Finding molecules that you randomly create using a modeling kit

Here's a list of all the stuff you can try:

For you, the common theme of all these steps would be to be able to correctly identify the skeleton of the molecule that you built yourself, and then being able to draw it on the online molecular editor.

Sign into Reaxys or SciFinder, draw your structure, and search for molecules or reactions. Note that these require institutional access (or a very hefty fee).
ChemSpider Structure search: It's a free tool, and my personal goto for getting access to several matching compounds. You can search for "exact matches", or partial matches like "tautomers", "same skeleton", or even "all isomers". There's search filters as well. But again, the first few options are sufficient for your snatoms kit.
eMolecules (in case you didn't notice, we also advertise it in our right sidebar) very similar to ChemSpider above.
PubChem Structure search - again, similar to the above two. Note that though this one has an odd (and old-looking UI) which you might not find intuitive.

The main difference between all four of these is primarily the level of detail and depth of their data. Functionally, they all do the same thing. But, they do so on different databases. That means for more complex molecules, their outputs may differ. However, this hopefully won't be a problem with your simple snatoms kit, so you can go ahead and try any of them. Good luck!

Addendum: generating a 3D structure from the molecule name

You may use this free online tool based on JSMol. Type the IUPAC name of the molecule (say (1-methylcyclobutyl)benzene) into the lower right box, and press Enter. Its 3D structure is then immediately transferred into the box above. You can zoom it in or out, or rotate it by 360deg. This would help when you're trying to create the 3D molecular structure given the IUPAC name.


Is trans-2-butene IR active with regard to its C=C bond?

The $\ce{C=C}$ stretch will not show in IR. The frequency is $\nu_{\ce{C=C}}=1700~\mathrm{cm^{-1}}$ and has A gerade symmetry (DF-BP86/def2-SVP). There is no change in the dipole moment.

However, some bending frequencies involving the $\ce{C=C}$ double bond will show up. For example $\nu_{\ce{HC=CH}}=963~\mathrm{cm^{-1}}$ with a medium intensity.


Will 2-methylbutanal undergo aldol condensation

There is no reason why the aldehyde cannot undergo an aldol reaction with itself. Enolates are not rendered unstable by the presence of additional alkyl substituents; in fact they are considered to be more stable. (See: formation of silyl enol ethers with $\ce{Et3N}$/$\ce{Me3SiCl}$.)

However, there is no way it can eliminate $\ce{H2O}$ to form the α,β-unsaturated ketone, since the aldol product does not have any α-hydrogens. Strictly you cannot call it an aldol condensation, although often no distinction is made in the literature between an aldol addition and a condensation.

The formation of the β-hydroxyketone above was reported as long ago as 1924, the reagent being potassium carbonate. More recently, in an attempt to make a 1,3-dioxolane from 2-methylbutanal with potassium t-butoxide and 2-chloroethanol, it was found that the enolisable aldehyde underwent "self-condensation" instead: Org. Lett., 2006, 8, 3745.


An exception to the IUPAC convention for italicisation (typography)

“The overall rule” [1, p. 7] in

The overall rule is that symbols representing physical quantities or variables are italic

a keyword here meaning that not every symbol of a physical quantity should be italicized.
Chemical element symbols are always upright unless they are used to denote a locant in a name of a chemical substance [1, pp. 8–9]:

Symbols for elements in the periodic system shall be roman. […]

Symbols for physical quantities are single, or exceptionally two letters of the Latin or Greek alphabet, but they are frequently supplemented with subscripts, superscripts or information in parentheses to specify further the quantity. Further symbols used in this way are either italic or roman depending on what they represent. […]

[…]

[…] Letter symbols for elements are italic when they are locants in chemical-compound names indicating attachments to heteroatoms, e.g. О-, N-, A-, and P-. The italic symbol H denotes indicated or added hydrogen.

For example, amount concentration of nitrate can be displayed as $c_\ce{NO3-}$, $c(\ce{NO3-})$ or $[\ce{NO3-}]$, where “$c$” is italicized as a symbols for physical quantity, whereas “$\ce{NO3-}$” is upright as it is a specification with its own typography rules.
As for the general use of brackets in written text, according to Wikipedia

In English, typographers mostly prefer not to set brackets in italics, even when the enclosed text is italic.

Further, from Bringhurst' The elements of typographic style [2, p. 85]:

The sloped square brackets usually found on italic fonts are, if anything, even less useful than sloped parentheses. If, perish the thought, there were a book or film entitled The View from My [sic] Bed, sloped brackets might be useful as a way of indicating that the brackets and their contents are actually part of the title.
Otherwise, vertical brackets should be used, no matter whether the text is roman or italic: “The View from My [sic] Bed” and "the view from my [sic] bed."

Similarly, from The Chicago Manual of Style (for earlier edition see this answer on English.SE):

6.5 Parentheses and brackets in relation to surrounding text. Parentheses
and brackets should appear in the same font—roman or italic—as the surrounding text, not in that of the material they enclose. This system, though it may occasionally cause typefitting problems when a slanting italic letter touches a nonslanting roman parenthesis or bracket, has two main virtues: it is easy to use, and it has long been practiced.

Specifically for use in technical documentation and in mathematical notations, brackets could be treated as a notation for operator (like matrix delimiters) [4, p. 316], yet another reason to use roman typeface:

6.2.5.3  Special Bracket Notation for Specific Operators
[…]

Brackets only should be used to indicate chemical concentrations:
$[\ce{S^2-}];$ $[\ce{Na+}];$ and isotopic prefixes: $[\ce{^{32}P}]\text{AMP}.$

The following comparison between the symbols fro amount concentration and specific optical rotatory power illustrates the problems arising from using italicized brackets, namely uneven spacing and clashing symbols:
$$
\begin{align}
&[\ce{^{7}NO3-}] &\quad &\textit{[}\ce{^{7}NO3-}\textit{]} \\
&[α]_λ^θ &\quad &\textit{[}α\textit{]}_λ^θ
\end{align}
$$
References

IUPAC “Green Book” Quantities, Units, and Symbols in Physical Chemistry, 3rd ed.; Cohen, R. E., Mills, I., Eds.; IUPAC Recommendations; RSC Pub: Cambridge, UK, 2007. ISBN 978-0-85404-433-7.
Bringhurst, R. The Elements of Typographic Style, 3rd ed.; Hartley & Marks, Publishers: Point Roberts, WA, 2004. ISBN 978-0-88179-206-5.
The Chicago Manual of Style, 17th ed.; University of Chicago Press: Chicago; London, 2017. ISBN 978-0-226-28705-8.
The Manual of Scientific Style: A Guide for Authors, Editors, and Researchers, 1st ed.; Rabinowitz, H., Vogel, S., Eds.; Elsevier/Academic Press: Amsterdam; Burlington, MA, 2009. ISBN 978-0-12-373980-3.


It's just a matter of definition. If a "layer" is understood as an infinitely thin plane passing through the centers of carbon atoms, then the electron density of the huge delocalized $\pi$ orbital is located above and below that plane, i.e., technically between layers. If a layer is defined so as to include whole atoms, then the said orbital is within layer.

In any case, the delocalized electrons are more or less "free" to move along the planes and not perpendicular to them, which explains the high anisotropy in electrical conductance and other properties of graphite.


Why do we do electrolysis and electroplating using warm electrolyte?

ringo makes good points in his answer. Additionally, though, the increased temperature enhances mass transfer of ions to/from the electrode surfaces by at least two mechanisms:

Higher temperature results in lower electrolyte viscosity, leading to a thinner fluid dynamic boundary layer and concomitant greater mass transfer to/from the electrode surfaces.
Higher temperature also increases the diffusivities of the electrolyte solutes, which also contributes to increased mass transfer to/from the electrodes.


Does higher Oxidation/ reduction potential translate to higher energy storage in battery?

Energy storage depends on the electromotive potential (i.e. difference between species in the electromotive series) and on the number of electrons available.

Li, for example, has an oxidation potential of ~3.04 V relative to hydrogen, but Al has one of 1.66 V, so Li has the greater potential. On the other hand, Li has only one freely available outer electron, but Al has three, so Al stores more charge per mol, giving three times the capacity in coulombs on that basis. On the gripping hand, however, Li has a mass of ~7 amu, and Al is ~27 amu, so an aluminum cell carries around the dead weight of unused neutrons and inner electron/protons.

So, depending on the needs, Li batteries weigh relatively little and provide comparatively high potential, aluminum batteries would hold more charge at a bit lower voltage, and even Zn is useful, though deficient compared to those others in volts and coulombs (it also has a bunch of inert neutrons, but there is no charge for them), because it is relatively inexpensive.


Can I use calcium chloride instead of sodium sulfate as drying agent?

Similar to @MathewMahindaratne's comment: if you know the drying agent suits your needs well enough, keep it.  In particular here, sodium sulfate is a neuter one (as is calcium sulfate, too) is a safe option because while drying the organic layer, the agent will neither protonate, nor deprotonate your product. Thus, if the protocol states $\ce{Na2SO4}$, stick to it.
This is in contrast to calcium chloride which not only will contain traces of calcium hydroxyde when taken from the jar, but upon contact with additional water (a.k.a. while drying your organic solution) will create new $\ce{Ca(OH)2}$ and thus is considered a basic (and potentially deprotonating) drying agent.  Note, $\ce{MgSO4}$ is considered as an acidic drying agent.  Using the wrong salt solutions or drying agents during the workup actually may cleave off (e.g., protecting) groups and thus complicate the isolation and purification of your product.


Is it true that all of the physical laws and mathematics needed to describe chemistry are known?

First, it is one thing to know the basis for a "law", and quite another to mathematically calculate the effects of that law. Consider just carbon, for example... chains may be made of thousands of atoms, with various functional groups attached to each. Though, as Dirac stated, it helps to have "shortcuts" to computation such as Fast Fourier transforms, there are still problems that cannot be solved in a "reasonable" time.

Second, if there are unknown laws, how would we know about it (not to quote Rumsfeld on unknown unknowns)?

And finally, even if all physical laws were known and understood, it would still be impossible to predict everything: Kurt Gödel's incompleteness theorems show that in a complex system (it does not have to be very complex; basic grammar-school mathematics qualifies), questions can be asked that cannot be proved true or false. This extends to chemistry and physics.


This is a really good question which I've never given much thought about.
I'll try to explain the reason as much as I know it. There may be a better explanation.

The reaction can be represented as -
$$\ce{2H_2 + O_2 <=> 2H_2O}$$

Now in O2 you have a structure like
$$\ce{O\dbond O}$$

And in H2O you have a structure like (ignoring the tetrahedral hybridisation and bent shape)

$$\ce{H\sbond O\sbond H}$$

We know that Oxygen is second most electronegative element after fluorine, which is because of it's small size and strong effective nuclear pull towards electrons.

Hydrogen on the other hand, is not as electronegative as Oxygen even after having a small size, because of very less effective nuclear pull.

When a bond between H and O is formed, O pulls the shared pair of electrons towards itself because it is more electronegative than H. This is the "gain" of electrons. The atoms do not actually take up electrons (when it is covalent bond) but pull it towards themselves, thus getting reduced. You will not find this "gain" or pull of electrons in the O-O bond because they have equal pulls, hence 0 oxidation state. But in water, Oxygen has +1 oxidation state.

Similarly, for hydrogen, it pushes the electrons towards Oxygen in water. Not exactly losing the electrons but density of electrons around Hydrogen atom decrease due to this push by hydrogen and pull by Oxygen. Thus it has "lost" it's electron, becoming positively charged which wasn't the case when it was in H2
Also, you'll notice that when more bonds are formed, there is more change in oxidation state of the atom.

Hope this helped.


Why is Density Functional Theory notoriously bad at describing oxygen molecules?

The answer is most certainly the Self interaction error of an electron with itself in the exchange term.

This would probably be more helpful than what I can write here, as I have little time to make this answer more complete.

A. Droghetti PRB 2008

Also have a look at
Klupfel2012


How to perform a total synthesis on paper ? (Think before to make it in lab)

Retrosynthesis, the way to determine which cuts to make and which steps to use in which order (at least in the general picture), is an art. It is not a science or a craft, it is an art. You learn it by doing it and there is only so much you can get taught by others.

Recently, a multi-group seminar I attended started a ‘molecule of the month’ challenge. We were given a molecule and divided into groups. Each group then had about six weeks to come up with a total synthesis plan, find literature precedence or good arguments for all steps and then present it. Our basic strategy was to draw substructures into SciFinder and rely heavily on its reaction search functionality. With more experience and having read more synthetic journal articles, we probably would have known what we would want to go for.

Finally, I feel like adding that sometimes it is not the molecule whose total synthesis you are publishing that piques your interest. There are also those that primarily do method development and then try to find out ‘which molecule could be a good target for the new and super-cool method I just established?’ Thereafter, the synthesis is modelled to fit the desired novel key step. This paragraph is just to show an alternative approach to total synthesis.


Rate Expression for Reactions Involving 2° Benzylic Alkyl Halides

My inclination is that you are correct and the reaction proceeds by an SN2 reaction. Secondary alkyl halides can react by either an SN1 or SN2 mechanism, so the other reaction conditions are important to make a decision of the predominant pathway. Since cyanide is a strong nucleophile (and often utilized as the prototypical SN2 nucleophile), that suggests that the reaction should go through an SN2 pathway. One other piece of information that we don't have is the solvent. If this is run in a protic solvent, that could tip the pathway toward SN1, while polar aprotic will favor SN2. I also would not be surprised if both pathways are operating.

It is true that the aromatic ring will stabilize a carbocation at the benzylic position, but it will also stabilize the transition state of the SN2 reaction.

A brief literature search has revealed less insight than I expected. The closest result I have been able to find is that an optically active benzylic bromide treated with trimethylsilyl cyanide in acetonitrile (polar aprotic solvent) does give an optically active benzylic cyanide, suggesting SN2.

DOI: 10.1055/s-0030-1260187


Why enzyme-catalyzed reactions mostly go one way?

Some enzyme-catalyzed reactions always go in the same direction in a live cell (e.g. reactions of citric acid cycle, RNA polymerase) and others go in either direction depending on circumstances (for example, some enzymes of glycolysis are also involved in the reverse pathway that makes glucose).

As for reasons why a reaction might go in one direction only, you are on the right track. You mentioned two reasons you could imaging for a preference in one direction:

  Does this mean that rate constants are very different in value depending on direction?

If a reaction is far from equilibrium (i.e. proceeds mainly in one direction), the forward rate and the reverse rate will be very different from each other. This could be because the rate constants a very different from each other (with similar concentrations of reactants and products) or because the concentrations of reactants and products are very different from each other, or both. So the safe thing to say is that the rates differ by a lot (rather than the rate constants).

  Or the reason is that reaction product molecules are always quickly consumed by the next step, and as a result the concentration of the product is always low?

This case also occurs. In RNA synthesis, for example, the reaction of a NTP (i.e. a triphosphate) with the growing RNA occurs near equilibrium, yielding longer RNA and diphosphate (pyrophosphate). In a second step, pyrophosphate hydrolyzes quickly, removing one of the products and making the first step virtually irreversible once the second step has occurred.

In the end, whether the reaction goes in both directions or just in one is not dependent on the enzyme: the enzyme is not capable of changing the position of an equilibrium. Whether the reaction goes forward or backward depends on the equilibrium constant of the net reaction (which the enzyme is not a part of) and the concentrations of the reactants and products.

You could envision a scenario where all enzyme-catalyzed reactions happen to be far from equilibrium under physiological conditions, but this is not the case, as the counterexamples mentioned above show.

Enzymes that catalyze reactions that are far from equilibrium are often regulated (turned on and off in various ways) to regulate entire pathways. If a reaction is near equilibrium and is used in different directions by distinct pathways, regulating the enzyme catalyzing the reaction would affect both pathways, so usually these enzymes are not regulated.


Reaction of furan with chloroform in base

The key to this question lies in realising that the Reimer–Tiemann reaction is not an electrophilic aromatic substitution but it is in fact

  addition of a carbanion to a strongly electron-deficient carbene.

This was discussed by @KlausWarzecha and @ron in answers to a previous question.

Therefore the major product will be the one where the attack comes from the carbon with the greatest electron density. Using MolCalc (partly developed by @JanJensen) I generated an electron density plot of furan which clearly shows that there is more electron density on the 3 and 4 carbons than on 2 and 5. This explains the given major product.


Action of KMno4/H+ and heat in this reaction

The $\ce{LiAlH4}$ will reduce the cyclic anhydride to the diol.

$\ce{KMnO4}$ re-oxidises both the alcohol groups (diol formed) to carboxylic acids and also oxidises the double bond to the diacid so the product you have drawn is correct, but it is an intermediate. You then have to consider the effect of heating this (to a pretty high temperature, I would assume).

Product A Heating causes a double dehydration to produce the di-anhydride shown but the right hand pair of acids do not recombine, instead the top acids combine and the bottom acids combine.

Product B Heating dehydrates the right hand diacid to the anhydride, the left hand diacid does a Claisen-ester-type thermal condensation followed by decarboxylation to give a ketone in  the left hand ring.


Receiving milk for working in harmful environment

I know that in India there was a very similar law extant at least back till the 1990's when I had encountered it. Not sure if it is still on the books.

The association with the USSR makes sense: during the cold war a whole bunch of Indian factories were constructed with Soviet aid. And the conventions, technology, etc. came along.

I had heard of several alternative explanations for the milk-law; not sure which (if any) were true. Maybe partly all were.

The Vitamin D or Calcium in milk competing against absorption of Lead or some other nasty element, I forget which the workers were exposed to in chemical factories.
The Milk just contributing to the general health of factory workers in an otherwise grossly malnourished nation (remember, this is cold war era India). Presumably due to the attendant health hazards of chemical factory work, keeping a worker there healthy was more crucial.
Just having the stomach full of a competing innocuous substance (e.g. milk) reduced the propensity of competitive absorption of other nasties in accidents, fugitive exposures etc.


Phosphorine (IUPAC: phosphinine) actually has aromatic character nearly as great (88%) as that of benzene.  According to the reference, phosphorine is sufficiently stable to be handled without air-free techniques; and it undergoes electrophilic substitution reactions similar to those of benzene.


I think your lecturer would have been more correct had he said that the carbon-sulfur bond reacts as if the sulfur is slightly $\delta^{-}$ and the carbon slightly $\delta^{+}$.

Sulphur is a larger atom so it has more, loosely held electrons than carbon.  This means that a sulfur atom is more polarizable than a carbon atom.  Although carbon and sulfur have similar electronegativities, the effect of this increased polarizability for sulfur is that sulfur behaves or reacts as if the bond were polar.  Take the case of nucleophilic attack on a $\ce{C-S}$ bond.  As the nucleophile approaches, the $\ce{C-S}$ bond will tend to polarize because of sulfur's polarizability.  And in which direction will the bond polarize?  Electron density will prefer to shift such that it resides on the larger atom (charge to size effect), sulfur.

This is the same argument used in the case of a carbon-iodine bond.  Again the two atoms have similar electronegativities (carbon = 2.55, iodine = 2.66), so although the bond is not polar based on electronegativity considerations, the polarizability of the iodine allows the bond to react as if it were polar when a nucleophile approaches.


Bonding in Square Antiprismatic Compounds

From what research I've done, there isn't a definitive answer, though people have ideas about what causes this phenomenon. Both sources here cited note the lack of adherence to the VSEPR model, and hypothesize that the steric effects of the eight fluorine atoms prevent the central unbonded lone pair from existing in a definite space, thereby distorting the antiprismatic shape.
To paraphrase an inorganic chemistry textbook:

$\ce{[XeF8]^{2-}}$ is difficult to interpret using VSEPR theory. There is a lone pair on the central xenon atom, which in the VSEPR model would be predicted to occupy a definite position about the central atom as do lone pairs in $\ce{NH3}$, $\ce{SF4}$, and $\ce{IF5}$. In $\ce{[XeF8]^{2-}}$, however, there is no definite location for this central lone pair.
One explanation is based on the degree of crowding around xenon—with eight fluorines attached, repulsions between the xenon-fluorine bonds are too strong to enable a lone pair to occupy a well-defined position. This gives $\ce{[XeF8]^{2-}}$ a very slightly distorted square anti prism structure, with one face being slightly larger than the opposite face. Although this distortion may be a result of the way in which these ions pack in a crystal structure, it is also possible that the distortion is caused by a lone pair exerting some influence on the size of the larger face.

From a piece of literature specifically analyzing the antiprismatic structure of xenon(VI):

We believe that the larger $\ce{Xe-F}$ separations are due to a weak interaction between some fluorine atoms and the neighboring cations, thus causing a slight but significant distortion of the antiprism. However, the distortion might also be explained as a manifestation of some steric activity of the xenon lone pair. Since the eightfold xenon coordination observed here provides no clearly defined ninth coordination position for the lone pair, it appears to offer little support for electron repulsion models which have been used with considerable success in rationalizing the structures of other noble gas compounds.


What is the reasoning behind the resonating structure of halo-benzene?

I think the misunderstanding here (it's a common one) is over the interpretation of resonance structures.

What you have to understand about resonance structures is that they do not actually exist in the molecule. The separate resonance structures are never formed on their own but they can be used to represent a delocalised bonding situation, more accurately described using molecular orbitals. So electrons do not actually move from the halogen to the $\alpha$ and $\beta$ carbons but what the resonance structures represent is that, only considering mesomeric effects, there is a greater electron density on the ortho and para carbons and than on the meta carbons, the carbon attatched to the halogen and the halogen itself.

However, this is not the full picture because halogens; especially fluorine and chlorine, have a very strong inductive effect and so there is actually considerably more electron density on the halogen than on any other atom.

Electrostatic potential map for chlorobenzene:


Writing Topology file for Ligand for Gromacs

First off, you should not be manually writing topology files whenever you can avoid it - it's far too prone to making silly errors. That's why there are so many programs out there that attempt to automate the process.

If however, you just want to understand HOW the topology file is structured, I recommend reading the GROMACS manual (chapter 5). These tutorials also do a good job of describing some of the key aspects of the topology files.


I think this Wikipedia article on hydrogen fluoride will answer your questions.  It discusses the acidity of $\ce{HF}$ from dilute to concentrated aqueous solution, and then pure $\ce{HF}$.  They present four equilibria to describe these situations

\begin{aligned}
\ce{H2O + HF &~<=> [H3O^+ {.} F^{-}]&&(I)}\\
\ce{[H3O^+ {.} F^{-}] &~<=> H3O^+ + F^{-}&&(II)}\\
\ce{[H3O^+ {.} F^{-}] + HF &~<=> H3O^+ + HF2^{-}&&(III)}\\
\ce{3HF &~<=> H2F^+ + HF2^{-}&&(IV)}\\
\end{aligned}

In dilute solution equilibrium (I) lies far to the right and the second equibrium (II) far to the left.  So even though the $\ce{HF}$ is extensively dissociated, the resultant hydrogen bonded ion pair is very stable and reluctant to undergo significant dissociation.  Hence the solution is weakly acidic.  As the concentration of $\ce{HF}$ is increased homodissociation comes into play [equilibrium (III)] producing the hydrogen difluoride ion as a discrete entity which is stabilized through intermolecular hydrogen bonding.  The increase in $\ce{H3O+}$ makes the resultant solution very acidic.  In pure $\ce{HF}$, autoionization occurs producing an exteremly acidic solution.  Page 4 of this link shows a graph of HF acidity from near zero to 50% concentration.


What is the reaction between plain soda and sugar?

Soda is carbonated water - it has carbon dioxide dissolved under high pressure. Carbon dioxide is a gas that would like to escape. Shaking the bottle of soda, for example, provides a suitable mechanism to allow release of the gas.

Adding dry sugar also provides a mechanism for this to occur through nucleation. The sugar crystals give the carbon dioxide a surface via which to escape the solution. This is not a chemical reaction per se, but a physical reaction. It is essentially the same process as the coke-mentos reaction. Sugar crystals are great for this, as they have a rough surface and therefor a very high surface area available.


How difficult is it to synthesize Novichok agents?

According to Fragmentation pathways and structural characterization of
organophosphorus compounds related to the Chemical Weapons
Convention by electron ionization and electrospray ionization
tandem mass spectrometry Rapid Commun. Mass Spectrom. 2016, 30, 2585–2593:

  N-[Bis(dimethylamino)methylidene]-P-methylphosphonamidic
  fluoride (3) was synthesized by the controlled
  addition of N,N,N′,N′-tetramethylguanidine to a solution
  of methylphosphonic difluoride (1). N,N,N′,N′-Tetramethylguanidine (0.60 mmol) and triethylamine (0.65 mmol) in dichloromethane (DCM) (500 μL) were
  added slowly into the solution of methylphosphonic difluoride
  (0.40 mmol) in DCM (500 μL), while stirring at 0–5 °C. After
  30 min, the resulting precipitate was filtered off and the solution was analyzed by GC/MS.

  The corresponding O-alkyl N-[bis(dimethylamino)-methylidene]-P-methylphosphonamidates 4 were synthesized by the addition of ROH to a solution of cpd 3. Sodium hydride powder (NaH, 60% in mineral oil, 0.5 mmol) was washed with dry hexane twice. The hexane was removed and DCM (500 μL) was added and the solution was kept stirring at 0–5 °C. The appropriate alcohol (0.5 mmol) was slowly added into the suspension and the reaction solution was stirred at 0–5 °C for 30 min. Compound 3 in DCM (200 μL) was then added dropwise to the solution, while stirring at 0–5 °C for 2 h. Any precipitate was filtered off and the resulting solutions of the end product were analyzed by GC/MS. It should be noted that, due to the extreme toxicity of these materials, the separation and purification of CWC-related chemical are very difficult and therefore should be carried out only by a trained professional in an efficient fume cupboard equipped with an active charcoal filtration system.


Is silicon dioxide more acidic than the oxides of nitrogen?

I think your question boils down to:

What is the meaning of acidic strength of oxides? How do we determine the acidic strength of oxides?

In general, the electropositive character of the oxide's central atom will determine whether the oxide is acidic or basic. The more electropositive the central atom, the more basic its oxide. The more electronegative the central atom, the more acidic its oxide.

Electropositive character decreases from left to right across the periodic table and increases down a group.
The trend of acid-base behaviour is from strongly basic oxides on the left-hand side to strongly acidic ones on the right, via an amphoteric oxide (for example, aluminium oxide) in the middle.

This trend applies only to the oxides of the individual elements  in the highest oxidation states for those elements. The pattern is somewhat less clear for other oxides.

A non-metal oxide's acidity is defined in terms of the acidic solutions formed as a result of reaction(s) with water. For example, sulfur trioxide reacts with water to forms sulfuric acid.

In general, acidic oxides are oxides of nonmetals and basic oxides are oxides of nonmetals.

Note: There are three nonmetal oxides from the upper right portion of the periodic table, $\ce{CO}$, $\ce{NO}$, and $\ce{N2O}$, which have low (not the highest possible) oxidation numbers for the central atom that they give neutral aqueous solutions.

Since the acidity of a cation rises rapidly with charge, d-block elements which exhibit a wide variety of oxidation numbers may have one or more oxides that exhibit only basic properties and one or more oxides that exhibit only acidic properties.

In general, the higher the oxidation number, the more acidic the corresponding oxide (similarly, reverse trend for basic character of oxides). Chromium makes a decent example here, as it exhibits a variety of oxidation states. $\ce{CrO}$ is basic, $\ce{Cr2O3}$ is amphoteric and $\ce{CrO3}$ is acidic.

P.S.
Let me introduce you to a theory developed by Hermann Lux and Håkon Flood, named Lux-Flood theory to explain the acidic or basic character of oxides. Here's what they had to say:

  An acid of Lux-Flood is an acceptor of $\ce{O^2-}$
  A base of Lux-Flood is a donor of $\ce{O^2-}$

For example:

$\ce{CaO}$ is a base of Lux-Flood because $\ce{CaO}=\ce{Ca^2+}+\ce{O^2-}$

$\ce{SiO2}$ is an acid of Lux Flood because $\ce{Si}$ has vacant d orbitals, allowing it to accept more $\ce{O^2-}$ ions.

The two compounds may react to give you $\ce{CaSiO3}$.


How are arsenic compounds removed from drinking water?

The US Environmental Protection Agency (EPA) has a page dedicated to arsenic in drinking water.  It contains a section that suggests ways to remove it from water.

Disclaimer: I have no idea if the arsenic removal methods on the linked page actually work.  You should consult someone trained in the science of water purification for definitive information.


What did I do wrong in my temperature change calculations?

You're not really wrong: the computer program is apparently expecting a float and not an integer, which is really stupid because the question is about temperature change and not the vagaries of significant figures.
Instead of you learning about heat capacity and the relation to temperature change, you're instead learning about significant figures first, and the chemistry/physics part second.
I find this sort of thing patently awful and a frustrating way for you or anyone else to "learn" anything about the principles.
You have to report your answer to 2 significant figures:
750 has 2 (the limiting number); 31500 has 3; and 4.184 has 4.
$$Q = m\times c\times \Delta T$$
or
$$\Delta T = {Q\over m\times c}$$
so
$$\Delta T = {31,500\ {\rm J}\over 750\ {\rm g} \times 4.184\ {\rm J}\cdot {\rm g}^{-1}\cdot {\rm ^{\circ}C}^{-1}}$$
which, to 2 significant figures, is:
$$\Delta T = 10.\ ^{\circ}{\rm C}$$
Notice I have included a decimal point, so the now ridiculous answer that should be accepted is 10. and not 10
Meanwhile, you've lost a lot of time wondering where your calculation went wrong because some programmer doesn't know how to cast. Or maybe you're really supposed to report numbers with a terminal decimal point, and I'll challenge anyone to find any respected publication where that happens:

"The temperature change, in degrees Celsius, is 10.."

Note that 10.0 has three significant figures, so that's also "wrong."
And that, my friend, is why "learning" from a computer is a lazy way out for the "teacher."


Why does the phosphorus trifluoride molecule have dipole moment?

Fluorine is more electronegative than phosphorus.
The phosphorus is at the apex of a pyramid, the base of the pyramid being an equilateral triangle with a fluorine atom at each vertex.
The F-P-F angles are 96 degrees.
Each F-P bond contributes to the net dipole moment, as a vector from the P to the F.  The net dipole moment is the vector sum of the vectors along the three P-F bonds.
If all four atoms were in a plane (a trigonal planar geometry), there would be no net dipole moment.  But because the geometry is pyramidal, the vector sum is not zero and there is a net dipole moment.
In some molecules lone pairs also contribute to the net dipole moment, in this particular case the lone pair of phosphorus is in a mostly s-like orbital, so in first approximation it can be neglected.


I myself was always confused why $\ce{H3O^+}$ is so well-known and yet almost nobody talks of $\ce{H4O^2+}$. I mean, $\ce{H3O^+}$ still has a lone pair, right? Why can't another proton just latch onto that? Adding to the confusion, $\ce{H4O^2+}$ is very similar to $\ce{NH4+}$, which again is extremely well-known. Even further, the methanium cation $\ce{CH5+}$ exists (admittedly not something you'll find on a shelf), and that doesn't even have an available lone pair!

It is very useful to rephrase the question "why is $\ce{H4O^2+}$ so rare?" into "why won't $\ce{H3O^+}$ accept another proton?". Now we can think of this in terms of an acid-base reaction:

$$\ce{H3O^+ + H+ -> H4O^2+}$$

Yes, that's right. In this reaction $\ce{H3O^+}$ is the base, and $\ce{H^+}$ is the acid. Because solvents can strongly influence the acidity of basicity of dissolved compounds, and because inclusion of solvent makes calculations tremendously more complicated, we will restrict ourselves to the gas phase (hence $\ce{(g)}$ next to all the formulas). This means we will be talking about proton affinities.

Before we get to business, though, let's start with something more familiar:

$$\ce{H2O(g) + H+(g) -> H3O^+(g)}$$

Because this is in the gas phase, we can visualise the process very simply. We start with a lone water molecule in a perfect vacuum. Then, from a very large distance away, a lone proton begins its approach. We can calculate the potential energy of the whole system as a function of the distance between the oxygen atom and the distant proton. We get a graph that looks something like this:

For convenience, we can set the potential energy of the system at 0 when the distance is infinite. At very large distances, the lone proton only very slightly tugs the electrons of the $\ce{H2O}$ molecule, but they attract and the system is slightly stabilised. The attraction gets stronger as the lone proton approaches. However, there is also a repulsive interaction, between the lone proton and the nuclei of the other atoms in the $\ce{H2O}$ molecule. At large distances, the attraction is stronger than the repulsion, but this flips around if the distance is too short. The happy medium is where the extra proton is close enough to dive into the molecule's electron cloud, but not close enough to experience severe repulsions with the other nuclei.

In short, a lone proton from infinity is attracted to a water molecule, and the potential energy decreases up to a critical value, the bond length. The amount of energy lost is the proton affinity: in this scenario, a mole of water molecules reacting with a mole of protons would release approximately $\mathrm{697\ kJ\ mol^{-1}}$ (values from this table). This reaction is highly exothermic

Alright, now for the next step:

$$\ce{H3O^+(g) + H+(g) -> H4O^2+(g)}$$

This should be similar, right? Actually, no. There is a very important difference between this reaction and the previous one; the reagents now both have a net positive charge. This means there is now a strong additional repulsive force between the two. In fact, the graph above changes completely. Starting from zero potential at infinity, instead of a slow decrease in potential energy, the lone proton has to climb uphill, fighting a net electrostatic repulsion. However, even more interestingly, if the proton does manage to get close enough, the electron cloud can abruptly envelop the additional proton and create a net attraction. The resulting graph now looks more like this:

Very interestingly, the bottom of the "pocket" on the left of the graph (the potential well) can have a higher potential energy than if the lone proton was infinitely far away. This means the reaction is endothermic, but with enough effort, an extra proton can be pushed into the molecule, and it gets trapped in the pocket. Indeed, according to Olah et al., J. Am. Chem. Soc. 1986,  108 (5), pp 1032-1035, the formation of $\ce{H4O^2+}$ in the gas phase was calculated to be endothermic by $\mathrm{248\ kJ\ mol^{-1}}$ (that is, the proton affinity of $\ce{H3O^+}$ is $\mathrm{-248\ kJ\ mol^{-1}}$), but once formed, it has a barrier towards decomposition (the activation energy towards release of a proton) of $\mathrm{184\ kJ\ mol^{-1}}$ (the potential well has a maximum depth of $\mathrm{184\ kJ\ mol^{-1}}$).

Due to the fact that $\ce{H4O^2+}$ was calculated to form a potential well, it can in principle exist. However, since it is the product of a highly endothermic reaction, unsurprisingly it is very hard to find. The reality in solution phase is more complicated, but its existence has been physically verified (if indirectly).

But why stop here? What about $\ce{H5O^3+}$?

$$\ce{H4O^2+(g) + H+(g) -> H5O^3+(g)}$$

I've run a rough calculation myself using computational chemistry software, and here it seems we really do reach a wall. It appears that $\ce{H5O^3+}$ is an unbound system, which is to say that its potential energy curve has no pocket like the ones above. $\ce{H5O^3+}$ could only ever be made transiently, and it would immediately spit out at least one proton. The reason here really is the massive amount of electrical repulsion, combined with the fact that the electron cloud can't reach out to the distance necessary to accommodate another atom.

You can make your own potential energy graphs here. Note how depending on the combination of parameters, the potential well can lie at negative potential energies (an exothermic reaction) or positive potential energies (an endothermic reaction). Alternatively, the pocket may not exist at all - these are the unbound systems.

EDIT: I've done some calculations of proton affinities/stabilities on several other simple molecules, for comparison. I do not claim the results to be quantitatively correct.

$$
\begin{array}{lllll}
\text{Species} & \ce{CH4}  & \ce{CH5+}    & \ce{CH6^2+}     & \ce{CH7^3+}     & \ce{CH8^4+}     \\
    \text{Stable in gas phase?} & \text{Yes} &  \text{Yes}   & \text{Yes} & \text{Yes}  & \text{No} \\
\text{Approximate proton affinity}\ (\mathrm{kJ\ mol^{-1}})  & 556 & -246 & -1020   & N/A   & N/A  \\
\end{array}
$$

Notes:

Even without a lone pair, methane ($\ce{CH4}$) protonates very exothermically in the gas phase. This is a testament to the enormous reactivity of a bare proton, and the huge difference it makes to not have push a proton into an already positively-charged ion.
For most of the seemingly hypercoordinate species in these tables (more than four bonds), the excess hydrogen atoms "pair up" such that it can be viewed as a $\ce{H2}$ molecule binding sideways to the central atom. See the methanium link at the start.

$$
\begin{array}{lllll}
\text{Species} & \ce{NH3}  & \ce{NH4+}    & \ce{NH5^2+}     & \ce{NH6^3+}      \\
    \text{Stable in gas phase?} & \text{Yes} &  \text{Yes}   & \text{Yes} & \text{No}  \\
\text{Approximate proton affinity}\ (\mathrm{kJ\ mol^{-1}})  & 896 & -410 & N/A   & N/A    \\
\end{array}
$$

Notes:

Even though the first protonation is easier relative to $\ce{CH4}$, the second one is harder. This is likely because increasing the electronegativity of the central atom makes the electron cloud "stiffer", and less accommodating to all those extra protons.
The $\ce{NH5^{2+}}$ ion, unlike other ions listed here with more than four hydrogens, appears to be a true hypercoordinate species.  Del Bene et al. indicate a five-coordinate square pyramidal structure with delocalized nitrogen-hydrogen bonds.

$$
\begin{array}{lllll}
\text{Species} & \ce{H2O}  & \ce{H3O+}    & \ce{H4O^2+}     & \ce{H5O^3+}      \\
    \text{Stable in gas phase?} & \text{Yes} &  \text{Yes}   & \text{Yes} & \text{No}  \\
\text{Approximate proton affinity}\ (\mathrm{kJ\ mol^{-1}})  & 722 & -236 & N/A   & N/A    \\
\end{array}
$$

Notes:

The first series which does not accommodate proton hypercoordination.
$\ce{H3O+}$ is easier to protonate than $\ce{NH4+}$, even though oxygen is more electronegative. This is because the $\ce{H4O^2+}$ nicely accommodates all protons, while one of the protons in  $\ce{NH5^2+}$ has to fight for its space.

$$
\begin{array}{lllll}
\text{Species} & \ce{HF}  & \ce{H2F+}    & \ce{H3F^2+}     & \ce{H4F^3+}     \\
    \text{Stable in gas phase?} & \text{Yes} &  \text{Yes}   & \text{Yes} & \text{No}  \\
\text{Approximate proton affinity}\ (\mathrm{kJ\ mol^{-1}})  & 501 & -459 & N/A   & N/A    \\
\end{array}
$$

Notes:

Even though $\ce{H3F^2+}$ still formally has a lone pair, its electron cloud is now so stiff that it cannot reach out to another proton even at normal bonding distance.

$$
\begin{array}{lllll}
\text{Species} & \ce{Ne}  & \ce{NeH+}    & \ce{NeH2^2+}    \\
    \text{Stable in gas phase?} & \text{Yes} &  \text{Yes}   & \text{No}   \\
\text{Approximate proton affinity}\ (\mathrm{kJ\ mol^{-1}})  & 204 & N/A   & N/A    \\
\end{array}
$$

Notes:

$\ce{Ne}$ is a notoriously unreactive noble gas, but it too will react exothermically with a bare proton in the gas phase.
Depending on the definition of electronegativity used, it is possible to determine an electronegativity for $\ce{Ne}$, which turns out to be even higher than $\ce{F}$. Accordingly, its electron cloud is even stiffer.

$$
\begin{array}{lllll}
\text{Species} & \ce{H2S}  & \ce{H3S+}    & \ce{H4S^2+}     & \ce{H5S^3+}     & \ce{H6S^4+}     \\
    \text{Stable in gas phase?} & \text{Yes} &  \text{Yes}   & \text{Yes} & \text{Yes}  & \text{No} \\
\text{Approximate proton affinity}\ (\mathrm{kJ\ mol^{-1}})  & 752 & -121 & -1080   & N/A   & N/A  \\
\end{array}
$$

Notes:

The lower electronegativity and larger size of $\ce{S}$ means its electrons can reach out further and accommodate protons at a larger distance, while reducing repulsions between the nuclei. Thus, in the gas phase, $\ce{H2S}$ is a stronger base than $\ce{H2O}$. The situation is inverted in aqueous solution due to uniquely strong intermolecular interactions (hydrogen bonding) which are much more important for $\ce{H2O}$.
$\ce{H3S+}$ also has an endothermic proton affinity, but it is lower than for $\ce{H3O+}$, and therefore $\ce{H4S^2+}$ is easier to make. Accordingly, $\ce{H4S^2+}$ has been detected in milder (though still superacidic!) conditions than $\ce{H4O^2+}$.
The larger size and lower electronegativity of $\ce{S}$ once again are shown to be important; the hypercoodinate $\ce{H5S^3+}$ appears to exist, while the oxygen analogue doesn't.


Why does the lone pair of phosphine not contribute to its dipole moment?

Let's see if we can use the following picture to help clarify things.

Let's look at the molecule $\ce{H3X-Z}$.  Let's say that the molecule is polarized such that the 2 electrons in the $\ce{X-Z}$ bond spend more time around $\ce{X}$ and less around $\ce{Z}$.  Let's also say that the $\ce{X-Z}$ bond is $\ce{sp^3}$ hybridized. Since the bond is polarized it will contribute to the molecular dipole moment.  Are you with me so far?

If we stretch that bond and look at the bond right before it breaks, it will still be polarized, but with a larger bond dipole moment because the electrons are still predominately located around $\ce{X}$ and the distance between those electrons and the electron depleted $\ce{Z}$ nucleus has increased.  If we stretch the bond a bit further it will break and we will wind up with two electrons around $\ce{X}$.  These electrons around $\ce{X}$ are still in an $\ce{sp^3}$ orbital as shown in the bottom part of the picture.  The orbital and the electrons in it still "point" towards where $\ce{Z}$ used to be.  The orbital and the electrons in it still have directionality.  They still contribute to a dipole moment.  Can you see this?

Let's continue assuming you can.  Let's repeat all that we've done above with one change.  Let's now say that the $\ce{X-Z}$ bond is $\ce{s}$ hybridized.  Now when we get to the last step and break the $\ce{X-Z}$ bond, the electrons remain in an $\ce{s}$ orbital.  An $\ce{s}$ orbital is spherically symmetric.  It has no directionality, it doesn't "point" in any specific direction because it is spherically symmetric.  So in this case we do not have a contribution to the molecule's dipole moment from the electrons in the spherically symmetric orbital.


Conformations of cyclohexane and flipping

Essentially, it is stating that a ring flip retains the absolute stereochemical configuration. Ring flipping is a physical phenomenon of 6-membered rings, not a chemical reaction that interconverts stereoisomers. No bonds are broken, though many do get strained in the process.

I'm not sure what you mean by your question about the importance of being in the axial/equatorial positions. Both the methyl and chloro groups are vying to reduce 1,3-diaxial groups by placing themselves in the equitorial position, but doing so for one places the other in the axial position. I think a natural question that arises "is which one wins?"


Extracting Zinc (oxide) from baby powder

This might be not worth it, because you will need special equipment.

Description totally theoretical.

Mixture consist of $Mg_3Si_4O_{10}(OH)_2$ - tale powder, $ZnO$ - zink oxide powder, and starch. 81% to 15% to 4% it's most common variation with 15% zink oxide, there also 10% mixtures, but process will be the same.

You need to get rid of starch. Best choice to convert it to dextrine, because starch in not fully soluble in water. You need to heat it in oven on a foil for about 2 hours, with temperature $200^{\circ}C$. It will have no effect for other components, zync oxide might colorize to orange, but when heat is gone it will be white again. Dextrine will be yellow color and it's fully soluble in water.
Mix new mixture with water, and pour it out. Now you have only $Mg_3Si_4O_{10}(OH)_2$ and $ZnO$ in you mixture. Both of them not soluble in water.
Then you need to obtain or build equipment for sublimation - sublimation apparatus. Because compounds have different melting point Talc - $800^{\circ}C$, zync oxide - $1975^{\circ}C$, you can get them one by one. You will need to freeze and dry your mixture first.
Heat over $800^{\circ}C$ to obtain talc, left over structure will be zinc oxide.

I'm not sure about sublimation from start, with starch inside, but this method in theory should work.


Sir Arthur7: From your comments you appear to know how to liberate CO2 and a proton by pushing arrows as illustrated by the blue arrows in generic α-oxoacid 1. You are correct in discrediting acyl anion 5 for two reasons. First, it is unstabilized and secondly, its protonation would provide an aldehyde, not a carboxylic acid. So why not push the red arrows in 1 to generate an acyl cation equivalent? The downside is that you don't want hydride as a leaving group. The solution is to replace the acidic hydrogen of 1 with a species that likes electrons, e.g., a high oxidation state metal or a hypervalent, non-metallic atom. Certainly, the nitrogen of HNO3 fits the bill where M = NO2 in structure 2. In this instance, nitrogen is n = 5 in structure 2 and 4. Bear in mind that for the purpose of electronic bookkeeping, transformation of the acyl anion 5 to a acyl cation 2 is a 2-electron oxidation while species M undergoes a 2-electron reduction. ---continued---

Moriarity, et. al.,1 have effected this oxidative decarboxylation with the hypervalent iodine species, iodosobenzene 7. The path of the red arrows within the brackets provides a step-by-step version of their mechanism using α-oxoglutaric acid 6.2 Notice that mixed "anhydride" 8 is the operational equivalent of species 2. The authors propose the formation of cyclic intermediate 9 that collapses to 12 with loss of CO2 and a proton. [Note: Structure 10 is drawn for clarity. Collapse of 9 to succinic acid 12 is proposed as concerted.] The collapse of structure 8 might be thought to follow the course of the green arrows but because the reaction is conducted in anhydrous dioxane, it is unlikely that an acylium ion is formed directly because it is apt to form succinic anhydride. See examples studied in the chart below.

The Moriarity paper describes the iodosobenzene oxidation of mandelic acid to benzaldehyde and CO2. See if you can now write a mechanism for this reaction.

1) R. M. Moriarty, S. C. Gupta, H. Hu, D. R. Berenschot, and K. B. White, J. Am. Chem. Soc., 1981, 103, 686. https://pubs.acs.org/doi/10.1021/ja00393a040
2) α-Oxoacid is preferred over α-ketoacid. The term keto (C=O) overcounts carbons. Oxo = (=O).


Prussian blue contains both $\ce{Fe(II)}$ and $\ce{Fe(III)}$. Mixed-valence compounds, such as the one in question, absorb certain wavelengths of visible light resulting due to the phenomenon of inter valence charge transfer (IVCT).
General Scheme for IVCT is:
$$\ce{L_nM^+-bridge-M'L_n + h\nu ->L_nM-bridge-M'+L_n}$$

In this case, orange-red light around $680\ \mathrm{nm}$ in wavelength is absorbed, and the reflected light appears blue as a result.

Additionally, changes colour from blue to colourless upon reduction. This change is caused due reduction of the $\ce{Fe(III)}$ to $\ce{Fe(II)}$, and thereby eliminating the IVCT responsible for Prussian blue's namesake blue colour. This is also why Prussian Blue exhibitis electrochromic behaviour.

For a more detailed description read this and this. Alternatively, you can look it up in a textbook.


Overpotential for photocatalytic water-splitting with TiO2

Photon absorption by titania results in an excitation of an electron from the valence band to the conduction band.  The difference in energy of an electron in the valence band and the conduction band is the band gap, which for titanium dioxide is about 3.3 eV.

What matters for electrochemistry, though, is the absolute voltage of (i) the excited electron, and (ii) the absolute voltage of the "hole" left in the valence band, not the just the difference between them (gap).  That is far more difficult.

The effective electrochemical voltages of the excited electron (or hole) can differ from what the desired electrochemical reaction is.  So there could still be an overpotential.  For titania, one paper I found estimates of the absolute energies of the valence band "hole" and the conduction band electron of -7.8 eV and -4.5 eV, respectively (at pH 0).  The thermodynamic potential for water oxidation is -5.8 eV, 2 eV higher than the hole energy.  That energy is lost if the hole is used to power water oxidation.  You could thus view excited titania having an overpotential of 2 eV for the water oxidation reaction.  In contrast, the theoretical potential for proton reduction is -4.5 eV, close to the conduction band electron energy.

Thus the energetics of bulk titania provide lots of overpotential for the water oxidation reaction but the tradeoff is that a high band gap means only high-energy photons can power the electrochemistry.  Lowering the band gap would make more and longer wavelengths of radiation available to power water photolysis, but the band gap can only be lowered in certain ways.  For titania, it would not be very desirable to lower the energy of the conduction band, because that would mean excited electrons would not be able to reduce protons to hydrogen .  Thus, the only way to lower the band gap is by raising the energy of the valence band, which would both lower the overpotential and and make more types of radiation accessible for powering water photolysis.


How do you explain very high vanadyl phosphate NMR chemical shifts?

$\ce{V^{4+}}$ (and $\ce{V^{3+}}$) is paramagnetic, and therefore you would expect a very large change in chemical shift. Paramagnetic shifts arise from the interaction of the observed nucleus with unpaired electron spin of the paramagnetic metal ion.
The observed isotropic shift comes from two interactions: the contact shift and pseudo-contact shift.

The contact shift comes from the direct interaction between the observed nucleus and the unpaired electron - a through bond effect. This gives very large shifts - many hundreds of ppm.

The indirect pseudocontact shifts essentially come from the interaction of the unpaired electron density with the electron density of the orbitals of the observed nucleus, and are effectively compensating for the nonzero magnetic field of the paramagnetic source - a through space effect. These are much smaller shifts, often within the usual observed chemical shift range.

Whilst the measurement of contact shift information is infrequently used for structural insight, pseudocontacts using paramagnetic lanthanides are used extensively for 3D solution structures of proteins.


How to calculate initial and final temperatures during supercritical steam expansion?

Since supercritical steam is not an ideal gas, you might want to use standard reference data (e.g. provided by NIST).

Initial state
mass m0 = 10 kg
pressure p0 = 500 bar = 50.0 MPa
volume V0 = 0.01 m3
density ρ0 = 1000 kg/m3
temperature T0 = 340.53 K
specific enthalpy h0 = 323.06 kJ/kg

Final state
mass m1 = 10 kg
pressure p1 = 500 bar = 50.0 MPa
volume V1 = 0.07 m3
density ρ1 = 142.86 kg/m3
temperature T1 = 925.86 K
specific enthalpy h1 = 3453.11 kJ/kg

Heat transfer
heat Q = ΔH = (h1 – h0) • m = 31.3 MJ


Finding the major contributing tautomer of 1-(propan-2-yl)-1,2,3,5-tetrahydro-6H-pyrazolo[3,4-b]pyridin-6-one

In keto-enol tautomerism, the equilibrium is mostly towards the keto form, as it should be, since $\ce{C=O}$ is a very strong and stable bond. In some cases however we find that the equilibrium sometimes shifts to enol form. This happens in cases where some additional benefit is done by the enol structure. here are two examples:
A) Removing anti-aromaticity or adding aromaticity to a ring.
B) Extended conjugation formation, providing additional stability.
In the structure you have shown, the enol form is more stable because it adds aromatic character to the ring. This additional stability drives the equilibrium
towards the enol form.


Is there a food-safe hydrophobic liquid with a substanially higher density than water?

Hello I know I went late to the party, but I have a solution to this. You all already have the right idea, but you need to put your ideas together.

Double layer your pool

Whether it's gallium, water, oil, or some other liquid it will all work and be safe under glass.

Their are hydrophobic resins that you can treat the glass with, and if you double paned the area that you're waterfall would be under, it would not cause a health problem because there would be no contact.

For example if you had made window panes in your pool (let's just say in a t-shaped stretching halfway under your pool), and you coated the window panes with a hydrophobic resin, and you kept it under let's just say a temperature controlled environment, that would one solve your pool problem of being boring and two, keep your waterfall from fogging up.

So in short
1 Removable hydrophobic panels for cleaning
2 Heated/ chilled pool or temperature controlled environment
3 Heated / cooling pump for waterfall
4 Well designed lighting ( you'd be surprised what a little bit of light in the right spot can do)

I do not know if it's available to the public yet but there is a coolant used for computers that is one consumable and two low maintenance called Coriant. It has a lighter density than water, if you're  panes are hydrophobate this would not present a problem.

Colored mineral oil or antifreeze would also work here.

If you use antifreeze and we're creative with its use, it may assist in some other mechanical uses as well as being a display piece.

Please note that any hydrophobic material or substance is going to be toxic to breathe or have extended contact with.

This should not be a problem if it is under a air / water tight encasing.

I apologize that I'm so late in responding to this but I hope someone reads this and finds it useful thank you for your time.


Will group 2 chlorides dissolve in water in water or react to produce distinct products? How does this differ by the type of salt?

Have you used anhydrous $\ce{CaCl2}$, dihydrate $\ce{CaCl2 . 2 H2O}$  or hexahydrate $\ce{CaCl2 . 6 H2O}$ ( 3 most common forms) ? It is a big difference in the resulting thermal effect.
Hexahydrate causes cooling down of the solution while being dissolved. If ice is used instead of water,  as the mixture hexahydrate : ice  2 : 1, it forms the famous freezing mixture, cooling itself down to $\pu{ -50 ^{\circ}C}$.
Anhydrous calcium chloride causes heating up while being dissolved. It is because of releasing energy of ion hydration. Rather then smoking, it was condensed water vapor.
Dihydrate is somewhere between, with roughly neutral thermal effect.
Interaction of the group 2 element chlorides with water is generally safe. There is very slight hydrolysis, growing progressively upwards along the group. The beryllium chloride hydrolysis is significant, releasing hydrogen chloride. Note that barium and especially beryllium are poisonous.


Why doesn't Epsom salt dissolved in water act like sulfuric acid?

Well, as you might have guessed, it's the $\ce{H+}$ in sulfuric that is dangerous, not the $\ce{SO4^2-}$. So, Epsom salts, being $\ce{MgSO4}$, aren't very dangerous.

In the same vein, table salt ($\ce{NaCl}$) is perfectly OK despite hydrochloric acid ($\ce{HCl}$) being rather nasty.

And $\ce{H+}$ is only corrosive in high concentrations, mind you - otherwise we'd be burning ourselves drinking orange juice, which itself has a small concentration of $\ce{H+}$. That also answers why water doesn't kill you: only a tiny fraction of it (roughly 1 in ten million molecules) actually undergoes autodissociation at any one time. It's not about the actual number of $\ce{H+}$ ions, but rather the concentration. If you dilute $\ce{HCl}$ enough you can pretty much just drink it. (Please don't be stupid and take this as a licence to do a home experiment and drink $\ce{HCl}$.)


Is temperature just the phenomenon of particles colliding against an object?

Heat transfer occurs when particles collide against an object (which is made of other particles) or when photons impact on the object.

Temperature is a measure of the average amount of kinetic energy of the particles (in a gas this will be simple related to how fast the particles of gas move but in a solid it will be about vibrations of the particles).

The sensation of heat is a product of both thermal radiation (infra-red photons are emitted by the hot object and absorbed by your hand) and contact with something that can transfer some of that energy directly (hot gas particles or a hot solid directly touching your hand). Living things have evolved to sense heat and cold because they only survive if they avoid extremes of both.


What is the precipitate of this reaction of magnesium?

With $\ce{(NH4)2HPO4}$, $\ce{Mg^2+}$ forms a white, crystalline precipitate of magnesium ammonium phosphate hexahydrate:

$$\ce{Mg^2+ + HPO4^2- + NH4^+ + OH- + 5H2O \rightleftharpoons Mg(NH4)PO4 \cdot 6H2O\downarrow}$$

This is a very sensitive test for $\ce{Mg^2+}$; however, calcium, strontium, barium and other heavy metal ions also form precipitates of phosphates under these conditions. $\ce{Mg(NH4)PO4 \cdot 6H2O}$ can be distinguished by the form of its crystals when they are examined under a microscope. It forms rhombic crystals with prismatic shape, and is less microcrystalline than other earth alkali metal phosphates (sources: 1, 2).


How to save the force on a molecule in GROMACS?

Gro output format does not support forces, so using it as an intermediate output format defeats your purpose.

Use

gmx traj -f your.trr -of

which runs the gmx wrapper binary, calling the traj module to do trajectory processing, names your input trajectory file after -f and triggers the output of forces with -of (optionally named after -of, defaulting to force.xvg), or similarly

gmx dump -f your.trr

to dump the entire contents of the file including forces) to the terminal.

See also

gmx help traj

or

gmx traj -h

or http://manual.gromacs.org/documentation/5.1/onlinehelp/gmx-traj.html

Note that I currently develop GROMACS.


When we blow on/stir a hot drink, why aren't we making it hotter?

This is related to a concept called "vapour pressure". Liquid water (and any other liquid for that matter) always has some amount of water vapour that has to be in equilibrium with it. For example, the air above water of 5 °C has about 1% of water vapour, 40 °C has about 7% of water vapour and at 90 °C it has about 70% of water vapour.

If there is too much vapour in the air for the temperature it is, then water condensates. That's why you get condensation on your glasses when you walk from a hot humid room outside into the cold at winter. If there is not enough vapour in the air, water evaporates to fill in for the missing vapour. That's why taking off your shirt after you've exercised and sweated makes you dry faster. The air trapped between your skin and clothes is very humid and in equilibrium with the sweat on your skin. Once you take it off, dry air comes into contact with the skin, forcing evaporation of your sweat (even though you're below the boiling point of water).

This process of evaporation is endothermic, meaning it consumes heat. Whenever water is evaporating, the temperature of the water goes down. When you're blowing air on a hot drink, you're basically replacing the humid air on top of the drink with dry air. This allows the water to evaporate more easily, reducing the overall temperature of the drink. This is also why you're feeling cold when you got wind blowing on your when you're sweating.


What mechanisms might make a substance travel different distances on the same TLC plate?

are there other reasons why the components of a mixture should have different Rf values when spotted separately?

One might speculate that two polar reactants form a less polar "adduct". This should however result in an additional spot, rather than in altered $\mathrm{R_f}$ values for the individual compounds.

But this is speculation. Admittedly, I have never used the cross spot method. Neither have I ever observed that spots of the reactants had different $\mathrm{R_f}$ values in the reaction mixture.

On a side note: What always worked for me was:

Don't apply spots but bands and overlap them.
"Sharpen" the bands on the base line by developping the TLC in methanol or another polar solvent first.

Together, these two tricks allow to destinguish between two compounds with very similar  $\mathrm{R_f}$ values.


Corrosion is a balancing game. In this case, by covering the beakers, you have limited the driving force for oxygen dissolution by limiting the amount of oxygen available to dissolve into solution and then the higher salt concentrated solution is able to further minimize the amount of oxygen that can be dissolved and participate in the redox reaction (as explained in the linked question's answer).


The short answer to this question is: yes.

The movement of particles in suspension like the powder in water would be a perfect example of Brownian motion as Mithoron already said. The particles are being kicked around by the water molecules, which are in constant motion as you already said.

This is also true for small airborne particles that collide with fast moving gas molecules.


How does the vapor pressure in a vessel affects the total pressure after a combustion reaction?

First, kudos for actually showing work on a problem!

Second, I'd solve this by inspection, rather than using the Ideal Gas Law, where time is critical on a test.

As you correctly calculated, the limiting reactant is $\ce{H2}$. Therefore the pressure of $\ce{H2}$ is nil; all the $\ce{H2}$ is in the water.

Half of the $\ce{O2}$ is used, reacting with the $\ce{H2}$ to make $\ce{H2O}$, so the remaining partial pressure of $\ce{O2}$ is ~157 torr.

The volume taken up by the liquid water is ~1/2,000 that of the original gases, so the volume left in the flask is still ~3 l, though slightly less. Since the answer choices are not close together, and the decrease in volume is negligible, I'd ignore it.

The vapor pressure of water at 25° C is ~24 torr, and at 30° C it's ~32 torr, so at 27° C, it would be about halfway, by linear interpolation, say ~28 torr.

The sum of those partial pressures is close to an answer choice.

Trivia: This question seems to be citing pi with the initial value, 3[.]14 torr. If the temperature had been -8° C (ice at that temperature has a v.p. ~2 torr), the answer would have been 159 torr, so the first six digits of pi could have been used; i.e. 314 ->159.


Is there a difference between NaCl or table salt (processed) and salt from the Himalayas?

No salt will be pure NaCl.  Each will have some degree of other elements.  The fact that the salt isn't white confirms there are other elements present.

See Analysis of Gourmet Salts for the
Presence of Heavy Metals which investigates 14 salts including two Himalayan salts.

See especially "Table 3. Comparison of toxic elements in Table Salt".

A "Himalayan Pink Fine Mineral Salt" was found to have the highest level of cadmium and tied for first for the highest level of nickel.

NaCl isn't molecular, but instead is a lattice of Na+ and Cl- ions.  Other ions can lie within this lattice as impurities.


I posted an answer here the other day which was completely wrong! I apologise and have left the incorrect text below for posterity to record how wrong people who should know better can be!

Looking on the NIST fluid properties database I see that the specific heat capacity (i.e. the heat capacity per unit mass) of D2O is just 1.5% higher than the specific heat capacity of H2O. So at 298 K this makes the molar heat capacity of D2O (84.963 J/K/mol) an astonishing 12.7% higher than H2O (75.38 J/K/mol). I am astounded and have no explanation.

Everything below this line is WRONG!

The molar heat capacity is almost exactly the same. i.e. it takes the same energy to heat the same number of molecules by one degree Celsius.
The difference in the *specific* heat is just because of the difference in density. D2O molecules have a relative molar mass of about 20 compared with 18 for H2O. Since the bond lengths are similar, the density is higher by a factor 20/18 - about 11% higher. So a given mass of D2O has about 11% fewer molecules in it.


Is a liquid in a container always in equilibrium with its vapour?

The definition of vapor pressure means that for any temperature and pressure within the limits of a liquid system, there is a certain amount of vapor present above the liquid. The liquid and vapor at this point are in equilibrium because the vapor pressure does not change over time for that given temperature and pressure. When the liquid starts boiling, the system is undergoing a phase transition and in not at equilibrium because the system volume is changing and energy is continually being added to raise the system's temperature. Of course, we know the temperature does not start to change until vaporization has completed.

Vapor pressure of any given pure liquid is normally a strong function of temperature and a weak function of pressure. If you took a beaker of a pure liquid and put it inside a container that could support a vacuum, then as the pressure external to the beaker starts to decrease, more molecules from the liquid phase have enough energy to escape the liquid phase and exist in the vapor phase. If you continued developing more vacuum, the liquid would boil. This is seen on the phase diagram as moving straight down from somewhere in the liquid region until you reach the gas-liquid equilibrium line.


Volume of dibasic acid required to neutralise 0.04 mole of OH-

Unfortunately you are missing something. The concentration is already given as an equivalence concentration, thus taking the dibasic nature of the acid into account. Therefore you have been given $c(\ce{H+})=0.1~\mathrm{mol\,L^{-1}}$, not $c(\ce{H2A})=0.1~\mathrm{mol\,L^{-1}}$. In the latter case your calculation would be correct.

The use of normality is ambiguous and it is not recommended to be used any more.


Charge-transfer absorption complex - solvent sensitivity

In a charge transfer DA complex the lowest excited state is the charge separated species $D^+ A^-$. The 'normal' lowest excited singlet states of D and A are each at higher energy. As the molecules D and A come together the interaction between them, although small, (the absorption spectrum is only marginally altered) is nevertheless enough to cause electron transfer upon photon absorption.

The dipolar excited state possesses its own electrostatic self-energy E, as does any dipole. This is the sum of the Born charges $\pm q$ at infinity less the Coulomb energy of bringing the charges together to from the dipole. This works out as $E=  q^2/(8\pi\epsilon_0 \epsilon a)$ where $\epsilon_0$ is the permitivitty of free space and $\epsilon$ the dielectric constant of the solvent. The charges have a combined radius of $2a$. The equation can be re-written as $E=\mu^2/(4\pi\epsilon_0\; \epsilon\; L^3)$ if the dipole is defined as $\mu=qL$. Thus you can understand why the energy is lowered, and hence CT transition red shifted in a polar solvent, e.g acetoninitrile $\epsilon =37$ vs hexane $\epsilon \approx 2$.

Notes: Electron transfer upon absorption is in contrast to photoexcited electron transfer where the excited state of the donor or acceptor, whichever is excited, undergoes a thermal reaction in the excited state to form $D^+ A^-$. In the CT complex the charge is transferred on absorption between one molecule and the other, rather than from one part of the molecule to another part as in a normal absorption process. The absorption process is so rapid that, there is plenty of time for this to occur whilst the D and A molecules are close to one another and in the correct relative orientation because molecular diffusion is a relatively slow process.


1) Yes and no ... the d-d orbital overlap will almost certainly have pi-orbital character (angular node along bond axis), and it could reasonably be considered "backbonding" because the electron density is going in the canonically "unexpected" direction (i.e. away from positively charged metal center).  On the other hand, chemists (particularly inorganic chemists) usually restrict the phrase "pi-backbonding" to refer to back donation from the d-orbitals of a neutral or ionic metal center into anti-bonding orbitals of a ligand, typically an olefin or a carbonyl ligand.

2) Yes .. that's essentially correct ... d-orbitals tend to be rather diffuse, and so can overlap with lobes of diffuse orbitals on ligands (or counter-ions, as with the metal sulfides).  In traditional metal-ligand pi-backbonding, lobes the d-orbitals overlap with lobes of an unoccupied anti-bonding orbital of the ligand.  In the example given above, they overlap with lobes of unoccupied d-orbitals on the sulfur.  For a given system, unoccupied, or virtual, orbitals tend to be much more diffuse than occupied orbitals.

3) Those diagrams are just intended to capture the qualitative aspects of the bonding interaction, such as the symmetries of the orbitals involved.  Remember, the orbitals are 3-dimensional shapes that describe the probability of finding an electron with a particular set of quantum numbers in a given region of space.  Depictions like the one in your post are helping to visualize the orbital by assuming a fairly tight "tolerance" for the probability. For example, those images might depict the surfaces that enclose 75% of the total probability density.  If you expand that to 90%, or even 99%, the lobes maintain the same basic shapes but get much larger, which might make the overlap seem more significant, but could also be confusing for other reasons.

4) I think it is quite well-written, but I suppose it would depend on the broader context in which it was being presented.  I guess it is presented at an inappropriately high level for an introductory chemistry textbook, for example.


So rotational spectroscopy is something one can spend a lot of time talking about so I'll try to keep this brief. Although this isn't a complete answer, it was too long for a comment, but hopefully this sheds some light on the topic.
For rotational transitions, one of the selection rules tells us that in order for EMR to induce a transition in the rotational energy levels, a permanent dipole moment must exist in the molecule (i.e. a single atom won't be observed in rotational spectra, nor would a molecule such as $\ce{H_2}$).
The rotational symmetry number is a variable that is used to account for the amount of rotations one can perform on a molecule that would make it indistinguishable from the starting point (i.e. a 90 degree rotational of $\ce{PCl_5}$ along the principal axis can be performed three times before returning each atom to its original position in space, however, each rotation produces an identical picture of the molecule each time).
Moment of inertia can be calculated from the rotational constant, however, in practice, one would usually calculate the moment of inertia with computational software and then use that to approximate the rotational constants which are employed to predict the energies and allowed transitions for the rotations of a molecule.
This is a link to an appendix with the equations used for calculating the moment of inertia in molecules based on the geometry (e.g. symmetric rotor, linear)
Here is an article that further discusses symmetry numbers and partition functions.
This one talks about rotational partition functions of diatomic gases.
This article talks about transition moment integrals and selection rules.


Do indigo and leuco-indigo have fully conjugated pi systems?

Let's just start with stating the very obvious: Figure 2.23 has more errors than there should be in an entire chapter.

Indigo

With the exception of bonds to hydrogen, all other bonds are nicely conjugated in indigo. The π molecular orbital show this very clearly.[1] The molecule has a fantastic C2h symmetry, which also means it is in one plane. Intramolecular hydrogen bonds certainly further stabilise this conformation. They can be visualised with the help of the Quantum Theory of Atoms in Molecules (QTAIM).[2]

 (click for full image)

Interestingly eleven out of twenty π orbitals are occupied, which leaves the HOMO being anti-bonding with respect to the central carbon-carbon bond, which is 138 pm long.

Going further I will first add electrons, then add protons according to the following scheme:

Indigo2-

I am happy to report, that the molecule stays flat. The central carbon-carbon bond elongates a little to 141 pm. It basically still has comparable orbitals, although the location of the nodal planes already accommodate the divide of the central bond. I am guessing that the intramolecular hydrogen bonds are accountable for this since the molecule is highly charged.

Now the basic question is how do the protons behave, where do they go, will they break the symmetry.

Leucoindigo

A bit to my surprise, the flat version of this compound is also most stable.

Preliminary conclusions

There are several points, where this project could have derailed. Level of theory is only one of them. The evidence presented here is not conclusive enough. In due time I might be able to do some more, but for now, at least the textbook picture is debunked.

Calculated at the DF-BP86/def2-SVP level of theory. Occupied orbitals are blue and orange, virtual orbitals (unoccupied) are red and yellow.
Analysis with MultiWFN 3.3.8, see http://sobereva.com/multiwfn
Nuclei are brown dots, bond critical points are blue dots, ring critical bonds are orange dots. Blue lines are zero-flux surfaces that seoarate the molecule into atomic basins.


For an example reaction $\ce{A + B <=> C + D}$

the forward rate is given by

$$r_\mathrm{f} = k_\mathrm{f}[\ce{A}][\ce{B}]$$

and the backward rate is given by

$$r_\mathrm{b} = k_\mathrm{b}[\ce{C}][\ce{D}]$$

For the rates to be equal one simply needs

$$k_\mathrm{f}[\ce{A}][\ce{B}] = r_\mathrm{f} = r_\mathrm{b} = k_\mathrm{b}[\ce{C}][\ce{D}]$$

Note that this does not imply that the two rate constants are the same

$$k_\mathrm{f} \neq k_\mathrm{b} \quad \text{(in general)}$$


Why do we use Ḡ ("G bar") as a notation for partial molar free energy?

Thermodynamic properties are functions of composition, which can be represented as $(\chi _1,...,\chi _N)$ where $\chi _i$ is the mole fraction of component $i$ present in the system. Now imagine a change in the composition such that the amount of component $i$ is increased by a very small quantity $dn _i$ (alternately, imagine the system to be infinitely large), and a thermodynamic property $X$ is changed by $dX$. Then (using the notation you describe) we define the partial molar property as $$\bar{X}=\left(\frac{\partial X}{\partial n_i}\right)_{p,T,n_j}$$ This notation means that we make a limiting, infinitely small change in $n_i$, such that the change in $X$ is for all practical purposes nil, ie $\Delta X=0$. The partial molar quantity describes the slope of the property $X$ (which can be represented as a surface in a multidimensional manifold) along the composition coordinate $n_i$, at a specific composition $(\chi _1,...,\chi _N)$.


What is antisymmetric exchange? What is J-strain? Where does it come from?

Maybe OP has already contacted the authors of this paper by now, but this was interesting. I'm no expert but I can Google things and this was too long for a comment.

Antisymmetric exchange:
At first I thought it was simply an exchange interaction where the wave function's sign is changed during exchange, now I don't think it's so simple.

Antisymmetric exchange is also known as DM-interaction (for Dzyaloshinskii-Moriya). I don't know exactly what it is, here is the original paper citation - can't find it anywhere though.

I.E. Dzyaloshinskii, Thermodynamic theory of weak ferromagnetism in antiferromagnetic substances, Sov. Phys. JETP 5(6), 1259-1272 (1957)

A definition

A slideshow, which contains some more info (slides 16-23)

J-Strain: Reading this, "J-strain" isn't the only "strain" that's referenced in this paper. Also referenced within is "G-Strain" which has a more elucidating Wikipedia article.

So, maybe it's safe to say, like G-Strain, J-strain is the broadening of the "distributed exchange coupling constants" (a.k.a. J-values). Inferred from a line in the paper: "Both proteins exhibit distributed J-values (J-strain)".

What did J-strain have to do with the experiment? "...The observed J-strain, which presumably represents a distribution of protein conformations “locked in” upon freezing the samples."

Where does the broadening of the J-values (exchange coupling constants) a.k.a. J-strain come from? Well apparently there's this thing called J-coupling that seems relevant. J-coupling is also known as indirect dipole-dipole coupling, and gives info about molecule interconnectivity. J-coupling relates nuclei to local electrons (All this is from the J-coupling article). The broadening of the lines may come from anisotropic interactions, which affect nuclear spin energy levels. Sure enough, the paper says it can't rule out anisotropy. An Alternative source? Non-homogeneous regions in the sample, which could go back to the frozen/locked conformations.

I still think it'd be great if someone doing physics research in this field could elucidate some more though.

Another J-coupling Webpage (U. Wisc)

Of course I should cite the original paper, "Evidence for Antisymmetric Exchange in Cuboidal [3Fe−4S]+ Clusters". Linked in question.


How does 1 mole of a compound = the molecular weight of the compound?

Well you can actually calculate it like this
$$1\ \mathrm{amu}=1.66\times10^{-27}\ \mathrm{kg}$$
Therefore
$$119\ \mathrm{amu}=119\times1.66\times10^{-27}\ \mathrm{kg}$$
But for 1 mole, multiplying by Avogadro number you get mass of one mole as $$\begin{align}
119\times1.66\times10^{-27}\times6.022\times10^{23}&=0.11899\ \mathrm{kg}\\
&=118.99\ \mathrm g\\
&\approx119\ \mathrm g
\end{align}$$


The concept of a mole(c. 1805), and Avogadro's number(1811) occurred after Dalton formulated his law (1801). So he would not have set out to measure a "mole of gas" as the modern formulation sets out. This history of Dalton may be useful to you: Google Book Search Most of his work centered around pressure readings, as his early interests were meterological.


Reasons behind the different colors of a solution containing Iodine

Iodine in an inert solvent is purple.  Electron donation alters that.  Linear starch (amylose)-iodine is dark brown.  Branched starch (amylopectin)-iodine is red.  The Wikipedia page shows iodine in chlorinated solvent (very weak electron donation) then addition of strong electron donor triphenylphosphine.

If an electron donor is reactive, then the interacting iodine molecule looks like iodonium (elecron accepter).  Initial interation is (4), post-mixing is (3).

Put a tiny iodine crystal into a melting point capillary, evacuate under vacuum, flame closed.  You will see a very pale purple vapor at equilibrium.  Crack the seal and the vapor is suddenly deeper purple.  Why?  Take a sealed cap and an evacuated sealed cap, and insert the vapor volumes into the beam of a green laser pointer. interesting. Take care that the optical geometry does not reflect back into your eyes.


How does boiling water affect its hardness?

You may need to think outside chemistry a little bit and pay a little bit attention to your everyday life. This is what happened when you boil water in a container:

Most hardness in water, calcium/magnesium, will precipitate onto the hot surface and is removed from water. Now if you can get more something by removing it, is it against the very fundamentals of science?


What type of material is this and how is it corroding?

Though a picture is not really enough information, approximate location at least and any simple tests on the materials shown would help a lot.

Its structural and ugly so its steel.
Its mostly likely painted steel. I'd say the paint flaked off so it has general uniform corrosion. Why it looks black and not the regular red-brown is probably combination of oxides and sulphides formed as the corrosion products. I don't see a hint that its galvanised steel.
Steel in exterior uses  rarely suffers any corrosion other than  general uniform corrosion.


Why does the pH before the equivalence point of a titration depend on the initial concentration only when the acid is strong?

To understand what is happening with the two sets of titration curves, start by considering the titration of acetic acid ("HOAc" for short) with sodium hydroxide solution. The titration curves are shown below:

In the figure, the mutual concentrations are simply the concentrations of the reactants. Thus, 0.1 M HOAc is titrated with 0.1 M NaOH, 0.01 M HOAc is titrated with 0.01 M NaOH, and so on.

Acetic acid is a monoprotic weak acid, with $\mathrm{p}K_\mathrm{a}=4.756$. The titration curves for the three highest mutual concentrations almost coincide at the half-equivalence points, i.e, at the center of the buffer region. This is as expected from the Henderson-Hasselbalch equation, which ignores the autoionization of water. However, the three lower concentrations demonstrate that the titration curves do not coincide at the half-equivalence points: as the concentrations decrease, the pH values at the half-equivalence pints move toward the assumed pH of 7 for pure water.

As the concentrations are lowered, the effect of water's auto-ionization can no longer be neglected and the Henderson-Hasselbalch approximation is no longer viable. The calculated pH values at the half-equivalence points are shown in the second column of the table below:

Note that for 0.1 M, $\mathrm{pH} = \mathrm{p}K_\mathrm{a} $, as expected. Even at 0.001 M, the approximation is not bad. However, as concentrations continue to decline, the pH at the half-equivalence points simply move even closer to that assumed for pure water.

Now consider the titration of hydrochloric acid with sodium hydroxide solution. The titration curves are shown below:

As shown in the figure and in the third column of the table, the pH values at the half-equivalence points are one pH unit apart until the concentration is down around $\mathrm{1 \mu M}$.  Then, as concentrations continue to decline, the pH at the half-equivalence points must invariably move toward that of pure water.

For hydrogen ion concentrations high enough to neglect the auto-ionization of water, the pH at the half-equivalence point is easy to calculate. First, divide the initial hydrogen ion concentration by 2, to account for the fact that half of the number of hydrogen ions, from the 50 mL of starting HCl solution, were neutralized by addition of 25 mL of equimolar NaOH solution. Then multiply the hydrogen ion concentration by 2/3, because the solution volume increased from 50 mL to 75 mL. The net result is that the hydrogen ion concentration, at the half-equivalence point, is 1/3 of the initial value. Then take the negative logarithm base 10, as usual:

$$\mathrm{pH} = -\log{({\ce{[H+] \times (1/2) \times (2/3)) = - \log{(\ce{[H+]/3)}}}} }$$

If $\ce{[H^+]}$ = 0.1 M, then pH = 1.477. N.B. Significant figures were ignored in this answer, so 0.1 M is treated as exact, etc.


What happens to an enzyme after it breaks down a starch molecule?

Enzymes are a form of catalyst. Catalysts do not get consumed in a chemical reaction. I believe in reality, enzymes do degrade overtime and you need to replenish them overtime to keep the reaction efficient.
Can one (1) enzyme break down multiple starches? Sure.
I did not find specific literature on fermentation, but I found one (1) carbonic anhydrase in our tissues and bloodstream can hydrate 1,000,000 carbon dioxide molecules per second. So your enzyme probably has a similar magnitude of catalytic rate.
Reference
Berg JM, Tymoczko JL, Stryer L. Biochemistry. 5th edition. New York: W H Freeman; 2002. Section 8.1, Enzymes Are Powerful and Highly Specific Catalysts. Available from: https://www.ncbi.nlm.nih.gov/books/NBK22380/


Volume of air required to oxidize glucose

As matt_black commented, there is no need to use ideal gas law as you are given the molar volume already. Volume of the air $V_\mathrm{air}$ can be found from the volume fraction $\varphi(\ce{O2})$:

$$V_\mathrm{air} = \frac{V(\ce{O2})}{\varphi(\ce{O2})}$$

Volume of the oxygen can be found from the molar volume $V_\mathrm{m}(\ce{O2})$ and the amount of oxygen $n(\ce{O2})$ required to oxidize given amount of glucose:

$$V(\ce{O2}) = n(\ce{O2})\cdot V_\mathrm{m}(\ce{O2})$$

$$n(\ce{O2}) = 6n(\ce{C6H12O6})$$

Getting things together:

$$V_\mathrm{air} = \frac{6n(\ce{C6H12O6})\cdot V_\mathrm{m}(\ce{O2})}{\varphi(\ce{O2})} = \frac{6\cdot\pu{2.47 mol}\cdot\pu{24.45 L mol-1}}{0.21} = \pu{1725 L}$$

Note that you have to use appropriate units, e.g. you cannot use all base SI units in the deal gas law and put the pressure in mm Hg! Also, never drop any units — they do participate in calculations as well.


Yes, you need additional quantities beyond the minimum necessary for calculating energies and the Hellmann-Feynman piece of the force when the wavefunction is not variational.

Here is a very rough sketch of why:

Puláy forces arise from applying the chain rule for calculating forces and was first discussed in the context of applying the Hellman-Feynman theorem.

Recall that the force is the change of energy with changes in (nuclear) coordinate and the energy is the expectation value of the (electronic) Hamiltonian, $E = \left<\psi\left|H\right|\psi\right>$. Applying the chain rule to this:

$$-F = \nabla E = \left<\psi \left\vert \nabla H \right\vert \psi\right> +  2\left< \nabla\psi \left\vert H \right\vert \psi\right>$$

The first term is what you get from Hellman-Feynman and is the expectation of a one-electron operator of the first you have listed. The second term goes away in Hellmann-Feynman only because it assumes the wavefunction is variational.

If you expand out the orbitals in the Slater determinant in some AO basis $\chi$:

$$ \psi(r_1, r_2, ..., r_N) = \phi(r_1) \wedge \phi(r_2) \wedge \dots \wedge \phi(r_N) $$
$$ \phi(r_1) = \sum_i c_i \chi_i(r_i) $$

then it is clear that $\nabla\psi$ generates terms which are MO coefficient derivatives (usually denoted $c_i^x$) and also AO derivatives ($\chi_i^x$). If the wavefunction were variational, then by definition all these derivatives are zero. If you work through the algebra, you will find that there are some new quantities that need to be calculated in order to work out the non-Hellmann-Feynman theorem, most notably the one-sided AO overlap derivative matrices $\left<\chi_i^x\vert\chi_j\right>$ which do not normally show up otherwise. These terms turn out to be critical for getting correct ab initio molecular dynamics since the wavefunction is rarely (if ever) variational in time.

There are tricks one can do to reduce the cost of calculating MO coefficient derivatives , but they cannot be obtained for free in general. And there's really no way around calculating the overlap derivative terms.


Why degree of dissociation/ionisation affected by dilution?

What motivates or pushes it to dissociate more?

Perhaps surprisingly, if you could look at an individual acid molecule, it does not (or hardly) changes its behavior upon dilution. It just dissociates with a certain rate, yielding the conjugate base and a hydrogen ion. There is no "extra motivation" to dissociate, and no "tapping on the shoulder" pushing it to react.

  Why is the degree of dissociation/ionisation affected by dilution?

The reverse reaction, where a dissociated acid (the conjugated base) reacts with a hydrogen ion to make the acid again, is affected by dilution. Here, two molecules have to come together, and this takes longer when the solution is more dilute.

This description is for the simplest case, where the net reaction not only describes the stoichiometry but also the mechanism of the reaction. However, even if there are intermediates, at equilibrium the forward and reverse rates are equal, and when you dilute solution, rates of the reverse steps would be significantly lowered, leading to a higher degree of dissociation.


Oxidation of Sodium Hydroxide in Ethanol?

[Answer based on comments from @KemonoChen, @IvanNeretin and @NilayGhosh. Thanks for your input :D]

  In the presence of alkali (or acid, for that matter) aldehyde will
  turn against itself, creating all matters of ugly polycondensate
  products. See Aldol
  condensation and
  note that it ends up in another aldehyde, so the process can and will
  be repeated

So yes it can happen.

And is is most likely ethanol oxidation to an aldehyde, which has (aldol) condensed into an 'ugly polycondensate'

It may also be Sodium Alkoxide, which is known to turn brown.


Since it is electrophilic addition, it can go via 2 ways. Either by forming classical carbocation or cyclic carbocation. Since the electrophile will be H+ it will go with classical carbocation. Therefore product will be 1,3 dibromo cyclopentane. Also taken in account the mesomeric effect of bromine atom


Chemical Kinetics - change to system conditions

As you said, your statement "If you add A and the volume is constant obviously the concentration of A increases" indeed does not apply.  In fact, the number of mols of A never changes.  Add a mol of A, and it reacts irreversibly with B and is gone forever as part of C.

This means that the total volume must decrease for the concentration to increase, which means that "addition of A at constant T and P" must be correct, as it is the only answer that allows for a decrease in volume and therefore an increase in the concentration of A.

An assumption made here that is not explicitly stated in the question, is that we never run out of B.


Let me approach this another way than the others: orbitals are NOT physical objects! They do  not exist in physical sense, they are theoretical constructs, chemical concepts that help understand / visualize / etc. mathematical solutions of Schrodinger / Dirac / Kohn–Sham / etc. equations.

Orbitals are not unique: given linear combinations are equivalent with each other, and there is no "correct orbitals", one can choose whichever they like. Canonical orbitals, natural orbitals etc are all good to go.

What is the evidence they exist? They do not exist, they are just mathematical solutions for given equations, and it is a purely mathematical question if they are good solutions for those equations or not. The theories themselves are consistent with experimental data, e.g. spectroscopic properties, geometries, reactivity.


This is not possible. The potentials are given only as references to the standard reduction potential for hydrogen. As such, they are not absolute.

You can convert between potential and free energy for a full redox reaction.

Actually, your statement has a deeper conceptual flaw. The standard formation free energy of the Fe(II) ion does not make any sense. The standard free energy of formation requires making something from its constituent elements in their standard states, but you cannot make an ion from its constituent neutral atoms. You'll need to account for the energy of the electrons and their binding energy to the ion. Therefore, the cation alone cannot have a standard free energy of formation.


How to unify all the subunits in a PDB file?

If you don't care about residue and atom numbering, which you probably do, then sed '/^TER/d' < 1A3N.pdb > 1A3N_combined.pdb will do. Since atom entries must be reordered, it looks like MDAnalysis will strip the TER entries automatically:

#!/usr/bin/env python2

import MDAnalysis

u = MDAnalysis.Universe('1A3N.pdb')

with MDAnalysis.Writer('1A3N_combined.pdb') as writer:
    writer.write(u)

So,

ATOM   1067  NH1 ARG A 141      26.176   8.362  17.810  1.00 11.11           N
ATOM   1068  NH2 ARG A 141      24.650   9.068  16.200  1.00 13.86           N
ATOM   1069  OXT ARG A 141      26.697  14.784  20.720  1.00 10.99           O
TER    1070      ARG A 141
ATOM   1071  N   HIS B   2       3.670 -13.643  19.447  1.00 38.58           N
ATOM   1072  CA  HIS B   2       2.695 -14.734  19.744  1.00 32.83           C
ATOM   1073  C   HIS B   2       1.379 -14.140  20.199  1.00 30.79           C

becomes

ATOM   1067  NH1 ARG A 141      26.176   8.362  17.810  1.00 11.11      A    N
ATOM   1068  NH2 ARG A 141      24.650   9.068  16.200  1.00 13.86      A    N
ATOM   1069  OXT ARG A 141      26.697  14.784  20.720  1.00 10.99      A    O
ATOM   1070  N   HIS B   2       3.670 -13.643  19.447  1.00 38.58      B    N
ATOM   1071  CA  HIS B   2       2.695 -14.734  19.744  1.00 32.83      B    C
ATOM   1072  C   HIS B   2       1.379 -14.140  20.199  1.00 30.79      B    C

where there are now chain IDs in the last column. See how this doesn't renumber residues? Modify the script; resids is a property, so the in-place object mutation works properly:

#!/usr/bin/env python2

import numpy as np
import MDAnalysis

u = MDAnalysis.Universe('1A3N.pdb')
u.residues.resids = np.arange(1, 1 + len(u.residues.resids))

with MDAnalysis.Writer('1A3N_combined.pdb') as writer:
    writer.write(u)

Finally:

ATOM   1067  NH1 ARG A 141      26.176   8.362  17.810  1.00 11.11      A    N
ATOM   1068  NH2 ARG A 141      24.650   9.068  16.200  1.00 13.86      A    N
ATOM   1069  OXT ARG A 141      26.697  14.784  20.720  1.00 10.99      A    O
ATOM   1070  N   HIS B 142       3.670 -13.643  19.447  1.00 38.58      B    N
ATOM   1071  CA  HIS B 142       2.695 -14.734  19.744  1.00 32.83      B    C
ATOM   1072  C   HIS B 142       1.379 -14.140  20.199  1.00 30.79      B    C