Where can I obtain tables of P-V-T/compression factor data?

One place to look for simple fluids is http://webbook.nist.gov/chemistry/fluid/.  But there are only a limited number of compounds treated.

Be warned that the data in the NIST Webbook is smoothed.  Collecting data of this sort is very difficult.  Many different sets of observations are combined and fitted to an appropriate equation.  What is at the URL given above is computed from that equation.

If you are looking for overall compression factors, this will do very nicely.  If you want five figure precision in compression factors, well, this won't do.

Another place to check is the  which is a primary source of raw data.


How did Halogens become known as Halogens? They are not the only elements that form salts!

This seems like a bit of a rhetorical question, so this isn't a terribly formal or authoritative answer, but anyhow - a lot of chemical nomenclature is like lava flow. It solidified and people just worked around it.

The halogens are so named because they have a rich chemistry of ionic compounds (fluorine through iodine, anyhow). However, both the halogens and the group I metals can form a wide range of things that aren't salts. The noble gases can form compounds with elements of low birth. The rare earth elements aren't particularly rare. Oxygen ('acid-former') is not a necessary component of acids. Technetium ('artificial' + ium) is produced in nature in significant quantities.

In addition, a lot of chemistry defies our human efforts to succintly categorise things, so at some point chemists have tried their best to find pragmatic general descriptors that unite groups of elements or molecules on the basis of the properties or constitution. The divide between organic and inorganic chemistry and the resulting exceptions and edge cases in classification (such as mellitic anhydride) is a good example of this. The map is not the territory.


How to differentiate magnetite from metallic iron?

These are the methods that come to mind:

LIBS
XRF
Streak color. When scratching on unglazed porcelain, magnetite leaves charcoal-black traces, whereas iron leaves shiny steel-gray streaks. Having a powdery mixture this test can be tricky, unless there are reasonably big shards left. Alternatively, one can try to assemble a bulky piece by gluing the powder together with silicate adhesive or cyanoacrylate (as paracetamol suggested in the comments, "rubbing some of the mysterious powder between two shards of porcelain" can also do).
Compare density of your sample with $\rho(\ce{Fe}) = \pu{7.874 g cm-3}$, $\rho(\ce{Fe3O4}) = \pu{4.9 .. 5.2 g cm-3}$ and, assuming this mechanical mixture consists of iron and magnetite exclusively, calculate the percentage of each phase.

Magnetite in its mineral form often contains $\ce{(Fe,Mg)O}$, $\ce{TiO2}$ and isomorphous impurities of $\ce{Mn}$, rarely $\ce{Cr}$, $\ce{V}$; therefore physical methods of analysis (1, 2) are more preferable. Due to the same reason I'm not suggesting treatment with $\ce{HCl}$ and other methods based on chemical transformations as primary for this analysis.


Why does brown sugar have a strong smell but white sugar doesn't?

Sugar is made by repeatedly boiling and cooling cane syrup (or sugar beet syrup).  After each cooling, the solution becomes supersaturated with respect to sucrose, causing sucrose to crystallize out of the "mother liquor" (the industrial term for the liquid solution from which crystals form). These crystals of "raw" sugar are heavily processed (washed in syrup, dissolved, treated with carbonate or phosphate minerals, recrystallized, and subject to ion exchange or charcoal color removal) to result in white sugar.

Nowadays, brown sugar is usually made by mixing refined white sugar with blackstrap molasses. In former times, the brown sugar was simply unrefined or unwashed crystal fraction from the 2nd or 3rd crystallization of cane juice. But precise control over the ratio of "contaminating" liquid to white crystalline sugar was difficult, which is why industry moved towards reconstituting brown sugar by mixing molasses and white sugar.

Blackstrap molasses is simply the spent "mother liquor" remaining after the third cycle of boiling and cooling to produce sugar crystals.

The smell of brown sugar thus comes primarily from molasses.  The smell of molasses is very complex.  Here is the abstract of a relevant paper:

  The characteristic sweet-aroma components in refinery final molasses were isolated by using a combination of sensory evaluation, column chromatography and gas chromatography, and were identified by infrared spectrometry, mass spectrometry and color reactions. Identified were ethyl n-hexanoate, ethyl phenylacetate, phenylacetic acid, n-butyl benzoate, isopropyl benzoate, l(+)-pantolactone, benzoic acid, o-toluic acid, m-toluic acid, β-phenylpropionic acid, succinic anhydride, maltol, isomaltol, and 2-methyl-5-hydroxy-6-ethyl-γ-pyrone.

  The aroma of the mixture of these compounds and vanillin and vanillic acid was similar to that of refinery molasses by sensory evaluation. The results indicate that these compounds are important in producing of sweet molasses aroma.


C-F bond lengths of octafluoropropane using Bent's rule

It is very much non-trivial to attempt estimations on a perfluorated alkane using Bent’s rule. In Bent’s rule, one should not only consider the effects of the electronegative substituents on the substituted atom itself, but also onto the next atom. Consider for example the following table taken from Bent’s review article:[1]

\begin{array}{lr}\hline
\text{Compound} & \angle(\ce{R^1-O-R^2}) \\ \hline
\ce{Me-O-Me} & 111^\circ \\
\ce{Me-O-H} & 107{-}109^\circ \\
\ce{H-O-H} & 104.5^\circ \\
\ce{F-O-F} & 103.8^\circ \\ \hline\end{array}

This very nicely corresponds to increasing group (!) electronegativities[2] in the series $\ce{Me} > \ce{H} > \ce{F}$. $\ce{CF3}$ should have a higher electronegativity than both $\ce{Me}$ and $\ce{H}$ (compare the acidities of formic acid, acetic acid and trifluoroacetic acid). Yet $\angle(\ce{C-O-C}) = 120^\circ$ in $\ce{F3C-O-CF3}$.[3] This corresponds to a higher $\mathrm s$ contribution than for the previous groups indicating other, more important contributions (here the anomeric effect of oxygen acting on the $\ce{CF3}$ group).[4]

Luckily, there are no such secondary effects in perfluoropropane. However, if you want to estimate all bond lengths compared to propane or 2,2-difluoropropane, these must be considered. Specifically, going from propane to 2,2-difluoropropane should not only reduce the $\ce{C-C}$ bond lengths due to increased $\mathrm s$ contribution but also due to partial double bond character due to a gauche-type effect which can be rationalised in Lewis formalism by the following resonance structure:

$$\ce{H-C-C-F <-> \overset{+}{H}\bond{...}C=C\bond{...}\overset{-}{F}}$$

Of course, the partial $\ce{C-C}$ double bond and the less-than-single $\ce{C-F}$ bonds also mess with the observed bond angles in 2,2-difluoropropane; they don’t only influence bond lengths.

So to attempt an estimation of the differences in bond length of the two $\ce{C-F}$ bonds in perfluoropropane, we would need to know how the group electronegativites of the $\ce{CF3}$ and $\ce{C2F5}$ groups compare to the electronegativity of fluorine and to each other. I don’t see any simple way to do this.

Mack et al. have published gas phase structures of 2,2-difluoropropane and perfluoropropane. These are their results (all values in picometres):[5]

\begin{array}{lcc}\hline
\text{compound} & d(\ce{C-C}) & d(\ce{F2RC-F}) & d(\ce{R2FC-F}) \\ \hline
\ce{Me2CF2} & 151.2 & \text{–/–} & 137.0 \\
\ce{(CF3)2CF2} & 154.6 & 132 & 134 \\
\ce{Me2CH2} & 153.1 & \text{–/–} & \text{–/–} \\ \hline \end{array}

As you can see, the experimental values just confirm your prediction of the central $\ce{CF2}$ having longer bond lengths than the terminal $\ce{CF3}$ groups. As expected, all $\ce{C-F}$ bonds in perfluoropropane are shorter than the bond lengths of 2,2-difluoropropane. And the $\ce{C-C}$ bond lengths increase from 2,2-difluoropropane to propane to perfluoropropane.

The conclusion is that indeed $\ce{CF3}$ is slightly less electronegative than fluorine, but the effect seems minimal.

References and Notes:

[1]: H. A. Bent, Chem. Rev. 1961, 61, 275–311. DOI: 10.1021/cr60211a005.

[2]: Bent explicitly considers group electronegativities. Only by that is the group electronegativity of a methyl group lower than the electronegativity of hydrogen.

[3]: A. H. Lowrey, C. George, P. D'Antanio, J. Karle, J. Mol. Struct. 1980, 63, 243–248. DOI: 10.1016/0022-2860(80)80330-9.

[4]: F. R. Leroux, B. Manteau, J.-P. Vors, S. Pazenok, Beilstein J. Org. Chem. 2008, 4, 13. DOI: 10.3762/bjoc.4.13.

[5]: H.-G. Mack, M. Dakkouri, H. Oberhammer, J. Phys. Chem. 1991, 95, 3136–3138. DOI: 10.1021/j100161a034.


Does a salt always have a neutral charge?

A salt is an ionic solid compound. As such, it needs to have a zero net charge.

However, that doesn't mean it should contain only one cation and one anion. The stoichiometry of the salt depends on the charges of the ions. For instance, iron (III) sulfate is $\ce{Fe2(SO4)3}$; this stoichiometry is required to balance the charges from $\ce{Fe^3+}$ and $\ce{SO4^2-}$.

Also, other types of salt exist, called mixed salts. They contain several different cations and/or anions (e. g. $\ce{KAl(SO4)2}$).

Compounds with a net charge exist, they are just not called salts (look up Complexes).

  In relation, are acids and bases charged?

Acids may (e. g. $\ce{NH4+}$) or may not (e. g. $\ce{H2SO4}$) be charged. Similarly, bases may (e. g. $\ce{SO4^2-}$) or may not (e. g. $\ce{NH3}$) be charged.


I mean doesn't every compound has its own different reduction and oxidation potential ? Or its just one potential that can be compared to understand which one will get reduced and which one will get oxidised ?

A compound has neither. You need to specify the half reaction. For example, take an aqueous solution of $\ce{Fe^2+}$ and hydrogen peroxide, among others. What are possible half reactions for $\ce{Fe^2+}$?

$$\ce{Fe^2+ -> Fe^3+ + e-}\tag{1}$$
$$\ce{Fe -> Fe^2+ + 2e-}\tag{2}$$

and maybe some more exotic ones. For hydrogen peroxide, you would have e.g.

$$\ce{H2O2 -> O2 + 2e- + 2H+}\tag{3}$$
$$\ce{2H2O -> H2O2 + 2e- + 2H+}\tag{4}$$

Notice that these are all written in the direction of reduction, and are associated with a reduction potential. To see which reactions are feasible, you would compare reduction potentials and come up with a redox reaction combining a half reaction as written with a half reaction (the oxidation half reaction) in the reverse direction. For example, hydrogen peroxide could undergo disproportionation:

$$\ce{2H2O2 -> O2 + 2H2O}$$

This is a slow reaction, but the enzyme catalase (which happens to have an iron(III) bound to its active site) increases the rate substantially, as you can see from the oxygen evolving when you put some hydrogen peroxide on a freshly cut slice of potato.

  Is redox potential a valid term?

Yes, under certain circumstances it makes sense to talk of redox potential as a property of an aqueous solution, similar to discussing its pH. For example, if a biological cell contains glutathione at fairly high concentration in the reduced and oxidized form (abbreviated as GSH and GSSG), this will determine the redox state of other substances in the cell. If a strong oxidation agent is introduced, it will react with GSH. If a strong reduction agent is introduced, it will react with GSSG. Like mainining a pH with a buffer, the redox potential of the cell is maintained through the ratio of GSH and GSSG (and we could call it a redox buffer).

The analogy between pH buffer and redox buffer only goes so far. The biggest difference is in the role of water. While water acts as an acid and base at any pH (i.e. whenever acids and bases are added), it does not undergo reduction and oxidation for many redox reactions in aqueous solution. Any pair of acid and base will react (because water provides the "reservoir" of hydrogen and hydroxide ions). Many pairs of possible oxidizing and reducing agents will not react, even when thermodynamically favored because there is no good mechanism for the electrons to be transferred (a question of kinetics).


Is the term "heteroatom" reserved for exclusive elements?

The IUPAC Goldbook provides no 'official' definition of the term heteroatom,and as such you're somewhat free to use (and abuse) the term as you see fit.

Since no IUPAC definition exists, it may be helpful for you to define what you plan on calling a heteroatom at the beginning of whatever you're writing, to inform the reader (common practice in scientific literature for poorly defined terms). Clearly an organic chemist would have a very different definition to someone involved in purely metallic synthesis.

I would however, point to several other IUPAC documents in which, whilst no 'definition' of a heteroatom is given, they do suggest lists of possible heteroatoms. For example, the following statement from the IUPAC nomenclature guidelines which includes many elements (including metallic ones, possibly going against your suggested definition).  This priority order of heteroatoms is repeated multiple times.

  P-22.2.2.1.3 If two or more kinds of heteroatoms occur in the same name, their order of citation follows the sequence: F, Cl, Br, I, O, S, Se, Te, N, P, As, Sb, Bi, Si, Ge, Sn, Pb, B, Al, Ga, In, Tl, Hg.


The rough texture is a physical catalyst (provides nucleation sites) for the evolution of carbon dioxide from dissolved carbonic acid. Pottery glaze fixes the foaming issue.


Test for diethylether or tertiary alcohol?

As mentioned in the comments, using Lucas' Reagent is the best option in this case because you're dealing with a tertiary alcohol vs an ether, but let's consider your proposal as well.

In principle, it should work because an ether will not react with ethanoic acid whereas an alcohol would. However, the formation of an ester simply by the addition of ethanoic acid to alcohol is not likely to complete (you generally need a dehydrating acid like $\ce{H2SO4}$ to drive the reaction). In other words, when you put in the bicarbonate you will most likely see effervescence.

If you're thinking of looking for effervescence, using thionyl chloride $\ce{SOCl2}$ might be an option since it will react with the alcohol to give a sulfur dioxide gas and HCl:

$$\ce{R-OH + SOCl2 -> R-Cl + SO2(g) + HCl}$$

You have to make sure the solvent is dry though.

In the specific case of t-Butanol vs diethyl ether, though, you could simply heat the solution slightly to about 40 degrees Celsius. If it boils, then you've got diethyl either. If it doesn't, it's t-Butanol.


Why does basic peroxide react via conjugate addition rather than through direct addition?

The reason that is given by Clayden et al. (chapter 23) is that because the peroxide has two adjecant lone pair on oxygen it exhibits an alpha effect, which is more or less a delocalization interaction between the lone pair orbitals. For reasons not entirely known (the basis of the alpha effect is an open problem in chemistry) this results in an increase in energy of the electrons in the resulting delocalized HOMO, making it more nucleophilic. You can imagine that this delocalized HOMO is also softer than the single localized oxygen lone pair and hence it tends to react at the soft double bond instead of the hard carbonyl site.

Excerpt from Clayden et al.


UV/Vis spectroscopy transition reference

Here are a number of resources that may be of use for your research.

The UCLA website presents tables and data in their webpage Theory of Ultraviolet-Visible (UV-Vis) Spectroscopy, including reference tables of common functional groups, aromatics and compounds often used in the laboratory.  (Many compounds you listed are in the tables).

A reference that may be of use is Spectroscopy of Organic Compounds (Kumar, 2006) which include several tables that include information of the compounds you list.

The book Spectroscopy of Organic Compounds (Kalsi, 2007), from page 12 provide several tables of information related to your listed organic compounds, including details of the electronic transitions. The chapter "Ultraviolet (UV) and Visible Spectroscopy" (starting page 9) has a subsection "The Designation of Various Transitions" which details the electronic transitions using examples in context (from page 11, summarised in a diagram on page 12).  Examples of solvents are discussed from page 23 onwards.  (All available online).

Transitions are explored in a bit of detail in chapter 4.3 of the UC Davis ChemWiki The Dynamic Chemistry E-textbook, Section 4.3: Ultraviolet and visible spectroscopy


What databases do you have access to? A Scifinder search would find literature references with no difficulty.
A Google search finds this technical report which has the synthesis of tetraphenylporphyrin and the preparation of Fe(II) and Zn(II) complexes report here.
The Cu(II) complex is prepared in this lab report here. The references in these reports should help further.
The references in this JACS paper here should lead you to other preps


What are the main differences of CTAB and DTAB surfactants? And what are practical applications of them?

CTAB, aka Hexadecyltrimethylammonium bromide

DTAB, aka Dodecyltrimethylammonium bromide

Sort of a vague question, don't know where to start. To begin, CTAB has a hydrocarbon chain length of 16 (as its name suggests) whereas DTAB has a -CH2- chain length of 12.

Otherwise they have the same hydrophilic/polar "head"; all this is obvious, you can tell just by the two compounds' names. They also have the aforementioned hydrophobic "tails", which refers to the hydrocarbon chains because they are non-polar.

CTAB and DTAB can be used to create micelles. It's like how a biological cell is separated from its environment with a phospholipid bilayer where the hydrophobic and hydrophilic components of the phospholipids align such that you get an enclosed structure that separates a non-polar "organic" region in a polar environment (e.g. water). The wikipedia link has good diagrams.
The reverse is called a reverse micelle, where the heads create a polar region separated from a non-polar environment (e.g. when the solvent is hexane).

For one example of practical usage, I did a lab just today using hexane as a solvent where I used CTAB to create an environment to limit the formation of CdS quantum dots. This worked by limiting the volumes where conditions aren't non-polar. In this case, reverse micelles were formed; tails pointing out, head surrounding to form a more polar center.

Another: In DNA extraction CTAB is used to pick up insoluble particles (e.g. polysaccharides) and exit the working solution.  You can relate surfactants to soap, which picks up insoluble (i.e. oily) substances and can easily be removed by water due to its hydrophilic outer layer, while containing the oily substances in its hydrophobic center.


How do the two half cells of Daniel cell know when they are connected?

No, electrodes do not know about each other.

When an electrode is inserted to an electrolyte, the electrochemical reaction is ongoing in both directions. If reduction direction overruns oxidation, the potential of the electrode is increasing ( or vice versa ) until the rate of both reaction gets equal, the net reaction rate is zero and the electrode reaches the equilibrium potential.

That may be quick or slow, depending on the electrode and respective reaction kinetics.

When both electrodes are galvanically connected, the electron exchange due the potential difference disbalances the equilibrium potentials of both electrodes and the net electrochemical reactions get the steady non zero rate.

Note that the electrode potential is conventionally referred to the Standard hydrogen electrode (SHE), which itself has potential $\pu{+4.44 \pm 0.02 V}$ ( Wikipedia ) wrt the potential of a free electron in vacuum. So saying $\ce{Zn}$ electrode potential is negative rather means it is less positive.

You may want to try searching various terms in wikipedia and following relevant internal/external links. The great source of compactly explaned topics is hyperphysics, but with limited scope. It is a kind of hyperlinked quick cards/cheat sheets, very handy, if you want to understand the basics.  This link is particularly for Electrochemistry


Why is this stereocenter by the Br an R? I thought it was S?

Both structure (a) and structure (b) have misassigned chirality centers. There should be only one $R$ assignment over all four structures. The correct assignments are as follows:

  Structure (a)

The misassigned chirality center is the one in the center. The carbon atom with the OH group has the highest priority (one C-O bond beats any number of C-C or C-N bonds). The second highest priority is the pyrrole ring (one C-N bond and one C=C bond beats one C-N, one C-C, and one C-H). The carbon atom with the primary amine is the third priority. The groups are arranged counter-clockwise and the lowest priority (H) is in the back.

  Structure (b)

The misassigned chirality center is the one with the bromo group. The bromine atom is the highest priority. The carbon atom bonded to chlorine is second, and the carbon atom bonded to fluorine is third. The groups are arranged clockwise, but the lowest priority (H) is forward.


Usage of ammonia as odorant in fuel gases

In principle, you could use ammonia, but there are practical disadvantages.

The sulfur-based odourants are some of the smelliest compounds known and are detectable in very low concentrations. In practice, they are used in concentrations of perhaps 30ppm. Several percents of ammonia would be thousands of times more than this concentration implying greater expense. At this level, the concentration of ammonia would be problematic because of the corrosive nature of certain types of pipe and would probably create other safety concerns (it might even be a direct safety hazard at the concentrations where it is readily detectable by smell).

Typical sulfur-based odourants are detectable at such low concentrations that they are not a direct safety hazard. Nor is there much worry about the combustion products at such low concentrations.

Ammonia is also much more reactive than the typical sulfur-based odourants and would be much more likely to be stripped from the gas during transmission which defeats the purpose.


Does Henry's law depend on curvature of gas/liquid interface?

No, but not because the vapour pressure of mixtures doesn't depend on curvature (it does), but because Henry is a narrowly-defined law that doesn't contemplate surface curvature.

Henry's law is, after all, a barely better description of vapour pressure in mixtures than Raoult, and that's because it introduces an experimental value (the Henry constant) for each solute/solvent pair. The only reason why it's (rarely) used is that it's extremely easy to operate with. And pretty much the only reason why it's taught is because it's a simple, mathematically trivial treatment that moves away from ideality; it's sort of the minimum, simplest step away from ideality, which is convenient for a classroom.

But if you attempt to combine Henry with the effect of surface curvature, say by combining it with Kelvin's equation, you run into a serious problem. The equilibrium difference of pressure between a liquid, $p^L$, and its vapour, $p^V$, when the liquid has a curvature of radius $r$ is

$$ p^L = p^V + 2\sigma/r$$

where $\sigma$ is the surface tension of the liquid. And that's when you run into a problem - surface tension of a liquid mixture also depends on the composition, so the entire point of Henry's law (that concentration of a species in solution is proportional to its partial pressure with a proportionality constant $H^{cp}$) is lost. Even if you accept the simplest case (that surface tension behaves ideally with composition following Gibb's isotherm) you are stuck with a $H^{cp}$ that is itself a function of $c$, and so the mathematical model has definitely moved away from Henry.

By which point you are better off using other, more realistic models for your non-ideal mixture. Note that even then, you are stuck with iteratively computing the crossed effects of surface tension on vapour pressure and vice versa, as the relationship can't be solved analytically as far as I know.


Here is my sage advice to you, Silvia95. Do not redraw structures!!! Use the method described here. Transcribing structures to place the lowest priority group in the rear will invariably lead to errors. Leave the structure where it is and use your hands as tools to determine chirality. Make a "thumbs up" gesture with either hand and point the thumb from the asymmetric carbon to the lowest priority group. Your fingers should point in descending order from highest priority to the next to lowest priority. If not, you are using the wrong hand. Right hand, R; left hand, S.C2 of dibromide a has the hydrogen in the rear. The fingers of your left hand show Br>C3>C1. C2 in dibromide b has the hydrogen in the front. Now the thumb of your left hand points toward you and Br>C3>C1. Using the same technique for C3 of a and b, your right hand fingers reveals Br>C2>C4. These two structures are identical, (2S,3R)-2,3-dibromopentane. If you rotate either structure by 180o about a vertical axis in the plane of the screen, the two structures will superimpose.


The naming conventions for atoms in PDB file require that at the C-terminus of a protein one of the carboxylate oxygens is called O and the other is called OXT.  But since it's a deprotonated carboxylate group they are both chemically equivalent, and you'd expect that bond lengths are equal.


Density usually specifies a ratio (or equivalent derivative) of the type "amount/unit dimension" where dimension is usually one of (distance, area, volume, hypervolume). Based on further examples of usage (see links in Todd Minehardt's answer) of the expression "density per unit dimension" where "dimension" might be "distance", "area", "volume" or other spatial dimension, the point with writing "density per unit volume" seems to be an attempt to be explicit about the dimension in the divisor (denominator). For instance, when writing  "density per unit dimension"  it is meant that "density=amount/dimension". Thus "density per unit volume" means "density=amount/volume". One could equally well say "volume density" or "area density" to be specific, but there might be ambiguity there as well, as one might confuse usage with expressions such as "number density" or "energy density" where the property in the numerator is being specified. Then "volume density"  might be interpreted as meaning "density of volume", even if farfetched. One could specify dimensions by placing the units in parentheses, for instance, as density (per unit volume) or density (per unit area) etc.


Why should alkylation of enols/carbonyl compounds be done at low
  temperature with a strong base?

If you start with an unsymmetrical ketone, then there are two different sides of the molecule you could remove a proton from. Let's use methyl ethyl ketone as an example.  Removing a proton from the methyl group leads to what is termed the kinetically favored anion (e.g. this hydrogen is the most acidic and the corresponding enolate anion is the fastest formed anion).  Removing a proton from the methylene group on the ethyl side leads to the thermodynamically favored anion (e.g the enolate anion with the most highly substituted [most stable] double bond).

Alkylation of carbonyls/enols is not always done using strong base and low temperature.  Strong base and low temperature generally leads to the kinetically favored enolate anion, while weaker bases and higher temperatures generally lead to the thermodynamically favored enolate anion.

In your case, acetone is symmetric so it doesn't really matter which side you alkylate.  But, once you form your desired product, methyl ethyl ketone, it can react further to give polyalkylated products.  These secondary products are more stable and thermodynamically preferred.  To reduce the likelihood of thermodynamic control leading to these undesired products, the reaction is run under kinetic control - strong base and low temperature - to maximize formation of the monoalkylated product.  Here is an earlier answer that further highlights kinetic vs. thermodynamic control in alkylation reactions.


The "rubber", or elastomer, likely has plasticizers to soften the material and make it more elastic and less brittle. These often migrate to the surface as parts age. Some of these plasticizers for elastomers include:

Sebacates, similar to animal fats
Adipates, based on adipic acid, which is found in beets
Phthalates, esters of phthalic anhydride
Some are innocuous, but others, such as bisphenol A (BPA) may have long-term toxic effects. Washing the parts, as you have done, should remove them from the surface. You might need to repeat washing, using different detergents.

Another possibility is depolymerization, actual breakdown of the plastic. At room temperature, this process is slow, though it could contribute to that sticky surface layer.
As for safety, you might check on rules at the time and place the toys were made. Modern toys are made with safety in mind, but plastics have had a bad rap in the past, such as causing liver damage to workers in vinyl production. Caveat emptor.


Mechanism of the reaction of cyclohexanone and nitric acid to form Adipic acid

It think this reaction goes via following mechanism.


Calculating equilibrium partial pressures of a gas given Kp

In general $k_{p}$ is not a dimensionless quantity. In the case of the reaction $$\ce{2NO + Br2 <=> 2NOBr}$$ the unit of measurement for $k_{p}$ is in fact $$\frac{pressure^2}{pressure^2 * pressure} = pressure^{-1}.$$ Thus the value 28.4 of the equilibrium constant given in the assignment is incomplete.

Given the correct answer, it turns out that $k_p$ is actually given in $\mathrm{atm^{-1}}$. Since $$\mathrm{28.4~atm^{-1}} = 28.4*(\mathrm{760~Torr})^{-1} = 0.0374~\mathrm{Torr}^{-1}$$ you should use this value for $k_p$ if you prefer to do your calculation with $\mathrm{Torr}$.


How to identify the structure of an organic acid based on its reactivity, mass and NMR spectrum?

I'm at a loss about how to deduce the molecular formula.

Compounds containing only C, H, and O have a certain number of hydrogens (h) compared to the number of carbons (c).

If there are no double bonds and no rings, h = 2c + 2. This is true for linear chain alkanes. Adding oxygens to those (hydroxyl groups, ethers) does not change the number of hydrogens. Turning the linear chain to a branched structure does not change the number of hydrogens.

For every double bond you add, you lose two hydrogens (think of it as turning sigma bonds to hydrogen into pi bonds between the heavier atoms). For every ring you close, you lose two hydrogens (one additional C-C bond instead of each carbon bound to a hydrogen, for example).

So for molecules with double bonds (d) and rings (r), you get:

h = 2c + 2 - 2d - 2r

  All you know is that D contains a C=C, benzene ring and most likely a -COOH group

So the benzene ring accounts for three double bonds and a ring, the C=C for a double bond and the -COOH for a double bond (C=O). At this point, we don't know if there are even more double bonds or rings, so we have d >= 5 and r >= 1 and the following inequality

h = 2c + 2 - 2d - 2r < 2c - 10

So the molecule will not have the maximum possible number of hydrogen atoms (2c + 2) but much less.

  $M_r(\ce{C_xH_y})= \pu{116 g mol^{-1}}$

Just like porphyrin said in his comment, there is only one solution to x and y that makes chemical sense: 9 carbons and 8 hydrogens. You don't even have to think about the rings and double bonds like I discussed above, but it is a nice check. For a hydrocarbon with 9 carbon atoms we would expect 20 hydrogen atoms in the absence of rings and double bonds. We have 12 less than that, so the number of rings and double bonds in the molecules is 6. That fits with the expectation of a benzene ring (accounting for 4), another C=C double bond (1 more) and a carboxylate group (1 more).


How to determine the magnetic character of heteroleptic complexes?

Based on the available information, we would start by figuring out the 3D geometry of the complex. $\ce{Cr^3+}$ should be octahedral, with two possible "combinations", cis or trans. Let's start with the trans version. The symmetry of this system is $D_{4h}$ so we call this "tetragonal".

With a tetragonal geometry, the MO's would be arranged in such a way that there will be two degenerate orbitals at the lowest energy, and three higher-energy orbitals. The MO diagram typically looks something like this:

Chromium(III) has an [Ar] $3d^3$ electronic configuration so we need to fill up three levels. Now the question is whether or not the splitting caused by the ligands are large enough to cause pairing.

We know for a fact that the trans version of the example is a green solution, so it absorbs red light (700 nm or ~14,300 $cm^{-1}$). Pairing energy is typically in the range of 20,000 - 30,000 $cm^{-1}$ so the 'gap' between yz and xy is probably about only ~14,000 $cm^{-1}$. As such, I would say the trans isomer is magnetic and not paired, because it is energetically more favorable to promote the 3rd electron to the higher energy level than to force it to pair.

Similarly, we know for a fact that the cis version is violet. This means it absorbs yellow light, about 570 nm or 17,000 $cm^{-1}$. This is still lower than the pairing energy, so I'd say that The tetraamminedichlorochromium(III) is magnetic.


Why don't alkynes undergo preferential anti bromination to the degree that alkenes do?

From what I know of the mechanism, Addition of Br2 takes place in the following mechanism

For an alkyne, the intermediate would not be stable due to the ring strain on 3 member ring with a double bond in it, thus reducing the rate of reaction.

Mechanism referred from: https://en.wikipedia.org/wiki/Halogen_addition_reaction


Good news is you are in right track and set up two equations correctly. However, I think the calculations would get easier if you have chosen your units of variables in  $\pu{mol}$. Also,  you may not needed to do more calculations since the answer is in $\pu{mol}$. Thus, your two equations should be:
$$32x + 46y = 2.16 \;\text{.... (1), and}$$
$$\frac{3}{2}x + 3y = 0.135 \;\text{.... (2)}$$
Simplify equation (2) as:
$$x + 2y = 0.090 \;\text{ and }\therefore x=0.090-2y $$

Substitute $x$ in equation (1):
$$32(0.090-2y) + 46y = 2.16 $$
$$(64-46)y = 32 \times 0.090 - 2.16 = 0.72$$
$$\therefore y = \frac{0.72}{18} = 0.04$$


pH of bleach solution slowly decreasing?

Looks like your bleach bleaches your indicator, just like it does pretty much everything else. After all, we don't call it "bleach" for nothing.

Sure, pH may be in fact slowly decreasing, and the bleach would eventually decompose and lose its properties, but these processes do not typically occur in a matter of minutes.

As to the mechanism, it is complicated, as are indicators themselves (usually, large organic molecules). One might think of a huge conjugated system of double bonds that produces color; all of a sudden, an oxidant appears and starts randomly snapping bonds in half. Soon the conjugated system is gone, and so is the color.


How to decide from the solubility product constants which hydroxide precipitates first?

If the question is about what compound would precipitate first, then there can be at most one answer.

Given equal stoichiometry  for the precipitation of $\ce{Fe(OH)3}$ and $\ce{Al(OH)3}$ and solubility product constants of $6 \cdot 10^{38}$ and $4 \cdot 10^{15}$ respectively, $\ce{Fe(OH)3}$ will be the correct answer.

$$\ce{Fe^{3+}_{(aq)} + 3 OH^{-}_{(aq)} -> Fe(OH)3_{(s)}}$$

However, in practice the precipitate will probably contain less water than stoichiometric $\ce{Fe(OH)3}$, and also contain small amounts of $\ce{Al(OH)3}$


How to remove pyridine from Reaction mixture containing chiral molecule with diacetal protection group?

Extract the organic soln of the product with dilute aq. Cu(II) until the characteristic blue of the Cu-pyridine complex is absent from the aq. phase.


The heat capacity is the slope of the plot of internal energy U with temperature T. The internal energy is energy that due to the rotation and vibrational energy a molecule possesses and as the temperature increases more rotational and vibrational energy levels becomes excited and so the internal energy increases.

Initially the internal energy is small and changes slowly as energy is added, mainly because there is not enough energy to excite many rotations or vibrations. Thus the slope of U vs. T is small and so is the heat capacity.  Then as temperature increases more of the energy levels become excited & the internal energy rises rapidly and so does the slope of U vs T and so heat capacity increases.

At very high temperatures  many, many energy levels are populated and so the rate of increase becomes constant as the temperature increases and so the heat capacity becomes constant, i.e. it levels out to a maximum value.  The plot shows the change of heat capacity with energy just for vibrational levels, but it has the same general shape when rotational and translational energy are added.

internal energy vs temperature

vibrational heat capacity vs temperature

low temperature region of heat capacity vs temperature.The 'bumpy' plot is due to the discrete values of vibrational energy levels


Universal symbols for molar mass, molarity, and molality

According to IUPAC, the official name for molarity is amount of substance concentration, and its symbol is $c$. The symbol for molality is $m$ or $b$, and the symbol for molar mass is $M$.

what are the meaningful differences between molarity and molality?

Molality does not change with temperature because mass and chemical amount is temperature-independent. The amount of substance concentration (molarity) does because substances expand (mostly) with in increasing temperature. So for most applications where the temperature changes, molality is the better choice.
On the other hand, for room temperature applications where volumetric measurements are convenient, it is common to use molarity as the measure of concentrations.


Calculating the equilibrium constants for parallel reactions

You can combine the two chemical reactions into a single global reaction scheme by adding them:
\begin{align}
\ce{A + B&<=>C + D}\tag{1}\\
\ce{A + B&<=>C + E}\tag{2} \\ \hline
\ce{2A + 2B &<=> 2C + D + E}   \tag{Global}
\end{align}
and the equilibrium constant for this overall reaction scheme is equal to the products of the two equilibrium constants:
\begin{align}
K_\mathrm{eq,global} &= K_\mathrm{eq,1} K_\mathrm{eq,2} \\
&= \frac{\ce{[C]^2[D][E]} } {\ce{[A]^2[B]^2}}
\end{align}
(where $\ce{[i]}$ denotes either molar concentration or partial pressure of component i). You do not need to adjust the final total pressure or mole amounts.


Alternative software for processing Molden files?

I am fairly sure Molden can be downloaded. However, programs like JMol, and Chemcraft are usually cross-compatiblr with molden.


Why is partial pressure used as a metric for oxygen in blood?

The equation P=CRT is the relationship for the oxygen concentration in the gas phase.  To get the mole fraction of oxygen the the liquid phase (blood), you first use the Henry's law constant.  Here is a link that can help:  https://www.google.com/webhp?sourceid=chrome-instant&ion=1&espv=2&ie=UTF-8#q=henry%27s%20law%20constand%20for%20oxygen%20in%20blood

If, after applying Henry's law, you end up with a concentration in units of x ml per unit volume, this result implies that x is the volume that the mass of dissolved gas (solute) would have it were an actual gas at the same pressure as that in the gas phase.


How to visualize or think about spin waves (magnons)?

In short this is the way I imagine them, as slight directional changes in spins propagating through the system.

The Heisenberg Hamiltonian for the exchange energy associated with magnetic coupling is a pairwise sum over an exchange integral $J_{ij}$ for two sites $i$ and $j$, with spin moments $\hat S_i$ and $\hat S_j$.
\begin{equation}
H_{ex}=-\sum _{i<j}J_{ij}\hat S_i\cdot \hat S_j
\end{equation}
Imagine a periodic array of lattice points each with a spin of two possible orientations $\pm\frac 12$, for a ferromagnetic ground state $(J_{ij}>0)$ with all the spins aligned, denoted as;
\begin{equation} |\uparrow \uparrow \uparrow \dots \uparrow\uparrow\rangle
\end{equation}
A low energy excitation would intuitively be a single flip of a spin somewhere in the lattice e.g.,
\begin{equation}
|\uparrow \downarrow\uparrow\dots\uparrow\uparrow\rangle
\end{equation}
However, this is not an eigenstate of the spin Hamiltonian. Instead low lying excitations have a complex wave character that is akin to a fractional displacement of a spin with respect to its neighbour. Such infinitesimal variation I personally imagine as a single reversed spin distributed over all the spins in the lattice.

Therefore you could view a spin wave as a superposition where all the spins contain a bit of this single inverted spin in the first excited state. It is easy to see that this is the case since we are using ladder operators to change our spins from up to down. Acting as a superposition over the lattice (spin wave) reduces the disturbance between any given adjacent pair, whereas the naive expected eigenstate would include a really big difference in exchange energies at a localised region.

As you may be thinking, how can I have an infinitesimally displaced spin when there are only two spin orientations? Well that is a bit of a misnomer as this oversimplified model has a discrete symmetry of 2. In reality there is a smooth symmetry associated with these spins so that the spin wave is actually a more realistic picture than the 2-fold orientation (Ising model).

Hope that helps :)


Can you theoretically hold more of a volume of water the colder it is?

Water reaches its maximum density at about $\pu{4 ^\circ C}$.  As its cooled from higher temperatures the density decreases as you would expect as the molecules move more slowly and they can thus "settle" in closer together.

But, as it approaches the freezing point, the molecules are moving slowly enough that they start combining together into a more crystalline, ice-like conformation.  It's still completely liquid, but is starting to organize into the ice structure, and ice is about $\pu{7\%}$ less dense than liquid water.

So in the case of water, you can "fit" the most water into a given volume at about  $\pu{4 ^\circ C}$.


What is the IUPAC name of this organic compound?

As of Loong's suggestion, I'd put the answer removing it from comment section: The name should be 4,4-dibromo-3,5-dichloro-2,5-difluoro-3,5-diiodo-pentan-2-ol as Maurice pointed out. However, keep in mind that there are 3 chiral centers within the molecule. But, the indicated name is okay since stereochemistry of the molecule is not shown.
If $\ce{OH}$ group is replaced by $\ce{H}$ atom, the name of new compound is 2,2-dibromo-1,3-dichloro-1,4-difluoro-1,3-diiodopentane. Again, there are still 3 chiral centers within the molecule.


Why does selenophene not undergo aromatic substitution?

The Se atom has such diffuse valence orbitals compared with carbon that there is not good pi overlap, so we don't really have a conjugated cycle.  By definition when there isn't a conjugated cycle, it's "not aromatic", not anti-aromatic or non-aromatic as they both require cyclic conjugation. Words involving 'aromatic' are tricky.


Molecular orbitals of allylic cations, anions and radicals

In allylic species 3 orbital wave functions are interfering to form 3 molecular orbitals among which the pi antibonding molecular orbital will have the max energy. The first two mo can accomodate 4 electrons. Since pi antibonding are at a higher energy level the first 4 electrons will remain in first two mo only. Similarly in cation and radical electrons will remain in these orbitals hence the pi antibonding will always remain empty.


I read that you are a computational chemist at university, so I assume you went through chemical training. This is a good thing. But please remember that cyanides are potent poisons and that protonated cyanide ($\mathrm{p}K_\mathrm{a}(\ce{HCN}) \approx 8$) aka hydrogen cyanide is a poisonous gas.

If you are not a trained chemist (or if you are one but do not feel $100~\%$ safe about using cyanides), stick away from them!

With that in mind this leaves two methods of acquiring $\ce{K[Au(CN)2]}$ — either buying it directly from YourSuppliers™ or by synthesising it. My preferred route of choice would definitely be to buy it from YourSuppliers™, as that is the safest way possible — and always always ensure that it remains in an alkaline solution.

I have no idea about the experimental procedure to synthesise $\ce{K[Au(CN)2]}$ according to the reaction you propose, but it didn’t escape my notice that you are generating $\ce{4KOH}$. This means, I’ll have to assume that the reaction runs under acidic conditions. If that is the case, do not use that method unless you have access to a proper and tested fume hood! If there are other procedures out there that employ alkaline reaction conditions you may (as a trained chemist!) decide to use them at your own risk.

Assuming you did find a method that employs alkaline conditions, the question remains where to get the cyanide from. Well, SigmaAldrich gives me a price of 38 € for $100~\mathrm{g}$ which means that the chemical is very cheap. Note that acquisition of potassium cyanide may well be regulated in your jurisdiction — and rightly so!

Whatever you do, do not attempt to synthesise $\ce{KCN}$ (or any other cyanide for that matter) at home!

Remember that a certain percentage of the population can sense $\ce{HCN}$ which they perceive as an almond-like smell. Don’t attempt to test which part you belong to. If the concentration of $\ce{HCN}$ is such that you can smell it, you are in severe danger!

This bears repeating for future visitors:
Do not do this unless you are a trained chemist and have access to a proper fume hood! Never, ever try this at home! Never work with cyanide in the open air or in any other place where unsuspecting passers-by may be endangered!


Where do the lone pairs go in the "true" resonance structure?

Resonance is a concept resulting from valence bond (VB) theory. Within this theory it is impossible to describe delocalisation. However, every configuration within this framework can be expressed as a Lewis structure. The congruent molecular orbital (MO) theory describes bonding in a delocalised way, therefore it is impossible to express it as a single Lewis structure.

Often we can find Lewis-like structures representing a resonance hybrid. For example from my own questions:

While it is simply impossible to write a valid Lewis-structure (localised electrons) and imply delocalisation, it is often used to denote how a molecule is "special" in a way. We make many assumptions here, and imply that everybody knows that all the electrons are delocalised. Therefore I omitted all electrons, implying the reader knows where to put them.

What you describe as the "true" resonance structure, is an attempt in the same manner to visualise a resonance hybrid (superposition of multiple valid Lewis structures) in a single Lewis-like fashion. The resonance hybrid, like it is used in LibreText, is an insufficient and inaccurate depiction. It is a futile attempt to jumble together VB and MO theory. Even though they are, at their respective theoretical limit equivalent, i.e. congruent theories, at that very limit they become so complex, that a visual depiction is no longer possible.

In this spirit, these depictions of resonance hybrids are overused in the same manner that resonance is often taught wrong and is an overused concept. (Unfortunately the linked LibreTexts article is point in case.) It is important to understand that these simple concepts have their limitations, and to always question the validity.
As you notice yourself in your question, the depiction of the resonance hybrid omits electrons, you have to be cautious to use these simplifications.

On a related, but slightly different note, I especially dislike the article you linked to on LibreTexts. Just to name a few errors:

it omits formal charges in the ozone example;
the delocalisation and resonance structure rules are complete made up, incomplete (1) or even wrong (4, 5);
in the concept of resonance it is inherently wrong to speak of the "correct"Lewis structure;
rules for estimating stability of resonance structures is complete garbage, there is no such thing as a most stable resonance structure;
Lewis dot structure for sulfate uses $\ce{S-O}$ double bonds, which is completely wrong;
it is not generally true, that the structure with fewer formal charges has a higher contribution to the wave function that one that has more charge separation;
etc..


Modelling pH as function of time and position

To answer my own question:
Using either the diffusion limited or the literature values of the acid-base neutralisation (Baldyga et al., Non-isothermal Micromixing in Turbulent Liquids: Theory and Experiment), I was able to simulate the $\mathrm{pH}$ as a function of time by making the simplifying assumption of equal diffusion coefficients of proton and hydroxide ions. I re-expressed the proton concentration in terms of $\mathrm{pH}$:
$$
\frac{\textrm{d} c_{\textrm{H$^+$}}}{\textrm{d} \mathrm{pH}} = -\ln (10) \cdot 10^{-\mathrm{pH}}
$$
plugged this into my original equation and made everything dimensionless.


What's the difference between perfect and ideal gas?

An ideal gas is the same as a perfect gas. Just different naming. The usual name for such gases (for which is assumed that the particles that make up the gas have no interaction with each other) is ideal gas, perfect gas is what such a gas is named in Atkins physical chemistry book. Personally I like the perfect gas naming better as it illustrates the perfect nature of the assumptions made about it.

For the simple systems (like monoatomic gases) when we can assume a perfect/ideal gas, $C_{V,m}$ is independent of temperature. For real gases this is certainly not the case.

Note that $C_{V,m}=\frac{3}{2}R$ only holds for monoatomic gases.


Thermodynamic quantities for the freezing of water

s is not correct because U is not just a function of temperature but also a function of chemical composition.  The internal energy of liquid water is higher than the internal energy of ice.  If you want to think of it in terms of temperature, then just view the heat capacity as being infinite at the phase transition (Dirac delta function of T), such that the integral over the phase change is equal to the heat of fusion.  As far as q is concerned, I guess they are assuming that the change in volume from solid to liquid is negligible.


Tools to draw molecular structure (wireframe) along with atom numbering

To answer the question, I assumed you would like to stick with Marvin if possible. Likely, the options described and found functional in version 5.2.4 met at this site are functional at later versions, too.

So I drew from scratch a similar representation of your molecule.  And indeed, there is overlap by labels and the drawing.

Yet the software allows you to print the drawings.  So, thanks to the installation of HP-1102P printer and printing into a file I got a postscript file.  This vector format is of course beneficial, you may import it in inkscape, for example.  Note two potential pitfalls: select "import" (instead of "open") and "import text as text".  Twice you have to ungroup the drawing before being able to move the labels around as you prefer.  Or to change their colour (vide infra), or substitute them.  Eventually, you may either save it in numerous file formats or like in this case, export the resulting bitmap.


Orientation in benzene rings with more than one substituent

I quote March's Advanced Organic Chemistry, Reactions, Mechanisms and Structure, 6th edition, page 666.

  Some substituents have a pair of electrons (usually unshared) that may be contributed
  toward the ring. The three arenium ions would then look like this:

  The stability of these two ions is increased by the extra form not only because it is another canonical form, but because it is more stable than the others and makes a greater contribution to the hybrid. Every atom (except of course hydrogen) in these forms (C and D) has a complete octet, while all the other forms have one carbon atom with a sextet. No corresponding form can be drawn for the meta isomer.The inclusion of this form in the hybrid lowers the energy , but also because it spreads the positive charge over a larger area—out onto the group Z. Groups with a pair of electrons (e.g., as the halogens) to contribute would be expected, then, in the absence of field effects, not only to direct ortho and para, but also to activate these positions for electrophilic attack. (my emphasis)

In your example, if chlorine electrophile attacks at C-4 or C-6, the resulting intermediate  is stabilized by an extra resonating structure, lowering the energy of this intermediate.Hence electrophillic aromatic substitution takes place.

If chlorine electrophile attacks at C-5 , the resulting intermediate is not stabilized compared to attack at C-6 .Hence the resulting electrophillic aromatic substitution is not seen.

This is the possible reason  that  the product with Cl at position 5 is not formed.

Quoting from chemistry.msu.edu:

  The influence a substituent exerts on the reactivity of a benzene ring may be explained by the interaction of two effects:

  The first is the inductive effect of the substituent. Most elements other than metals and carbon have a significantly greater electronegativity than hydrogen. Consequently, substituents in which nitrogen, oxygen and halogen atoms form sigma-bonds to the aromatic ring exert an inductive electron withdrawal, which deactivates the ring.

  The second effect is the result of conjugation of a substituent function with the aromatic ring. This conjugative interaction facilitates electron pair donation or withdrawal, to or from the benzene ring, in a manner different from the inductive shift. If the atom bonded to the ring has one or more non-bonding valence shell electron pairs, as do nitrogen, oxygen and the halogens, electrons may flow into the aromatic ring by p-π conjugation (resonance).

  The second factor that becomes important in reactions of substituted benzenes concerns the site at which electrophilic substitution occurs.

  The exact influence of a given substituent is best seen by looking at its interactions with the delocalized positive charge on the benzenonium intermediates generated by bonding to the electrophile at each of the three substitution sites.

Therefore chlorine is an ortho para directing group with minor meta product.

References:

Advanced Organic Chemistry 5th edition Part A: Structure and Mechanisms, Francis A. Carey and Richard J. Sundberg
March's Advanced Organic Chemistry, Reactions, Mechanisms and Structure, 6th edition


What program could I use to create a protein model from scratch?

You have several options to generate smaller peptides using the peptide builders in

Avogadro or
PyMol


Is it known how this cleavage occurs?

The mechanism proposed for the cleavage reaction involves an initial attack by the donor on one boron atom in diborane, leading to cleavage of one $\ce{B-H-B}$ bridge.

This is followed by the attack of a second donor molecule, cleaving the remaining $\ce{B-H-B}$ bridge. If the boron atom in the pendant $\ce{BH3}$ group is attacked, two moles of adduct result, whereas attack on the atom already carrying a donor produces cation and borohydride:

Support for this view comes from the fact that the intermediate ($\text{I}$) is detectable by tensimetric titration of amine-borane with diborane and by $\ce{^{11}_{}B}$ nmr.

Source: Boron Hydride Chemistry, Earl Muetterties


Why does NOCl dissociate as +NO and Cl-?

Firstly, the correct comparison to make to determine which is more likely of $\ce{NO+}+\ce{Cl-}$ and $\ce{NO-}+\ce{Cl+}$ is the ionization potential vs. the electron affinity of the the two species. The electronegativities are correlated with these other metrics, but it is important to emphasize that those are really the important things.

Relatedly, nitrosonium $\ce{NO+}$ is stabilized by resonance and so this is a considerably stabilized cation (i.e. IP not too high). On the other hand, the IP of $\ce{Cl}$ is huge while the EA is small; it would much rather get an electron than lose one. The hyponitrite anion $\ce{NO-}$ actually usually dimerizes for reasons outside the bounds of the question. So $\ce{NO+}+\ce{Cl-}$ is way lower in energy.

In general, a red-flag should go up when you see halogens with positive charges. They do happen, but very rarely, especially at the level of chemistry in an intro organic course.


Why the atomic (metallic) radius of gallium is less than that of aluminium?

First, you have to look at the definition of metallic radius, which is the half distance between two atoms in a lattice. It has a significant dependence on crystal structure.

  Tanget relevant to the question and other answers:  Gallium has an orthorhombic crystal structure (CN = 6) whereas as aluminum has a face-centered cubic crystal structure (CN = 12).  This difference in coordination requires a goldschmidt correction to compare values as if gallium were 12-coordinated which makes the gallium's corrected metallic radius actually even smaller compared to aluminum at about $\mathrm{130pm}$.

Gallium exists as a diatomic solid in the lattice, which causes the atoms to be closer together on the  a axis because they are covalently bonded. Gallium's covalent radius is about $\mathrm{122pm}$ which is very much smaller than the metallic radii of either element. This value between the bonded atoms lowers the average distance between atoms giving gallium a lower metallic radius.

If you look at the Van der Waals radii of the elements, (more representative of a monoatomic gaseous atom) you will see that aluminum is actually smaller though not by a lot, giving some credence to the d-electrons explanations, but not a complete explanation in regard to the metallic radius.


Solvents imply dissolving process as intermolecular interactions between solvent and solute, leading to disperging the solute from the condensed form.

"Permanent" gases near normal conditions do not provide such interactions. Volatile compounds evaporate with or without presence of gas. Gas is rather a bystander/spectator of such evaporation.

Fluids near critical conditions like supercritical CO2 share many properties of liquids and gases, so such interactions are present, so in some sense and with both eyes closed, one can consider it as solvent and gas.


Basicity has everything to do with electron density on the electron-pair donor.  Electronegative subsituents actually decrease basicity because they draw electrons towards themselves, though nitrogen and chlorine have comparable electronegativities (3.04 and 3.16 respectively), so I'm not sure you can use that argument here. Instead of inductive effects, let's consider the resonance structures of our molecules:

$\hspace{5.6cm}$

A positive charge on a nitrogen is always more favorable than a positive charge on a chlorine. From this we know nitrogen is donating its electrons into the ring to a much greater extent than chlorine. Because the electron density is much greater on the heterocyclic nitrogen in B, it will therefore be the more basic of the two.


Why $\psi(r_e, r_n) = \sum \psi_e^i(r_e|r_n) \chi^i(r_n)$ in Born-Huang expansion?

After reading this lecture note I have a better understanding.

The reason I'm so confused is that the form
$$\psi(r_e, r_n) = \sum_{i = 1}^{\infty} \psi_e^i(r_e|r_n) \chi^i(r_N)$$
seems to be a linear combination with coefficients that are all $1$. And this is very importand for the derivation of the set of entangled nucleus time-independent SEs since there is no arbitrary coefficients in the set of equations.

After I read the lecture note, here is my understanding:

For a fixed atomic configuration $R^{\alpha}$, the electronic state can be represented by a linear combination of the eigenstates of $\hat H_e(R^{\alpha})$
$$\psi(r_e| R^{\alpha}) = \sum_{i = 1}^{\infty} c_i(R^{\alpha})\psi^{i}_e(r_e|R^{\alpha})$$
although $\psi(r_e| R^{\alpha})$ is an electronic state, this is essentially a solution to $\hat H \psi = E \psi$ at a fixed atomic configuration.

Since by varying $R^{\alpha}$, the set of linear coefficients {$c_i$} will change. So the general $\psi(r_e, r_n)$ can be represented as
$$\psi(r_e, r_n) = \sum_{i = 1}^{\infty} \chi^i(r_N) \psi_e^i(r_e|r_n)$$
in which the set of {$\chi^i(r_n)$} is just the set {$c_i({R^{\alpha}})$}.

Essentially, the $\psi(r_e, r_n)$ that we plug back into $\hat H \psi = E \psi$ is still an electron wavefunction.


How do I compare the C=C bond lengths in this cumulene?

Label the double bonds (a), (b), and (c), in correspondence with the image provided in the question. Label the carbons in the cyclopropane ring $\ce{A}$, $\ce{B}$, and $\ce{C}$, with $\ce{A}$ being the carbon with the double bond (c).

At $\ce{A}$, the usual $\ce{sp^2}$ hybridization would lead to bond angles of $120^\circ$, whereas a lack of hybridization (i.e. simple $\ce{p}$ orbitals) would lead to bond angles of $90^\circ$. The latter is not usually favored, but the high ring strain in cyclopropane leads to carbon-carbon bonds $\ce{A-B}$ and $\ce{A-C}$ having greater $\ce{p}$ character than is usual. The double bond (c) at $\ce{A}$ thus receives more $\ce{s}$ character, and is stronger than a usual $\ce{sp^2{-}sp}$ bond would be (i.e. (a)).

We conclude, therefore, that the bond strengths are (b) > (c) > (a), and hence that the bond lengths are (a) > (c) > (b).


How can every atom have unique spectral lines?

Do all noble gases or alkaline earth metals have similar spectral
  lines considering the above points?

The question is interesting after you modified it. The basic set of reasoning you provided is the main story. Each element has a different nuclear charge and the outermost electron(s) is responsible for the atomic emission spectrum. Since the nuclear charge is different, those outermost electrons experience a different potential energy. Their kinetic energy is also different from element to element.

The key question is what is meant by similarity? The atomic spectra of all the elements is visualized as bright lines on a dark background. The reason they appear as lines is just because of the instrument used to observe the atomic spectrum. There is nothing fundamental in the "line"spectrum. The atomic emission appears as lines because the slit in the monochromator is shaped like a very narrow rectangle. This is the image of the slit. If I made a very narrow circular opening, the images will appear as bright points rather than lines. Consider that line spectrum is the conventional way of looking at the atomic spectrum.

For the visible region, let us look at the visible spectrum of alkali and alkaline earth metals.

or look at the visible spectrum of noble gases

Are they similar in any way? Visually there is no similarity at all in the spectrum. Look at the spectrum. The reason we cannot see any similarity is because the dispersion of the lines is on the "wavelength" scale. However, very bright, I mean really genius spectroscopists of the 18th and the 19th century were able to find out a patterns (or call it as mathematical series). They figured out that each set of lines forms a pattern as Sharp series, Principal series, Diffuse Series and Fundamental series and there they found similarities in patterns-in terms of mathematical series. In short, our eyes cannot interpret those series. Another complication arises because we are just looking at the visible region. There is a ultraviolet region and then there is an infrared region. We cannot see it without using more sophisticated instruments.


How does one tell if a specific molecule is acidic or basic?

It is sometimes challenging to determine if a molecule is going to be acidic or basic if the system in which it is reacting is not considered.  An important point to consider when dealing with acids and bases is that acid/base strength is inherently tied to the solvent.  For this answer, I'm going to limit the discussion to acids and bases in an aqueous environment.
It is helpful to consider the terms acid and base as a means to classify substances.  This way, Chemists can explain chemical reactivity and structure-function relationships of substances.  Very early classification systems depended on our senses (acids are sour, bases are slippery to the touch) and more recent classification systems utilize structural characterization tools such as NMR or crystallography.  Many classification systems have been proposed over the years, and only a few of them have found sufficiently widespread use to end up in textbooks used in the standard Chemistry curriculum.  Below are a few systems, taken from Miessler & Tarr's Inorganic Chemistry textbook.  The 2nd through 4th entries are those in common use today.

NAME (YEAR) acid definition [example]; base definition [example]
Liebig (~1776) Acid: an oxide of N, P, S [$\ce{SO_3}$]; Base: Reacts with
acid [$\ce{NaOH}$]
Arrhenius (1894) Acid: Forms hydronium ion [$\ce{HNO_3}$]; Base: Forms hydroxide ion [$\ce{NaOH}$]
Brønsted (1923) Acid: Proton donor [$\ce{HCl}$]; Base: Proton acceptor [$\ce{NaOH}$]
Lewis (1923) Acid: Electron-pair acceptor [$\ce{Ag^+}$]; Base: Electron-pair donor [$\ce{NH_3}$]
Ingold-Robinson (1932) Acid: Electrophile [$\ce{BF_3}$]; Base: Nucleophile [$\ce{NH_3}$]
Lux-Flood (1939) Acid: Oxide ion acceptor [$\ce{SiO_2}$]; Base: Oxide ion donor [$\ce{CaO}$]
Usanovich (1939) Acid: Electron acceptor [$\ce{Cl_2}$]; Base: Electron donor [$\ce{Na}$]
Solvent system (1950s) Acid: Solvent cation [$\ce{BrF_2^+}$]; Base: Solvent anion [$\ce{BrF_4^-}$]
Frontier Orbitals (1960s) Acid: LUMO of acceptor [$\ce{BrF_3}$]; Base: HOMO of donor [$\ce{NH_3}$]

I find the various acid/base systems very enlightening.  Note how From Arrhenius through Lewis there was a broadening of the acid/base classification system; Arrhenius can't be used to describe non-aqueous acids and bases and Brønsted can't be used with aprotic substances.  Yet, the Lewis definition incorporates the previous two (a Brønsted acid is also a Lewis acid; an Arrhenius base is also a Lewis base).  One then may ask, what's up with the Lux-Flood definition then?  This definition is counter to the trend of broadening the classification system, and yet it is useful in describing anhydrous solid-state chemistry and is used to describe geochemical reactions as well as the chemistry of high-temperature melts.  My point being: classification of substances into acids and bases is only meaningful if it helps explain chemical phenomena.
Which brings us to the Usanovich definition, which essentially states that every reaction is an acid-base reaction.  Such a broad definition is not overly helpful, and to those of us with an affinity towards electrochemistry (ahem), is somewhat arrogant :-)
I do believe that determining if a substance will behave as an acid or a base requires a bit of chemical intuition (or a Socratic method).  For example, you may have performed an experiment in which $\ce{KOH}$ served as a base; you know that, like potassium, sodium is an alkali metal; therefore you presume that $\ce{NaOH}$ would be a base as well.
Back to the question at hand
So how do I suggest one use this information to predict whether a substance will behave as an acid or a base using its structure alone?  Personally, I find the Lewis theory as the most useful classification system in answering this type of question.  If the structure of a compound is set before me and I am to predict its acid/base chemistry, I will ask two questions:

Are there any lone pair electrons that can be donated?
Are there any electron-deficient atoms that could serve as electron pair acceptors?

If the answer to question 1 is yes, then the molecule can behave as a base.  If the answer to question 2 is yes, then the molecule is an acid.  If both are true, then I have an amphoteric substance.
In proofreading this answer, I realize I said I would restrict myself to aqueous systems, in which case using the Brønsted system may be more helpful.  In this case, the questions become:

Is there a hydrogen that can be donated? (You'll be right more often than you are wrong if you rephrase this question as "Is there a hydrogen that is attached to something other than carbon?").
Is there a lone pair that can accept a proton?


Rotational degrees of freedom (3N-5 and 3N-6)

Rotations around bonds are typically termed "internal rotations", and represent one of the most common problematic cases for the rigid-rotor-harmonic-oscillator (RRHO) model of internal molecular motion.  This is because RRHO assumes that any vibrational amplitudes are "small," and internal rotations are most definitely not small!  Such rotations still involve the internal degrees of freedom of the molecule, though, and thus (as Jan notes) are considered as part of the 'vibration' of the molecule, not its 'rotation.'

Internal rotations typically have some energy cost involved (as in your pentane example), and so cannot be treated as "free" rotations. They are usually termed "hindered rotations," and there is a great deal of literature studying them.  Some citations I know offhand:

Classic Pitzer (J Chem Phys 5 469, 1937), and Pitzer & Gwinn (J Chem Phys 10 428, 1942)
Independent-rotations approximation for treating hindered internal rotation: Pfaendtner (Theor Chem Acc 118 881, 2007)
Other / more-advanced treatments of internal rotation:

Hecht (J Chem Phys 26 31, 1957)
Strekalov (Chem Phys 355 62, 2009)
Vansteenkiste (J Chem Phys 124 044314, 2006)
McClurg (J Chem Phys 106 6675, 1997)

One species of particular interest to me on this topic has been nitromethane, which has an extraordinarily small barrier to rotation around the $\ce{C}-\ce{N}$ bond (see Strekalov, above), such that at ambient temperature it can be considered essentially a free rotation.


You totally can't have a gas (or any other state of matter) made of positive ions alone.

That's not what you asked, though. The question was: what if we theoretically could have it. Well, then in a pretty short instant you will stop having it. The said instant would be rather violent, too(*).

To put things into perspective, imagine an indestructible spherical vessel about as big as to contain 1 mole of gas at normal conditions, which implies the radius of 17.5 cm. Now let's magically put 1 mole of $\ce{O+}$ inside. That would give it an electric charge of about 96500 coulombs, which is quite a lot. The electric field outside the sphere would be $$|\vec E\,|={1\over4\pi\varepsilon_0}\cdot{q\over r^2}=2.836\cdot10^{16}\;\rm V/m$$

Do you know how big is this? Wrong! it is bigger than you think. Mere $3\cdot10^6\;\rm V/m$ would cause a breakdown in dry air. Good solid insulators can handle more, but I think we may safely assume that $10^{10}$ is enough to break anything that is made of atoms. Mind you, we have six orders of magnitude on top of that(†). The conductor will be instantly vaporized and thrown out with immense force, enough to start many nuclear reactions and create a bunch of weird elementary particles. What metal it was made of would be the least of your concerns.

So it goes.

(*) Reminds me of a certain what-if episode, but they used a bigger sphere.
(†) Thought we would break the Schwinger limit, but we didn't.


Hydrogen, having very small and fast moving molecules,  diffuses faster and through smaller holes than air. Your balloons seem to be permeable for hydrogen. Try a different brand, if it helps. But as others have mentioned, rubber is not the best material to keep hydrogen. Better is a metallised plastic.
Pure hydrogen just burns in a flame.  If you have ever  watched the film footage (1937)  of the catastrophic accident of the airship Hindenburg , it was just a huge scale of burning in a flame.
What makes an explosive "bang"is a hydrogen mixture with air, even better with oxygen. Because  hydrogen does not explode, until is contains >6% v/v of oxygen.
Any binary chemical explosion of gases requires a self-sustaining chain reaction. That does not happen if a mixture contains too little of one of mutually reacting gases.


Why do methylamine and ethylamine have nearly the same basicity constant?

Ethylamine has a pKb slightly greater than that of methylamine.

As you are saying

  It is known that ethyl is better than methyl for inductive effect. It's positive inductive effect is greater than methyl.

is absolutely true. But the basicity in aqueous medium is not only determined by +I effect but also solvation of substituted ammonium cation.

Even though +I effect in ethylamine is more than that in methylamine, ethylamine  is more bulkier and sterically hinders the formation of intermolecular H−bonds with water.


Help to kick out the anomaly in proving $Q$ = $K_\text{eq}$

In order to compare $K$ and $Q$, you have to define them in terms of the same measures of concentration and the same standard state. The only difference in the definition should be that in order to determine $K$, you are required to use the concentrations or activities of an equilibrium state, while $Q$ may be determined for states that are not necessarily at equilibrium.

please help me to understand why at chemical equilibrium $Q=K_{eq}$ holds

Once you have the definitions match, it becomes fairly trivial. The equilibrium constant is what you observe whenever the reaction has had enough time to go to equilibrium. The individual equilibrium concentrations might be distinct for different starting concentrations, but $K$ comes out the same (if it does not, it means the reaction is not reaching equilibrium). If $Q$ is not equal to $K$, you are not at equilibrium.
I realize this is a bit of a tautology. The explanation why the particular form of the equilibrium expression leads to a value that is a constant requires a knowledge of how chemical potentials (or $\Delta_r G$) vary with concentration. The reaction will proceed in the direction that minimizes $G$, so at equilibrium $\Delta_r G = \frac{dG}{d \xi}$ will be zero.


What is the entalphy change with respect to pressure at Isothermal Process for *Real Gases*?

The equation for the partial derivative of enthalpy with respect to pressure at constant temperature is given by:  $$\left(\frac{\partial H}{\partial P}\right)_T=V-T\left(\frac{\partial V}{\partial T}\right)_P$$From the equation of state for your gas, you can evaluate the rhs of this equation.


Sand is a mixture. Sand is classified as a heterogeneous mixture because it does not have the same properties, composition and appearance throughout the mixture. A homogeneous mixture has a uniform mix throughout.


I'm not sure I completely understand what exactly your problem is, but:
You want to resolve to signals at approx. m/z = 1060.2 and m/z = 1061.2.
So $\Delta m = 1$ The required resolution is: $\frac{m}{\Delta m} = \frac{1061}{1} \approx 1060$: The spectrometer in (a) has only 1000 which is too low. The spectrometer in (b) has 5000 which is quite more than needed.
Maybe it is easier to think of the reciprocal: $\frac{\Delta m}{m}$ which tells you how far two signal need to be apart so your spectrometer can resolve them.
In (a) this is $\frac{\Delta m}{m} = \frac{1}{1000} = \frac{1.06}{1060}$ two signals can be resolved if they are at least 1.06 m/z apart at m = 1060. I.e. if they were slightly further apart than they are.
The spectrometer in (b) can resolve signals that are at least  $\Delta m/z =  0.21$ apart at  $m/z = 1060$.
FWHM and resolution
The FWHM is basically part of the resolution. The $\Delta m$ is the m/z difference between two signals that are just resolved.
Have a look at the IUPAC's definitions of resolution in mass spec. There are several approaches to specify what "just resolved" means.
The FWHM can be used directly as the $\Delta m$ of the "peak width definition".
If instrument A has $5\times$ the FWHM than instrument B, then resolution of A  $\approx \frac{1}{5}$ resolution of B.
I'm more used to the "10 % valley" which roughly equals the full with at 5 % height: you specify how much signal is allowed at the minimum between the two peaks, e.g. 10 %.
It is important to realize that these two definitions differ by almost a factor 2!
Here are some results of "simulated" peaks at different resolutions and $\Delta m/z$ (all gaussian, which of course doesn't need to be the case in reality). I use the full width at 5% = 10 % valley definitions for the resolution.
The first row is resolution 5000 with the, second row is resolution 1000. You see that the two signals are much better resolved than in your example picture above. Which means that above another resolution definition was used, possibly $\Delta m$ = FWHM. In any case, a resolution of 1000 should be almost sufficient to resolve m/z 1060.2 from 1061.2.
The third row shows two signals of equal height with $\Delta m/z $ = 1 = FWHM. According to the resolution definitions above, this is a resolution of ca. 520. Note that while you can say that there are at least two peaks, you cannot take the two maxima as the m/z of the two underlying signals (grey).
The fourth row is almost the same, but I multiplied the second signal by 56% to account for the frequency of having bradykinin with one $^{13}$C. Looks quite similar to your picture.

Also, have a look at M. P. Balogh: Debating Resolution and Mass Accuracy, LC•GC Europe, 17(3), 152–159 (2004).


Oxalate ion dentiticity - why is it bidentate and not tetradentate?

The oxalate anion cannot be tetradentate. While there are four oxygen atoms, the ion is planar and the oxygen atoms are pointing in different directions.

The best any metal can do is bond to two of them, although there are two bonding modes possible:

Perhaps you were envisioning some $\kappa^4$-bonding mode that is
"side-on". However, most of the nonbonding electrons (orbitals) are also in the plane.

This is not to say that oxalate cannot form four bonds to metal centers; it just cannot do so with only one metal center. Oxalate can act as a bridging ligand


Does resonance affect the polarity of the molecule?

When talking about resonance it is important to know that this is only a concept to visualise certain bonding features. It is very important to understand, that the actual bonding situation is a mixture of all possible resonance structures. The Molecule itself exists at all times as this mixture, hence neither of these resonance representations can be observed. One can regard it also as a superposition of these structures.

Resonance structures had to be introduced into the Lewis formula context, when it was obvious, that one formula can no longer describe a bonding pattern remotely accurate. While the Lewis formalism is great for many things, as it keeps chemistry visualisable and simple, it unfortunately has its boundaries. Only for molecules with very simple bonding patterns it is somehow possible to guess some physical properties (qualitatively).

Formal charges are also only a concept, or a result from the limited depiction of the Lewis representation - hence the name formal. They do not usually represent any physical charge. However, sometimes, the physical charge will coincide with the formal charge. Therefore it is usually not possible to derive any actual dipole moment from them.

Charge separation within molecules depend of the geometric arrangements of the nuclei and the electrons around them, i.e. the electron density. It is a physical property which can be observed and also predicted by means of quantum chemical considerations. Molecular Orbital Theory is a very fundamental concept which allows to derive these properties (at least on a qualitative level).

The bent arrangement of ozone is a result of its electronic structure. And the dipole moment is a result coming from that.

Upon the extension of the question and some comments:

Like Thomij already stated, the formal charges are only for bookkeeping. They always have an integer value. A partial charge, however, is a physical observable, resulting from the difference from the electron density and the field of the nuclei.

In all cases the overall formal charge has to add up to the overall physical charge, which can also be calculated from all partial charges. These partial charges can also be calculated from quantum chemical calculations, since you are approximately solving for the optimal electron density at a certain nuclear arrangement.

There is absolutely no possibility to predict a dipole moment without knowing the electron density (or wavefunction) at a given nuclear geometry. A Lewis formula can be generated from these facts, showing the simplest but still well matching bond description. This representation has to sometimes include formal charges. The linked wikipedia page does not employ this concept to explain the polarity, they rather use partial charges.

The partial charges in ozone result from a superposition of the molecular orbitals (click here to enlarge). These are the valence orbitals obtained from BP86/cc-VPDZ:

Especially the HOMO (very right) and the HOMO-3 (4th from right) and HOMO -6 (4th from left) contribute massively to a slight charge separation in ozone.


Can I convert from an emission amount (in tonnes per area) to a concentration (in ppbv)?

What you are doing seems reasonable but there are many hidden assumptions in your technique:

The data source you are using actually provides a rate, i.e. metric tons per area per year.
Thus your calculation is also implicitly a rate: the results will be in tons per volume per year.
You are assuming that the only source of atmospheric $\ce{SO2}$ is the emissions quantified by your data source.  It may well be that there are other, natural sources of $\ce{SO2}$, or more likely, natural sinks of $\ce{SO2}$.
If you want to know actual concentrations, you will need to quantify all the other sources and sinks of $\ce{SO2}$, and integrate all sources and sinks over time.


How does the electronic configuration for this acid oxide correspond to its structure?

The compound is $\ce{SeO3}$, and here is my reasoning:

From the electronic configuration given in the question, the central atom is selenium (atomic weight 34).

As you can see from its electronic configuration, the valence shell has 6 electrons and it goes on to form 4 covalent bonds with oxygen, of which 3 are $\sigma$ bonds and 1 is a $\pi$ bond. Therefore, it shares 4 electrons and has a lone pair. Its steric number is 4 and hence its hybridisation is $sp^3$.

With the tetrahedral geometry, the lone pair will occupy the uppermost position (which is shown by a dash) and the three oxygen atoms will occupy the base of the distorted tetrahedron (due to the lone pair and arising from VSEPR theory). Further, there would be a negative charge on two of the oxygen atoms.

The central atom is not gallium since its electronic configuration is $(4s)^2(4p)^1$. And krypton doesn't react with oxygen as far as I know.


What do D and L stand for in the Fischer-Rosanoff convention?

Dextro (right) and levo (left). (I learned it as "laevo", but I don't see that spelling in a quick Google search.)


Writing Ksp when the ions are on the left of the equilibrium equation?

By convention, the solid is on the left of the side dissolving reaction. If you do it the $K_{sp}$ has the correct form.

The value of the $K_{sp}$ given above suggests that it conforms to this convention.


The amylose present in starch is responsible for the deep blue colouration.  The structure of amylose consists of long polymer chains of glucose units connected by an alpha acetal linkages. All of the monomer units are $\alpha$ -D-glucose, and all the alpha acetal links connect $C1$ of one glucose and to $C4$ of the next glucose.

As a result of the bond angles in the acetal linkage, amylose actually forms a spiral structure.

X-ray diffraction studies$^{[1]}$, demonstrate that the amylose is organized as left-handed helices, which have outer diameters of 13 Å and a pitch of 8 Å, with each turn of the helix corresponding to six 1,4-anhydroglucose units. The iodine components are linearly arranged in the 5 Å wide inner cavity of the helices with an $\ce{I–I}$ distance of approximately 3.1 Å.

Starch-iodine complexes show dichroism of flow$^{[2]}$, i.e light with its electric vector parallel to the flow lines is more strongly absorbed than light with its electric vector normal to the flow lines. The dichroism of flow is shown to require that the long axes of the iodine molecules in the
complex be parallel to the long axis of the starch iodine complex.

The exact structure of the polyiodide chain remains contentious, however, the following species are commonly implicated as key elements of the substructure:

$$\ce{I2 + I^- -> I3^-}$$

So, ignoring the structural specifics of the polyiodide chain, one notes that if an iodine molecule enters the internal channel of the amylose
helix, it may be surrounded by about six glucose residues. The following image is from reference (3):

Bear this image in mind, and note that the hydroxyl oxygen is pojected into the helix cavity.

Also, when iodine is dissolved in ether or alcohol, a new strong absorption band appears in the near ultraviolet region. In relation to the study of the benzene-iodine complex, Mulliken postulated the complex formation
between iodine and ether or alcohol, and explained this new band by the concept of intermolecular charge transfer spectra$^{[3]}$.

The "simple" idea behind a charge transfer complex is given below:

$$\ce{I2 + D -> I^{\delta -}...D^{\delta +}}$$ Reference 4 gives more details. (It also likens this bonding to a "hydrogen bond" if that helps, you understand). In this case, D is the donor atom (electronegative atoms like nitrogen, oxygen etc.)

The iodine-donor distance is usually in 2~3 Å range. Thus, based on all of this one can conclude that possible charge transfer interactions between oxygen atoms that are projecting in towards the helical cavity and the iodine molecules in the polyiiodide chain are responsible for the colouration. You can view how the oxygen molecules are pointed towards the chain of iodine here. (I recommend rotating the diagram so that the axis of the cavity is towards you, and you are essentially looking down the tube)

References

1.https://www.researchgate.net/publication/244335699_The_complex_of_amylose_and_iodine

http://pubs.acs.org/doi/abs/10.1021/ja01244a017
http://scitation.aip.org/content/aip/journal/jcp/22/3/10.1063/1.1740076 (I recommend reading this if you are not satisfied with the brief, qualitative discussion I have provided here)
Inorganic Chemistry (Nils Wiberg) (google books, relevant portion is part of the preview)


this paper stated that $\ce{CO2}$   could be formed as well, how?

Usually when we consider permanganate oxidations we think of

oxidation of alkyl benzenes to benzoic acid
cleavage and oxidation of a double bond to an acid, through the aldehyde, if there is one hydrogen on the double bond
cleavage and oxidation of a double bond to a ketone if there are no hydrogens on the double bond

However, there are a few special cases such as formic acid and oxalic acid, or their corresponding aldehydes.  These compounds are oxidized by permanganate to $\ce{CO2}$ (for an idea as to why these compounds react differently than, say, acetic acid, see Scheme 1 here).

With this information in hand we can re-examine the permanganate oxidation of piperic acid.

We see that oxalic acid is one of the primary products produced in this reaction.  As discussed above, oxalic acid can then undergo further reaction to produce carbon dioxide.


Why are synthetic pH indicators used over natural indicators?

In acid-base titrations, synthetic indicators are exclusively used to find accurate end-point determinations because they always have a highly defined color change at certain pHs. For example, phenolphthalein ($\mathrm{p}K_\mathrm{a} = 9.7$ at $\pu{25 ^{\circ}C}$) is colorless in acidic solutions (precisely $0 \lt \mathrm{pH} \lt 8.2$), but it is pink in basic conditions when the pH of the solution goes above 8.2 (precisely $8.2 \lt \mathrm{pH} \lt 12.0$), which makes it ideal for strong acid-strong base titrations (Wikipedia).  On the other hand, bromothymol blue ($\mathrm{p}K_\mathrm{a} = 7.1$ at $\pu{25 ^{\circ}C}$) is yellow in acidic solutions (precisely $0 \lt \mathrm{pH} \lt 6.0$), but it is blue in basic conditions when the pH of the solution goes above 7.6 ($7.6 \lt \mathrm{pH}$). However, it shows greenish-blue color in a neutral solution (precisely $6.0 \lt \mathrm{pH} \lt 7.6$; see the picture below).

It also change color from yellow to pink in high acidic conditions such in concentrated hydrochloric acid ($0 \gt \mathrm{pH}$; see most left test tube in the picture below), thus it is ideal to use as a universal indicator (Wikipedia). These two examples would clearly show you, as you correctly put it, synthetic indicators are much more accurate and reliable since they are specifically designed for the task.

Sure, natural indicators could be easily prepared from easily attainable row materials, and hence, could be regularly used as a cost-effective and environmentally friendly alternative to synthetic pH indicators, but I regret to say that it can be used only in high school science activities to fascinate science-loving kids. Since those natural sources contain several color changing substances (e.g., flavanones, flavones, flavonols, and anthocyanidins), the color change at the end-points is not sharp, perhaps, due to the difficulty of determining $\mathrm{p}K_\mathrm{a}$ of relevant indicator(s). For example, although boiled red cabbage water can be used as a universal indicator in high school projects and show the end point of an acid-base titration (see the following figure), it cannot be used in a day-to-day analytical lab to do accurate calculations.

Also, another reason for not using natural indicators is that most of these dyes tend to decompose in higher pH values.  The best example is most common natural color indicator, anthocyanidins. The stability of anthocyanidins is dependent on pH. At a low pH (acidic conditions), colored anthocyanidins are present, whereas at a higher pH (basic conditions) the colorless chalcones forms are present (see scheme below) (Wikipedia).

Most importantly, aside from the above, flavonols including anthocyanidins have been studied extensively in drug development due to their plethora of biological activities (Ref.1). Thus they have been targetted in much more important tasks than a usage as pH indicators.

Reference:

M. Rahnasto-Rilla, J. Tyni, M. Huovinen, E. Jarho, T. Kulikowicz, S. Ravichandran, V. A. Bohr, L. Ferrucci, M. Lahtela-Kakkonen, R. Moaddel, “Natural polyphenols as sirtuin 6 modulators,” Scientific Reports 2018, 8, 4163 (11 pages) (DOI:10.1038/s41598-018-22388-5).


How is change in entropy of universe related to the energy released by system that can be used to do work?

Well a reaction which is spontaneous has negative value of Gibbs or Helmholtz free energy. This free energy is the energy which is made available to be used for work.

How does that relate to your question? Both Gibbs and Helmholtz free energy are derived from the argument that change in entropy of the universe is greater than zero. Physically, that is an argument that no chemical system will react in such a way that decreases the number of available microstates unless it increases the number of available microstates for the surroundings by a greater amount.


Ionic Character, on which factors do it depend

When you are considering lattice energy as the parameter for judging the ionic character of compounds, you are assuming two things.

The constituting elements of this alkali metal halides are purely ionic,and thus the Lattice energy (U) is the released energy when those two isolated gaseous ions forms the crystal lattice. But, in practice the constituting ions are never purely ionic (especially smaller alkali metal halide such as $\ce{LiF} $ or $\ce{NaF}$.)
The ions constituting the crystal lattice are hard non-elastic spheres. They hardly (in general) obey this postulate. This postulate is again not totally valid for smaller alkali metal halides due to distortion of their electron cloud due to large polarization.

So, it is seen that if you just theoretically say that the electrostatic force as simply Coulomb's law and thus lattice energy decreases, it doesn't conclude anything as it is seen that the Lattice energy is not a proper parameter for judging ionic character.

The correct approach in judging the covalent character present in an ionic compound is to follow Fajans' rules. According to this rule, $\phi _{cation}$ is inversely proportional to $r_{cation}$ and directly proportional to $Z_{cation}$ Here, $Z_+$ for all the cation is same. But radius of $\ce{Li^+}$ is smallest and ionic radius gradually increases from $\ce{Li^+}%$ to $\ce{Cs^+}$. So, $$\ce{$\phi$_{Li^+} > $\phi$_{Na^+}  > $\phi$_{K^+}  > $\phi$_{Rb^+}  > $\phi$_{Cs^+} }$$ Here  $\phi$ is the measure of polarising power of cation. As, the value of $\phi$ increases, the polarising power increase and thus the anion's electron cloud gets attracted towards cation nucleus which tends to form a kind of overlap between the orbitals of cation and anion, which increase covalency and decreases ionic character.

So, according to Fajans' rules, increasing ionic character (i.e. more covalency to less covalency) is, $$\ce{LiF < NaF < KF < RbF < CsF}$$


Titration with EDTA and Eriochrome black T

Assuming you've used Erichrome Black-T as an indicator, the second your whole solution changes from wine red to blue, as in the moment you realise there is a change in colour, however light it is, that's your end point and you should stop.

Make sure you use a white tile under your conical flask and remember to pause the titration and swirl the flask little bit every now and then as you near the end point. That should even out the colour and make the change noticeable before it's too late.


Just a quick answer without looking too much into it.
You have been using a lot of polar solvents without any luck. The reason is that the molecule no longer has the groups that make it water-soluble because of the thermal decomposition. Only hydrocarbons chain remain. You need to use an organic solvent (toluene, benzene, hexane). Since you probably don't have access to a lab, I would try with gasoline.
Let me know if it works.
Cheers.


Is chlorophenylmethane the correct IUPAC name rather than chloromethylbenzene?

The preferred IUPAC name is, indeed, (chloromethyl)benzene, built from substitutive prefix chloromethyl. The parentheses around "chloromethyl" are required because it is a compound substituent (a simple substituent attached to another simple substituent, specifically, methyl and chloro in this case). The Blue Book [1, p. 66] has plenty of examples similar to yours:

  $$
\begin{array}{lll}
\ce{C6H5-CH2-NCS} &\quad &\text{benzyl isothiocyanate} \\
  & &\text{(isothiocyanatomethyl)benzene (PIN)} \\
  & & \\
\ce{C6H5-NC} &\quad &\text{phenyl isocyanide} \\
  & &\text{isocyanobenzene (PIN)}
\end{array}
$$

Note that common names for this molecule also include: benzyl chloride and α-chlorotoluene.

References

Favre, H. A.; Powell, W. H. IUPAC “Blue Book” Nomenclature of Organic Chemistry: IUPAC Recommendations and Preferred Names; IUPAC Recommendations; Royal Society of Chemistry: Cambridge, England, 2013. ISBN 978-0-85404-182-4.


[…] one professor I know who insists that it should be typeset with an en dash.

I disagree! A standard minus sign should be

on the same height as the horizontal line of a plus sign;
as long as an equal sign.

The en dash is

lower than the horizontal bar of the plus sign;
slightly longer than the equal sign.

Unicode has dedicated characters for minus and superscript minus

standard minus: U+2212: −
superscript minus: U+207B: Cl⁻

See also:
The situation, and how to realize the correct typography using $\mathrm\LaTeX$ has been described in the German article Feinheiten bei wissenschaftlichen Publikationen – Mikrotypographie-Regeln, Teil II by Marion Neubauer in Die TeXnische Kommödie, 1997, 1, 25-44 (PDF).


Bond order for carbonate ion for resonance

There definitely is an easy way to do this.

One of the easiest examples that comes to mind is benzene.  If we look at the structure:

we see that each of the carbon bonds are either single bonds to the adjacent carbon or double bonds to the carbon in the other direction.  If we draw the other resonance structure these bonds shift by 1 atom to look like the following:

Thus, the overall structure can be obtained by looking at those two structure and averaging the bond order between them.  For benzene one resonance form for each bond will be single and the other double $(2 + 1)/2$. The "$2$" on the bottom of the equation is the number of resonance structure being considered.  Thus, for benzene the bond order is $1.5$.

Applying this same method to the carbonate ion, we have 3 resonance structures with bond orders of 2, 1, 1 when considering the bond between carbon and a single oxygen.  Since there are 3 structures the answer is $(2+1+1)/3 = 1.33333$ or $4/3$.