text:
Is there a chemical that will etch antimony but not stainless steel?
I am working with ultra high vacuum equipment made out of stainless steel that has been contaminated with a number of metals, but in particular Antimony.  We'd like to do some machining of the equipment but this requires cleaning prior.  The amount of contamination is only a few hundred nanometers, but needs to be removed.  Is there a wet chemical etch that would selectively etch antimony (and antimonides) but not stainless steel?
I had initially considered $\ce{HNO3}$ as a possibility, but it appears without $\ce{HCl}$ or $\ce{HF}$ (which will make the solution etch Stainless) its etch rate of antimony is either slow/negligible or unreported.  I'm currently trying an experiment etching a small antimony pellet, but the initial results are not promising.
Any possible solutions would be preferably easy to mix and not require overly strict ventilation.  The etch will be contained and ventilated to an exhaust; however, doing it in a fume hood is not practical due to the size of the equipment.  Additionally, this means that highly toxic/corrosive chemicals are undesirable (e.g. $\ce{HF}$).

Some remarks:

Antimony and antimonides suggest that you need an oxidizing agent.
The presence of chloride ions might cause pitting corrosion to the steel.
As a model for testing you could probably use antimony deposited on a surface through a Marsh test.
I would try if NaOH/H2O2 is able to dissolve the antimony deposit (ammonium hydroxide/H2O2 won't do it).


text:
How to explain the discrepancy between the calculated absorption wavelength within the particle in the box theory and the observed wavelength?
The experimental observed wavelength for this molecule is 425 nm.
We are supposed to estimate this wavelength theoretically using this equation (energy change in a 1D box), where $N$ is the number of pi electrons.
$$\Delta E=\frac{h^2(N+1)}{8m_\mathrm{e}L^2}$$
$L$ is given as 11.84 angstroms.
I'm getting a wavelength of 200 nm. It seems to be too far off.
Molecule:

I'm counting 22 pi electrons and using that as N
This question is different from How to identify the number of pi electrons in a conjugated system to calculate the HOMO-LUMO gap with the particle in the box approach? in that this question is specifically addressing the discrepancy between the theoretical and experimental wavelengths, whereas the previous question was specific to the number to use for $N$.
Also how does this even make sense? As conjugation increases the energy gap between HOMO and LUMO should decrease. With this relationship as $N$ increases as does the energy gap. To me it makes no sense, how can it be explained?

Possible duplicate of What pi electrons to include when calculating Energy change

No the previous question was specifically addressing what to use for N.

Why did you post it again?! You can edit and any additional info

This question is specifically asking why there is such a large discrepancy between the theoretical and experimental wavelengths. The previous question was asking only about how many pi electrons to include for N

Well you have wrong number for starters and you should simply edit old stuff to include it. This is still about the same task.

Are you missing $\pi ^2$ in the denominator?

no according to the equation we're given there's no $$\pi ^2$$ in the denominator. And @Mithoron not sure what you mean by the wrong number

@Mithoron is saying that your value of N is not correct. There are more than 16 pi electrons.

Oh ok is that because the S and N lone pairs will rehybridize to sp2 with a p orbital open? In that case I'd say there are 22 pi electrons. That gives a theoretical wavelength of 210 nm which is even further off. Also Not understanding because as N increases as will the energy gap, but I thought that the energy gap will decrease as N increases.

See this PDF describing a problem like yours. It's about computing absorption spectra for conjugated dyes similar to yours and contains hints about the number of electrons and the model.

Thanks @ToddMinehardt. I don't see where that article explains how to calculate N. I only see it explaining what n is. I know that N=8 for this molecule: https://www.dartmouth.edu/~chemlab/chem6/dyes/graphics/chemistry/chemistry-img14.gif , but I don't see how! I only see 6 electrons participating in the conjugation

Nevermind N=6 for that molecule. It was labelled confusingly. N=22 for my molecule I'm quite sure of that. Calculated wavelength is 200 nm, which is off by 225 nm. Something's wrong!

Luckily I found a source containing your molecule as an example.
You need your equation for the energy difference between the HOMO and the LUMO, which is (nearly) defined as you wrote:
$$\Delta E = \frac{(N+1)h^2}{8 m_e L^2}$$
As you want a wavelength, you need to convert the energy difference, which yields then:
$$\lambda = \frac{8m_ecL^2}{h(N+1)}$$
Now they state, that the length of the box can be guessed by the amount of carbons (p) between both nitrogens by using this formula:
$$L = (p+3+\alpha)l$$
... with $l$ being the length of typical benzene-C-C-bonds, thus being 0.139 nm and $\alpha$ is an empirical correction factor for polarizable end groups.
The amount of pi-electrons (N) that you need to inlude can be calculated by a similar equation:
$$N = p+3$$
... where the additional "+3" come from both nitrogens (would be actually +4, but it's +3 because one nitrogen atom had to form an ion).
This leads to their equation 4:
$$\lambda = 63.7\frac{(p+3+\alpha)^2}{p+4}$$
... where all the physical constants have been combined to give 63.7.
As you have three carbon atoms between both nitrogens, you come up with 328 nm and when you include their "correction factor for polarizable end groups" of $\alpha = 0.675$, you come up with 405 nm, which is more than close to the experimental result.

When I use your box length and use 12 electrons (3 from the "left" benzene ring, 4 from the two double bonds to the "right" nitrogen" and 1 from the lone pair of this nitrogen), I come up with about 356 nm, which is not that far away as it sounds. In eV this is an error of 20% (the 405 nm are wrong by about 5%), whereas your 200 nm would have been an error of 120%!

This is awesome, but it really feels a lot like tinkered to me. Make the numbers work. I will add an answer later better explaining what I actually mean, but suffice to say that estimating the box as going from nitrogen to nitrogen (and a little bit further) is sketchy at best and does not resemble the HOMO/LUMO distribution at all. However, I think this is a standard approach and hence you get my thumbs up for finding it. (The cited J Ed Chem 1985, 62, 351-352 is a better source though: mirror)

You can google it as I did and you will find many sources that state exactly that. There is even a Paper in J Chem Educ about why this model works surprisingly well.

You mean the paper I linked? I am about to destroy it.

J. Autschbach; Why the Particle-in-a-Box Model Works Well for Cyanine Dyes but Not for Conjugated Polyenes. J. Chem. Educ. 2007, 84 (11), 1840

The energy level of the particle in a one-dimensional box are given by
$$E_n = \frac{n^2\hbar^2 \pi ^2}{2mL^2} = \frac{n^2 h^2}{8mL^2}.\tag1$$
We see that the energy is quantised and dependent on the quantum number $n$. We also see that the energy difference between those levels gets larger with $n$. The only variable in this equation is the length of the box $L$. The larger the box the denser the states.
For an extended pi system like the ones in the dyes we make various simplification. First and foremost we reduce a multidimensional problem to one dimension. We also treat the barriers of the box as infinitely large potentials. Then we can only estimate the length of the pox. We are completely ignoring the fact that the molecule will have vibrations. And there are a couple of more assumption we make. All of them will influence the accuracy of the calculation. You should discuss these issues thoroughly and come to your own conclusions.
pH13 outlined a procedure that has been established in the literature, which neatly correlates the experimental findings with the calculations.[1] I personally find this method a little bit sketchy and not very intuitive at all. My first issue is the choice of the box itself; it seems rather arbitrary to me, as they decide to ignore the phenyl rings, add half a length of a bond, choose the average bond distance to be the one of benzene.

In this case, the box extends from nitrogen to nitrogen plus one-half of a bond length beyond each nitrogen. That is, $L = (p + 3)l$, where $p =$ the number of carbon atoms in the chain and $l$ is the average bond distance in the chain. The number of electrons in the chain, $N$, is $p + 3$.
  [...]
  [...] we let $l = 1.39 \times 10^{-8}~\mathrm{cm}$ (the bond distance in benzene) [...]

For a slightly simpler model I have estimated the length of the box to be $L=1.20~\mathrm{nm}=1.2\times10^{-9}$ and the number of π electrons to be $N= 22$.
With the reasonable assumption that orbitals are doubly occupied, we can hence find the the quantum number of the energy level of the HOMO $E_\mathrm{HOMO}$ to be $n_\mathrm{HOMO}=N/2$ and in analogy to that for $E_\mathrm{LUMO}$ as $n_\mathrm{LUMO}=N/2+1$. The energy of the photon to excite one electron from the HOMO to the LUMO is in first approximation it's difference $\Delta E_\mathrm{gap}$. Therefore we will find the formula you used after some transformations.
\begin{align}
\Delta E &= E_\mathrm{LUMO} - E_\mathrm{HOMO}\\
  &= \frac{n_\mathrm{LUMO}^2 h^2}{8m_\mathrm{e}L^2}
   - \frac{n_\mathrm{HOMO}^2 h^2}{8m_\mathrm{e}L^2}\\
  &= \frac{h^2}{8m_\mathrm{e}L^2}\left[n_\mathrm{LUMO}^2
       -n_\mathrm{HOMO}^2\right]\\
  &= \frac{h^2}{8m_\mathrm{e}L^2}\left[\left(\frac{N}{2}\right)^2
       -\left(\frac{N}{2}+1\right)^2 \right]\\
  &= \frac{h^2}{8m_\mathrm{e}L^2}\left[N+1\right]\tag2\\
\Delta E &= \frac{hc}{\lambda}\\
\lambda &= \frac{8m_\mathrm{e}cL^2}{h(N+1)}\tag3
\end{align}
Plugging in the data, we find $\Delta E = 5.965~\mathrm{eV}$ and $\lambda=207~\mathrm{nm}$. Well, that's almost the same as you got, and that is very disappointing. Could it be, that there are a couple of approximations, that do not work well?
Farrell also admits that his approach is flawed:

Agreement between predicted transition energies and experimental energies, Table 1, is not good (although the agreement is rather surprising in view of the approximations made).

Therefore a parameter $\gamma$ which is adjusted to give a good fit to the experimental results is introduced. The length of his box becomes $L=p+3+\gamma$.
Elongating the conjugated bridge between the nitrogen atoms, lengthening the box, will result in a better fit, even without that adjustment. In other words, the less important the end groups (phenyl moieties) become, the better the simple model performs. That is not surprising at all.
The following table is reproduced from that paper (with slight modification). The number of carbons in the chain is $p$.
\begin{array}{lcccc}
p & \lambda_\mathrm{exp}/\mathrm{nm}
  & \Delta E_{\mathrm{exp}}/\mathrm{eV}
  & \Delta E_{\mathrm{th.}\, \gamma=0}/\mathrm{eV}
  & \Delta E_{\mathrm{th.}\, \gamma=0.68}/\mathrm{eV}\\\hline
3 &  422 & 2.94 & 3.78 & 3.05\\
5 &  555 & 2.23 & 2.74 & 2.32\\
7 &  650 & 1.91 & 2.14 & 1.88\\
9 &  760 & 1.63 & 1.76 & 1.57\\
\end{array}
Conclusion
The failure of the model for short chains is not surprising. It will get better the longer the box is and the less influential the end moieties are. This system especially is well suited for discussions of such approximations and gain a basic understanding of quantum chemistry. Stay sceptical.
Reference

John J. Farrel J Ed Chem 1985, 62, 351-352 [mirror]

Addendum

As conjugation increases the energy gap between HOMO and LUMO should decrease. With this relationship as N increases as does the energy gap. Makes no sense

It actually does. The more particles you fill in such a well, the higher the energy of the same ensemble gets. What makes the energy gap smaller and the states more dense is the extension of the actual box. The larger $L$ the closer are the states together. If the box size grows larger faster than the electrons you fill in it, then the gap shrinks. The gap is linear dependent on $N$, but inverse quadratic in $L$.

When one "arbitrarily" insists that the box is of a certain length, that is where any method fails.  The molecular geometry provides the boundary conditions on the standing waves for the electrons, so people fix things by adding an extra length (fudge factor) to account for the discrepancy in the model.  So, the Kekule structure(s) (Lewis structure(s)) people start with to model the box length are very approximate.
So, can we make a better approximation without having to account for every electron?  Yes, but we need to include those electrons in the molecule that are leading to the measured "characteristic length". What is that thing, but the "conjugation length".  You are gonna say, "the whole thing is conjugated", but look closer.
We all learn that conjugation is alternating double and single bonds right, but what is that here and how does it help?  Although the conjugation method is limited too, it does offer a better approximation to a "length" since it starts with a good fragment (stable "vinylic like CH2=CH-" chunks).
Use what you know about MO's from organic chemistry and try assigning phases to the Kekule structures and see what you need to do to make the longest length of alternating (++)_(--) units.  That is, try alternating phase pairs of the pi orbitals and see which configuration(s) gives the longest length.  You may need to change the number of alpha and beta phases that appear on one side of the molecule  to get the longest length (flip a pi-bond over to get its phase out of the way of conjugation path).  How many bonds long is that, and is that an integer multiple of the measured length (or a reasonably close approximation to an integer multiple of a bond length)?  Compare that best guess and shape to what the HOMO and LUMO look like.  What do you think conjugation is now and how does the end group affect conjugation?  Cheers.

I appreciate that you are trying to get the OP to think through what the answer would be, but it would be preferable to add a definitive answer at the end so that others could benefit from this in the future.  Thanks!

Better than presenting the wrong answer such as that with pure fudge factors that are erroneously repeated as some kind of understanding or knowledge; L=(p+3+α)l...This equation is totally arbitrary and without any physical significance as it is completely contrived. (tinkered)  The fact that it approximates the solution is what the student should be thinking about.    Eventually the student should consider Huckel MO theory and phases of p orbitals.  Thank you.

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How are the hydrogens attached to the nitrogen atom in the ammonium ion?
Naively attaching hydrogens to a nitrogen atom's electron orbitals results in a picture like this:

However, there are two problems with this drawing that I can see.
First, I have heard that the ammonium ion has a tetrahedral geometry so the hydrogens would have to be moved around and the orbitals would have to be stretched around or distorted for this picture to make sense.
Second, while I have drawn each of the hydrogens attaching to different external orbital I don't actually know which orbitals the hydrogens would attach to or why.
How are the hydrogens actually attached to the nitrogen atom in the ammonium ion?
I suppose one possibility might be to have the molecule be a hybrid of various $\ce{NH3 + H^+}$ configurations. I guess this could be tested by analyzing the charge distribution of ammonium and to see if the positive charge is more distributed around the outside or the inside.

This is because atomic orbital methods like you've described here don't accurately reflect how bonding takes place in real molecules. You can google "VSEPR theory" to learn more, or someone should come along in time and do a proper explanation in the answers section. Great question, I wish I saw more of this in chemistry students.

You heard about hybridisation?

The nitrogen atom, on both ammonium ion ($\ce{NH4+}$) and ammonia ($\ce{NH3}$), is $\ce{sp^3}$ hybridized. That's because we have 4 sigma bonds on ammonium ion and 3 sigma bonds + 1 electron pair in ammonia. The most stable orbital configuration for this set up is a tetrahedral shape.

Nitrogen can have a $\ce{sp^2}$ configuration if it has 3 sigma bonds and a pi bond, like in $\ce{H2C=NH2}$. The $\ce{sp^2}$ orbitals are "used" in the sigma bonds and the remainder $2p_z$ orbital is "used" in the pi bond. This gives a triangular plane shape to the molecule.

Finally, nitrogen can also have a $\ce{sp}$ configuration if it has 1 sigma bond and 2 pi bonds, in $\ce{N2}$ for example. In this case, the $\ce{sp}$ orbital is "used" in the sigma bond and the remainder $2p_y$ and $2p_z$ orbitals are "used" in the pi bonds. This gives $\ce{N2}$ a linear geometry.

Molecular orbital theory and hybridization are a little hard to process (at least for me it was and it's still being), so take your time to read some more about it, but at least with those examples you should have an idea of what to expect from some molecules geometries.

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Is there a reaction between ethanol and aluminum?
Is there a potential reaction that would occur between ethanol and aluminum? If so, would it potentially create anything toxic?

Yes it is, it is an industrial way to prepare aluminiuum isopropoxide.

In what situation? Normally there's no significant reaction.

Mercury(II) chloride is a necessary condition for this synthesis to occur, because usual aluminum is covered by a thin waterproof and continuous protective layer of aluminum oxide (alumina). This layer prevents the metal aluminum to get in touch with any liquid. In the presence of dissolved mercury chloride, the layer is not continuous anymore. So aluminum can react with water, ethanol, or isopropanol, producing gaseous hydrogen and aluminum hydroxide (or ethoxide, or isopropoxide).

@Maurice, For rapid synthesis, a way of breaking the Al2O3 coating is needed (might Ga would work, too?), but for long-term storage, aluminum is attacked by the lower alcohols, as I've observed by using a container with (unanodized) Al top for storing isopropanol. After a year or so, the top was covered with a white crust and badly corroded. No Hg was involved in this inadvertent test!

Thank you for the information, Dr Pippik. I did not know about this extremely slow reaction. And of course, gallium works as well.

The reaction of aluminum and ethanol forms aluminum triethoxide,  $\ce{ Al(OC2H5)3}$, but the reaction is fairly slow at room temperature, particularly if the aluminum is anodized. In the presence of any moisture it decomposes into aluminum hydroxide and ethanol.
The Safety Data Sheet indicates that aluminum triethoxide is both flammable and can cause "severe skin burns and eye damage", but in atmospheric conditions it likely decomposes as quickly as it forms, so the main adverse effect will be the corrosion of the aluminum, covering it with a white crust of $\ce{Al2O3}$ (I've seen this happen to an aluminum lid on a container used for ethanol).
UNH has posted a chart that shows that anodized aluminum is resistant to ethanol attack, but does not state at what temperature, so you may have to experiment to see if it suitable in your application.
BTW, many other alcohols react in a similar manner, such as isopropanol, which forms aluminium isopropoxide with aluminum.

I have seen the same:  the main adverse effect will be the corrosion of the aluminum, covering it with a white crust of Al2O3
A
l
2
O
3
(I've seen this happen to an aluminum lid on a container used for ethanol).


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When is the Assumption that Enthalpy is Constant over Temperature Change Invalid?
In a question, I was asked to integrate the Gibbs-Helmholtz equation to derive some formula. Then the next question asked:

Identify one assumption that you have made in deriving the above equation and identify one
  scenario where it is a very bad assumption, and explain why the assumption fails

Now the assumption that was made was the $\Delta H_{\mathrm{fus}}$ is constant over the change in temperature. However, could someone please explain when and why this assumption fails.
Thanks.

The $G{-}H$ equation describes the effect of $T$ on $\Delta G$ at constant pressure. But, along the fusion contour of $T$ vs $p$, $T$ is a unique function of $p$. So, along this contour, $p$ is changing. So the $G{-}H$ equation is not valid to use for a change of phase such as fusion. The $G{-}H$ equation was developed to describe the effect of $T$ on $\Delta G$ for chemical reactions at constant pressure, usually for the change from separate pure reactants to separate pure products at the standard pressure of $1\ \mathrm{bar}$.

Thanks for your great answer. Out curiosity, would assuming the enthalpy of fusion being constant also be a bad assumption in certain scenarios?

If both the temperature and the pressure change (i.e., in tandum along the equilibrium contour), the change in enthalpy between the solid and the liquid in equilibrium with one another  also changes.  But usually, the effect of pressure on the heat of fusion is going to be very weak.


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Test for Fe2+ ion in salt using potassium hexacyanidoferrate(III)
It's given in my book that $\ce{Fe^3+}$ ion reacts with potassium hexacyanidoferrate(III) to give brown colour.
$$\ce{Fe^3+ + [Fe(CN)6]^3- -> Fe[Fe(CN)6]}$$
Both the reactants have the same oxidation states.
My question is, can I use the same test for $\ce{Fe^2+}$?
First, the hexacyanidoferrate(III) will oxidize $\ce{Fe^2+}$ to $\ce{Fe^3+}$ like this:
$$\ce{Fe^2+ + [Fe(CN)6]^3- -> Fe^3+ + [Fe(CN)6]^4-}$$
Then the reaction will continue to give brown colour.

You kinda can, but the product would be different: the Prussian blue, famous in its own right.

@IvanNeretin I thought prussian blue is obtained when $\ce{Fe^3+} $ reacts with hexacyanidoferrate(II) or $\ce{[Fe(CN) 6]^4-} $. I am talking about $\ce{[Fe(CN)6]^3-}$.

for prussian blue, $\ce{Fe^3+ + [Fe(CN)6]^4- -> Fe4[Fe(CN)6]3}$

$\ce{Fe^3+ + [Fe(CN)6]^4-}$ is exactly what you have on the right side of your own equation.

@IvanNeretin But can't the reactant $\ce{Fe^2+}$ combine with the product ferricyanide to give Turnbull's blue?

It's pretty much the same thing.

$\ce{Fe^2+}$ is oxidised to $\ce{Fe^3+}$ while $\ce{[Fe(CN)_6]^{3-}}$ is reduced to $\ce{[Fe(CN)_6]^{4-}}$. These then form the prussian blue precipitate.


text:
Isomers of coordination compounds
Does [Pt(CH3-CH(NH2)(COO-))2] show geometrical and optical isomerism?
According to me Pt should be dsp2 hybridised giving a square planar geometry. So it is of the form M(aa)2 type. The ligand itself has a chiral carbon. Hence it can show optical isomerism. But can it show geometrical isomerism?

First of all, the compound is not M(aa)2 type.M(aa)2 shows no kind of isomerism (with normal ligands). Since the donor atoms on a ligand are different, it would be M(ab)2. And square planar structures show only geometrical isomerism and not optical isomers, as you think. Why?
Optical isomerism is when the compounds can have non super-imposable mirror image like structure. They should not have a plane (or axis) of symmetry. However, in a square complex, it is not possible. Since it will always have a plane of symmetry or a mirror image which can be superimposed on the previous compound, even if it is of type  M(abcd) (Try this out for yourself)
But, you are right, carbon on your compound is chiral, so it can have different configuration of methyl and hydrogen groups in space so as to remove that plane of symmetry factor and make them optical isomers.
Geometrical isomers, however are variations of a compound about the central metal atom atom or ring or a double bond, in which the arrangement of groups is different. So square planar complexes can show geometrical isomers. The one is your question can be depicted as -
If you include the configuration of the chiral carbon of the ligand, there can be a few more structures made.


text:
Electrochromic properties of ethyl viologen
I came across this article on electrochomic displays. It mentions Ethyl Viologen as an electochromic chemical. I would like to attempt an experiment to showcase it's electrochomic properties but don't really know where to start.
First off, where to get some. Googling I came across these, three, pages. Will one of these work? If not where do I get some? Or do I need to synthesize it (if so how)?
Second, what to do with it once I have it? I take it there's more to it than just sticking some electrodes in a beaker of it.
The article states, "If the EV-molecule absorbs a second electron, EV turns pale blue. This reduction however is irreversible and definitely not wanted." What care (if any) do I need to take to avoid this?
Sorry if I ask something obvious, I'm very inexperienced with chemistry.
Thank you.

I know it's been a while since you asked this question, but just remember to be careful with the viologens, as they are fairly toxic. Also, if you still want an answer I can try to help. What equipment do you have access too? Do you have fume hoods, a potentiostat, etc..?

[…] I'm very inexperienced with chemistry.

Have a look the material safety datasheet (MSDS) of ethyl viologen dibromide and pay attention to the H-statements:

H302 Harmful if swallowed.
  H312 Harmful in contact with skin.
  H315 Causes skin irritation.
  H319 Causes serious eye irritation.
  H332 Harmful if inhaled.
  H335 May cause respiratory irritation.

Can you handle this?
You asked:

what to do with it once I have it? I take it there's more to it than just sticking some electrodes in a beaker of it.

It is not that simple.
It is true that 4,4'-dialkyl pyridinium dications undergo a reversible one-electron reduction. This can be achieved by applying a voltage or by electronic excitation in the presence of an electron donor.
However, the resulting radical cations are very sensitive to oxygen and they have been used as probes to monitor oxygen permeation in polymers. With other words, you would have to exclude oxygen from your setup for a lasting colour effect.
In summary, I'd rather advise against the use of ethyl viologen for showcasing, unless the experiments are performed by professionals in a safe environment.


text:
How to convert an oxirane to an alkanol with one additional carbon atom?
This is a question given in our exam.
In the reaction,

Find the reagents X and Y required for steps(ii) and (iii) are _______ and ________ respectively.
I have tried in this way but I have stuck in second step. The given compound reacts with PhCO3H to form an epoxide.
Am i correct? Then please help me continue the reaction.
The answer given is CH3MgBr and H3O+.

This seems like 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.

You're right: The reaction of the alkene with perbenzoic acid forms an oxirane.
My suggestion for the next step is:

Draw the epoxide.
Draw the final product.
Map the atoms: Count and assign the carbon atoms. Do you spot any difference?

Supposed that the oxygen atom of $\ce{OH}$ group in the final alkanol is the same atom introduced by the action of perbenzoic acid, the oxirane was apparently opened.
Which general ways to open an oxirane exist?
Try to think in general synthons first.
One of the two steps in the sequence might just be a neutralization, quenching or workup step.

UPDATE
With a bit of drawing, atom counting and mapping, you will have realized  (by now) that:

The final alkanol has one carbon atom more than the oxirane intermediate.
The additional methyl group was added to the less substituted side of the oxirane.

Now, try to think in synthons, such as "$\ce{CH3+}$" and "$\ce{CH3-}$", rather than in reagents.
Where would a "$\ce{CH3+}$", thrown at the oxirane, add? Probabaly at the centre of high(est) electron density. This is the oxygen atom, and thus the wrong position. Apparently, "$\ce{CH3+}$" is not your synthon.
What about "$\ce{CH3-}$"? It's a strong nucleophile that would preferably attack centres of low(er) electron density, such as carbon atoms with electron withdrawing substituents. You have two of these, namely the two carbon atoms that constituted the $\ce{C=C}$ double bond and which are now in your oxirane. If "$\ce{CH3-}$" attacks one of these carbon atoms, the oxirane is cleaved in such a way that one $\ce{C-O}$ bond is broken and the oxygen atom gets the negative charge. Note that sterical hendrance comes into play here too.
In summary,"$\ce{CH3-}$" seems to be the right synthon. Now, you have to translate the synthon, "$\ce{CH3-}$", into a reagent that you can either find in the lab or make in situ. In order to get the negative charge on the carbon atom, it might be a good idea to attach an electropositive to it.
Consequently, a Grignard reagent, $\ce{CH3MgBr}$ is a good choice.
As I mentioned initially, while trying not to spoil the fun of solving, the last step is just an aqueous, acidic workup to release the alkanol ;-)


text:
CuCl2 has colour but CuCl does not
$\ce{CuCl2}$ is green in colour while $\ce{CuCl}$ is white. What is the reason for this? Is it because the first compound exists as a complex ion and has d-d transitions?

Both should be considered as complex ions in solid state.

The reason is indeed $\text d$-to-$\text d$ transitions. The d orbitals of both $\ce{CuCl}$ and $\ce{CuCl2}$ have two possible energy levels (based on crystal field theory).
$\ce{Cu(II)}$ is a $\text{d}^{9}$ metal, and $\ce{CuCl2}$ exists as a (distorted) octahedral around the copper atom. Based on crystal field theory, the d orbitals of an octahedral metal center can be represented as shown below. In this case, because $\ce{Cu(II)}$ is $\text{d}^{9}$, the three lower energy suborbitals are fully filled in the ground state, and but the top two suborbitals only contain a combined three electrons. Therefore, the energy required to excite that electron is $\Delta _o$. When that electron returns to the ground state, it gives off a different wavelength of light than it absorbed (in this case, the light emitted is apparently not visible light).

$\ce{Cu(I)}$, on the other hand, is a $\text{d}^{10}$ metal ion, so each of the d suborbitals is fully filled. In order for an electron to be excited in this situation, it would have to leave the 3d orbital (and the $n=3$ shell altogether) and enter the $n=4$ shell. This is rather large  energy jump compared to the intraorbital jump an electron in $\ce{CuCl2}$ has to make. Therefore, $\ce{CuCl}$ does not readily absorb visible light as $\ce{CuCl2}$ does, so it appears colorless. That is, until it see moist air.


text:
Stereoisomers of octahedral co-ordination compounds of the form [Ma3b3]
Does $\ce{[Co(H2O)3Br3]}$ show geometrical and optical isomerism?
According to me $\ce{Co}$ should be $\mathrm{sp^3d^2}$ hybridised giving an octahedral geometry. It is of the form $\ce{[Ma3b3]}$ type. I know it can show geometrical isomerism, namely fac and mer isomers. But can it show optical isomerism?

Is there any way you could position the ligands such that it has a non-superimposable mirror image?

@DGS yeah I think so. When similar ligands are all adjacent to each other. But I think they will have a centre of symmetry.

I don't believe either fac or mer isomers result in non-superimposable mirror images.

This very much looks like a [tag:homework] question. Therefore I have given it the corresponding tag. As per our homework policy, please show your thoughts on this or the question may get closed. Note that you need to consider both fac and mer separately whether they can (or can not) exhibit optical isomerism. Hint: It may help to draw the isomers on paper. Also note that you should remove hybridisation from your mind for transition metal coordination compounds; they are much better described without it.

@Jan No, this is not a homework question. I am asking if anyone of the fac or mer isomers can show optical activity.

According to me $\ce{Co}$ should be $\mathrm{sp^3d^2}$ hybridised giving an octahedral geometry.

No, please don’t do this. The most accurate description of cobalt here would be $\mathrm{d^5}$, maybe $\mathrm{d^5s}$. But don’t extend the hybridisation concept from main group elements to the transition metals; it doesn’t belong and it doesn’t help here. Rather, there are six ligands around cobalt, so you have an extremely high chance for octahedral geometry. Which is also the case.

I know it can show geometrical isomerism, fac and mer.

That’s a good start.

But can it show optical isomerism?

When asking about optical isomerism, ask yourself, if the compound is superimposable on its mirror image. One method to check this is to draw the compound, draw its mirror image and rotate either until they overlap — or until you think they cannot. The second method is much more efficient though: Still draw them, but check whether you can find a centre of inversion symmetry, a plane of symmetry or an axis of improper rotation (turn by $\frac{360^\circ}{n}$, where $n$ is a natural number and then mirror the result through a plane perpendicular to the rotation axis). If you can find any of these, the compound is automatically identical with its mirror image, and if there are none of these, then it displays optical isomerism. Remember to consider 3D structures.
It should be rather simple to determine this for the mer isomer. For the fac, consider a view where three ligands of the same type are above the paper plane, the central cobalt occupies the paper plane and the three ligands of the other type are below it. For both considerations, you can ignore the two hydrogens on each oxygen.
The results are given in a spoiler (mouse over to see).

 The mer isomer has planes of symmetry that transect three meridional atoms of one kind and the middle atom of the other three.

 The fac isomer has planes of symmetry that transect the cobalt, one ligand of each kind and the middle between the other two ligands of each type.

 Therefore, neither displays optical isomerism.

pls explain how do you highlight certain part of question while answering. It will be helpful for me while answering questions so that it will be clear for the viewer.

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


text:
Anomalous Electronic Configuration of Thorium
The electronic configuration of thorium ($Z=90$) is $5\mathrm f^0 6\mathrm d^2 7\mathrm s^2$. But, according to the aufbau principle, the electrons should first enter the $\mathrm f$ subshell and not the $\mathrm d$ subshell. This is not a special case of a half-filled subshell or a fully-filled subshell like we see in the respective cases of chromium and copper. Then why do we write the electronic configuration like this, first filling the $\mathrm d$ subshell and not the $\mathrm f$ subshell?
P.s I came across the anomalous configuration of thorium here: https://en.wikipedia.org/wiki/Thorium#cite_ref-Wickleder5960_19-0
But it doesn't state the reason for the anomaly.

Welcome to Chem.SE! Thanks for posting your question here. For completeness, please add a citation to where you found that configuration for $\ce{Th}$. If you haven't already, check out the tour and help pages for more information. Enjoy!

Other related questions here and here.

For first four actinide elements $\ce{Th}$, $\ce{Pa}$, $\ce{U}$ and $\ce{Np}$, the difference in energy between $\mathrm{5f}$ and $\mathrm{6d}$ orbitals is small. Thus in elements electrons may occupy the $\mathrm{5f}$ or $\mathrm{6d}$ levels or sometimes both.
Later in actinide series the $\mathrm{5f}$ do become appreciable lower in energy, thus from $\ce{Pa}$ onward the $\mathrm{5f}$ shells fill in regular way.


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How to easily distinguish salts physically? Edit
I have basically 12 salts to identify in exams. They are ***

Ammonium Carbonate, Ammonium Sulphate, Ammonium chloride, Aluminium
  Nitrate , Aluminium Sulphate, Barium chloride, Calcium Carbonate,
  Magnesium Carbonate, Magnesium Sulphate, Lead Nitrate, Zinc Sulphide,
  Zinc Sulphate

***. I will be given one of them. I know all the tests I have to undertake to identify these salts (such as ash test, flame test, sodium carbonate extract tests etc). But I would like to know if there is any quick way to easily identify my salt on looking it or some easy tests so that I get a leverage over time and could finish my tests easier. I would like to know ASAP, as I have my exams tomorrow.
Thanks in advance

You just need to know the test for anion and cation

I know the tests, but it is practically time consuming.... If only i had an alternative way..

Then prioritize your step on the sequence for your test

@MukilanSC None of these tests are particularly time consuming. How fast are you meant to be doing this??

I have been given 3 hours for a volumetric analysis as well as this qualitative analysis and also this time includes summarizing the result of each individual test.

That is actually way more than what is required. I had to do a titration, a salt analysis and prepare a colloidal solution all in 2 hours. If you try to find the salt by blindly guessing, you will waste a lot of time. Start with dil. H2SO4 test, then do flame test, then do ash test. By now, you would've identified the radicals.

Draw a matrix of cations vs anions, mark the possible combinations, and calculate the probabilities.
The chances are 4/12 for a sulphate, 3/12 for a carbonate, etc.
I wouldn't start testing for sulphides ;-)


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How would I name O-methylations on sugars while retaining the sugar’s common name?
Suppose I had a monosaccharide, e.g. D-arabinose. Suppose I do a few synthetic steps with it, and end up with the compound methyl (2S,3R,4R)-3,5-dihydroxy-2,4-dimethoxypentanoate. (as labelled by ChemDraw. If that is not the PIN, please correct.)

However, when writing my paper/thesis/lab report, I would love to derive the base name from D-arabinose for simplicity and consistency. The monosaccharide derivatives with a carboxylic acid in place of an aldehyde are -onic acids so my base name would be methyl D-arabinonate. But how would I correctly and unambiguously label the methoxy groups?
If I used 2,4-dimethyl, that would sound like this second compound:

And if I used 2,4-dimethoxy, wouldn’t that make it this one:

Is there a trick I am missing for clearly naming the compound as above with methyl D-arabinonate as a base name?

Did you have a look at Pure Appl. Chem., 1996, 68, 1919-2008 IUPAC?

@KlausWarzecha No, because I didn’t know it existed ^^'

How about methyl 2O, 4O-dimethylarabinoate?

@CurtF. After skimming across Klaus’ source, yeah, something like that. If someone who had previous experience in doing so could answer, that would be great ^^

It is possible to name such compounds based on retained carbohydrate names in accordance with the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) as follows.

P-102.5.6 Monosaccharide derivatives
P-102.5.6.1 O-Substitution
In order to maintain the integrity of structures and take advantage of retained names to imply the absolute configuration, O-substitution is allowed in carbohydrate nomenclature. Substituents replacing the hydrogen atom of an alcoholic hydroxy group of a monosaccharide or monosaccharide derivative are denoted as O-substituents. The substitution of an anomeric hydroxy group is discussed in P-102.5.6.3.2. The O-locant is not repeated for multiple substitution by the same atom or group. Number locants are used as necessary to specify the positions of substituents; they are not required for compounds fully substituted by identical atoms or groups.
P-102.5.6.1.1 O-Acetyl and O-alkyl functionalization.
For O-acyl derivatives, names with the acid component cited as a separate word ending in ‘ate’ after the monosaccharide name are preferred to names using O-acyl group prefixes. However when the ose ending is changed (e.g. to denote a glycosyl or an acid function having seniority over an ester) O-acyl prefixes are required. O-Alkyl derivatives are always expressed by prefixes.

Therefore, the name for the corrsponding D-arabinose derivative with methyl substituents replacing the hydrogen atoms of two alcoholic hydroxy group is 2,4-di-O-methyl-D-arabinose.


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determine density of oxygen gas using modified ideal gas equation
I am trying to determine the density of a gas (oxygen). As the gas is a compressible fluid a modified version of the ideal gas equation is required. I have been given the following equation and I am struggling to interpret it. I would like advice on the unit conversions and whether this is the correct equation at all.
$$\rho=\frac{(P+1.01325).10^5.MW}{z.8314.(T+273)}$$
where:
$\rho$ = density (Kg/m3)
P = pressure (bar(g))
MW = molecular weight (in kg/kmol)
z = compressibility factor (dimensionless)
T = temperature  (deg. C)
I think that the 1.01325 converts the bar.g. to bar.a. then to pascals using the 100,000 multiplier?
Has the universal gas constant $8.314 J/(k.mol)$ been converted to $J/(K.kmol)$ ?

I can't see why you bother with it. It's simply pV=zRT transformed so you could easily calculate density with data given.

Z been compressibility in your equation? Apparently I need to consider compressibility. Something to do with pressure drop over pipe lengths. I will comment further soon

This has nothing to do with flow in pipes.  The z in this equation is the compressibility (correction) factor for real gases compared to ideal gases (not the distance along a pipe).  You need to determine the reduced temperature and pressure of oxygen, scaling by the critical properties.  You have a graph that gives z as a function of reduced temperature and pressure.  Once you know z, you can calculate the density.  Isn't this being covered in your course?

I am not on a course, unfortunately, and I am having to learn this as I go. My intention was to use this equation to obtain the density, which is used in the Darcy equation to calculate pressure drop per unit length of straight pipe.

@ChesterMiller In addition, if you have any recommended literature that would help me in the area that you have mentioned; that would be greatly appreciated and thanks for your input.

See any Thermodynamics book, but the best one that I am aware of is Fundamentals of Engineering Thermodynamics by Moran et al.  To get an extended treatment of chemical thermodynamics (chemical reactions and phase equilibrium), I suggest Introduction to Chemical Engineering Thermodynamics by Smith and Van Ness.

What an awkward, ugly equation that is! Mithoron is correct about the traditional, elegant and understandable form.
Yes, (P+1.01325)*10^5 converts gauge pressure in atmospheres to absolute pressure in Pascals.
z is compressibility - look up the Nelson-Obert generalized compressibility charts.
As for the 8314, the units here should be Pa*m^3/kmol-K. Dividing it by the molecular weight in kg/kmol will give you the specific gas constant for oxygen in Pa-m^3/kg-K. That's already been done in this equation.
Look up compressibility factor on wikipedia then find the Nelson Obert charts online, also critical data for oxygen, and use them to determine whether your oxygen is ideal in the P, T range you are dealing with (z factor between 1.01 and 0.99).  The specific volume of the gas depends on z as well as on P and T, but you can neglect the z correction if it's within a percent or two of the ideal value of 1.
Responding to your subsequent comment:
P, T and z determine the gas's specific volume in m^3/kg, which along with mass flow rate (kg/s) and pipe cross-sectional area (m^2) determine flow velocity (m/s), which along with pipe roughness will determine friction factor, which along with gas viscosity, pipe length and the presence of fittings like elbows and open valves, will determine pressure drop.


text:
Bioplastics from eggshells
There are many resources online which talk about projects making bioplastic from egg shells. However none of these actually show or describe the procedure needed to make it. Does anyone know a site/paper which one could use to figure out how to do this for a science fair project?

Hi and welcome to chemistry.stackexchange.com. Feel free to take a [tour] of this site. For any questions about how it works, consult the [help].

What kinds of bioplastic? Bones are made of calcium phosphate, so egg shells won't do.

Eggshell is 95% calcium carbonate. It seems more likely to be used as an additive in a composite plastic.  Testing composite plastics produced from adding ground eggshell to common plastics may make for an interesting science project.
References:
Dissertation on meat and bone meal derived plastics
Patricio Toroa et. al. Eggshell, a new bio-filler for polypropylene composites

And I don't relay see the benefit of the calcium carbonate coming from eggshells other than to advertise with "bio" buzzwords..

Eggshell is a waste product poultry producers have to pay to landfill.  The goal in using egg shell in plastics is to avoid paying money to send egg shells to the landfill and instead use the shells to save money on plastic packaging for eggs.

https://en.wikipedia.org/wiki/Eggshell_and_protein_membrane_separation -

@_Agriculturist I can't completely follow your argument about using egg shells in plastic surgery (in small quantum) to avoid having to pay for depositing them, as there are many clay soils that would profit form egg shells.

@Gyro  You seem confused about the term "bioplastics".  This article should help..   https://en.wikipedia.org/wiki/Bioplastic     ---

Yes, I read bioplastics as some kind of bone-replacement, or something like that, within our bodies.

Gyro - You have a good point, IUPAC discourages the use of the term bioplastic.

Btw, where do the hens get all that calcium carbonate they need for laying eggs? Wouldn't egg shells be a perfect recycling food supplement?

Please post this as a separate question.

I don't think this appropriate to chemistry.stackexchange. And I would say, YES it is, but I can't cite resources.


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How to apply oxidation numbers in the reaction to form hydrogen peroxide from a superoxide?
In the following reaction:
$$\ce{2O2- (aq) + 2H+ (aq) <=> H2O2 (aq) + O2(g)}$$
I see where some oxygen atoms have lost electrons, but I do not see what has gained electrons.
On the reactant side, the superoxide has a $-1$ charge per oxygen atom. On the product side, half the oxygen atoms now have a $0$ charge, so they lost an electron each (right?). For it to be a true redox reaction, does not a species/atom have to gain electron(s)?

On the reactant side, the superoxide has a $-1$ charge per oxygen atom.

No, the superoxide has a $-\frac{1}{2}$ charge per oxygen atom.
The reaction is better thought of as two separate reactions:
\begin{align}
\ce{O2- &<=> O2 + e-}\tag1\\
\ce{O2- + 2H+ + e- &<=> H2O2}\tag2
\end{align}
This demarcation is particularly relevant in a biological context, because there is a class of enzymes Superoxide reductase (SOR) that only catalyze the reaction forming $\ce{H2O2}$ and another class of enzymes Superoxide dismutase (SOD) that catalyzes both.
Here, oxygen is the element which is simultaneously undergoing oxidation (equation (1)) and reduction (equation (2)). This is termed a disproportionation reaction.
See also: A special case of disproportionation (or 'dismutation') is 'radical disproportionation'


text:
What are the bubbles and crystals found in sodium acetate hot packs after use and storage in the solid form?
When sodium acetate trihydrate hot packs are regenerated soon after activation, they regenerate to a clear liquid upon boiling (typically with an air bubble that was not present in the new pack)
On the other hand, when activated hot packs are stored in the crystallized form for a while, regeneration by boiling in water yields a mixture of clear liquid and very thin clear glossy plate-like crystals almost like mica sheets, plus large air (gas) pockets.
I suspect that the bubbles arise due to diffusion of room air through the flexible plastic envelope of the pad. When a new pad is activated and sodium acetate trihydrate crystals form, small gas spaces are left between the crystals which are necessarily filled only with water or acetic acid vapors which have low vapor pressure at room temperature. The low gas pressure in between the crystals (I estimate < 3 kPa) would create a large driving force for gas diffusion through the polymer envelope.
The crystals could be sodium acetate anhydrous formed because of water vapor loss by diffusion through the polymer envelope, but I'm not as sure on this. Perhaps they are sodium hydroxide formed because of acetic acid loss by diffusion? The liquid remaining around the crystals can still be activated to crystallize like a new hot pack, so the crystals are definitely not the trihydrate.
I'm looking for a reality check on these speculations, ideas on how to find out what the crystals are (ideally without needing sophisticated lab equipment), and advice for restoring hot packs containing these flat crystals to the initial clear liquid state. Photos below. Many thanks!

Hi random geologist here with relatives having a heat pack factory
1; bubbles form most prominently the longer you leave the pouch in crystallized state
2; if heated sodium acetate dissolves in its own crystal water (water that is built in as H2O molecule into the crystal structure)
so if you heat your pouch above 58°C it dissolves and if you do it correctly you dissolve every crystal nucleus (otherwise the pouch crystallizes again whilst cooling) so with 0 nucleii to form crystals from your pouch cools storing the latent heat of crystallization (264 kJ/kg) in it (heavily undercooled liquid) when you activate your pouch acicular crystal aggregates sprout out of your metal plate. (the more your liquid is undercooled the less it cares in what form it crystallizes, first its skeletal crystals then dendritic (snowflakes) then acicular). I hope this helps


text:
Can carbon subnitride be used as a fuel with similar properties to gasoline?
I am trying to figure out if Dicyanoacetylene, aka carbon subnitride, aka $\ce{C4N2}$, can have a similar reaction with air as gasoline, or if it only creates a lot of heat but not much expansion?  As I understand it, the thing that makes gasoline useful is that when it combusts, the product has a much lower density than the reactants, causing a huge expansion to reduce pressure (which drives pistons in an engine).  Does $\ce{C4N2}$ have a similar effect, or does it have only comparatively negligible expansion compared to gasoline and simply generates more heat? Or might it produce a similar expansion or more than gasoline?

Hi and welcome to chemistry.stackexchange.com. Feel free to take a [tour] of the site. All reactions leading to gaseous products will have a much greater desire to expand than liquid ones, so at first sight the effect should be similar. I’ll leave it to someone closer to the field to actually answer, though.

I was pretty sure it would have the expansion effect, but I also want to know how it compares to the expansion of the gasoline reaction.

Why are you even using strange "expansion effect" expression? It's not important at all what, say, molar volume would have products in room temp. Important is that reaction generates a lot of heat, which leads to pressure rise.

I guess it does come out a bit strange.  I think what I was going for when I asked was what volume does it's by product occupy ultimately and whether the pressure produced is greater than that of gasoline.

What matters in combustion reactions in engines is not the relative volume of the reactants and products but the amount of energy produced. And why would any highly unstable, potentially poisonous and spontaneously explosive compound be good for anything as a fuel?

No
You wouldn't want to burn this thing anywhere, it burns like, 9010 degrees... Fahrenheit. It's nearly the temperature of the surface of the sun (around 900 degrees off). Not much can contain it. So unless you had some kind of super-metal, than you are out of luck trying to use it as fuel.

Wow!  A chemical that burns as hot as the surface of the Sun!?!   PLEASE give a reference, I am intrigued.


text:
Fractional distillation of a product differs from the components listed in the SDS. What am I doing wrong?
I recently bought a product called Prestone De-Icer for Windows and Wipers which claimed to contain Methanol. I was puzzled when using it to note that it did not smell much like Methanol so I looked up the SDS for the product which stated that it contained 50-100% Methanol, 1-10% Ethylene Glycol & 1-5% Carbon Dioxide.
Out of curiosity I decided to carry out a fractional distillation to see just what the constituents were and was puzzled to see the thermometer shoot up through the Methanol BP at 66C mark without stopping and the first stop was about 76C which would suggest Carbon Tetrachloride. It then stopped at about the 79C mark which would indicate Ethanol and then carried on up to about 98C which would suggest Heptane.
So my question is this - Is there something wrong with my methodology or interpretation of what I am seeing or is it possible that the manufacturer is in error in the SDS. The latter would be very serious if true since the SDS would be the resource used should it be suspected that someone has been poisoned by the product.
My instinct is to suspect that I am doing something wrong but for the life of me I can't see what it could be. The thermometers I have used ( three in all ) are calibrated to 0.1 degree and were one mercury, one alcohol and one digital and all agreed to within 0.4C max deviation.
I should point out that I am not a professional chemist, just an amateur who has taken the hobby up in retirement so I am quite prepared to be told that I have got it wrong. It's all part of learning in my opinion.
Here is a link to the SDS -
http://www.homedepot.com/catalog/pdfImages/2c/2cd5847a-9dd0-4cf0-a5ef-83efbd532f4c.pdf
I am including a photo of the setup I am using to do the separation. The Vigreaux column is actually wrapped in aluminium foil.

I would be stunned if there was carbon tet in antifreeze. It's very toxic and dangerous to the environment.

I can't imagine heptane either. In the rain heptane would separate out and make a mess on the windscreen.

The product is for de-icing frozen windscreens so all it has to do is lower the freezing point of water so the ice melts.

My question boils down to this - if you have a substance which is at least 50% Methanol, then when heated the thermometer at the top of the Vigreaux column should rise to the BP temperature and remain steady until most of the Methanol has boiled off, then carry on up to the next highest product. If it is not stopping at the BP of Methanol is there anything that can explain this other than the absence of Methanol?

Maybe they use ethanol or isopropanol as a safer alternative to methanol and haven't updated the SDS? Isopropanol has a characteristic (pleasant) smell easy to recognise. Also, it should be an aqueous solution which explains the $\mathrm {98  \ ^\circ C}$ fraction.

Thanks @K_P
That is certainly possible I suppose but uber-naughty of them if that's the case. The MSDS is the first port of call should someone require emergency treatment.
I don't think they are using ethanol or isopropanol though as the smell is more like ammonia. It is definitely a suffocant even with just a very slight inhalation.

It looks like there are a bunch of Prestone deicers with varying compositions. Some contain ethanol, methanol, propylene glycol, ethylene glycol, water etc. and MSDS only need to have hazardous compounds listed. Distillation of multicomponent mixtures is not always as easy as the textbooks say - you may want to consider - are there components that might form an azeotrope? Have you tried redistilling the fractions and seeing what happens? What speed are you distilling at, and how rapidly do the temperatures change? (For closer-to-ideal separation, go as slow as possible with heat changes!)

Two thoughts. 1: GCMS is clearly the best option here (though likely not the most practical for at-home chemistry). 2. Did you consider azeotroping between two solvents?

Best Answer is this:
Your observations are consistent with the SDS for the reasons below.
A couple of issues see below.

Fractional distillation is much more complicated than you suppose. You have no way to control reflux ratio, (amount of distillate vs reflux) which right away will limit your separation. The column is not insulated or temperature controlled.
Vigreux columns are good for distilling one component from non-volatiles, but are not made for fractional distillations. You will need a much more efficient column, the cheapest choice being a vacuum jacketed Hempel column with some sort of packing, ie Raschig rings.
With the equipment you are using, a range of 10 C between ideal boiling points would be considered normal, you will not see the sharp cut offs you are expecting.
The first component is likely methanol, and the boiling point range you are seeing is due to the above reasons.
Ethylene glycol does not boil to almost 200 C, when you end the distillation, is there any liquid left in the distilling flask?
As far as I could quickly check, methanol and ethylene glycol do not form an azeotrope, but the tables I checked were on Wikipedia and not thorough, you might want to check that. The formation of an azeotrope definitely explains the observations especially if it boils in the range of 98C +/- 20 C
Not sure on the exact conditions or if the MSDS is wrong, but it is conceivable dimethoxyethane (dimethyl glycol) formed or was the original constituent, this has a boiling point of 85 C, it could have even formed in the distilling flask depending on conditions.
Note the MSDS or SDSs are they are now called are not complete, and there are usually other minor constituents that might appear and could possible catalyze the formation of dimethoxyethane, assuming the SDS was correct in the first place.
Distillation is not the appropriate method here, GC-MS would be the way to go.

Hope that helps! I suggest you purchase Handbook of Laboratory Distillation, it is the fundamental text for distillation in theory and practice, the price is well worth it.
In conclusion your observations could be explained by: azeotrope formation, incorrect SDS, formation of dimethoxyethane, and incorrect set up.

A very comprehensive answer. I'm not sure that I fully understand all of it but it has given me some useful insights. The book recommendation is a good starting point. Thanks again.


text:
Is there a material that turns from being electrically conductive to non-conductive irreversably upon exposure to some form of radiation?
For my project I am looking for a material which can turn from being electrically conductive to nonconductive (has to be irreversible) upon exposure to electromagnetic radiation (say UV). I tried googling for my question but found no relevant answer. A short explanation or a link would be greatly appreciated.

I figure this is the case with most conducting polymers, especially if irradiation is performed in an oxygen environment.


text:
Non-Ideal Gas Behavior
I obtained measurements of temperature and pressure at various temperatures ($-10, 0, 20, \text{ and } 50^\circ \mathrm{C}$) and plotted them on a graph. Projecting backwards, the best-fit line predicts 0 pressure at significantly above $-273.15^\circ \mathrm{C}$(0 Kelvin). I am wondering why this is.
My understanding is that a true linear relation assumes "Ideal Gas Behavior", which is predicated on the false assumptions that molecules have no attraction for each other and that they take up no space. I know that at high/low temperatures and pressures the realities about volume and attraction begin to make a difference. But I don't understand how these account for my results.
The pressure for all measurements fell between $680$ and $820$ torr. My understanding is that that is not a significant amount of pressure. In terms of predicting pressures at lower temperatures, I understand that the less kinetic energy of the gas the more relevant the attractive forces are...but I would think that would lower the pressure which makes things worse--if the pressure actually falls faster and faster as temperature drops, then the predicted zero-pressure point would be even higher than the predicted values I got. I need it to be lower!
I am under the (possibly incorrect) impression that both the realities of volume and attraction are supposed to be causing my incorrect prediction...but I don't understand how either does...
Thanks for any help!

That's a pretty narrow range of temperatures and pressures.  Maybe extrapolating all the say back to 0 K is too inaccurate.  Show us your data.

Yeah, I guess it is too narrow. But that's the point of the experiment I think...to get data that doesn't extrapolate properly and then have to explain the deviations :)

There is little reality to such an extrapolation. Attaching much significance will just lead you astray. There are a number of real gas equations which purport to correct for this or that "problem" with the ideal gas equation. The greater truth is that the equations just add extra fitting coefficients which make the equation more accurate. Different equations will work for different gases in different ranges.


text:
How do I figure out the amount of water (mL) and amount of Aluminum (g)?
We have a project where we have to extract copper from copper sulfate. But my group is given a powder version ($5~\mathrm{g}$), so we have to first dilute it with water and then combine it with aluminum to extract the copper. If this is the equation:
$$\ce{3Cu(SO4) + 2Al + H2O + salt -> Al2 (SO4)3 + 3Cu}$$
Salt is acting as a catalyst for the experiment.

Hi and welcome to chemistry.stackexchange.com. Feel free to take a [tour] of the site. I edited your post to include MathJax for chemical and mathematical expressions. For more information on how to do so yourself, check out the [help], this meta-post or this one. What actually is your question?

The first step you need to take is to determine the moles of your copper sulphate.
This is simple arithmetic. $$moles = \frac{mass}{molar.weight}$$
The mass of the substance is your measured 5 grams. The molar mass of your compound can be obtained from Wikipedia. In this case, anhydrous copper sulphate has a molar mass of $159.61\frac{g}{mol}$.
Plugging into the equation: 0.0031 $moles = \frac{5}{159.61}$
Now that you know the moles of your copper sulphate, you should be able to work out the the needed amounts of your other compounds using stoichometry.

No issues with your method, but it is very likely that she was using the pentahydrate, so the molar mass would need to be adjusted.  If the powder was white, it was anhydrous, if it was a nice blue color, it was hydrated.

This is why I made sure to specify that the anhydrous species was used for calculations.

Indeed you did.  :-)  I was trying to make it more clear, since this was a basic question and it seems reasonable that she would not immediately consider the possibility of a hydrate and its effect on the solution.


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Why does calcium carbonate decompose in sulphuric acid?
Here's the reaction:
$\ce{H2SO4(aq) +  CaCO3(s)  ->   CaSO4(s)  + H2O(l)  +   CO2(g)}$
I don't understand why calcium carbonate decomposes in sulphuric acid. It's not soluble in water. I could understand the hydrogen ions oxidizing the calcium, yet the reaction isn't a redox reaction; the oxidation states remain the same!

Very related: http://chemistry.stackexchange.com/questions/47049/why-does-caco3-react-with-hcl-but-not-with-h2so4/47050#47050

I don't understand why calcium carbonate decomposes... It's not soluble in water.

It's more soluble in water than you might think.  If it weren't soluble, we wouldn't have awesome things such as stalagmites in caves.
Calcium sulfate is more soluble in water than the carbonate, but that isn't saying much.  It's solubility is only about $\pu{2 g/L}$. So writing the product as $\ce{CaSO4(s)}$ is only correct if you are talking about dilute solutions.  Anything too concentrated would decompose to form precipitated calcium sulfate.

I could understand the hydrogen ions oxidizing the calcium...

Now that would be a very curious reaction.  The calcium in calcium carbonate is already in the +2 oxidation state. Calcium has only two valence electrons and further oxidation would require disrupting an intact octet.  It would be very unfavorable.  $\ce{Ca^{2+} -> Ca^{3+} + e-}$ would be far too unfavorable a reaction for protons to carry out. $\ce{Ca^{3+}}$ would be a much stronger oxidant than aqueous protons.

Why does calcium carbonate decompose in sulphuric acid?

The decomposition can still happen even with insoluble, solid-phase calcium carbonate because reactions can happen at surfaces. An acidic proton could be attacked by a solid-phase carbonate ion at the crystal surface, forming bicarbonate.  The bicarbonate could attack another acidic proton, forming a surface-bound $\ce{H2CO3}$ molecule, which could spontaneously transform into $\ce{CO2}$ and water, which would be very loosely bound to the surface, allowing them to escape and exposing fresh not-yet-reacted calcium carbonate molecules to the aqueous surface, and the process begins anew...

Since carbon dioxide is released out, leading to decrease  in concentration of product ,hence the reaction goes in forward direction to attain equilibrium leading to breakdown of calcium carbonate.


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Which acid to choose when designing a world with acid seas?
I'm currently writing a narrative and hoping to include a world which is home to a sea of some form of 'nasty' substance (doesn't have to be acid).
The varieties of inhabitants of this world have evolved mostly on the lands and eventually built their cities right up to the shorelines of this acidic sea. (In some cases I may even have cities that float and migrate across it).
I'm imagining some kind of scientific or industrial endeavor they perform whereby they've created ships that can withstand the chemicals (at least in short term) by lining their hulls with a type of ablative shielding.
Also desirable (although not essential) would be for the composition to not be immediately fatal. A carbon life form (or would they likely be silicate?) could fall in and maybe swim for a few minutes and leave with only some semi minor chemical burns. Yet prolonged exposure would most definitely be fatal.
I'm also wondering if the inhabitants could harness this chemical in some imaginative way such as for propulsion or agricultural/industrial means?
My question is, what kind of chemical/cocktail could I use to fit these kinds of parameters?

As any "rocky planet" is made of metal oxides, Fe oxides, Si oxides, Ca oxides that would counter any real acidic ocean, "best" you can hope for are organic acids from an abundance of carbon that has not yet met the abundance of (bound) oxygen. So if this should be a "hard fiction" you will have to make due with $\ce{H2CO3}$ and acetic acid. Or, you need some special polt to explain it.

Hi and welcome to chemistry.stackexchange.com. Feel free to take a [tour] of the site. Your question may be a bit broad, but that’ll be up to the community to decide. As for the usage part: Most certainly, all acids that the chemical industry produces are useful for something (and most of the time, one just uses sulphuric acid because it is the cheapest).

@GyroGearloose thanks, that's helpful. What if the size of the body was reduced to that of a lake? These certainly occur at least here on Earth: https://www.google.co.uk/webhp?sourceid=chrome-instant&ion=1&espv=2&ie=UTF-8#q=acidic%20lake

Local conditions could be very, very different. Decaying pyrite can release sulfuric acid. (I'm now only a hobbyist on chemistry, so I can't be really sure.) Search the internet for ecological disasters with mining Fe, and be sure to verify (there's a lot of myths out there lacking any science).

Get realistic. Venerean clouds are made of sulfuric acid, they say.

@IvanNeretin yes, but there is no ocean. Sulfuric acid aggressively binds any water, so terraforming Venus  by cooling its atmosphere down and adding lost hydrogen from outside would certainly result in highly acidic oceans, at first. Until the temperature has decreased to a human-bearable temperature. Then, the acid would dissolve the soil and (mostly) neutralize.

That's quite possible. Now what if we imagine a world which is mostly like Venus, but has less metals (say, not enough to neutralize all that acid)?

If your story can accommodate a basic ocean or sea, I can suggest a really good one.

Basic is easy to get: water+ammonia. This has been explored before.

Sure @ShankRam. What are you thinking?

exactly what at @IvanNeretin told. Water + ammonia fits in beautifully. At high conc. it is corrosive, it can be used to make fertilizers, prolonged exposure causes burning sensations and so on...

Coca Cola or a mixture of Coca Cola and Pepsi Cola would probably be a good choice.
It's not immediately fatal for carbon life forms that fall into it. Only prolonged exposure will cause digestion and damage to bones and teeth.
Due to the phosphoric acid ships with iron hulls would withstand corrosion due to passivation through the formation of iron phosphate (also a molluscicide to protect from shipworms).
In agriculture phosphate is also a potent fertilizer.
For propulsion purposes, Cola together with Mentos is a well-known hybrid rocket propellant.

Great answer. I'm now wondering how I could account for the occurrence of phosphate and how to keep the acid concentration up through some kind of phosphorous cycle? (phosphorous rocks/rain?)

I think that it is really hard to consistently model a habitable world with acidic oceans since acids would tend to be neutralized by many minerals in such a world. Phosphoric acid could be formed by hypothetic vulcanic exhalations of P4O10 (fumaroles, hot springs, or, under the sea, black smokers). However, afaik this does not occur on earth. Here you rather have CO2 or SO2. The latter forms sulphuric acid which is more aggressive than H3PO4.


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Are there IP limitations on (M)SDS documents?
Much effort is put in putting together (M)SDS documents:

Many companies offer the service of collecting, or writing and
  revising, data sheets to ensure they are up to date and available for
  their subscribers or users. Some jurisdictions impose an explicit
  duty of care that each SDS be regularly updated, usually every three
  to five years. However, when new information becomes available, the SDS must be revised without delay.

Are there any intellectual property (IP) limitations on MSDS documents? i.e. are they copyrighted?

A really nice question. First, let's point out that specifics about these issues will vary from country to country, as they all relate to the laws of that country. Secondly, let's clarify that copyright and IP aren't exactly the same. Copyright covers the exclusive right of that owner to make copies, license and otherwise exploit their works (printed, artistic etc). Intellectual property is property that results from creative thought, and can include patents, copyrighted material and trademarks. So, IP is a bit like an umbrella that encompasses copyright material.
The specifics relating to copyright and IP differ slightly from country to country, but it is safe to say that the (M)SDS document is a copyrighted document. Most SDS documents will have this specified, usually at the very end of the document - I just checked some of the ones I have on file from commercial suppliers, and they all have a copyright statement. This means that you cannot make copies etc under the usual copyright limitations.
There are requirements about what information goes into a SDS. Again, there will be differences from country to country, but for the US, for example, I think all hazardous components above 1% must be specified, and known carcinogens above 0.1% must also be listed. The exact numbers might be a little different, but the general gist is that hazardous components must be listed.
HOWEVER, every country has a clause in their code governing SDS requirements that specifies, essentially, an exemption from listing ingredients that may be regarded as a 'trade secret', or puts at risk an identified Intellectual Property claim. How this is expressed varies; in some countries, a trade name may be used to 'hide' the true identity of a IP-protected ingredient. In other countries, a summary of functional groups can be used to represent that key ingredient. Usually, though, the nature of the hazard must be represented somehow on the SDS document. I know in some countries, even if an ingredient is protected by an IP clause, its true identity must be revealed to health practitioners if requested. I'd have to dig around for the exact laws for each country/zone if you needed more specific answers.

Would it be considered "fair use" or so if I reformat an existing sds? For example mass-converting sigma's pdf files into a publicly available html format?

@Sparkler  (caveat) I am not a lawyer, but I would say no. By Sigma, do you mean Sigma-Aldrich? They have at the bottom of each SDS the copyright declaration: "Copyright XXXX Sigma-Aldrich Co. LLC. License granted to make unlimited paper copies for internal use only." Sigma chemicals, Australia, do not have this copyright statement on their SDS, but in most parts of the world copyright does not have to be explicitly stated.


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Aromaticity of 1,1'-bi(cycloprop-2-en-1-ylidene) and 1,4-dioxine
Are these Anti aromatic/aromatic or non aromatic?
1.

Here, if we were to consider the resonating structure(s) with charge separation

One ring is aromatic and other is anti aromatic, and secondly there is not continuous delocalization of pi electrons between those rings {if we don't include no resonating structures with no formal charges, as their contribution would not be equalish to normal form} what category would this go in?
2.

Like before, there can be no continuous electron delocalization like stated in previous example , only other ones the diff forms of

If we consider it to have enough delocalization, it would have 4n electrons and hence antiaromatic {although all the orbitals wont be mixed as 3 on each side twice with oxygen's lone pair being common between all of them}

As we can see in pyrene, if we consider central double bond to be non participating, huckle rule works, is there a better (and just by observation by eye) method that we could use for multi ring systems? what is the basic condition for aromaticity besides planar,cyclic,sp2 {nitrogen and oxygen in general change their hybridisation as to make it aromatic with their lone pair(s)}

What is the question?

is the second one aromatic or not? and if there is a better way to evaluate aromaticity

The second one is definitely not aromatic.  It would have 4n electrons if planar; the oxygens would need to be sp2 hybridized to make this molecule planar, meaning there would be a lone pair in a p-orbital each

i mean anti aromatic or not

Antiaromaticity is a fictional state with real consequences,  that any molecule will distort to avoid.
In the case of molecule 1, the primary state will be a triplet (unpaired pi-electrons), with the central bond lengthening to decrease the pairing of the pi-electrons.  This will lead to 3 electrons in each pi-system, and hence preventing antiaromaticity, because there are no longer 4n electrons in the rings.  A fictional molecule, with the constituency shown for molecule 1 that is in a singlet state would be antiaromatic.
In the case of molecule 2, the molecule will distort itself to make the at least one of the oxygens non-coplanaer with the pi-system, and no longer being co-planar, the system will not be antiaormatic.  A fictional molecule with the constituency shown for molecule 2 that is co-planar, would be antiaromatic.
It is unclear what you are asking about for pyrene.

ok, understood, would the same happen  here?

Yes, and the net effect is that the 2-proton in 2-H pyran is basically undeprotonateable.

Wouldn’t molecule 1 also distort so that the two rings are perpendicular to each other?

@lighthart so the 2-"proton" in 2H-pyran could in fact be hydridic! In the case of 4H-pyran, which is better known, the CH2 hydrogen does in fact transfer as a hydride ion when it dispropotionates.


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What is insoluble in drinking alcohol, but edible?
What substances are edible for human consumption, but will not mix with alcohol? What is a substance that would block alcohol from mixing with its counterparts in a product?
For example: You make a cupcake and want an alcohol shot in that cupcake. What substance could the alcohol shot be contained in so it wouldn't disperse into the cupcake, but still be edible?

When you say edible, you mean non-poisonous and human-stomach-degradable?

A sugar hull with a chocolate coating to prevent evaporation does a great job https://en.wikipedia.org/wiki/Praline .

In a cupcake? So, edible AND able to withstand a cooking process to 180oC??

AND able to withstand the pressure of alcohol vapour at 180oC.

Ethanol does not dissolve chocolate, which consists mostly of butter and sugar. These mini alcohol filled chocolates are very common party candies. You could try inserting one of these into your cupcakes after you have baked them.
You could also try making a mold for a gelatin-based shot cup, as gelatin is also insoluble in ethanol. Gelatin is capable of absorbing alcohol, however, so I do not recommend making these too far in advance, lest they turn soggy.

In first order approxiamtion: Ethanol and similar substances solvable in water cant be mixed with unpolar substances. You could carry alcohol in a bowl of solid fat for some time.
But to solve your problem of sweet drunkness have a look in a candy shop: You can fill chocolate (made up of fat and sugar primary) with liquor. Liquid alcohol will barely survive head of a stove in same time biscuits need, as well as solid fat.


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How do partially protonated sulfate ligands affect a metal's charge?
I'm currently reviewing the Periana-Catalytica system and I have a question about one of the intermediates, namely $\ce{[(bpym)Pt(OSO3H)2]^+}$, which is found in the following tutorial review. How do the $\ce{OSO3H}$ ligands affect the oxidation state of the metal? If I count each of these ligands as a monoanionic ligand, and also factor in the positive charge of the complex, wouldn't the oxidation state of the metal be +3? This seems a bit strange to me, seeing as Pt typically has oxidation states of +2 or +4, but never +3.

http://www.wag.caltech.edu/publications/sup/pdf/512.pdf

So judging from the complexes I see in Figure 3, I've deduced that $OSO_{3}H$ is a monoanionic ligand. Am I correct?

Yeah, your right. And there's no problem with intermediate on +3 oxidation.

Thank you for clearing that up. So is the +3 oxidation state not really a problem for Pt, per se--it's only not really seen because it's an intermediate and is therefore very transitory?

There are also Pt (III) compounds, at least dinuclear ones http://pubs.acs.org/doi/abs/10.1021/ic00137a074


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the longevity of amorphous metal
I gathered that many of amorphous solids are made by quickly cooling the liquid material, thereby giving it no time to form any long-range order (crystallinity). So it makes a perfect sense that when high heat is applied to amorphous solid, it will regain its crystallinity, and will no longer be glassy.
But is this behavior non-continuous? That is, can sustained low thermal excitation give rise to the crystallinity in amorphous solids given enough time? Or is there a threshold temperature (or a level of thermal/mechanical excitation) below with no excitation can bring back the crystallinity?

Compounding wins - Better living though chemistry. I bought a ton of coal for my 401(K) plan. By the time I'm ready to retire it will have been transformed to diamonds. ;-)

There is no threshold, but the time required varies dramatically. What takes seconds at high temperature, without heating could easily take millenia.

The answer to your question is no in general. Because amorphous metal is usually an alloy rather than a pure metal. These alloys are made of atoms of largely different sizes, which results in low free volume and higher viscosity (compared with other metal and alloys) in molten state. The viscosity hinders the atoms movement to form a crystal.
So, the significantly different atomic radii of the components that constitute the amorphous metal (about 12%) favors the high packing density and low free volume. This leads to a complex crystal units with higher potential energy and lower chance of formation.
Accordingly, I think that even if we heat an amorphous metal above a threshold temperature  and let it cool slowly, we won't have a conventional crystalline structure. At best, we get regions of some order.


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Hybridisation of hexaamminecobalt(III) complex
Does $\ce{[Cr(NH3)6]^3+}$ have $sp^3d^2$ or $d^2sp^3$ hybridisation?
Oxidation state of $\ce{Cr}$ is $+3$. The electronic configuration is $3d^3$ and the $d$ electrons occupy $t_{2g}$ orbitals. If the electrons from the ligands were to occupy the $e_g$ orbitals the stability of the complex decreases. So shouldn't the hybridisation be $sp^3d^2$?

How are $sp^3d^2$ and $d^2sp^3$ different?

In d2sp3, inner orbitals (in this case, 3d) will be used. In sp3d2, the outer 4s, 4p, 4d is used.

"If the electrons from the ligands were to occupy the $\ce{e_{g}}$ orbitals the stability of the complex decreases.". Why would the compound want to be unstable? The $\ce{d}$ electrons occupy $\ce{t_2 g}$ orbitals according to Hund's rule

@NilayGhosh the t2g orbitals are already filled. The ligand electrons have to occupy eg orbitals.

Reason for the above duplicate vote: The octahedral geometry is the same, the oxidation state of the metal is the same, in one case the ligands are the same and the question is equally wrong in its premises.

@Jan seems like you had an argument with a user. Even my teacher told that $\ce{NH_3}$ acts as a strong field ligand for certain metals. It's not specified in standard books like JD Lee or Atkins. But for the purpose of getting the correct answer, we are taught like that. This question comes up in my exam frequently and since I am in high school, the only way to get the correct answer is by learning it by heart. (I hate doing it).

@AdityaDev: Are you referring to $\ce{NH3}$ being strong field? Well … I can understand where you’re coming from; if you define water to be a neutral ligand and anything weaker being weak-field and anything stronger being strong-field then hmm … but at least it’s consequent. However, I was taught to distinguish strong-field ligands like $\ce{CO}$ or $\ce{CN}$ from medium-field ligands like most nitrogen-ligands $\ce{NH3, en}$, porphyrins and group in water as the weakest medium-field ligand. I agree that $\ce{NH3}$ is not weak-field though.

@Jan is it okay if I memorize that the complex is d2sp3? I am still at high school and there is only one month for my exam. (I have physics and maths to study as well).

@AdityaDev Read the second-last paragraph in my answer to the dupe question ;)

It has d2sp3 configuration since NH3 is a strong field ligand which causes the electron to pair up and use the inner d orbitals(3d) .

Can you elaborate your answer with a help of a hybridisation diagram? It looks like a comment rather than an answer right now.

With 40 reputation I cannot comment.

"Can you elaborate your answer with a help of a hybridisation diagram? "

I thought NH3 was a weak field ligand.

Calling $\ce{NH3}$ a strong-field ligand is really pushing it: It’s probably the weakest of all ligands stronger than water. Also, the answer is wrong because it is wrong to think of hybridisation here.

@NilayGhosh look at jan's comment

@Jan I read your answer in the above question. I find it correct logically. actually on elementary level we are taught that NH3 is a SFL , and rather we should taught on the basis of MOT , but it was not in our syllabus! So I gave that reason.


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Why isn't Graham's Law working here?
In graham's law of diffusion most numericals I have seen assumes the formula
$$\frac{r_1}{r_2}=\frac{P_1}{P_2}\sqrt{ \frac {M_2}{M_1}}$$
Where they say $P_1,P_2$  are the partial pressures of the two gases receptively inside the container.But my question is should'nt it be the difference in partial pressures inside and outside the orifice?
Because say if the pressure outside equals pressure inside then the rate of diffusion becomes $0$ .Then graham's law fails in such a case.
Please tell me if I am going wrong.

Well, yes, it should be the difference of pressures inside and outside; it's just that we assume the gas is escaping into a kind of void, so the outside pressure can be ignored.


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Raoult's Law and Mole Fraction
Raoult's Law states that the partial vapor pressure of each component of an ideal mixture of liquids is equal to the vapor pressure of the pure component multiplied by its mole fraction in the mixture.
As we know that the liquid and vapour in a container are at dynamic equilibrium,
So the mole fraction which is mentioned in Raoult's Law is measured at equilibrium or initially?

Raoults Law says that, for an ideal liquid mixture, the partial pressure of each component in the gas phase is equal to the vapor pressure of the pure component (at the temperature of the system) times the  mole fraction of that component in the liquid phase.  If the system is at equilibrium, the mole fraction to use in the equation is the current mole fraction.

So you want to say that if we have to calculate partial pressure at any time then we must use mole fraction of the component in solution at that time in the equation?

Yes, that is what I’m saying.

But that creates a problem, Initially partial pressure is zero but the product vapor pressure  as a pure component and mole fraction is not equal to zero, leading to a contradiction to your statement

Why is the partial pressure zero initially?  Do you now have vapor and liquid initially at equilibrium?

Initially partial pressure will be zero as nothing will have been evaporated from solution

This is a non-equilibriium situation, and Raoult's law applies only to equilibrium.  However, at the interface between the liquid and gas phases, the local partial pressures equilibrate instantaneously with the liquid at the interface, so it always applies at the interface.  It's just that, if the global system is not at equilibrium, the partial pressure in the gas phase vary with distance from the interface, and diffusion is occurring in the gas phase.

But as you say that to calculate partial pressure at any time then we must use mole fraction of the component in solution at that time in the equation, then that statement is also incorrect since we will not have equilibrium every time

In a transient situation, the equilibrium relationship applies only at the interface.

Ok So When we have reached saturated conditions, we take the mole fraction components of solution of that time i.e when liquid-vapour equilibrium is achieved

Globally, yes.  At the interface, the local concentrations in the liquid and vapor phases are always at instantaneous equilibrium.

Can I have a discussion with you in chat to tell you the question  related to this which caused all confusion?


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Is water "hardness" dependent more on calcium or magnesium?
If one sample of water has a certain concentration (g/L) of calcium ions, and another sample as that same concentration of magnesium ions, will one solution have a greater "hardness than the other"? In other words, does one ion have more influence over water hardness than the other?

There are other ions that contribute to hardness, particularly iron and aluminum, just FYI.

The one with the magnesium would be harder (assuming no other hardness-inducing ions) because magnesium has a lower molecular mass, so the molarity of magnesium would be higher. Assuming a 1:1 molar ratio of magnesium:calcium would give the same hardness (i.e. the same $ppm\ \ce{CaCO3}$ rating).

The permanent hardness of water is typically given in one of three types of measurements: grains per gallon, milligrams per liter (mg/L), or parts per million (ppm) of "calcium carbonate" in the water. Since calcium carbonate has a mw of 100 g/mole the equivalents of calcium carbonate would be: $$ \text{g }\ce{CaCO3} = 100 \text{ g/mole }\times \text{(}\ce{[Ca^{2+}] +[Mg^{2+}]}\text{)}$$ where $\ce{[Ca2+]}$ is the molarity of calcium and $\ce{[Mg2+]}$ is the molarity of magnesium. So in the sense of molarity, calcium and magnesium are equal.
However you could also measure calcium and magnesium as mass of the cation per volume in which case $$ \text{mass of }\ce{CaCO3} = 2.5 \times \text{(mass of }\ce{Ca^{2+}} \text{)} + 4.1 \times \text{(mass of }\ce{Mg^{2+}} \text{)}$$  so in the sense of the mass of cations, then 1 g/l of magnesium is harder than 1 g/l of calcium.


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Hydroxide storage
I am figuring out what chemicals to store with Ammonium hydroxide. At the moment a 2-shelf corrosive resistant fume hood cabinet will be used.
I have in mind storing the Ammonium hydroxide with Potassium hydroxide, Sodium hydroxide, and Calcium hydroxide on the lower shelf of the cabinet. Would this be okay?
And would it be okay to store carbonates and bicarbonates on the upper shelf? Any other suggestions as to what would be okay to store together with the chemicals listed?
Many thanks!

Welcome to Chemistry.SE! Take the [tour] to learn more about this site. Enjoy!

I would not store carbonates or bicarbonates with hydroxides. I would restrict that cabinet to alkali materials only. A general chemical safety principal is to store bases separately, acids separately, corrosives, etc. I'd store the carbonates and bicarbonates in a separate cabinet for inorganic solids.

@GeoffHutchison Thanks, Geoff! I am following you advice and will store the hydroxides on their own.  The cabinet we have is under the fume hood but does not have any sort of exhaust from the cabinet to the fume hood and out. Would this be okay to store the hydroxides? Everything in the cabinet is corrosion-resistant.

My lab doesn't have separate venting for hydroxides. At least in the US, I don't believe this is required for acid/base cabinets. It's very common here to have an acid or base storage under a hood but not vented.

On a personal note, I think that the somewhat popular term "ammonium hydroxide" is awfully misleading and should be avoided.
National regulations concerning occupational safety and environmental protection might vary. My suggestions are based German requirements.
According to TRGS 510, aqueous ammonia and the solid hydroxides mentioned in the question fall in the same storage class (8B Non-combustible corrosive substances) and may be stored together.
However, aqueous ammonia is a caustic aqueous solution of a gas. At elevated temperature, pressure may build up.

Make sure that the bottle can not break under pressure. For glass bottles, use PE stoppers. Alternatively, use plasic bottles.
Make sure that the faces and eyes of lab workers aren't directly hit when something goes wrong. Store bottles with aquous ammonia below hip level.

If you take a look at the GHS hazard statements for aqueous ammonia, you'll see H400: Very toxic to aquatic life. According to German regulations, aqueous ammonia is filed under "Wassergefährdungsklasse (WGK) 2"). You might want to make sure that no aqueous ammonia is released to the water in case of a leakage. Use suitable retention and/or make sure that all drains in the lab are connected to a central netralisation.


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How to remove salt from powdered soup mix?
Sometimes I carry powdered soup mix (Knorr/Maggi) when I travel but they are way too salty. Is there an efficient and simple chemical process to lower the salt content?
I know sodium is super soluble and is hard to remove by precipitation, but is there a work around to perhaps only dissolving the salt and take them out? Thanks!

I think this question belongs in LifeHacks SE.

This might get good answers at LHSE, but it might also get good answers here.  It's sad to me that over time this community seems to focus on helping only academic chemists and spurning those who want to use a little chemistry in their daily lives.

Heard that a little egg's white could help in case of too salty soups. It gets salted out binding salt.

There isn't any magic way to get rid of excess salt in food.

The simplest solution is to add more water. If the soup becomes too bland just add some more other spices.

I think this question should stay. I've often had the same question!!

Peel and thickly slice several potatoes and cook these in your soup. Remove them when they have softened. They will absorb the salt while cooking.
You can use this method if you accidently add too much salt to stews or sauces (normally this becomes apparent when reducing.)
The removed potatoes can be discarded or used in another meal, maybe mashed.

Do you have a reference for this? I am interested in learning the mechanism behind this.

Can it be any type of potato? I tried this once with a high starch potato and the potato fell apart during cooking. Maybe the lower starch, so called "waxy potatoes" are better suited for this application.

The consensus I've seen from sources that actually go to the bother of testing this (rather than blindly repeating the old wives tale) is that it doesn't work, or if it does work it's only to the extent that the water content of the potatoes dilutes the soup. - If you have a reference which actually tests and verifies this, I'd also be interested in seeing it.

I think that using chemistry is the hard way to solve this problem. In my country, all foods have nutrition information which includes sodium levels. So either select low sodium dried mixes or take dried food to make your own soup.

Is there an easy way? Yes. But it will make the soup toxic, so I will not explain the details. I'm sure there are some ion exchange columns which have been manufactured so that they are suitable for pharmaceutical or food use. So, what you'd have to do is first filter the soup to remove particulates and colloids (such as oil globules) and then pass the clear liquid through the ion exchange bed(s). I'd estimate it would cost less than 100 $ per bowl of soup. As far as potatoes, they do have a water content. Perhaps cooking them with your soup would eventually decrease the salt in the soup and increase it in the potato, that sounds reasonable. I'd suggest you try low-sodium powdered mashed potato mix (if you can find it) and you could also try rice or pasta to absorb some of the salt water and improve the taste. Just some wild guesses. You really don't want to stray from various cooking or food additive processes when it comes to stuff you eat (or drink).