Calculating the pH of a mixture of Na2HPO4 and Na3PO4?

I'll give a round about answer based on significant figures.

The whole truth is that any time that you add any phosphate ion into an aqueous solution, then you will have all four phosphate species ($\ce{H3PO4}$, $\ce{H2PO4^-}$, $\ce{HPO4^{2-}}$, and $\ce{PO4^{3-}}$) in solution. The relative amounts depend on the 3 equilibrium equations, and the total concentration of all of the phosphate species.

For phosphoric acid the three pKa's are different enough so that only two phosphate species will have a "significant" concentration at whatever pH the solution is at. Here is where the answer gets fuzzy. What is "significant"? To account for 99% of the species (2 significant figures) is typically good enough and at most two species would need to be considered. However if you want to account for 99.99999999999% of the species (12 significant figures), then you're going to have to consider all four phosphate species. So the gist is how many significant figures do you need to consider in the calculations?

What is typically done is to simply the four equilibrium equations to the two "significant" ones (maybe only 1 species at high or low pH's), and then calculate the concentrations of the last two species using the found concentrations of the first two.

In reality there is another consideration. If you look at the pKa values for phosphoric acid, Wikipedia lists pKa1 = 2.148, pKa2 = 7.198, and pKa3 = 12.319. There are only three significant figures in each of these equilibrium constants. (Only the mantissa counts, not the characteristic.) So you can only have three significant figures for any given phosphate species.

So, to three significant figures, for any sort of mixture of $\ce{Na2HPO4}$ and $\ce{Na3PO4}$ salts you'll need to consider both $pKa_2$ and $pKa_3$ and you'll end up with a quadratic equation to solve. The simplifying assumption is that $$\ce{[H3PO4] + [H2PO4^-] << [HPO4^{2-}] + [PO4^{3-}]}$$

An "exact" iterative solution, considering all four species, can easily be solved via a computer program, but it is really messy to do such a calculation by hand.

As a final check you should calculate the concentrations for $\ce{H3PO4}$ and $\ce{H2PO4^-}$ and verify that the assumption holds. (For phosphoric acid if the assumption doesn't hold to three significant figures, then normalizing the calculated values would yield a "good enough" result to 3 significant figures.)


$\mathrm{pH}$ is a measure of acidity but defining acidity as the concentration of dissoluted protons in solution. This is quite a bit different from defining acidity as the amount of acid in solution.

Taking your example, the appropriate concentration of $\ce{HCl}$ would be something like $1~\mathrm{mM}$ while the corresponding acetic acid concentration is maybe $1~\mathrm{M}$. (These numbers are most certainly not correct, and maybe even the dimension is off by one or two orders of magnitute, but they can still serve well as an example.) So you could say ‘but there’s one thousand times as many acid molecules in the acetic acid solution!’

When considering the actual effects of an acid, the concentration of the original acidic species is irrelevant. Vinegar is, for example, usually $5~\%$ acetic acid and can be added to food (tough guys might even drink it) — I wouldn’t try that with $5~\%~\ce{HCl}$ since that contains a much higher active proton concentration.

What you can do, though, is consider the higher concentration of weak acids a type of buffer: Dilute acetic acid and the $\mathrm{pH}$ will hardly change while diluting $\ce{HCl}$ will have an immediate effect on $\mathrm{pH}$. So a weak acid can supply the same weak source of protons for a longer time.

And even though real-world analogies should be avoided in any case since they often give a wrong impression: If you jump down the staircase of a high building two steps at a time there will be likely no effect (small effect but buffered — acetic acid) just like if you only jump the two steps once (small effect unbuffered — strongly diluted $\ce{HCl}$), but if you try to jump the whole staircase (large effect unbuffered — $\ce{HCl}$ in the same concentration as acetic acid) you won’t remain unaffected.


Unfortunately GAMESS does not support any other than the named population analyses.

You will need to use a different program. For that matter I suggest MultiWFN, which comes with a variety of different population analyses and is still in active development. From the website:

  Population analysis. Hirshfeld, VDD, Mulliken, Löwdin, Modified MPA (including three methods: SCPA, Stout & Politzer, Bickelhaupt), Becke, ADCH (Atomic dipole moment corrected Hirshfeld), CHELPG, Merz-Kollmann and AIM methods are supported.

The currently developed version MultiWFN 3.4 should be able to even read GAMESS output directly as stated in the update log:

  GAMESS-US output file now can be used as input file (not comprehensively tested, currently only single point task at HF/DFT level is formally supported). The suffix of output file should be changed to .gms so that Multiwfn can properly recognize it

As a workaround for older versions, and a reliable way to import this data, you can export a .wfn file with GAMESS. I will give a brief summary for this, since I am still using MultiWFN 3.3.8. From the corresponding manual:

  For example, the wavefunction represented by GTFs is enough for Hirshfeld population, so you can use .fch/.molden/.31~.40/.wfn/.wfx file as input, but .pdb, .xyz, .chg, .cub and .grd files do not carry any wavefunction information hence cannot be used.

As I have previously stated, .wfn files are technically outdated, especially when it comes to larger basis sets. However, for the time being, they are our only option to make it work.

I am using GAMESS version % DEC 2014 (R1) and the following input to demonstrate the procedure. I am using a Linux environment, so you might have to adjust this guide accordingly.

 $CONTRL SCFTYP=RHF            ! Restricted calculation.
         RUNTYP=OPTIMIZE       ! Geometry optimisation.
         COORD=ZMT             ! Z-matix specifies coordinates.
         AIMPAC=.TRUE.         ! Requests wfn file to be written.
         $END
 $BASIS  GBASIS=STO NGAUSS=3   ! STO-3G minimal basis
         $END
 $GUESS  GUESS=HUCKEL          ! Start with Hückel MO.
         $END
! Following group contains molecule specification.
! Blank lines are important.
 $DATA
Water
Cnv  2

O
H  1 rOH
H  1 rOH  2 aHOH

rOH=1.09
aHOH=110.0
 $END

Note that the indentations before the keywords are important. ! indicates a comment. Any additional output will be written to the punch file. This is dependent on how you set up GAMESS, but it normally is located in ~/scr/<rootfilename>.dat.

To extract the .wfn file you can use the following little script, or extract it manually from the .dat file.

#Script to extract wfn file from GAMESS dat (punch) file
#! /bin/bash

[ ! -z $1 ] && inputfile="$1" || exit 1
outputfile="${inputfile%.*}.wfn"

startpattern="----- TOP OF INPUT FILE FOR BADER'S AIMPAC PROGRAM -----"
endpattern="----- END OF INPUT FILE FOR BADER'S AIMPAC PROGRAM -----"

sed "/$startpattern/,/$endpattern/!d;//d" "$inputfile" > "$outputfile"

This should give you the following .wfn file for analysis:

Water
GAUSSIAN              5 MOL ORBITALS     21 PRIMITIVES        3 NUCLEI
  O    1    (CENTRE  1)   0.00000000  0.00000000 -0.14560943  CHARGE =  8.0
  H    2    (CENTRE  2)  -1.43257773  0.00000000  1.05592484  CHARGE =  1.0
  H    3    (CENTRE  3)   1.43257773  0.00000000  1.05592484  CHARGE =  1.0
CENTRE ASSIGNMENTS    1  1  1  1  1  1  1  1  1  1  1  1  1  1  1  2  2  2  3  3
CENTRE ASSIGNMENTS    3
TYPE ASSIGNMENTS      1  1  1  1  1  1  2  2  2  3  3  3  4  4  4  1  1  1  1  1
TYPE ASSIGNMENTS      1
EXPONENTS  1.3070932E+02 2.3808866E+01 6.4436083E+00 5.0331513E+00 1.1695961E+00
EXPONENTS  3.8038896E-01 5.0331513E+00 1.1695961E+00 3.8038896E-01 5.0331513E+00
EXPONENTS  1.1695961E+00 3.8038896E-01 5.0331513E+00 1.1695961E+00 3.8038896E-01
EXPONENTS  3.4252509E+00 6.2391373E-01 1.6885540E-01 3.4252509E+00 6.2391373E-01
EXPONENTS  1.6885540E-01
MO  1                     OCC NO =   2.00000000 ORB. ENERGY = -20.25158070
  4.22735195E+00  4.08851077E+00  1.27421022E+00 -6.18804850E-03  8.27701479E-03
  6.24678653E-03  0.00000000E+00  0.00000000E+00  0.00000000E+00  0.00000000E+00
  0.00000000E+00  0.00000000E+00  6.97684832E-03  4.38722945E-03  6.95010854E-04
 -1.54630881E-03 -1.49552256E-03 -4.66089059E-04 -1.54630881E-03 -1.49552256E-03
 -4.66089059E-04
MO  2                     OCC NO =   2.00000000 ORB. ENERGY =  -1.25754111
 -9.93961353E-01 -9.61316148E-01 -2.99600254E-01 -2.02173727E-01  2.70423693E-01
  2.04092795E-01  0.00000000E+00  0.00000000E+00  0.00000000E+00  0.00000000E+00
  0.00000000E+00  0.00000000E+00  2.05800516E-01  1.29412887E-01  2.05011755E-02
  4.30889918E-02  4.16737970E-02  1.29879022E-02  4.30889918E-02  4.16737970E-02
  1.29879022E-02
MO  3                     OCC NO =   2.00000000 ORB. ENERGY =  -0.59384261
  0.00000000E+00  0.00000000E+00  0.00000000E+00 -0.00000000E+00  0.00000000E+00
  0.00000000E+00  1.02654522E+00  6.45519180E-01  1.02261083E-01  0.00000000E+00
  0.00000000E+00  0.00000000E+00  0.00000000E+00  0.00000000E+00  0.00000000E+00
 -1.24405583E-01 -1.20319664E-01 -3.74983838E-02  1.24405583E-01  1.20319664E-01
  3.74983838E-02
MO  4                     OCC NO =   2.00000000 ORB. ENERGY =  -0.45973131
 -4.42352599E-01 -4.27824176E-01 -1.33334109E-01 -1.28843153E-01  1.72338126E-01
  1.30066155E-01  0.00000000E+00  0.00000000E+00  0.00000000E+00  0.00000000E+00
  0.00000000E+00  0.00000000E+00 -1.26642108E+00 -7.96359559E-01 -1.26156734E-01
 -8.17262213E-02 -7.90420432E-02 -2.46339524E-02 -8.17262213E-02 -7.90420432E-02
 -2.46339524E-02
MO  5                     OCC NO =   2.00000000 ORB. ENERGY =  -0.39261686
  0.00000000E+00  0.00000000E+00  0.00000000E+00 -0.00000000E+00  0.00000000E+00
  0.00000000E+00  0.00000000E+00  0.00000000E+00  0.00000000E+00  1.67545020E+00
  1.05356804E+00  1.66902879E-01  0.00000000E+00  0.00000000E+00  0.00000000E+00
  0.00000000E+00  0.00000000E+00  0.00000000E+00  0.00000000E+00  0.00000000E+00
  0.00000000E+00
END DATA
RHF      ENERGY =      -74.9659012162   VIRIAL(-V/T)  =   2.00600297

You can now use this file and load it with MultiWFN. After the start-up process you should find the key values printed to the screen like this (I abridged it a little bit):

 Multiwfn -- A Multifunctional Wavefunction Analyzer (for Linux)
 Version 3.3.8, release date: 2015-Dec-1
[...]
 System energy:     -74.965901216200 Hartree,   Virial ratio:  2.00600297
 Total/Alpha/Beta electrons:     10.0000      5.0000      5.0000
 Net charge:     0.00000    Expected multiplicity:    1
 The number of orbitals:     5,  Atoms:      3,  GTFs:     21
 This is restricted close-shell single-determinant wavefunction
 Title line of this file: Water
 Formula: H2 O1
 Molecule weight:        18.01528

 Loaded water-sto3g.wfn successfully!

                    ------------ Main function menu ------------
[...]
 7 Population analysis
[...]

You may now choose 7 and enter the following menu:

                 ============== Population analysis ==============
 0 Return
 1 Hirshfeld population
 2 Voronoi deformation density (VDD) population
 10 Becke atomic charge with atomic dipole moment correction
 11 Atomic dipole corrected Hirshfeld population (ADCH)
 12 CHELPG ESP fitting charge
 13 Merz-Kollmann (MK) ESP fitting charge
 14 AIM charge

Follow the guidance of the manual and the on-screen instructions to produce the required charges. For comparison here are the computed Hirshfeld charges, written to water-sto3g.chg with the following command sequence 7; 1; 1; y.

  O     0.000000    0.000000   -0.077053   -0.296017
  H    -0.758088    0.000000    0.558771    0.147959
  H     0.758088    0.000000    0.558771    0.147959

The element identifier is followed by the Cartesian coordinates, which are followed by the charges.

In principle all other implemented charges are equally as easy to compute. Keep in mind, that basin based charges (e.g. AIM) need more computing power and depend heavily on molecule and basis set size.

Apart from this procedure, you can obtain natural charges with the NBO6 program, which conveniently interfaces with GAMESS. Unfortunately you have to buy it.


Reaction of phenol with concentrated sulfuric acid

It's a reversible reaction, so at higher temperature the thermodynamic para product is formed from the kinetic ortho product.

Further reading here 1


A primary standard is critical for an analytical result. If we dissolve a known mass of a material into a known volume, we must be able to know its concentration with confidence. Such a standard must be easily made very pure (> 99.95%), it should not decompose, and it must be fairly robust to say heating or vacuum. We use such standard solutions as, for example, titrants. This is surprisingly hard to come by, and so we say that secondary standards are materials that do not meet these strict criteria.

Say you want to titrate a sulfuric acid solution for quantitative work, you could use a known concentration of NaOH or KOH and volumetrically determine the concentration of the sulfuric acid solution. In fact, dissolving pellets of these particular bases in a known volume of water is not sufficient to calculate concentration with any confidence in analytical work because the solids are contaminated with carbonate from atmospheric CO$_2$. In general these carbonates contaminate alkaline solutions quite readily through this reaction: $\ce{OH- + CO2 -> HCO3-}.$ This is why we would have to titrate the concentration of $\ce{NaOH} $ solution with a primary standard before using it as a secondary standard.

Now there are acid and base primary standards available, so why would one ever need to use a secondary standard? Usually it’s a matter of specific needs and convenience. Whenever possible, use a primary standard if your results actually matter.

Let us look at a specific case where a secondary standard is useful: potassium permanganate is exceptionally useful as a titrant because it serves as its own indicator where the presence of a purple color signifies the end point, but it is unstable in light or heat due to this reaction: $\ce{4 MnO4^- +2H2O -> 4MnO2 + 3O2 + 4OH-}$. This is why solutions of it must be stored in a dark cool place; even still, we must standardize the concentration of such a solution and we can only trust its measured concentration for a short time thereafter.

A secondary standard solution of $\ce{KMnO4}$ is usually titrated by oxalic acid, a primary standard, and then used to determine the concentration of a number of other analytes.

The reaction with oxalic acid can be summarized as $\ce{5 H2C2O4 + 2 MnO4- + 6H+ -> 10CO2 + 2 Mn^{2+} + 8H2O}$ where everything except $\ce{MnO4-}$ is colorless. The end point of this titration is a marked by a purple color due to unconsumed $\ce{MnO4-}$. This allows you to make a standardized solution of potassium permanganate that can be used for a short while.

Now, how would we use this? Solutions containing $\ce{Ca^2+, Pb^2+, Ag+, Mg^2+, Co^2+, Zn^2+}$ and etc. can be precipitated as a metal oxalate. You would then digest this ppt in an acidic solution and titrate the released $\ce{H2C2O4}$, where the end point is marked by a purple color, allowing you to infer the original concentration of the metal ion in the original solution.

$\ce{KMnO4}$ was used in mineralogical surveys of uranium prior to the Manhattan project, where better techniques were developed to quantitate.


Is liquid water in equilibrium with water vapor at room temperature?

It depends on the circumstances, but mostly not

Under most normal atmospheric circumstances water in an open vessel is not in equilibrium with water vapour in the atmosphere. We know this because water usually evaporates, albeit slowly.

One way to measure whether water is in equilibrium is to measure the humidity which is a measure of the maximum concentration of water vapour under the prevailing temperature and pressure. Humidity is mostly less than 100% which means that the air could have more water vapour in it before the vapour spontaneously condenses. This isn't always true, which is why we get fog and why some countries are unbearable to live in.

At normal temperatures the equilibrium between water and water vapour is established slowly. Any vessel containing water that is open to the atmosphere will try to establish that equilibrium, but that won't be complete until the air is 100% saturated with vapour (humidity=100%). And you'd have to saturate the entire atmosphere in the room or, if the room is open, the entire planet's atmosphere to achieve equilibrium.

That this doesn't happen quickly is obvious. But it can happen locally (we call that weather). It it obviously not true across the whole planet at the same time or the planet would be called "sauna" and not earth.

The point is that the atmosphere is rarely 100% saturated with water vapour, so nothing open to the atmosphere will normally be in equilibrium with liquid water. A close bottle left long enough might establish equilibrium (with 100% humidity) inside the closed bottle, but any exchange with the surrounding air will lower the humidity and the equilibrium might take some time to re-establish.

The key to understanding this is that the kinetics of establishing the equilibrium are slow and the normal atmosphere is not saturated with water vapour. This means that the air above a glass containing open water will have a slightly higher concentration of vapour than the air surrounding the glass but with not be fully in equilibrium because the room surrounding the glass is big. And the atmosphere around the room is even bigger and clearly isn't in equilibrium (we have weather and wind to demonstrate that). The only systems where equilibrium will be established are closed ones which have been left a reasonably long time. atmospheric exchange causes the first; slow kinetics the second of these constraints.


How do I make a ZMAT or XYZ file from scratch?

I know of two three ways to do this, but in either case it's quite a bit of work since I don't know any software that can read molecular structures from pictures. The second and third way use freeware only.

Draw the structure in Chem3D. For whatever stupid reason Chem3D doesn't support xyz files but you can save as *.pdb. If you need *.xyz then:

save as Gaussian input
delete the first 3 lines
Put the total number of atoms in the resulting first line
remove any "atoms" labeled as LP (lone pair, Chem3D uses them for force field optimization)
remove the second column next to the elemental symbol (if you sue Notepad++ you can select that by using ALT)
change file extension

Use openbabel 2.2.0 or higher to convert SMILES into *.xyz or *.pdb using the "gen3d" keyword. Of course for this you need the SMILES so you either have to produce that yourself or you need to draw it in a software that is able to produce SMILES like, again, Chem3D or this free website
It turns out this NIH website is also able to produce 3D structures, also in pdb file format, directly from SMILES or drawing the structure in their editor.


Is it possible to solve the equation for the diffusion of gas through a sphere analytically by applying e.g. "combination of variables"?

Yes. Let's improve the wording here: that is not just an expression, it's the definition of a dimensionless variable that eases the understanding of the solution of the transient diffusion is a semi-infinite medium. Because there is no clear characteristic length as in a finite diffusion path problem, Buckingham π theorem will tell us the two dimensionless quantities we need are
$$\Pi_1 = \frac{z}{\sqrt{D t}}$$ (Or any function of it, like squaring it or dividing it by 4, such as your $\sigma$)

And
$$\Pi_2 = \frac{y-y_{A\infty}}{y_{As} - y_{A\infty}}$$

With $s$ being the surface index and $\infty$ indicating a position "at infinity" (often we use $y_{A\infty} = 0$).

Then we conclude $\Pi_2 =\textrm{function}(\Pi_1)$. This particular function can found by solving Fick's second law with the appropriate boundary and initial conditions. In the most common case, it yields a Gauss error function.

Had we be initially interested in a sphere instead of a planar surface, the only major difference would be the use of spherical coordinates instead of a Cartesian frame, which would result in a different functional solution. But the problem of defining the dimensionless variables would be the same, given we assume perfect angular symmetry. The $z$ position would just be substituted by a radial position $r$.


Selecting parent chain considering substituent (prefix) functional groups

According to the usual nomenclature rules, two equivalent names are possible for the compound that is given in the question:

6-amino-4-(2-hydroxyethyl)hexanoic acid
4-(2-aminoethyl)-6-hydroxyhexanoic acid

In order to identify the preferred name, we have to dig deeper in the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), where we finally find two potentially relevant rules:

  P-45.2.3 The preferred IUPAC name is based on the senior parent structure that has the lower locant or set of locants for substituents cited as prefixes to the parent structure (other than ‘hydro/dehydro’ prefixes) in their order of citation in the name.

and

  P-45.5.1 The preferred IUPAC name is the name that is earlier in alphanumerical order (see P-14.5). Alphabetic letters are considered first in the order that they appear in the name; all Roman letters are considered before any italic letters, unless the latter are used as locants or are a part of a compound or composite locant, for example, ‘N’ and ‘4a’. Then, if still there is a choice, numerical locants are considered in the order of their appearance in the name.

Since P-45.2.3 comes first, the preferred name is 4-(2-aminoethyl)-6-hydroxyhexanoic acid and not 6-amino-4-(2-hydroxyethyl)hexanoic acid because the locant set ‘4,6’ is lower than ‘6,4’.

By way of comparison, a similar example given in the (corrected) Blue Book is 2-(2-bromoethyl)-4-chlorobutan-1-ol and not 4-bromo-2-(2-chloroethyl)butan-1-ol since the locant set ‘2,4’ is lower than ‘4,2’.

Remarkably, also according to the subordinate rule P-45.5.1, the preferred name would be 4-(2-aminoethyl)-6-hydroxyhexanoic acid and not 6-amino-4-(2-hydroxyethyl)hexanoic acid because ‘aminoethyl-hydroxy’ is earlier alphabetically than ‘amino-hydroxyethyl’. However, the preferred name for the example given in the Blue Book would have been 4-bromo-2-(2-chloroethyl)butan-1-ol and not 2-(2-bromoethyl)-4-chlorobutan-1-ol since ‘bromo-chloroethyl’ is earlier alphabetically than ‘bromoethyl-chloro’. This example was actually given in the Blue Book by mistake but it has been corrected as part of the official Corrections to Nomenclature of Organic Chemistry. IUPAC Recommendations and Preferred Names 2013.


How to explain (non-/anti-) aromaticity in fulvene with the help of resonance structures?

TL;DR: You cannot assign aromaticity based on a couple of resonance structures. Penta-fulvene has negligible (anti)aromatic character, which is supported by computational and experimental investigations.

Introduction

Aromaticity is a complex and still not fully understood phenomenon. Active investigations are experimentally and computationally challenging. Unfortunately in schools and university it is often taught as something quite simple to fathom, which can be explained by looking at Lewis structures and counting electrons. This maybe true for many common compounds, but when you dig deeper you'll very soon find the limitations. (See the notes below.) It is certainly not helpful in the case of fulvenes.

Penta-fulvene's resonance and (anti)aromaticity

The resonance structures you have drawn are correct, but the set is missing one member, coincidentally the more important one. (Please see the notes below on resonance.) There are more, but those are with more charge separation and likely have only little contribution.

In general you cannot judge one resonance structure on its own. In this case it is not helpful at all. In all resonance structures, the π-system is fully conjugated, and delocalised over the whole molecule.
Penta-fulvene has C2v symmetry, and we see deviations in the single and double bond lengths. The values are from a quite extensive study on substituted fulvenes: K. Najafian, P. von Rague Schleyer, and T. T. Tidwell, Org. Biomol. Chem. 2003, 1, 3410-3417 (DOI: 10.1039/B304718K). Unfortunately, they use the unsubstituted fulvenes as comparison. From the abstract:

  The fulvenes (1a–4a) have modest aromatic or antiaromatic character, and are used as standards for comparison.

Another study basically comes to the same conclusion, see E. Kleinpeter, and A. Fettke, Tetrahedron Lett. 2008, 49 (17), 2776-2781 (DOI: 10.1016/j.tetlet.2008.02.137). Citing quite liberally from various parts and omitting any literature references:

  Fulvenes 1–4 have previously been synthesized (triafulvene 1, pentafulvene 2, heptafulvene 3 and nonafulvene 4), and were studied with respect to their dipole moments and NMR spectra. The 1H and 13C NMR spectra of triafulvene 1 (both protons and carbon atoms of the 3-membered ring moiety display resonances in the region of aromatic compounds) evidence a significant contribution of the resonance form 1b [aromatic charge separation]; the corresponding NMR spectra of 2–4, however, display typical olefinic compounds with strongly alternating bond lengths and only a small extent of charge separation (corroborated by the relatively small dipole moments).

  [...]

  Dependent on the criterion employed, 1–4 were reported as partially aromatic, non- or even antiaromatic.

  [...]

  [...] However, the expected partial aromaticity of the 3-membered ring moiety of 1 was not observed (vide supra).
  Similar conclusions can be drawn for the presence of partial aromaticity in 2: even if the occupation of πC=C of the exocyclic double bond is lowest in the series (which can be realized with the participation of 2a, corroborated by the correct direction of the dipole moment), both ICSSs [iso-chemical-shielding surfaces] at ±0.1 ppm [2: ICSS = −0.1 ppm (5.0); ICSS = +0.1 ppm (6.2)] are far away from reference benzene 7 [7: ICSS = −0.1 ppm (7.2); ICSS = +0.1 ppm (8.9)] or even from cyclopropenylium cation 6 [6: ICSS = −0.1 ppm (5.9); ICSS = +0.1 ppm (7.2)]—pointing to 2π electron aromaticity. Again, if there is partial 6π electron aromaticity in 2, due to the contribution of 2a, then it is only very small.

  [...]

  Compared with the corresponding fulvalenes, studied previously, which are genuine push–pull olefins and exhibit partial (anti)aromaticity in the corresponding 3-, 5- and 7-membered ring moieties (in the latter if structurally planar), the 3-, 5- and 7-membered ring moieties in fulvenes 1–4 reveal very small, if not negligible (anti)aromaticity only.

From all of the above I hope I was able to make clear how complex the concept of aromaticity is. Only because of thoughtful investigation and interplay between experiment and theory, penta-fulvene can be described as having negligible (anto)aromatic character.

Notes on aromaticity

The original definition of aromatic (gold book) only states is very broad and may include any and no compounds:

  In the traditional sense, 'having a chemistry typified by benzene'.
  A cyclically conjugated molecular entity with a stability (due to delocalization ) significantly greater than that of a hypothetical localized structure (e.g. Kekulé structure ) is said to possess aromatic character. If the structure is of higher energy (less stable) than such a hypothetical classical structure, the molecular entity is 'antiaromatic'. The most widely used method for determining aromaticity is the observation of diatropicity in the 1HNMR spectrum.
  See also: Hückel (4n + 2) rule, Möbius aromaticity
  The terms aromatic and antiaromatic have been extended to describe the stabilization or destabilization of transition states of pericyclic reactions The hypothetical reference structure is here less clearly defined, and use of the term is based on application of the Hückel (4n + 2) rule and on consideration of the topology of orbital overlap in the transition state. Reactions of molecules in the ground state involving antiaromatic transition states proceed, if at all, much less easily than those involving aromatic transition states.

Much more rigorous is Hückel's (4n + 2) rule and therefore includes much less compounds. The main problem here is, that its application is often taught carelessly or even wrong. When one considers whether a compound is aromatic or not, it is probably one of the worst rules to follow. For fulvenes it certainly leads to the wrong conclusions.
The main problem is that this rule often gets reduced to counting π-electrons, but that is only a small part of it. Even if we include more recent developments and extensions of the rule, there is a lot more to it. (Originally only valid for a couple of hydrocarbons from which it was derived.) I like to encourage you to read up on the full definition (and links within) in the gold book:

  Monocyclic planar (or almost planar) systems of trigonally (or sometimes digonally) hybridized atoms that contain (4n + 2) π-electrons (where n is a non-negative integer) will exhibit aromatic character. The rule is generally limited to n = 0–5. This rule is derived from the Hückel MO calculation on planar monocyclic conjugated hydrocarbons (CH)m where m is an integer equal to or greater than 3 according to which (4n + 2) π-electrons are contained in a closed-shell system. [...]

There is a more updated version on aromaticity in the gold book, which allows a more rigorous approach to the whole subject. Unfortunately it is not as simple as what was there before. You will need to understand a lot more about quantum chemistry, especially how to construct molecular orbitals. While Hückel MO calculations (which you could probably still do with a pencil and a [few] paper[s]) still provide a good entry point and approximation, it is more convenient using modern electronic structure programs and density functional theory (or similar) to elucidate aromaticity.
For the sake of completeness, here is the newer definition:

  The concept of spatial and electronic structure of cyclic molecular systems displaying the effects of cyclic electron delocalization which provide for their enhanced thermodynamic stability (relative to acyclic structural analogues) and tendency to retain the structural type in the course of chemical transformations. A quantitative assessment of the degree of aromaticity is given by the value of the resonance energy. It may also be evaluated by the energies of relevant isodesmic and homodesmotic reactions. Along with energetic criteria of aromaticity, important and complementary are also a structural criterion (the lesser the alternation of bond lengths in the rings, the greater is the aromaticity of the molecule) and a magnetic criterion (existence of the diamagnetic ring current induced in a conjugated cyclic molecule by an external magnetic field and manifested by an exaltation and anisotropy of magnetic susceptibility). Although originally introduced for characterization of peculiar properties of cyclic conjugated hydrocarbons and their ions, the concept of aromaticity has been extended to their homoderivatives (see homoaromaticity), conjugated heterocyclic compounds (heteroaromaticity), saturated cyclic compounds (σ-aromaticity) as well as to three-dimensional organic and organometallic compounds (three-dimensional aromaticity). A common feature of the electronic structure inherent in all aromatic molecules is the close nature of their valence electron shells, i.e., double electron occupation of all bonding MOs with all antibonding and delocalized nonbonding MOs unfilled. The notion of aromaticity is applied also to transition states.

Notes on resonance

I will not go into a lot of detail here, because bon did an excellent job explaining it in: What is resonance, and are resonance structures real? However, allow me to make one point very clear: You cannot treat resonance structures on their own. You always have to treat them as a set, a superposition. There is no such thing as a most stable resonance structure, as well as there is no such thing as one of these structures dictating the reactivity. From a pencil and paper approach you hardly ever can judge which structure is most important for the description of the total bonding. Also, from a simple Lewis-type drawing you can almost never judge the properties of the compound.


Molecular orbitals of heteronuclear diatomic molecules

The rule as generally stated in textbooks is that “atomic orbitals of the more electronegative element are lower in energy than the corresponding orbitals of the less electronegative element”.

Now, MO diagrams are only simple for elements of the second row of the periodic table ($\ce{Li}$ through $\ce{Ne}$). Involving heavier atoms makes it harder to guess at molecular orbital diagrams, and there is need for quantum chemistry calculations. Thus, the rule becomes: the further to the right your element is, the lower its energy levels are. The reason for this is that the charge of the nucleus increases with its atomic number, and thus lowers the energy levels.


Kolmogorov backward equations in active biological systems

UPDATED: I wrote a first answer assuming that $t>0$ which got close to what's in the paper, but not quite the same. Thanks go to Karsten Theis for pointing out that $t<0$ in these "backward equations". Here's a corrected explanation:

The scenario is that we have a system that can be in any of seven consecutive states (numbered 0 through 6). At some time $t<0$, we want to know something about the probability of a system in each state reaching state 6 by time $t=0$, which is expressed as $F_i(t)$ where $i$ indicates the state number.

We start with a system in state 6 at time $t$. Since reversal back to state 5 is not permitted and there is no state beyond state 6, any system that is in state 6 will remain in state 6 up to (and past) time $t=0$. Thus the probability $F_6(t)=1$. In the paper, they use a Dirac $\delta$ function to indicate this, because that is necessary when $F_6(t)$ is used in the other equations.

For the other five states, the equations do not give explicit expressions for the probabilities, but instead express the change in probability over time as a function of the probability in a differential equation.

Let's consider a system that at time $t$ is in state $5$. During a time interval $\Delta t$, two things can happen: Either (1) the system remains in state $5$ or (2) the system switches to state $6$.

One assumption they seem to make is that the only way the probability can change is if the system changes state. So if the system remains in state 5, its probability $F_5(t+\Delta t)$ remains equal to $F_5(t)$. If it changes to state 6, then its probability increases to $F_6(t)$. The change is thus a weighted combination of these two.

In more detail, if we consider the probability of switching from state 5 to state 6 during the interval $\Delta t$ to be $P_{56}$, we can say that our new probability is the weighted sum

$F_5(t+\Delta t)=P_{56}F_6(t)+(1-P_{56})F_5(t)$,

so our change in probability is

$\Delta F_5=P_{56}F_6(t)+(1-P_{56})F_5(t)-F_5(t)=P_{56}[F_6(t)-F_5(t)]$.

Our last step is to figure out what $P_{56}$ is, and it is given by the rate constant for the change from state $5$ to state $6$. The rate constant is the average frequency of the reaction, so the probability of the reaction occurring in a given time interval is the rate constant times the length of time. So we replace $P_{56}$ with $k_{-P}\Delta t$ and get

$\Delta F_5(t)=k_{-P}\Delta t[F_6(t)-F_5(t)]$.

Dividing by $\Delta t$, we have

$\dfrac{\Delta F_5(t)}{\Delta t}=k_{-P}[F_6(t)-F_5(t)]$.

As $\Delta t$ goes to $0$, this becomes the derivative:

$\dfrac{dF_5}{dt}=k_{-P}[F_6(t)-F_5(t)]$.

You can repeat this for the other five states to generate the other equations in the paper. The key point is that even when back reactions are permitted (such as from state 1 to state 0), the probability function is only for the systems that began in the indicated state, so the rate constant for a system coming back to that state does not enter into the equation. An implicit assumption is that the system does not make two moves during the time interval $dt$. That is, it cannot move from state 0 to state 1 and back to state 0.


What causes the smell of something being "musty"?

The "musty smell" is from mold, sometimes bacteria. Different compounds are responsible depending on the species. Some examples: trichloroanisole, tribromoanisole, dimethylmethoxypyrazine. Search "mold smell molecules" for many links.

Anisole is also called methoxybenzene, structure here: https://en.wikipedia.org/wiki/Anisole


How does the number of electrons increase in beta decay?

The two commenters have pretty much answered your question: after beta emission, which is a sufficiently energetic process to tear apart a molecule, the resulting fragments/ions would likely undergo a series of additional reactions with any nearby molecules to finally produce stable species. The emitted electron also would be captured by nearby molecules, leading to a cascade of reactions, also eventually producing stable species. In any case you do indeed end up with a nitrogen atom and an extra electron.

Example: if decay occurred in a bottle of liquid methane, you would get ammonium cation (isoelectronic with methane) and an extra electron loosely attached to methane - a methane radical anion. These could recombine to give neutral ammonia and a hydrogen atom, and then further reactions are possible.

I would also point out that carbon 14 has 6 protons and 8 (not 12) neutrons, but in any case your question remains valid.


3-step process to make propyl ethanoate from propene

My proposal is hydroboration followed by Fischer esterification. It's 2 or 3 steps depending on how many you count hydroboration to be.

Alternatively if you are intent on going through a haloalkane pathway, you could try anti-Markovnikov addition of $\ce{HBr}$ by reacting it in the presence of $\ce{H2O2}$ to prepare 1-bromopropane. Then you could do nucleophilic substitution of $\ce{OH-}$ followed by Fischer esterification. This method would have a lower yield though due to the competing E2 reaction.


Contradiction of general rules of bond character

Although atomic bond in a compound such as $\ce{M+X-}$ (e.g., $\ce{Na+Cl-}$) is considered to be 100% ionic, in reality, it also has some covalent character. An explanation for the partial covalent character of an ionic bond has been given by Kazimierz Fajans in 1923 (Ref.1). According to Fajans, if two oppositely charged ions (say $\ce{M+}$ and $\ce{X-}$) are brought together, the nature of the bond between them depends upon the effect of one ion on the other. In inorganic chemistry, these Fajans' rules are used to predict whether a chemical bond will be covalent or ionic, and depend on the charge on the cation and the relative sizes of the cation and anion, summarized in the following table:
$$
\begin{array}{c|c}
\hline
\bf{\text{Ionic}} & \bf{\text{Covalent}} \\
\hline
\text{Low positive charge }(\text{e.g., } \ce{Na+}) & \text{High positive charge }(\text{e.g., } \ce{Al^3+})  \\
\text{Large cation }(\text{e.g., } \ce{K+}) & \text{Small cation }(\text{e.g., } \ce{Li+})  \\
\text{Small anion }(\text{e.g., } \ce{F-}) & \text{Large anion }(\text{e.g., } \ce{I-})\\
\hline
\end{array}
$$
When the two oppositely charged ions, $\ce{M+}$ and $\ce{X-}$, approach each other, the positive ion ($\ce{M+}$) attracts electrons on the outermost shell of the anion ($\ce{X-}$) and repels its ($\ce{X-}$) positively charged nucleus. This results in the distortion, deformation, or polarization of the anion. If the polarization is quite small, the forming bond between them would be more ionic, while that would be more covalent, if the degree of polarization is large.
The power of an ion (cation) to distort the other ion is known as its polarization power and the tendency of an ion (anion) to get polarized by the other ion is known as its polarisability. Based on the (cation’s) polarizing power or (anion’s) polarisability, the formation of covalent bond is favored by the following factors:

Small positive ion (Cation): The smaller the size of cation, the greater is its polarising power. Due to greater concentration of positive charge on a small area, the smaller cation has high polarising power. Thus, the greater will be the covalent nature of its bond with an anion. This explains why $\ce{LiCl}$ is more covalent than $\ce{KCl}$.
Large negative ion (Anion): The larger the size of anion, the greater is its polarizability, i.e. susceptibility to get polarized. It is due to the fact that the outer electrons of a large anion are loosely held and hence can be more easily pulled out by the cation. Thus the greater will be the covalent nature of the bond with a cation. This explains why iodides (e.g., $\ce{Kl}$), among halides (e.g., $\ce{KF}$ and $\ce{KCl}$), are most covalent in nature.
Large charge on either of the two ions: As the charge on an ion increases, the electrostatic attraction of the cation for the outer electrons of the anion also increases. As a result, the cation’s ability for forming the covalent bond increases (see the similarity with second point). For example, the covalency increases in the chlorides of $\ce{Na+, Mg^2+,}$ and $\ce{Al^3+}$ in the order of $\ce{NaCl < MgCl2 < AlCl3}$.
Electronic configuration of the cation:  For the two ions of the same size and charge, one with a pseudo-noble gas configuration (i.e. 18 electrons in outer-most shell) than a cation with noble gas configuration (i.e. 8 electrons in outermost shell) will be more polarizing. Thus copper(I) chloride is more covalent than sodium chloride although $\ce{Cu+}$ ion ($\pu{1.04 \mathring A}$) and $\ce{Na+}$ ion ($\pu{1.05 \mathring A}$) have similar size and charge.

References:

Kazimierz Fajans, “Struktur und Deformation der Elektronenhüllen in ihrer Bedeutung für die chemischen und optischen Eigenschaften anorganischer Verbindungen,” Naturwissenschaften 1923, 11, 165–172 (https://doi.org/10.1007/BF01552365).
Noam Agmon, “Isoelectronic Theory for Cationic Radii,” J. Am. Chem. Soc. 2017, 139(42), 15068–15073 (https://doi.org/10.1021/jacs.7b07882).


What is the derivation for the expression of depression in freezing point?

To derive it, you have to know how an equilibrium depends on concentration and on temperature.
At the freezing point, pure liquid and the solid are at equilibrium. If you lower to concentration of the liquid by adding solute, you disturb the equilibrium. If you lower the temperature by the right amount, you are back at equilibrium.
The relationship you get is:
$$x_{A}=\frac{\Delta H_{m}}{RT^{\text{2}}}\ \Delta T$$
This is from a General Chemistry text that also does not derive it. With some algebra, you can express $K_f$ in terms of the enthalpy of fusion comparing this and the equation given by the OP.


Calculating the energy used during electrolysis of NaCl

Let's reformulate your equations a little bit.
The ideal gas law was just fine:
$$n = \frac{p V}{RT}$$
A problem lies in your formulation of Faradays law.
We are not interested in masses, but in moles. Of course it is interchangeable by using the molar mass, but why make things complicated. As a rule of thumb you can remember for such tasks: Convert from mass in the beginning and convert to mass in the end (if needed). Nearly all equations in "the middle" are based on the concept of fixed relationsships between the number of involved particles and all equations are much easier to understand if you use $n$.

Faradays law means that the overall charge is just the product of charge per particle times the number of particles:
$$
Q = nzF
$$

The energy is just the product of $Q$ and $U$ divided by $\mu$. You don't need time and amperage.
This gives:
$$
E = \frac{QU}{\mu} = \frac{zFpVU}{RT\mu}
$$

Unfortunately when putting values into these equations you get another result. The following python code will give 448.4 kWh as result. I think that your result is correct and our difference results from differently used constants...:

constants = {}
constants['F'] = 96485.33289
constants['R'] = 8.314
conversion = {}
conversion['to kWh'] = 2.7777e-07

n = lambda p, V, T : p * V / (constants['R'] * T)
Q = lambda n, z : n * z * constants['F']
E = lambda Q, U, mu : Q * U / mu
# Using p=1.01325*10^5 Pa, V=100m^3, T=298K, z=1, U=3.6V, mu=0.88
print(E(Q(n(1.01325e5, 100, 298), 1), 3.6, 0.88) * conversion['to kWh'], 'kWh')


Are these statements about change in enthalpy contradictory?

They are basically same statements. We can use Hess law to elaborate these two statements. For example, we can use the following reaction,
                                 CH4=C+2H2
Here, Delta reaction = Enthalpy of CH4 - Enthalpy of C + 2(Enthalpy of H2).......(1)
We can break the reaction events as,
1) Breaking of 4 C-H bonds
2) Formation of 2 H-H bonds
So, delta H = Energy required to break 4 C-H bonds + Energy required to forming two H-H bonds.......(2)
Now we always take the reactor to be the system and that's why heat input to the reactor is positive and heat going out of the reactor is negative. In the formation of bonds, the energy are released, so the formation energy of H-H is negative with respect to our thermodynamic system.From 2
So,  delta H = Energy required to break 4 C-H bonds - Energy released in the formation of two H-H bonds.......(3)
Now, energy required to break 4 C-H bond= Energy of C + Enrgy of 4 H - energy of CH4
                                        =-Energy of reactants...(4)
Similarly,  energy required to form 2 H-H bond= Energy of 2 H2 - energy of 4 H
                                              =Energy of products (As C is in reference state).........(5)
Adding 4 and 5 yields, equation (1), which proves that both statements are same.


Why does $\text{Cr}^{3+}$ not have the same electron configuration as $\text{Sc}$?

The electron configuration of chromium is : $\ce{[Ar] 4s^1 3d^5 }$ (chromium is one of those examples of special electron configurations : the electron configuration is not $\ce{[Ar] 4s^2 3d^4 }$ as many people might suspect, an electron is moved to the 3d orbital because this configuration is more stable  --> that is the reason why chromium has this special electron configuration)

So the $\ce{Cr^3+}$ ion loses an electron from $\ce{4s}$ and loses 2 electrons from $\ce{3d}$ to give it an electron configuration of $\ce{[Ar] 3d^3 }$

Scandium has an electron configuration of $\ce{[Ar] 4s^2 3d^1  }$ and isn't one of those exceptions.

These are the reasons why the chromium(+III) ion doesn't have the same electron configuration as scandium, despite the fact that they have equal amounts of electrons.

As for why the 4s orbital is emptied first:

  In all the chemistry of the transition elements, the 4s orbital
  behaves as the outermost, highest energy orbital. The reversed order
  of the 3d and 4s orbitals only seems to apply to building the atom up
  in the first place. (Using the Aufbau principle) In all other respects, the 4s electrons are always
  the electrons you need to think about first.

(this link is interesting for you in case you wanted a more detailed explanation about the 4s / 3d problem :http://www.chemguide.co.uk/atoms/properties/3d4sproblem.html )

EDIT : Chromium, just like all other elements, tries to obtain more stable configurations by losing certain amounts of electrons. Now, in the case of $\ce{Cr^3+}$ half/fully filled subshells can't be achieved ( you get $\ce{[Ar] 3d^3}$ ) but that doesn't necessarily mean that it is not stable. But chromium does have other ions in which it does have more stable electron configurations. Take a look at the $\ce{Cr^6+}$ ion : here the ion has an electron configuration that is equal to the one of argon, which does have completely filled valence subshells. ($\ce{[Ne] 3s^2 3p^6}$)

So ions can't always achieve these more stable half/fully filled subshells, but that doesn't necessarily mean that they are not stable.


Test for glyphosate in water, boolean style

The proposed method above most likely uses ELISA (enzyme-linked immunoabsorbent assay). In most modern labs, ELISA is quite a standard procedure, however, each assay is easily worth a couple hundred dollars.

Another method of detection is using HPLC (high-performance liquid chromatography), however those machines can run upwards of a couple hundred thousand dollars and each test is expensive as well.

The EPA has a protocol which they used to detect glyophosphate contamination in water:
https://www.bucksci.com/catalogs/EPA%20Method%20-%20547%20-%20GLYPHOSATE%20IN%20DRINKING%20water%20by%20isocratic%20HPLC%20with%20flourecense%20detector.pdf

What may be of interest to you is an article from the journal of the Royal Chemical Society in which they published an article where they used gold nanoparticles to detect presence of glyphosphate using the naked eye or UV light.

http://pubs.rsc.org/en/content/articlelanding/2013/AY/c2ay26391b#!divAbstract


The IUPAC name of sodium nitroprusside, and back-bonding in it

Nitrosyl is the textbook example for a non-innocent ligand — that is, one that can appear in different oxidation states with different numbers of donated electrons and can make determining the bonding situation rather complicated. It can appear as the nitrosyl cation $\ce{NO+}$ (isosteric to $\ce{CO}$ or $\ce{CN-}$), the nitrous oxide radical $\ce{NO^{.}}$ or the ‘hyponitrite’ anion $\ce{NO-}$ (isosteric to either $\ce{{}^1O2}$ or $\ce{{}^3O2}$). I’m not going to start talking about possible donated electron counts yet.

Nomenclature

For nomenclature purposes, any form of $\ce{NO}$ bound to a metal is always treated as a neutral three electron donor and is always termed nitrosyl. This is because IUPAC wants rules that can be applied blindly without having to perform complicated experiments to determine how exactly the ligand is bound. And we shall see in a second that it is very non-trivial.

Considering that and checking the $\ce{[Fe(CN)5NO]^2-}$ fragment, we take away 5 $\ce{CN-}$ leaving us with $\ce{[Fe(...)NO]^3+}$ which indicates that the correct IUPAC name for sodium nitroprusside is sodium pentacyanidonitrosylferrate(III). (Not (II).)

The product formed would be, by the same approach, sodium pentacyanidonitrosylsulfidoferrate(III). But we’ll come back to that.

But what is the actual ligand?

As I already stated, nitrosyl is a non-innocent ligand. It can be $\ce{NO+}$ and allow significant π-backbonding (metal to ligand) as exemplified in the tricarbonylnitrosylferrate($\mathrm{-I}$) system $\ce{[Fe(CO)3NO]-}$. By formality, this is considered a neutral donor giving iron a formal oxidation state of $\mathrm{-I}$. But in reality the ligand is $\ce{NO+}$ meaning that we are dealing with iron($\mathrm{-II}$) — analogous and isoelectronic to the tetracarbonylferrate($\mathrm{-II}$) complex.

On the other side of the spectrum we have the complex formed in one nitrate detection method: $\ce{[Fe(H2O)5NO]^2+}$ — pentaaquanitrosyliron(II) according to IUPAC. Approaching this complex is difficult, however most evidence points towards some structure between $\ce{[Fe^{II}NO^{.}]^2+}$ and $\ce{[Fe^{III}NO- ]^2+}$. The $\ce{Fe-N-O}$ unit is still linear (VSEPR would predict $120^\circ$ for  a hyponitrite). This can be explained by considering the $\ce{NO-}$ (or $\ce{NO^{.}}$) in a triplet state much like $\ce{^3O2}$ (or a corresponding dublet state in case of $\ce{NO^{.}}$). There is a slight tendency for the unpaired electrons in iron and $\ce{NO}$ to couple; this is usually understood as antiferromagnetic coupling and results in an overall observed spin of $+\frac{3}{2}$.

In the $\ce{NO+}$ case, the bonding picture is rather clear: $\ce{NO+}$ behaves exactly like $\ce{CO}$. We have a σ donating (forwards) bond and significant π back bonding. The other case is rather hazy except for the σ bonding, which is of course ligand to metal. As I wrote, the spin coupling is considered antiferromagnetic coupling and not actual bonding usually, and if anything, it would be a covalent type of bond (each fragment donating one electron). Thus it would be wrong to speak about back-bonding — especially since metals in high oxidation states usually don’t back-bond significantly.

Removing the five cyanido ligands in nitroprusside leaves us with a $+3$ charge as shown above. Thus we are either dealing with $\ce{[Fe^{II}NO+]}$ or $\ce{[Fe^{III}NO^{.}]}$. The extremely short $\ce{N#O}$ distance points strongly towards an $\ce{NO+}$ cation. $\ce{Fe^{II}}$ is again not good at back-bonding and no π forward bonding is to be expected due to the mismatched energy of the π electrons of $\ce{NO}$. Therefore, the $\ce{NO}$ group is rather labile and can be displaced at ambient temperature and pH. This is the mode of action of nitroprusside as a drug: It releases $\ce{NO}$.

And what about the sulfide test?

I didn’t dig into it deeply, but it seems like you are mistaken in the product formed. What I found on the web[1] implies (I couldn’t read the actual article, only the abstract) that sulfur does not coordinate additionally to the iron centre (there would not be much space anyway) but rather attacs the $\ce{N#O}$ triple bond nucleophilicly on the nitrogen atom. Therefore, the final complex is better written as $\ce{[Fe(CN)5(SNO{-}\kappa N)]^4-}$ and better named pentacyanidothionitrito-κN-ferrate(III).

References

[1] S. L. Quiroga, A. E. Almaraz, V. T. Amorebieta, L. L. Perissinotti, J. A. Olabe, Chem. Eur. J. 2011, 17, 4145. DOI: 10.1002/chem.201002322

Background from Prof. Klüfers’ internet scriptum to his coordination chemistry II course.


How to tell whether lone pairs in heterocycles contribute to aromaticity?

For the most common situations use this rule: a conjugated ring atom needs either a ligand or a localized electron pair to replace a missing ligand.  Any remaining electron pairs are then part of the conjugated pi system.

Thus pyridine needs a localized lone pair on nitrogen to replace a missing ligand, and that leaves only three bonding pairs for the pi system.  Pyrrole has a ligand on nitrogen so the lone pair is conjugated into the pi system along with two carbon-carbon bonds.  Furan has two lone pairs on oxygen but a missing ligand, therefore oxygen contributes one pair to the pi system.


Is it possible to make a drug that liquefies heart plaque to treat heart disease?

Plaques are continually being removed from arteries by natural mechanisms within the body.  Statin use, when combined with aggressive dietary changes, can slow down the rate of plaque deposition to the point that the rate of plaque removal is actually higher than the rate of plaque deposition, and, as a consequence, plaque thinning can be observed.

Reducing plaques faster (plaque liquefication) has some potential, life threatening drawbacks.

We've all seen how ice cubes, when frozen together into a clump and then placed in a water bath, dissolve.  Some ice cubes break off from the main structure and float away to dissolve separately.  The same thing can happen with a large plaque deposit.  If a piece of the plaque breaks off during the liquefication process and continues to travel through the blood stream it can lead to downstream blockages, particularly in the brain (stroke).  Rapid dissolution of plaques would increase the risk for such adverse events.
While plaques form more or less everywhere as relatively uniform deposits, the real concern is areas where plaques have grown to large size and begin to block arteries.  It has been found that these larger deposits are initiated by events related to damage of the arterial wall at that specific location. The damage leads to infection (that's why highly sensitive C-reactive protein [hs-crp] measurements are routinely made in conjunction with cholesterol measurements).  The body mobilizes the immune system to fight the infection; microphages and white blood cells are directed to these infection sites.  Once this happens deposition of other materials (lipids, calcium, etc.) start at the infection site and abnormal plaque growth results.  Because of the 1) initial damage, 2) the infection and 3) infection fighting, the arterial wall becomes abnormally thinned at the site.  To some degree, the plaque actually serves to stabilize the thin wall preventing wall rupture and leaks.  Rapid removal of the plaque would increase the chances for the wall to rupture,  producing adverse events and increasing mortality.

So while it is likely that drugs could be designed to remove the plaque extremely fast (e.g. liquefication), the risks might far outweigh the benefits.

  would the liquefied plaque be eliminated as regular fluid is through
  the kidneys?

A major component of plaque is lipoprotein.  Lipoprotein is transported by blood as insoluble particles of varying density and size.  Lipoprotein itself is an assembled complex of proteins and fats.  The fats are various esters of glycerol including esterified cholesterol.  The lipoprotein transports cholesterol and fats, via the bloodstream to cells where these compounds are needed for biosynthesis. After the lipoprotein has released the transported molecules it returns to the liver where it is "reloaded" and sent out again. Lipoproteins are not eliminated (unless damaged) from the body, but are rather recycled.

If a large amount of lipoprotein were suddenly released into the body, I'm sure some would be eliminated through the kidneys.  Certainly the lipoprotein receptors would sense this event and stop synthesis of new lipoprotein until systemic balance was restored. it seems likely that adverse events like kidney damage, gallstones, bile duct clogging, etc. would occur.

Since statins were first released onto the market, drug companies have been considering where to go next.  I have to believe that plaque liquefication was among the options considered.  While the reduction of plaque build up (statins) have their share of problems, I would guess that these problems are minor compared to the possible problems that can result from liquefication (discussed above).  Indeed, with the recent announcements (see here for example) of monoclonal antibodies achieving plaque reductions superior to statins, it seems that drug companies are continuing to focus on avoiding the problem in the first place, rather than fixing things (liquefication) down the road.


How to balance the reaction between ethanol and chlorine using oxidation number method?

The key  is to remembering that  $n \cdot m = m \cdot n$.
It seems obvious, but you would not believe how much struggles people suffer when trying to apply this trivial equation to chemical reactions.
In this special case, The total change of oxidation numbers is zero, otherwise, it would mean we have lost some electrons somewhere.
$$\sum{n_i \cdot \Delta ON_i } = 0$$.
The stoichiometric coefficients are in the reciprocal ratio to ( absolute value of ) the oxidation number change ratio.
So the sum of the oxidation number change of carbons in 1 ethanol molecule is $( (+2 - -3) + ( +2 - -1) ) = +5 + +3 = +8$
The sum of the oxidation number change of chlorine in 1 chlorine molecule is $2 \cdot (-1 - 0) = -2$
We have then:
$$n_{\ce{C2H5OH}} \cdot 8 + n_{\ce{Cl2}} \cdot -2 = 0$$
$$n_{\ce{Cl2}} = 4 \cdot  n_{\ce{C2H5OH}} $$
Note that you are missing negative ions of the left side for charge balance, also overall mass balance is missing.

Enumeration of stoichiometric coefficients have also a general, analytical mathematical solution, but personally, I never used it, as it would take much longer time than just enumerating it on the fly by the chemist's heart:

You are solving a set of linear algebraic equations for multiple variables.
Each variable is a stoichiometric coefficient for a reagent or a product.
Each equation represents some kind of balance.
1 equation for the atom count balance for each of the elements
1 equation for the charge balance, if applies.
In most cases there will remain 1 degree of freedom, so 1 stoichiometric coefficient must be chosen and some others will depend on the choice.
E.g. there can be $\ce{n H2 + n F2 -> 2n HF}$, so we choose $\ce{n=1}$ for $\ce{H2}$  and rest will follow.


How many significant digits are appropriate when one or more digits are eliminated (during subtraction)?

There's an easy way to look at this.

Lets say the value $50.0$ refers to $\pu{50.0 cm}$ measured accurately to $\pu{0.1 cm}$, and that $49.6$ refers to $\pu{49.6 cm}$ measured accurately to $\pu{0.1 cm}$. The difference would be, as you've said, $\pu{0.4 cm}$ measured to $\pu{0.1 cm}$ accuracy.

So, yes, the answer has only one significant digit.

Your initial measurements aren't more accurate than $0.1$, so adding two extra significant digits is incorrect.

I hope this helps.


Developing Computational Chemistry Software

How do I get into method development?

As was said in the comments, one would join a group that works in this field and learn from them. From up close, there is some derivation from previous results and some trial and error involved. And yes, a PhD-sized investment is typically required.

How do I get into method implementation?

It depends on how deep you wish to go. There are open-source projects to look at and contribute to. There are publicly available exercises on implementing HF and beyond. I would not advise to implement DFT from scratch because the numerical integration grid is both critical and difficult. Same is true for the integrals, use a library for them.

How do I learn to use Program X?

It is helpful to learn one program suite well and learn others later. The reason is that many concepts are similar between the programs. Learning one program comes down to locating good worked examples, reading the manual, attending a workshop if possible. One may for instance attempt to reproduce data from a DFT benchmark, since the supporting material usually has a detailed listing, possibly down to absolute energies.


Is my work correct so far?

Yes, your conversions to the units for the inputs to the equation (#1 to #3) look correct to me.

How do I simplify the units in my last step in order to then convert to a value with the desired units...?

You don't.
This sort of expression embodies a purely empirical relationship -- it has no particular basis of connection to the underlying physical relationships between these various quantities.  The reason the units of each of the input quantities are specified ($^\circ\mathrm{F}$, $\mathrm{\frac{ft}{min}}$, etc.) is because you're supposed to use just the numerical values of those quantities, after conversion to the indicated units, and substitute those into the equation. So, instead of the equation you have in #4, it would just be:
$$
H = 0.0061 \cdot \left(0.03^{0.6} + 0.038 \cdot 86\right) \cdot 246^{0.35}
$$
Then, you take the purely numerical result you obtain for $H$ from this equation and tack on the indicated units; in this case, $\mathrm{BTU \cdot {hr}^{-1} \cdot {ft}^{-2} \cdot \ {^\circ F}^{-1}}$. From there, you can carry out a straightforward conversion to the metric units asked for in the final answer.
These sorts of empirical correlations are extremely common in chemical engineering practice, where one doesn't especially care if an equation has solid theoretical underpinnings, as long as it performs reasonably well in predicting quantities of interest.
To note, while you can construct internally consistent units for the various constants in the formula, they usually don't end up having any particular useful physical interpretation. For example, since it's only possible to add two quantities together if they have the same units, the $0.6$ exponent on $X$ means that the $0.038$ coefficient on $T$ would have to have a dimensional value of $0.038\ \mathrm{{in}^{0.6} \over {^\circ F}}$, which has no particular physical significance that I can determine. I'm not even sure what sense there is to be made of units such as "$\mathrm{{in}^{0.6}}$", "$\left(\mathrm{ft\over min}\right)^{0.35}$", etc.


Clear definition of a 'non-oxidizing acid'?

A non-oxidising acid is an acid in which there is no stronger oxidant than $\ce{H+}$ at $\mathrm{pH}\ 0$.

Or, in a slightly different wording, a non-oxidising acid would be one that cannot dissolve any noble metal at $1~\mathrm{M}$ concentration, but should (in the absence of effects like overpotential) dissolve any non-noble metal at that concentration.

Nitric acid has the strong oxidation agent $\ce{NO3-}$ present, which will reduce to $\ce{NO}$ or $\ce{NO2}$ depending on the concentration. It is able to oxidise noble metals such as silver.

Hydrochloric acid, on the other hand, has no other oxidising agent. The chloride ion is already fully reduced. Therefore, it is not able to oxidise noble metals such as copper.

Some acids are more strongly oxidising if they are concentrated. Sulphuric acid is one. Only in higher concentrations will reduction of sulphate to $\ce{SO2}$ take place and make the acid oxidising. In lower concentrations, only $\ce{H+}$ takes part in oxidations.


How to correctly define anode and cathode?

Cathode is the electrode where the Reduction reaction takes place.

Anode is the electrode where the Oxidation reaction takes place


Among thioester and ester which has more electrophilic carbonyl carbon?

Reactivity of thioesters with respect to nucleophilic attack

Why is the thioester bond weaker than a regular ester bond?

Let's consider a reaction mechanism like this:

A steady-state analysis will show that the rate of formation of product is

$$r = \frac{k_1k_2[\text{ester}][\text{Nu}]}{k_{-1} + k_2}$$

$k_1$ for the thioester is larger, because the sulfur 3p lone pair overlaps poorly with the C=O π* orbital (which is formed from 2p orbitals). Recall that esters are less electrophilic than ketones because there is an ester oxygen which has a lone pair capable of overlapping with the C=O π*. In this case, since this overlap is diminished, it turns out that thioesters are typically roughly as reactive towards nucleophilic attack as ketones.
$k_2$ for the thioester is also larger because thiolates are better leaving groups than alkoxides.

Depending on the relative rates of $k_2$ and $k_{-1}$, though, the second point may or may not be relevant. For example if $k_2 \gg k_{-1}$, then the rate equation simplifies to $r \approx k_1[\text{ester}][\text{Nu}]$, in which case the comparison of $k_2$ is no longer useful.


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

  P-14.4 NUMBERING

  When several structural features appear in cyclic and acyclic compounds, low locants
  are assigned to them in the following decreasing order of seniority:

  (…)

  (j) When there is a choice for lower locants related to the presence of stereogenic centers or stereoisomers, the lower locant is assigned to CIP stereodescriptors Z, R, M, and r (pseudoasymmetry) that are preferred to E, S, P, and s, respectively, which are preferred to the non-CIP stereodescriptors cis, trans, or r (reference), c, and t (…).

  (…)

Therefore, the chirality center R receives the lowest locant and thus a systematic name is (2R,3S)-2,3-dibromobutane and not (2S,3R)‑2,3‑dibromobutane.

However, your drawing shows (2R,3R)-2,3-dibromobutane.


Difference between precipitation and crystallisation

The two words are often used interchangeably (perhaps incorrectly), though there is a slight nuance.
IUPAC defines the following:

Crystallisation: The formation of a crystalline solid from a solution, melt vapour or a different solid phase, generally by the lowering of the temperature or by evaporation of a solvent.
Precipitation: The sedimentation of a solid material (a precipitate) from a liquid solution in which the material is present in amounts greater than its solubility in the liquid.

In essence, a crystallisation is the formation of a crystalline solid (i.e. not amorphous), for instance leaving a beaker of $\ce{NaCl}$ in water out for a long time will eventually grow crystals as the water evaporates.
Precipitation can be thought of a something 'crashing out', for instance if you have certain kinds of amines, you can add $\ce{HCl}$ in diethyl ether to precipitate out the $\ce{HCl}$ salt of the amine as they are not soluble.
There are of course in-between cases. Attempts at 'crystallisation' can yield amorphous solids if the conditions are not right, and, equally, precipitation can sometimes give crystalline solids.
The choice of words is therefore always ambiguous, but in general the IUPAC nomenclature should be followed where appropriate (eg. you can never crystallise to give an amorphous solid).
To answer the other part of your question about chemical vs. physical changes, you are correct that crystallisation is always a physical change (different isomorphs of the same material: amorphous, crystalline needles etc). Precipitation can be either, for instance in the $\ce{PPh3}$ case described by Jan, or in the $\ce{HCl}$ amine salt I previously described.


A chemistry approach to environmentally friendly yoga mats… NBR, NR, TPE, PER

You will need a natural woven fiber. Any film will not help the skin in contact with it.


How to determine the molecular formula of a compound given its molecular mass and mass percentage contribution of the involved elements?

The percentages are based on the molar mass. You have to multiply the percentages with the molar mass and divide by the mass of the atoms.

\begin{align}
\ce{C}:&& \frac{0.5999 \cdot 160.178}{12.011} &\approx 8\\
\ce{H}:&& \frac{0.0503 \cdot 160.178}{1.01} &\approx 8\\
\ce{N}:&& \frac{0.3498 \cdot 160.178}{14.007} &\approx 4\\
       && \implies \ce{C8H8N4}\\
\end{align}


Are there any amphoteric substances other than water and ammonia?

First, it may be worth elucidating the nomenclature a bit and distinguishing between amphiprotic and amphoteric  substances. The former are those that can react as either acids or bases in the Brønsted-Lowry sense, while the latter are those that react in either manner per the broader Lewis definition. As you mention, water and ammonia are amphiprotic (and thus also amphoteric by definition), which is evident even from the fact that both undergo auto-ionization:

$$
\ce{H2O + H2O <=> H3O+ + OH-}\\
\ce{NH3 + NH3 <=> NH4+ + NH2-}
$$

In fact, a large number of protic substances exhibit this type of behavior to an appreciable degree, including concentrated solutions of various acids (e.g., sulfuric, hydrofluoric, etc.). Another canonical example of an extremely common amphiprotic molecule prevalent in nature is bicarbonate, which you find in countless consumer products, as well as within the human body as a physiologic buffering agent.

Under the right conditions, a very wide array of molecules of diverse types can be amphoteric. The two necessary and sufficient criteria for amphoterism are:

The molecule possesses lone pairs, so that it can act as a base.
The molecule can accept electron pairs, so that it can act an acid.

Those are not rare properties, hence amphoterism isn't really an exotic phenomenon.


Would dissolving a salt and letting it sit for 48 hours have an effect on the solution?

The only reactions that I would expect to occur to any significant extent with pure water are acid-base reactions between the citrate and water, where water, being amphiprotic, could act as either an acid or a base:

$$\ce{C6H6O7^2- + H2O <=> C6H7O7- + OH-}$$
$$\ce{C6H6O7^2- + H2O <=> C6H5O7^3- + H3O+}$$

Citrate, being the conjugate base of a weak acid, is itself a weak base (according to the citric acid Wikipedia page, the $\mathrm pK_\mathrm a$ for the dissociation of the second proton is 4.75, and for the third between 5.41 and 6.40). Citrate is tripotic, and in the magnesium salt citrate possesses one undissociated acidic proton, which makes it an amphiprotic molecule as well. Given that the $\mathrm pK_\mathrm a$ of water is approximately 15.7, and the $\mathrm pK_\mathrm a$ of hydronium ($\ce{H3O+}$) is −1.74, both reactions above favor the reactants over the products by many orders of magnitude, though the second reaction is favored above the first (i.e., the dibasic magnesium citrate salt should still be very slightly acidic, overall).

Acid-base reactions typically reach equilibrium quite quickly, so there should be very little change in pH and no pH-attributable differences in taste over time. If any change does occur, I would expect it to be the result of a change in concentration as the water gradually evaporates. Another possibility is that, if the water is impure (which tap water certainly is), additional side-reactions might occur. What those are would depend on the exact composition of the water as well as environmental conditions. It's possible for the magnesium to form a precipitate, for example, in combination with other anions dissolved in the water (unlikely given the quantity in your example, but possible). If the magnesium citrate is part of a commercial product containing other ingredients, then it may well be that other agents are responsible for any effects you observe.


Is it possible to separate salt from food, or at least convert it into something that doesn't taste salty?

Even if there is a method to extract salt from food, specially from apple crumble, it’ll probably ruin it.

The simplest way I can think of doing this is dissolving the apple crumble in water and blending it. Filter it and then boil the filtrate.

Given that every solid ingredient but salt is organic, when all the water boils off you should be left with salt and burnt carbohydrates, since NaCl boiling point is way above 1000 °C and a stove will reach 400 °C tops. Now, if you ever burnt sugar you know the caramel sticks to the pan, so if you’re lucky just add water again to separate the salt crystals from the burnt carbs and transfer it to a different recipient. Boil the water to get the salt.

You can add the burnt carbs back to the apple crumble but I don’t think it’ll taste like apple crumble again.


When 3,4-dimethyl-2,4-hexadiene reacts with HBr what is the major product?

Conjugated dienes can undergo two types of additions with HX, they can undergo 1,2-additions and 1,4-additions.  A 1,2-addition occurs when HX adds to only one of the double bonds, not interacting with the other.  a 1,4-addition occurs when HX adds X to one of the double bonds, and because the system is conjugated, the hydrogen will be added three carbons away from X.  The 1,2-adduct and 1,4-adduct are either thermodynamically or kinetically stable.  The kinetic product will form quicker, but is less stable, it will dominate at low temperatures and short reaction times.  The thermodynamic product is more stable but takes more time to form.  It will be the major product at high temperature and long reaction times.  In this case, the kinetic product would be 2-bromo-3,4-dimethyl-3-hexene, the thermodynamic product would be 4-bromo-3,4-dimethyl-2-hexene, because the thermodynamic product has the bromine on the most substituted carbon.  Depending on the reaction conditions, either could form.  At low temps. and short reaction times, 2-bromo-3,4-dimethyl-3-hexene will form, at high temps. and long reaction times, 4-bromo-3,4-dimethyl-2-hexene will form.


How important is it that geometry be optimized at a high level of theory?

Method

Most of the time CCSD(T) would indeed be a huge overkill for geometry optimisation. DFT and MP2 have way better performance/cost ratio. Note that I said DFT and MP2 above, not or MP2; this is well known procedure to compare DFT geometries with MP2 ones to be on the safe side: there should be no big differences.
On the DFT side hybrids might be overkill as well. Try simple pure GGA functionals like BP86 and PBE as well meta-GGA functionals like TPSS first. There might be almost no advantages in using hybrids for geometries.
Include empirical dispersion corrections in DFT calculations even if you do not suspect dispersion forces to be decisive. Just in case.
Employ the density fitting (also known as the resolution of identity) for DFT as well as MP2 to significantly accelerate calculations.

Basis set

With respect to the basis set convergence of geometries is typically reached at the triple-zeta level. This is fairly well-known from excessive benchmark studies for correlation consistent1 and polarisation consistent basis sets2, but Turbomole def2 basis sets should have similar behaviour as well. References are exclusively about DFT, but the same is true for WFT as well.
In practice, though, for DFT calculations (but not necessarily for MP2) a triple-zeta basis might also be an overkill. Try double-zeta def2-SVP first (augmented with diffuse and polarisation functions to your taste). Even Turbomole developers themselves describe DFT/SV(P) and MP2/TZVPP levels of theory as "almost quantitative".

Benchmark

Compare the calculated geometries with experimental data if such are available for your (or at least similar) compounds. At the end of the day, how well does a particular level of theory reproduces experimental geometries is the main criteria of choice.

1) Wang, N. X., & Wilson, A. K. (2004). The behavior of density functionals with respect to basis set. I. The correlation consistent basis sets. The Journal of Chemical Physics, 121(16), 7632–46. http://doi.org/10.1063/1.1792071

2) Wang, N. X., & Wilson, A. K. (2005). Behaviour of density functionals with respect to basis set: II. Polarization consistent basis sets. Molecular Physics, 103(2-3), 345–358. http://doi.org/10.1080/00268970512331317264


Why should all the acid be used up when making zinc sulphate crystals?

I don't think you have really explained the experiment correctly. I'd guess that you're using a diluted solution of sulfuric acid.

$\ce{Zn + H2SO4 ->[H2O] ZnSO4 + H2 ^}$

It would be difficult to have exactly the right amounts of zinc and sulfuric acid, so one or the other will be in excess.

If zinc is in excess the solution can be filter to easily remove the
excess Zn.
If Sulfuric acid is in excess then when all the water is gone you get zinc
sulfate which is wet with sulfuric acid. The sulfuric acid won't evaporate
on a water bath like water.


What are the active ingredients of shampoo, and how do they work?

Shampoo has a few common components: Surfactants, preservatives and a number of additives meant to give the product desirable properties.
When you wash your hair, you're removing debris and buildup in two steps; the polar components (such as salts and aerosols that land in your hair) are washed away by the water, and the non-polar by the surfactants.
Surfactants are molecules where one side is polar and the other is non-polar. Things with similar polarity like to group together, and the opposite happens with polar/non-polar pairs (I'll call this the 'Like attracts Like rule, it's very important to this particular problem).
These surfactants can group together with  each other to form little spheres known as a micelle if there are enough around. This level is known as the critical micelle concentration.

(source: ekshiksha.org.in)
Because the non-polar parts of the surfactants come together, the center of the micelle is also non-polar and can hold small droplets and pieces of other non-polar substances like the sebum that builds up on your hair naturally over time.
Almost all shampoos use a nearly identical set of surfactants. Common examples are sulfates (ammonium lauryl sulfate, Sodium laureth sulfate, sodium pareth sulfate, sodium lauryl sulfate) and/or polysorbates (PEG20 and PEG80).
There are also additives that thicken it up, keep it from going bad and make that rich wonderful lather that we've all come to know and love.
What you are paying for aside from the above is mainly botanical or synthetic additives meant to soothe or beautify. The best way is to look up these ingredients on a case by case basis and decide if they would benefit you. I would say "be wary"; a large number of ingredients provide virtually zero added benefit for your money. Some work extremely well for certain people; you never know until you try.
Always read your labels! If you don't know what something is, google it!


Mechanism of acid-catalyzed Robinson annulation?

As I mentioned in my Comment the only serious issue was the formation of an allylic anion in acid medium. You seem to know how to form enols from ketones and how to form ketones from enols. I will ignore those steps in the diagram. Protonation of the carbonyl in 1 can cause ring fragmentation to form transient enol 2 that rapidly tautomerizes to diketone 3. Cyclization via enol 4 leads to β-hydroxyketone 5. Although loss of water from 5 as you have described is viable, I prefer formation of the enol 6 because loss of water is facilitated over a tertiary cation given the allylic nature of the species 7. Loss of a proton from carbocation 7 leads to 3-methylcyclohex-2-en-1-one (8).
Addendum: The process is not a Robinson annulation, which involves an initial Michael addition followed by an aldol condensation. The latter reaction includes the transformation 3 $\rightarrow$ 8. The transformation 1 $\rightarrow$ 3 may described as a retro-aldol reaction.
You may want to see the formation of mesitylene (1,3,5-trimethylbenzene) from acetone under acid conditions.


How to determine order of nucleophilicity among commonly used nucleophiles?

The relative strengths of nucleophiles can be correlated with three structural features:

A negatively charged nucleophile is always a more reactive nucleophile than its conjugate
acid. Thus $\ce{HO^-}$ is a better nucleophile than $\ce{H_2O}$ and $\ce{RO^-}$ is better than $\ce{ROH}$.
In a group of nucleophiles in which the nucleophilic atom is the same, nucleophilicities parallel basicities. Oxygen compounds, for example, show the following order of reactivity:
$$\ce{RO^- > HO^- >> RCO_2^- > ROH > H2O}$$
This is also their order of basicity. An alkoxide ion ($\ce{RO-}$) is a slightly stronger base than a hydroxide ion ($\ce{HO^-}$), a hydroxide ion is a much stronger base than a carboxylate ion ($\ce{RCO2^-}$), and so on.
When the nucleophilic atoms are different, nucleophilicities may not parallel
basicities. For example, in protic solvents $\ce{HS, CN}$, and $\ce{I}$ are all weaker bases than $\ce{HO^-}$, yet they are stronger nucleophiles than $\ce{HO^-}$.
$$\ce{HS^- > CN^- >  I^- > HO^-}$$

Nucleophilicity versus Basicity: While nucleophilicity and basicity are related, they are not measured in the same way. Basicity, as expressed by $\mathrm{p}K_\mathrm{a}$, is measured by the position of an equilibrium involving an electron pair donor (base), a proton, the conjugate acid, and the conjugate base. Nucleophilicity is measured by relative rates of reaction, by how rapidly an electron pair donor reacts at an atom (usually carbon) bearing a leaving group.

For example, the hydroxide ion $\ce{OH^-}$ is a stronger base than a cyanide ion $\ce{CN^-}$; at equilibrium it has the greater affinity for a proton ($\mathrm{p}K_\mathrm{a}(\ce{H_2O})= 16$, while $\mathrm{p}K_\mathrm{a}(\ce{HCN})= 10$). Nevertheless, cyanide ion is a stronger nucleophile; it reacts more rapidly with a carbon bearing a leaving group than does hydroxide ion.


Instead of glassware, you might have to think about metallic containers. A 1 M NaOH solution should not be that harsh for stainless steel for a short period of time. Imagine, in the chlor-alkali industry, nothing is made of glass. The electrodes are made of titanium as far as I remember. If we were rich, a platinum lined crucible would have worked! Titanium might work too, very resistant to bases.

Here is a useful link: Selecting Stainless Steel for Caustic Soda

Regarding silanes, the Achilles heel is not the perfluoro functional group on the silane but the siloxane bond itself- i.e., the Si-O-Si ---- (functional group). Base will chew this bond apart into pieces even at pH 8-9. This is the biggest drawback of HPLC silica columns. We have not been able to synthesize any base stable silica bonding.


To predict if a substance is deliquescent

There is no general rule, just partial empirical rules for particular anions, which can be memorized from textbooks.

There is theoretical reasoning why salts  are soluble or not, concerning the energy of the crystal lattice, energy of ion hydration, entropy change etc. But easier way is to go for empirical rules and "chemical sense".


What aspects of a macromolecule/protein do the various contributions to its entropy relate to?

I've only really understood entropy to be made up of rotational, vibrational and translational components

There is also electronic entropy, but it's a fun exercise to show that (at least for the hydrogen atom) the corresponding partition function $q_{\text{elec}} = 1$ at room temperature, and for all intents and purposes there is only population of a single state. So, the other 3 kinds of entropy (translational, rotational, vibrational) are those directly from a statistical mechanical derivation that are relevant for most (bio)molecules.

  I've come across a few different contributions to entropy in macromolecules such as proteins: configurational, conformational and vibrational.

Hopefully you agree that vibrational entropy in the context of proteins is the same as in the context of any other system in that vibrations can't be decomposed into other kinds of excitations (translational or rotational), regardless of whether or not the vibration is localized to a bond, highly delocalized, or quaternary structure subunits moving with a regular period.

Configurational and conformational entropy are being used slightly differently. From Wikipedia,

  Conformational entropy – is the entropy associated with the physical arrangement of a polymer chain that assumes a compact or globular state in solution.
  Configuration entropy - is the portion of a system's entropy that is related to the position of its constituent particles rather than to their velocity or momentum.

For proteins, I think the distinction between conformation and configuration is less clear than it is for more molecular systems. In my limited experience, a configuration of a protein is similar to a macrostate in statistical mechanics, which is distinguished from other possible configurations by some criteria related to the position of a protein's primary or secondary structure. More specifically, different techniques can be used for clustering the different protein conformations together into distinct configurations.

In these cases, the greater the number of conformations a protein can take, or the number of configurations it samples based on some metric, the greater the entropy. For something like an intrinsically disordered protein, its conformational and configurational entropy are likely to be higher than for a globular protein.


What is relationship between molecular weight distribution of PE and its strength?

I have found an old paper that might answer your question, which is:

"The Effect of Molecular Weight Distribution on Polyethylene Film Properties"
P. J. Perrono, P. B. Lederman, Polymer Egineering and Science, 1972, 12, 5, 340-345 (DOI: 10.1002/pen.760120505).

The properties of polyethylene evaluated in this paper as a function of the molecular weight include film gloss, haze, tear resistance and film impact strength.
This study shows that broadening the molecular weight distribution increases haze and reduces film gloss. Further, it was shown that a linear relationship exists between film gloss and external haze. Both values are measures of surface irregularities in the film which are affected by the drawing characteristics of the polymer. A broader molecular weight distribution results in increased impact strength as measured by the Dart Drop Impact Test. This is, it is believed, a result of the increase in long chain branching of the higher molecular weight fractions of the polymer which cause a higher degree of molecular weight entanglement at the branch sites. In contrast the tear strength is reduced as the molecular weight distribution broadens because of the low molecular weight fraction in the broad spectrum material which tend to decrease resistance to tear.

Hope this helps.


Do we consider the dissociation of water when dealing with ion concentrations of a solution?

It dissociates into 7 parts; copper and sulfate ions and five moles of water.

Water is also the solvent, so these don't count.

However the answer "two" is indeed not perfectly right, because the sulfate ion would (partially) react with water, to form

$$\ce{SO4^2- + H2O <=> HSO4- + OH-}$$

A sulfate ion is so to say the opposite of sulfuric acid, which reacts

$$\ce{H2SO4 + 2 H2O -> HSO4- + H3O+ + H2O <=> SO4^2- + 2 H3O+}$$

$\ce{HSO4-}$ is not strongly acidic anymore $(\mathrm{p}K_\mathrm{a} = 2),$ so the sulfate ion logically must react somewhat basic. (OK, $\mathrm{p}K_\mathrm{a} = 2$ is quite acidic, acetic acid has $\mathrm{p}K_\mathrm{a} = 4.8.)$

So the correct answer is slightly larger than two. But from the given possibilities in this multiple-choice test, two is by far the most reasonable.

(Update: The hydrated $\ce{Cu^2+}$ ion might well be acidic enough to overcompensate the basicity of the sulfate. In which case you'd still have more than two equivalents of ions, just it'd be a small amount of $\ce{H3O^+},$ not $\ce{OH^-}.$ And of course there is always the autoprotolysis of water.)


Why objects usually don't exibit fluorescence/phosphorescence?

You are correct in most of what you write. Absorption in the visible (and uv) parts of the spectrum are of far larger energy that IR and microwave and the lowest electronic transition involves promoting an electron from the highest populated orbital into the lowest unoccupied one. Once the excited state is formed it can emit via fluorescence and phosphorescence, which are both at a slightly lower energy that the light absorbed and this is because higher energy vibrational levels can be populated in the ground state. (Look for picture of a Jablonski diagram).

However, fluorescence is emitted into all directions ($4\pi$ radians due to molecular motion between absorption and emission) and only a small part of this could enter the eyes but also and most importantly there are non-radiative processes that competitively remove the energy from the excited state ultimately populating high vibrational levels in the ground state from which energy is rapidly lost into the surroundings. These non-radiative events mean that for most types of molecules the fluorescence yield is small, which means that only a few % of the excited state's energy ends up as photons. The non radiative processes are called intersystem crossing between singlet and triplet excited state and internal conversion between states of the same spin state. The processes causing these (besides and chemical reaction) are anharmonicity in molecule vibrations and particularly heavy atoms or unpaired electrons as part of the molecule itself or in nearby molecules such as in a solvent. Some molecules may also have symmetry forbidden optical transitions in emission but not in absorption: carotenes are of this type.


Because "finding two spots" was difficult to understand, my answer to this includes an idealized graphical representation of a TLC.  It consists of two lanes A, and B for two starting materials. There is a reference lane (D) of the already known, confirmed and purified product -- for example obtained by a previous batch of synthesis, or (as met once in lab classes) the TA offers to spot it, using his/her known reference.
If lane C was used to co-spot both the starting material(s) as well the reaction product, yet you do not detect (by UV-Vis, or derivatization) a spot corresponding in $R_f$-value to your reference (in lane D), than your analyte analyzed does not contain significant amounts of your intended product.

Depending on how / if at all a work-up prior to spotting the sample of the unknown was executed, this observation indeed may refer to "the chosen conditions did not yield significant amounts of the intended product".
Yet if you forgot to quench/neutralize (and subsequently extracted) your analyte sample prior to spotting, your product may be still at the ground line -- esp. for salty samples on a silica TLC -- and hence not elute at $R_f = 0.75$. Second potential pitfall I imagine: you quench your reaction with water, and extract at different pH values (separation of strongly acidic / acidic / basic compounds) -- your intended product simply is in the subsequent extract that accidentally was not spot on lane C.

The TLC was idealized that the compounds would nicely separate from each other, without any smearing/fronting/tailing either individually, nor in presence of each other (+ other components of the reaction mixture not detected by the TLC [e.g. solvent(s)]).  Simultaneously, it is assumed there is no decompostion of any of the products while eluting the TLC, that could be checked by 2D-TLC.


Concentrating Isopropyl alchohol using non-iodized salt

Common table salt can have additives, so I would recommend working with high purity Epsom Salts (magnesium sulfate hydrate). Heat to remove water of crystallization at 100 degrees Celsius for two hours.

Add your isopropyl alcohol with the MgSO4 in excess and wait until there is a separation. I would decant and allow for a safety margin above the visible separation line.

Here is a reference.


Is there a difference between parchment paper for writing on and the type you use in cooking?

Parchment paper for cooking is coated with silicone.  Therefore, most traditional writing implements would not even be able to leave a permanent mark on it.

In addition to the coating, at least one brand says it uses additional manufacturing steps to strengthen its cooking parchment paper:

  A spokesperson for Reynolds Consumer Products, says that the company’s
  parchment paper is pressed into a sheet, then dipped into an acid
  bath, washed, and “passed over a series of hot rotating drums that
  realign the fibers and give the paper its strength,” before the
  silicone coating is applied.


First, remember that electrons cannot exist between principle energy levels (distances). In a flame, electrons from lower principle levels (n=1,2,3... etc) will be knocked up to higher principle levels from the absorbed energy. Electrons do not like to stay in this high energy state and will travel back down to their ground state, and in the process release that stored energy in the form of radiation (often visible light). The frequency of the radiation (color of the light) depends upon how big the fall back to ground state was. Different metals have different attainable energy levels, and so will display different colors when burned.

Some elements only have a few possible emission colors, while other elements are a mixture of dozens of colors ! This principle helps us tell what the composition of stars are, as well as allows us to test for metal contents in food products or other applications.


Why does trihaloacetaldehyde formed during the haloform reaction not undergo a Cannizzaro reaction?

Albeit haloform reaction is primarily for methyloketons, aldehydes which have relatively good leaving groups connected with carbonyl (like chloral), or get it during halogenation, indeed also react.

Here we have another factor limiting usage of Cannizzaro reactions, as although intermediate tetrahedral anion (6) could in principle undergo this reaction if $\ce{R = H-}$, it doesn't if there's a better leaving group then $\ce{H-}$. The point of haloform reactions is to make them if they aren't present already.

(source of picture)


Are there alternative ways to find n or number of e- in problems like the one pictured?

In this case, we've to calculate the oxidation states of carbon in the reactant and product side separately, the difference between those two and finally multiply it with the number of carbon atoms. While this is may be a seemlingly simple calculation, things can get uber-tricky because of the mammoth number of exceptional cases. Let's start with the basics; here's a gift for you:

$${\begin{array}{|l|c|} \hline
\pu{Element} & \pu{Oxidation Number} & \pu{Remarks} \\\hline
\pu{Alkali Metals} & \pu{+1} & \pu{Always} \\
\pu{Alkali Earth Metals} & \pu{+2} & \pu{Always} \\
\pu{Flourine} & \pu{-1} & \pu{Always} \\
\pu{Halogen Family} & \pu{-1} & \pu{Generally} \\
\pu{Oxygen Family} & \pu{-2} & \pu{Generally} \\
\pu{Hydrogen} & \pu{+1} & \pu{Generally} \\\hline
\end{array}}$$

Whoa! I see three exceptional cases already (remarked "generally" in the table) and things haven't started yet. What's up with them?

Oxygen: Oxygen doesn't always exist as an oxide ion $(\ce{O^2-})$ in its compounds, it can often-times exist as

Peroxide $(\ce{O_2^2-})$: Oxidation state in this case is $-1$. Popular examples include $\ce{H2O2}$, $\ce{Na2O2}$, $\ce{BaO2}$, etc.
Superoxide $(\ce{O_2^-})$: Oxidation state is $-0.5$. Common examples $\ce{KO2}$, $\ce{RbO2}$, $\ce{CrO2}$
Ozonide ($\ce{O3-}$): $\frac{-1}{3}$ oxidation state; for instance in $\ce{KO3}$.

Halogen atoms (except flourine): When $\ce{Cl, Br or I}$ combine with elements of higher electronegativity ($\ce{O and F}$), then these halogen atoms can show positive oxidation numbers; commonly $\pu{+1, +3, +5, +7}$. In $\ce{BrF5}$, that's an inter-halogen compound, $\ce{Br}$ is in $+5$ oxidation state.

Hydrogen: In case of metal hydrides,  it can show $-1$ oxidation state. As in $\ce{NaH}$.

Phew, er any more exceptions? You dare ask? Metals. Wait! Aren't metals electro-positive!? Yes, they are, but in some cases they can show zero or even negative oxidation numbers. For example in $\ce{CsAu}$, gold shows a $-1$ oxidation state. Let's bring in coordination-compounds: nickel in $\ce{[Ni(CO)4]}$ finds itself in zero oxidation number.

There are plenty of them, I'm saving them for another day.

Let's try calculating things you've asked.

For ethanol ($\ce{C2H6O}$) we don't know the oxidation state of the carbon atom. Why don't we assume it to be $x$. Now, the oxidation state of a compound is either zero (if it's a neutral species) or it's equal to the charge it bears. Is ethanol neutral? Yes it is. So the individual oxidation numbers of elements (times the number of elements bearing that oxidation state) should sum up to be zero.

$$2x+6(+1)+1(-2)=0$$

Solving, we get $x=-2$. Doing the same thing for carbon-dioxide, we can get $x=+4$. So a carbon atom changed its oxidation state from $-2$ to $+4$, that's a jump of $+6$. And how many carbon atoms were doing this? Two carbons. That gives us $n=12$.