Very likely, the title of your question already hits the nail.

Since mobile phone tend to be "smart", filled to the rim with computing power and data, users wish to protect their companions with a coating that repells water.  Often, the devices are not only coated on the front side, which would at least hamper accumulation of residues of gliding fingers, but all around, including the back.  As a result, you might get a cell phone that may drop into a bowl of water, and still remain functional.

Among the different technologies available (a short presentation by CNN, back 2013 is found here), there are polymer coatings, or a mixture of anastase / $\ce{TiO2}$ and PFTE / teflon (like the coating of your frying pan) particles with some perfluorinated liquid (example).  For some, you need a larger installation that appears like a scilab, for others, you may prepare it at home with a spray can (example).  Hence the results in terms of performance, maintenance, and longevity may vary, too.  As a result, if applied both on front and back of the phone, not only water, but the components of "normal glue" (referring to what backs a velcro tape) may be repelled more or less efficiently, too.

Your observation that the coating appears little to non-affected both by water as well as iso-propanol (which for some coatings is a no-go, as it would tear the layer off), as well as for WD 40 (which may be summerised as a blend of apolar hydrocarbons (ref)) allows the speculation that the coating might be either polymer-based, or the triple mix of anastase + teflon + perfluorinated binder.  To discern between the two, remotely, is not possible.

If a mechanical, pocket-shaped mount like

(source)

or a pouch, to be bound around your upper arm while riding a bicycle, like

(source)

is not what you may deploy (or alter a little to fit your needs), and if the coating consists of the "polymer-only" variant, then maybe it is possible to establish a connection between the back plate of the phone, and either the velcro tape or something already connected to the velcro tape with the help of cyan-acrylate based gorrilla super glue.  These glues often bind very strong and to polymers, however partially dissole the polymer a little.  This chemical approach however comes with many "ifs" and uncertain outcome.


Temperature dependence in absorption spectroscopy

I am aware of one effect: doppler broadening. When the temperature increases, the speed distribution spreads out. That spreads out the distribution of the velocity component in the direction of your light source. The light the particles see from the light source then get red- or blue-shifted relative to what actually comes out of the source. So the particle population ends up absorbing a broader range of wavelengths than it did at a lower temperature. The effect is  $\Delta\lambda/\lambda_0 \propto \sqrt{T/m}$


What is true about this reaction involving hydroxy carboxlic acid?

I think your guess for P is correct, we just don't know whether it is 2- or 3-hydroxy-butanoic acid.  Reaction of thionyl chloride with the hydroxyl group in the alkyl chain can proceed with retention or inversion of configuration depending on reaction conditions, but the point is, if P is optically active, then Q will be optically active too (so A seems correct).  Reaction of Q, an acid chloride, with ethyl amine will produce the corresponding ethyl amide, R.  Since amides are basic (they can form salts with acids), I think B and C are true too.  An optically inactive isomer of P would be 4-hydroxy-butanoic acid.  On heating it would dehydrate and form a stable lactone (S) that would not react with thionyl chloride, so I think D is true also.  I just wonder if we're missing something that would allow us to determine if P is the 2- or 3- isomer?

Update:  To be complete and accurate let's update.  Based on the comments below (particularly JA's) I was right about "B" (but for the wrong reason, I thought the amide would be basic enough, but it is not; B is true because, as JA points out, heating with excess amine will convert P into an amino amide (I've labeled it as X) which is basic).  I was wrong about C, R is not basic enough to form a salt.  I still don't know if P is 2- or 3-hydroxy-butanoic acid, but I've added a picture based on the 3-hydroxy isomer to try and summarize the chemistry.


Why is MP2 failing to calculate proper IR frequencies for [Co(CO)4]- and [Fe(CO)4]2-?

Some months later but here we go...

I agree with most of the comments that multiconfigurational effects should be present for this system, and maybe they are very strong. But I tend to think that this is not the cause of the huge difference with experimental values. The fact that B3LYP returns reasonable values should be taking into account. Surely it "could be" just a coincidence but let's face it, I is very unprovable that all the results are reasonable due to coincidence (it is very simple to check it out, just try some others well proved functionals).

My guess is that you found the correct nuclear coordinates for the minimum of the PES, but the SCF part of the calculation have serious problems and it converged to some unreasonable electronic "state". This would imply that the MP2 results are ridiculous.
As HF and B3LYP are very different in the SCF part, B3LYP converged correctly.

You could try different initial guesses (Hückel, PA, a previous converged calculation) and different methods (like RHF, OHF, UHF etc.). Other basis sets can also help (a little larger if possible). I would try to avoid effective core potentials! (I understand that this can be a problem for your teaching purposes). Convergence method is of course of main importance (consider quadratic or very damped convergence) . I found almost impossible that calculating inner electrons correlation helps in this case.

Of course that the refinements that you tried are useful, but I think that just to improve quantitatively the results.

I hope that the ideas above help you. I would like to know if you effectively solved your problem.

Regards


What is the liquid in Mathmos lava lamps?

The company is not very forthcoming with the ingredients in their FAQ; however, they do state in regards for disposal:

  Q. How should I dispose of my lava lamp bottle when it reaches the end of its life?

  A. Please make a hole in the metal cap with a sharp point. Then pour
  away the liquid. The bottle may then be placed in the glass
  recycling.

However, personally, I go by a general rule that if I do not know what substance is, I do not pour it down the sink.  The authors of the HowStuffWorks webpage How to Dispose of Lava Lamps state the actual ingredients are a trade secret, but the first patent for lava lamps states the ingredients as being:

  "solidified globule of mineral oil, paraffin and a dye as well as paraffin wax or petroleum jell, preferably Ondina 17 with a light paraffin, carbontetrachoride, a dye and the paraffin wax or petroleum jelly."

as for disposal, the author of that website made contact with a company, who advised that

  They told me that I could wrap the lava lamp in newspaper and throw it in the trash. I was told that it was non-toxic. They warned me not to pour the liquid down the sink because it contained wax.

The wax could make a mess of the pipes from your sink if it were to accumulate and solidify, and there is more than enough rubbish and pollutants in the water ways as it is.

Best to consult your local area's rules and regulations as to disposal.


Measuring distance in a conductive liquid

(Initial, preliminary answer, here.)
There are several possible reasons for your unstable results:

Possibility 1:  Wrong electrodes

Ag|AgCl pellets seem like an odd choice of electrodes for this application.  What are the reasons for using them as electrodes?  What is their surface area?  Their complex shape means the electrical current maybe isn't uniform between the two electrodes.  Also, the fact that you're drawing current through an electrode with an AgCl coating may mean the coating is building-up/breaking-down (which could mess up your measurements) ...  In your current application, it seems like you'd want to use two, big, parallel stainless steel plates (with a lot of corrosion protection!) as your electrodes.

Possibility 2:  Changing salinity

Depending on the local conditions, the conductivity of the seawater may be changing.  For example, if you are by a river delta, the tides may be washing fresh water between your plates.  Or, if the electrodes are at the water surface in really sunny location, day-time evaporation may increase salinity ...  If either of those are the case, install a second set of plates with a known, fixed electrode spacing to measure salinity, and use that to modify the results from the original set of electrodes.

Possibility 3:  Convection Currents

Is the seawater between the two plates moving?  If so, how fast?  Does the speed change over time?  All of these factors could contribute to unstable electrochemical readings, too ... If so, you may add mesh or baffles to create more-consistent flow conditions.

(Follow-up answer, here)

As far as I can tell, you're method sounds alright from an electrochemical perspective.  Cathodic protection is fine for minimizing corrosion.  Having "known distance sensors" will avoid problems with changing salinity, etc (as you mentioned).


Why is Ionization Energy definition restricted to gaseous atoms?

In fact, it is not restricted at all. You may compare the energies needed to remove one electron out of a solid; these are also meaningful, albeit in a different way, and known for a wide range of substances. But when you are talking about atoms, you want to measure atoms, and the only way to have an undisturbed lone atom is to put it in a gaseous state. And of course, the resulting ion should be in that state, too; after all, you can't make a solid out of positive ions alone, and if you want to add negative ions to counterbalance them, an ambiguity arises as to which ions should you use.


Determining enthalpy and internal energy of formation for potassium superoxide

Now, when we say that the temperature rose by $\pu{2.62K}$, this is the temperature of the calorimeter and the water together. We are assuming that they have been given time to reach thermal equilibrium (and generally water and copper calorimeters take relatively less time to reach equilibrium, it's pretty safe to neglect the time). Calorimeters can be filled with varied amounts of water, so the calorimeter's constant is for just the calorimeter, not the water in it.

Note that if we were to assume that the specific heat of potassium/oxide were not negligible, we would have to know the specific heats of the two, as well as the temperature at the start of the reaction and the temperature at which they want $\Delta H^o$ to be calculated at. "Standard state" doesn't fix temperature, though when you say $\Delta H^o$, you generally mean $\Delta H^o_{\pu{298K}}$. Using these, you can form a reaction pathand calculate the answer (I can elaborate on this if you want, it's pretty interesting)

With these assumptions, one gets the heat released to the order of 10 kJ, and the energy is to the order of 100 kJ/mol. Not too extravagant, remember that a joule is a rather tiny unit of energy. And formation enthalpies/energies are generally to the order of 100 or 1000 kJ.

Finding the enthalpy change now is pretty easy. Since $\Delta H=\Delta U+ \Delta(PV)=\Delta U+\Delta(nRT)$. Using this and looking at the overall reaction, it shouldn't be too hard :)


Generally, energy optimization is done as iterative process with following steps repeating

calculate forces of nuclei
move nuclei according forces known so final structure were less tense.

Most focus in science is focused on how first step is made, as choice of the second step does not affect final structure except pathological cases.

The main difference is usually tied to computational capabilities of 'silica' available. Complexity of quantum mechanical computations usually grows as $O(n^3)$, $O(n^4)$ or even $O(e^n)$ where n is number of atoms of same type in system. So, the bigger system is, the 'worse' approximation is used. Biological systems are usually so large, that any sort of quantum mechanical computation is not an option at all, so "molecular mechanics/dynamics" approach is used, with empiric formulas of forces based on relative positions of atoms


Getting the pressure using the equation of states - will the answer be the same?

In general no, the pressure will not be the same, although there will be intersection points where (by coincidence) they will be the same, and there will be regions of the phase diagram for some equations of state (EOS) and for some substances that might be close enough that it doesn't matter.

For example, the van der Waals EOS gives pressure values at high temperature and low pressure for noble gases that are close to what you would obtain with the ideal gas law.

Since it is typically easy to calculate pressure from an EOS given temperature and volume, you can try this yourself and see how much of an impact the choice of EOS makes for your particular system at a particular point in phase-space.


Experimenting with electrochromic prussian blue

You pose a series of questions. Let me attempt to answer them all:

  The instructions say to provide 40µA per cm2, where does that number
  come from?

This comes from an article of Itaya and coworkers where they studied the electrochemical synthesis of Prussian Blue. (http://pubs.acs.org/doi/abs/10.1021/ja00382a006) In fact; the website in your question refers to an article in the Journal of Chemical Education (http://pubs.acs.org/doi/abs/10.1021/ed075p881) which in turn refers to the JACS article above.

From my understanding, this number should be seen as an experimental condition. It is given so that the experiment is reproducible.

  The TIO material I'm using has a different resistance than the glass
  listed, will that change anything?

The material you are using has roughly twice the resistivity as compared to the one used in the experiment. In principle, this would not pose too much of a problem. You can tune your electric system using the variable resistor. Just make sure you stick to 40µA per cm2.

  All the pt wire I could find was very pricy so I though I would use a
  graphite electrode instead, but all I can find is carbon electrodes.
  Is that close enough?

This has already been answered by A.K. The carbon flat electrode should work fine. Though I am not sure whether you can use it repetitively (in multiple experiments).

As a boost of confidence. The procedure you are using is both reported in JACS as well as in J. Chem. Ed.; it should work. If you are conducting this experiment at your university or high school, perhaps someone can give you a copy of the articles to which the website refers. The experimental procedures in these articles can be really helpful in preparing your experiment. Best of luck!


How useful is the assumption that water doesn't auto-ionize in the following?

First, while it is useful to know approximate methods of solving aqueous solution equilibrium problems, you have to realize that in a professional context, these methods are not used other than for back-of-the-envelope calculations. There exist software that perform that kind of calculations without introducing approximations, and they are widely used. So, to answer the question in your title: such approximations are useful for simple calculations, and for those only.

Now, let’s try to evaluate the effect of neglecting auto-ionization of water on your problem.
You can write the full system of equations describing the aqueous equilibrium:

conservation of mass: $[\ce{HA}] + [\ce{A-}]=c_\ce{A}$, where $c_\ce{A}$ is the analytical concentration in chlorobenzoic acid (0.1 mol/L).
electroneutrality: $[\ce{H+}]=[\ce{A-}]+[\ce{HO-}]$
water autoionization equilibrium: $[\ce{H+}][\ce{HO-}]=K_\text{e}$
acid-base equilibrium: $\displaystyle K_\text{a}=\frac{[\ce{H+}][\ce{A-}]}{[\ce{HA}]}$

You can solve this system of equations, as you have four equations and four unknowns ($K_\text{a}$, $[\ce{A-}]$, $[\ce{HA}]$ and $[\ce{HO-}]$). The answer gives $\mathrm{p}K_\text{a} = 9.19$ (given the precision given in experimental measurements).

You don’t say what figure you obtain, but I can take the same system above and simulate “neglecting auto-ionization” by setting $K_\text{e}=0$. Doing this, I obtain $\mathrm{p}K_\text{a} = 9.18$, which means the error involved in your approximation is 0.01 on the $\mathrm{p}K_\text{a}$. So, this is indeed a good approximation.


Assigning E–Z configuration for alkenes with cyclic substituent groups

The compound is a tetra-substituted alkene with four different groups. Thus, it could be a (E)- or (Z)-ethene. To find the priority of the groups, you must follow CIP rules. The four groups are cyclobut-1-enyl, cyclobut-2-enyl, 2-methylcyclobutyl, and 1-methylcyclobutyl. Since it is an alkene (planner sp2–sp2 bond), we don't have to consider all four groups at once. What you have to do is consider two groups at once on each side of the double bond (say they are a and b):

On side a the first carbon of each cyclobutyl group (cyclobut-1-enyl and 1-methylcyclobutyl) has equal priority ({C,C,C} and {C,C,C}) as indicated in the image. However, the second carbon of cyclobut-1-enyl group has the priority {C,C,H}, while the second carbon of 1-methylcyclobutyl group has the lower priority {C,H,H} compared to that in cyclobut-1-enyl group. Therefore, between these two groups, cyclobut-1-enyl group has the higher priority.
On side b the first two carbons of each cyclobutyl group (cyclobut-2-enyl and 2-methylcyclobutyl) have equal priority: First carbons of both rings have {C,C,H} and the second carbons of both rings have {C,C,H} as well (see the image). Yet, the third carbon of cyclobut-2-enyl group has the priority {C,C,H} while the that of 2-methylcyclobutyl group has the lower priority {C,H,H} compared to cyclobut-1-enyl group. Therefore, between these two groups, the cyclobut-2-enyl group has a higher priority.
Now, you see higher priority groups of a and b sides are oriented on the same side of the double bond. Therefore, the compound is (Z)-isomer. The name is (Z)-1-(cyclobut-1-enyl)-2-(cyclobut-2-enyl)-1-(1-methylcyclobutyl)-2-(2-methylcyclobutyl)ethene.
Also, note that this molecule also has three chiral centers as indicated in the image. However, their stereochemistry is not given, so I neglect them.


Potassium sulfate as a byproduct of manganese heptoxide synthesis?

The reaction you mentioned shows the formation of manganese heptoxide and potassium bisulfate. But if you alter the stoichiometry of the reaction, i.e. change the moles of the reactant, you will obviously get different products like manganese(II or III)sulfate, manganese oxide, potasium sulfate etc. For this, the reaction conditions should be changed.

$$\ce{2KMnO4 + 2H2SO4 ⟶ Mn2O7 + H2O + 2KHSO4}$$

  Potassium permanganate react with sulfuric acid to produce
  manganese(VII) oxide, potassium bisulfate and water. Potassium
  permanganate - solid. Sulfuric acid - 98% solution. The reaction
  proceeds at room temperature. Impurities: manganese(IV) oxide
  $\ce{MnO2}$(source)

$$\ce{2KMnO4 + H2SO4 → Mn2O7 + K2SO4 + H2O}$$

  Potassium permanganate react with sulfuric acid to produce
  manganese(VII) oxide, potassium sulfate and water.(source)

$$\ce{4KMnO4 + 6H2SO4 → 4MnSO4 + 2K2SO4 + 5O2 + 6H2O}$$

  Potassium permanganate react with sulfuric acid to produce
  manganese(II) sulfate, potassium sulfate, oxygen and water. Sulfuric
  acid - 60% solution. Impurities: ozone $\ce{O3}$.(source)

$$\ce{2KMnO4 + 4H2SO4 ->[70-75°C] Mn2(SO4)3 + K2SO4 + 4H2O + 2O2}$$

  Potassium permanganate react with sulfuric acid to produce
  manganese(III) sulfate, potassium sulfate, water and oxygen. Sulfuric
  acid - 98% solution. This reaction takes place at a temperature of
  70-75°C.(source)


Do changing opt=modredundant to opt in Gaussian makes geometry optimization not to take into account frozen angles?


Does NaCl reduce the surface tension of water?

Gibbs adsorption isotherm tells us that (roughly speaking) the decrease in surface tension is proportional to the surface excess concentration of the solute.

Positive surface excess means that the solute tends to concentrate on the surface rather than stay in the solution. The surface concentration has no physical limit and can go arbitrarily high. In soap and other surfactants, it is hundreds of times higher than the concentration in the solution, therefore their effect on surface tension is negative and very strong.

Negative surface excess means that the compound prefers to stay in the solution rather than on the surface. It has a physical limit, though: the concentration can't be lower than 0. Therefore the effect of such compounds on surface tension is positive, but small (not noticeable with naked eye). This is the case with $\ce{NaCl}$ and pretty much all "normal" inorganic salts.


H1 NMR Spectra of inorganic complex, Cu4OCl6(2-mepy)4

What you have there is a typical example of an H NMR spectrum of a paramagnetic compound. The formal oxidation state of the Cu here is Cu(II) resulting in d9 configuration. This means that there will be an unpaired d electron making the Cu atoms paramagnetic.

The lone electron in the Cu atoms disturbs the magnetic field the Hs' feel resulting in extreme chemical shifts and also provide a very efficient relaxation mechanism resulting in extreme peak broadening (in some cases you cant even see the resonances at all). The highest paramagnetic shift I have encountered so far was around 250 ppm for an Fe(II) complex I made. In another occasion we had resonances at 1200 ppm!

As for the peak broadening I had peaks of ν1/2 of a few thousands of Hz. The integration wont always make much sense with paramagnetic compounds either.


Nucleophilicity with respect to the solvent

Actually nucleophilicity is inversely proportional to electronegativity when comparing compounds across the period, and inversely proportional to size when comparing in the same group. This condition is meant in aprotic solvent.

In protic solvents  the more electronegative ions gets easily solvated to a larger extent (i.e., like a ball with the ion in the centre and the water molecules around it) so the order across the period remains same since electronegativity is not effected but the order about the group changes since the now smaller ions become the larger ones.

For example, the trend for nucleophilicity is  $\ce{F- > Cl- > Br- > I-}$ in aprotic solvent but in protic solvent it is
$\ce{I- > Br- > Cl- > F-}$    ($\ce{F-}$ is now larger than others in size because it is solvated.)


Magnetic moments of tetrahedral Cobalt (II) and (III) complexes

The question isn't really clear exactly what complexes you are referring to.  However, cyanide is quite a high field ligand, so perhaps the Co(II) cyanide compound is low spin  while the 3.87 value is for high spin (3/2).  I don't see how it could be low spin if it is really tetrahedral, but isn't $\ce {Co(CN)_4^2-}$ square planar?

For Co(III) alkyl, alkyl ligands can also be relatively high field, and it is at least theoretically possible to have low spin (s=1) Co(III) tetrahedral complex. However, it is rare for tetrahedral complexes to be low spin because the ligand field splitting energy is lower than for octahedral.

If those are not the reasons, see Iron, Cobalt, and Nickel Complexes having Anomalous Magnetic Moments Quarterly Review Chemical Society volume 22, pp 457-498.


Reaction with tin(IV) iodide and hydrogen peroxide

Note: this is a preliminary description, which I think answers the question, but disagrees in one detail (see asterisk).

Tin(IV) iodide is completely hydrolyzed by water [1, p. 120] to colorless hexahydroxostannic acid:

$$\ce{SnI4 + 6 H2O <=> H2[Sn(OH)6] + 4 HI}$$

Hydrogen peroxide can not oxidize $\ce{Sn^4+}$ further (in fact, hexahydroxostannate(IV) is used as a stabiliser for hydrogen peroxide), and is likely going to oxidize hydroiodic acid to iodine (which attributes to the brown color) instead:

$$\ce{H2O2 + 2 HI -> I2 + 2 H2O}$$

Hexahydroxostannic acid only exists in solution and tends to form cloudy/milky gel over time [2, p. 1021]; this can be simplified as formation of colloidal particles of tin(IV) oxide [hydrates]:

$$\ce{H2[Sn(OH)6] <=> SnO2 · n H2O + $(4-n)$ H2O}$$

When petroleum ether is added, iodine is extracted to organic phase (on top due to lower density of the ether). Tin(IV) oxide sol remains in aqueous phase and attributes to its slightly pink color due to Mie scattering of colloidal solution.*

* Phase order doesn't match OP's description which suggests opposite coloring of the phases (additional clarification on the type of petroleum spirit or a photo of the binary phase mixture would help).

References

Inorganic Syntheses; Bailar, J. C., Ed.; Inorganic Syntheses; John Wiley & Sons, Inc.: Hoboken, NJ, USA, 1953; Vol. 4. https://doi.org/10.1002/9780470132357.
Holleman, A. F.; Wiberg, E.; Wiberg, N. Lehrbuch der anorganischen Chemie, 102nd ed.; Fischer, G., Ed.; Walter de Gruyter: Berlin New York, 2007. ISBN 978-3-11-017770-1. (in German)


How to explain the structure of HNO3 with a Molecular Orbital diagram?

You have 24 valence electrons. Counting the electrons: 6 of these will go into the $σ$ framework and there are 14 electrons in non-bonding, either $sp^3$ or $sp^2$ hybrids, on the oxygens. This leaves 4 electrons for the $π$ system.

We can think of the Nitrogen and Oxygen in the double bond being as both being $sp^2$ hybridised, with vacant $p$ orbitals orthogonal to the plane of the molecule. These 4 remaining electrons will then go into the $π$ system formed by the overlap of these 2 orbitals, which overlap to form 2 molecular orbitals, the $1π$ and the $2π$ MOs.

We expect the $1π$ to have greater contribution from the Nitrogen due to the lower orbital energy of the $p$ in N compared to O, and the $2π$ to have greater contribution from the O instead of the N.

A full MO calculation of the molecule would also take into account the non bonding $sp^3$ electrons on the oxygen, of which one lobe will be able to interact slightly with the $ p$ orbitals, giving a delocalised system over the Nitrogen double and single bond, exactly as we see drawing resonance structure for the molecule.


Comparing the bond strength between a polar bond and non-polar bond

In polar covalent bonds, say $A–X$, assume electronegativity of $X$ is greater than that of $A$ then their is polarisation of bond $A–X$, $X$ gains partial negative charge $\delta–$ and $A$ gains partial positive charge $\delta+$ but this charge is very less (as electrons of sigma bond are held tightly) to account for any significant increase in bond strength of $A–X$ bond but the $–X$ group in $A–X$ molecule tends to attract $A–$ group of another $A–X$ molecule in this way there is electrostatic interaction between two different $A–X$ molecules giving rise to intermolecular forces (Dipole-Dipole interaction, hydrogen bonding etc).

DO NOT CONFUSE
FORCES BETWEEN TWO DIFFERENT MOLECULES AND COVALENT BONDS WHICH EXIST IN A SINGLE MOLECULE. THEY ARE TWO COMPLETELY DIFFERENT THINGS

However, attraction or repulsion may exist between two parts of a single molecule (generally happens for organic  molecules) but doesn't significantly affects bond strength between them rather they tends to change orientation of those groups or may rotate about the sigma bond.

And for $ICl$ and $I_2$, bond of iodine and chlorine is weak due to polarisation. And bond between iodine and iodine is strong covalent bond. I hope you know that polarisation of electron cloud of sigma bond decrease overlapping between orbitals.


Why can't the strength of superacids be measured in water?

Any acid-base reaction is always an equilibrium:

$$\ce{HA^1 + (A^2)- <=> (A^1)- + HA2}\tag{1}$$

and for each pair of acids $\ce{HA^1}$ and $\ce{HA^2}$ you could calculate a $K_\mathrm{a}$ value to determine one acid’s strength with respect to the other. This $K_\mathrm{a}$ value is typically calculated according to equation $(2)$ if $\ce{(A^2)-}$ (which does not have to feature a negative charge; I just wanted to avoid different descriptions for the two acids) is the solvent.

$$K_\mathrm{a} = \frac{[\ce{HA^2}][\ce{(A^1)-}]}{[\ce{HA^1}]}\tag{2}$$

If $\ce{(A^2)-}$ is $\ce{H2O}$ (i.e. $\ce{HA^2}$ is $\ce{H3O+}$) and the acid $\ce{HA^1}$ is a strong acid or a superacid, then this equation is of little use; the equilibrium $(1)$ will be strongly shifted towards the product side, the concentration $[\ce{HA^1}]$ will be very small and thus $K_\mathrm{a}$ very large. With ‘very small’, I mean that $[\ce{HA^1}]\approx10^{-10}$ — acid $\ce{HA^1}$ is said to be fully dissociated.

Exact measurements of $K_\mathrm{a}$ become very hard if not impossible because you would need to determine extremely minute concentrations. The concentrations $[\ce{HA^2}]$ and $[\ce{(A^1)-}]$ are of little help since:

$$[\ce{HA^2}] = [\ce{(A^1)-}] \approx c_0(\ce{HA^1})\tag{3}$$

This is the levelling effect of water: no matter how strong the acid, there is no acidic species stronger than $\ce{H3O+}$ that can survive in aqueous solution for extended time. Most importantly, consider two very strong acids with different $K_\mathrm{a}$ — e.g. one with $\mathrm{p}K_\mathrm{a} = -15$, the other with $-10$. They have a very different acidity, yet the concentrations of all relevant species in aqueous solutions will be nigh identical.

Therefore, a new, more expandable way to measure and compare the strengths of very strong acids was required.


When writing the net ionic equation, if one of the products ionizes, what is the most appropriate way to account for this in the answer?

We consider mixing of $\ce{H2SO4(aq)}$ and $\ce{Ca(OH)2}$ in molar ratio 2:1.
Molecular equation version:
$$\ce{2 H2SO4 + Ca(OH)2 -> Ca(HSO4)2(aq) + 2 H2O}$$
Eventually if $[\ce{Ca^2+}][\ce{SO4^2-}] = [\ce{Ca^2+}] \cdot K_\mathrm{a2} \cdot \frac {[\ce{HSO4-}]}{[\ce{H3O+}]}> K_{\mathrm{sp,}\ce{ CaSO4}}$ ($\mathrm{p}K_\mathrm{a2}=1.99$):
$$\ce{Ca(HSO4)2(aq)  <=> CaSO4(s) + H2SO4(aq)}$$
Ionic equation version:
Separated solutions:
\begin{align}
  \ce{H2SO4(aq) + H2O &-> H3O+(aq) + HSO4-(aq)}\\
  \ce{HSO4-(aq) + H2O &<<=> H3O+(aq) + SO4^2-(aq)}\\
  \ce{Ca(OH)2(aq) &-> Ca^2+(aq) + 2 OH-(aq)}
\end{align}
More exactly:
\begin{align}
  \ce{Ca(OH)2(aq) &<=>> Ca(OH)+(aq) + OH-(aq)}\\
  \ce{Ca(OH)+(aq) <=>> Ca^2+(aq) + OH-(aq)}\\
\end{align}
Mixing solutions with 2:1 molar ratio:
$$\ce{2 H3O+(aq) + 2 HSO4-(aq) + Ca^2+(aq) + 2 OH-(aq)-> 4 H2O + 2 HSO4-(aq) + Ca^2+(aq) }$$
And after elimination of spectator ions:
$$\ce{2 H3O+(aq) +  2 OH-(aq)-> 4 H2O }$$
Then, further ionizations of the bisulphate, as $\ce{[H3O+]}$ decreased:
$$\ce{ HSO4-(aq) <=> H3O+(aq) + SO4^2-(aq)}$$
Again, eventually if $[\ce{Ca^2+}][\ce{SO4^2-}] = [\ce{Ca^2+}] \cdot K_\mathrm{a2} \cdot \frac {[\ce{HSO4-}]}{[\ce{H3O+}]}> K_{\mathrm{sp, }\ce{CaSO4}}$ ($\mathrm{p}K_\mathrm{a2}=1.99$):
$$\ce{Ca^2+(aq) + SO4^2-(aq)  <=> CaSO4(s)}$$


Separation of a solution of n-butanol and nitroethane

Use counter-current extraction.
I'm not sure about solvents, but let's say that I can separate nitromethane from water by distillation. Also that butanol can be separated from ether by distillation. (as I remember ether and water aren't miscible...)
So I line up 7 sepatory funnels.
Into sepatory funnel #4 I put my mixture and 50 ml of water and 50 ml of ether. Equilibrate then water phase goes to sepatory funnel #5 and ether phase goes to sepatory funnel #3. Sepatory funnel #4 is now empty.

In sepatory funnel #5 I add 50 ml ether, Equilibrate then water phase goes to sepatory funnel #6 and ether phase goes to sepatory funnel #4
In sepatory funnel #3 I add 50 ml water, Equilibrate, then water phase goes to sepatory funnel #4 and ether phase goes to sepatory funnel #2

So basically I have the water extracts going one way and the ether extracts the other way.


Decrease in molar conductivity with increase in concentration

The equation you quote is the empirical relationship discovered by Kohlrausch (1907) for the conductivity of strong electrolytes, and is valid for concentrations up to $\sqrt{c} \approx 0.3$. $\Lambda_m$ is the molar conductivity usually given with units $\mathrm{\Omega^{-1}\,cm^2\,mol^{-1}}$ ($\Omega$ = ohm).  The $\Lambda_m^0$ is the molar conductivity at infinite dilution which effectively means that of pure water.

The reason that the conductivity falls with concentration is due to inter-ionic forces on the ion mobilities. There are two related effects, the Electrophoretic and Relaxation Field effects.

Each ion is surrounded by a 'cloud' of oppositely charged ions and when a field is applied there is simultaneous movement of an ion in one direction and of the opposite movement of the ionic atmosphere; the electrophoretic effect. Both the central ion and its atmosphere drag solvent molecules along with them  and this results in a retardation of the movement of the central ion. The effect can be evaluated using Stoke's Law $v=f/6\pi\eta L$, where velocity is $v$ and force $f=zeE$ for an ion of charge $z$ in field $E$. The solution viscosity is $\eta$ and $L$ the Debye length which is the effective radius of the ionic atmosphere, typically a few nm in millimolar solution. The velocity is identified with the deceleration of the ion caused by the ionic atmosphere moving in the opposite direction.

The second effect is a relaxation effect. Suppose the central ion is randomly moved, the overall spherical symmetry that existed before any movement occurred will be broken, i.e. the central ion is asymmetrically located with respect to its atmosphere, and because of this a 'relaxation field' is set up. The symmetry is restored by the effect of the relaxation field which takes only ten to hundreds of nanoseconds.

When the ion is under the influence of the applied field the time average of the forces on the ion are therefore no longer zero. The external field driving the ions through solution is reduced by the continual effect of the relaxation field. This seems strange since this field relaxes so very rapidly compared to motion of the central ion through its atmosphere, $\approx 10^{-3}$ sec, however, its magnitude is huge $10^5$ V/cm compared to the applied field and even small changes in symmetry do have a measurable effect in acting against the external field.


Why is this a Lewis structure of ethyl acetate?

The two rules you cite are guidelines for the 'best' Lewis structure.  However, in the example provided, all electrons are accounted for and all atoms are connected making it a 'valid' Lewis structure.

In the context of resonance forms, this is likely a minor contributor.

However the electrophilic center in ethyl acetate is the carbon you marked as positive, this form has some descriptive  and predictive power in discussing reactions of ethyl acetate.  In general, for $\ce{R2C=O}$ molecules (esters, aldehydes, ketones, etc), The carbocation (sextet) carbon has some relevance.


Why do alkyl chlorides have lower boiling points than the corresponding alkyl bromides and iodides?

Your knowledge of the bond strengths is correct, but your understanding of boiling is flawed.

As orthocresol has mentioned in his comment, boiling is not about breaking the carbon-halogen bonds in your tetrahalomethanes.

Boiling does not mean decomposition!

You only transfer your material from the liquid to the gaseous state. Here, the molecules are still intact. But upon heating to the the boiling point, you have overcome the interactions between the molecules that existed before and held the whole bunch of molecules together as a lquid.

So, why is there the observed trend in the boiling points? Did you have a look at the molecular weights? ;)

What about the boiling points of alkyl halides, $\ce{R-Hal}$?

Let's have a look at the data for haloethanes, $\ce{C2H5-Hal}$:

$$\begin{array}{cr}
\mathrm{Hal} &\mathrm{bp~[°C]}\\
\hline
\ce{F} & -37 \\
\ce{Cl} & 13 \\
\ce{Br} & 38 \\
\ce{I} & 71 \\
\end{array}$$

Again, the trend is similar: Fluoroethane has the lowest, iodoethane has the highest boiling point.

Apparently, compounds with a higher molecular mass have a boiling point.

As a rule of thumb, the heavier the molecule, the more kinetical energy is needed to transfer it to the vapour phase. However, please note that this is a simplified model which might break when strong interactions between molecules, such as hydrogen bonds are involved!


Heisenberg uncertainty principle and lifetime of the excited state

For practical purposes the relevant aspect of the relation between the uncertainties in time and energy is that it is a Fourier relationship that explains how peak widths in a spectrum (with linewidths in frequency units) relate to the decay rates of time-dependent signals. A decay can be considered the statistical result of many independent events with an average probability of occurring described by a time constant $\tau$. It can further be shown via the Fourier transform that when a signal decays exponentially with time constant $\tau$, this leads to a Lorentzian peak shape with a width at half height proportional to $\tau^{-1}$. It is even simpler to show this in the case of a Gaussian shaped decay with linewidth ~$\sigma_t$. The resulting frequency peak is then Gaussian with linewidth ~$\sigma_t^{-1}$. When you multiply the linewidth parameters of the Gaussian functions you get:

$$\sigma_f\sigma_t=1$$

If you use Planck's relation $E=h\nu$ if follows that

$$\sigma_E\sigma_t=h$$

Now - as mentioned earlier - you can think of the width of the Gaussian or Lorentzian time-dependent decay as proportional (or described by) a time constant $\tau$ (which is the default description in the case of exponential decays). It follows that

$$\sigma_E~\tau \approx  h$$


Is there any difference between table salt and common salt?

They are both the same. If you look at this wiki page, it uses common salt and table salt as the same name. It's mainly just household terms, like people calling hydrochloric acid muriatic acid etc.


The important aspects of reagent purity are the level of purity with respect to compound(s) X, and the purity with respect to compound(s) X required for your particular use case.

For example, if you are conducting a reaction that requires anhydrous reagents but is unaffected by trace metals, then you would want to see a certificate of analysis or equivalent showing an acceptable limit for water in the reagent. Whereas if you require sub-ppm levels of trace metals, then that's the parameter of interest to you and you probably don't care if there is 0.5% water present.

In other words, a 99.5% pure ethanol product and a 99.8% ethanol product could be drastically different in their composition with respect to your needs; i.e. the apparently slightly more pure compound could have orders of magnitude more trace metals present than the slightly lower overall purity product. Furthermore, terminology regarding reagent purity is often similar between different vendors (i.e. pure, ultrapure, milipure, etc.), but frequently has very different implications on the aspect of purity that is important to you.

Summary, TL;DR:
Although you can frequently make generalizations about the meaning of "ACS Reagent" grade or "[pick-your-prefix]pure" grade, you need to inspect either the specifications on the bottle or certificate of analysis for each reagent to know for certainty that it meets your needs. The simplest reliable source for such information is the manufacturer, most of which have online resources providing this information for their products.


Why is the pure tungsten (W) melting point higher than that of its oxides (either WO3 or WO2)?

As far as I know,

$\ce{WO2}$ and $\ce{WO3}$ have a covalent characteristic, and thus lower melting point than the metal.

while, $\ce{Al2O3}$ have more ionic characteristic, and thus higher melting point than the metal.

$\ce{W}$ is a d-block metal from the $\ce{3^{rd}}$ row which make it possible to have covalent materials rather than ionic materials.

this article might prove this point
http://pubs.acs.org/doi/abs/10.1021/jp963724z?journalCode=jpcbfk


How can rotamers of alpha-keto esters be named to distinguish between different geometries?

The IUPAC recommends s-cis and s-trans for the rotamers of conjugated dienes.

The nomenclature derives from having a "cis-like" or "trans-like" geometry about a sigma bond.

I feel like something similar should be appropriate, however that same IUPAC link recommends E/Z or sp/ap (syn-periplanar/anti-periplanar) for the N-alkyl amides (5,6).

Amide 5 would be Z (but ap w/r/t alkyl groups) and amide 6 would be E (and sp). E/Z is less ambiguous in my opinion.

Ketoester 1 is E, and ketoester 2 is Z. I suppose that ketoester 1 is also ap and ketoester 2 is also sp.


It is a good question, it is a gemstone hiding in the mud.

I have searched the Cambridge database, all the bis-acetylacetonatonickel complexes which have four coordinate nickel centres are square planar. There are three common coordination geometries for nickel(II) which we need to consider.

Tetrahedral, square plannar and octahedral. Here is a diagram (not to scale) of the d oribital energies in the three types of complex.

Nickel(II) has eight d electrons, if we pack them into the octahedral or tetrahedral complex we will end up with two unpaired electrons. But if we stick the electrons into a square planar complex then if the difference in the energy between the two uppermost d orbitals in energy is sufficient then we will get a low spin $\mathrm d^8$ complex. To get spin-pairing, the energy difference between the orbitals must be greater than the energy required to spin pair two electrons.

There is a whole continuum of complexes between perfect octahedral and perfect square planar. If the distance between the metal and the two ligands on one axis is increased (tetragonal distortion of an octahedral complex) then the complex is part way between the two extremes. There is a good undergraduate lab experiment based on the early work of David Goodgame which is all about the nickel(II) complexes of N,N-diethylethylenediamine. I did this experiment as an undergraduate, I supervised students doing it as a PhD student and as an academic I have used it in a lab class. Trust me, it is good!

What the person asking the original question should have been thinking about is the question of square planar vs octahedral (or tetragonally distorted octahedral).

If I was to take an acac-type ligand with bulky groups in place of the methyls such as tert-butyls then I am sure that I would have an isolated nickel. This nickel could well be tetrahedral. But in fact it is square planar (F.A.Cotton, J.J.Wise, Inorganic Chemistry, 1966, 5, 1200, CCDC code DPIMNI)

If we look at a nickel(II) complex of an acac where the methyl groups have been replaced with phenyl groups, the closest oxygens to the nickel other than the four we expect are $\pu{4.4 Å}$ from the nickel. They are a long way in terms of angle away from the $z$ axis of the nickel. I think it is safe to say that they are not binding to the nickel. Here is a picture of the nickel complex. (D.V.Soldatov, A.T.Henegouwen, G.D.Enright, C.I.Ratcliffe, J.A.Ripmeester, Inorganic Chemistry, 2001, 40, 1626.)

Just in case anyone is not sure it is very much a square planar complex.

But if you look at the crystallographic literature you will see G.J.Bullen, R.Mason, P.Pauling, Inorganic Chemistry, 1965, 4, 456. Here is the answer: what happens is that the nickel acac complex forms a trimer and the bridging oxygens make all three nickels have an octahedral environment. The CCDC code is ACACNI01 for the best determination which was M.B.Hursthouse, M.A.Laffey, P.T.Moore, D.B.New, P.R.Raithby, P.Thornton, Dalton, 1982, 307.

Also look out for hydrated complexes such as AQACNI (R.L.Harlow, C.E.Pfluger, J. Coord. Chem., 1973, 2, 289).

By the way one of the lectures I had as an undergrad mentioned this problem of waters adding to the coordination sphere or oligomers forming, I think it was in a lecture by David Goodgame.

While I have never seen a 4-coordinate nickel bis-acac complex which is anything but square planar, if you look at the coordination chemistry of a related phosphorus(V) ligand (dithioimidodiphosphinate) you can see both tetrahedral and square planar nickels.

If we look at YUSFIB which is for the tetraphenyl ligand (P.Bhattacharyya, J.Novosad, J.Phillips, A.M.Z.Slawin, D.J.Williams, J.D.Woollins, Dalton, 1995, 1607) we can see that the nickel is tetrahedral. Also in the tetraisopropyl ligands complex (D.Cupertino, R.Keyte, A.M.Z.Slawin, D.J.Williams, J.D.Woollins, Inorganic Chemistry, 1996, 35, 2695) it is tetrahedral as can be seen in the following two pictures.

But the THF solvate of the dppaS2 complex of nickel is square plannar. (E.Simon-Manso, M.Valderrama, D.Boys, Inorganic Chemistry, 2001, 40, 3647.)


Number of collisions of particles in a box (or a sphere)

If you want a number of collisions in a given time $\Delta t$, then simply from a units perspective, the other answer doesn't make sense, as it produces a number with dimensions of length. You want something that is dimensionless.

Often, physical chemistry textbooks will provide a derivation for collisions per unit volume per unit time. The proofs are a little bit involved and I'd rather refer you to a textbook instead of typing it all out. However, the bottom line is that the collision frequency is given by

$$Z_{AB} = \pi(r_A + r_B)^2\left[\langle v_A \rangle^2 + \langle v_B \rangle^2 \right]^{1/2} \left(\frac{N_A}{V}\frac{N_B}{V}\right)$$

$Z_{AB}$ is the number of collisions between A and B per unit time per unit volume
$r_A$ and $r_B$ are the radii of A and B
$\langle v_A \rangle^2$ and $\langle v_B \rangle^2$ are the mean square velocities of A and B
$N_A$ and $N_B$ are the number of particles of A and B
$V$ is the total volume

For a derivation of this I suggest reading Levine's Physical Chemistry, 6th ed, section 14.7.

For your specific case you've assumed that $r_B = 0$, $v_A = 0$, $v_B = s$. Therefore this simplifies to

$$Z_{AB} = \pi r_A^2 \cdot s \cdot \left(\frac{N_A}{V}\frac{N_B}{V}\right)$$

Obviously, the radius of A, $r_A$ is related to its volume $V_A$ by $V_A = 4\pi r_A^3 /3$. To obtain the number of collisions in a given time you just need to multiply by the total volume and total time.

$$f =  \pi r_A^2 \cdot s\Delta t \cdot \left(\frac{N_A N_B}{V}\right)$$

As you can see, the form of this equation is not all too different from the one in the other answer. The key difference is that the distance travelled $s\Delta t$ must be multiplied by a cross-sectional collision area, equal to $\pi r_A^2$, and not by a volume. The textbooks will explain this well, I will try to elaborate on this later if I have time.


Overlapping oxygens in crystal structure -- what's going on?

It's a bit hard to answer without having coordinates file (*.res or *.xyz), but the simplest answer that I would expect is that these oxygen atoms belong to two different molecules that are disordered over the same position (i.e. they have only partial occupation on the site that in sum is 1). It's very common for f.e. thf coordinated to alkali-metals.

If you have shelx input for the structure look for the line that defines one of oxygens and see if 6th parameter is 11.0

O1    5   -0.180348    0.245831    0.298215    11.00000    0.02208    0.02526 =
                                               ^^

This means fixed (10) occupancy of 1.

If you have cif-file look for the atom loop

loop_
 _atom_site_label
 _atom_site_type_symbol
 _atom_site_fract_x
 _atom_site_fract_y
 _atom_site_fract_z
 _atom_site_U_iso_or_equiv
 _atom_site_adp_type
 _atom_site_occupancy
 _atom_site_symmetry_multiplicity
 _atom_site_calc_flag
 _atom_site_refinement_flags
 _atom_site_disorder_assembly
 _atom_site_disorder_group
O1 O -0.18035(6) 0.24583(6) 0.29822(5) 0.02794(8) Uani 1 1 d . . .

8th parameter is occupancy (_atom_site_occupancy).

These both examples were for not disordered atoms. If they are disordered then it looks like this:

shelx-ins/res

O1    4    0.297097   -0.198418   -0.064164    10.50000    0.34896    0.34935 =
                                                  ^ occup. 1/2

or

FVAR       0.22777   0.51340   0.51266
...
O1    4    0.551971    0.536866    0.295322    21.00000    0.03100    0.03031 =
                                                ^ occup. bound to 2nd free variable
                                                  here 51%

cif-file:

O1_20 O 0.6673(9) 0.3140(7) 0.1724(3) 0.0434(14) Uani 0.513(3) 1 d PDU C 1
                                                      ^


This all looks correct.  It's worth noting that the voltage generated by this cell occurs through the flow of electrons from half cell A to half cell B as the system reaches equilibrium.  It's also worth noting that the final concentrations of ions should eventually become nearly identical given enough time and free interaction between the two half cells (at which point the voltage becomes 0).

For further reading, this site has some good examples:
http://chemwiki.ucdavis.edu/Core/Analytical_Chemistry/Electrochemistry/Voltaic_Cells/Electrochemical_Cells_under_Nonstandard_Conditions/Concentration_Cell

While wikipedia has a (brief) overview of the theory:
https://en.wikipedia.org/wiki/Concentration_cell


Why does dimethyl ether have a bond angle bigger than 109.5 degrees?

In case of $\ce{Me2O}$, the positive deviation of the bond angle can be explained on the basis of steric crowding of two methyl groups.

There are other contradiction too like this:

In case of $\ce{(SiH3)2O}$, the huge bond angle increase is mainly due to back bonding and partly due to steric repulsion.


Before getting to the azeotropes, it's worth reminding ourselves of the relevant assumption that we make about ideal mixtures: that the intermolecular interactions have constant energy regardless of the mixture.

Given the important role that intermolecular interactions have in determining the boiling point of a pure liquid, it's surprising that so many mixtures behave approximately ideally, in line with this assumption. There are two main reasons for this.

First, liquids are only miscible if they have somewhat similar types of intermolecular interactions with similar magnitudes of energy. For example, water and hexane have very different intermolecular interactions, so surrounding water with hexane or hexane with water is quite unfavorable, and the two do not mix. Thus, any actual mixture has interactions between the two types of molecules that are not hugely different from the interactions within the pure liquids.

Second, substances that are liquid are room temperature generally do not have especially strong intermolecular forces, so the small differences in the intermolecular interactions of two miscible liquids is usually dwarfed by the impact of the difference in molecular weight that is a major determinant of boiling point. Water is quite unusual in having such a low molecular weight and still being a liquid at room temperature, a symptom of the exceptionally strong role of intermolecular interactions in determining its boiling point.

With that in mind, we can talk about azeotropes. To get a minimum boiling azeotrope, the deviation from Raoult's Law requires that the vapor pressure of the lower boiling component be reduced from what is predicted for ideal behavior and that the vapor pressure of the higher boiling component be increased. In order for that to happen, we need two liquids with substantially different intermolecular interactions whose mixing changes those interactions in opposite directions.

In the example of the water:ethanol azeotrope, the water is able to increase the amount of hydrogen bonding between ethanol molecules by increasing the density of hydrogen bond donors and acceptors per unit volume. At a low water concentration, the water also does not interfere greatly with the hydrophobic interactions between the ethyl ends of the molecules. As a result, the average intermolecular interaction for the ethanol molecules is stronger than in pure ethanol, and we have the necessary decrease in the vapor pressure of ethanol. For the water, when it is at low concentration in mixture, each water molecule is surrounded by ethanol molecules, so it has fewer hydrogen bonding interactions than in pure water. As a result, its average intermolecular interactions are weaker than in pure water, and we have the necessary increase in vapor pressure. Many low molecular weight alcohols form azeotropes with water for the same reasons.

When the water concentration gets below what is required for the azeotrope, enough of the alcohol molecules interact only with other alcohol molecules that the deviation from ideality decreases.

For the maximum-boiling azeotropes, the requirement is that the vapor pressure of both components is decreased. This is quite common in acid:water mixtures where the degree of ionization of the acid in the mixture is higher than when it is in pure form, which means that there are more ionic interactions, which are quite strong. If the water concentration gets below what is required for the azeotrope, the amount of ionization decreases and the deviation from ideality decreases with it. When the water concentration is higher than in the azeotrope, there may be a great deal of ionization of the acid, but there is enough water that is not interacting with the acid (and conjugate base) molecules that the deviation from ideality is less.


Esterification and dehydration synthesis

The reverse of esterification is called "hydrolysis."  The hydrolysis reaction can be performed under acidic or basic conditions.  This link shows the mechanisms involved for both the acidic and basic hydrolysis of an ester.  Each step in the reaction is explained in detail and accompanied by a picture.


Why is potassium ferrocyanide considered safe for consumption, when it is just one reaction away from the highly toxic potassium cyanide?

Under biological conditions it is almost impossible to release HCN

Free cyanide can be released from potassium ferrocyanide by heating or by strongly acidic conditions (and some heat). Neither of these conditions is possible in a living organism.

The reason why this is so is because the cyanide ligands are very strongly bound to the iron (in slightly more technical terms the dissociation constant is very, very small (ca 10-24)). Most of the mechanisms that generate cyanide require much more violent chemical attack: waiting for the cyanide to dissociate and then protonating it to give deadly HCN will never result in a notable amount of NCH. This means that, in biological systems, ferrocyanide is effectively inert.

This explains why the LD50 in mice, for example, is about 6g/kg which is a lot ([see the safety info here]2 for both ferri- and ferrocyanides).

It is also worth noting that food safety authorities have studied the compounds and agree that:

  The Panel concluded that ferrocyanides (E 535–538) are of no safety concern in these current authorised use and use levels.

So, while it might look like you are "one reaction away from cyanide" in practical terms this is not remotely true.


What are the preferred IUPAC names of these two compounds?

We can only speculate about the various automatic algorithms that are available to generate names for chemical structures. Anyway, all known programs have limitations or shortcomings. In particular, the algorithms may not include all rules or they may use obsolete rules. Note that the IUPAC recommendations have changed several times and that the current (2013) recommendations have more than 1500 pages.

In my experience and also in this case, ACD/ChemSketch is more reliable than MarvinSketch. However, such programs are typically updated frequently, and some issues might be solved in a new version.

The compounds that are given in the question are not trivial cases. In order to find the preferred IUPAC names, we have to dig a bit deeper and consider the rules given in Section P-45 of the current recommendations. These rules are applied if two or more names may result according to the basic nomenclature rules, because of different substitution patterns or multiple occurrences of the same senior parent structure. Apparently, these rules are taken into account in the present version of ACD/ChemSketch but not in the present version of MarvinSketch.

Thus, ACD/ChemSketch correctly gives the preferred IUPAC name for the first compound as 4-(2-bromobutyl)-6-methyloctan-3-ol.

The name ‘6-bromo-4-(2-methylbutyl)octan-3-ol’, which was proposed by MarvinSketch, is not wrong, it is just not the preferred name. Remarkably, this name was in accordance with obsolete IUPAC recommendations from 1979 (retained in 1993). Therefore, some homework exercises might still expect this answer. However, this does not necessarily mean that MarvinSketch is using old IUPAC recommendations from 1979 or 1993; it could be a mere coincidence as the second example below shows.

ACD/ChemSketch also correctly gives the preferred IUPAC name for the second compound as 4-(2-bromobutyl)-6,6-dimethyloctan-3-ol.

The name ‘6-bromo-4-(2,2-dimethylbutyl)octan-3-ol’, which was proposed by MarvinSketch, is not in accordance with IUPAC recommendations (neither 1979 nor 1993 nor 2013). Apparently, the relevant (past or present) tie-breaking rules for such cases are not included at all in the present version of MarvinSketch.


You are correct in your ideas. The effect is important in the van-der-waals interaction and appears as an additional  $1/r^7$ term compared to the usual $1/r^6$ term. (This is sometimes called a retardation effect). It is present only at larger distances (a few nm) and arises only in induced-dipole interactions (dispersion interaction) because of the time it takes for the interaction to propagate between molecules. During this finite time (governed by the speed of light in the medium) the first dipole's direction has changed during the time it takes to cover distance $2r$ so that the inducing and induced dipoles become less correlated the farther the two dipoles are apart. The equation describing this effect is called the Casimir-Polder eqn. $W(r)=-23\alpha_0^3hc/(8\pi^2(4\pi\epsilon_0)^2r^7)$ where $\alpha_0 $ is the polarisability and $\epsilon_0$ the permittivity of free space.


Why does pKa of a acid-base indicator equal to the pH when the equivalence point is reached?

Look at the following equation:

  $$\mathrm{pH} = \mathrm pK_{\mathrm a} + \log\frac{[\ce{A^-}]}{[\ce{HA}]}$$

At the half equivalence point, say we have 10 moles of $\ce{WA},$ and so there will be 5 moles of $\ce{SB}$ as the name suggests ("half equivalence"). So we know that the 10 moles will be neutralized to 5 moles of $\ce{WA}$ by the $\ce{SB}.$ If we do this then the $\ce{CB}$ is also now equal to the $\ce{WA}$ since it will increase by 5 moles.

$$\ce{[WA]~ =~ [CB]}.$$

Thus,

\begin{align}\mathrm{pH} &= \mathrm pK_{\mathrm a} + \log 1\\  &=\mathrm pK_{\mathrm a}\;.\end{align}

So this is why the the $\mathrm{pH} = \mathrm pK_{\mathrm a}$ at the Half-Equivalence Point.

  Vocabulary:

  SB : Strong Base. (eg. $\ce{NaOH}$)
  CB : Conjugate Base. (eg. $\ce{Ac^{-}}$)
  WA : Weak Acid. (eg. $\ce{HAc}$)

NOTE: We can never be 100% accurate here, this is just used to derive that $\mathrm{pH} = \mathrm pK_{\mathrm a}$


Why boron trihydride is weaker lewis acid than boron tribromide and boron tri-iodide?

Predicting the acidic strengths of $\ce{BX_3}$ using back-bonding is one plausible theory. This article http://pubs.acs.org/doi/abs/10.1021/ic990713m hints us to use bond-strength data to predict the same:

  The weaker Lewis acid strength of $\ce{BF3}$ compared to $\ce{BCl3}$ with respect to bases such as $\ce{NMe3}$ is explained in terms of the ligand close-packing (LCP) model. The halogen ligands remain close-packed during the formation of the complex leading to an increase in the $\pu{B−Hal}$ bond length. Because a $\ce{B−F}$ bond is stronger than a $\ce{B−Cl}$ bond, more energy is required to stretch a $\ce{B−F}$ bond than a $\ce{B−Cl}$ bond. Hence $\ce{BF3}$ is a weaker acid than $\ce{BCl3}$.

Pulling up the bond strength data, I get:

Bond |  Strength (KJ/mole)
--------------------------
B-H  |  389
B-F  |  613
B-Cl |  456
B-Br |  377
B-I  |   ?

Building up on these two pieces of information, it can be said that since $\ce{B-H}$ bond is stronger than $\ce{B-Br}$ and $\ce{B-I}$ (see below), $\ce{BH3}$ is a weaker acid than $\ce{BBr3}$ and $\ce{BI3}$.

I'm afraid I couldn't find the data for $\ce{B-I}$ bond, but since we can compare the bond-strengths of boron trifluorides on the basis of back-bonding theory, we can be sure that the strength of $\ce{B-I}$ bond will be $\lt  \pu{377 kJ mol^-1}$.


Ion selective electrodes are a group of electrodes that are very different from each other.

The common bits are, a selective membrane, a "sensing" electrode and a reference electrode to measure against.

The details of the membrane and sensing electrode depend highly on what selective electrode you are trying to build.

Here is an old patent on a Cl selective electrode for reference:
http://www.google.com/patents/US3740326

Regarding getting accurate results with a DIY: this really depends on which electrode you are trying to build. In the Cl example, I don't see any reason why a DIY would perform worse.


Why do they flush people with water after chemical attack?

A standard protocol after unintended chemical exposure is to quickly douse the victim in large quantities of water to wash away chemicals that could remain and cause further harm, such as powders and other types of solids that could spread over your body as you moved. Also, the water will dilute any chemicals to the point of least harm that soak into clothing (provide a large amount of water is used). However, as Ivan Neretin pointed out in the comment, sarin gas does not do physical damage to the external parts of the body, but enters the body and causes nerve damage extremely quickly. In this type of situation, the water dousing probably did little to nothing to help the victims.


I think this could be a counter answer. In contrast, I found a literature evidence that disproportionation of $\ce{Ag(II)}$ to $\ce{Ag(I)}$ and $\ce{Ag(III)}$ has had happens Ref.1). The abstract states that:

  Interaction of $\ce{Ag+}$ salts in anhydrous liquid hydrogen fluoride, $\ce{aHF}$, with $\ce{AgF4-}$ salts gives amorphous red-brown diamagnetic $\ce{Ag^IAg^{III}F4}$, which transforms exothermally to brown, paramagnetic, microcrystalline $\ce{Ag^{II}F2}$ below $\pu{0 ^\circ C}$. $\ce{Ag^IAu^{III}F4}$ prepared from $\ce{Ag+}$ and $\ce{AuF4-}$ in $\ce{aHF}$ has a tetragonal unit cell and a $\ce{KBrF4}$ type lattice, with $a = \pu{5.788(1) Å}$, $c = \pu{10.806(2) Å}$, and $Z = 4$. Blue-green $\ce{Ag^{II}FAsF6}$ disproportionates in $\ce{aHF}$ (in the absence of $\ce{F-}$ acceptors) to colorless $\ce{Ag^IAsF6}$ and a black pseudotrifluoride, $\ce{(Ag^{II}F+)2Ag^{III}F4^-AsF6-}$. The latter and other $\ce{(AgF)2AgF4MF6}$ salts are also generated by oxidation of $\ce{AgF2}$ or $\ce{AgF+}$ salts in $\ce{aHF}$ with $\ce{F2}$ or in solutions of $\ce{O2+MF6-}$ salts ($\ce{M = As, Sb, Pt, Au, Ru}$). Single crystals of $\ce{(AgF)2AgF4AsF6}$ were grown from an $\ce{AgFAsF6/AsF5}$ solution in $\ce{aHF}$ standing over $\ce{AgF2}$ or $\ce{AgFBF4}$, with $\ce{F2}$ as the oxidant. They are monoclinic, $P2/c$, at $\pu{20 ^\circ C}$, with $a = \pu{5.6045(6) Å}$, $b = \pu{5.2567(6) Å}$, $c = \pu{7.8061(8) Å}$, $β = 96.594(9)°$, and $Z = 1$. The structure consists of $\ce{(AgF)_n^{n+}}$ chains ($\ce{F−Ag−F} = 180^\circ $, $\ce{Ag−F−Ag} = 153.9(11)^\circ $, $\ce{Ag−F} = \pu{2.003(4) Å}$), parallel to $c$, that enclose stacks of alternating $\ce{AgF4-}$ and $\ce{AsF6-}$, each anion making bridging contact with four $\ce{Ag(II)}$ cations of the four surrounding chains “caging” them. There is no registry between the ordered array in one “cage” and that in any neighboring “cage”. The $\ce{F-}$ ligand anion bridges between the anions and, with the $\ce{Ag(II)}$ of the chains, generates a trifluoride-like structure. $\ce{(AgF)_2AgF4AsF6}$ [like other $\ce{(AgF)_n^{n+}}$ salts] is a temperature-independent paramagnet except for a Curie “tail” below $\pu{50 K}$.

However, I cannot find a counter example for $\ce{Cu(II)}$ complexes. Yet, $\ce{Cu(I)}$ is known to disproportionate to $\ce{Cu(II)}$ and $\ce{Cu(0)}$, which has limited $\ce{Cu(I)}$ chemistry. For example, if you react $\ce{Cu(I)}$ oxide ($\ce{Cu2O}$) with dilute sulfuric acid (hot), you'd get a brown precipitate of $\ce{Cu(0)}$ and a blue solution of $\ce{CuSO4}$ (a $\ce{Cu(II)}$ salt), because the disproportionation reaction dominates. You would not get the expected copper(I) sulfate solution from this reaction.

$$\ce{Cu2O + H2SO4 -> Cu + CuSO4 + H2O}$$

However, if you use certain ligands to form $\ce{Cu(I)}$ complexes, they would stabilizes the $\ce{Cu(I)}$ oxidation state. For example, both $\ce{[Cu(NH3)2]+}$ and $\ce{[CuCl2]-}$ are $\ce{Cu(I)}$ complexes, which don't disproportionate.

References:

Ciping Shen, Boris Žemva, George M. Lucier, Oliver Graudejus, John A. Allman, Neil Bartlett, "Disproportionation of$\ce{Ag(II)}$ to $\ce{Ag(I)}$ and $\ce{Ag(III)}$ in Fluoride Systems and Syntheses and Structures of $\ce{(AgF+)_2AgF4^-MF6-}$ Salts ($\ce{M = As, Sb, Pt, Au, Ru}$)," Inorg. Chem. 1999, 38(20), 4570-4577 (https://doi.org/10.1021/ic9905603).
Also read for the Disproportionation of$\ce{Au(II)}$: Khaldoon A. Barakat, Thomas R. Cundari, Hassan Rabaâ, Mohammad A. Omary, "Disproportionation of Gold(II) Complexes. A Density Functional Study of Ligand and Solvent Effects," J. Phys. Chem. B 2006, 110(30), 14645-14651 (https://doi.org/10.1021/jp062501y).


Confusion regarding the mechanism of acidic/basic buffer solutions

While learning about acidic and basic buffer solutions we were told that they can resist pH change if we add small amounts of acid or base by neutralizing the $\ce{H+/OH−}$ ions from the added acid or base.

Yes, the pH changes less when you add small amounts of acid or base to a buffer, and more if you add it to pure water or an unbuffered aqueous solution.

But I am confused about the fact that if we add a small amount of base to an acidic buffer then after neutralization of the excess $\ce{OH−}$ ions from the base by the H+ ions from the buffer, shouldn't the concentration of $\ce{H+}$ ions in the solution decrease,

Yes, that is what happens. As a consequence, the weak acid/base reaction of the buffer  substances is no longer at equilibrium, and there will be a net reaction to replenish some of the $\ce{H+}$ that was lost.

thereby making the solution acidic and reducing the pH. (vice-versa for the addition of acid to a basic buffer)

This is probably a typo. When you add a small amount of base, the pH would go up, buffered or not. In the presence of buffer, it goes up a little, and in the absence of buffer, a lot.

(vice-versa for the addition of acid to a basic buffer)

Buffers stabilize the pH near the $\mathrm{p}K_\mathrm{a}$ of the conjugate weak acid in "both directions", so even if the pH of a solution is slightly acidic, it is able to buffer against addition of acid or base.


Do electrons only fill 'spin up' first? Or could it start filling 'down spins' first?

No it does not matter whether you start with 'spin up' or 'spin down' electrons as long as you are consistent: If you start filling some orbitals with 'spin up' electrons you will have to keep doing this for all the orbitals you want to fill.

But: The convention is that you start with 'spin up' as this usually is taken to represent the $\frac{1}{2} \! \hbar$ eigenvalue of the $z$ component of the spin angular momentum operator and you should adapt to that in order to avoid confusion.


these type of things are possible,

esterification is possible, check the reference.


I think what is given here is $K_p= p(\ce{NH3}). p(\ce{HI})$ and not $K_c$. So, What you have here are the partial pressures of the gases. As you have a sealed vessel, so the sum of the partial pressures of the gases gives the total pressure, 8.5 atm


It is interesting to consider that either outcome is possible, depending on the order of addition.

If a solution containing 0.5 moles of HCl is added to a solution of 1 mole of K2CO3, slowly, with mild stirring, cold, the products will be 0.5 KCl + 0.5 KHCO3. Nothing is lost to the atmosphere.

However, if a solution of 1 mole of K2CO3 is added to a solution of 0.5 moles of HCl, it would be possible to get 0.5 KCl + 0.5 K2CO3 with slow addition, rapid stirring, warm/hot. The difference here is that the initial K2CO3 is completely overwhelmed by the HCl and produces KCl + H2CO3, which can be degraded to H2O + CO2. By the time half of the K2CO3 is added, the products are 0.5 KCl + CO2, and the CO2 could be mostly evolved as gas and disappears. The last half of the K2CO3 just comes in and sits there.


Finding order of acidic strength for given compounds

Here, this was a simple question of checking the most acidic compound out of the four and this would have given you the answer. $\ce{H2O}$ will be most acidic as the negative charge will be formed on an oxygen compared to nitrogen and carbon in the other two cases.
However, the interesting question here is the part in the middle, what about $\ce{C2H2}$ and $\ce{Ph3CH}$? Which is more acidic?
If you take a look at the $\mathrm p K_\mathrm a$ values of the given compounds, you would notice that the values would come out to be:
\begin{array}{l|c}
\hline
\text{Compound} &\mathrm {p}K_\mathrm {a} \\
\hline
\ce{H2O} & 15.7 \\
\ce{NH3} & 38\\
\ce{C2H2} & 24\\
\ce{Ph3CH} & 31.5 \\
\hline
\end{array}
Now, this is the definitive set of values which states the order should be  $\ce{H2O > C2H2 > Ph3CH > NH3}$.

Note: Next part is speculative as I do not have a solid reason for $\mathrm pK_\mathrm a$ values being so for acetylene and the triphenyl system.
A question that may arise at this point is why is this so? Why would a triphenyl system with $3$ phenyl rings be less acidic than acetylene? What about the three rings providing resonance?
This is because in reality, only one phenyl ring provides resonance at a time. Due to the size of phenyl groups, the structure is not fully planar. Another reasoning is that the negative charge lies on an $\mathrm{sp^3}$ carbon whereas in acetylene it is on an $\mathrm{sp}$ carbon.


Why does Finkelstein reaction only work with R-Cl and NaI?

The classic Finkelstein reaction entails the conversion of an alkyl chloride or an alkyl bromide to an alkyl iodide by treatment with a solution of sodium iodide in anhydrouss acetone (Wikipedia). You can find here why this reaction works in this manner.

The reaction is essentially driven toward products by mass action due to the precipitation of the poorly soluble $\ce{NaCl}$ or $\ce{NaBr}$. Solubility of one specific salt (only $\ce{NaI}$) is important to the reaction. I emphasize above that your acetone has to be very dry, otherwise $\ce{NaCl}$ and $\ce{NaBr}$ are also going to be soluble large scale in acetone and reaction won't go to completion, and thus, won't be useful. For example, the solubility of $\ce{NaI}$ in acetone is found to be 29.2% by weight at $\pu{25 ^\circ C}$ but it is increased to 40.5% when 6.1% (by weight) of water added to the solution (Ref.1). Similar effect should be expected for the solubility of
$\ce{NaCl}$ and $\ce{NaBr}$ (Ref.2).

To answer your particular question about $\ce{KI}$, you need to look at the solubility of it in acetone: Unlike $\ce{NaI}$, which has maximum solubility in acetone at $\pu{25.7 ^\circ C}$, maximum solubility of $\ce{KI}$ in acetone is at $\pu{-57.5 ^\circ C}$ (Ref.3):

  [....]. In this paper are listed the solubilities of potassium iodide in acetone from $+54.5$ to $\pu{-78.5 ^\circ C}$. While, in general, the solubility curves of sodium and potassium iodides are similar in form, the maximum solubility of potassium iodide is about threefold less than that of sodium iodide, and it occurs at $\pu{-57.5 ^\circ C}$, instead of $\pu{+25.7 ^\circ C}$ as was reported for the sodium salt. The solid phase which is in equilibrium with the solution below $\pu{-57.5 ^\circ C}$ is a solvated form containing five (or six?) molecules of acetone per molecule of potassium iodide.

As stated in above abstract, $\ce{KI}$ molecule is surrounded by 5 to 6 acetone molecule, instead of 3 in $\ce{NaI}$ case. That makes it crowded (steric effect) as well. Its solubility is reported as 1.330% by weight at $\pu{+25 ^\circ C}$ (Ref.3), compared 29.2% of $\ce{NaI}$ under the same conditions. These facts make $\ce{KI}$ unsuitable for used in Finkelstein reaction. Yet, there are several instants of using $\ce{KI}$ in this type of reactions, e.g., see Ref.4.

There are also instances of using substrates other than $\ce{R-Cl}$ or $\ce{R-Br}$, e.g., see Ref. 5.

References:

R. Macy, E. W. Thomas, “The System: Sodium Iodide-Acetone-Water,” J. Am. Chem. Soc. 1926, 48(6), 1547–1550 (https://doi.org/10.1021/ja01417a015).
M. Li, D. Constantinescu, L. Wang, A. Mohs, J. Gmehling, “Solubilities of $\ce{NaCl, KCl, LiCl}$, and $\ce{LiBr}$ in Methanol, Ethanol, Acetone, and Mixed Solvents and Correlation Using the LIQUAC Model,” Ind. Eng. Chem. Res. 2010, 49(10), 4981–4988 (https://doi.org/10.1021/ie100027c).
R. Livingston, R. R. Halverson, “Solubility of Potassium Iodide in Acetone,” J. Phys. Chem. 1946, 50(1), 1–6 (https://doi.org/10.1021/j150445a001).
F. G. Bordwell, W. T. Brannen, “The Effect of the Carbonyl and Related Groups on the Reactivity of Halides in $\mathrm{S_N2}$ Reactions,” J. Am. Chem. Soc. 1964, 86(21), 4645–4650 (https://doi.org/10.1021/ja01075a025).
D. J. Maloney, S. M. Hecht, “A Stereocontrolled Synthesis of $\delta$-trans-Tocotrienoloic Acid,” Org. Lett. 2005, 7(19), 4297–4300 (https://doi.org/10.1021/ol051849t).


Can we manipulate the rate of water's autoionization?

First of all, I'm assuming you mean the equilibrium position. This is not the same as the rate of the reaction.

Sure there is. The easiest way to increase the equilibrium position water's ionization is by heating it.

To find out how much increasing the temperature of water will affect $K_w$, the Arrhenius equation will work for values of $T$ close to $0 °C$. $K_w=Ae^{\frac{-E_a}{RT}}$.

Of course, you'll have to find $A$ from experimental data.

Here is a graph of how changing the temperature affects $pK_w$. For the most part, as temperature increases, you will see an increase in K.


As we go down a fluoride-alkali metal series, why doesn't the boiling point decrease?

Polarizability of the fluoride ion by the alkali metal ion decreases going down the group, this decreases the covalent character in each ionic bond in question.

See the related questions and answers here:

Does covalent character increase or decrease bond strength?

Melting points of alkali metal halides


I don't have any journal article citations to back me up, but I've always understood that this distinct smell is from water aerosol.  You can detect the same smell during a fine misting rain as well as near water sprinklers (if the sprinklers produce a very fine mist of water).  But you don't get this smell from water vapor (steam) or bulk water, so there is clearly a specific range of water droplet size that triggers the olfactory response.

I've personally always wanted to know what the specific size range of these aerosol droplets needs to be to trigger the olfactory response.  I'm confident someone has studied this, but I don't know which journals to search to find the answer.


Effect of increasing nitrogen concentration in the synthesis of ammonia

The final concentration of H$_2$ is lower than the concentration of NH$_3$ because for each NH$_3$ you produce, you need 1.5 H$_2$.  So the rate of consumption of hydrogen is higher than the rate of production of ammonia.

Here you can read about the multiple steps in the synthesis of ammonia, also called the Haber-Bosch Process.  As you mention, by increasing the concentration of nitrogen, you can increase the rate.  The rate limiting step is separation of the nitrogen atoms in N$_2$


Method and Tips for Inorganic Qualitative Analysis

This is a rough draft of the complete answer. A lot more will be added to the answer very soon. At present, it includes the common tests for sulfite and bisulfite ions. Please do not downvote just because it is not a complete answer yet.

Identification Of Sulfite Anion: ($\ce{SO3^{2-}}$)

All sulfites are insoluble except "group IA sulfites " and$\ce{(NH4)2SO4}$. Sulfite ppt. is always white in color.

Test With Acid:

$\ce{SO3^{2-} + H^{+} -> SO2 (g)}  $

Sulfur dioxide gas is evolved which gives lime water test as well as baryta water test:

$\ce{SO2 + Ca(OH)2 -> Ba(HSO3)2}[\text{soluble}]$

$\ce{SO2 + Ba(OH)2 -> Ca(HSO3)2}[\text{soluble}]$

Test with $\ce{BaCl2/CaCl2}$:

$\ce{SO3^{2-} + CaCl2 -> CaSO3 + 2Cl^{-}}$

$\ce{SO3^{2-} + BaCl2 -> BaSO3 + 2Cl^{-}}$

Note that $\ce{BaSO3}$ is slowly oxidized by atmospheric oxygen to form heavy white ppt of $\ce{BaSO4}$. Similarly:

$\ce{BaSO3 + Br2 -> BaSO4 + Br-}$

$\ce{BaSO3 + H2O2 -> BaSO4 + H2O}$

$\ce{BaSO3 + HNO3 -> BaSO4 + NO}$

Test With $\ce{Pb(NO3)2}$:

$\ce{Pb(NO3)2 + SO3^{2-} -> PbSO3 [\text{white ppt.}]}$

On boiling $\ce{PbSO3}$ in presence of $\ce{O2}$, $\ce{PbSO4}$ is produced.

(On boiling $\ce{PbS2O3}$ we get $\ce{PbS}[\text{black ppt}]$ and $\ce{H2SO4}$. Hence, this fact can be used to distinguish sulfites from thiosulfates).

Test With $\ce{AgNO3}$:

Schiff's Test/Fuschin's Test/Magenta Reagent Test:

Both $\ce{SO3^{2-}}$ and $\ce{SO2}$ are decolorized in this test.

Distinguishing b/w $\ce{SO3^{2-}}$ and $\ce{CO3^{2-}}$ ion:

Some Important Redox Reactions:

$\ce{MnO4- + SO2/H2S -> Mn^{2+} + SO4^{2-}/S}$

$\ce{Cr2O7- + SO2/H2S -> Cr^{3+} + SO4^{2-}/S}$

$\ce{Br2 + H2O + SO2/H2S -> Br- + SO4^{2-}/S}$

$\ce{Fe^{3+} + SO2/H2S -> Fe^{2+} + SO4^{2-}/S}$

$\ce{IO3- + SO2/H2S -> I2 [\text{gives deep blue color in starch}] + SO4^{2-}/S}$

$\ce{SO2}$ can either be oxidized to $\ce{SO4^{2-}}$ or reduced to $\ce{S^{2-}}$. But, $\ce{H2S}$ can only be oxidized to $\ce{S}$.

Notable Reaction (Comproportionation) : $\ce{\ce{SO4^{2-}} + \ce{H2S} -> S + H2O}$

Points to remember:

$\ce{CO2/CO/CO3^{2-}/HCO3^{-}}$ do not show any change on treatment with with oxidizers like $\ce{KMnO4/K2Cr2O7/Br2\text{ water}/Fe^{3+}/IO3^{-}}$
$\ce{S^{2-},SO3^{2-}/HSO3^{-}}$ do show change on treatment with with oxidizers like $\ce{KMnO4/K2Cr2O7/Br2\text{ water}/Fe^{3+}/IO3^{-}}$.
$\ce{SO3^{2-} + H2O2 -> SO4^{2-} + H2O}$ $ [\text{no visible change is observed}]$

Dvarda Alloy ($\ce{Zn}/\ce{Al}/\ce{Cu}$):

$\ce{Zn + NaOH -> Na2ZnO2 + 2H}$

In presence of dvarda alloy the following reactions can take place:

$\ce{SO3^{2-} -> S^{2-}}$

$\ce{SO2 -> H2S}$

$\ce{NO2^{-} -> NH3}$

$\ce{NO^{3-} -> NH3}$

$\ce{Zn}$ reacts with dil. $\ce{H2SO4}$ to produce $\ce{H2}$ gas which can also reduce $\ce{SO4^{2-}}$ to $\ce{S^{2-}}$.

Test with Sodium Nitroprusside:

$\ce{Na2[Fe(CN)5NO] + ZnSO4 -> Zn[Fe(CN)5NO] [\text{salmon pink ppt }]}$

$\ce{Zn[Fe(CN)5NO] + \text{moist} SO2 -> \text{red color compound}}$

$\ce{S2O3^{2-}}$ and $\ce{S^{2-}}$ ions also give the test. So they can be removed on treating with $\ce{HgCl2}$ as follows:

$\ce{Hg^{2+} + S^{2-} -> HgS}$

$\ce{Hg^{2+} + S2O3^{2-} -> HgS2O3}$

$\ce{HgS2O3 [\text{on heating}]-> HgS [\text{black ppt}] + H2SO4}$

Identification Of Bisulfite Anion ($\ce{HSO3^{-}}$):

Test With Acid:

$\ce{HSO3^{-} + H^{+} -> SO2 (g)}$

Test With $\ce{H2O2}$:

$\ce{SO3^{2-} + H2O2 -> SO4^{2-} + H2O} $

$\ce{HSO3^{-} + H2O2 -> H^{+} + SO4^{2-} + H2O} $ (acidic)


Why can zwitterions be difficult detect by HPLC-MS?

Zwitterionic amino acids are routinely detected by HPLC and by MS.  The only challenge is to find an appropriate stationary phase.  The most common stationary phases for this type of separation would include octadecyl-derivatized silica (ODS, also known as C18) and a HILIC stationary phase.  Many column vendors make columns explicitly marketed for separation of amino acids.

Examples:

Sciex markets an LC-MS method using a C18 column.
Agilent markets an LC-MS method using a HILIC column.
A decently-cited paper from 2016 describes a 3-minute LC-MS method for amino acid separation using HILIC columns.

What needs to happen is that analytes need to reversibly equilibrate between the stationary and the mobile phase via a single "mode".  If there are multiple modes by which analytes can interact with the column, then peak shapes will be highly distorted and/or flattened into noise.  An example of this would be if the analyte can both interact hydrophobically with the C18 moiety of ODS and via Lewis acid / Lewis base interactions with bare silanols (Si~OH) groups on the surface of exposed, underivatized silica.  These days, column manufacturers go through great lengths to insure that very few if any exposed silanols remain exposed on derivatized silica resin used for analytical LC columns.  Zwitterions are really no different than other analytes in this regard.

At the MS, zwitterions will need to be transferred to the gas phase as charged ions for detection.  But this isn't really a problem either, at least for e.g. electrospray mass spectrometry.


In water, will sucrose be hydrolysed by water to form glucose and fructose molecules?

Not without a catalyst present. The process is usually performed using invertase, but some other enzymes with similar activity exist and addition of acid also can catalyse the reaction.


I would suggest adding anhydrous $\ce{MgSO4}$ to your distillate, like Nicolau Saker Neto already mentioned in his comments. Mix the suspension thoroughly (e.g. by stirring) and look how long the salt remains dispersed. When it clumps up and settles quickly due to the formation of hydrates, there is still some water in your mixture, and you need to repeat the procedure until the added $\ce{MgSO4}$ remains dispersed as fine particles for a longer time (it should look similar to a snow storm). After the salt has settled, you can recover the dried oil with a pipette. Because magnesium sulfate is not listed as toxic or environmentally hazardous, it can be considered a "greener" alternative to, for example, copper(II) sulfate or diethyl ether (the latter is additionally highly flammable).


Internal energy change for reactions at constant volume vs constant pressure

The key point to keep in mind in this type of problem is that the energy of an ideal gas is only a function of temperature, not of volume or pressure$^\ast$. We also assume that the dependence of the energy of the condensed phases on p and V is negligible. Therefore the final pressure or volume is not going to affect $\Delta U$ for the reaction, provided n and T are constant. Since here the reaction refers to a conversion of a stoichiometric amount of reactants into products at the specified T, we therefore do not expect much of a change in $\Delta U$ with change in p or V.

$^\ast$Note that the total differential for U can be written as
$$\begin{align} dU &= \left\{ T \left( \frac{\partial p}{\partial T} \right)_V -p\right\} dV + C_VdT \end{align}$$ Therefore at constant T,
$$\begin{align} dU = &= \left\{ T \left( \frac{\partial p}{\partial T} \right)_V -p\right\} dV  \end{align}$$
The term in parentheses is zero for an ideal gas.


Why does cinnamaldehyde pass the ferric chloride test?

I think the cinnamaldehyde gives $\ce{FeCl3}$ test due the similar reason why acetic acid can give $\ce{FeCl3}$ test as $\ce{Fe^3+}$ can act as oxidizing agent.

As in case of acetic acid $\ce{Fe(CH3COO)3}$ (red colour) is formed, a similar complex may be produced.


The thread indicated in the comment by @andselisk dates back to July 2014. It may be condensed to

  "A substructure querying with openbabel requires a good SMARTS string. In case you don't have one, often a good SMILES string may be used instead of a SMARTS string, too."

A check for a rather simple molecular structure of (benzylsulfinyl)benzene

there is more than one way to describe it in either form of a SMARTS, or a SMILES string.  Consequently, the strings per se sometimes differ from each other.  Consequently, recognizing the relevant entries in the collection of .mol2 with openbabel sometimes failed.

Below I summerize (status: 2019-May-15) a few experiences: the string offered by the software, its type, and source; eventually, if it worked (i.e., entries retrieved), or not (i.e., entries not retrieved) in a small test set of 419 model data and 7 relevant hits.

|-------------------------------------+--------+--------------------+--------|
| entry                               | type   | source             | status |
|-------------------------------------+--------+--------------------+--------|
| c1ccccc1-[#6]-[#16](-c2ccccc2)=[#8] | SMARTS | pubChemSketcher    | works  |
| C1=CC=CC=C1C[S](C2=CC=CC=C2)=O      | SMILES | pubChemSketcher    | fails  |
|-------------------------------------+--------+--------------------+--------|
| O=S(Cc1ccccc1)c2ccccc2              | SMILES | ACD ChemSketch     | works  |
|-------------------------------------+--------+--------------------+--------|
| C1=CC=C(C=C1)CS(C(C=CC1)=CC=1)=O    | SMILES | ChemDoodle         | fails  |
|-------------------------------------+--------+--------------------+--------|
| O=S(C1=CC=CC=C1)CC2=CC=CC=C2        | SMILES | ChemDraw Pro12     | fails  |
|-------------------------------------+--------+--------------------+--------|
| O=S(CC1=CC=CC=C1)C1=CC=CC=C1        | SMILES | Marvin JS          | fails  |
|-------------------------------------+--------+--------------------+--------|
| O=S(Cc1ccccc1)c2ccccc2              | SMILES | Peter Ertl JSME    | works  |
|-------------------------------------+--------+--------------------+--------|
| c1cccc(c1)CS(=O)c1ccccc1            | SMILES | openbabel, default | works  |
| c1ccc(cc1)CS(=O)c1ccccc1            | SMILES | dito, universal    | works  |
| c1ccc(cc1)CS(=O)c1ccccc1            | SMILES | dito, Inchified    | works  |
| O=S(c1ccccc1)Cc1ccccc1              | SMILES | dito, canonical    | works  |
|-------------------------------------+--------+--------------------+--------|

PubChem Sketcher: version 2.4, https://pubchem.ncbi.nlm.nih.gov/edit2/index.html
ACD ChemSketch:   Freeeware version 2015
ChemDoodle:       version 8.1.0, 2019 and webcomponents,
                  https://web.chemdoodle.com/demos/sketcher/
ChemDraw:         version 12 Pro, (then) CambridgeSoft, 2009
Marvin JS:        ChemAxon, v19.11.0, on-line test at
                  https://marvinjs-demo.chemaxon.com/latest/
                  Its SMARTS (very differently written) equally failed:
                  O=S([#6]-[#6]-1=[#6]-[#6]=[#6]-[#6]=[#6]-1)[#6]-1=[#6]-[#6]=[#6]-[#6]=[#6]-1
Peter Ertl JSME:  https://peter-ertl.com/jsme/JSME_2017-02-26/

Addition:
The approach mentioned in the question basically calls babel (or òpenbabel) as an iteration per file.  Especially if the data set is accessed more than once to identify different structures, and / or consists of multiple thousand models, this may be tediously slow.  Instead, it is worth to "bundle" the individual information

babel *.mol2 mymols.sdf --unique

creating a container like a multi-model .mol2 (this even works with .xyz files as input).  The .sdf may be indexed

babel mymols.sdf -ofs

by fingerprints (cf. chapter 5 of the manual).  Now, very quickly the names of the files about a strucure matching the query substructure are identifed, e.g.

babel mymols.fs -ifs -s"O=S(c1ccccc1)Cc1ccccc1" results.smi -xt

and stored in results.smi.  Consequently, any further work may focus on them.


Which plastic can be partially broken by pressure easily?

If you need good control, put a spring-loaded or cantilever-loaded valve at the tip.  Rupture diaphragms typically have a needle point at the far end. The bubble deforms with pressure, contacting the point that then ruptures the diaphragm that is under tension.  It is much easier to control and reproduce deformation than it is to engineer catastrophic failure at maximum deformation.


Can I normalize my data in order to compare it?

I assume you have a UV/Vis absorbance based measurement. Then the absorbance should follow Beer´s Law which states that absorbance is proportional to concentration. Therefore you should be fine with your normalization approach. You could check if you are in the linear range by performing a calibration.


The attacking nucleophile does not just shove its electron pair to the carbon from some random direction. No, it must donate that pair to the antibonding orbital ($\sigma^*$) of the $\ce{C-X}$ bond, thus weakening and eventually breaking that bond. And it just so happens that the antibonding orbital is mostly located on the side opposite to X.

Upd. Thought I'd better look up a nice picture. Here's one:

(from this page)


Your description matches fluorite, you can see many pictures in Internet, and compare with your stone. To determine what it is precisely, you have to post a picture.

EDIT. After the pictures are posted, I take back my suggestion: the key difference
is that fluorite is not transparent, at least those pieces that I possess, and those which I can see on Internet. But in the dark, it looks exactly of the right color. There  are very many fluorescent stones, look for example, here:
 http://www.galleries.com/minerals/property/fluotabl.htm


Why is tetrathiocyanatocobaltate(II) blue?

Let us consider this qualitatively with crystal field theory (CFT). Co is a transition metal with $d^7$ configuration. CFT says that depending on the geometry of the metal complex, the d-orbital energies will split into different energy levels. If the ligands (in this case, the four $\ce{SCN-}$ ligands) align with certain orbitals, then the electrons in these orbitals will feel an increased repulsion due to the closeness of the ligands, and increase in energy. The orbitals in-between the aligned axes, will not feel this repulsion, and will not increase in energy. The splitting diagram might look something like this

There is an equation that relates the (splitting)
 energy to wavelength:

$$
\Delta E = \frac{hc}{\lambda}
$$

From this equation, we see that if the splitting energy increases, wavelength will have to decrease, i.e. be shorter than before. This means a shift towards the violet region. As you say, $\ce{SCN-}$ is a strong ligand, meaning it leads to a rather high $\Delta E$. So we could say that $\ce{Co(SCN)_4^{2-}}$ is blue due to the high energy gap between the d-orbitals. Electrons are excited from one of the lower orbitals up to one of the higher orbitals. Upon de-exciting, light is emitted with an energy equal to $\frac{hc}{\lambda}$. This happens to be in the blue/violet region.

Weaker ligands would split the energy levels less, and the complex's color would therefore have a color closer to the red part of the visible spectrum.

Did this make it a little clearer?


What happens when FeF3 is reacted with benzene?

Iron catalyst will not be needed as Fluorine will react directly with benzene explosively,  the probable product formed would be hexaflurocyclohexane , which can predicted by looking at similar reaction of chlorine with benzene in presence of UV rays which is mentioned in NCERT class 11 chemistry  Textbook.

(Replace $Cl$ with $F$  and since fluorine is more reactive UV light is not needed.)


How to identify a compound as chiral or achiral

Look for carbons with four different groups attached to identify potential chiral centers.

Draw your molecule with wedges and dashes and then draw a mirror image of the molecule. If the molecule in the mirror image is the same molecule, it is achiral. If they are different molecules, then it is chiral.

Here you're particularly interested in the C in the CH. However, take a look at the C in CH3. Why isn't it a chiral carbon?


Why is the letter J omitted in the spdf... sequence?

Omitting j when alphabetically enumerating things has a long tradition.

First of all, the alphabet did not always exist in the form we know it today.
Quoting Wikipedia:

After [...] the 1st century BC, Latin adopted the Greek letters ⟨Y⟩ and ⟨Z⟩ [...] Thus it was during the classical Latin period that the Latin alphabet contained 23 letters:
[no J, V, W]
[...]
It was not until the Middle Ages that the letter ⟨W⟩ [...] was added
[...]
only after the Renaissance did the convention of treating ⟨I⟩ and ⟨U⟩ as vowels, and ⟨J⟩ and ⟨V⟩ as consonants, become established. Prior to that, the former had been merely allographs of the latter.

In some books, this has consequences till today.

The footnotes to the Confession and Catechisms [of the Presbyterian Church], containing the proof texts, are enumerated in the traditional manner, that is, by letters of the alphabet (omitting j and v, as alternative forms for i and u in the Latin alphabet). Source

But even later, long after the letters I and J were considered distinct in terms of proper spelling, their alphabetical order (I preceding J) was not firmly established.
The New General English Dictionary of 1768 had a combined section for I and J, treating both equal with respect to alphabetical order (but not regarding spelling).
Same with Handwörterbuch der allgemeinen Chemie, a German chemistry book printed in 1818.
The latest book, I could find, is the Handwörterbuch der reinen und angewandten Chemie of 1850. This is pretty close to when the letters spdefghik.. must have been defined! (Does anyone know when exactly this was?)
And now, it perfectly makes sense to me. Even if i and j were distinct letters at the time and their order should have been commonly established in the early 20th century, the possibility that some readers could still be confused about what comes first, must have lead to the decision to leave j out.

Update (2017-12)
Concerning the spelling and order of names, I was able to find evidence that is even 100 years younger.
The Berlin telephone book contained a spelling table. The first book of 1890 contained a spelling table that assigned numbers to each letter, omitting the letter J, i.e. I=9 and K=10. That is pretty similar to our orbital labels, isn't it? In the 1903 printing, words were assigned to the letters, but J was still left out. The 1905 printing was the first to include J into the spelling table. (source)
Even the Berlin address book of 1943 did not distinguish between I and J. For instance Jutta is listed before Iwanski. (Interestingly, this book doesn't even use different glyphs for I and J in the Fraktur font.)
Of course, there are many books around that time, and earlier, that sort I before J and that use different glyphs, even if printed in Fraktur (example). Nevertheless, this shows that the convention "I before J", as we know it today, was not firmly established in early 20th century Germany.


Confusion about comparing the charge densities of benzene and fluorobenzene

This is the way of directing the electrophiles to the ortho and para places  of the benzene cycle. As you see the ortho and para places are more negatively charged.That's true. It is due to the ability to donate its electron pair to the benzene cycle.

But an electronegative element can attract the electron cloud to it. Fluorine is the most electonegative element, so it has a greater ability to attract the electron cloud of the benzene cycle. So that makes it relatively  resistant to an incoming electrophile.

This does not say that it is not a  ortho-para director. It's a ortho-para director but it deactivates the ortho and para places of the benzene cycle.


Converting units and sig figs (finding molarity)?

In terms of significant figures, you always use the number with least amount of sig figs to determine how many sig figs your answer has. In this case, one number has 3 sig figs, and the other has 1. Therefore your answer should have 1 sig fig and be 1 mol/L, as you said. To make the zeroes in the 500 mL become significant it would need to have a period afterwards like this: 500. mL. Perhaps the writer forgot to do that, however as it is written the answer would be 1 sig fig: 1 mol/L


Use household tools!

Can you bend or fold the note? Is the layer brittle and does break off?
Can you scratch the layer off with a sharp blade?
Place the note between paper (kitchen roll) and use a hot flat iron. Does the layer melt?
Can you wash the layer off using a cotton bud and

water
methylated spirit
nail polish remover