text:
Effect of concentration on molar conductance
It has been stated that the molar conductance($\Lambda_m$) of strong electrolytes is not affected to a greater extent on dilution and so to find the limiting value of molar concentration($\Lambda_m^0$) we can extrapolate the graph of molar conductance vs the concentration ($\sqrt c$). But as the concentration would near zero shouldn't the molar conductance decrease sharply to the range of say $10^{-8}$ which is the conductance of distilled water as the amount of solute that is the strong electrolyte would be negligible to consider?Also in this case extrapolating the graph should not be possible?

The molar conductivity of a strong electrolyte $(\Lambda_m)$ is given as the ratio of measured conductivity $(\kappa)$ to the molar concentration $(c)$:
$$\Lambda_m=\frac{\kappa}{c}$$
However, as you note, the value of $\Lambda_m$ is not independent of concentration. The reason for this is that $\kappa$ does not scale linearly with concentration.
$$\Lambda_m = \Lambda_m^\circ -K\sqrt{c}$$
In this equation, $\Lambda_m^\circ$ is the limiting or intrinsic conductivity of the electrolyte and $K$ is an empirical constant. The interpretation of $\Lambda_m^\circ$ is the molar conductivity at infinite dilution as $c\rightarrow0$.
$$\lim_{c\rightarrow 0}{\Lambda_m}=\Lambda_m^\circ$$
The infinite dilution scenario is different from the scenario when there were no ions in solution to begin with, just as mathematically $\lim_{x \rightarrow 0}{f(x)}$ does not always equal $f(0)$ (often because $f(0)$ is undefined).
In the limiting case of $c\rightarrow 0$, the ion concentration never reaches zero, but it does keep getting close. You can never dilute a solution containing a solute to zero concentration (i.e. pure solvent).
If you graph $\Lambda_m$ as a function of $\sqrt{c}$, you should get a straight line with a non-zero intercept $(\Lambda_m^\circ)$. This value is an empirical parameter used to model data. It can also tell you something about the intrinsic conductivity of ions (see this Wikipedia article). However, as you say, at exactly $c=0$, he conductivity should be negligible. At any other value of $c$, however the conductivity should follow from the equation given. Thus, the conductivity is not necessary linear in the infinitesimal range around $c=0$:
$$\Lambda_m=\begin{cases}
10^{-8}\approx0, &c=0\\
\Lambda_m^\circ - K\sqrt{c}, &c>0 \end{cases}$$

On increasing the dilution the conductivity($\kappa$) does decrease as it is proportional to the number of ions in a unit volume. But the molar conductance is the product of $\Lambda_m=\kappa V$ where $V$ is the volume containing one mole of the dissolved electrolyte.It is proportional to the number of ions in $V$ volume of the electrolytic solution which includes all the ions in one mole of the electrolyte plus the additional contribution from water. Hence, on increasing the dilution $\kappa$ goes down but $V$ goes up in such a way that its product goes up. Therefore in increasing the dilution, the conductivity will fall to the level of distilled water but the molar conductance still includes the volume containing one mole electrolyte and hence will have a value higher than that of distilled water and the concentrated solution of the electrolyte.
[The general term Conductance (as defined by $C=\frac{1}{R}$) is different from the quantity molar conductance. The former, deals with a electrolytic setup with a definite cell constant and fixed condition. The other is equal to the Conductance of $V$ volume of electrolyte between two electrodes placed 1 UNIT APART.]


text:
Increase in bond length of BF₄⁻ compared to BF₃
Why does $\ce{BF4^-}$ have a larger bond length than $\ce{BF3}?$ Usually in molecules the bond angle increases to reduce repulsion but, here the bond length increases. Why is it so? Does the bond length increase with the bond angle in other cases of such kinds of molecules?

Isn't this then a simple case of the higher bond length of $Sp^3$ hybrid orbitals in $\ce {BF_4^-}$ than the $Sp^2$ hybrid orbital in boron trifluoride?

@SatwikPasani hybrid orbitals don't make for good answers in general since they don't actually exist.  The fact that they remain in the chemistry vernacular has been discussed recently

@bobthechemist as much as I personally dislike the overuse of hybrid orbitals in chemical education, they have their use… (and lots of limitations, yes)

There seems to be a bit of confusion between bond length and bond angle here.  It is true that VSEPR theory predicts that bonding pairs of electrons (and lone pairs, for that matter) will repel one another as much as possible, thereby increasing bond angles, but this has little to do with bond length.
One way to explain the different bond lengths in $\ce{BF_3}$ and $\ce{BF_4^-}$ is by looking at the molecular orbital diagram of $\ce{BF_3}$:

The linked diagram isn't great since it does not show the filled orbitals: the $1a_2'$ $1e''$ and $4e'$ orbitals are all filled, making the lowest unoccupied orbital (LUMO) the $2a_2''$.  If you haven't wrapped your head around molecular orbital diagrams of this complexity, don't sweat it.  The bottom line is, the $2a_2''$ orbital has antibodning interactions with the $2p_z$ orbitals of boron and the $2p_z$ orbitals of the surrounding fluorines.  Thus, adding electrons to this orbital makes the $\ce{B-F}$ bonds weaker, and therefore longer.

Sorry if I didn't understand well. but I imagine all orbital symmetries and energies are going to shift around once a fluoride ion is added to $\ce{BF3}$, so to me it seems a bit strange to argue about the bonding properties of $\ce{BF4^{-}}$ by considering a simple electron addition to $\ce{BF3}$ MOs.  Is this really safe?

@NicolauSakerNeto You are correct that the orbital symmetries will change and the energies will shift.  We'd need better computational methods to get at quantitative explanations.  I'm making an assumption that the HOMO of $BF_{4}^{-}$ will still be higher in energy (and somewhat antibonding) even in a tetrahedral conformation; however all I've seen is MO diagrams for methane or tetrahedral d-metal complexes, so I can't be certain.

$\ce{B}$ in $\ce{BF3}$ is $sp^2$ hybridized (since the molecular geometry is trigonal planar), and therefore the unhybridized $p$ orbital in $\ce{B}$ can overlap with the $p$ orbitals in $\ce{F}$. You can draw resonance structures, and from the valence bond theory perspective, the bond order of $\ce{B-F}$ in $\ce{BF3}$ will be approximately 1.33 due to the partial $\pi$ bond character in the $\ce{B-F}$ bond in $\ce{BF3}$.
For $\ce{BF4-}$, the $\ce{B}$ is $sp^3$ hybridized and hence p orbital overlap is not possible. Hence, the $\ce{B-F}$ bond is simply a conventional 2-centred, 2 electron bond with bond order 1. Comparing bond orders, it becomes clear that $\ce{BF4}$ has a longer bond.
If you wish to, you can draw a simplified MO diagram showing stabilization due to overlap of the $\ce{B}$ $2p$ and $\ce{F}$ $2p$ orbitals, but the valence bond theory argument suffices.

Can we explain this  on the basis of charge density?
Like, in $\ce{BF3}$, the number of $\ce{e-}$  around $\ce{B}$ are 6 while in $\ce{BF4-}$ there are more number of $\ce{e-}$ around $\ce{B}$.
Since we know that $\ce{B}$ has small atomic size, so more number of $\ce{e-}$ means reduced effective nuclear charge on the valence $\ce{e-}$, due to which the bond length increases.
Thus, $\ce{BF4-}$ has longer bond length than that in $\ce{BF3}$.

In $\ce{BF3}$ there is backbonding from $\ce{F},$ which increases the double bond character of the $\ce{B-F}$ bond. There is no backbonding in $\ce{BF4-}.$ This causes shorter bond length in $\ce{BF3}$ compared to $\ce{BF4-}.$

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Is there a correlation between the susceptibility of a polymer to undergo hydrolysis and its melting point?
As per topic, I was wondering why the Tm of the polymers PGA, PLA and PLGA are in the following order. Does it correlate to their ability to undergo hydrolysis?
PGA>PLA>PLGA

I don't have the information available to do more than comment, but rates of hydrolysis (generally) are PGA > PLGA > PLA. The hydrolysis rates are controlled by the sequence, e.g. GG hydrolyzes faster than GL or lG, which hydrolyze faster than LL (all for steric reasons). Melting point has to do with the overall order of the polymer. Tm of PLGA depends on composition, sequence, polymerization method, and control of stereochemistry.

Off the cuff answer: in solution Tm would have no effect.  Out of solution (ie caustic solution applied to solid polymer), surface wetability for the solvent carrying base would be a much stronger component than Tm.  Well underneath the Tms, no change would be observed.  However, lower Tm at a temperature near the Tm would result in better adsorption by solution all other things being equal, so the 'ability' to undergo hydrolysis (generally described as a rate) would be higher.  All of that being said, I think Tg would be more important than Tm, but less important than wetability.

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How do I make a paper white again?
Suppose I have a white paper originally. Then it is placed in a trunk, where soda spills occasionally and gasoline gas/gasoline also spills due to no gas tank cover. Over time, this paper becomes discoloured.
How do I make it white again? What equipment do I use?

The Library of Congress experimented with using diethyl zinc to preserve and deacidify old documents without damaging the ink used to write them. It apparently did a pretty good job, provided the reactor chamber was thoroughly cleared of reactant before the hatch was opened and oxygen was admitted; otherwise the processed papers would spontaneously burst into ravenous flames.

One approach would be to use various solvents. You would require solvents in which your various stains were soluble in. Alcohols / water would be suitable to dissolve the soda. Whereas an organic solvent would be suitable for washing away the gasoline. Though gasoline is typically colourless.
Another approach would be to expose the paper to copious UV irradiation, materials are coloured due to absorptions in the visible spectrum which trigger electronic excitation. These absorptions bias the transmission and reflection of otherwise white light and give the materials their colour. UV radiation is of higher energy than visible light radiation, and therefore the chemical excitations involved can prompt reactions which result in the destruction of the materials themselves.
A similar approach to invoke this chemical degradation of the coloured materials could be achieved through using a strong acid or base. This would be akin to the effect of bleach on household surfaces. Again, exposure to acids and bases can prompt reactions which result in the degradation of coloured materials.
It is worth noting that all of these methods will have some affect on the paper. Continued washing / drying of the paper with solvents will likely cause the paper to become brittle and affect the absorption of the paper, which may affect your ability to write / print on it. A similar effect may be noticed under UV irradiation, though I expect it wont be to the same extent.
A sufficiently strong acid or base may be able to completely degrade your paper as well as your coloured contaminants.

one more category of chemical reagents Horba misses is the oxidizers . sodium hypochlorite , or ozone , can kill many colour compounds .

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text:
Balancing a redox reaction with only one product
Consider the following comproportionation reaction:
$$\ce{NH4NO3 -> N2}$$
How would I go about balancing this using the half reaction/ion-electron method?

You should use the the same product species in both half -reactions. In the example you gave, this is $\ce{N2}$.
$$\ce{NH4NO3-> N2}$$
The half-reactions are constructed from the ammonium and nitrate ions separately, since these feature nitrogen in different oxidation states:
\begin{align}
\ce{NH4+ &-> N2} \tag{1} \\
\ce{NO3- &-> N2} \tag{2}
\end{align}
Balancing the reduction half-equation $(1)$:
\begin{align}
\ce{2NO3- &-> N2} & & \text{(balance N)} \\
\ce{2NO3- &-> N2 \color{red}{+ 6H2O}} & & \text{(add } \ce{H2O} \text{ to balance O)} \\
\ce{2NO3- \color{red}{+ 12H+} &-> N2 + 6H2O} & & \text{(add } \ce{H+} \text{ to balance H)} \\
\ce{2NO3- + 12H+ \color{red}{+ 10e-} &-> N2 + 6H2O} & & \text{(add } \ce{e-} \text{ to balance charge / oxidation state)}
\end{align}
And balancing the oxidation half-reaction:
\begin{align}
\ce{2NH4+ &-> N2} & & \text{(balance N)} \\
\ce{2NH4+ &-> N2 \color{red}{+ 8H+}} & & \text{(add } \ce{H+} \text{ to balance H)} \\
\ce{2NH4+ &-> N2 + 8H+ \color{red}{+ 8e-}} & & \text{(add } \ce{e-} \text{ to balance charge / oxidation state)} \\
\end{align}
In order to combine the two balanced half-equations, we need to multiply the first by $4$ and the second by $5$:
\begin{align}
\ce{8NO3- + 48H+ + 40e- &-> 4N2 + 24H2O} \\
\ce{10NH4+ &-> 5N2 + 40H+ + e-} \\ \hline
\ce{8NO3- + 10NH4+ + 48H+ + 40e- + 10H2O &-> 9N2 + 24H2O + 40H+ + 40e-}
\end{align}
which finally reduces to
$$\ce{8NO3- + 10NH4+ + 8H+ -> 9N2 + 14H2O}.$$


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Shelf-life of solutions
If I prepare the silver nitrate solution or any other solution, how can I know the shelf life of this solution? I know that it depends on the stability of the substance, but maybe there is some document which give information and is interesning how realy it was found.

You won't find a general formula for the shelf life of a solution.  A silver nitrate solution will have a very short shelf life if it is kept in the sunlight, whereas potassium nitrate could last for years.  The shelf life is also going to be very strongly dependent on the quality of the container.  Think about what happens to the aging of a wine bottle if the cork does not provide a sufficient seal .  Interactions between the solution and the container have an impact on the solution shelf life as well.  To stick with the wine theme, corked wine is due to contamination of the wine with 2,4,6-trichloroanisole which is generated by natural fungi in the cork that have been exposed to chlorine.
So it is important to understand the chemistry of the solution with which you are working in order to determine its shelf life.


text:
Would Chlorine or Chloramine gas show up on a VOC test?
Would Chlorine or Chloramine gas show up on a VOC test?  My apartment was cleaned using chlorox and lysol, and there has been a smell that's been lingering for about 10 days.  The air was tested with a VOC meter but nothing came up.  Would Chlorine and Chloramine have shown up on this test?

Chlorine, no. Chloramine, maybe.
Monochloramine, $\ce{NH2Cl}$, is an inorganic compound, and not very volatile (when dissolved in water it forms more or less of an azeotrope, evaporating slightly faster than the water itself). So, not a VOC.
However, the way it does its job as a disinfectant is to basically shred organic compounds, like cell walls, key enzymes, etc, forming "organic chloramines" where one or more of the remaining hydrogens is replaced with an organic functional group. These now-organic compounds have a higher vapor pressure, so by design, chloramine is used to create VOCs. The idea, when used in tap water or pool/spa chlorination, is to allow the organic chloramines to boil off and disperse in the air, leaving clean water behind. That "chlorine smell" of a chlorinated pool is actually mostly organic chloramines evaporating out. In a more enclosed space like an interior room, these compounds can't disperse and so the smell builds.

Very helpful, thank you. What are some definite ways to test for Chlorine and Chloramine in the air?


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Hybridisation of the negatively charged carbon atom in a carbanion
Given the carbanion, $\ce{R3C-}$, the carbon is $\mathrm{sp^3}$ hybridized, unless it is participating in resonance. This is clear from its steric number.
However, I am having trouble in drawing its orbital diagram. I know that if this was a neutral carbon atom, it would look like this:

$\mathrm{3s}$: _
$\mathrm{2p}$: ↑ ↑ _
$\mathrm{2s}$: ↑↓
$\mathrm{1s}$: ↑↓

Which would hybridize to this:

$\mathrm{3s}$: _
$\mathrm{2sp^3}$: ↑ ↑ ↑ ↑
$\mathrm{1s}$: ↑↓

Now if I was to examine the carbanionic state, would I simply be adding one electron to the top most diagram, then re-hybridize that? Or would I add two electrons because the carbon is not sharing the electrons with any other atom? That's another question I have with orbital diagrams. Do we break each bond homolytically, and base our diagram off of that? This would give:

$\mathrm{3s}$: _
$\mathrm{2p}$: ↑ ↑ ↑
$\mathrm{2s}$: ↑↓
$\mathrm{1s}$: ↑↓

I am not sure how I would rehybridize that. Have I drawn the radical form?

You are asking for $\ce{R_3C^-}$ and then you draw orbital diagram for isolated carbon, that does not mix well. Additionally, that carbanions are planar, so adjust your hybridization properly. And finally, it's you who has to decide, if you talk about radical anion or closed shell anion.

I don't agree about planarity, but inversion is an option.  To flesh out Ssavec's answer slightly: you can't really consider sp3 hybirdization in the context of just the carbon atom.  You have to mix in hydrogen orbitals to form a  molecular orbital diagram.

Hybridization is a concept, which is often misconceived. One has to understand, that it is a mathematical concept, which explains a certain bonding situation (in an intuitive fashion). In a molecule the equilibrium geometry will be formed respecting steric and electronic interactions and possibly interactions with a medium or external field. The geometric arrangement will not be formed because a molecule is hybridized in a certain way.
Hybridisation is therefore only a result of a certain geometric arrangement of atoms in a molecule. A linear combination of all available (atomic) orbitals will form molecular orbitals (MO). These can then be rearranged (recombined) in a certain way to form localised molecular orbitals (LMO). These can then be used to interpret a bonding situation in a simpler theory.
Each LMO is expressed as a linear combination of the atomic orbitals, hence it is possible to calculate contributions of the atomic orbitals and describe these also as hybrid orbitals.
In a very simple case methane, the $\ce{C-H}$ bonds are formed from contributions of carbon's $\ce{2s, 2p}$ and hydrogen's $\ce{1s}$. For the carbon this may be interpreted as a $\ce{2sp^3}$ orbital.
In the simple case of $\ce{{}^{-}CH3}$, as Uncle Al already pointed out is isoelectronic to ammonia, the bonding situation may be also described in terms of $\ce{2sp^3}$ orbitals. As well as ammonia (or amines in general) it may also undergo (nitrogen) inversion.
(source)
In this inversion scenario the hybridisation of the carbon (nitrogen) changes from $\ce{2sp^3}$ to $\ce{2sp^2}$ to $\ce{2sp^3}$.
With larger carbanions this becomes a little more complicated. As long as $\ce{R=Alkyl}$, hyper conjugation is not strong enough to prevent pyramidalisation. However, the inversion berrier is furthermore reduced.
If any of the $\ce{R}$ groups are different from alkyl substituents, then I would expect that the environment around the carbon becomes (more) planar, i.e. the contribution of carbon's $\ce{2s}$ orbital increases. That may have various reasons, but some kind of conjugation or hyperconjugation will be involved. When these $\ce{R}$ groups become pretty bulky, I would expect the same, due to steric interactions.

Summary
Please understand hybridisation as a tool to understand bonding situations. Therefore it is a result of a geometric alignment of atoms in a molecule. It is not the driving force to form a certain geometric constellation.

$\ce{H3C^-}$ is isoelectronic with$\ce{NH3}$, as $\ce{CH4}$ and $\ce{NH4^+}$ correspond. Each of the four $\ce{sp^3}$ hybrid orbitals contain a pair of electrons, in all cases shown.  Nuclear charge versus electron count and the source of the four electrons added to the naked $\ce{sp^3}$ hybrid orbitals are accessories.


text:
Yellow precipitate in the presence of ammonium molybdate?
We had analytical chemistry experiments to determine acidic radicals from a mixture of two salts. After the chloroform-chlorine water test for distinction between halides $(\ce{salt\ solution + H2SO4 + CHCl3 + Cl2_{(aq)}})$ when I emptied the test-tube in the sink, all of it suddenly turned yellow.
What reaction could have occurred to cause this?
The sink mostly contained diluted $\ce{H2SO4}$ and $\ce{HNO3}$ along with small quantities of $\ce{AgNO3, BaCl2, NH4OH}$, lead acetate and ammonium molybdate.
All of these would mostly be in small quantities since I had rinsed the sink with water before emptying the test-tube.

Off-topic, but important: why do you poor chemicals like these in the sink and not in the appropriate chemical waste container?!

We don't have an "appropriate chemical waste container". Though I hope all the sink outlets are connected to a separate tank.

@Michiel Safety and environmental concern is never off-topic.

@Martin That's why I still mentioned it, but with the 'disclaimer'

[...] diluted $\ce{H2SO4}$ and $\ce{HNO3}$ [...] $\ce{NH4OH}$ [...] and ammonium molybdate.

The yellow precipitate indicates that phosphate was present too. Under these conditions a yellow ammonium molybdatophosphate is formed:
$\ce{H2PO4^- + 22 H+ + 12 MoO4^2- +3 NH4+ -> \underset{yellow}{(NH4)3[P(Mo3O10)4]} +12H2O}$

$\ce{PbMoO4}$ has solubility 0.00001161 g/ 100 g and is yellow. I believe, that with diluting acidity of the solution fell down, leading to significant deprotonation of $\ce{H2MoO4}$ and subsequent sedimentation of $\ce{PbMoO4}$.

Wulfenite is a lead molybdate mineral with the formula PbMoO4. It can be most often found as thin tabular crystals with a bright orange-red to yellow-orange color, sometimes brown, although the color can be highly variable. In its yellow form it is sometimes called "yellow lead ore".

Permeakra already wrote about that, years ago...


text:
What substance is used for staining brass?
While working at CNC engraving, I had at my disposal a small bottle of liquid used to stain brass for creating signs etc - engrave a writing, stain the engraving, then grind/polish the flat topmost layer to remove the colored layer and leave it only in engraved depressions.
The bottle didn't reveal much about composition of the liquid, beyond some warnings about toxicity. It was a brass staining liquid, period.
The coloration it created was dark brown, that seemed black by comparison with the surrounding metal. The reaction took only a few seconds until full pitch of coloration was achieved and the liquid itself was transparent with very mild cyan-blue hue. The coloration was quite persistent; even roughly 10 years afterwards it's still good as new on the pieces I kept and required rather deep brushing to remove from the topmost surfaces.
Can you tell me what substance it was?
(below a sample with dark parts achieved through that liquid)

Edit: I found a very similar liquid on sale, but the page is equally devoid of composition details. The instructions say: [the liquid] is to be used in room temperature. It allows the finish in tones between black and light brown. The time of reaction is almost immediate. The color and speed can be controlled through dilution in water up to 50%.

Could you also tell us on which metals this is used?

@Martin: The label reads "for Brass and Copper".

This site regarding patina forming solutions might help:  http://www.sciencecompany.com/-W160.aspx#1

Due to link rot, it makes your answer better if you can make some general statements or summary comments, even if the page linked to is a full and detailed answer itself.

It doesn't seem to be any of these. It was a cold process on cold metal, and the solution wasn't light-sensitive (plain transparent flask), didn't have any sediment and was kept in liquid form with years of shelf life, so the only cold "to brown" process on the list doesn't match - the rest require the solution to be hot, or produce other colors.

Sulfurated potash (which is on that list under brown to black) does come in a gel form, it could be that.

It was definitely not gel- it behaved completely like water, and it had pretty much indefinite or at least very long shelf life. Also, the color doesn't seem to match (I've updated the question with a photo of such a liquid.)

The colour could be a colouring agent rather than the natural colour of the active ingredient.


text:
Is sublimation caused by an environmental difference or an element's composition?
I was having a conversation when the topic of sublimation came up. Prior to the discussion, I had always thought that sublimation was caused by a great pressure/temperature difference, but another person stated that whether or not an element sublimates depends on its composition.
Will any element sublimate under the right conditions? Or are there elements which will always pass through a liquid state?

To a first approximation, any pure substance can sublimate. However, complex substances may decompose or change their structure before any appropriate sublimation conditions are reached.
The only (and very remarkable) exception is helium (in fact, helium is so special that helium-3 and helium-4 have quite different phase diagrams even though they're isotopes!). Note that due to the presence of superfluid phases, there is no direct transition between the solid and gas phases in either isotope.
As an aside, there is nothing prohibiting a substance to exhibit a solid-gas transition curve crossing a region of mild temperature and pressure conditions. Though unusual, an interesting example is iodine heptafluoride. It has a very narrow liquid range at ambient pressures (meaning a small decrease in pressure is enough to go below the triple point pressure), and the liquid range happens to fall near very reasonable temperatures.

Depending on the field of study sublimation is either always possible, or dependent upon the substance.
In Organic chemistry, sublimation only occurs if the phase diagram (wikipedia, phase diagram) would allow it. If a phase change from solid towards gas is possible in the phase diagram, sublimation is possible when changing temperature and pressure in the right way. When no transition can be made between the two, sublimation is not possible. It all depends upon thermodynamics.
In the field of inorganic chemistry however, sublimation-like techniques are almost alway possible. If I'm not mistaken, with low pressure and high temperature, sputtering (wikipedia, sputtering) can be used to deposit single elements in a film.
I am an organic chemist myself, so the details of the sputtering I don't particularly know, sorry.
Hopefully this does answer your question.


text:
Lone pair electrons on secondary carbon vs primary carbon
I know that a molecule is more stable when a free radical is on a secondary carbon over a primary carbon, but how about when there is a lone pair of elections that can be placed on a secondary or primary? Is there any difference in the stability of the molecule or does that only apply to free radicals?

Welcome to Chemistry.SX! Please consider reading our [about] page if you haven't done so already in order to get acquainted with the customs of this site.

Could you clarify what you mean by having a lone pair? Are you interested in carbene stability, carbanion, or both?

I am not sure if you are referring to carbanions or carbenes (I suspect the latter), because both contain a carbon with a lone pair, so I will give explanations on both in the following.
Carbanions are destabilized by the exact same effect that stabilizes carbocations and radicals: hyperconjugation. Donation of electron density from an adjacent $\sigma$ bond into the empty/half-filled $p_z$ orbital stabilizes the cation/radical, but not a carbanion center with its negative charge in the fully-filled $sp^3$ orbital lobe. Therefore, a primary carbanion is more stable than a secondary one.
The inverse is true for carbenes: Secondary carbenes are way more stable than primary ones. This is because carbenes are per se a very reactive species, and because of this property they are often used in organic synthesis as intermediates which are generated in situ. To obtain a reasonably stable carbene, the reactivity of the lone pair needs to be somewhat reduced. Two substituents at the carbene carbon are needed which stabilize the carbene either by steric hindrance (i.e. making the lone pair difficult to reach for other reactants) or mesomeric effects (for example, lone pair donors which alleviate electron lack of the "sextet" carbon, delocalization of the carbene lone pair into an aromatic system). The Arduengo carbene (named after its discoverer) is a good example for this. The stability of this compound is such that it can be isolated as a crystalline solid at room temperature. The carbene lone pair can be delocalized into the five-membered ring which then contains 6 $\pi$ electrons (2 double bonds + lone pair of uncharged nitrogen) and is aromatic. Bulky $\mathrm{R}^1$ groups like adamantyl further contribute to stability by shielding the lone pair from potential reactants (source).

Arduengo carbenes are not the only persistent (especially stable) carbenes, and the understanding of their stability is still a subject of ongoing research, because structural elements (bulky substituents, aromaticity) which had been regarded as necessary for stability were often disproved as such by the synthesis of carbenes which did not contain them. The compounds do not need to be cyclic, as shown by bis(diisopropylamino) carbene, and even heteroatoms next to the carbene center do not need to be present, like in the case of bis(diisopropylamino)cyclopropenylidene.

Yes, there is a difference.  You should look into "carbene" stability: http://en.wikipedia.org/wiki/Carbene
A secondary carbene will be more stable.
Some other interesting carbenes that demonstrate carbene stability; called "persistent carbenes", http://en.wikipedia.org/wiki/Persistent_carbene

This answer would be better if you summarized some of the content from those Wikipedia articles so that we don't have to go somewhere else to learn why secondary carbenes are more stable than primary carbenes.


text:
References to draw 3D molecules with directionality of non-bonding electron pairs and p-orbitals
My instructor has been drawing 3D molecules that show the directionality of non-bonding electron pairs and p-orbitals. I've been trying to find references online that show this process, but I'm having a hard time finding anything useful.
Can someone please explain the process or link to a source that does? I've included two pictures of molecules that show what I'm looking for.

Welcome to Chemistry.SX! I'll take the opportunity to point you to our [about] which contains some useful information. Do I understand you correctly: You want to know why the p orbitals and free electron pair orbitals are drawn the way they are in your instructor's Lewis structure pictures?

Thank you! I'm curious as to how these structures are drawn. For example, given some completely "random" molecule, how can I know how to draw the orbitals (and fill the electrons) the way they're presented here?

In order to give a helpful answer it might be good to know a little bit about your chemistry backgroud. Are you still in High School or at university (undergrad, grad student)? Do you know how to draw Lewis structures? Do you know what orbitals are (do you know basic quantum mechanics)? Do you know about conjugated bonds and such things?

I'm an undegrad student. I know how to draw Lewis structures with nonbonding electrons, but I am specifically confused about the p orbits surrounding the carbon/oxygen atoms. Why does the bottom picture have two p orbitals (does this represent a triple bond?).

Regarding the lower picture, the right half does show a C-C-triple bond.  What is their hybridisation?  What does it imply on the geometrical arrangement between the atoms?  Later, the horizontal dash between the carbon atoms, this symbolizes the sigma bond of rotational symmetry along the C-C direction.  Black "lobes" perpendicular are for one set of pi-bonds, the curved line between them may remember you about their nickname, "banana bond". The blue lobes again perpendicular to these is the second set of pi-bond.  On the left side, the hybridisation clearly differs.

As it is already pointed out in the comments, what you are learning for is hybridisation and some basic MO theory. Do not go too far into the MO direction, because you will get into somewhere completely different place, and you will hard time to make the connections.
So aftre you reviewed hybridization: Once you can identify in a molecule the sp1, sp2 and sp3 carbons, you should be able to orient the hybrid orbitals. For example in the case of sp2 carbon of the lower image, it is in the HHCC plane. In case of sp1 there are two, for sp2 one un-hybridized p orbitals. These unhybridized orbitals should be perpendicular to the hybridized ones: perpendicular to the axis of sp1 or the plane of sp2.


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Why does ammonia act as an acid in only the first of these two reactions?
$$\ce{2Na(l) +2NH3 (g) ->2NaNH2(s) + H2(g)}$$
$$\ce{CH3Cl (alc) +NH3 (alc) -> CH3NH2 (alc) +HCl(alc)}$$
Doesn't ammonia donate a proton in both cases?

The first reaction is a redox reaction where you can't give $\ce{NH3}$ the role of a proton donor in the same sense as an acid is a proton donor. In the second reaction, you have to look at the reaction mechanism: it's a $\mathrm{S}_{\mathrm{N}}2$ reaction. Here, the $\ce{NH3}$ does not donate a proton directly but it first substitutes the $\ce{Cl}$ on methylchloride, thus forming an (much more acidic) ammonium salt, and then the ammonium salt donates the proton acting as an acid.

@Philipp Thanks for the comment. Unfortunately, I don't understand how it forms an ammonium salt. Doesn't ammonium have to be $\ce{NH4+}$? Where does the extra proton come from? Also, why is it that in a redox reaction $\ce{NH3}$ cannot act as a Bronsted-Lowry proton donor?

Ammonium doesn't mean $\ce{NH4^{+}}$ in general. There can be any organic group attached to $\ce{N}$ in an ammonium salt. In this case you have $\ce{CH3NH3^{+}}$, which is formed when $\ce{NH3}$ kicks out the $\ce{Cl^{-}}$ group in the first step of the $\mathrm{S}_{\mathrm{N}}2$ reaction thus leading to a positively charged ammonium cation. As for your second question: In the first reaction $\ce{NH3}$ doesn't dissociate like a normal acid, i.e. there is no reaction like $\ce{NH3 <=> NH2^{-} + H^{+}}$, instead the $\ce{H}$ that is lost ends up with the oxidation number $0$, i.e. not a proton.

@Philipp: That's probably enough to form a decent proper answer rather than a comment.

Yes. I agree. Thanks @Philipp! It sure answered my question.

@Aesin Ok, I added some arguments involving $\mathrm{p}K_{\mathrm{a}}$ values and hope that the answer is coherent.

Ammonia has a $\mathrm{p}K_{\mathrm{a}}$ value (a description what that means can be found in this answer of mine) of 33 and is thus not acidic at all - you'd need a $\mathrm{pH}$ value around 30 to deprotonate $\ce{NH3}$ to any appreciable amount! This means that $\ce{NH3}$ very rarely acts as a proton donor itself.
So, what happens in your first reaction is the following: It is widely known that when you dissolve $\ce{Na}$ in liquid ammonia you get solvated electrons (in the form of electride complexes). But electrons have of course very strong reducing power, so after a while they attack $\ce{NH3}$ abstracting one of its $\ce{H}$ atoms under the liberation of hydrogen gas. The important point here is that there is no dissociation of $\ce{NH3}$ here, i.e. no reaction $\ce{NH3 <=> NH2^{-} + H^{+}}$, so $\ce{NH3}$ does not act as a proton donor in the same sense a Bronsted acid would.
Your second reaction is an $\mathrm{S}_{\mathrm{N}}2$ reaction where $\ce{NH3}$ acts as a nucleophile. So, it doesn't act as a proton donor in this reaction neither. What it does is that it replaces (substitutes) the $\ce{Cl}$ in $\ce{CH3Cl}$, thus forming a methylammonium salt, $\ce{CH3NH3^{+}Cl^{-}}$.

Now, the $\mathrm{p}K_{\mathrm{a}}$ of an alkylammonium cation lies somewhere around 10 (the $\mathrm{p}K_{\mathrm{a}}$ of $\ce{NH4^{+}}$ is 9.24). So, it is much more acidic than $\ce{NH3}$ and also more acidic than the surrounding alcohol (I assume that "(alc)" means that the reaction is done in alcohol), which has a $\mathrm{p}K_{\mathrm{a}}$ between 15 and 17, and thus it (the ammonium salt) acts as a proton donor.


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References for learning about acid and bases in organic chemistry
Can someone help me find references to help solve these problems. These are problems suggested by my instructor.
I'm specifically interested in learning how to solve question A.(b). How can I know whether a reaction will occur between the two products and further, how can I predict what the product will be if a reaction does occur (what atoms bond together, and how can I know which atoms bond together in any organic reaction)?

Concerning your first question: I found that the  book "Organic Chemistry" by Clayden, Warren, Wothers and Greeves and here especially the 8th chapter contains a very good and readable introduction to acidity and basicity. After you have worked through this chapter you should be able to explain why certain molecules are as basic/acidic as they are and you see a lot of typical $\mathrm{p}K_{\mathrm{a}}$ values for a lot of substance classes so that you get some knowledge about what $\mathrm{p}K_{\mathrm{a}}$ values you can expect for other substances. As for your second question: If you want to know why reactions occur in the way they do you'll have to take courses in Organic Chemistry. Apart from that, I wholeheartedly recommend the aforementioned book again. If you work through this book (this will take you a while: it has 1200 pages... but then again, that is 1200 very readable pages :)) you'll have a very solid basis in organic reaction mechanisms and organic chemistry in general. If you want to dig a little deeper this book by Bruckner is very thorough but a little dry.
One particular perspective that I found very helpful in explaining why reactions happen in the way they do was the frontier orbital theory. A very good and readable book about its use in organic chemistry is this one by Fleming. The book by Clayden et al. also uses this theory a lot in explaining reaction mechanisms.

+1, I echo your sentiments re: the Clayden, et al. and (especially) Fleming books. The latter is really excellent in dealing with an intrinsically abstruse subject. On the introductory level, I think Klein's Organic Chemistry is also fairly good, though a bit handwavy w/r/t certain complexities, and with disappointingly meager coverage of MO theory. At a higher level, I thought Carey & Sundberg's two-volume Advanced Organic Chemistry was highly instructive and lucid.


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Gaussian vs ABINIT for solids
I see that Gaussian has feature to set periodic boundary conditions specifying the parameter Tv in the input file. Does it do it via the plane wave basis set? Also, the question is whether it is efficient to use Gaussian to compute the relaxed lattice constant, band structure and DOS for solids or it is better to use programs like ABINIT.

Even if it is possible to calculate periodic structures in Gaussian, I would not suggest it.
Better option is to use codes devoted to calculation of solids, among others the ABINIT you mentioned, or e.g. Quantum Espresso.
Gaussian does not use plane wave basis set, will be slower and in general less powerful than codes devoted to solids.


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On nomenclature of certain compounds
I have some questions on the nomenclature-
(1) In $H_2O$ we have $H$ written first and $O$ is second but when it comes to Hydroxyl group $OH$ the $O$ comes first, Why?

(2)Why is Propanal called Propion-aldehyde and not Propane- aldehyde?

The answer to your first question is given here: http://chemistry.stackexchange.com/questions/974/why-is-methanes-molecular-formula-conventionally-ch4-while-water-is-h20-a.

To answer (2) first, propionaldehyde is the common (Trivial) name and hence is derived and held-onto only due to historical reasons and now obsolete nomenclature reasons. The 3-C aliphatic saturated qroup was historically called Propion- and hence propanoic acid corresponds to propionic acid and propanal corresponds to propionaldehyde, and propanoate(ester) corresponds to propionate in trivial nomenclature. It is a consequence of the history of the original naming of the 3-C aliphatic group at the time of its discovery.
As for (1), it is a consequence of pure convention; just as water is not called hydrogen hydroxide!

Note that because "propion-" was first, propanoic acid did not become propionic acid. It was the other way around. The "prop-" in "propion-" is where our IUPAC "prop-" prefix comes from.

@BenNorris That is true. I should've written "corresponds to" instead of "became". Corrected in the answer.


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Clarifications on thermodynamic cell efficiency
The thermodynamic efficiency of any cell (especially Fuel cells) is given as $$\frac{\Delta G}{\Delta H} \times 100$$
I understood this partly, that since $\Delta G$ is the useful work obtained in the ideal case, the efficiency expression must have it in the numerator. But I fail to see the intuition (or maths) behind the $\Delta H$ term in the denominator. I have two questions regarding this:

On applying this equation to ordinary (not fuel cells), for a general reaction $$\ce{A^+ {(aq)} + B {(s)} -> A {(s)} + B^+ {(aq)}}$$ the $\Delta G$ term is given as $-nF E_{\text{cell}}$ where $E_{\text{cell}}$ depends on the concentration of the two ionic species as given by the Nernst equation $\left(E=E^\circ-\frac{RT}{nF}\ln\frac{[\ce B^+]}{[\ce A^+]}\right)$. On the other hand, is the $\Delta H$ term independent of the concentration? and if not, how would it (and consequently efficiency) depend on the ionic concentrations?
For fuel cells, suppose the cell reaction is such that the $\Delta S$ is positive and since $\Delta G=\Delta H -T\Delta S$, therefore $|\Delta G|>|\Delta H|$ since the second term in the equation for $\Delta G$ becomes negative due to positive entropy change. Would this predict an efficiency greater than 100%?

${\Delta G}$ and ${\Delta H}$ are normally expressed on a per mole basis.  Your assertion that ${\Delta G}$ is related to concentration is not precise.  It is related to the ratio of concentrations, and as such the incongruity with ${\Delta H}$ is not troubling.
If you found a fuel cell with a gain in entropy, according to your equation you would have more than 100% efficiency.  I seriously doubt such a situation exists.

I think - correct me if I'm wrong - the first point is slightly incorrect. If the given equation involved n moles, then both ΔG and ΔH would have scaled n times, since they are extensive properties. Though, their ratio would again scale back to 1. So, the result is correct (efficiency doesn't depend on the concentrations), just that the derivation seems a bit incorrect


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How to properly work with potassium HMDS
What is the proper procedure for working with KHMDS?  I would like to use it to deprotonate a compound and have a bottle of KHMDS in toluene.  Does it need to be handled under inert atmosphere?  And is it soluble in halogenated solvents?

Did you look at the MSDS or any other literature pertaining to KHMDS or would you rather have someone else give you the answer without you doing the work?

SDS's are nice but they tend to give safety instructions appropriate for tanker rail car volumes of material. (i.e. why do I need a respirator, face shield, apron, and gloves for everclear and table salt salt?)

If you would look up the properties of KHMDS from a chemical supplier, you will find it will react violently with with water. So do keep it under inert atmosphere.
As for solubility, it is a non-nucleophilic base, so probable no substitution and addition like reactions will occur. I wont suspect it will be a problem there.
A comment on the use though. If you plan to use this, explore the dangers and possibilities of a substance like this a bit further than asking it here. For safety considerations. Look up the MSDS sheets from your supplier and be sure to only use this in a properly equipped lab.
Hope this helped.

I checked the msds right after posting the question and read up on it and was able to carry the reaction out successfully.  Thanks for your response.


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How to calculate the probability of finding an electron in a box between 0.25L and 0.75L?
If the wavefunction of the electron in a box of length $L$ is $$\psi{(x)}=\sqrt{\frac{2}{L}}\cdot\sin{\frac{n\pi x}{L}}$$
What would be the probability of finding the electron between $0.25L$ and $0.75L$?.
From postulate 1 of quantum mechanics I know that $$1=\int_{0}^{L}{\psi(x) \psi^{*}(x)\,\mathrm{d}x}.$$
Trying out a similar reasoning leads me to think that the required probability is the integral $$ \int_{0.25L}^{0.75L}{\psi(x) \psi^{*}(x)\,\mathrm{d}x}$$ which gives the answer as $0.5$. But the book gives the answer as $0.82$.
What is the problem with my method and can anyone explain me why the answer is $0.82$?

Have you tried to calculate the integral by finding the Antiderivative of $\psi(x) \psi^{*} (x)$ and then substituting the integration limits in?

You have the reasoning correct, so I suspect you made a mistake in your integral calculation. The probably of finding the particle between $a$ and $b$, for the $n$-th state, is indeed given by:
$$P_n(a,b) = \int_a^b \left|\psi_n(x)\right|^2\, \mathrm{d}x $$
but if I do the calculation for the ground state ($n = 1$), I find the same results as your textbook:
$$ P_1\left(\frac{L}{4},\frac{3L}{4}\right) = \frac{1}{2} + \frac{1}{\pi} \approx 0.818 $$

Here is a more complete solution

Welcome to chemistry.se! If you have questions about how to beautify your posts, have a look at the [help]. Do you want to know more about this site, please take the [tour]. While this answer might be correct, it would certainly profit from some more explanation. Also it would be better if it was typeset with the tools available on this site.


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Why was anhydrous magnesium carbonate “upsalite” discovered so late?
What complicated the synthesis of upsalite (anhydrous $\ce{MgCO3}$), for it to be synthesized and published in 2013 as opposed to the 20th century, as one would expect for such a simple salt?

Did you try looking up the paper and maybe reading the introduction?

@Nope I didn't realize it was published in PLOS. Nice to see that it was. My interest was piqued by a popular-science magazine.

The paper describing the synthesis of Upsalite has a clear introduction explaining the problem:

Magnesium is the eighth most abundant element in the earth’s crust and essential to most living species. It can form several structures of hydrated carbonates such as nesquehonite ($\ce{MgCO3·3H2O}$), and lansfordite ($\ce{MgCO3·5H2O}$), a number of basic carbonates such as hydromagnesite ($\ce{4 MgCO3·Mg(OH)2·4 H2O}$), and dypingite ($\ce{4 MgCO3·Mg(OH)2·5 H2O}$), as well as the anhydrous and rarely encountered magnesite (MgCO3) [5]. In contrast to other alkali earth metal carbonates, chemists have found anhydrous magnesium carbonate difficult to produce, particularly at low temperatures. Above 100°C, magnesite (crystalline $\ce{MgCO3}$) can be obtained from $\ce{Mg(HCO3)2}$ solutions by precipitation. However, at lower temperatures, hydrated magnesium carbonates tend to form, giving rise to what has been referred to as “the magnesite problem” [6].Yet, not only chemists have been intrigued by magnesium carbonates. Although abundant in nature, where crystalline forms exist as traces in most geological structures, pure magnesium carbonate is seldom found on its own in larger deposits, a fact that has puzzled geologist for more than a century [7].

As you can see, anhydrous crystalline magnesium carbonate has been synthesized for a long time (and occurs in nature), in the form of a dense phase: magnesite. The new material reported under the name of upsalite is simply an porous amorphous polymorph of magnesite. New polymorphs of inorganic structures are regularly discovered/synthesized by researchers, and reported in the literature. The most striking example may be the family of zeolites, with more than 200 polymorphs of aluminosilicate.

Thanks to F'x for this fine answer and to Lo Sauer for the question.  Cool paper that I hadn't seen before!


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Oxidation state of group 2 elements
I have came across the following statement : "the oxidation state $+II$ is stable for the elements of the alkalin earth metals" , the same is true for the elements of the first group , i've looked for a possible explanation and heard that it has a relation with the charge densities of the elements , if that's true what is that relation ? otherwise , what could explain the oxidation states for each of the groups in the periodic table ?

Can you clarify "the same is true for elements of the first group"? If you mean "the oxidation state +2 is stable for the elements of the alkali metals" then the statement is not correct.

@bobthechemist no , i mean that the oxidation states +1 and + 2 are stable for the alkali and the alkali earth metals respectively .

The 'preferred' oxidation states of s-block metals can be explained by looking at their electron configurations:

Note that for the alkali metals, each element has one more electron than a noble-gas configuration.  This electron is very weakly bound to the nucleus as evidenced by the ver low ionization energy:

It is much harder to remove the 2nd electron, since it is contained within a completely filled shell.  (Please ignore the 2nd IE for Fr, as it doesn't exist in my data source.)
Therefore, the alkali metals have a tendency to lose just one electron, forming the +1 oxidation state.
The same thought process can be applied to the alkaline earth metals.  Here are the first three ionization eneriges for each:

Hopefully, it is clear that in the case of the alkaline earth metals, it is the 3rd electron that is in the completely filled shell, and that electron is extremely difficult to remove.  Thus, the metals in this group tend to stay in the +2 oxidation state.


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does Sodium persulfate evaporate at room temperature
I use Sodium persulfate to etch a copper board to manufacture home made Printed Circuit Board (PCB), and I use the same etching tank on the below photo which has no cover on the top. I wonder if it is okay to keep it like that at the room temperature or would it evaporate?

Regardless of evaporation, it is an oxidizer and unhealthy so it is recommended to keep it in a closed container.
With that being said, you most likely have a solution of sodium persulfate in water.  Water will evaporate so again, yes, keep it closed.

Sodium persulphate solutions slowly decompose while stored. The decomposition generates gaseous oxygen. Make sure that there is a way for the gas to excape or the bottle might rupture.
The storage temperature should be low. The decomposition rate of persulphate solutions depends strongly on temperature. For a 10 % solution of sodium persulphate 10% of the active material will decompose in about 30...40 hours at 50C temperature. At 25 degrees centigrade the same amount of persulphate is lost in 30 days.
If you need to store a persulphate solution please keep the temperature very low. Storing in less than 8 degrees of Centigrade will probably keep the solution in working condition for at least half a year or more.
http://www.peroxychem.com/media/90826/AOD_Brochure_Persulfate.pdf
Storing PCB chemicals in the same fridge with food is not recommended.

Storing any chemicals in the same fridge as food is a stupid idea.


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Is it possible to save an insoluble snowflake?
In The Big Bang Theory Season 3 Episode 1 "The Electric Can Opener Fluctuation"
Leonard gives Penny a snowflake that will never melt. He said some type of chemical gas name when he try to explain, but I don't understand what he said exactly so I can't do any research about that. You can watch it on this YouTube video as well.
I used Google for some searches like doesn't melt snowflake and insoluble snowflake but results gave me nothing.
So, is this possible to save a snowflake preserved in glass with a chemical element, gas or something?

This link looks promising: http://www.instructables.com/id/Catch-a-snow-flake-and-keep-it-forever/

By the way Leonard used a 1 % solution of polyvinyl acetal resin.

@Philipp You have a reference or did you review your DVD collection?

@bobthechemist If you mean my statement concerning "polyvinyl acetal resin": I watched the scene from Big Bang Theory again and that's what Leonard says (at least it's what I understand). To back it up: here is another link which looks relatively trustworthy and where "polyvinyl acetal resin" is mentioned in the context of preserving snowflakes: http://www.its.caltech.edu/~atomic/snowcrystals/preserve/preserve.htm

@Philipp Yup, I was just curious.

With gas no absolutely. To preserve ice at room temperature using the pressure of a gas you should reach a pressure of one billion Pascals. Too dangerous for Penny safe...
In fact there are some imprecisions in the video Leonard cites two different methods to "preserve" snow flakes with an organic polymer. In fact these methods create a snow flakes replicas, they are called snow crystal fossils, because there is no more ice inside them. Both methods are describe here, have a look!
Formvar Replicas

Leonard: "it will last forever I preserve it in a 1% solution of
polyvinyl acetal resin"

This is the Schaefer and Day technique [1] the solution Leonard is talking about is made from 1% of polyvinyl acetal resin in ethylene dichloride.
Superglue Replicas

Leonard: "in fact is a pretty simple process, you see, Cyanoacrylate
are monomers which polymerize mmm... "

Cyanoacrylate are the main components of Super Glue  this is easy method propose by the chemist Tryggvi Emilsson that you can do at home.
You have to choose one of these methods, you can't do both of them as Leonard (it seems) did... Take in account that Leonard is a physicist he doesn't know much about this stuff...;-)
[1] V. J. Schaefer and J. A. Day, Peterson Field Guides: Atmosphere (Houghton Mifflin, 1981).


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Number of pi-electrons in an energy state
I am learning quantum chemistry and I am trying to understand the statement in a .pdf I'm reading which states
"The energy transition associated with the spectral transitions observed is:
$$\Delta E = E_{LUMO} - E_{HOMO} = \frac{(N+1)h^2}{8mL^2}$$
where $L$ is the length of the "box" (it's talking about particle in a box), $m$ is the mass of the electron, and $N$ is the number of $\pi$-electrons in the system. I know due to the solution of the Schrodinger equation that $$E = \frac{n^2 h^2}{8mL}$$
So, $$\Delta E = E_{HOMO} - E_{LUMO} = E_{k+1} - E_{k} = \frac{(2k+1)h^2}{8mL^2}$$
So, I want to figure out why $2k = N$?

Welcome to Chemistry.SE. Is there any possibility that you could link to the pdf in question?

Your calculation is indeed correct. However, because each electron in the $\pi$ system can have two spin states, there are two electrons per energy level of the “particle in a box” Hamiltonian. Hence, $N=2k$.

I've said that $N$ is the number of $\pi$-electrons. That's all my source says.

@SamuelReid sorry, I read too fast… answer amended


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Predicting electronic geometry, observable geometry, and hybridization for any atom in a molecule
Can someone please explain the intuition around the answers to this problem:

I'm finding it very confusing because it seems like you have to nitpick between two resonance structures to get the desired results. Do we just always look for the "lowest" possible VSEPR in any resonance structure? Why isn't the observable geometry for "X" linear instead of none?

I'm not sure if it matters, but this really isn't a homework problem

But it looks like a homework problem or a problem from a text book or study guide. Even if this is for self-study, our approach is still the same. We want to help you get to the answer and not give you answer. For more info, see: How do I ask homework questions on Chemistry Stack Exchange

There's something funky going on in this example, because resonance does not require translational motion of nuclei.  can you provide the reference to this VSEPR vs. observable geometry?

Do we just always look for the "lowest" possible VSEPR in any resonance structure?

Yes. Consider the two resonance contributors below for a N,N-dimethylacetamide, which has a functional group in common with your molecule. The rule of thumb I teach is: **if an atom appears to be $sp^2$-hybridized in one resonance contributor and $sp^3$ in another, then that atom is $sp^2$-hybridized always.

In the figure above, I have used the curved arrow notation to represent how the lone pair on the nitrogen atom is delocalized into the $\pi$-system. In order for this delocalization to occur, that lone pair needs to be in a $p$ orbital. In order for that nitrogen atom to have a $p$ orbital, it must be $sp^2$-hybridized (or possibly $sp$-hybridized if applicable). All three of the atoms (N, C, and O) in the delocalized system need a p orbital, and thus must be $sp^2$-hybridized.
Once you have hybridization, then you have limited your choices of geometry:

$sp^3 \implies$ tetrahedral geometry group: tetrahedral, trigonal pyramidal, bent, "none"
$sp^2 \implies$ trigonal planar group: trigonal planar, bent, "none"
$sp^{\ \ \  }  \implies$ linear group: linear, "none"

Why isn't the observable geometry for "X" linear instead of none?

You are confusing observable geometry at an atom with molecular geometry.
Observable geometries at an atom are a relationship describing the relative geometric positions of other atoms bonded to that central atom. The central atom is not counted.A linear arrangement means that the atom in question is bonded to two other atoms or groups and those two atoms or groups are $180^\circ$ apart. If there is only one atom or group bonded to the atom in question, we cannot define the geometric relationship between that atom or group and the other atoms or groups bonded to that atom because there are not other atoms or groups. Consider the following examples from the tetrahedral group. The last case is "none" or "undefined, because you need at least two things to have a relationship.

This is different than molecular geometry which deals with the shape of the entire molecule. Let's look at HF. The shape of the molecule is linear, because the two points define a line (segment). The geometry around F is "none" or undefined. We cannot describe the arrangement in space of the hydrogen atom relative to other atoms bonded to fluorine because there are none.

I've never seen this distinction made between "VSEPR geometry" and "observed geometry", and it does not make much sense to me. It would seem more correct that what they call "VSEPR geometry" includes the lone pairs, while their "observed geometry" is only the arrangement of atoms, ignoring the lone pairs. Thus, geometry around X is ill-defined ("none"), because it has only one neighbor (linear geometry require two neighbors, like the central C in O=C=O).


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How will citric acid be deprotonated by pyridine?

Why do we deprotonate that specific hydrogen as opposed to any other one? What information does the table on the right give us? It seems unrelated to citric acid - where does the fluorine come from?

Welcome to Chemistry.SE. This feels like a homework-style question, so I'll go ahead and give it the [tag:homework] tag. I'm also adding [tag:acid-base] tag, since this is what the question is about. The tags [tag:bond], [tag:hydrogen], and [tag:molecules] do not really match the question.

The table on the right looks like a standard group of acids shown to illustrate the effect of inductive withdrawing on carboxylic acid pKa. The first two structures are relevant to the question. The table may have been copied from somewhere else because it had both structures on it. Two of the pKa values in the table are every similar to two of the pKa values given in the question. The fluoroacetic acid is a fine distraction; you are correct in thinking that it is not involved.

As for figuring out which of citric acid's three carboxylic acids is deprotonated first...

Citric acid has three carboxylic acid groups of two different types. Can you see that? One of the acids is closer to the OH group than the other two. Which acid or acids look more like the first entry in the table (except jammed onto the citric acid structure)? Which acid or acids look more like the second entry on the table? Once you have made that distinction, you should be able to use the pKa data on the table to determine which of the three carboxylic acid groups in citric acid is the most acidic (goes with the 3.13 pKa).

I see what you mean. So the table on the right just provides us with information to get a close approximation to the most acidic group in the molecule?

@BobDylan That is correct

It will deprotonate the acid group in the middle as is more acid the base can easily take it´s proton away, rathe than having problems with the protons in the other acid groups.

Welcome to Chemistry.SE! Would you be willing to elaborate? What types of problems would pyridine have with the other protons?

It helps here if you explain the why of your answer rather than just stating it.


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Chelation and its relation to transition metal ligands
Can someone explain chelation and its relation to monodentate and tridentate ligands? I dont understand what they mean by chelate effect and chelating rings.

Would you be willing to provide some more information, for example some specific compounds that seem to be giving you trouble?

I'm giving the answer thinking that you are familiar with ligands and Denticity.
Polydentate ligands form two or more coordinate bond with the central metal atom. For eg:EDTA can form up-to 6 coordinate bond with the central atom. In these situations a single ligand encloses the metal ion like enclosing it in a cage. This is called chelation. Co-ordination compounds with chelated metal ions are more stable then those in which the metal ions are surrounded by many monodentate ligands.If you want more information read the Wikipedia article on this http://en.wikipedia.org/wiki/Chelation.


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Is the nomenclature "primary", "secondary", "tertiary", and "quaternary" carbon atoms limited to sp3 carbon atoms?
In organic chemistry we can classify carbon atoms as primary, secondary, tertiary, or quaternary based on the number of additional carbon atoms bonded to the carbon atom of interest. The definitions in the textbook I teach from (Klein, 1st Ed. Wiley, 2011), as well as in several others I have, and from from several locations around the web, are:

Primary $(1^\circ)$ carbon atom - bonded to one other carbon atom, e.g. $\ce{HO} {\bf \color{red}{\ce{C}}}\ce{H2CH3}$
Secondary $(2^\circ)$ carbon atom - bonded to two other carbon atoms, e.g. $\ce{HO} {\bf \color{red}{\ce{C}}}\ce{H(CH3)2}$
Tertiary $(3^\circ)$ carbon atom - bonded to three other carbon atoms, e.g. $\ce{HO} {\bf \color{red}{\ce{C}}}\ce{(CH3)3}$
Quaternary $(4^\circ)$ carbon atom - bonded to four other carbon atoms, e.g. ${\bf \color{red}{\ce{C}}}\ce{(CH3)4}$

Is this labeling system limited to $sp^3$-hybridized carbon atoms? I do not see anything in the definitions that exclude $sp^2$- or $sp$-hybridized carbon atoms. Nor have I seen or read anything that explicitly excludes these atoms. However, the images in textbooks and around the web all show and discuss only $sp^3$-hybridized carbon atoms in this context. Are the following valid?
The central atom in propene $\ce{CH3 -}{\bf\color{red}{\ce{C}}}\ce{H=CH2}$ is a $2^\circ$ carbon atom because it is bonded to two other carbon atoms.
The carbon atoms in acetylene $\ce{HC#CH}$ are $1^\circ$ carbon atoms because they each are bonded to one other carbon atom.
The ipso carbon in toluene $\ce{C6H5CH3}$ (the carbon atom where the $\ce{CH3}$ is attached) is a $3^\circ$ carbon atom because it is bonded to three other carbon atoms.

To best of my knowledge, the classification primary to quaternary is used for discussing reactivity of hydrocarbons, i.e. alkanes. There the number of substituents matters, as it governs carbocation stability, steric hindrance, etc...
In all other cases, the electronic structure of the molecule rules the reactivity. As you mentioned alkenes, alkynes or aromates, there it makes no sense to discuss such subtle effect as number of neighbouring carbons, when you have $\pi$-system at hand.
The IUPAC Gold Book entry for alkyl groups supports this distinction:

The groups $\ce{RCH2}$, $\ce{R2CH}\ (\ce{R} ≠ \ce{H})$, and $\ce{R3C}\ (\ce{R} ≠ \ce{H})$ are primary, secondary and tertiary alkyl groups, respectively.

Thanks for your answer. However, I was not asking whether they were useful labels for these systems. As you point out, they clearly are not helpful in relating structure to properties. I was wondering whether the extension of the definition is valid under the formalism of the nomenclature.

According to IUPAC Gold Book http://goldbook.iupac.org/A00228.html, I would say, that it is meant only for alkanes and alkyl groups.

the correct definition is
"a tertiary carbon atom has 3 bonds to ANY number of carbon atoms"
the carbon atoms in acetylene are tertiary, not primary (this came as a shock to me too the first time i heard it).
the reason for this is, you can further improve the classification by adding the hybridisation of the C atom, so in acetylene, the carbon atoms are both SP tertiary, while in isobutane, the central carbon atom is sp3 Tertiary.
if we would follow the flawed definition "a tertiary carbon is bound to 3 other carbon atoms" , then adding the hibridisation would be useless, as it would not alter the classification, which is why this definition is flawed.

Can you provide a source for your definition? It appears to contradict the IUPAC Gold Book.


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Why is aluminium carbonate unstable?
Can it be isolated at room temperature?  Or even at any temperature?

Aluminium carbonate reacts easily with water/water vapour to form aluminium hydroxide as you can see on the Wikipedia page.
$$\ce{Al2(CO3)3 + 3 H2O → 2 Al(OH)3 + 3 CO2}$$
So, if you want to store aluminium carbonate, you have to make sure it is free from moisture.

Why it is unstable:
Smaller and more charged metal cations (i.e. metal cations with high charge density) tend to distort the electron cloud of the carbonate ions (polarisation).
Polarisation eventually leads to abstraction of oxygen from the carbonate ion, making one of the $\ce{C-O}$ bond in the carbonate weak, which makes it very prone to attack by water and heat (similar to the decomposition of copper (II) carbonate when compared to alkali metal carbonates).
The reaction itself doesn't release a lot of energy; hence why it is rather unstable and not reactive in chemistry terms.

So if it can be synthesised from the reaction on the Wiki page, how is it isolated?  I've searched quite a bit, and can find almost nothing about it on the web, nor in any of my inorganic textbooks.  Lots of reagents are sensitive to water, but that doesn't make them unstable, just reactive.  Fluorine, for instance.  You wouldn't think of that as unstable, although it's exceptionally reactive.

@ChrisA I added some more information to my answer. I tend to miss the title of the question once I read the body :( For the production of Aluminium Carbonate, I can't say for sure, but that could be prepared without the presence of water. Both $\ce{Al2(SO4)3}$ and $\ce{Na2CO3}$ are anhydrous solids; the former is acidic and the latter basic, the reaction should occur without much difficulty even without water involved.

Aluminum carbonate, if existed, would decompose by the following reaction:
$$\ce{Al2(CO3)3 → Al2O3 + 3 CO2}$$
to $\ce{CO2}$ and $\ce{Al2O3}$, with a release of energy, even in the absence of water. Therefore, it is unstable, and probably could not be isolated under any conditions.

This answer seems to be the only one so far that directly answers the question of instability even in isolation as opposed to reactivity (which could be viewed as instability in the presence of other compounds like water).

Aluminium carbonate is a salt of a very weak acid and a weak base, so it tends to hydrolyze. Similar may be said for a much easier available aluminium sulfide: it reacts with water in the air, so it is a common tool for rebellious teenagers if they can get the materials.
Moreover, aluminium cation is extremely small. It means, that it forms very strong bonds with small anions (oxygen). Since carbonate is a rather big cation, it is not a best partner for aluminium, so it should release carbon dioxide very easily.
For a better understanding consider second group carbonates. Barium carbonate is very stable, calcium carbonate dissociates around $1000^oC$ and berrilum carbonate does not form. Aluminium is an even smaller cation, than berillium.

It does not exist in nature, but in theory you could make it in the lab by adding pressure.
From a traditional pencil and paper calculation of Gibbs energy I found the pressure needed for
Al2/3O (s) + CO2 (g) = Al2/3CO3 (s)
is about 10 GPa.
Or you could make an Al2(CO3)3 (aq) solution with concentration of $10^{-5}$ M.
Well, still not impossible. And there should be some special solvents and/or complex structures that easily allow this reaction (postulation).


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How can I get ChemSketch to get N and H atoms closer while still operating in ACS style and without adjusting atom sizes?
A person at wikipedia is getting on my case for creating this image because the N and H symbols are so separated even though this seems to be the default of ChemSketch (one of the softwares that's recommended for chem structure drawing in the Wikipedia manual of style) and I'd like to know how to make them happy and get the separation between the N and H shortened without changing the style and letter sizes.

You could explicitly draw in the N-H bond.


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Naming of a molecule
$\ce{H2C=C=O}$ what is the group of the  molecule? How to characterize it and how  is this dissimilar to ketone?
If we add $\ce{O2}$ with it then will we get $\ce{CO2}$?

en.wikipedia.org/wiki/Ketene

Your question is unclear - the body does not seem to go with the title. Do you want to know the name of this molecule (ketene)?  or what the functional group is (ketene) and how it is different from the ketone group (complex)? or what happens when you burn it (the same thing that happens when you burn all organic compounds)? Could you focus your question?

As, your question isn't much clear and it needs to be focused little more.
From current state of this question, I can provide you the answers for specific questions that I get to understand from its current state.

$\ce{H_2 \ \ C=C=O}$
What is the group of the  molecule?

Well this is known as ketene.

how to characterize it and how this is dissimilar to ketone?

Ketenes are generally very reactive, and participate in various cycloadditions. They will also undergo [2+2] cycloaddition reactions with electron-rich alkynes to form cyclobutenones, or carbonyl groups to form beta-lactones.

Ketones engage in many organic reactions. The most important reactions follow from the susceptibility of the carbonyl carbon toward nucleophilic addition and the tendency for the enolates to add to electrophiles.

See this and this

If we add $\ce{O2}$ with it then will we get $\ce{CO2}$?

No! I hope structure of ketene and wikipedia articles make it clear to you why. :)
Also see this.


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Why do pharmaceutical drugs have a expiration date?
Every now and then you reach into the bathroom cabinet to find that some pill or ointment you where planning to use has gone past its expiration date.
I am interested whether this expiration date has any value. In a short search I found this article from Harvard medical school stating that some test found that 90% of all drugs where still perfectly active 15(!) years after the expiration date. The article then continues to give some vague statements on possible reasons for the expiration date (including financial reasons).
What I am wondering is whether there would be any chemical reasons for drugs to become less effective or even become harmful over time, which could warrant the existence of an expiration date?
As a specific example, in case it is difficult to answer this question in general, let's take an ointment meant to treat muscle pains which use Benzydaminehydrochloride as an active component.

It's a good question, and I cannot answer your specific example but can give some generic reasons why pharmaceuticals have an expiration date:

Degradation of the compound/formulation over time. This obviously comes to mind, but there is more to it than meets the eye. For example, you gave the example of an ointment and cited the active ingredient: however, the product you buy (and use) is more than just the active molecule. The whole formulation has to be stable, not only the benzydamine. The ointment is likely to be a complex emulsion, designed to target many different physical and chemical properties, including chemical stability, thermal stability, but also viscosity (so it's easy to apply), color, etc. I suppose in that particular case, the emulsion might be less stable than the active ingredient taken alone.
Also to note: many pharmaceuticals are sensitive to air, in many cases because they are unstable in the long term in presence of oxygen, or moisture. Such products are sold enclosed in a plastic, metallic or composite packaging that protects them. There, the limited stability in time might simply be due to the limit of stability of the (cheap) packaging itself: e.g., after some time of possible light exposure, plastic will become porous and let moisture in.

Being conservative, because we're talking about health.

Being conservative, because the company selling the product doesn't want to be liable.

Economics: it takes money to make absolutely sure the drugs are safe for a longer period of time. Are customers willing to pay more? Most people don't store drugs for use years later (nor is it recommended), so they won't see the point in paying more for longer expiration.

Being cynical - to add to the economics - so that people will throw them out and buy more. The degradation point is important - if there is a drug that is supplied as a pure enantiomer, but in time undergoes racemization, the unwanted enantiomer could be toxic and/or have very serious side effects. On the other hand, some drugs can be kept for ages, but since the average consumer can't detect the degradation or non-degradation, there has to be an expiry date.

@Epicentre I am trying to think of some hypothetical mechanism by which an enantiomer of a pharmaceutical could be toxic, or have any side effects beyond that of the therapeutic enantiomer.  Can you provide any example?  Or even a hypothetical by which that might occur?

The classic example of a toxic enantiomer was thalidomide

@Epicentre can you provide a reference?  It's not clear that the bioactive enantiomers of thalidomide aren't also the toxic ones. In general one would expect that, to the extent a molecule has any isomer-specific bioactive properties, it would be the same enantiomer that produces both "therapy" and "toxicity."

@feetwet: Another example of isomer-specific toxicity/side-effects (although not one susceptible to racemization in storage, to my knowledge) would be levomethamphetamine/dextromethamphetamine; both enantiomers are TAAR1 agonists promoting norepinephrine release, but the dextro form has additional psychoactive and toxic effects (mediated by its causing the release of several other neurotransmitters as well) which render it less-clinically-useful than the levo form.

A good question, I'd say!
As for the reason drugs 'go bad', it is true, that if the molecules are stable, they could last for a long, long time and in some cases it is perfectly possible that they would not harm anyone.
On the other hand, molecules used as medication etc. are often highly functional. This means, they contains parts that perform a reaction, bind to proteins in your body or something like that. The reason the drugs work, is this functionality. Is is, however a very good possibility for a functional group to perform a reaction that is not desired. The reason for performing the wrong reaction, is that the functional groups of the molecules are not 100% selective for their use in the body. So if any other reaction goes on with the drug, it loses/changes functionality and could even become harmful. Another term used for this process is 'degrading'. Part of the active medicine in the drug has changed into something else.
With a wide range of drugs available, it is not possible to always know which degradation product will form. Therefore, it is also not possible to ensure the drugs work, and are safe, for years and years. For health and safety reasons then, a 'best before date' or expiration date is added. This date works both ways, warns you that the drugs might be harmful, and protects the manufacturer of the drugs against misuse and lawsuits.
Hope this helped!


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Why do people often capitalize element names?
According to IUPAC rules, names of chemical elements should not be capitalized. See Wikipedia’s take on the issue:

According to IUPAC, chemical elements are not proper nouns in English; consequently, the full name of an element is not routinely capitalized in English

However, I commonly see people (even in research articles) capitalize the element names, and I wonder why. Is there a historical trend of doing so? Did the rule change at some point? Are element names capitalized in other languages?

I often wonder about this myself. Some people also capitalize most compound names (including water). However, I'm not reading research articles; I'm reading lab reports.

@BenNorris it happens it articles, but it's not common (and it would mean the typesetter has done a bad job). It's more common on the internet at large (including many questions on this site)

Agreed. I would assume the same misunderstanding that leads to capitalization errors for elements and compounds on the internet leads to the same errors in students' work.

How strange! I worked and taught chemistry (Chemistry) for 20 years, 20 years ago, and have no idea if I capitalized sodium fluoride for example! When I asked students to name NaF was I expecting Sodium Fluoride? It is not in a sentence, so does that matter?

Perhaps they recently took biology, and were unconsciously influenced by the initial capitalization of scientific names in binomial nomenclature, e.g., Lepomis cyanellus (green sunfish).

These people are either writing in German, or weren't educated properly.

This is a Google ngram of the usage of the words: iodine, carbon, nitrogen, zinc and Iodine, Carbon, Nitrogen, Zinc (not at the beginning of the phrase).
There is no historical trend, at least not an usage of an archaic form (we can see that the maximum  usage of the capitalized element name is from 1900 to 1960 but normalized to the overall usage is steady).

We can make some hypothesis about why someone could choose to write capitalized element names:

If we must write element abbreviation capitalized (Zn, Cu, C, N) I  can think I have to write the element capitalized even if is not abbreviated.
Someone (me too) may think that if I write Zinc it is more unambiguous and means I'm talking about the concept of element "all atoms with the same number of protons in the atomic nucleus" and not about the metal zinc.  If you write report with some elemental techniques (XRF, AES) sometimes you feel the urge to specify to the reader (that maybe is not a chemist): "when I write that I've found zinc I mean that in the sample could be present $\ce{Zn}$ but also $\ce{ZnO}$, $\ce{Zn+}$, $\ce{Zn^{2+}}$".
The fact the atoms are identical and the definition of a proper noun is: "a noun that in its primary application refers to a unique entity" could induce to think that elements may fit the definition, and so gain the privilege to be capitalized.
In alchemy elements such water and fire were capitalized.

However up to my knowledge the form not capitalized is preferred even in the others languages, in Italian I'm sure.

I wrote an answer on English.SE about capitalization: http://english.stackexchange.com/questions/329217/in-english-are-words-like-english-monday-and-january-considered-common/329850#329850. Since any piece of "gold" has unique properties that identify it as gold and not silver, it isn't a proper noun, just like how any tree that bears apples is an apple tree, not an *Apple Tree.

The argument about unicity implying being a proper noun is specious. I have only one nose, but that doesn't make it my Nose.

@DavidRicherby I could argue that your nose is not an unique entity, something that has separate and distinct existence and objective or conceptual reality, but simply but simply a part of an unique entity yourself. But of course is only an hypothesis I am not saying that is the reason...

Which elements are Jupiter and Venus?


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Do lone pairs contribute towards the pi electron count of heterocycles?
I'm not able to understand how the following compounds are aromatic.
When should the lone pairs on heteroatoms be taken into consideration when counting the number of π electrons?

Only count the lone pairs/ pi-bonds/ groups which are participating in conjugation and ignore them in all other cases.
For example, in compound 2 (thiophene), there are two lone pairs on sulfur.

One lone pair (brown) is in a p-orbital, and hence participates in conjugation with the two π-bonds. The other lone pair (blue) is pointing outwards from the ring in an $\mathrm{sp^2}$ orbital. This lone pair is orthogonal, or perpendicular, to the π-system and hence cannot take part in conjugation. In total there are six π electrons, and the compound is therefore aromatic.
See also: Conjugated system on Wikipedia.

The Nitrogen atom is already an $\mathrm{sp^2}$ hybridized atom, so its lone pair does not take part in conjugation
The sulfur atom is $\mathrm{sp^3}$ hybridized, it has two lone pairs, so only one of its lone pair will take part in conjugation for it to be $\mathrm{sp^2}$ hybridized atom
$\ce{N}$ bonded to $\ce{H}$ is $\mathrm{sp^3}$ hybridized and has its lone pair taking part in conjugation, thus its lone pair counted as 2 pi electrons. the lone pair of the $\mathrm{sp^2}$ hybridized $\ce{N}$ atom is not pi electrons
Lone pair on $\ce{S}$: one pair is pi electrons, lone pair on $\ce{N}$: not pi electrons

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text:
Coarse-grained force field for ion–water interactions
I want to simulate, at a very coarse grained level, the interaction of water molecules with metal ions, specifically Calcium and Potassium. Is there a standard force field for this. I was told to use to the Martini potential but it doesn't seem to have metal ions...

I want a non atomistic (i.e. molecular) model of water and a compatible Ca and K model. I was thinking of using the Tip3p water model and trying to extend it for Ca and K....

TIP3P is not a coarse-grained model, as it has one point charge per atom… a coarse-grained force field has beads, i.e. each force center represents multiple atoms.

Forcefields for alkali ions in water have been developed for decades now, so the literature is rich with examples… and yet there still are significant developments (see this very new paper for only one such example).
If you want to have a forcefield for alkali ions that mix well with the TIP3P water model, you better look at what existing studies on such systems have done… this is an example, but you can find many others with searching through the literature.


text:
Extraction of Sn from SnO2
In the extraction of $\ce{Sn}$ from $\ce{SnO2}$ by carbon reduction method, why is  $\ce{CO}$ formed instead of $\ce{CO2}$ as per the following reaction?
$\ce{SnO2 + C -> Sn + CO}$

Above a $1000\ \mathrm K$, the $\Delta_\mathrm fG$ of $\ce {CO}$ from $\ce C$ is more negative than that of $\ce{CO2}$ formation from $\ce C$. Therefore, during smelting, when coke ($\ce C$) reacts with $\ce {SnO2}$, the formation of $\ce {CO}$ rather than $\ce {CO2}$ is thermodynamically preferable.
Below that temperature, the reduction with carbon is not possible since $\Delta_\mathrm fG(\ce{SnO2})\lt\Delta_\mathrm fG(\ce{CO2})\lt\Delta_\mathrm fG(\ce{CO})$. Therefore at all temperatures where this reduction is possible, thermodynamically carbon monoxide formation would be more favourable as compared to carbon dioxide formation. To see for yourself, the variation of $\Delta_\mathrm fG$ of various compounds, have a look at Ellingham Diagrams, a link to a comprehensive set of which is Given Here.
$\Delta_\mathrm fG$ is the Gibbs free energy of formation.

Nice answer! I wonder if we can talk of Boudard reaction for this case too....


text:
How to make artificial sea water
My 7th grader is embarking on a long-term science project and I'm on tap to "help." He is showing a strong interest into finding ways to get drinking water to the masses. His idea is that we need to find inexpensive and efficient desalination processes. So, we are exploring current desalination processes to see what makes it "expensive" and "inefficient."
However, we live in the Rocky Mountains and don't have access to proper sea water. We started to explore simply adding salt to tap water and researching how much. Then, we discovered "Artificial Sea Water" or "ASW" is a common creation in science. Because sea water contains more than just salt. It contains chemicals like fluoride, strontium, boron, bromide, inorganic carbon, potassium, calcium, sulfate, magnesium, chlorine and of course, sodium. I am not a chemist, so this is starting to look daunting.
So, here are my questions to counsel my son:

To experiment with desalination, is adding salt to tap water
sufficient or do we need to investigate finding official ASW?
How do we test the salinity of water? Do I really need to invest
hundreds of dollars on a salinity meter?

If you look at the (average) composition of seawater:

you will find that NaCl is the main solute of this complex solution. Its concentration is 10 times higher than that of the next constituents (magnesium and sulfate). So, it seems to me like a simple NaCl solution is a good approximation for your experiments.
Regarding testing salinity of water, I don't have a definite answer on that, I do have two suggestions to offer:

Measure salinity by electric conductance of the solution, e.g. by making a simple conductometer and testing it with solutions of known concentration of salt.
Measure salinity by the quantity of salt (solid residue) once all water is evaporated. You can speed the evaporation by heating up.

Thank you for your answer. Can you provide some guidance of building a "conductometer?" I have some familiarity with electronics. Am I measuring resistance (ohms)? Can I just use an ohmmeter? Also, It seems that temperature will affect density and thus "conductance"? Will I need to ensure that all measurements are at the same temperature?

@mawcsco you want to measure resistance between two electrodes in solution, with fixed geometry and (yes) fixed temperature. You can find DIY guides on the internet for such devices, e.g. here


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How to check for geometrical isomerism in cyclic compounds?
I have no problem with identifying whether a given molecule has geometrical isomerism if it's not cyclic. But cyclic compounds are confusing for me. For example, how to check whether the following molecule posses geometrical isomerism?

For a molecule to show geometrical isomerism different molecules should be attached to the double bonded carbons. So, I'm sure that geometrical isomerism (if it is present) is along the double bonded carbons. But how to check whether different groups are attached to it when the double bonded carbons are the part of a cycle?

I hope the statutes of limitation haven't run out. Structure 1 is a (Z)-alkene. Using Cahn-Ingold-Prelog (CIP) protocol, the rings are deconstructed as shown in diagram 2. The larger red arrow marks the higher priority carbon (C,C,C). The smaller red arrow points to the lower priority[carbon (C,H,H). For the blue arrows, sulfur takes priority over carbon. [Addendum: Strictly speaking, the two blue arrows should point to the methylene groups next to the blue sp2 carbon such that C(S,H,H) > C(C,H,H).] The two larger arrows establish a (Z)-configuration. See: What are the CIP rules for cyclic substituents?

Two ways to think about it:

Intuitively: Take one side of the double bond (say, the sulfur-containing one) and imagine rotating it by 180° (so what was up is now down, in your drawing). Is this molecule the same? If it is, then you don't have isomerism. If the procedure gives a new, different molecule, then you have isomerism.
More formal: use the Cahn–Ingold–Prelog rules to rank both sides of the two carbon atoms of the double bond. If priorities are different, then you have different isomers (and you can actually assign them a name.

In your particular example, the compound has another isomer, and the one you have drawn has the (E) configuration.

I just read the Wikipedia article on Cahn–Ingold–Prelog rules. Can you give me an example on how to apply those in a cyclic compound? They say making a tree like structure and introducing ghost atoms. Can you please show an example?

@RajathKrishnaR sorry, this is really undergrad-level chemistry, and you will find plenty of explanations in any textbook… the site's goal is to help with specific questions, not to write courseware!


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How to quickly cool a test tube without cracking it?
Is there a way to safely cool a heated test tube quickly without cracking it due to sudden thermal stress in the glass? One thing that I tried was blowing on it, cooling it in slightly lukewarm water followed by cold water but this method kind of involves a "hope for the best factor". Is there a better technique?

I searched a lot on how to prevent shattering of glass but everywhere it is advised to use other material than glass..!! See this : http://www.studyhelpfox.com/3107678/How-Can-You-Prevent-A-Test-Tube-In-Water-From-Shattering-From-Thermal-Shock%3F

Are you trying to cool the tube itself or can the contents be equivalently cooled as well?  I'd just dump a little bit of ice into your solution.  Probably the simplest way to cool the tube down.

Rather than working on your technique, check whether other test tubes are available.
Apparently, a glass-like material with a very low thermal expansion coefficent would be a good choice.
The glassblowing shop at the University of Delaware provides a nice overview of the possible materials and their properties, and there are definitely additional online resources, e.g. by manufacturers like Schott, Corning, etc.
Here's a summary (coefficients are given in $cm/cm/°C$):
soda lime glass $89 \times 10^{-7}$
Pyrex, Duran $32.5 \times 10^{-7}$
Vycor  $7.5 \times 10^{-7}$
fused quarz $5.5 \times 10^{-7}$
Please note that quarz is both pretty expensive and demands special care! Alkaline solutions and quarz are a bad conbination!

Regular test tube is not really a good choice for rapid cooling for two reasons:

it is often cheap glass,
it has a relatively thin wall, and large space for the sample inside.

This later is important, because if you want to cool down something really quick, it should be relatively small, and have good contact with the wall. I assume you want to cool something really quick or else the question has not much point.

Have you tried better quality tubes, e.g quarz, Pyrex?
Have you tried transfer the sample to a cooled place, instead of cooling the whole thing down? E.g. rapid freeze quench techniques are often based on squirting the sample into a large volume of cooler liquid or on a cooled surface.


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What are identical molecules?
I don't know what identical molecules are but these are by thoughts.
Since from the word identical the molecules should have the same number and kind of atoms attached in the same way. Since in stereo-chemistry the isomerism is due to spacial orientation all stereo-chemical isomers are identical.
Is my thinking correct? What factors determine whether two molecules are identical or not?

Identical compounds have the same chemical and physical properties (and other properties too).
Of the two types of stereoisomers, diastereomers (non-mirror image stereoisomers) are clearly nonidentical. Consider the isomers of tartaric acid. Meso-tartaric acid and (+)-tartaric acid have different melting points (165 $^\circ$C and 171 $^\circ$C, respectively). They also have different acidities.
The (+) and (-) isomers (enantiomers) are more challenging, but they are distinguishable by one physical property - the way they rotate plane-polarized light. If the two molecules were identical, they would not be isomers - we would not be able to distinguish them. If they are distinguishable, they are not identical.


text:
How to change CO₂ to a less polluting gas?
I want to change $\ce{CO2}$ coming out of the tailpipe of a car to some other compound that is considerably less harmful to the environment with a filter of some sort.
What are some ways that this is possible?
$\ce{CO2}$ to carbon and oxygen?

Actually, $\ce{CO2}$ is already very less "polluting" as compared to other gases in the exhaust like $\ce{NO_x,SO_x,CO}$ and unburnt hydrocarbons. To eliminate $\ce{CO2}$ while still using a combustion engine will not be practically feasible, since $\ce{CO2}$ is the final oxidation product for all "organic" fuels. You can shift to an alternative power source (Hybrid, Electric) to eliminate $\ce{CO2}$. Otherwise, an engine which emits only $\ce{CO2}$ and $\ce{H2O}$ is pretty non-polluting engine by practical standards.

@SatwikPasani Thanks for your edit! I have reverted the title formatting back due to this issue.  It's not a big problem, just wanted to let you know for future edits.

Since cars are inherently going to pollute would it then be a better idea to filter out all carbon particulate matter? even from non-diesel engined cars? Thank you both for the answers by the way.

@DerekSmith A pragmatic approach would be to eliminate automobiles altogether.

It makes no difference where you grab the $\ce{CO2}$.  Don't burden the car.
Plant mammoth sterile forests with gene-gineered high lignin wood for building.  Or, deep bury the trunks in the Sahara Desert of the Empty Quarter.  RuBisCO is the slowest, least efficient enzyme on Earth, the first carbon fixation step in photosynthesis.  Gene-gineer a faster, more efficient (e.g, not poisoned by oxygen) RuBisCO for the forests.
http://www.cea-jaea-collaboration.net/literature/cea/ICCAP%20Nice%202007%20NaCO2finalpaper.pdf
A reactor for your tailpipe.