to know if it would be a valid answer to give Ba2S2, instead of BaS. Since they both give the same ratio of Ba : S.

@Dan I would say no, not normally. As Ivan has pointed out it is true, but it is like saying the answer to 20 divided by 4 is $10/2$. This is correct, but not simplified.


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Should you work in metres-cubed or litres in chemistry?
Generally, is it better to work in metres-cubed or litres in chemistry?
My undergraduate degree was in civil engineering and we always worked in $\mathrm{m^3}$.
However, I notice the gas constant $R$ is used in a few different forms:
$$R=0.0821\mathrm{\frac{L\cdot atm}{mol\cdot K}}$$
$$R=8.3145\mathrm{\frac{J}{mol\cdot K}}$$
$$R=8.2057\mathrm{\frac{m^3\cdot atm}{mol\cdot K}}$$
among others.
I guess it depends on the application but is it generally easier to work in litres or metres-cubed?

In chemistry, I have only ever used $\mathrm{\frac{J}{K \cdot mol}}$ for $R$ and only ever litres for volumes.

Generally for calculations involving R where I'm given a table of R values, I use the units that are the easiest to cancel out. I agree with Jan that litres are typically the only thing I use practically; the only real reason for metres-cubed are those general chemistry unit conversion problems.

In chemistry, unlike engineering, volumes are usually reported in dm$^3$, rather than m$^3$ such as mol.dm$^{-3} $ for concentration and dm$^3 $.mol$^{-1}$. s$^{-1}$ for second order rate constants.

Values for the gas constant are tabulated for various different units as a matter of convenience, since when carrying out calculations, the units of the gas constant need to match out with the other values being used (which in turn depends on the expression being used in the calculation) and its often easier to look up a value of the gas constant than to convert everything else into a specific set of units.
Generally, chemists aim for SI units, however there is a great deal of discrepancy between, for example, chemists in the USA vs. chemists in Europe. This is especially apparent in physical chemistry.
For example, the SI unit for energy (+work, +heat) is joules, such that the 'correct' way would be to express energy is terms of kJ/mol or J/mol, however, for historical reasons, many american chemists still work in calories (ending up with kcal/mol etc).
In terms of the volume question, we have equally confusion.
For small quantities, volumes are measured in units like millilitres, micro litres etc, though cubic centimetres (CC) do occasionally get used. For calculations involving moles/concentrations however, we default to decimetres and metres cubed. (you should note that 1 decimetre cubed == 1 litre)
In conclusion, in an ideal world we'd all use SI units, but various quirks have meant that we don't. If in doubt, look at what other people in the literature are doing, and, whatever units you choose, make sure that the units in your calculation match up.

Of importance is the fact that the litre is a more convenient unit for the bench chemist whose daily work typically involves smaller scale solutions and reactions. The cubic meter (= 1000 L) is more convenient for the chemical engineer because it likely more closely matches the scale.

The units depend on the context that you are using at the present moment.
We generally perform most calculations in SI units hence, cubic meters is preferred,  unless everything that you are dealing with is not in SI units.


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How to generate fully coloured electrostatic potential maps in Avogadro?
I want to generate electrostatic potential maps of molecules using Avogadro, that show the colour scheme ranging from blue through green and yellow to red.
In this Youtube video, at 1:32, the video author does just that.

However, when I try the same thing on my Avogadro (version 1.2.0), it shows this:

How do i display the colourful electrostatic potential map on my computer? My computer has a 4 GB RAM and an NVIDIA GeForce 210 Graphics Card.
Thank you!

Did you ask the Avogadro support forum? That seems a more appropriate venue for this question.

The answer is that the color scheme changed, due to user (and developer) feedback. More people preferred the red to white to blue scheme for electrostatics.
In Avogadro 2, we hope to have a full-featured gradient editor, allowing you to set the colors you wish.

5 months since this post and situation seems the same

@TonyM - as indicated in the comment above, there's a support forum for Avogadro and you can post requests on GitHub - while I use Chem.SE, it's no substitute for a GitHub issue tracker.


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Volume increase when air is enriched with water
When air is enriched with water, how much volume increase can be expected?
I am aware that the volume increase will not be proportional (i.e. $\pu{1L}$ air + $\pu{10mL}$ gasified water will result in a volume of less than $\pu{1.01L}$).
So how big is the volume increase actually?

At normal pressures, water vapor is pretty much an ideal gas, as is air.

The short answer is that the behavior of the system you describe should very well approximate that of an ideal gas at room temperature and 1 atmosphere, and that the addition of $\pu{0.01L}$ of water vapor to $\pu{1.00L}$ of air will give you a total volume very close to $\pu{1.01L}$ (assuming constant temperature and pressure).

The departure from ideal behavior for water vapor strongly depends on relative humidity. While the absolute humidity given in the question is only $1\%$, this would correspond to $100\%$ relative humidity at $\pu{6^oC}$, and greater than $100\%$ relative humidity at lower temperatures. Gas molecule interactions so strong that condensation occurs essentially represents the opposite extreme of the assumptions made for an ideal gas.
So, at the extreme of very low temperatures, essentially all of the water vapor would condense and your calculation would have a high bias of 1% if you were to assume ideal gas behavior.  However, at room temperature (i.e. $\pu{20^oC - 25^oC}$) this will not be the case and the assumption of ideal behavior would introduce negligible error.


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Is isobutanol miscible with ethanol?
I can only find info on sec-butanol (aka butan-2-ol) and ethanol. I need to know whether isobutanol forms an azeotrope with ethanol. Can you direct me to a legitimate info source too, please?

http://www.sciencedirect.com/science/article/pii/037838129180013L

From PubChem, 2 butanol is soluble in water, very soluble in acetone and miscible in ethanol and ethyl ether. If acetone is not a problem you can  dissolve in a little acetone then add to ethanol.

Unlikely. Compounds with similar polarity/chemical structure do not form aceotropes with each other. https://en.wikipedia.org/wiki/Azeotrope#Mechanism

I've come to the conclusion the EtOH and iBuOH are miscible and will therefore not form an azeotrope based on the info provided here and elsewhere.

That's a non sequitur. Compounds may be miscible and still form an azeotrope, like water and ethanol do.

According to the Merck Index${^{[1]}}$

isobutanol is miscible with ethanol

1) O'Neil, M.J. (ed.). The Merck Index - An Encyclopedia of Chemicals, Drugs, and Biologicals. Cambridge, UK: Royal Society of Chemistry, 2013., p. 953


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Claisen Condensation vs. Trans-Esterification
In the Claisen condensation, an alkoxide $\ce{RO-}$ acts as a base. However, in trans-esterification, an alkoxide $\ce{R'O-}$ attacks the carbonyl carbon of an ester as a nucleophile. Is this because a different 'R' group is used in trans-esterification?

RO- doesn't act as a nucleophile in the claisen condensation. It's like an Aldol but the alkoxy group leaves giving a beta keto ester.

my bad, I meant it acted as a base

In the Claisen condensation, notice that the alkoxide base usually matches the ester given. For example, if an ethyl ester is used, the base will be ethoxide. The reason for this is that trans-esterification is occuring, except that the product is the same as the reactant.
In general, the Claisen condensation is slow compared to trans-esterification. The rate limiting step of the Claisen is the C-C bond formation. Keep in mind that there will be very little of the active ester enolate nucleophile present, because the equilibrium of alkoxide deprotonating an ester's alpha-position will lie dramatically to the ester (by ~ 10 billion fold).
If one tried to run a Claisen condensation where the base does not match the ester (e.g., an ethyl ester with methoxide as the base), I would expect a mixture of methyl and ethyl esters as the products.

Figures to be added when I get a chance later.


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Carbocation Rearrangement - Till which neighbouring carbon is it possible?

In the above reaction, clearly the OH will be protonated followed H2O leaving, seeing that it's a good leaving group.
Now there can be two possibilities either a alkene is formed as is, or the carbocation rearranges and ends up opening the ring, thus increasing the final stability of molecule, followed by alkene formation. (5 carbon ring is more stable)
But as per some texts, the Carbocation only rearranges until the neighbouring carbon and then alkene is formed without ring opening. Is that correct, if yes, why?

Consider that in the situation you present, it would be much easier to lose the proton to form the alkene immediately, than to migrate the positive charge until a position where ring-opening is viable.
Not to say it is impossible, but the required 'steps' to get to that point are very improbable to the point that it doesn't happen.

I think that this certainly is the correct approach. Also, my professor mentioned that carbocation rearranges to the adjacent carbon only if the new intermediate carbocation is more stable. Clearly, here, after one rearrangement, no more stability can be achieved by rearranging to neighbouring carbon, and hence it forms an alkene.


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Does boiling water alter the chemical properties of dishwashing liquid?
An ex-housemate of mine claimed to be a former chemist, and once told me not to use boiling water (or recently boiled water) when leaving dishes to soak, as the temperature would chemically alter the dishwashing liquid, rendering it useless or at least significantly less effective. However I've never found anything in internet searches to support this assertion. Can you lovely peeps help give me a definitive answer?

It depends rather strongly on your brand.
Soap, by itself, is unlikely to be affected by the boiling water. They're relatively stable, especially for an organic molecule. However, many of us use detergents that include enzymes that may denature in the boiling water, which is what your friend may have been referring to: A denatured enzyme is not very effective at cleaning at all.
So, look on your bottle! If it mentions enzymes, it probably won't go over well with boiling water. If it's just soap, and, say, bleach, boil away!


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Ferric Chloride solution weight
When I notice that ferric chloride is sold as $40\%$ solution w/v, does it mean weight / volume?
Is it a matter of for example adding $40$g of ferric chloride crystals to a water solution where the total volume will then be $100$ml?
Or do I have to workout via the molar mass of $\ce{FeCl3}$ to water? How do I calculate the weight of crystals necessary to create a $40\%$ solution of $\ce{FeCl3}$?
Thanks
EDIT:
The system seems to think this is a homework question (whatever that is). If I divide $100$ml of water (so $100$g) by $40\%$, I get $40$g of $\ce{FeCl3}$. If I dissolve $40$g of $\ce{FeCl3}$ in $100$ml of water, will I get a $40\%$ solution of $\ce{FeCl3}$? And a total volume of $114$ml? ($\ce{FeCl3}$ is $2.9\frac{\mathrm g}{\mathrm{cm^3}}$, so $40$g is approx $13.8$ml)?

Welcome to Chemistry! This is a homework question. We have a policy which states that you should show your thoughts and/or efforts into solving the problem. It'll make us certain that we aren't doing your homework for you. Otherwise, this question may get closed. Please [edit] in your full reasoning or thoughts on this.

What? That doesn't make any sense?

You're right in that 40% weight/volume means that you take 40 grams of ferric chloride and dissolve it in 100 mL of water. This is the definition of w/v(%) and there is no need to work out the molar concentration - the main benefit of this method of stating concentration.
The total volume will not be (volume of liquid + volume of solid) in most cases of dissolution. In most cases, you will observe very little change in the volume of the resulting solution as the solid molecules separate and fit into the spaces between the liquid molecules. You may even occasionally observe a decrease in volume as shown in this experiment.

My scientist friend explained to me that a 40% solution means that in 1 liter of solution there are 400 grams of solute. In your case you would add 40 grams to some water, dissolve it, then top it up to 100ml total solution.
If you add the solid to the liquid you have don't know what the total volume will be - therefore you don't know what the %-age will be. That's why you make it up to the final specified volume.


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What happens when you mix milk and coke?
Add small amounts of milk to a bottle of coke, and you end up with a white-ish precipitate with a clear liquid on top. "The Internet" seems to agree that phosphoric acid in coke reacts with something in the milk (possibly proteins, it seems to me). The Telegraph even wrote that it is the "milk molecules" that react.
So what is the reaction(s) that takes place?  I can't find any proper sources that explain this in chemical terms.
Telegraph's milk-molecule analysis:
 This is what happens when you add milk to coke
or via the Internet Archive (without pictures).

The phosphoric acid causes denaturation of milk proteins which precipitate out of the solution (milk is an emulsion, to be exact). The flock that is formed also adsorbs dye molecules (caramel) which is why the solution become clear.
There is nothing coke-specific about it. Any acid would denature milk just like that. And nearly any huge organic molecule would get absorbed, just like caramel dye.

This is a bit more complicated than it seems. Milk contains about 3% protein. Most proteins are not soluble -- otherwise our bodies would dissolve when it rained! The most common type that are soluble are the albumins, which rely on their tertiary structure for solubility. Hence anything that disrupts the tertiary structure -- denatures it -- can expose hydrophobic parts of the chain, causing them to agglomerate and form a solid precipitate. In the case of milk, this process is called "curdling" and the soft solid is called "curds." It is perfectly safe to eat, in fact quite nutritious. It is the first step in making cheese, and in any case happens as soon as the milk hits your stomach acids.
Now, typical things that cause albumins to denature include heat, acidity, and concentrated salts. So, soft drinks are usually quite acidic, they cause curdling, problem solved. Hmm, almost. Actually the main protein in milk is casein, and casein is not an albumin. Its method of maintaining suspension in milk is much more complex: the casein micelle. If you are familiar with the micelles formed by detergent: these are different. They share the name because they look similar at high magnification, but definitely don't work the same way. In fact the question of how casein remains suspended in milk is an open research topic. Suffice to say, your denaturant (acid, in this case) denatures something, this disrupts the casein micelles, and they agglomerate to form delicious, nutritious, slightly not-good-looking curds.

very well written

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


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Why is the boiling point of a solution higher than that of a pure solvent?
Why is the boiling point of a solution higher than that of a pure solvent?
I think when a non-volatile solid solute is added to a solvent, some of the liquid solvent molecules are replaced by solute molecules at the surface. The lesser the number of solvent molecules at the surface that fewer molecules evaporate, and thus the vapor pressure of the solution will be lower than the pure solvent.
Is this correct?

Can a solute affect the entropy of the system? Google  colligative properties.

The change in vapour pressure with added solute  is called Raoults law. Other colligative properties are elevation of boiling point, depression of freezing point and Osmotic pressure. These depend only on the number of solute molecules in a fixed volume of solvent, not their nature.

I've  given a rather comprehensive answer about this here. I hope it is helpful.

The rate at which molecules evaporate has nothing to do with where the thermodynamic equilibrium will be.

The change in vapour pressure with added solute is called Raoults law. The vapour pressure of a species i above a solution is proportional to mole fraction of this species (solute) $x_i$ and its pure vapour pressure $p_i^*$ or $p_i=x_i.p_i^*$.
As the solution now contains a fraction of a low vapour pressure solute then this component will not contribute much to the overall vapour pressure which will be less (in this case) than pure solvent.
It is not correct to say that solvent molecules are replaced at the surface as both solute and solvent are distributed uniformly in the solution. Thermal diffusion moves the solute and solvent around in a random manner. It is that there are now fewer solvent molecules as some have been replaced by low vapour pressure solute (in this example) and because they are of low volatility they need more energy to evaporate than the solvent does. As the temperature is constant, by the Boltzmann (energy) distribution fewer solute molecules populate higher energy levels compared to solvent and so have a lower vapour pressure.

YES the explanation is correct.
In fact it is a qualitative property of solutions.
Elevation of boiling point /lowering of vapor pressure/lowering of freezing point all depend on same reason.

No, that explanation is not thermodynamically correct.


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Trans-hydrogenation of alkenes
Both this and that question seem to indicate that the hydrogenation of alkenes to alkanes or alkynes to alkenes is always a syn process.
However since hydrogenation is not a concerted process, I do not see why we could not have a non-negligible (although small) portion of the trans product.
Following this reaction mechanism for syn additions :
before the penultimate step (the addition of the second hydrogen), can't the molecule simply rotate around the $\ce{C-C}$ axis (I expect this should be possible with small enough alkenes) ?
The Wikibooks article on alkenes does indicate that typically the hydrogenation of alkenes leads to only $98\%$ of the syn product, but I have been unable to find more precise information elsewhere.

Yes, the trans product probably will be formed in small quantities. Given the very small size of a hydrogen atom I would guess that the second hydrogen transfer step is very fast, so rotation is unlikely, given that it will be sterically hindered.

Is there some experimental data with concrete examples, and which maybe shows how it evolves with the size of the alkene ? I haven't been able to find any @bon

I don't have the time or resources to search for it now but maybe someone else will turn up something.

This link has a good summary of hydrogenation. Syn addition is very much preferred because of the way the molecule 'adheres' to the surface. As you note, however, there are cases where anti-addition is observed - particularly for sterically hindered molecules such as 1,2-dimethylcyclohex-1-ene which shows up to 30% of the trans/anti-addition product (which also so happens to be the thermodynamic product).
The reason we see the syn-addition more often is because the Hydrogen atom attack happens really quickly (it's the kinetic product). If you pump more heat energy into the reaction though, you could expect to see more trans product - but this is dangerous as hydrogen is explosive.


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Aqua Regia Coloring
I was working with Aqua Regia in the lab today and was impressed by how quickly the solution (HCl and HNO3) changed colors from a colorless liquid to orange to a deep red.  I didn't time it, but I would estimate that the change from clear to orange took under a minute while the color change from orange to red took about an hour.
Why does this color change occur?  Why does the first stage (clear to orange) occur so quickly without mixing?  I'm sure this information's out somewhere, but I couldn't find an answer with searching.

The initial yellow/orange colour is mainly dissolved chlorine ($\ce{Cl2}$) and notrosyl chloride, ($\ce{ClNO}$), the latter affording Aqua Regia much of its potency.
The reds and browns are almost certainly due to its subsequent decomposition into nitric oxide ($\ce{NO}$) and nitrogen dioxide ($\ce{NO_2}$) as $\ce{NO_2}$ is strongly coloured.

Awesome, thanks!  That explains why the change of color from orange to red takes longer as well.

Nitric oxide is colorless.

Good point, @vapid . I'll reword it to make that clear.


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What are the order of the reaction under the following conditions?

The experimental rate of a reaction
$$\ce{A + 2B -> C + D}$$
$$\frac{\mathrm dx}{\mathrm dt} = K[\ce{A}][\ce{B}]^2$$
Calculate the order of the reaction when (i) $[\ce{A}]\gg[\ce{B}]$  (ii) $[\ce{B}]\gg[\ce{A}]$

My efforts :
Should we ignore the concentration of the reactant that is present negligibly? And thus the orders will be:
(i) Order = 1
(ii) Order =2
Please guide.

When you have something in large amount the effect of that species becomes negligible. You are limited by your limiting reactant.

@mamun So it should be the reverse
(i) In case I since B is the limiting reactant the order of the reaction will depend on it and the order will be 2
(ii) In the 2nd case it is A that acts as the limiting reactant and the order of the reaction is 1

Your latest comment is spot on.  Nice job.

Let's compare the rate when $A$ and $B$ are present at their initial amounts, say $A_0$ and $B_0$, to when the reaction is 99% complete.
When $[\ce{A}]\gg[\ce{B}]$, the reaction will be 99% complete when 99% of $B$ is gone, i.e. when $B = 0.01B_0$.
At that point, $$\frac{dx}{dt} = k[A][B]^2 = k(A_0 - \frac{0.99}{2}B_0)(0.01B_0)^2 $$
Now since $A\gg B$ also means $A_0\gg B_0$, then it follows that $(A_0 - \frac{0.99}{2}B_0) \approx A_0$.  Thus, when the reaction is 99% complete and $B$ is limiting,
$$\frac{dx}{dt} \approx kA_0(0.01B_0)^2 = kA_0 (0.0001)B_0^2$$
If we lump $k$ and $A_0$ into a single parameter, $k_{app}$, then we can write $\frac{dx}{dt} = k_{app}B^2$, which shows that when $B$ is limiting, the reaction is 2nd order.
The case for $A$ being limiting shows the reverse.

The order overall is 3 , first in A and second in B.
When [A]>>[B], A can be considered as constant, because all of B can be consumed but [A] is hardly changed. In this situation the concentration of A ([A]) can be lumped in with the 3rd order rate constant $k_3$ to give a pseudo-second order reaction, $ k_{psuedo}=k_3[A]$. This has 2nd order units $dm^3mol^{-1}s^{-1}$. A similar argument applies when [B]>>[A] but with a different pseudo order rate constant.


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Why are aqueous solutions included in the equillibrium constant but liquids are not?
I understand solids are insoluble and will just sit at the bottom of the vessel so we don't include them.
But I am confused as to why aqueous solutions are used in the equilibrium equation but liquids are not.
E.g. for the reaction:

The equilibrium constant is:

Why isn't H20 (liquid) included in the equilibrium constant yet KCl (aq) is?
Thank you

An example would help.

In the example you give, water would be excluded because it is the solvent of HCl and KOH (both are aqueous). Any change in water then, from the equilibrium, would be insignificant.


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Water side-reactions in organic electrochemistry (redux)
(note: this is a reposting of part of my original post which I was told had too many questions in one post)
Please bear with me as I am a chemistry newbie, but I am autistic and have recently developed a fixation on electrosynthesis and must understand it.
I'm interested in split reaction systems, where a salt bridge connects two chambers, and an aqueous media is used. I'll try to split my questions about this into two sections; water alone, and organic synthesis that take place in water.
In the anode chamber, with the use of a catalytic anode (platinum, platinized titanium, palladium, etc), will the free radicals form hydroxides, peroxide, etc. just build up if the water is stirred properly? Or will ions coming across the salt bridge balance it out somehow? I am confused about that because I thought the point of the salt bridge is to isolate half reactions. So I assume you get unbalanced results isolated in each chamber?
When you add a complex organic molecule, how do the electrochemists keep the side-reactions from trashing their molecules? Do they have to factor in the water reactions as part of their strategy to reduce or oxidize the compound?

You lost me in "will the free radicals form hydroxides, peroxide, etc. just build up if the water is stirred properly". Also, "when you add a complex organic molecule"? Add to what?

As potential is developed between the electrodes, the water itself forms acids and alkali at the cathode and anode respectively. These in situ reagents are what I am talking about. They only form near the electrodes, so the media(water), must be stirred somehow. At high enough voltage the bubbling of escaping hydrogen and the formation of the aforementioned reagents stirs for you.
In regard to the complex organic molecule, I am talking about adding it to the water. I'm just restricting it to water to keep it simple, but I do know that redox reagents are used to facilitate some reactions.

How do I get this question taken off of hold? Everything I have posted so far has been appropriate and friendly(though apparently unclear), but I have not had a very welcoming experience here so far.

It would have to be heavily edited.

I'm trying to figure out how to just get rid of my account, but its not nearly as easy as it should be.

You're right, a salt bridge is used to isolate the reactants but still to allow charge to flow. For example, in the case of a Sodium Chloride Salt Bridge, Sodium ions will flow into one chamber while the Chloride will flow into the other. Overall the system must stay charge neutral.
If the system contains water only, on both sides, then what will happen is that you will electrolyze the water and get $\ce{H2}$ and $\ce{O2}$ gas.
As for complex organic molecules, I believe when oxidizing or reducing them the preference is to use chemical reagents instead of electrolysis.

Thanks for your response. That is much the way I was imagining it, but I am still a bit unclear on how this doesn't result in sodium, potassium, or whatever from becoming part of your finished product.
It seems like there is a lot of research going on in Asia, and they even sell machines for complex electrosynthesis that they claim are capable of working CO2 up to complex hydrocarbons. I read a book about this subject that was written in the late 1800s by Walther Löb, and it seems to me like this area of research may have been neglected a bit during a time of cheap petrochemicals...

@DangusMcFinghin Book from 19th century? Get some textbook not ancient stuff.


text:
Is there safe chemicals for making voice pitch or bass?
I'm looking for safe experiment that can pitch and bass my voice.finally I find a video that do it but he said "not try in home,ok..?".Is there safe tool or chemicals for Manipulation of the larynx is used to generate a source sound with a particular fundamental frequency.
other my question is about below viedo link:How these chemicals make voice pitch or bass!(Helium and Sulfur Hexafluoride)
Pitch is a perceptual property of sounds that allows their ordering on a frequency-related scale,1 or more commonly, pitch is the quality that makes it possible to judge sounds as "higher" and "lower" in the sense associated with musical melodies.
video link

There are no safe chemicals except oxygen, nitrogen, and water; and even those may kill if used improperly. If you know what are you doing, go on, breathe helium, it's fun. If you don't, don't.

What if you inhaled both (He + SF6) at the same time?

This is actually a problem in dealing with undersea habitats for divers: special gas breathing mixtures are used, to reduce or eliminate nitrogen narcosis, “the bends”, etc. The U.S. Navy has naturally done lots of research in this area. Look up gas mixtures for deep sea divers or something along those lines. The problem is that communication is weird: divers sound like Donald Duck and electronic voice correction may be necessary.

Assuming that helium and sulfur hexafluoride (as shown in the video) will give you the correct sounding "pitch" and "base", it would be fine if done correctly and safely.
Helium is incredibly inert and so will not cause any sort of toxicity. It is also a very light gas, so it will exit your lungs quickly. Just do not breath only helium, or for a long time or you could suffocate. Make sure to get lots of normal air.
Sulfur Hexafluoride is also rather unreactive, but more dangerous. This is because it is heavy and will tend to stay in your lungs. Only use it for a short period of time, or you can asphyxiate. While the same goes for helium, it tends to exit your lungs more easily.
The actual mechanism that causes them to work is based of how dense they are. It does not change the pitch, however. Instead, it changes something called timbre. This is basically the fact the helium allows sound to travel much faster through it, so it is a similar affect to playing a video at a faster speed.
Sulfur hexafluoride does the exact same thing, except it is much denser and does not allow sound to travel through it very quickly.
If you want to do this, be careful! You can suffocate, especially with sulfur hexafluoride.

Unless you're upside down?

Ehm, well I guess so.

Xenon will also work, but it can act as an anaesthetic so would not be safe if used unsupervised. Short exposure is probably OK.

A  possible safe way to inhale helium is to mix it with 20% oxygen.
Likewise, mix sulfur hexafluoride with 20% oxygen.

Possible safe is not good enough.

You can't just have it be possible,it has to be confirmed to be safe otherwise accidents are going to happen.


text:
What is an active centre of a catalyst?
I was studying catalytic poisoning and read that in temporary poisoning, the poisoners are held at active centres by weak forces.
What is an 'active centre' in this context?

Same as "active sites" or where the catalyzed reaction takes place.

The active centre is the site where the substrate binds to the catalyst in order for catalysis to take place. On a molecular level, most catalysts (or most substances really) are not just flat surfaces, they have complex three dimensional shapes. The substrate often has to bind to the catalyst in a very specific way in order for the catalysed reaction to take place - this is the active centre.


text:
Can a system of 2 consecutive biphenyl systems be optically active , even if any adjacent biphenyl systems aren't optically active?

This question came in an online test, where they asked whether or not 22,26-dibromo-12,32-dichloro-23,25-diiodo-16,36-dimethyl-11,21:24,31-terphenyl was optically active.
The answer is that, it is optically active.
According to Solomon and Fryhle:

"Biphenyl is an asymmetric compound so it must not have a plane of symmetry or center of symmetry to be optically active".
  Hence compounds which have 2 same adjacent groups on a single phenyl in any biphenyl system cant be considered optically active .

Applying that concept above as same groups (Br and I) are attached to the middle phenyl system, the individual biphenyl systems on both sides are optically inactive, hence the compound should be optically inactive.
Please tell me where I am going wrong in my interpretation.

The key point you have to remember is that optical activity isn't decided by a single section of the molecule but rather it is a property of the entire molecule - and while not being able to find a mirror plane is almost a guarantee, the nail in the coffin to decide that a molecule is optically active is if you cannot find an improper rotation axis (reflect and rotate).
There is no mirror plane in the tri-phenyl molecule you show (cut it into left and right the Br and I don't overlap; cut it up and down the methyl and chloride won't overlap; and so on and so forth). Hence, it is optically active.

This is not entirely true, the condition for chirality is that the molecule must not possess an improper rotation axis (of which mirror planes are a subset of)

The presented compounds are typical examples of  atropisomerism.

Here is the correct representation of possible forms of this compound.

Let central ring be in the plain. Then other two rings are twisted. The darker line on a diagram indicates a group pointing towards us. It is important to remember that the angle between the planes of the rings is neither 0$^o$ nor 90$^o$ (otherwise your question would be valid).
Note that -CClBrI radical is not optically active. However, combining two such radicals you can have an optically active molecule IBrClC-CClBrI.

why wouldnt the angle between the planes of the ring be 90 degrees ? due to steric hindrance isn't that the best configuration for them to minimize repulsion ?

@IshitaGupta You would guess it is, but it is not a case. In the case of unsubstituted biphenyl, the equilibrium torsional angle is 44.4° (from wiki about biphenyl).


text:
How do non-sublimable solids exist in equilibrium with vapour?

How do non-sublimable solids exist in equilibrium with vapour?

All substances exist in equilibrium with their vapour, something I learned from Jan's comment. But my question is how does that happen with solids that are non-sublimable, like ice, piece of iron, or, a piece of cotton?

Actually, ice sublimes quite easily. But cotton or iron have so low vapor pressure that I think it is unmeasurable (iron is non-volatile because of a high crystal lattice energy, and cotton because it is made of very large molecules that are bound strongly together).

@vapid So could we say that though verys less but iron and cotton both sublime?

No, I wouldn't say that. I think that even after billion years you wouldn't be able to observe any loss of mass due to sublimation of these substances. So at any reasonable timescale it is safe to say that iron or cotton doesn't sublime.

Thermodynamically, even cotton exists in equilibrium with its vapour, even though this is completely silly in any practical sense. The reason is that thermodynamics does not involve time; it may take $10^9$ years, but so what! The equilibrium may only consists of one molecule / km$^3$ but $technically$ its still equilibrium! This is because you can't arbitrarily pick and choose the cases you want thermodynamics to apply to, but you can apply reasonable physical insight based on observation rather than applying theory blindly.

There is no such thing as 'cotton vapor.'  Thermodynamically, cotton fibers decompose.

I agree with porphyrin and disagree with Agriculturist's comment.   For a given cellulose chain length in a given crystal structure, there will be some theoretical vapor pressure of that cellulose chain.  It might be lower than the "vacuum" of intergalactic space, but there is still some non-zero value.

So a shirt converts to its 'shirt vapor' if thrown into a theoretical vacuum.  While this sounds absurd,  a shirt is only slightly more complex than a cotton fiber.  A living cotton fiber is covered in pectin and other compounds.  The inside has a lumen which dies off when the bowl opens.  The middle layers contain mostly cellulose with varying chemical properties.  This answer needs refinement; for instance, it may make sense to talk about a homogeneous compound like cellulose along with which theory, body of literature, or viewpoint this answer refers.

This questions seems based on out-dated theory of olefaction (which is the science of smells.)
In the old theory, the nose detects minute particles of a substance in air by receptors keyed for specific particles.  These particles are assumed to be in the air because we can smell them.
This old theory can be refuted based on three observations:

Compounds with different chemical compositions can smell
exactly the same. (creosol and eugenol)
Compounds of different isotopes can smell different. (musk)
Concentrations of compounds can change the odor completely. (phenol)

Since the advent of genetic technologies, a new theory is on the horizon is the underlying mechanism for smell is based on quantum oscillations.  This theory is derived from information gathered from studying the genome which encodes the proteins responsible for the sense of smell.
The metallic smell does not come from metal.  One current theory  is the metal causes lipids on the skin to oxidize.  The oxidized lipids are what travels to the nose.  The nose is actually picking up a human scent of blood at a very small concentration.  Even at small concentrations, blood is something the which the nose is extremely sensitive.

This reply does not answer this question. The idea of molecular vibrations of one type of molecules vs. another causing us to perceive different smells, 'quantum oscillations' as you call them, is still very speculative.  I am very surprised that isotopic substitution say H for D, such as in naphthalene (moth ball smell), causes a difference. If it did it would be a plus for  a molecular vibration to smell model. I don't think that its true that we are $very sensitive$ to smell of blood, taste maybe, but if I have small cut I don't smell  blood.

@porphyrin - The question is based on a premise derived from outdated  science which has been recently called into question.   The odor detection threshold for blood is typically in the parts per trillion range; however,  your threshold could be different based on both your geneotype, phenotype, alleotype, health, and psychological biases.

Looks you wanted to recycle your answer, but it didn't go well. Question is simply about non-zero vapor pressure not smell.

@Mithoron -  The question references 'Jan's answer' .  The answer to a different question is what created the confusion, so I answered the question here then backtracked to the older answer and included the information there as well.   Looks like you were jumping to conclusions without studying the question.

It references Jan's comment http://chemistry.stackexchange.com/questions/39540/is-every-substance-in-equilibrium-with-its-vapours#comment62324_39540

One more http://chemistry.stackexchange.com/questions/38465/how-are-odor-and-electronic-structure-related-to-aromaticity


text:
Is it possible to observe ATP energy transport in table experiment?
Is it possible to take AMP or ADP, then phosphorylate it convert to ATP to charge with energy, then transport to other place an then observe energy freeing?
In table / home / school experiment?


text:
Benzalkonium chloride breakdown time

I wonder if and how long (approximately) it will take for benzalkonium chloride (shown above; $n = 8, 10, 12, 14, 16, 18$) to breakdown (degrade fully or partly) in a closed water loop with continuous flow and a volume of approximately $1.5\ \mathrm{L}$ distilled water.
The concentration will be around $2\ \mathrm{ppm}$ to $10\ \mathrm{ppm}$.
UV and general light will be kept to a minimum but can't be fully shut out. The temperature will flucturate between $19\ ^\circ\mathrm{C}$ to $50\ ^\circ\mathrm{C}$.
Thank You!

Are you trying to break it down asap or keep it active for as long as possible? It is very stable under neutral or slightly pH, but it can decompose in 2M NaOH. See the following reference for details (Journal of Liquid Chromatography & Related Technologies, 2012, 35:2, 231-239, DOI)

I'm sorry for the late reply, didn't notice the question had been answered until now. I am trying to keep it active as long as possible to use it as an biocide in water. I don't know if or when to add more to compensate for degradation.


text:
Oxygen and Hydrogen
I wonder if two gases , Hydrogen and Oxygen (product of electrolysis, HHO) are placed in a glass jar and let it seat for a few hours, will they separate (Oxygen to the bottom and Hydrogen to the top) by it’s own weight difference ? Thanks

possible duplicates:  http://chemistry.stackexchange.com/questions/43118/can-gases-be-immiscible/43124#43124  http://chemistry.stackexchange.com/questions/43320/why-do-we-consider-all-gasses-to-be-in-one-phase/43321#43321

At room temperature, gas molecules are in random motion, bouncing off each other and occupying the available space. Gases would diffuse and form a homogeneous mixture, just like air. However a gas centrifuge can generate enough differential acceleration to separate them based on their molecular masses.
[ A gas centrifuge is a device that performs isotope separation of gases. A centrifuge relies on the principles of centrifugal force accelerating molecules so that particles of different masses are physically separated in a gradient along the radius of a rotating container. - Source:https://en.wikipedia.org/wiki/Gas_centrifuge]

A few hours is not long enough for the two gases to separate.
This experiment http://lss.fnal.gov%2farchive%2f2005%2fconf%2ffermilab-conf-05-635-a  conducted at Fermilab to verify safety procedures for the release of asphyxiant gases, suggests that gases of sufficiently different densities will stay striated for some time if the release of the heavier or lighter gas is slow enough as not to disturb the present atmosphere, but that gases that are initially mixed will remain mixed for much longer than several hours.


text:
Why are Earth's oceans salty?
As we all know water is $\ce{H2O}$, it does not contain sodium chloride. The ocean is a solution composed of water and a solute which is salt. I was wondering how the massive amounts of $\ce{H2O}$ in our oceans first mixed with salt? I'm sure the water didn't contain salt when it first got to the oceans so how did it get mixed with salt? Thanks.

duplicate http://earthscience.stackexchange.com/questions/2956/what-is-the-origin-of-the-oceans-salt

But it is  not in Chemistry Stack Exchange

Early history of Earth is a question to paleohydrology and paleochemistry. Briefly, water wasn't pure on the early stages of Earth.
Brief history of water on Earth. The planet was hot and seismoactive (lots of vulcanos) in the beginning. Water (as steam) escaped from magma, cooled down and precipitated on the Earth's surface. The surface already had salts on it, so it became saline from the very beginning. A better question is: how come we have fresh water in rivers and saline water in oceans? This separation is due to water cycle on our planet. Rivers bring dissolved salts to the ocean, then evaporates and precipitates over the land. Salts cannot evaporate (at any practical rate), so they stay in the ocean. This is similar to washing dishes. Consider this model: dish gets cleaner - that is a river; drain water gets dirtier - that is an ocean; drain water goes through purification and becomes clean before going to the supply pipe - this is evaporation of water from the surface of an ocean and precipitation over the land in our model.

The most common answar is-
Water is a good solvent. When it flows, many kind of salts released by rocks dissolves in it. All these salty water goes to oceans and make them salty.

This reminds me of my school days when we had to watch this movie:
Hemo the magnificant
It's an old movie but worth the time to watch it.
Actually sea water contains traces of all known natural elements including sodium and chlorine ions and our blood is an analog of the sea water of which it evolved.  Salt does not exist in water until the water is evaporated away.

I'm completely baffled by your last sentence. It seems to contradict itself.

It's not salt unless it is a solid so evaporation must take place to crystallize to salt. We call it salt water but it is the sodium and chlorine ions that make it "salty".


text:
Is this chemical name wrong? 2-bromobutane vs 1-bromobutane
Well, technically it's already wrong because it should end in "-propane" but my textbook is so crappy let's just ignore that for a second. The naming convention goes that substituent groups should be named so that they have the lowest number possible. Then, in both of the bromopropanes below, shouldn't they both be 1-bromopropane because there are no other substituent groups besides the bromine?

I completely don't get your question.

@Mithoron since we're supposed to number the substituent with the smallest one we can give it (whether we read the carbon chain's numbering left-to-right or right-to-left, in the above example), can't we write 1-bromobutane as the name for the 2-bromobutane if we just read the carbon chain from right-to-left?

@MaxLi No. The bromine is not on the first carbon, regardless of which direction you number it from.

The numbering of locants in the question is correct. In acyclic hydrocarbons, the longest chain is numbered from one end to the other.

The corresponding subsection of the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) reads as follows:

P-15.1.7.1.4 The longest chain is numbered from one end to the other by arabic numbers, the direction being chosen so as to give the lower locants to the substituent groups (side chains) [see P-14.4(f)]. (…)


text:
Do 2-hexyne and hexa-2-yne mean exactly the same thing?
Do 2-hexyne and hexa-2-yne mean exactly the same thing? If so, what is the purpose of having two naming conventions?

These are simply old and new nomenclature.

The latter, more systematic nomenclature variant is necessary for more complex compounds, but complete overkill for simple, monofunctional chemicals. "Two-hexyne" is a lot easier to get than "hexatwoyne".

For 2-hexyne, the extra naming possibility isn't really a help.
However, consider a more complex molecule, say, a hexynol.  It has both an alkyne and an alcohol.  A name like 2,4-hexynol would be very confusing.  Which group is at carbon 2 and which is at carbon 4?  That is why IUPAC doesn't provide for that possibility.  Instead, Writing it as hex-4-yn-2-ol is actually clearer and more understandable.

The names “2-hexyne” and “hex-2-yne” describe the same compound. The name “2-hexyne” is in accordance with obsolete IUPAC Recommendations. The relevant rule of the IUPAC Nomenclature was changed in the 1993 IUPAC Recommendations:

R-0.1.2 Position of locants
Locants (numerals and/or letters) are placed immediately before the part of the name to which they relate, except in the case of traditional contracted forms (see R-2.5).

The corresponding subsection of the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) reads as follows:

P-14.3.2 Position of locants
Locants (numerals and/or letters) are placed immediately before that part of the name to which they relate, except in the case of the traditional contracted names when locants are placed at the front of their names.

Therefore, the preferred name in accordance with current IUPAC Recommendations is “hex-2-yne”.


text:
Will a Dole Pineapple juice can made out of tin leach the tin into the juice & cause harm?
I drink a lot of Dole Pineapple juice from the can, because it has no added sugar or ingredients. Can it be harmful because of the tin in the can? I called Dole and the can is not lined.

No, otherwise they would get sued. If their can contained something that is toxic and leaches out, they would have legal problems galore.
Tin is very low in its toxicity, so you can eat and drink out of it and be fine. While the pineapple contains some acids, it is unlikely that they will react very vigorously with anything in the can.
As a side note, however, some "tin" cans are actually steel, which is also not very toxic.
TL;DR You are fine, enjoy your pineapple juice.


text:
Work done on a container, resulting in pressure change
Why does pumping gas into a rigid container (causing the pressure and temperature to increase), not mean that work has been done on the container?
Work is given by the formula :  W = -P ΔV
If I go off the formula alone, ΔV=0 (as the container is rigid) and hence W=0
But if I ignore the formula and take the definition of work as "the energy it takes to move an object against a force", wouldn't pumping gas into the container require work in itself, due to the effort required to get the gas molecules into the container e.g. like pumping a bike tyre, the gas is not going to go into the container unless I pump it in?

For reference:
Question from textbook: A rigid container of constant volume is used to store
compressed gas. When gas is pumped into the container,
the pressure of the gas inside the container is increased and
the temperature of the container also increases. Which
statement is true of the work done on the container?
(a) The work is equal to the increase in the pressure
inside the container.
(b) The work is equal to the increase in the temperature inside the container.
(c) The work is equal to the sum of the pressure and temperature increases.
(d) There is no work done on the container.
Answer from textbook:  (d) the container does not expand, so as there is no
change in volume of the container, no work is done on the
container.

Which textbook is this?

Yes, the explanation

the container does not expand, so as there is no change in volume of the container, no work is done on the container.

is right as it takes into account that the container is rigid, so its volume can not be changed. Thus, there is no work done by the gas on the container.
Also, if you say

" pumping gas into the container require work in itself, due to the effort required to get the gas molecules into the container "

Yes, there is work done on the gas, but it is by the piston(or the pump).
You should see how the laws of Thermodynamics make clear about the boundary between system and its surroundings, so work has to be done by "something" on "something".
Also, In the question, it is clearly asked "Which statement is true of the work done on the container?" So, as the container's volume is unaltered, there is no work on it.

Thank you for the response ABC Dexter.
That makes sense.

In the open system version of the first law, the work W is divided into two distinct parts:  the work involved in pushing material into and out of the system, and everything else $W_{shaft}$.  The latter is loosely referred to at the "shaft work" and the former is always conveniently lumped in as part of the enthalpy of the streams entering an leaving the system.  In your problem, because the container is rigid and there is no rotating shaft present, the shaft work is zero.  However, there is still the work involved in pushing material into the system, which is lumped in as part of the inlet enthalpy.


text:
How to Make Ferrofluid in a Bottle
I am trying to accomplish the very famous magnetic Ferrofluid in a bottle aka office desk toy.
Before anyone suggest to buy online, I would like to make it myself and learn the whole process VS store bought item.
here is a short video of it
Grade 11 chemistry was the last time I studied chemistry.
I would like to create the similar/same toy
My first question is: is ferrofliud thick liquid you can just buy online or do you have to make it yourself using ink toner or magnetic powder etc
I tried making it  using ink toner powder mixing it with vegetable oil. The results were not the best but worth a shot, it ended up thick blob and alot of particles it was not clear at all and very hard to see through it.
for the clear liquid I used 50% cleaning alcohol,
i tried using water before and it just ended up making the whole glass black.    So in short this is the measurements I used and once again I apologize if I forgot to mention anything.  I used black ink toner mixed it with vegetable oil, ink toner 10 grams vegetable oil 20ml, and 100ml was cleaning alcohol
Alot of people suggest use magnetic powder instead of ink toner what do you recommend? What different liquids should I be using so it does not mix and stays clear.  Some people suggest online make sure you only use water without minerals and iron and many other liquids

related http://chemistry.stackexchange.com/questions/409/what-kind-of-suspension-liquid-should-be-used-with-ferrofluid-so-it-does-not-st

It is always easier to buy ferrofluid than to make it. Some of the steps don't work well unless the conditions are just right and it is hard to get them right on your first attempt. Luckily there are plenty of suppliers (this UK supplier is top of my Google search for "ferrofluid sales" and they also supply lots of cool magnets).
Ferrofluid is a suspension of small magnetic particles (often magnetite, the magnetic oxide of iron) in a fluid. To avoid the heavy and magnetic particles clumping together or precipitating, they need to be coated in a surfactant. And this only works if the particles are the right size (very small) and you have the right mix of surfactant that works in the carrier fluid (the fluid is usually mineral oil or vegetable oil; alcohols will be a poor choice).
Laserprinter toner isn't usually a mix of suspended magnetic particles: only specialty toner inks are (the ones used for machine readable printing). If you can get the right sort, it will save a lot of effort, but the wrong toner won't work at all and will just make a mess.
There are quite a few recipes on places such as Youtube, but some of them don't work well or don't explain some of the important details. So, if you don't want to spend a lot of time experimenting to get the synthesis right, just buy the ferrofluid.

You have to be sure to use the type of toner powder that is made to print "machine readable type" like on checks.  It is magnetic, the regular toner powder is not magnetic and does not work. It makes a mess.

I have not yet tried it but there is a youtuber who has made excellent ferrofluid at home. The broad steps are as follows :

Making the magnetic nanoparticles (Fe3O4)
Coating them with a surfactant (Oleic acid)
Dissolving in suitable carrier solution (petrol/kerosene)

Making the magnetic nanoparticles
Add FeCl2 and FeCl3 solution in 1:2 ratio in a beaker and continue mixing it while adding NH3 drop wise. First the solution will turn brown. Keep adding until the solution turns black
$$3NH_3+3H_2O+FeCl_3→3NH_4Cl+Fe(OH)_3$$
$$2NH_3+2H_2O+FeCl_2→2NH_4Cl+Fe(OH)_2$$
$$Fe(OH)_2+2Fe(OH)_3→4H_2O+Fe_3O_4$$
This leaves us with Fe3O4 in NH4Cl solution.
You can also heat the solution for sometime at about 80 ℃
Note - some people suggest adding the iron solution to ammonia and not he other way around.

Separating out the Fe3O4 from the solution
To the solution add small amount of Oleic acid and some excess of ammonia which will form ammonium oleate which binds to the magnetic particles and is soluble
$$OleicAcid+NH_3→AmmoniumOleate$$
heat the solution to about 80 ℃ – 90 ℃ . The ammonium oleate will decompose to again form oleic acid (which is hydrophobic) and it will separate out the magnetic particles from the solution.

Purification
We don't want

Particles that are too big
Particles that are not properly coated with Oleic acid

To deal with them follow these steps

Dissolve the Oleic acid coated particles in petrol/kerosene and then keep the beaker over a big magnet to precipitate the unwanted big particles. Then Decant the useful Oleic acid coated particles which stayed in the solution
Now add double the volume of Acetone to extract these particle out of the solution. They stay coated with Oleic acid.

This was the purification process. Note that you can also use paint thinner in place of petrol and isopropyl alcohol in place of Acetone.

Dissolving in carrier fluid
Now let the particles dry out and then add a few drops of kerosene/petrol to make a concentrated solution. The ferrofluid should be ready now.

Additional information
If you don't have oleic acid then you can obtain it by saponification of olive oil.
$$OliveOil+CausticSoda/NaOH \xrightarrow{\text{boil}} soap$$
Dissolve the soap in water with some acid (sulphuric). The oily substance collected at the top is mostly Oleic acid with stearic acid and linoleic acid. This can be further purified by refrigeration.
Melting point of oleic acid is 13–14℃ while that of stearic acid is 68℃ and linoleic acid is -12℃
You can also make ferrous chloride from ferric chloride by adding irin nails to it.


text:
Why is $^3P_1$ an atomic term symbol for carbon?
I'm using these rules, $^{2S+1}L_J$ with $J=L+S$ and it seems to imply that $^3P_1$ means $S=1$, $L=1$, and $J=1$ but clearly $1+1 \ne 1$ so according to me this shouldn't be an atomic term symbol but it is. Why?

This is not the fundamental one. But it not means it is not one.

You are correct in saying that the total spin for two electrons is S = 1 as its spin multiplicity is 3 and that L =1. The J value is found in the same way as S and L values were using the Clebsch-Gordon series, $$J= |L+S|...|L-S| $$ This means find the maximum value, then the minimum make it positive, and separate any other values by 1. Sometimes there is only 1 value so be careful. We use this type of series as a short-cut for vector addition of angular momentum which is why you question '1+1 = 1'.
In your case J = 1+1 to 0 , which gives three values J = 2, 1 and 0 so three possible term symbols.  There will also be singlet terms where the total spin S = 0.
The triplet term means that electrons are in different orbitals, generally this lowers the energy, and also means that there is no Pauli principle to consider. Hund's rules are used to determine which levels and states are lowest in energy. (See for example Steinfeld, 'Molecules & radiation')

Ah ok, so it seems I was just told some quick short cut for some actual process. What's the actual process called if I want to do it "the real way"; Clebsch-Gordon addition of angular momentum or something I guess?

I'm not sure it has a name as such, but its based on the Russell Saunders coupling scheme and vector addition of angular momentum.  This is described in most undergraduate phys chem text books.

Yeah, for reference I was reading McQuarrie Physical Chemistry and this seemed to be just thrown at me without justification so I guess I need to refer to a different book for the explanation on that.

with $J=L+S$

Nope. In general, it can be shown that the permitted values of the $J$ quantum number for the total angular momentum that arise from two sources characterized by quantum numbers $J_1$ and $J_2$ are given by
$$
    J = J_1 + J_2, J_1 + J_2 - 1, \dotsc, |J_1 - J_2| \, .
$$
In this particular case, $J_1 = L$ and $J_2 = S$, so $J$ ranges from $L + S$ to $|L - S|$ in integral steps and not necessarily equal to $L + S$.