text:
Lewis structure for NCO+
I proposed the following lewis structure for $\rm NCO^\oplus$

The problem is that this structure is linear, while my book tells that the structure is not linear. What will be correct canonical structure?
Also is it correct to call the following as resonance structures?

Your structure has two unpaired electrons, which looks pretty unstable. Try making that go away for starters.

@Variax Can I transfer one electron from N to O?

It's also not a very good idea to have the most electronegative element carry the positive charge.

@Martin-マーチン so should I put the electron of N on O?

@MaxPayne Yes, always try to have your electrons paired. Now, in order so make your molecule angular you'll need to sacrifice one of the two double bonds. I'm thinking that the positive charge might sit more comfortably on the carbon atom if that helps.

@Variax But that would be a minor contributor, since no. of bonds would be 3, and octet of 2 atoms will also be sacrificed. Also, can we simply hop unpaired electrons anywhere in the molecule?

As always, writing a Lewis structure seems to be a trivial task, but sometimes it just isn't. The smaller the molecule becomes, more often than not the bonding situation becomes more complicated and complex. This can often be attributed to the high symmetry of such molecules. Prime examples are $\ce{O2}$, $\ce{CO}$, $\ce{CO2}$, $\ce{N_xO_y}$, etc..
I found no evidence anywhere that $\mathbf{\ce{NCO+}}$ is bent. I have tried calculations in the singlet and triplet states and all of them lead back to the molecule being linear. It would be nice to know which textbook you are using for general reference.
To be perfectly candid here, it is not even a trivial task to set up the right calculations. At the DF-BP86/def2-SVP level of theory, the triplet state is 151.8 kJ/mol lower in energy than the singlet state. A broken symmetry singlet (a biradicaloid electronic state) is still 111.3 kJ/mol lower in energy.
On the CASSCF(6,6)/def2-SVP level of theory the triplet wave function is 143.3 kJ/mol lower in energy than the singlet wave function.
Lewis structures were not meant to accommodate complicated bonding situations. Instead they are meant to be easily understood. Even the concept of resonance can be quite tricky in certain situations. Without knowing that the ground state of dioxygen is a triplet state, we would interpret its common Lewis structure wrong.
The same shall apply here. We need to treat the concept of Lewis structures within its limitations. Based on the triplet DF-U-BP86/def2-SVP calculation the best representation is probably a bit surprising: The positive charge would be located at the most electronegative atom, while the spin density is located at the nitrogen (1). The broken symmetry solution gives us instead two double bonds with the positive charge at the nitrogen (2).
In all cases we shall not forget, that the most electropositive element is carbon in this molecule, so the actual charge is most likely located there (3). Since the π-orbitals are stretched over the whole molecule, the spin density will not entirely be located at nitrogen alone (4). A meaningful interpretation in terms of resonance structures (NRT of NBO) is, however, almost impossible. I tried it based on the closed shell singlet DF-BP86/def2-SVP calculation and there are three contributing main structures (5).

For what it's worth, the structures you have proposed are valid resonance structures. They are not all of them, but they are a good start. As you can see, any of them capture only a tiny glimpse in the complicated nature of this molecule.

Very nice answer, although I couldn't understand some complicated parts of it. I think I got too serious about lewis structures.
This question is actually from a high school level MCQ, and the book plainly stated that the molecule was non-linear.

a bent structure is discussed here: http://dx.doi.org/10.1063/1.449527 , but while they seem to think it is important in a reaction mechanism, I don't think they mean it is the ground state


text:
Portable library to render 2D structural formulas as vector graphics from SMILES or InChI
For a web-based project, I am looking for a piece of software to create a 2D vector graphics structural formula from a textual structure encoding scheme such as SMILES or InChI, running on the server or client, and callable via some kind of API.
I’m imagining a client- or server-side library that you could just use like this:
svg_output = lib.render_from_smiles(smiles_input);

Ideally, it would have the following properties:

Take a SMILES or InChI representation of the compound as input
Compute the structure graph and 2D positional data (positions of atoms, types of bonds, etc.)
Render the structure in some kind of vector graphics representation (SVG would be perfect, but any other open format would do)
Or, alternatively, just output the computed drawing parameters, to be passed on to a separate renderer (which would need to be written, but that is the less challenging part)
Written in a language that is easy to use and deploy in web environments (PHP, JavaScript, Python, Ruby, Go, etc.)
If a renderer is included, allow me to control the rendering style (stroke width/colours, text attributes etc.)
Open source or with license terms that are affordable to a non-profit project :)

The application just needs to render individual structures. No reaction mechanisms or anything fancy. Small organic molecules only.
I know there are various standalone software packages (ChemDraw, MarvinSketch etc.) that do this kind of thing, but I did not find anything that can be used as a library like this.
Any suggestions or ideas? Thank you!

If you can't find anything you like, I would like to know about it; I'm a coder and a chemist, I can definitely set this up if it isn't out there.

@sqykly - I think you're far better contributing to RDKit or Open Babel. Setting this up requires lots of twisty passages, all alike.

You might want to have a look at Open Babel.

It is licensed under GNU GPLv2
It has bindings for Python
It has bindings for PHP
It has bindings for Ruby
It can read (and write) SMILES
It can read (and write) InChI
It   can write SVG
It can write a list of painter commands

smiles shortcut: obabel -:"O=C(O)c1ccccc1OC(=O)C" -osvg -O asp.svg src Of course OB is not without bugs and may fail for 3D type structures, but that shouldn't be a problem for this simple conversion

@Martin Yes, the shortcut is a neat feature in the shell. OTOH, even ghosts are cool in the shell ;)

@Martin-マーチン- "may fail for 3D type structures"? Any particular examples?

You should also look at RDKit, which is at its core is C++ code for manipulating molecular structures, but which also has Python and Java bindings.  Most people use it via its Python bindings.
Here's some example code.
# import rdkit components
from rdkit import Chem
from rdkit.Chem import Draw
from rdkit.Chem.Draw import IPythonConsole
IPythonConsole.ipython_useSVG=True

# displaying an SVG structure (in a Jupyter notebook)
mol = Chem.MolFromSmiles('O=C(O)c1ccccc1OC(=O)C')
mol

Inside of a Jupyter notebook, that results in a nice SVG-formated image.  .  You can get a file of this image using the standard way to get SVG out of Jupyter notebooks, which I can't remember right now.
# 2D positional data for atoms
for atom in mol.GetAtoms():
    print (atom.GetIdx(), atom.GetSymbol(), atom.GetExplicitValence(), atom.GetIsAromatic())

# 2D positional data for bonds
for bond in mol.GetBonds():
    print (bond.GetIdx(), bond.GetBeginAtomIdx(), bond.GetEndAtomIdx(), bond.GetBondType())

The code above gives you just a sense of the flavor of how things are done, and just a few examples of properties available for bonds and atoms.  It certainly isn't an exhaustive list of bond/atom properties.  The output is:
(0, 'O', 2, False)
(1, 'C', 4, False)
(2, 'O', 1, False)
(3, 'C', 4, True)
(4, 'C', 3, True)
(5, 'C', 3, True)
(6, 'C', 3, True)
(7, 'C', 3, True)
(8, 'C', 4, True)
(9, 'O', 2, False)
(10, 'C', 4, False)
(11, 'O', 2, False)
(12, 'C', 1, False)
(0, 0, 1, rdkit.Chem.rdchem.BondType.DOUBLE)
(1, 1, 2, rdkit.Chem.rdchem.BondType.SINGLE)
(2, 1, 3, rdkit.Chem.rdchem.BondType.SINGLE)
(3, 3, 4, rdkit.Chem.rdchem.BondType.AROMATIC)
(4, 4, 5, rdkit.Chem.rdchem.BondType.AROMATIC)
(5, 5, 6, rdkit.Chem.rdchem.BondType.AROMATIC)
(6, 6, 7, rdkit.Chem.rdchem.BondType.AROMATIC)
(7, 7, 8, rdkit.Chem.rdchem.BondType.AROMATIC)
(8, 8, 9, rdkit.Chem.rdchem.BondType.SINGLE)
(9, 9, 10, rdkit.Chem.rdchem.BondType.SINGLE)
(10, 10, 11, rdkit.Chem.rdchem.BondType.DOUBLE)
(11, 10, 12, rdkit.Chem.rdchem.BondType.SINGLE)
(12, 8, 3, rdkit.Chem.rdchem.BondType.AROMATIC)

Converting between/from/to InChI and SMILES is easy.  Here's an example of InChI output.
    # output InChI
    Chem.MolToInchi(mol)

'InChI=1S/C9H8O4/c1-6(10)13-8-5-3-2-4-7(8)9(11)12/h2-5H,1H3,(H,11,12)'

You can learn a bit more about how to do common tasks in RDKit via this GitHub repo of example Jupyter Notebooks.

openchemlib is a  library that can convert a smiles into a SVG.
It can be used on the the server (Java) or on the client side (JavaScript).
https://github.com/Actelion/openchemlib
https://github.com/cheminfo/openchemlib-js

source: Stack Exchange

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text:
What does the "N" in N-nitrosoamine mean/stand for?
What does the "N" in N-nitrosoamine mean/stand for?  Also, if the functional group is connected to a benzene ring what do I call it?

An older review article: http://pubs.acs.org/doi/pdf/10.1021/bk-1979-0101.ch001

http://chemistry.stackexchange.com/questions/44349/amine-nomenclature-when-to-use-n

@Mithoron Related, agreed, but not duplicate.

The "N" means that the nitroso group ($\ce{-NO}$) is attached to the amine nitrogen.
So, for example, in N-nitrosomorpholine the nitroso group is attached to the nitrogen atom of the saturated, heterocyclic morpholine ring:

You could never have an "N-nitroso" group attached to a plain benzene ring, since there is no nitrogen atom in a benzene ring for the "N-" prefix to refer to.
In general, though, you would name aromatic nitroso compounds in the same way as you would those with any other substituent like 'chloro-' or 'nitro-'. For example, the following compound would is named nitrosobenzene:

A nitroso group is a nitrogen double bonded to an oxygen (think carbonyl). When that nitroso group is attached to an amine, it is an N-nitrosoamine. We add the N- designation so we know that the nitroso group is bonded to the nitrogen of the amino group as opposed to somewhere else on what is presumably an organic molecule.

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text:
Is it possible to create a solution with superoxide?
I am not a chemist, and somehow I cannot find any material on this topic, so I decided to ask here.
I am looking to create a solution with a specified amount/concentration of superoxide ($\ce{O2^.-}$). Can you suggest some compounds that will enable me to "isolate" superoxide in a solution?

Potassium superoxide (Aldrich) and an equimolar
  amount of 18-crown-6 ether used to increase solubility were
  dissolved in dry DMSO.

Role of Hydrogen Bonding in the Active Site of Human Manganese Superoxide
Dismutase Biochemistry 2004, 43, 7038-7045

I took a read through this paper. I don't understand how this has anything to do with producing superoxide in stable amounts.

@A.I. They create superoxide solutions in DMSO, which is then added to aqueous solutions.  The superoxide is relatively stable in the DMSO solution.

OK, so what you're suggesting is to mix $KO_2$ and $C_{12}H_{24}O_6$ in $(CH_3)_2SO$, then the whole thing is added to an acqueous solution, which should have $O_2^{\bar{\cdot}}$. Correct?

OK, this seem s like a good path. I'll accept this as an answer. This article might also be useful for others reading this.

@A.I. once it's added to aqueous solution,  the superoxide reacts quickly.  See "uncatalyzed, second-order rate constant of 440
M-1 s-1." in caption of Fig. 2.  In the DMSO, superoxide is dissolved as a separate ion and is relatively stable.  In aqueous solution, it is still dissolved as a separate ion, but is unstable, which is why the stock solutions are made in DMSO.


text:
Can I solve the partial derivative (dG/dH)s ? Does it make sense?
It is possible to express   ,   in terms of Cp, Cv, V, P, kt etc. ?
I know that G and H are both derived from the Legandre transformation of U, so they are the same equivalent representation of internal energy but expressed like G=G(T,P) and H=H(S,P).
So I'm wondering if this partial derivative makes sense...

Is this a homework problem, or is it just something you dreamed up?

No, it's not an homework problem. It is something I ask to myself, and I'm quiete curious to know the answer, because I search but didn't find nothing about.

$$dH=TdS+VdP$$
At constant S, $$dH=VdP$$
$$dG=-SdT+VdP$$
If S is constant, then $$\left(\frac{\partial S}{\partial T}\right)_PdT+\left(\frac{\partial S}{\partial P}\right)_TdP=0$$But,$$\left(\frac{\partial S}{\partial T}\right)_P=\frac{C_p}{T}$$and$$\left(\frac{\partial S}{\partial P}\right)_T=-\left(\frac{\partial V}{\partial T}\right)_P$$So, $$dT=\frac{T}{C_p}\left(\frac{\partial V}{\partial T}\right)_PdP$$and$$dG=\left[V-S\frac{T}{C_p}\left(\frac{\partial V}{\partial T}\right)_P\right]dP$$Dividing by dH then gives:
$$\left(\frac{\partial G}{\partial H}\right)_S=1-\frac{\alpha T}{C_p}S=1+\frac{(G-H)\alpha}{C_p}=1+\frac{\alpha T}{C_p}\left(\frac{\partial G}{\partial T}\right)_P$$

It might be worth pointing out that the $\alpha $ in the last equation is the coefficient of thermal expansion $\alpha  = V^{-1} (\frac{\partial V}{\partial T})_{P} $

Thanks.  I meant to.  Now that you have done it, there is no need for me to do it.

I'm generally cautious about subtracting something in the numerator of a fraction. Are you sure that it's G - H in the parentheses on the last line, or is it H - G? Also, isn't the partial of G wrt T at const P equal to -S, not S?

@timaeus222  Ooops.  Good catch.  Sorry everyone.  I'm going to go back and correct it.  It also looks like I made a sign error in the last term.


text:
Is 70% isopropyl alcohol still inflammable after 24h?
I was wondering if Isopropyl can still produce flames after being vaporised on wood and having a day to evaporate plus rinsed with hot water afterwards? If I put heat on the wood (around 50 degree Celsius) after all this, do I still have a change of burning it or should it be fine?

The autoignition temperature of isopropyl alcohol is about 400°C. So given that there's no flame or spark, a temperature of 50°C would not suffice to ignite it.


text:
Why is the calomel electrode widely preferred despite its toxicity?
The following pagargraph is from the Wikipedia page of calomel:

Mercurous chloride is employed extensively in electrochemistry, taking
  advantage of the ease of its oxidation and reduction reactions. [...] Over the past 50 years, it has been superseded by the
  silver/silver chloride (Ag/AgCl) electrode. Although the mercury
  electrodes have been widely abandoned due to the dangerous nature of
  mercury, many chemists believe they are still more accurate and are
  not dangerous as long as they are handled properly.

Yes, the calomel electrode is a breakthrough in electrochemical industry due to its ease of its oxidation and reduction reactions. But, the nature of mercury is such that every chemist fears it. There has been cases of mercury poisoning and deaths related to it. Despite its toxicity, electrochemist prefer the SCE
over the silver/silver cloride electrode. Why?
Due to its toxicity, the SCE has almost been replaced by the silver/silver chloride electrode. But still, the SCE is used as a reference electrode. Why? Why is the silver/silver chloride electrode not as good as the SCE?
Also, what are the safety measures/protocols undertaken by electrochemists while performing experiments using the calomel electrode?

It was replaced by Ag not the other way round.

I’m not scared of mercury. None of my colleagues are, either.

[…] Why and how chemist are still able to perform experiments using Calomel electrodes despite its toxicity? […]

I mostly used non-aqueous silver electrodes as a reference for my measurements in acetonitrile.
But I wouldn't be afraid to use a saturated calomel electrode (SCE) when necessary. We have recently discussed the toxicity of mercury metal and I think it is largely overrated.
As far as $\ce{Hg2Cl2}$ (calomel), the other mercury component in the SCE is concerned, have a look as the wikipedia article that you have cited yourself. The solubility is extremely low ($\mathrm{2.3\ mg\cdot l^{-1}}$) and the ${\mathrm{LD_{50}}}$ is $\mathrm{210\ mg\cdot kg^{-1}}$ (rat, oral uptake). This isn't overly dramatic either, when compared to the infamous potassium cyanide (${\mathrm{LD_{50}}}$ $\mathrm{5\ mg\cdot kg^{-1}}$, rat, oral uptake).
Actually, calomel had medical uses :)
In an article on Venereal disease prevention, the author writes:

[…] the absence of therapeutic modalities spurred the United States Armed Services to utilize Credé's solution for urethral irrigation and locally applied calomel ointment for prophylactic purposes in World War I […]

As far as the toxicity is concerned, both mercury metal and mercury(I)chloride are by no means comparable with dimethylmercury.


text:
Why is hexafluoridonickelate(IV) diamagnetic?
As fluoride is a weak-field ligand, it should not pair up the electrons. So why is $\ce{[NiF6]^2-}$ diamagnetic?

@Close voters: please state your reasoning why you consider this a close-worthy homework question. OP has used the knowledge available to them to draw a conclusion, finds out that this conclusion does not fit with the experiment and questions the fact. It does relate to a specific context OP is unclear with.

This is a common mistake to make when analysing transition metal compounds. The undergraduate or the non-coordination chemist looks at a complex, identifies a ligand, puts it into their box of weak-field or strong-field ligands, deduces an electronic structure and is then completely perplexed if something does not work the way the simple picture would assume.
It is then helpful to remember what the molecular orbital scheme of am octahedral complex looks like; refer to scheme 1. Note that scheme 1 ignores π contributions of the ligands but fret not for I shall briefly explain where (and how) they come in.

Scheme 1: Octahedral $\ce{[ML6]}$ complex with no π interactions. Image copied from this answer and originally taken from Professor Klüfers’ internet scriptum to his coordination chemistry course.
On the left-hand side of the scheme you can see the metal’s orbitals, from top to bottom they are (in nickel’s case) $\mathrm{4p\ (t_{1u}), 4s\ (a_{1g})}$ and $\mathrm{3d\ (e_g + t_{2g})}$. On the right-hand side, we have the six ligand orbitals which transform as $\mathrm{a_{1g} + e_g + t_{1u}}$. Only orbitals with identical irreducible representation interact constructively (or destructively) leading to the corresponding molecular orbitals shown in the middle.
What is commonly termed field split, $10~\mathrm{Dq}$ or $\Delta_\mathrm{o}$ is the difference between $\mathrm{t_{2g}}$ and $\mathrm{e_g^*}$ — the former being entirely metal-centred if π interactions are not considered, the latter being formed by both the metal and the ligand in a bonding-antibonding interaction.
Adding the ligands’ π orbitals, realise that they would transform as $\mathrm{t_{1u} + t_{1g} + t_{2u} + t_{2g}}$. Most of these interactions are either strictly non-bonding or contribute only weakly to bonding since the corresponding symmetry $\mathrm{t_{1u}}$ is already being used to make σ interactions. Only the $\mathrm{t_{2g}}$ p orbitals introduce a new interaction with the central metal by turning what are nonbonding orbitals in scheme 1 into a bonding/antibonding pair of orbitals. Since the interaction is π type, it will be a weak splitting, but it contributes to the decrease of $10~\mathrm{Dq}$ by raising the bottom layer.
I have talked a lot but so far I did not mention the point why hexafluoridonickelate(IV) is a low-spin complex. However, the basis has been laid. It is important to visualise that the field split is — because it derives from the strength (or weakness) of bonding/antibonding interactions in the molecular orbital scheme — dependent both on the metal’s and the ligands’ properties. This can be visualised in the very simple (and thus not satisfying; when approaching a higher level) equation $(1)$:
$$10~\mathrm{Dq} = g_\mathrm{M} \times f_\mathrm{L}\tag{1}$$
In this formula, $g_\mathrm{M}$ represents the contribution of the central metal while $f_\mathrm{L}$ that of the ligands. Obviously, if we modify the metal in a certain way, we can significantly influence the value of $10~\mathrm{Dq}$, i.e. the field split, i.e. whether the complex prefers a high or a low-spin state. Enough circling the issue, let’s get to the point.
The preference for high or low spin can be greatly influenced by the metal’s oxidation state. Remember that a positive charge attracts electrons, so that their respective orbitals will (all!) be lowered in energy by oxidation. Thus, the energies of the metal’s d orbitals and those of the ligands’ σ-symmetric orbitals will move closer together creating a larger stabilisation/destabilisation. At the same time, the π symmetric ligand orbitals will be more removed energetically and thus the weak π interaction will be further diminished. These effects in combination lead to a greater field split for high central metal oxidation states. In hexafluoridonickelate(IV), the oxidation state is $\mathrm{+IV}$ — pretty high for nickel which is probably most commonly found as nickel(II) in coordination complexes. Thus, it is not surprising that this higher oxidation state leads to a low-spin complex. Low-spin $\mathrm{d^6}$ is, of course, diamagnetic.
For further exemplification, I quote two values from a table of $g_\mathrm{M}$ values Professor Klüfers included in the linked internet scriptum (in German):
$$\begin{array}{cc}\hline
\ce{M^n+} & g_\mathrm{M}/ 1000~\mathrm{cm^{-1}}\\ \hline
\ce{Co^2+} & 9 \\
\ce{Co^3+} & 18.2 \\ \hline\end{array}$$
As can be seen from these values, increasing cobalt’s oxidation state from $\mathrm{+II}$ to $\mathrm{+III}$ doubles the metal’s contribution to $10~\mathrm{Dq}$. And that was only an increase of one.


text:
How to calculate the charge distribution at the ortho/para or meta sites of a substituted benzene ring
I've modelled a few substituted benzene rings in avogadro and retrieved the partial charges of each atom; I've tried to sum the partial charges of the carbon and hydrogen atoms at the ortho/para and meta sites to see the overall partial charge at that site but results don't make sense as the para site always ends up being 0.
How can I calculate the charge at the sites?

Summary: The method used does not handle π-conjugation
The partial charges in Avogadro (and many other software programs) are assigned by default by the Gasteiger-Marsili scheme†
Importantly, while the Gastieger charges are fairly good for atomic partial charges of organic-ish molecules, they're unlikely to reproduce effects in conjugated systems very well.

A method is presented for the rapid calculation of atomic charges in σ-bonded and nonconjugated π-systems. Atoms are characterized by their orbital electronegativities. In the calculation only the connectivities of the atoms are considered. Thus only the topology of a molecule is of importance.

My suggestion is to use some type of quantum chemical calculation, e.g.:

NWChem
OpenMOPAC
ORCA
(etc.)

Even semiempirical quantum chemical methods (e.g. PM7) will allow you to assign charges based on the quantum electrostatic potential around the molecule. They'll be much more accurate at the trends you want.
† "Iterative Partial Equalization of Orbital Electronegativity - A Rapid Access to Atomic Charges," Tetrahedron, Vol. 36, pp. 3219-3228, 1980.


text:
Confirming the presence of 4-bromoacetanilide during bromination of acetanilide
In my lab, acetanilide was added to bromine in aqueous acetic acid, and discolouration of the reaction mixture was obtained, indicating that bromine was consumed.
After some discussion, it was concluded that an electrophilic aromatic substitution might be taking place. This was backed up by a procedure we found online. What was left to do was to confirm the presence of the product, 4-bromoacetanilide.
The challenge is that the reaction mixture contains unreacted acetanilide as well as 4-bromoacetanilide, acetic acid, and possibly some polyhalogenated byproducts. Any conditions that could result in amide hydrolysis, e.g. heating with base, should also be avoided.
No residue was observed in the reaction mixture.
So, what chemical property can be used to determine the presence of 4-bromoacetanilide in the reaction mixture?

Did you try to isolate and purify the product?

@KlausWarzecha Actually, no. I'm quite sure that spectroscopy would quickly yield the answer, but as of now, both those techniques are out-of-bounds. This is why the question focuses on chemical properties.

@EashaanGodbole It's always easier when your material is purified. The most simple method to check whether bromination took place in the para position is to measure the melting point. The melting points of the bromoacetanilides strongly depend on the position of the bromo substituent, and they are all different from the parent acetanilide.

I advise against fishing in the mixture in this case. Try to isolate the product, purify it and determine its properties.
Have a look at the melting points of the starting material and the possible monobromination products:

acetanilide 114 °C
2-bromoacetanilide 96.5–100.5 °C
3-bromoacetanilide 81–85 °C
4-bromoacetanilide 165–169 °C

This promising! You can of course use further spectroscopical methods (NMR), but I'd start with the melting point.
You mentioned possible complications for the workup, but I disagree.
The reaction is typically performed in glacial acetic acid with a slight excess of bromine to make sure that there's (almost) no starting material left.
For workup, you pour the whole mixture into a beaker with water that has been placed in a bowl with ice. If you want/need to destroy an excess of bromine, add some bisulfite solution until the aqueous phase is colourless.
There should be some crystals. Filter them off, and recrystallize the material from ethanol.

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text:
Which component acts as the electrophile in the synthesis of the Sudan-1 azo dye?
In the two-pot synthesis, which acts as the nucleophile, 2-naphthol or phenyldiazonium?
Solomons et al. (2014) says that phenyldiazoium is weakly electrophilic. Yet in the questions to be answered in our laboratory manual, it asks for resonance structures on the attack of phenyldiazonium on carbon 1 and carbon 3 of beta-naphthol.

I'm confused. I can't seem to find a mechanism wherein phenyldiazonium acts as a nucleophile in the internet or in my books.

Maybe you can draw a mechanism yourself and add it to your question? If you're looking for a free molecule editor:  I'm using BKChem for my posts on Chem.SE.

Even though arrows are always drawn to represent nucleophilic attack, the word "attack" can in general refer to electrophilic attack as well, so there might not be any real contradiction here.

Following up on @orthocresol's comment, we typically learn the term 'attack' with respect to nucleophiles (e.g., in SN2 reaction), but 'attack' can be used for electrophiles as well.

Some pedantry: your second arrow should really be a $\ce{<->}$, not a $\ce{<=>}$. Those are two resonance forms of naphthol which are not being rapidly interconverted.

It was mentioned in the comments but I shall mention it again.
Attack does not have to mean nucleophilic attack. In your case, the attack is electrophilic, the electrophile phenyldiazonium attacking the naphtholate.


text:
Where to find a look-up table for mechanical properties of thermoplastics?
Does anybody know of a good compilation, regarding mechanical properties of thermoplastics or in general polymers?

I suggest MatWeb:

MatWeb's searchable database of material properties includes data sheets of thermoplastic and thermoset polymers such as ABS, nylon, polycarbonate, polyester, polyethylene and polypropylene; metals such as aluminum, cobalt, copper, lead, magnesium, nickel, steel, superalloys, titanium and zinc alloys; ceramics; plus semiconductors, fibers, and other engineering materials.

There are 84,173 entries in the Polymers tab in the physical properties search page at the time of my writing this. Note that you can select up to 3 properties at a time to narrow your search (or 10 if you pay for the service: MatWeb is a freemium deal, so be sure to look into whether or not your needs are satisfied by the free/pay versions.)


text:
Avogadro Conformer Properties with Linux Version
My Windows version of Avogadro has the following menu point:
View > Properties > Conformer Properties

My vanilla Avogadro version under Ubuntu does not even have the menu. What do I need to do, to either activate the menu or to get to this option under Linux?

I can't reproduce the problem.

I'm using Ubuntu 16.04 and installed Avogadro from the repository. The version is 1.1.1, just as described on Launchpad.
This version does have the menu entry in question:
Ansicht > Eigenschaften > Konformereneigenschaften
After a conformation search, it does open a table with the found conformers and their energies.


text:
Is the compound CM-7 in Red Comet fire grenades azeotropic dichloromethane?
Red Comet Manufacturing changed the liquid in their fire grenades from carbon tetrachloride to something they referred to as CM-7 in 1955.
Is CM-7 the same thing as azeotropic dichloromethane? If not, what is the composition of CM-7? What are the gases given off by CM-7 when it is exposed to heat?

Nope. I've run a distillation of the CM-7 fluid and I got roughly four different fractions across 50-70 C (83 kpa ambient pressure where I am). Too high to be dichloromethane or trichlorotrifluoroethane, too low to be carbon tet, if there's chloroform in there it's not much, I've trawled the internet as hard as google will allow and I can't find a good answer for what liquid I have three liters of.

Passive fire suppression systems by Red Comet included a glass bulb with colorless transparent liquid, and had a sticker:

MC-7 This fire extinguisher contains azeotropic chlormethanes®

Registered trademark symbol already signifies that marketing already prevailed at that time. Indeed, among chlormethanes $\ce{CH_xCl_{4-x}}$ chloromethane $\ce{CH3Cl}$ is flammable. Dichloromethane $\ce{CH2Cl2}$ is not combustible itself, but supports burning and flashes. Both chloroform $\ce{CHCl3}$ and $\ce{CCl4}$ are not flammable, but don't form azeotrope and still would produce phosgene $\ce{COCl2}$ when react with atmospheric oxygen. Also, the company denied presence of $\ce{CCl4}$ in MC-7 (Kiplinger's Personal Finance, Oct 1956, p. 48):

CM-7, which has replaced carbon tetrachloride in all our grenade extinguishers. CM-7 does not produce toxic gases. It converts toxic gases created by the fire itself into additional fire-extinguishing chemicals.

As it seems like Red Comet doesn't hold any patents for inventions in chemistry, most likely it's a hydrochlorofluorocarbon that's been known at that time.
These days MC-7 extinguishers are sold as vintage items, and uniformly it's claimed (e.g. here and there) that the liquid component is 1,1,2-Trichloro-1,2,2-trifluoroethane, which at that time indeed has been studied as a very unreactive (even upon heating) vaporizable liquid agent (e.g. Fire Research Abstracts and Reviews, Vol. 3
1961, p. 148).
Starting 1960s, several acts in Montana, Chicago, and later across US no longer permitted usage of both tetrachloromethane and MC-7 as fire-extinguishing chemicals. Halons took over.


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Which Phenols give the colours blue, green or violet in the neutral FeCl3 test?
I have read about the neutral FeCl3 test for Phenols, where a positive test is indicated by the colour of the solution which ranges from green to blue to violet. But which specific derivatives of phenol give a specific colour? Or is it random?

The color of the solution will depend on what substituent is attached to phenol. The following is an excerpt from an e-book:

Ferric Chloride Test: To 2 ml aqueous or alcoholic solution of compound, add 5-6 drops of freshly prepared neutral ferric chloride solution. Colored complexes of phenols are produced
\begin{array}{c|c}
\mathbf{Compounds} & \mathbf{Color} \\\hline
\text{phenol, resorcinol, o-cresol, p-cresol} & \text{Violet or blue}\\
\text{catechol} & \text{Green} \\
\text{hydroquinone} & \text{Violet or transient blue which changes to yellow on excess addition of FeCl3}\\
\text{pyrogallol} & \text{Blue rapidly changing to red}\end{array}
Aqueous solution of 1 and 2-naphthol however do not give any characteristic coloration but their alcoholic solution give violet-blue and green coloration respectively due to formation of dinaphthols.

Ref.: Systematic Lab Experiments in Organic Chemistry by Arun Sethi, New Age International, 2006

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


text:
Addition of cyclohexene with itself?
So I was practicing for exams, and this is the specific question.
I get that structure 2 is possible through polymerization and 3 doesn't fit to be the polymer of cyclohexene. I'm kinda consfused about 1 tho, I see that the double bonds have broken and joined with another cyclohexene, but I'm not sure about how that works. Like, is it similar to the addition of electrophiles..if so why isn't this seen with non-cyclic alkenes?

You're going to need an initiator of some kind. Probably a radical generator. That radical will begin a chain reaction that propagates by adding into the double bond to produce a new secondary radical.

1 is essentially a cyclic form of 2 with $n = 1.5$. So theoreticly, it could form.

$n = 0.75$ of course. Not sure where my brain was in November 2016 …

I'm fairly sure this question is merely testing one on their ability to count atoms. If the structure has a chemical formula of $\ce{(C6H10)_m}$ where $m \in \mathbb{Z}^+$, then it is – at least in principle – a possible polymerisation product of cyclohexene $\ce{C6H10}$. Whether it is formed in real life is a different question, but my belief is that that is not part of the question, so you're probably overthinking this if you're wondering about how exactly 1 is formed.
1 ($\ce{C18H30} \equiv \ce{(C6H10)3}$) and 2 ($\ce{(C24H40)_n} \equiv \ce{(C6H10)_{4n}}$) fulfill the above criterion, but 3 ($\ce{(C21H36)_n}$) does not.

I think OP went further from his answer - this isn't homework or overthinking, but curiosity about actual trimerisation reactions.

It's overthinking for the purpose of answering that question, so I decided not to talk about it - however, if we don't restrict ourselves to that question, it's certainl ynot a bad thing to be curious.

This is a less mathematical, and thus slightly simpler, alternative to the user orthocresol's answer.
There's absolutely no need to get confused when coming to addition polymerisation questions. When I learnt about this reaction in my chemistry class this year, my chemistry taught us that a very simple way of handling the question is using this converting to "H form" strategy: Break the $ \pi $ bond and draw the substituents and the two carbons in the shape of the letter H, as shown in the image below. Notice that each of the two carbons have only formed three bonds and each can make another single bond to another atom.

Image source: https://getrevising.co.uk/https_proxy/747

For your question, the 'H form' of cyclohexene is shown above. Now, it is very easy to see that 1 is in fact possible because you are connecting three of these 'H units' in a circular fashion.
Actually, for this particular question, this conversion to the "H form" strategy may seem trivial but for alkenes where the substituents may be "ugly"-looking carbon chains, this strategy can be prove to be very useful.
I hope you find my answer insightful.


text:
Which lone pairs/bonding pairs go into which pi molecular orbitals?
I was just doing some practice midterms and I came across this question. The answers were posted, but there are no explanations for anything. I wasn't able to find anything about this online, either.
Here are pictures of the question.
Here's the first part:

Here's the second part:

Part e) is what is throwing me off. I was flitting back and forth between O and N, and O, N, and C, but in the end got it wrong. It says to base the answer off of part d), but there are two filled orbitals! How do I know which pair of electrons occupies which orbital?
Does the sp3 lone pair of the nitrogen occupy the lowest occupied MO or the highest occupied MO? Because it appears from the correct answer that it occupies the higher one - but why is that? The other delocalized pair of electrons is the pi electrons in the C=O double bond, and I get that maybe they have different energies, but I can't think of any reason one pair should occupy one particular orbital over the other.
Furthermore, according to the movement of electron pairs in the resonance structures of f) (which were all correct according to the answer key), the third structure suggests the lone N pair is shared between the carbon and the nitrogen, not the oxygen and the nitrogen. What gives?
Also, as a side question - the reason why the solvent is polar aprotic, despite having the capacity to shed a hydrogen from the carbon to form an induction-stabilized carbanion, is because protic solvents have the labile hydrogen attached exclusively to an oxygen or a nitrogen, correct?

You can add images to your post with this button.

I think what the (bad) test question is suggesting is that HOMO electrons are delocalized and the bulk of the density is on O and N because there is a node at C. I don't think they should refer to it as a "lone pair" for the purposes of this question...

So it is inaccurate altogether to think that particular delocalized electrons present within the LOMO and HOMO are sourced from particular regions of the molecule? That any electron could occupy those orbitals, and that statistically speaking, it just would largely spatially correspond to the areas around O and N?

Thank you for your help - I will use the image button next time. I would ask my professor, but he has some policy against answering questions... Nonetheless, maybe I should suck it up and risk drawing his ire anyway. I really want to know the answer to this. I will leave this question as unanswered for now though, if anyone can chime in with a more concrete answer.

This is a common attempt to visualise what π systems actually mean for the overall Lewis structure. But let me note a couple of things first:

Obviously, there is no $\mathrm{sp^3}$ lone pair on DMF’s nitrogen. Instead, the atom is $\mathrm{sp^2}$ hybridised and the lone pair (if it is there) occupies a p-type orbital.
Remember that the π analysis only deals with the π orbitals (obviously); there is an underlying σ system which is implied but hardly mentioned. Thus, if ‘one bond’ is created, it is actually a double bond, while ‘no bond’ means ‘only the underlying single bond’.

If you want, in simplistic terms, you can add the π orbitals one after the other. The first is fully bonding and has contributions by all three atoms. Since it is fully bonding across two bonds but only contains two electrons, we can say that it contributes $\frac22$ bonding electrons to each bond or a bond order of ½.
The second orbital is not really bonding as it has a perpendicular nodal plane through the central atom (carbon). Thus, we cannot really analyse it in terms of the $\ce{C-O}$ or $\ce{C-N}$ bond order directly. However we can say that the electron density will be stronger at the ends, i.e. on nitrogen and oxygen. Each of these atoms practically experiences $\frac22$ lone pair electrons (do not participate in bonding) or ½ a lone pair.
Consider what would happen if you had filled the third orbital. It would have been antibonding with respect to both the $\ce{C-O}$ and the $\ce{C-N}$ bond. It would have contributed the same bond order of ½ to both — but this time in negative. That would contribute to an overall bond order of zero or the bonding electrons being on either side of the bond as lone pairs.
For your question, this means:

the fully bonding orbital cannot contribute to any lone pairs since it is not compensated for by an antibonding one
the nonbonding orbital contributes the lone pair. Since it only experiences contributions from oxygen and nitrogen, this is where the lone pair will be located.
it would be wrong to look at the Lewis structure, see a lone pair on nitrogen and immediately associate that with the ‘lone pair’ of the question. If anything, consider all relevant resonance forms and all their lone pairs. Also, as soon as it has resonanted to form a bond, it is no longer a lone pair.
finally, as per your side question, note that the $\ce{C-H}$ bond in DMF has a very, very low tendency to deprotonate. That proton is not even mentioned on both the Evans table and the Bordwell data. While it might lead to a carbene, the carbene is not exactly stable. A much more likely deprotonation would be that of a methyl group which would extend the π system. Note that the sequential treatment with butyllithium and DMF is a method to convert electron-rich aromatic rings (e.g. thiophene) into their corresponding carbaldehydes — in this reaction, the proton of DMF is less acidic than that of a thiophene or proton exchange would occur in place of nucleophilic attacks.

I understood the first part about the respective electron densities of each molecular orbital. But when you say that the fully bonding orbital cannot contribute to any lone pairs since it is not compensated for by an antibonding one - what does it mean to be compensated for?

Is the idea that since its asking about a lone pair, you need to pick the MO that even allows for such a configuration? And that since the bonding MO involves electrons shared among all three atoms, it couldn't possibly house the electrons of the lone pair?

@leadgame Well, think linear combinations. Consider a hydrogen molecule. $\sigma_1$ is the bonding MO and $\sigma_2$ is the antibonding MO. If you linearcombine $\sigma_1$ and $\sigma_2$ you get $\Psi_3 = \sigma_1 + \sigma_2$ and $\Psi_4 = \sigma_1 - \sigma_2$. $\Psi_3$ is localised only on one hydrogen, $\Psi_4$ on the other. If both $\sigma$ orbitals were filled, then we can consider it the same as a lone pair on both sides. That’s also the formal explanation for why number of lone pairs increases when going from $\ce{N2}$ over $\ce{O2}$ to $\ce{F2}$ and finally $\ce{[Ne2]}$.

(Note that $\ce{Ne2}$ doesn’t exist, hence the square brackets.)


text:
How to remove fluorine from 2-[2-(3-bromophenyl)ethenyl]-5-fluoro-1,3-benzoxazole?
I tried to develop a synthesis strategy for 2-[(E)-2-(3-bromophenyl)ethenyl]-1,3-benzoxazole below. I was able to find the synthesis plan shown on the picture below, however my target molecule is one without a fluorine on the benzoxazole. Hence, my questions is: would this reaction go the same way if I were to react the benzoxazole without the halogen (F)? If not, is there a way to remove it from the target molecule?
Here's a link to the synthesis plan I currently have:

And here's my target molecule:

edit:
Here's the reagents and the procedure as stated on reaxys:

With  N-benzyl-N,N,N-triethylammonium chloride; sodium hydroxide in  >dichloromethane; water
Inert atmosphere;
2.1. Synthesis
General procedure: The SBOs were prepared by the base-catalysed >condensation of the appropriate 5-halogeno-2-methylbenzoxazole with the >requisite aromatic aldehyde under phase transfer conditions. In a typical >experiment, equimolar quantities (5 mmol) of the starting materials were >dissolved in dichloromethane (20–50 ml) in the presence of >benzyltriethylammonium chloride (3 mmol) and stirred magnetically under a >nitrogen atmosphere as an aqueous solution of sodium hydroxide (50percent, >w/v, 5 ml) was added dropwise over a period of 10 min. After being stirred >for 2–36 h until analytical thin layer chromatography indicated that the >reaction was complete, the mixture was diluted with water (50 ml) and the >SBO was extracted with dichloromethane (3×20 ml), dried (MgSO4), filtered, >evaporated under reduced pressure and recrystallized from aqueous methanol >or ethanol.

Something that would remove $\ce{F}$ but not $\ce{Br}$? I doubt it.

@ringo You might be right. That's why I'm asking whether this reaction could be done with benzoxazole that has no halo grp.

I guess it's follow up of http://chemistry.stackexchange.com/questions/61854/development-of-retrosynthesis-plan

@Mithoron Indeed it is.

Oh, so it's basically an aldol condensation (the base deprotonates the methyl group on the benzoxazole, you delocalise the negative charge onto nitrogen, and then that anion attacks the aldehyde. The quaternary ammonium salt is a phase transfer catalyst to bring the NaOH into the organic phase). Well, I don't think the C-F should affect anything really. Pretty smart reaction.

How to remove fluorine from this compound?

There's no need to try that!
Just use  the commercially available 2-methyl-1,3-benzoxazole and try to react it with 3-bromobenzaldehyde in N,N-dimethylformamide under reflux for 2-4 h using self-prepared $\ce{KF-Al2O3}$ as a catalyst.
The reaction has been described (DOI) for several benzaldehydes, such as $\ce{4-H, 4-Cl, 4-Me, 4-OCH3, 4-N(CH3)2}$, with yields > 80%.

Thank you! This is exactly what I wanted to know :) Oh and this would be a KF/AL2O3 catalysed condensation reaction, right? And if so, would you mind showing me what would the reaction mechanism for this look like? Anywho. Thank you again!

I've done some research and it seems that I can make Potassium fluoride on alumina simply by mixing KF with Al2O3. Is that correct?


text:
Emission and Absorption Spectroscopy?
We're learning about emission and absorbtion spectroscopy, and how an emission spectrum is produced by passing a high voltage through a gas, and an absorbtion spectrum is produced by shining white light on a gas. Since both techniques seem to be inputting energy into the sample, why don't both produce an emission spectrum (as the electrons fall back down to their 'natural' energy state)? Also, why do the electrons even fall back to their original energy state - since a constant stream of energy is being delivered, why don't they stay at an elevated state until that stream of energy stops?

After absorption there are quite many pathways by which the molecule goes back to the ground state; not all of them involve radiative decay i.e. decay with emission of a photon. 2. Absorption of energy isn't classical - it's quantum. So it's not the same as a ping pong ball with a hairdryer underneath. If you excite a molecule with light $\ce{M + $h\nu$ -> M^*}$, shining more photons with energy $h\nu$ on the excited state doesn't "prop" the excited state up. It could decay back down to the ground state and get re-excited, but it will decay, purely because it likes having lower energy,

Thanks for the clarification and the help! I didn't realize that I was thinking of electrons in the wrong way the whole time.

To emphasize the point that @orthocresol made: suppose you have a transition between states $A$ and $B$ with energy difference $E$. If introduce photos with energy $E$, you will promote both the transition $A\rightarrow B$ and $B\rightarrow A$. This process is governed by the Einstein coefficients for absorption and stimulated emission.


text:
Way to express charges?
I would like to ask a question regarding electric charge. I see that charge has $2$ units.

Elementary charge
Coulomb

For example
A proton has a charge of $+1e$ or of $+1.602  * 10^{-19}$ Coulombs.
But there is another way but I don't have the name of the expression way or method. I see people say a proton has a charge that of +1 ( within out any unit). This is a way I assume is originated from the elementary under t but I don't know what's the name. I would be grateful if someone could help me.

Simply if you talk about particles it's obvious that you're using elementary charges and there's no need to write "e" What's here to ask about?

There are 3 units that are used to express charge.

Coulomb (C)

This is most commonly used in macroscopic situations.
1 coulomb of charge is that which, if placed at a separation of 1 metre from another charge of magnitude 1 coulomb, experiences a force of $8.99\times 10^9$N along the line joining the charges.
This definition results from the observation-based Coulomb's Law of Electrostatic Forces:
$F_e=k_e\frac{q_1 q_2}{r^2}$

e

This is a unit of charge most commonly used in microscopic situations due to its convenience.
It is the magnitude of charge present on 1 electron, or 1 proton - it is equal in magnitude.
$e\approx 1.602\times 10^{-19}$

Electrostatic unit of charge (esu)

The least used of the three, it is the CGS unit of charge. It is also known as franklin (Fr), or statcoulomb (statC).
1 esu of charge is that which, if placed at a separation of 1 centimetre from another charge of magnitude 1 esu, experiences a force of magnitude 1 dyne along the line joining the two charges.
Again, from Coulomb's Law of Electrostatic Forces, we can determine that
$1 esu\approx 3.335\times 10^{-10} C$

Oh my god thank you. Now I know one more unit (which is 1 esu). I would like to ask in terms of why sometimes people use: a proton has "a charge of $+1e$" and if this kind of expressions has a label for it !?


text:
Salting out constant in activity coefficient
Based on this Wikipedia page for activity coefficients, I know that the activity coefficient for a nonelectrolyte can be determined using a "salting-out" model with the following equation:
$$\begin{aligned}\log_{10}(\gamma_0) &= bI\\
\gamma_0 &= \text{activity coefficient}\\
b &= \text{empirical constant}\\
I &= \text{ionic strength}
\end{aligned}$$
This equation and the accompanying wiki leaves three related questions about its use:

How would I determine that empirical constant $b$?
How do I interpret the ionic strength of a nonelectrolyte solution?
What does this equation have anything to do with "salting out" - reducing the solubility of nonelectrolytes in aqueous solution by increasing the ionic strength?

https://en.wikipedia.org/wiki/Salting_out

@ringo - your link does not include an equation for determining activity coefficients based on the "salting-out" model. The wikipedia page for activity coefficient does give the formula, but does not provide an explanation of how to determine the empirical constant (or honestly how the formula is related to salting out).


text:
What would happen to feces when set on fire?
What would happen, if, under safe conditions, I took my feces, doused it with lighter fluid, and set it on fire? What the neck would burn? And what would be left? What if I threw in some wood twigs?

According to the German nursing profession wiki, the contents of human feces are as follows:

75 % water
  10 % (non-digestable) remainder of food (e.g. cellulose)
  8 % enteric bacteria
  7 % epithelium (cells from your own gastrointestinal tract)
  rest = mucus and salts

Given that, initially only the lighter fluid will burn. This will heat up the surface of the mass and evaporate some water from it. As the mass reaches a sufficiently high temperature, organic material with a low boiling point might be set free as well and possible burn.
On a closer look, it turns that the VOCs (volatile organic compounds) in human feces constitute an interesting matter (yes, pun intended).
In a freely available study titled Volatile organic compounds from feces and their potential for diagnosis of gastrointestinal disease, which was published in FASEB J, 2007, 21, 1675-1688 (DOI), the authors identified a large number of of VOCs from asymptomatic donors and patients with different diseases by gas chromatography. Among the compounds found in at least 60% of the asymptomatic donors than were

3-carene, limonene, $\alpha$-pinene and other terpenes
ethanal, acetone, 2-butanal , 2-heptanone and other carbonyl compounds
toluene, styrene, phenylethyl alcohol
heterocyclic compounds, such as indole, 3-methylfuran

But given that you chose pyrotechnics, you might not be interested in these details and were looking for less pleasant and productive experiments ;)
If you are simply interested to incinerate and burn feces to leave as little "solid" remains as possible, remember that in some parts of the world (animal) feces are commonly used as fuel. Consequently, you might be better off with dried feces.


text:
Manufacturing coloured crystal glass using gold nanoparticles
I came across this video which shows how a crystal chandelier is made. I recommend that anyone reading this post should watch it; it is quite fascinating.
Anyway, at about the 2:15 mark, the video talks about how ruby red crystal is produced by adding 24-karat gold powder to the glass. However, they say the colour emerges after reheating in an oven at $1000\ ^\circ C$.
My question is, why is the reheating step necessary ?

To nucleate and grow the appropriate sized gold nanoparticles. Both gold and silver were used in medieval cathedral stained glass windows.


text:
How was the dissociation constant of water determined?
I was studying about pH and pOH of different acids and bases and I read about the dissociation constant of water ($\ce{H2O}$).
It turned out to be $1.00 \cdot 10^{-14}\ \mathrm{mol^2\cdot L^{-2}}$ at $25~\mathrm{^\circ C}$.
Could I get some insights to how was this specific value determined?

http://chemistry.stackexchange.com/questions/52001/why-is-the-water-dissociation-constant-valid-for-acid-base-solutions http://chemistry.stackexchange.com/questions/23693/ph-poh-autoionization-of-water

@Mithoron NO....I was asking how was it determined that the concentration of hydroxide and hydronium ions is $ 10^{-14} $  M and not something else......How was this specific value determined?

You can do it using a pehameter and pure water... Are you asking about history or what? BTW you're mistaking concentration with constant in your comment.

Thanks for pointing that out......but could you please tell me what is a pehameter...I googled it but couldn't find anything useful... I know how dissociation constant is calculated...but for that don't we need concentration of ions ...How was this concentration calculated? or is there some other way?

I meant pH meter

but a pH meter needs a solution of known pH to calibrate it.....how was the pH of that solution determined? Anyway...Thanks for your help

I'd dare to say that it was first determined in experiments on electrical conductivity around 1890 by the physical chemist Friedrich Kohlrausch.
In Die Wasserstoffionenkonzentration. Ihre Bedeutung für die Biologie und die Methoden ihrer Messung (Leonor Michaelis, 1922, ISBN: 978-3-642-88800-7) the author writes

F. Kohlrausch und A. Heydweiller kamen durch wiederholte Destillation des Wassers unter ganz besonderen Kautelen zu einem Grenzwert der Leitfähigkeit des Wassers, der sich durch weitere Reinigungsversuche nicht weiter herabdrücken ließ und daher als die eigene wahre Leitfähigkeit des Wassers gedeutet werden mußte. Sie stellten auf diese Weise die Dissoziation des Wassers zum erstenmal zahlenmaßig fest. Später wurden dann mehrere andere Methoden zur Dissoziation des Wassers gefunden, die wir bald kennenlernen werden, und diese bestätigten die Resultate von Kohlrausch.

This roughly translates to:
Upon repeated destillation of water under extreme precautions, F. Kohlrausch and A. Heydweiller eventually reached a limit for the conductivity […]. This way, they were the first to quantitatively determine the dissociation of water[…]
In a footnote to this paragraph a reference is given:
F. Kohlrausch und A. Heydweiller, Zeitschr. f. physikal. Chemie, 14, 317 (1894).

I have a last question......I understand how water was distilled but still you just couldn't count the concentration of ions...how was concentration of ions determined?

I love the old language … ‘the own and true conductivity of water’ =D


text:
Peculiar behaviour of tea boiling as milk+Water
While making tea last night I forgot all about it and left the kitchen. When I did remember about it, I ran back and I was surprised to see that it was still in the vessel (though it formed a peculiar foam-like mass on the top).

This pushed me to do the same with water and milk (as these were constituents this tea).
I was not so lucky in case of milk, When I boiled the milk it rose too much and overflowed.

Boiling water alone didn't help either. The level of water in the vessel rose a little but it was far from overflowing (I provided the highest temperature as I could provide to vessel).

So I observed three different phenomena for three different liquids:

Tea (having water and milk in proportion of 3 and 1) rose till the height of container.

Milk overflows.

Water doesn't even remotely show anything that would indicate it was responsible for this peculiar behaviour of tea.

I can't seem to figure out why this happens. Would anyone know why?Thanks for any help
EDIT
After listning to people saying that they need more experimental data I decided to go further.
I boiled the milk along with water in a proportion of 1 by 1 (without tea leaves and sugar to it) and did the same with different proportions of milk and water.
After doing so I got that As I increased the amount of water (in combo of milk and water) the level of solution started decreasing and ultimately at the ratio of 1 and 5 (for milk and water) the solution didn't overflow.
The results were as expected but I thought these can be helpful for you guys in answering the question.

Related: Why does milk overflow when boiled?

There is one thing common. Should i remove the milk from my question @Loong??...:p

You rightly mention that milk and water are components of tea. But perhaps I should direct your attention to the fact that they aren't the only components of tea. (There is of course, the powdered tea leaves you stir in)

@AaronAbraham should i try to boil tea leaves and sugar too?/ :) :)

-_- Do I detect an undertone of sarcasm?

@AaronAbraham No No No No.........

What I implied, is that the powdered tea leaves clump together at the surface, forming that mass on top of the tea. This of course, wouldn't happen if you tried using a certain device we call a tea bag ;)

Okay so you mean that i should boil a mixture of milk and water only and then check the result.

Thought this was a lame question at first but because of the effort the OP put in with the photos it along with the response became quite interesting.  Thanks for the effort.  ps do the suggested experiments and report back.

Original Hypothesis
As mentioned in this post, milk will overflow thanks to the effect of surfactants, which allow for the formation of stable bubbles. Since tea contains both milk and water, it contains surfactants from the milk, but at a lower concentration thanks to the water. This would result in some bubble formation, but not as much as you would find in pure milk, explaining why the tea filled up the container, but did not overflow. Either it was a coincidence that the bubbling stopped right at the top of the container, or it relates to the fact that more surface tension is required to create bubbles without the support of the container walls. In addition, other components in the tea could be contributing to this effect, but I think it is a result of the fact that milk and water are mixed. If you want to experiment with this, I would try a different volume of your water/milk mixture and see if that also makes it to the top of the container without overflowing.
Results of Additional Experiments
The results of your additional experiments seem to indicate that something else is going on. It is interesting that you found a different ratio that filled the container when there was no tea involved. This seems to indicate that the tea is at least somewhat involved in this interaction. Also, the fact that past the ratio of 1:5 (milk:water) the container did not fill up with bubbles seems to indicate that my hypothesis about surface tension was wrong, and that there is nothing special about filling the container. In this case, you probably just got lucky that your original tea did not overflow.
These results are dependent on the fact that you used the same vessel for your new experiments and allowed adequate time for bubbles to form. More information on your experimental procedures will help me narrow down on a possible explanation.

I have updated my question by adding new info to it.Hoping for a better answer.

Did you preform these experiments in the same vessel? If so, it is somewhat surprising that the first milk:water ratio to not cause overflow was 1:5 while it was 1:3 with the tea. I will look into this more if it is the case. Also, did the 1:5 ratio stop right at the top of the vessel? If so, did a ratio like 1:6 also stop at the top, or did that just fill partway?

Ratio of 1:5 reached too the top but as i started decreasing amount of milk,the height of solution get lower and lower

@THELONEWOLF. I added additional information based off of your experimental procedure. I can narrow down on an explanation if you could write down more details about your procedure.


text:
What is the standard way of reporting/interpreting measurement uncertainty in scientific publications?
As an undergraduate student I was often told in lab classes to report every single measured or calculated quantity with its associated uncertainty in lab classes, in the form $(10.1 \pm 0.1 )~\mathrm{g}$ for instance, where 0.1 is the standard deviation or estimated uncertainty. I'm starting to find, however, that this practice is seldom followed in scientific publications.
When reporting uncertainty in a scientific publication, is it good practice to report every measurement in the form $(300.1 \pm 0.1 )~\mathrm{K}$ ? I sometimes see this done in the body of scientific papers, but usually never in tables/lists of data. This worries me because, in my understanding, all measurements have an uncertainty associated with them, so I find these data hard to interpret unambiguously.
Is it usually assumed that numbers written without an explicit uncertainty are accurate to the last expressed digit? (i.e. that $10.5~\mathrm{g}$ implies $(10.5 \pm 0.1)~\mathrm{g}$ ?) sometimes this seems to be the case, but doesn't this negate the difference between $ \pm 0.1$ and $\pm 0.5$ for instance? Is this difference usually considered unimportant?
As a side question: Regardless of the usual practice in academic writing, do you consider that demanding every single number to be reported with an associated uncertainty is a good idea if teaching a lab class?

The international reference standard is the so-called GUM: JCGM 100:2008 Evaluation of measurement data – Guide to the expression of uncertainty in measurement. The same content can also be found in an ISO/IEC standard ISO/IEC Guide 98-3:2008 Uncertainty of measurement   Part 3: Guide to the expression of uncertainty in measurement (GUM:1995)

In particular, see Subsection 7.2.2 of the GUM.

Thanks, I had read it prior to asking the question, (not back to back, some sections) but I believe that is a sort of "idealized way" in which a measurement uncertainty should be reported, as I said, I have seldom seen even a ±0.1 on a table. When I asked about the "standard way" I didn't have in mind what the ISO et al people say but what scientists actually do. Having said that, I have seen m = 100,021 47(35) g or similar things, but only in NIST values, etc.

Should I edit the question to make that more clear?

In my experience, this varies greatly with your field of study.
In organic chemistry, I have yet to experience proper uncertainty reporting in journal articles or lab reports with the notable exception of physical organic chemistry. Once in my undergraduate years, I noted while writing a report that $130~\mathrm{mmol}$ where actually $1.3 \times 10^2~\mathrm{mmol}$ and wrote it that way. Okay, I should have written $0.13~\mathrm{mol}$ in a more correct notation to signify the significant digits. But the way the TA marked the report clearly told me ‘I don’t give a damn about correct error margins or significant digits, just God damn write $130~\mathrm{mmol}$!’ Lesson learnt from that anecdote: synthetic organic chemistry does not care about minor details such as significant figures or uncertainty.
The closer you get to physical chemistry, the more your measurements actually represent data values to be cited in a table and thus the more exact the values determined become, the more people care about reporting error margins and uncertainty. And funnily enough, in biology this is also done much more regularly. I suspect that there are different underlying reasons: One side (physical chemistry/physics) does not want the negative outcome of ‘you pretended your results are more accurate than they are!’, while the other side, biology, knows that it is intrinsically dealing with very error-prone systems that require large repetition counts which can then be analysed statistically.
Although synthetic organic chemistry practically also deals with rather error prone systems, either the idea is that there are too many uncontrollable variables or the idea is $n=1$ (always!); in any case, error bars or uncertainty are deemed superfluous.

Seems to me synthetic organic chemistry is more focused on the "ooh, look what neat thingy I made!" rather than the "I know to three significant figures the mass of the pile of the neat thingy I made."

@hBy2Py Thanks, now I have the image of a cute, cuddly, pink, fluffy little thingy in my flask =^,…,^=


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Is polymeric ethylene glycol conductive?
The other day, when we were doing the chapter Polymer Chemistry my teacher mentioned that Polymeric Ethylene Glycol (PEG) is conductive.
[EDIT- By 'conductive', I guess my teacher meant, "sufficiently conductive, so that it can permit a significant amount of current"...Now I really don't want to go and elaborate on "significant"]
Naturally I was taken aback, having lived all my life believing anything even remotely related to plastics could possibly conduct electricity. So I Googled it, but that endeavor wasn't particularly fruitful. I ended up getting links to articles like these: http://www.sciencedirect.com/science/article/pii/S001346861101190X
As I wasn't able to get my hands on any piece of scientific literature, that definitively puts my question to rest, I resolved to have a look at the chemical structure of PEG:

But I wasn't able to identify anything that could possibly impart conductor-like properties to PEG. No free ions or any delocalized electrons that could facilitate conduction. Nothing ._.
However, I was able to discern something about possibly 'altering' PEG, via a method that's akin to doping. At any rate, I'm getting no where here.
Would anyone happen to know if PEG really is conductive in the pure state? Or is it somehow 'modified' to impart conductive properties to it? If so, then how is it modified?
A good reference would be appreciated.

Depends on what someone means as conductive.

Roger that! I'll edit it.

Uptake of water from the air might be one factor. See if you can get a hand on this article: http://scitation.aip.org/content/aip/proceeding/aipcp/10.1063/1.2733226

Good link Herr @Klaus! But it mentions that the water serves to increase the conduction. Would you happen to know any link that covers that?

@AaronAbraham Unfortunately not. I was just curious and ran a cursory search myself ;)

@KlausWarzecha The other link was quite interesting at any rate. So danke! ^_^

Though not exactly common, there are still many conductive polymers. Some are only conductive when adulterated (e.g. iodine-doped polyacetylene, which led to a Nobel prize in chemistry),  but others are conductive even when pure. See https://en.wikipedia.org/wiki/Conductive_polymer


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Manifolds in quantum chemistry
Sometimes the concept "manifold" is used in quantum chemistry, for instance,
in Calculation of size‐intensive transition moments from the coupled cluster singles and doubles linear response function:

where $\tau_{2s}$ and $\tau_{2t}$ denote the singlet-singlet and triplet-triplet spin coupled double excitation manifolds where spin couplings initially are carried out on the occupied-occupied and unoccupied-unoccupied orbital indices.

However, in general, manifold refers to a geometric object that is locally Euclidian. This is not essential to the coupled cluster method.
My question is, what is the motivation to introduce manifolds in the coupled cluster method?

In general, the term "manifold" might refer to about a dozen of different and unrelated things (this, for example). Using it without any context is a great way to get misunderstood.

The use of the word "manifold" in most of the quantum chemistry literature is purely as a synonym for multiple or different. Usually it appears when the number may be very large. Nothing more complicated than that.
For a simpler example than coupled cluster, consider configuration interaction with only single excitations (CIS). One of the key quantities is the set of single excitation amplitudes, $t_{ia}$ or $t_{i}^{a}$ or $k_{ai}\hat{a}^{\dagger}\hat{i}$ depending on your preference for notation. $i$ is an occupied MO index, and $a$ is a virtual MO index. We might say that the set $\{t\}$ forms a manifold, and the number quickly becomes large as you increase the basis set size: there will be $N_{ov} = N_{\text{occ}} \times N_{\text{virt}}$ single excitation amplitudes.


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Coating applied to plastic (polymer suitable for straws) to repel carbon dioxide to prevent "floating"
We are four students currently finalizing our last exam-project before our bachelor.
We confirmed our hypothesis that straws floating to the top of your drink is a "problem" from our survey.
Our question here would be: Is there any coating or similar scientific method that can be applied to any suitable plastic material, (for the use of consuming beverage) that can prevent the carbon dioxide from sticking to the plastic material, and thereby lifting it to the surface?


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What is the function of tosylchloride in the synthesis of an ether?
Considering the following reaction:

Today my teacher explained this reaction as follows:
The electron pair of $\ce{OH}$ (in the first compound) will "attack" the $\ce{S}$, making the $\ce{Cl}$ "jump out". This is something I didn't understand clearly:
Why does the electron pair attack the $\ce{S}$ and not one of the oxygens, or even the ethanol? Is there any particular reason?
I'm not a chemical engineering student (I study engineering physics) so I don't have a great background in organic chemistry.

Sorry about the title, this is actually a Sn2 reaction..

No worries, I changed the title to something more specific and replaced your drawing (at least you tried!). Feel free to revert if it violated your intention.

It won't do anything with ethanol because that's only added in step 2.

There are probably only two principal mechanisms in organic chemistry, one being electrocyclic reactions (not applicable here) and one being nucleophile/electrophile reactions. In the latter, which is what is happening in your case, you must always look for the group with either the highest electron density or the ‘best’ (I know, a hand-wavery term) nucleophilic properties — this group is the electrophile — and for the group with the lowest electron density or the ‘best’ (same hand-waveryness) electrophilic properties — termed the nucleophile.[1]
The first step to determining this would be to check electronegativities and oxidation states. Groups with high electronegativities and low oxidation states are typically nucleophiles and vice-versa. In your example, all alcoholic oxygens have a much higher electronegativity than all atoms they are connected to (carbon and hydrogen), are in the $\mathrm{-II}$ oxidation state and would thus act as nucleophiles. The oxygens and the chloride of tosyl chloride also have a larger electron density than their surroundings. For reasons outlined above that I cannot go into details with, they will not react nucleophilically, though.
The carbon atoms which are directly attached to the alcoholic oxygens all lack electron density due to the neighboring electronegative oxygen. They are electrophiles. And also the central sulfur of tosyl chloride — which not only features a $\mathrm{+VI}$ oxidation state but should also be drawn with a double positive formal charge to satisfy the octet rule[2] — is an electrophile.
To synthesise the ether from the two alcohols, we would now somehow need either of the two alcoholic nucleophiles to attack the other carbon electrophile and displace the other alcoholic hydroxy group — this reaction is termed $\mathrm{S_N2}$ reaction or bimolecular nucleophilic substitution. However, it is not likely to occur directly: a direct side-product would be the removal of a hydroxide ion, which is termed a bad leaving group.[3] We need to turn it into a better leaving group.
This is where tosyl chloride comes in. But first another little excursion: Notice the little 1. and 2. in your reaction scheme? They stand for first and second, implying that we are dealing with two separate reactions interceded by a workup and typically purification of an intermediate. So first we react the 1-phenylpropan-2-ol with tosyl chloride, collect the product and then react said product with ethanol. As I outlined before, a pretty good electrophile in tosyl chloride is the sulfur atom, and with the adjacent chloride, we have a good leaving group. So under a nucleophilic substitution,[4] this chloride can be displaced (along with a proton) leading to our intermediate 1-phenylprop-2-yl tosylate.
Technically we would now need to make a fresh analysis of nucleophiles and electrophiles. But it may suffice to note the key differences: there is no longer a good leaving group on sulfur (the alcoholate which would be a leaving group again is the conjugate base of a very weak acid), but if ethanol now attacks the carbon adjacent to what is now a tosylate group we have the tosylate anion as a good leaving group. In a second bimolecular nucleophilic substitution, tosylate and a proton are displaced and the ether is formed.

Notes:
[1]: Actually, which group is nucleophilic and which is electrophilic depends on the frontier orbitals, and that can also be used to unify both reaction types into one general scheme, but that analysis is very hard to do a priori.
[2]: Organic chemistry’s preference to write sulfonic sulfur atoms as doubly bonded to two oxygens, creating a dodectet gained popularity probably because of one of Pauling’s not-so-bright moments. However, evidence points towards d-orbitals (which would be required for the dodectet) not taking part in bonding for non-metals and thus the charge separation depiction being a more accurate description of the truth. This is documented a lot across this site.
[3]: Again, I won’t go into detail here when leaving groups are good or bad, but just leave as one principal pointer that conjugate bases of strong acids are bad leaving groups. The conjugate base of $\ce{OH-}$ is $\ce{H2O}$ which is a very weak acid; thus, hydroxide is a very bad leaving group.
[4]: Technically, this is no longer an $\mathrm{S_N2}$ reaction. Since sulfur is participating as the central, electrophilic atom, the transition state is more long-lived involving a four-electron three-centre bond. The difference is marked enough to call this bimolecular nucleophilic heavy atom substitution or something along the lines. It will usually not be referred to as an $\mathrm{S_N2}$ reaction.


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Regioselectivity in Carbometallation
I am a student of organic chemistry and I frequently watch online open course lectures by various professors around the globe. Of these, in https://www.youtube.com/watch?v=OYAMls5x4EI, at about 27:10, I had a question, as mentioned in the image below.

Mechanism

What factors determine where the palladium will attach?
My initial thoughts were to find out the electron-rich regions of the $\pi$ bond, and the palladium atom in the $\mathrm{+II}$ oxidation state will act as an electrophile. But I wasn't able to figure it out exactly.
I understand that the ratio of products shown in the figure can be controlled by using various ligands, especially based on their steric effects, in this case.

@EashaanGodbole One thing that is important to note is that the hydroxyl group acts as a hydride source. This is only possible for the above product (see my answer).

Your mechanism is missing an important step: between the oxidative addition of palladium into the $\ce{C-Br}$ bond and the alkene insertion there should be an alkene annealing forming a palladium-alkene complex. This is displayed in the mechanism scheme of a generic Heck reaction below, which I sheepishly borrowed from Wikipedia.

Scheme 1: Catalytic cycle of a Heck reaction. Image taken from Wikipedia, where a full list of authors is available. Note that the β hydrode elimination in step C should be syn and not anti or gauche as displayed.
The annealing step and the π complex are important because they influence the Heck reaction’s selectivity. Due to the avoidance of steric stress, both residues will adopt an anti-type conformation as depicted in structure 3. On the (syn-selective) insertion of the alkene into the $\ce{C-Pd}$ bond, this geometry is kept resulting in the palladium catalyst ending up on the higher-substituted alkene carbon. For the inverse insertion, both organic residues — in your case two phenyl groups — would have to occupy the limited space on one side of the catalyst.
The subsequent β hydride elimination can theoreticly occur in both directions in your molecule. However, eliminating towards the hydroxy group allows keto-enol tautomerism with further stabilisation. Since β hydride elimination is a reversible process, the reaction will select the ketone over the allylic alcohol that would be formed upon opposite β hydride elimination.
The alkene insertion into the $\ce{Pd-C}$ bond, is not reversible. Two $\ce{C-C}$ single bonds are formed at the expense of a palladium-π interaction resulting in a notably downhill process.

Just have a look at the reductive elimination step of this Heck reaction:

In case of the second reaction the resulting allylalcohol is far less stable than the enol tautomerizes to the ketone. The other thing are sterics aspects that make the migratory insertion step far less favourable for the second product as well.

No. The selectivity is determined when the $\ce{C-C}$ bond is formed. Ease or unease of the following beta hydride elimination is irrelevant for the proceeding of the reaction; if the $\ce{C-C}$ bond formed a branched compound, the final product will be branched.

@Jan You sound so definite in what you are saying. Then could you please provide proof of what you are saying, i.e. that the C-C bond formation in a Heck reaction is irreversible?

@Jan Looking at e.g. the pinacol-pinacolon rearrangement, benzoin condensation and so on it seems entirely possible that a C-C bond formation is reversible, and I am actually pretty sure that all steps of the Heck reaction are reversible...

Of course, reversible $\ce{C-C}$ bond formations are possible; they are a key point of the mechanism of aldol condensations or Horner-Wadsworth-Emmons reactions. However, in all of those cases we are dealing with somewhat activated bonds where bond dissociation enthalpy ($\approx 400~\mathrm{kJ \cdot mol^{-1}}$!) is partially regained by other features of the system. Unfortunately, the Kürti-Czakó does not state the irreversibility of the $\ce{C-C}$ bond formation in a Heck reaction, but it also does not call it reversible.

@ketbra the answer that you provide is based on the thermodynamics of the system, as Jan mentioned. But for the product you say it less favourable, a readdition is certainly

Certainly possible.also, the final product in that case seems more stable. So, shouldn't the basis be kinetically controlled?

@EashaanGodbole I really don't think so. Typically, all steps of cross coupling reactions are reversible. In fact, the outcome of the Heck reaction is in most cases controlled by the final beta hydride elimination (so the reductive elimination). So I dont see the problem with my 'product controlled reaction' argument. But sure, the formation of the first Intermediate after carbometalation will also be far less favorable for the second reaction. However, this has nothing to do with kinetic or thermodynamic control its only thermodynamic control.

@EashaanGodbole Looking at both arguments either arguing about the disfavoured carbometalation and/or the disfavoured beta hydride elimination: Even if the reaction would be allowed to react for 10 billion years the first product would still be the only product, as the other product is just too unstable to be formed anyway. It is just a thermodynamic argument. By the way, the lectures of Dyker are really nice : )


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Why does a 1:1 mixture of Polyvinyl Alcohol (PVA):H2O decompose/carbonize during heating?
I bought a powder of Polyvinyl Alcohol (PVA from Sigma Aldrich). I mixed the powder with water in a mixture 1:1 to place a droplet on a hot plate. To figure out the melting temperature of the PVA, I heated it, meanwhile the water should be evaporated. According to Wikipedia or other chemical databases, partially hydrolysed grades of PVA melt between $150-180\ ^\circ C$
.
My questions:

Why did my mixture decomposed/carbonized above $120\ ^\circ C$
(its color changed from transparent to brown, so I guess it decomposed)?
The mixture did not melt between $150-180\ ^\circ C$
. Why?
Did I make any general mistake at all?

I experienced similar phenomena in my experiments.
In my case, I put PVA pallets (M.W. is 84000~12500, 87~89% hydrolyze degree) in the oven at 120∘C. After over 12 hours, the color became yellowish (partially brown). But after dissolving the pallets in the water, there was no difference between control sample(without heat treatment).
There are several papers about formation of polyene in PVA chain under acid catalyst. And, almost commericial PVA product has always small amount of acetic acid which hydrolyzed from PVAc spontaneously. Therefore partial polyene-nization (semicarbonization) was possible with heating under melting temperature.
Also, I made acidic PVA film by solution casting (10% PVA, pH 2 by adding HCl). Then, the color of PVA film was changed to black at 120∘C in the oven.
So color change of PVA is common while experiments. Maybe it is related with carbonization.

maybe polyenenization or carbonization.
I remembered that PVA becomes viscous near 150-180∘C. Measuring Tm of polymer exactly is not easy. Especially hot plate is not suitable tool to analysis physical properties of polymers. Maybe due to cooling by air, melting of PVA is indistinguishable to glass-transition. I recommand to use DMA instead of your experiment.
To measure Tm, you don't have to mix PVA with water. DSC or DMA is better.


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Emulsion formation during extraction
While extracting benzoic acid from ether with $\ce{NaOH}$ solution, an emulsion forms between the layers. On the other hand, when aniline is extracted from ether with $\ce{HCl}$ the separation proceeds smoothly. Why does the emulsion form in the former case?


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Trend in atomic radius for noble gases
In an exam, we were given the following graph and asked to explain why the slope of the change in r vs. Z changes dramatically at a point along the curve. I understand that the d block causes the change in slope for all of the series, but why is the slope from $\ce{He}$ to $\ce{Ne}$ less than that from $\ce{Ne}$ to $\ce{Ar}$? Why is this trend not present for any of the other series? I initially thought that this might have to do with the introduction of the p-block, but that would not explain the absence of the trend for the other series.

I suspect that there exists an argument along the lines of the 2p orbital not having a radial node, therefore rapid stabilisation going across Period 2, leading to Ne being a tiny bit smaller than expected. But I'm not willing to flesh it out now (rather late and I want to sleep).

Isn't this because He is the only element that has 1s orbital in this chart? Maybe if we included hydrogen, we would see a trend? All other elements start from the 2nd row and jump to the 3rd row except for Helium and Neon.

@QuantumCAPUCCINO, $\ce{Li+}$, $\ce{Be2+}$, $\ce{B3+}$, and $\ce{C4+}$ all have the same electron configuration as He, so I don't think that is the case.

@Benzene Uh oh I missed that. Sorry~

They should’ve included hydride in the series beginning with fluoride …

I have a theory for the question.
First lets look at what the graph is telling us about the radius and the atomic number. For the Noble gases, as the atomic number is increasing the radius increases. It is because of the addition of more electron shells (simplification for now).
Helium has a full 1s orbital. Whereas Neon has a configuration of $1s^2,2s^2,2p^6$. Therefore the electronstatic force occurring on the electrons is slightly bigger than the electron on the helium atoms. Whereas Argon has a configuration of $[Ne]3s^23p^6$.
Now my theory. In He, we only had $1s^2$ and in Neon we added the extra shell and we went up til the 2p. Now the shells also act as barriers sometimes, therefore we have something called an effective nuclear charge. Now, since He has 1s2 and we added an extra shell, well the $1s2$ does not act much of a barrier and the nucleus is still able to attract the electrons in the second shell ($2s^2 2p^6$) electrons and the radii does not change drastically. But when you add a full shell on top of that, well then we have a problem. Because then the second shell electrons can act as a barrier in the electrostatic force, and the effective nuclear charge decreases and the repulsion between the electrons in the third shell will be greater, hence a greater atomic radii occurs between Ne and Ar, compared to He and Ne.
Now this trend will not be seen in the other ones. It is because the electron configuration is one thing but is not the only thing that is being shown on the graph. All of the other ones are ions. When something is an ion, for example the positive ones, then there are two things to look at, electron configuration and the charge. Meaning in a positive ion, you have more protons than electrons, meaning that the electrostatic force will much more greater in $Li^+$ than the one in $He$, even though they have the same electron configuration.
Whereas the trend in the noble gases is based on the fact that they all have $e^- = p^+$ and are neutral, whereas the others are not.
I hope I make sense here. But like I said, its just my theory for the question and I can be wrong :)


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Why is chemical accuracy defined as 1 kcal/mol?
"Chemical accuracy" in computational chemistry, is commonly understood to be $1~\mathrm{kcal\over mol}$, or about $4~\mathrm{kJ\over mol}$. Spectroscopic accuracy is $1~\mathrm{kJ\over mol}$, and that definition has intuitive sense. However, where does the $1~\mathrm{kcal\over mol}$ quantity come from?
From Wikipedia:

A particularly important objective, called computational thermochemistry, is to calculate thermochemical quantities such as the enthalpy of formation to chemical accuracy. Chemical accuracy is the accuracy required to make realistic chemical predictions and is generally considered to be 1 kcal/mol or 4 kJ/mol.

Short answer: the goal of "thermochemical accuracy" for computational chemistry is to match or exceed experimental accuracy. Thus, ~1 kcal/mol comes from the typical error in thermochemical experiments.
The drive began with John Pople, who begin the modern effort to consider "Model Chemistries," comparing the accuracy of different methods across many molecules and often multiple properties. He realized that for thermodynamic properties, one could approach the accuracy of experiments. (See, for example his Nobel lecture).

As the model becomes quantitative, the target should be that data is reproduced and predicted within experimental accuracy. For energies, such as heats of formation or ionization potentials, a global accuracy of 1 kcal/mole would be appropriate.

He then started work on composite methods like G1, G2, G3, etc. that could approach predicting many chemical properties to this accuracy.

I tried fairly quickly to see if the 1 kcal/mol number is solely due to Pople or others suggested it previously. Certainly he pushed this threshold in his widely-used Gn methods.

I see, so in short, "thermochemical accuracy" became shortened to "chemical accuracy" over time, since that was the main area of chemistry quantum chemistry aimed to match in accuracy.

Exactly. Also, if you can reach 1 kcal/mol accuracy, that's ~0.05 eV, which would be fairly good for ionization potentials or electron affinities. (We're not there yet.) So it was a good target for other chemical properties.

Certainly experimental accuracy is a major factor. However, another argument for this rule of thumb is that at room temperature, a difference in free energy of 1.4 kca/mol results in a  equilibrium/rate constant with a relative change of ~10. Therefore, in order to be within an order of magnitude of experiment, you need to have an accuracy of ~1 kcal/mol.


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Do electrons in an atom revolve around the the nucleus clockwise or counterclockwise?
Do electrons in an atom revolve around the the nucleus clockwise or counterclockwise?
Is there any rule to determine?

No. Electrons don't revolve around anything at all. You have a choice of two inappropriately-termed words to describe the electronic state you're interested in: spin up and spin down. Neither term refers to spin in any sense of the word, not to mention up or down, which are even less appropriate. In short: electrons have a property called spin which can have values of up or down (more confusingly notated as 1/2 or -1/2).

What would be the spin dependence of time?

Come to think of it, even ordinary things in 3D do not rotate clockwise or counterclockwise. (Also, electrons are not ordinary things, and they do not rotate at all.)

Even in an orbital the electrons in an atom so not rotate, they just have a probability of being at some position. The spin of an electrons as mentioned above does not literally mean that they 'spin' but rather that they have a property that obeys equations describing angular momentum. Yes it is all very odd compared to everyday experience, but its what experiments teach us.

http://chemistry.stackexchange.com/questions/36987/does-the-shared-electron-in-covalent-bonds-revolve-around-nucleus

Lets make this clear:

Electron is elementary particle.
Like other particles it can be also described as wave (wave–particle duality).
It may be a point particle, but it possibly has finite, but very small radius.
Electrons do move around nuclei. Corpuscular nature of electron completely excludes the possibility that electrons in atoms don't move. They also may or may not have true orbits depending on which interpretation of quantum mechanics is really "true". So, they move and are very fast; their velocities can be even relativistic (see this answer) and in such cases even Schrödinger equation doesn't describe atoms properly.

As Ivan pointed out in comments, orbits of electrons in atom, which wasn't disturbed in any way, can't be seen, which is true for any system (observer effect). Still even single electron during ionisation of an atom was observed using attosecond laser; a film of movement of electron cloud was created (see also).
As for OP's question "clockwise" or otherwise is relative to observer and therefore there's no point in asking that. Circular motion in original Bohr model was only an approximation and electrons move in unpredictable, chaotic way. Even on short timescales single atoms can be considered spherically symmetrical - they don't have left side or up side.

The question does pose itself (and I am in no way qualified to understand that paper … I gave up quickly) whether electrons can be seen to circulate the nucleus or whether they just move randomly in its vicinity.

@Jan Bohr modelled orbits with circels Sommerfield improved with ellipses, but the movement is more or less chaotic https://en.wikipedia.org/wiki/Old_quantum_theory#Limitations_of_the_old_quantum_theory

@Mithoron You are confusing people for no reason. These papers do not say that an electron residing on an orbital moves in any literal sense. The second paper describes an excited electron which oscillates between certain states (each state being a probability cloud), and the first one deals with an ionized atom and separated electron. By skipping these technicalities, you fuel the false impression which many students already have, that electrons on (say) 1s orbital sort of fly around the nucleus, like planets around the Sun. This is quite wrong. Sorry, but you aren't making this clear.

@IvanNeretin Well, for starters electrons do not reside on orbitals. Orbitals are only models, by no means a perfect ones. I felt the need to make this post because it's you who is wrong and I'll try to convince you.

Yeah, sure, if we look deeper, the orbitals are but abstractions and all that. Now what's about the electrons which are "moving"? In one case, an electron is oscillating between two probability clouds, which probably can count as "moving", but mentioning that without further explanation still does create a wrong impression. In the other case, an electron gets wrestled free from an atom and is moving in a somewhat classic fashion; that's OK. Now to the point: what about an electron in a ground-state atom, is it "moving"?

@IvanNeretin See Dave's answer - http://chemistry.stackexchange.com/questions/13581/i-am-trying-to-picture-how-electrons-move-around-in-atomic-orbitals

The problem with the interpretations of quantum mechanics, and also the reason why there are so many of them, is that they are all equivalent when it comes to the observables. Bohmian trajectories, even if they exist, aren't and can't be seen. Your answer implies that the true trajectories of electrons in undisturbed atoms were actually observed. This is quite an overstatement.


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Dichromate in acidic medium
We were taught that $\ce{Cr_2O_7^{2-}}$ (dichromate) in $\ce{H_2SO_4}$ gives $\ce{CrO_3}$
$$\ce{Cr_2O_7^{2-} +H_2SO_4 -> CrO_3 + ...}$$
Link: wikipedia
But we know that dichromate in acidic medium converts to $\rm Cr(III)$.
$$\ce{ Cr_2O^{2−}_7 + 14 H3O+ + 6 e− → 2 Cr^{3+} + 7 H2O}$$
Link: wikipedia
How is this possible?

Your question answers itself! With $H_2 SO4$, we get $CrO_3$. With $H_3 O^+$, we get $Cr^{3+}$.

It's a matter of concentration

The first reaction is in anhydrous $\ce{H2SO4}$. The second reaction is in an aqueous acidic solution.

Both statements on the behaviour of $\ce{Cr2O7^{2-}}$ are true, but you are confusing two different reactions.
The formation of chromium trioxide from (sodium) dichromate in sulfuric acd is not a redox reaction. There is no partner that could be oxidized by $\ce{Cr2O7^{2-}}$. Concentrated sulfuric acid is an oxidant itself and metals like copper are dissolved (oxidized to $\ce{Cu^{2+}}$) while not hydrogen is generated.
Remember also that sulfuric acid is a strong, oxidizing and dehydrating acid. This effect is used in the production of $\ce{CrO3}$, which can be considered the anhydride of chromic acid ($\ce{H2CrO4}$). Note that chromium in $\ce{Cr2O7^{2-}}$, $\ce{H2CrO4}$, and $\ce{CrO3}$ always has the same oxidation state.
The second reaction that you mentioned does happen too, and this is indeed part of a redox reaction in which $\ce{Cr(VI)}$ is reduced to $\ce{Cr(III)}$. However, this can only happen when a suitable reaction partner is present, which in turn is oxidized by $\ce{Cr2O7^{2-}}$.

Sorry I was not taught about any details of the reaction, only the reactions were taught. Very nice explanation! Thank you!

@MaxPayne No worries, that's what we are here for :)


text:
K°P vs KP in equilibrium thermodynamics
I'm having a little trouble with definitions.
Here's a copy and paste from my hwk:
...K°P, and K°c have no units...Please enter the KP in 1/bar in  3 significant figures...
So these values even have different units, it's not a typo or anything.
Given: aA+bB $=>$ cC+dD
Is $$K°P=\frac {[C]^c[D]^d}{[A]^a[B]^b}$$
or is
$$KP=\frac {[C]^c[D]^d}{[A]^a[B]^b}$$
the question stated that K°P had no units, and the equation could be unitless if a+b=c+d, but that may not always be the case.
Can somebody tell me which "KP" matches the definition, and also provide, or link the formula for the other "KP"?
Sorry for asking such a trivial question but I'm having alot of trouble finding these definitions.
Thanks all.

The equilibrium constant should be dimensionless, it has to be to be used in equations such as the free energy for reaction at equilibrium
$\Delta G^0_R = -RT\ln(K_p)$.
If the reaction is
$\ce{\alpha A + \beta B \<=>\ \gamma C + \delta D} $
 if we write
$$K_p= \frac{P_A^{\alpha} P_B^{\beta}}{P_C^{\gamma}P_D^{\delta}} $$
where $P_a$ etc are partial pressures at equilibrium then $K_p$ is going to  have dimensions in general depending ion the values of $\alpha , \beta $ etc. as you state in your question.
In the past it was sort of implicitly assumed that each pressure was divided by 1 atm so then $K_p$ becomes dimensionless (or the units were simply ignored and numerical values used) but nowadays this is explicitly added so we obtain
$$K_p= \frac{\left(\frac{P_A}{P^0}\right)^{\alpha}  \left(\frac{P_B}{P^0}\right)^{\beta}}{\left(\frac{P_C}{P^0}\right)^{\gamma}\left(\frac{P_D}{P^0}\right)^{\delta}} $$
where $P^0 = 1 \pu{ atm}$  and so makes things dimensionless no matter what the values of $\alpha, \beta $ etc are but does not change any numerical value.
Although I have used partial pressures the same applies to concentrations $\ce{[A], [B]}$ etc where now each concentration is divided by  $\ce{[C^0]}$ where this is  $ \pu{1 mol/dm^3}$.
I'm not quite sure what your notation $K^0P$ etc. means but I suspect it refers to the two cases I have described. Hope this helps.