text:
Correct equation for Ionic Conductivity (λ) in Solutions?
We haven't started on Electrochemistry at school yet, but I did manage to find some time to read up on the topic. One thing I've noticed from when I started, is that different books and sites use different terms for the same symbols (For example: One book says Molar conductivity is represented by $\Lambda$, but another says it's Molar Conductance, and another book makes use of both Molar Conductance and Molar Conductivity, albeit inconsistently).
Now, my school textbook, as well as another book, Physical Chemistry by Wallwork and Grant employ the following equation for ionic conductivity:
$$\lambda=F\mu$$
Where $F$ is the Faraday constant and $\mu$ is the ionic mobility of the ion under consideration.
But I saw another equation on Wikipedia

$$\lambda=zF\mu$$
Where $z$ is the ionic charge of the ion being considered.

So my question is, which of these equations are right?

Short answer: $\lambda = z \mu F$ is correct. The link  provided by @Loong confirms this (page 73 of the document). To add to that I have provided you with a short derivation.
Prerequisite reading (so to speak) for definitions of a few terms/notation would be my answers to this question, and this question
Let me establish an important distinction (or definition) which can (possibly) be a source of confusion. The two relevant equations from my previous answer  to the first question are:

$$\Lambda _{m}=\Lambda _{m}^{0}-K{\sqrt  {c}}$$

and,

$$ \Lambda_{m}^{0}= \nu_{+}\lambda_{+}^{0}+\nu _{-}\lambda _{-}^{0}$$

Before I begin my discussion, I would like to note that "$\Lambda_m$" (uppercase ) is the molar conductivity of the solution, and "$\lambda$" is the molar conductivity of the ion (ionic conductivity). The second equation has a "special name", that I did not use in my previous answer-"Kohlrausch's law of independent migration of ions".
In general for dilute solutions of strong electrolytes, molar conductivity can be decomposed into contributions from the different ions in solution (this is the spirit behind the previously quoted equation):
$$ \Lambda_m^0 = \Sigma \nu_i \lambda^0_i$$
Now, let's attempt to relate the ionic conductivity to the ionic mobility (it clearly makes sense to do so). So I am assuming we have a dilute solution of completely disassociated strong electrolyte with molar concentration $c$. Each formula unit of this electrolyte gives rise to $\nu_+$ cations of charge $z_+e$ and $\nu_-$ anions of charge $z_-e$.
The concentration of each type of ion is therefore, $\nu c$ ($\nu$ could either mean $\nu_+$, or $\nu_-$ to avoid unnecessary notational complexity).
the number density of each type of ions is:
$$\text{No. density} = \nu c N_\mathrm{A}$$
Now picture a small window of area $A$ in your solution. The number of ions that can pass through said window in a time interval $\Delta t$ is equal to the number of ions within the distance $s \Delta t$ (where $s$ is the drift speed of the ions)
$s \Delta t A$, happens to define a volume element and the number of ions in this interval is given by: $ V \times \text{no. density}$, i.e
$$ s \Delta t A \nu c N_\mathrm{A}$$
Thus, the flux of ions is
$$J_\mathrm{(ions)} = \frac{s \Delta t A \nu c N_\mathrm{A}}{ \Delta t A} = s\nu c N_\mathrm{A}$$
remember flux always has dimensions of "something" per area per time
Flux of charge is $$J_\mathrm{(charge)} = ze s\nu c N_A = z s \nu c F $$
again, I am omitting the +/- signs for the sake of clarity. Also, note $F := e \times N_\mathrm{A}$
Now, drift speed is $s = \mu E$, so the flux of charge (henceforth, simply J) is
$$J = z \mu \nu c F E$$
The current is charge flux times area, so
$$I = (z \mu \nu c F E)A$$
Additionally, $E = \frac{\Delta \phi}{l}$ (i.e the potential gradient)
$$I = (z \mu \nu c FA) \frac{\Delta \phi}{l} \tag{I}$$
Using Ohm's Law (i.e $\Delta \phi = IR$}, one obtains
$$I = \frac{\Delta \phi}{R} = G \Delta \phi = \frac{\kappa A \Delta \phi}{l} \tag{II}$$
Comparing the (I) and (II) one is lead to $$ \kappa = z \mu \nu cF$$
This is the ionic conductivity. We need to divide by concentration of ions to the get the molar quantity. The concentration of ions however is not $c$, but rather $\nu c$ and our final result is
$$\lambda = z \mu F$$

I might be wrong, but I would suggest checking for the measurement units. Molar conductivity is measured in
$$C^2 s^{-1} m^{-2} N^{-1} \text{mol}^{-1}$$
so somewhere $F$ (and $z$) should be squared.
I think the possible inaccuracy with the previous answer is that the current (density) is the flux multiplied by the charge (per mole - i.e. $zF$).


text:
Is it possible that in a water soluble polamerization, by injecting some additional initiator, the monomer concentration increases?
I have a simple polymerization process with the following kinetic mechanism:
$$I_2 \stackrel{k_d}{\rightarrow}2\dot{I}$$
$$\dot{I}+M\stackrel{k_p}{\rightarrow}\dot{P_1}$$
$$\dot{P}_i + M\stackrel{k_p}{\rightarrow}~~\dot{P}_{i+1}$$
$$\dot{P}_i + \dot{P}_j\stackrel{\frac{1}{2}k_{tc}}{\rightarrow} D_{i+j}$$
This polymerization is taking place in a tubular reactor.
By default, the initiator is injected at the inlet of the reactor. Now, in addition to this injection, I have added another injection of the initiator at the middle of the reactor. I am expecting that the monomer concentration at the reactor outlet reduces, but this does not happen. Indeed, the monomer concentration at the reactor outlet increases.
I wanted to ask you guys, if such a behaviour is possible?

source: Stack Exchange

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/

sort: votes


text:
What is lithium pyroborate?
This component is mentioned in an article called "Novel geopolymer materials containing borate and phosphate structural units".
I've never heard about this before, and a google search didn't really give an answer it seems.

google for it. If you have a problem with guessing structure of a compounds, but have  a name, google usually helps. Sometimes you need enclose it in quotes.

Also try googling the words individually.  If you did that for pyroborate, you would find definitions that would solve the problem for you.

Lithium pyroborate ($\ce{Li2B4O7}$) is the salt of lithium ($\ce{Li}$) and pyroboric acid ($\ce{H2B4O7}$).

Nomenclature
The prefix ortho- designates an acid with the maximum number of hydroxyl ($\ce{OH}$) groups [reference]. For boron ($\ce{B}$), orthoboric acid is $\ce{H3BO3}$ (also written as $\ce{B(OH)3}$ to show that there are $3$ hydroxyl groups).
The prefix pyro- here seems to not match with the usual usage, so I will not be explaining the usual usage here. One common feature between this and the usual usage is that their molecular formulas are multiples of the ortho acid's molecular formulas, minus some water molecules:
$$\ce{4H3BO3 -> H2B4O7 + 5H2O}$$

Gallery
Some proposed structures of lithium pyroborate:

Some proposed structures of pyroboric acid:

The above are modified from two proposed structures of sodium pyroborate, with the first one referenced here and the second one referenced here.

Why the downvote???

Because, obviously, the structure is completely wrong?

@permeakra There are at least 4 structures of tetraborate that I found online.

@user34388 - The formula doesn't include hydrogen, so your proposed structures that do include hydrogen are incorrect. The first image you have is correct. See ChemSpider.

@ToddMinehardt My proposed structures that do include hydrogen are pyroboic acid structures.

@user34388 - I stand edified. Retracting down-vote.

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/


text:
Which electronic effects are responsible for the reduction of nucleophilicity of a hydroxy group orthogonal to an ester?
I am facing a similar synthetic problem as Boekman Jr. et al noted in their synthesis of Tetronolide.[1] I will quote their’s rather than disclosing my actual synthetic problem, but note that the key element — an α-hydroxy-γ-lactone is identical.
The authors’ intended convergent synthesis involves coupling the two compounds 1 and 2 shown in scheme 1 in a ‘tandem ketene-trapping/[4+2] cycloaddition’, which requires the free hydroxy group in 1 to nucleophilicly attack the ketene.

Scheme 1: Tandem reaction attempted by Boekman Jr. et al; would include a nucleophilic attack of 1’s hydroxide onto 2’s ketene motif.
However, as indicated this reaction did not produce 3. The authors note:

However, despite considerable experimentation, all attempts to couple lactone [1] with [2] to afford β-ketoester lactone [3] failed. We surmised that the failure stemmed from impaired nucleophilicity of the tertiary OH group owing to a stereoelectronic effect resulting from the enforced alignment of the lactone carbonyl π-orbitals and the adjacent CO bond. (Compound numbers adapted to fit this scheme.)

Instead, the authors used alcohol 4, which readily reacted with the masked ketene 2 to give the ester 5 as shown in scheme 2.

Scheme 2: Successful tandem reaction attempted by Boekman Jr. et al; included a nucleophilic attack of 4’s hydroxide onto 2’s ketene motif.
My experimental findings have confirmed a similar effect for my compound. However, I have a bit of trouble discerning which electronic effects exactly cause this unreactivity. My first assumption is that the conformation of the five-membered ring forces the hydroxy group into an angle of approximately $90^\circ$ with respect to the carbonyl-π system. This should allow mixing of σ and π orbitals of whichever kind.
My second assumption was tracing the problem back to the anomeric effect. In that effect, a nonbonding electron pair interacts with an adjacent σ* orbital, lowering its own energy but reducing the σ bond strength. The required electrons could be supplied by the π system, resulting in the effect being π and σ*. However, I do not quite see yet, how this kind of mixing would reduce the nucleophilicity of the oxygen atom.
Thus, the questions are:

Which electronic interactions exactly are responsible for this effect?
Has this been the topic of a research paper?

Reference:
R. K. Boekman Jr., P. Shao, S. T. Wrobleski, D. J. Boehmler, G. R. Heintzelman, A. J. Barbosa, J. Am. Chem. Soc. 2006, 128, 10572. DOI: 10.1021/ja0581346.

Do you have a tertiary alcohol as well? My first thought is that sterics are the dominant factor. What is the track record of alpha-hydroxy esters (hindered and unhindered) adding to ketenes in general? I'd want to know that before speculating on a stereoelectronic effect.

@jerepierre The paper goes on to state that they decided to couple a different alcohol with the ketene. The carboxyl was isopropyl-protected, the alcohol that would form the lactone vinyl-protected. In my case, the alcohol is secondary, but there are very bulky groups in beta- and gamma-position probably fixing the conformation of the gamma-lactone.

I am extremely dubious of the stereoelectronic effect. I can't recall ever seeing an example where an antibonding orbital aligns with a CO sigma, and that reduces nucleophilicity. I would be much more inclined to think that this is steric in nature. Have you modeled what the two substrates looks like (or your reactant)? You'd have to pretty convincingly rule out sterics for this kind of heavily substituted system before you want to make a claim of lone pair stabilization...

One more thought. When we open the lactone in compound 4, is the hydroxyl group even pseudo-axial? It might be pseudo-equatorial and less sterically encumbered.

I'd like to second the concerns of Zhe and jerepierre, an electronic explanation would certainly explain a slow reaction. If that reaction doesn't go at 140°C (certainly enough to disrupt a hydrogen bond) I think the steric explanation fits better.

If we focus on the 2-oxo-3s-hydroxy-tetrahydrofuran the electronic  effects present are the hyperconjugation and the dipole minimization. "Both effects contribute to the preferred (Z)-conformation of esters over the (E)-conformation. In the (Z) conformation, the lone pair of electrons in the alpha oxygen can donate into the neighboring σ* C-O orbital. In addition, the dipole is minimized in the (Z)-conformation and maximized in the (E)-conformation. But this electronic effects occur between the carbonyl carbon and the oxygen of the ring but the hydroxyl group." [a]

In contrast, I calculated the PES of hydroxy group in the following two systems:

2-oxo-3s-hydroxy-tetrahydrofuran [S1]
(1S,2S)-isopropyl 1-hydroxy-2-((vinyloxy)methyl)cyclohexanecarboxylate

The latter one is:

The oxygen of iprO-(C=O)- group was pointed away from [S2] and towards to [S3] cyclohexane. The PES is: (It was calculated with am1 due to computational resources, it was done with my laptop.)

It can be observed that the intramolecular hydrogen bond (IHB) in [S3] is more stable than in the other systems. The order of intramolecular hydrogen bond preference is [S3]>[S2]>[S1]. In other words, the IHB formed with the oxygen SP$^3$ of iprO-(C=O)- group is preferred over the carbonyl oxygen. [b] This can be seen on the rotational barrier between both minima. This conformational preference allows the lone pair of hydroxyl oxygen to be more nucleophilic than in the other systems. As it can be seen in the following figures:
[S1]

[S2]

[S3]

Related to the electronic effects present in this IHB, it can be complemented that "this is a consequence of interacting the lone pair on the accepting oxygen with an empty hydride antibond of the donating O-H atoms, $n_O$ → $\sigma^*OH$. Occupancy of the antibonding orbital raises the energy of the O-H species while stabilizing the oxygen." As Albrecht et. al. concluded in her paper.[c]
Bibliography

[a] Cuevas, Eusebio Juaristi, Gabriel (1995). The anomeric effect. Boca Raton: CRC Press. ISBN 0-8493-8941-0.
[b] Rincon et al. J Solution Chem (2011) 40: 656.
[c] Albrecht et al. J. Phys. Chem. A. (2012) 116: 3946

By the way; how did you define the angle in the graph? I.e. what corresponds to $0^\circ$, what corresponds to $90^\circ$?

Additionally, scanning through Albrecht’s paper it appears your quotation is not a literal one. You should use quotation markes only for fully literal quotes.


text:
What are bonds in molten NaCl?
I know there are ionic bonds in solid NaCl. But what bonds are present in liquid (molten) NaCl? I believe there should be some kind of attraction there, because particles (ions) are pretty close to each other. What are these attractions?

Those are ionic bonds.

You mean, ionic bonds do not get broken when NaCl gets molten?

I did not say that. Rather, they break and form readily, which is why they can move past each other, which is why it is called a liquid.

Energy should be released at the moment of formation a new ionic bond, so in this way this process should be exothermic. As we heat NaCl at that time (in order to make it molten), so we supply energy, and temperature is high, it is difficult to imagine successful exothermic process under this conditions.

View it as a dynamic equilibrium, where the bond-forming exothermic processes and the bond-breaking endothermic processes balance out each other.

You could ask same thing about water; there's h-bonding in ice and liquid, what else would there be?

So what is the difference between inter-molecular bonding in ice and water? Apparently they look different: solid vs. liquid.


text:
Borax bead test technique
In the observation tables provided in books and websites two different colors are assigned to a specific borax+salt bead in both oxidising and reducing flames. Not all basic radicals show this trend but some like Cu$^+$$^/$$^+$$^2$ does. It appears green in hot condition and blue in cold condition when made using oxidising flame.

My question is what is done to actually reduce the temperature of the borax bead? Is it held under running tap of water or just outside the flame?

"In hot" and "In cold"  are really weird column headings to me. It would seem better to use "While Hot" for column and move that column to the left. Use "Cold" for the second column. You of course have to fuse the bead before you can see the "Cold" color.

The table is from ncert. By 'in cold' and 'in hot' they must have meant in hot condition and cold condition respectively.

I know what Borax bead tests are so I understand what the author meant. I just think the notion was phrased poorly. Also since we read left to right there is the nuance of time. I think having the hot column first then the cold column would imply time sequence better.

@tytoalba where did you find this table? I couldn't find it in ncert class 11

"Hot" refers to the bead when just taken out of the flame of the Bunsen burner, while "cold" refers to the state when the bead cooled to room temperature. Provided the small size these beeds typically have (a toothpick like rod of magnesium oxide to carry them), this is just a matter of about a minute of exposure to ambient air. Duing this process, you may spot differences, too.  Conversely, intentionally cooling the hot glass-like beads under tap water often shatters them into pieces.
Side note #1:  Testing with a bead of Borax is a one quick pretest, to be confirmed by subsequent experiments.  And as a good habit: once you found the ions, compare your bead made with the analyte with a reference, a bead made with a sample of known composition.
Side note #2: Of course I miss one, an entry for cobalt ... regardless of the flame (oxidation or reduction), hot or cold: an intense dark blue. Almost Yale blue.


text:
Can we oxidize oxygen?
I have encountered this equation:
$$\ce{SO4^2- -> S2O8^2-}$$
In this reaction 2 of the oxygen ions change from $-2$ to $-1$ so they are getting oxidized.
It is because two oxygen ions are connected together.
A question rises: why can't we oxidize oxygen ions in $\ce{NO3-}$ or any other?

$\ce{NO3-}$ can become $\ce{NO2}$, which according to this post, has a canonical form containing a peroxide bridge causing the oxygen to have oxidation states of $-1$.

Oxygen can be oxidized. However, it is uncommon due to its high electronegativity. In fact, it is highest second only to fluorine. This makes it extremely difficult to remove electrons from oxygen, thus making it hard to oxidize.
One definitive example of oxygen being oxidized is oxygen difluoride. This exception is even listed on many lists of oxidation state rules (see: oxidation states).
And of course, if you go beyond stable molecules, it is possible to ionize oxygen with an electrical charge.

What I basically meant is oxygen in ions, In NO3 nitrogen cant oxidizes (reached its highest oxidation state) but Oxygen is not.
So from your answer, We can force oxygen to be oxidized, right?

Of course we can. Just electrolyze nearly any salt with oxygen-containing anion, and that would be it.

@Biker I don't know if I understand your follow up. This might be interesting to you, though: But see: https://en.wikipedia.org/wiki/Fluorine_nitrate

It is just because our exams gives us a multiple choice:
Which of the following cant be oxidized?
and the answer is NO3- because N is +5
They seem to exclude O-2 for a reason but it can be oxidized under certain cases like Fluorine or electrolysis. We can exclude electrolysis because it requires energy and maybe they are referencing that it must happen spontaneous so we are left with fluorine. That is why I asked if we can oxidize oxygen ion or no.

@Biker Based on that, it sounds like they're asking about the molecule as a whole, not the atoms in it. Although it is a bit of a suspicious question if that's the exact wording, because it does sound confusing. It might just be a case of poor wording; you could ask the teacher for an explanation of what was intended.

If they meant molecule would that make difference?
When I asked the teacher, he simply replied that nitrogen has +5 oxidation state which means it cant get oxidized more. I asked him what about oxygen in the molecule. He said we ignore it.

Oxygen is also oxidized by platinum hexafluoride, where the platinum (in a high oxidation state) rather than the fluorine is the oxidizing agent, see https://en.wikipedia.org/wiki/Platinum_hexafluoride. The reaction is represented as
$\ce{O2 +PtF6 -> O2+PtF6-}.$
This reaction heralded the possibility of similarly oxidizing noble gases, which busted the "inert gas" myth (at least for the heavier noble gases).

In the equation $\ce{2SO4^2- -> S2O8^2- + 2e-}$ sulfate is being oxidized.
This is different from oxygen being oxidized.
The statement in the OP "In this reaction 2 of the oxygen ions change from $-2$ to $-1$ " is incorrect. Instead, each sulfate ion, taken as a whole, loses 1 electron.  There is no reason to ascribe the loss to oxygen, rather than sulfur.
The analogy for nitrate would be $\ce{2NO3- -> N2O6 + 2e-}$ , which is discussed on page 86 here

Umm because Sulfur can't be oxidized? It has reached its highest oxidation state and if the ion loses electrons then it must be the oxygen not the sulfur that was my reasoning.

@Biker in some artificial book-keeping system we can pretend that in sulfate S is +6 and each O is -2, but what is it in the real world? "In
$\ce{SO4 ^{2-}}$ there is 0.5 of negative charge on each oxygen
and no charge on sulfur." http://scitation.aip.org/content/aip/journal/jcp/45/6/10.1063/1.1727864

@Biker and "the peroxide O-O bond in $\ce{S2O8 ^{-2}}$ is nonpolar with a negative Mulliken charges of 0.65 on both peroxide oxygen
atoms, as determined by our AM1/COSMO computations." http://pubs.acs.org/doi/abs/10.1021/jp2041845 So O going from -2 to -1 is just imagination, just as the IUPAC definition of oxidation state is "the charge an atom might be imagined to have" http://goldbook.iupac.org/O04365.html

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


text:
Volume expansion from phosphate buffer solution
I want to mix a phosphate buffer at $\mathrm{pH\ 7.2}$ at $20~\mathrm{^\circ C}$. I do this by dissolving $0.0036\ \mathrm{mol}$ of sodium dihydrogen phosphate monohydrate ($\ce{NaH2PO4*H2O}$) and $0.0063\ \mathrm{mol}$ of di-sodium hydrogen phosphate dihydrate ($\ce{Na2HPO4*2H2O}$) in water per liter of buffer. I add $0.15~\mathrm{M}\ \ce{NaCl}$. How much water do I add if I want $1~\mathrm{L}$ of buffer? Do I just use $1~\mathrm{L}$ of water — will the salts not expand the volume? Is it insignificant and at what point do I need take into account the volume expansion of the solution?
Notes:

$$M(\ce{NaH2PO4 . H2O}) = 137.99~\mathrm{g/mol} \Longrightarrow 137.99~\mathrm{g/mol} \times 0.0036~\mathrm{mol} = 0.496764~\mathrm{g}$$
$$M(\ce{Na2HPO4 . 2 H2O}) = 177.99~\mathrm{g/mol} \Longrightarrow 177.99~\mathrm{g/mol} \times 0.0063~\mathrm{mol} = 1.121337~\mathrm{g}$$
$$M(\ce{NaCl}) = 58.44\ \mathrm{g/mol} \Longrightarrow 58.44\ \mathrm{g/mol} \times 0.15~\mathrm{mol/L} = 8.766~\mathrm{g/L}$$

Welcome to chemistry.stackexchange.com. Feel free to take a [tour] of the site. Visit the [help] for unanswered questions about how it works.

There is an exact way of doing this and there is a practical way of doing this.
The exact way involves a volumetric $1~\mathrm{l}$ flask. You would first fill the salts into the volumetric flash, add some water to dissolve them and then carefully add more water until the flask contains exactly $1~\mathrm{l}$ of solution. That removes the error of the salts increasing the overall volume.
However, it is also part of the exact way to add a pH-monitoring device before you have finised to make sure the pH is exactly at $7.2$ — and to carefully add phosphoric acid or sodium hydrogen phosphate if the pH is off. (A better solution would be to make sure the pH is over the desired value so phosphoric acid can simply be added; as it is a liquid it can be added and dissolved easier.) For most applications outside of thermodynamics, this is an overkill.
The practical way involves one of the following:

using an Erlenmeyer flask, beaker or similar with a $1~\mathrm{l}$ mark; filling the salts first and then the solution up to the mark
simply measuring $1~\mathrm{l}$ of water and adding that.

Why are the practical solutions okay? In most cases, it does not matter whether the molar concentration of, say $\ce{NaCl}$ is exactly $0.15~\mathrm{M}$ or whether it is $0.14~\mathrm{M}$. This rather unimportant experimental error is simply accepted. It also does not really matter whether the pH is exactly $7.2$ or whether it is, say $7.5$. Therefore, one often decides to go the not-so-exact but easier and faster way.

Thanks for the help! Do I add NaCl after i mixed the salts and dH2O? Do you think a magnetic stirrer is necessary?

@Lennart No, add NaCl with the phosphates. A stirrer is only permitted if you are using the practical way and actually add a physical litre of water, otherwise you are introducing additional error ;)

Alright. I think I'll go with the practical way :) Regarding the pH.. is it still OK to be not-so-exact when using it in HPLC as an eluent? I'm using SEC- size exclution chromatography

@Lennart For that it is definitely okay ;) It’s how we make our HPLC-buffers, too ;)

Ah, ended up with 6.98 pH wanted about 7.2 pH hope it doesnt affect too much..


text:
Energy comparison between an atom and its ion
Is it possible to compare energy(enthalpy or similar) between an atom and its ion, possibly in the same state(gas)?
For example, $\ce{O}$ and $\ce{O^+}$(intentionally cation), $\ce{Na}$ and $\ce{Na^+}$.
At first I though of ionization energy, but its RHS(products) has electron, so I thought it can't be used to directly compare them.

The electron is assumed to have $0$ energy...

@user34388 How about electron's kinetic energy?

Assume it's zero.

Give us some context please. There are some good answers below, but I think you are aiming for something else. As a general rule, don't compare total energies of different compounds directly. I don't think it's really clear that you want the ionization energy but rather something more dubious

Ionization energy is, exactly, a comparison of the energy of (i) a given species and (ii) that same species with one electron removed.
Ionization energies can be thought of as a reaction like the following, using the oxygen atom as an example:
$$
\ce{O -> O+} + e^-
$$
When evaluating quantities like the ionization energy, "external" energies (this is my term, not an official/formal one!) like kinetic and gravitational potential energy are disregarded. Further, the reference energy is generally defined to be zero for an isolated electron at "infinite separation".
Thus, the energy of the above reaction is:
$$
E_\mathrm{rxn} = \left(E_{\ce{O+}} + E_{e^-}\right) - E_{\ce O}
$$
But, since the electron is taken as being in its reference state, this is simply:
$$
E_\mathrm{rxn} = E_{\ce{O+}} - E_{\ce O} \equiv IE
$$
PLEASE NOTE that the above energies are only the internal/electronic energies of the $\ce O$ and $\ce{O+}$ species, and not their enthalpies $\left(H_{\ce{O+}},\,H_\ce{O}\right)$, free energies $\left(G_{\ce{O+}},\,G_\ce{O}\right)$, etc.

If you want to compare the energy between an atom and its parent ion you basically are indeed interested in the adiabatic ionization energy of your atom (i.e., the difference in energy between the atom and ion in their lowest energy state). This is the energy needed to remove the electron away from the ion core.
It might be instructive consider the spectrum of the hydrogen atom. You might know that the energy levels of the H atom w.r.t. its ground state are given by
$$E_n=E_\text{IP}-\frac{\mathcal{R}}{n^2}$$
where $E_n$ is the energy of the state with principal quantum number $n$, $E_\text{IP}$ is the adiabatic ionization energy, $\mathcal{R}$ is the Rydberg constant. In case of hydrogen, $E_\text{IP}=\mathcal{R}$. The hydrogen atom thus has an infinite amount of energy levels that become more closely spaced as $n$ increases and converges on the ionization energy. This corresponds to an electron that is so far away from the proton, that the system can be considered to consist of an isolated electron and ion. For Na and O, the formula for the energies is not so simple anymore, but you'll still find that the energy levels converge on the ionization energy of the atom.
Another way to look at it is by considering an low-energy collision between an ion and an electron.


text:
Is it more economical to liquify natural gas to ship overseas or convert it to propane (C3H8) and ship it overseas?
Problem statement: Companies want to frack oil wells and then export natural gas overseas using transport via high pressure pipeline followed by liquefaction at a terminal port.
If the methane component in natural gas ($\ce{CH4}$) can be converted to propane $\ce{C3H8}$, that would leave $\ce{H2}$ which can be combined with nitrogen to create ammonia-based fertilizer.
If the process is done at the well, propane could also be reverse-fed to wells in liquid form to use in place of water for fracking, after which the propane could be drawn off by lowering the pressure (such that it is then in the gas phase) and recompressed to liquid at $160 \ \mathrm{psi}$. This reduce water contamination and propane would not be lost.
In the liquid phase at $160 \ \mathrm{psi}$, propane has a volume $\frac{1}{250}$ as it does in the gas phase, and consequently propane would require a pipeline much smaller than natural gas for same volume sent.
On the receiving end, propane has a higher value for its ease of transport and storage as well as larger consumer base.

One might suspect that the oil and gas companies have calculated out the economics pretty well, including all the handling and converting of LNG to LPG. (And the required costs of capital investments - a point often missed).

I'm voting to close this question as off-topic because there's no question there, just a bunch of statements.

Gas to liquid technology is known, but propane isn't a good choice of end product
The idea of converting natural gas (which is mostly methane) into liquids is widely used not least because the earth has, inconveniently, located the best natural gas fields some way from the places that most want to use natural gas.
Right now (in 2016), though, it seems it is still not desperately uneconomical to convert the methane into a liquid by refrigeration and ship it to customers in the largest known thermos flasks (LNG tankers). Both the ships for transporting liquefied natural gas (LNG) and the plants to create it are very expensive (tens of billions of dollars for the minimal economic scale). There are something like 410 of these ships and there seem to be dozens of new units being built every year (see this report on the industry for more detail: warning pdf).
But there are some examples of plants that convert the local gas supply directly to more transposable liquids. Shell, for example, have developed a process that converts gas directly into clean liquid fuels (the initial plants produce very clean diesel, but other fuels are possible). See this description of the Shell process. Some other firms are operating similar processes.
So it seems that, right now, LNG is the best economical option. But if you are going to use a technology to go directly to something more transportable than methane, it is better to go the whole hog to a widely used fuel (diesel and petrol are both basically longer chain hydrocarbons than propane).
This may, of course, change.

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


text:
How do I dissolve sodium oleate in distilled water
I wanted to know how to dissolve sodium oleate powder in distilled water?  Does it have to be at a certain temp?  I need a 2% solution

A product information by Sigma-Aldrich, a chemical supplier, cites Merck Index regarding the solubility of sodium oleate.  And states "It is
also soluble in water (approximately 100 mg/ml)", in other words, 10 mass%.  Since this value is larger than the 2 mass% requested by you, and provided sodium oleate is a surfactant anyway, just go ahead.  A gentle warming, like 30 to 40 Celsius, may accelerate the process, though.


text:
Mustard gas hydrolysis (at a very high rate)

Explain mustard gas hydrolysis at a very high rate:
$\ce{Cl-CH2-CH2-S-CH2-CH2-Cl}$

I tried searching online but did not find this reaction anywhere. Can someone explain me the steps/mechanism of this reaction? Even some hints might help.
My attempt:
According to Wikipedia

Solvolysis is a special type of nucleophilic substitution ($\mathrm{S_N1}$) or
  elimination where the nucleophile is a solvent molecule. For certain
  nucleophiles, there are specific terms for the type of solvolysis
  reaction. For water, the term is hydrolysis; for alcohols, it is
  alcoholysis; for ammonia, it is ammonolysis; for glycols, it is
  glycolysis; for amines, it is aminolysis.

It clearly states that hydrolysis/solvolysis can only be $\mathrm{S_N1}$ or elimination. However I'm wondering how can $\mathrm{S_N1}$ occur on a primary carbon! Morever water doesnt seem to be a strong enough base for elimination. So what will be the preferred reaction mechanism?

The cation is somehow stabilized via S. Without S, it would be just ethyl chloride which hydrolyzes pretty slowly, as expected.

You mean SN1 occurs?What will be the end product?Can you link the mechanism?

Luckily I got the mechanism via Google Image Search.

Internal attack by sulphur atom takes place during the reaction. This is called anchimeric assistance, neighbouring group effect or $\mathrm{S_N{NGP}}$. Wikipedia describes it well.

Yes, an $S_N1$ reaction takes place. As, @Ivan pointed out, if it weren't for the it would just be like ethyl chloride. The key to this whole process is the presence of lone pair of electrons on sulfur which it uses to stabilise the primary carbocation (by forming the triangular cation depicted).

@getafix I would rather say it an extended version of $\ce{S_N1}$ i.e. $\ce{SN_{NGP}}$.It's not strictly $\ce{S_N1}$.

Of course I agree. There is clear neighbouring group participation. Sorry, my comment did not make it explicit. However, I never knew $SN_{NGP}$ was a term..learn something new everyday

@getafix Never mind.We're all here to learn :-)!

Indeed. What I meant is that the rate determine step is unimolecular, i.e $S_N1$


text:
Comparing rates of hydrolysis in cyclohexane ring (involving neighbouring group participation)

I need to compare the rates of hydrolysis of these three compounds.

MY ATTEMPT:

By definition hydrolysis is either $\ce{S_N1}$ or elimination.But water being a weak base (and weak nucleophile) $\ce{S_N1}$ is preferable.
$\ce{PhS^-}$ is a better leaving group than $\ce{Cl-}$ due to mesomeric effects.
Comparing with the middle compound the leftmost and rightmost compound will certainly have higher rate of hydrolysis due to presence of good leaving group $\ce{PhS^-}$
But how to compare between the leftmost and rightmost compound's rates?

You asked this before. http://chemistry.stackexchange.com/questions/58888/is-the-sn1-reaction-faster-with-an-axial-or-equatorial-leaving-group

In fact, no, this is slightly more complicated. I would expect there to be some degree of neighbouring group participation by the SPh (forming a sulfonium ion, followed by water attacking) This can only happen in the trans configuration, however it requires ring flip into the higher-energy trans-diaxial conformation. Complicated.

@orthocresol We don't need to worry about the ring flip I guess as it will take place automatically.

Ring flip definitely happens. The only issue is that the trans-diaxial conformation will be less populated, which will affect the rate negatively. I can't determine just from this alone which isomer is going to react faster. In the absence of NGP the cis isomer is faster, but NGP speeds up the trans isomer reaction. Is it enough to make it faster than the cis? I would guess yes, but I don't know.

@orthocresol I know it might be less populated.However internal attack mechanism rate supersedes all the external attack mechanisms.So $rate_3$ must be highest by any logic...

"When NGP is in operation it is normal for the reaction rate to be increased." -Wiki https://en.wikipedia.org/wiki/Neighbouring_group_participation

@orthocresol If you want more proof here you go http://meskvmcollege.org/upload/study_material/30275.pdf . See page 10 of the pdf. Reaction rate in trans is 670 times greater than cis (in that case).

The third reaction infact occurs via neighbouring group
  participation.

(https://en.wikipedia.org/wiki/Neighbouring_group_participation)

In the third compound $\ce{S}$ atom attacks carbon containing $\ce{Cl}$ by $\ce{SN_{NGP}}$ forming sulphonium ion. An attack by $\ce{H2O}$ on the least hindered carbon of the three member ring occurs. In this case both the carbopns in three member ring will be similar though. So the third compound has highest rate of reaction.

(I'd like to add "This can only happen in the trans configuration, however it requires ring flip into the higher-energy trans-diaxial conformation." as pointed out by - @orthocresol in the comments to the orginal question.)

Adding as a proof that trans reacts faster than cis form during NGP in cyclohexane ring.
  http://meskvmcollege.org/upload/study_material/30275.pdf (See page 10
  of the pdf)

The first compound has second highest rate. Anchimeric assistance cannot occur in cis form. Negative inductive effect by $\ce{Cl}$ enhances the rate. Reaction mechanism is $\ce{S_N1}$ and water itself acts as nucleophile.
The central compound has the lowest rate among the three. $\ce{S_N1}$ mechanism occurs and $\ce{Cl}$ is replaced by $\ce{OH}$ where water itself acts as nucleophile.

So summing up: $$\ce{rate_3>rate_1>rate_2}$$


text:
How to calculate pH of solution made of H3PO4 and Na3PO4?
I am unable to calculate the $\mathrm{pH}$ of a solution made of $\pu{0.1 M}$ $\ce{H3PO4}$ and $\pu{0.05 M}$ $\ce{Na3PO4}$.
How to set the approximation and which ions are present at the equilibrium?

Hint here.

Hint 2 - Phosphoric acid can lose three protons. Write each of the reactions down with pKa's. For such polyprotic acids (in general...) only two species will be important. so it's either (1) H3PO4 & H2PO4- (2) H2PO4- & HPO4-2 or (3) HPO4-2 & PO4-3  see http://ion.chem.usu.edu/~sbialkow/Classes/3600/Overheads/H3A/Image133.gif

@Andreade Welcome to ChemistrySE. This appears to be a homework question, and you must demonstrate some effort towards solving the problem. You have already received 2 hints, that should get you started. Feel free to take the tour and get familiar with the site.

Thank you for the hints, I know it is a polyprothic acid but I can't manage to understand which ions leave in solution. I think due to the acid(h3po4) base(na3po4) reaction, only H2PO4- and HPO4-- leave and make a buffer solution, but I'm not sure about that.

See: https://chemistry.stackexchange.com/q/8254/95526

You have the the following six independent equations in order to specify the unknown concentrations of six species as present in an aqueous solution consisting of $\pu{0.1 M}$ $\ce{H3PO4}$ and $\pu{0.05M}$ $\ce{Na3PO4}$:
\begin{align}
K_1 &= \frac{[\ce{H+}][\ce{H2PO4-}]}{[\ce{H3PO4}]}
     = 7.5 \times 10^{−3} \tag{1}\\
K_2 &= \frac{[\ce{H+}][\ce{HPO4^2-}]}{[\ce{H2PO4-}]}
     = 6.2 \times 10^{−8} \tag{2}\\
K_3 &= \frac{[\ce{H+}][\ce{PO4^3-}]}{[\ce{HPO4^2-}]}
     = 3.6 \times 10^{−13} \tag{3}\\
K_\mathrm{w}
    &= [\ce{H+}][\ce{OH-}]
     = 1 \times 10^{−14} \tag{4}\\
0.15
    &=[\ce{PO4^3-}] + [\ce{HPO4^2-}] + [\ce{HPO4-}] + [\ce{H3PO4}]\tag{5}\\
[\ce{H+}]
    &= 3[\ce{PO4^3-}] + 2[\ce{HPO4^2-}] + [\ce{HPO4-}] + [\ce{OH-}] -0.15\tag{6}
\end{align}


text:
How does the wax layer of a leaf get damaged by acid rain?
I know that wax is not soluble in water so I think it doesn't react with or dissolve in water-based solvent. Some sources I have consulted say that it does react but slowly. So getting confused by different opinion, I have no idea what the truth is.

Being insoluble in water does not mean something cannot react with water or water-based solvents.
Plant waxes are made up of a number of constituents; amoung them esters, alcohols, alkenes, fatty acids and carbonyls. Especially the esters can be hydrolysed by acidic conditions, creating a more hydrophilic area which water (and more acid) can use to penetrate the layer.
Another probable reason is the fact that acidic proton transfers often involve the release of heat which will make the wax layer less viscous. All these things taken together mean that slowly but surely, the wax layer will be destroyed given enough acid rain.


text:
Ions and Ionic Bonds
I've recently just started learning about Chemistry, and more specifically, Ions.
In class, we learned about how Valence Electrons are electrons located in the outer-most orbitals and how every atom (with the exception of Hydrogen and Helium), want $8$ valence electrons.
When possible, an atom will attempt to get rid, or add, an electron to get $8$ on its outermost shells. For example, Sodium and Chlorine.
We learned that it's easier for Sodium ($\ce{Na}$) to lose an electron to get the $8$ outer electrons it always wanted, and for Chlorine to receive that $1$ refuted electron. Then, they form an Ionic bond and you get Table salt!
But now, I'm wondering: what if that's not possible? What if you have an element such as Beryllium and Lithium? Would they simply not combine? Or would something else happen? I am confused.

I'm not sure what you're really asking. But in general there are three types of bonds - ionic, covalent and metallic. Only the noble gases are relatively inert. // The octet in the Lewis dot model is a very limited way to show bonding, but it does work for a lot of molecules/compounds.

I'm just wondering if its possible to have Beryllium and Lithium to have an Ionic bond. I haven't learned about Covalent and Metallic though...

Yes, both form ionic bonds. For example Beryllium sulfate https://en.wikipedia.org/wiki/Beryllium_sulfate and Lithium chloride https://en.wikipedia.org/wiki/Lithium_chloride

Are you talking about an ion bond between Be and Li? or between Be/Li and something else? I'm not sure what you're really asking

@getafix Between Be and Li.

The variation from ionic to covalent to metallic bonding is not clear-cut, but a matter of degree. For example, the bond between hydrogen and oxygen in water is mostly covalent (sharing electrons), but at room temperature, about one in 10,000,000 bonds break, with the hydrogen losing an electron to oxygen -- the ionic bond you discussed.
However, in the case lithium and beryllium, there is not enough difference between them to favor one getting an electron instead of the other. However, something more interesting happens: both are willing to let electrons go, so there is a sea of electrons, unattached to a particular atom! This is called a metallic bond, and it enables many metals to mingle to form alloys, such as bronze, amalgams and stainless steel.

I did not know that!


text:
Can an elimination reaction be visualised as the combination of the loss of a leaving group and proton transfer?
Specifically in the conversion of a haloalkane to an alkene, i believe this is the reaction mechanism.
Extrapolating on this, do all elimination reactions function as such, or is the term elimination reaction a reference to the ends rather than the means?

Elimination refers to the ends rather than the means. On top of that, it does not necessarily refer to the loss of a proton + a leaving group; it can in general refer to the loss of any small molecule. So, for example, a loss of CO2 is also considered an elimination, even though there is no proton there.
As far as proton + leaving group eliminations go (elimination of HX), the order in which the leaving group leaves and the proton falls off determines what type of elimination it is.
The leaving group can leave first - that is called an E1 elimination.
The proton can fall off first - that is called an E1cb elimination.
And if both of them happen at the same time, it is an E2 elimination.


text:
Calculate the amount of substance of an isotope based on the percent composition

Calculate the amount of substance of $\ce{^37Cl}$ in a $\pu{42.3 g}$ sample of $\ce{Cl}$ containing $24.00\%~\ce{^37Cl}$ and $76.00\%~ \ce{^35Cl}.$
  Given the atomic masses $m(\ce{Cl}) =\pu{35.45 amu}$, $m(\ce{^35Cl}) =\pu{34.97 amu}$, and $m(\ce{^37Cl}) =\pu{36.97 amu}$.

I know you would multiply $\pu{42.3 g}$ by $0.240$ to get $\pu{10.2 g}$ of $\ce{^37Cl}$, but my teacher says you then divide by the atomic mass of $\ce{Cl}$ rather than $\ce{^37Cl}$. That is the part that confuses me, she says it is because

You don't know what those $\pu{10.2 g}$ are made of, so you need to use the atomic mass of $\ce{Cl}$.

Your teacher  is wrong. You know precisely that 24% are $\ce{^37 Cl}$, as this is given in the question. Also not that amu is a deprecated unit and it should be replaced by u, the unified atomic mass unit.

If you have given an accurate account then your teacher is mistaken. We've all gotten confused while trying to talk and think at the same time.
\begin{align}
\pu{42.3 g} \ce{Cl}\cdot 0.24\frac{m(\ce{^37Cl})}{m(\ce{Cl})}
&= \pu{10.2 g} \ce{^37Cl} \\
\implies
\frac{\pu{10.2 g}~ \ce{^37Cl}}{\pu{36.97 g//mol}\ce{^37Cl}} &= \pu{0.276 mol}
\end{align}


text:
What is the right way to express pH and concentrations?
Which of the following expressions are used in English to express the pH of solutions and the concentration of solutions?  I think they are all clear to someone with knowledge of chemistry, but I want to avoid awkwardness in my English.

A. I prepared a solution such as below:

a solution of pH 3
a solution with pH 3
a solution with a pH of 3
a solution of a pH of 3

And

B. I prepared a solution such as below:

a 3 nM buffer solution
a buffer solution of 3 nM
a buffer solution of 3 nM concentration
a buffer solution with a concentration of 3 nM
a buffer solution of a concentration of 3 nM

I'm not a native speaker and confused. Especially, can I use A-1, B-2 or 3 expressions?

FWIW, the only ones that sound wrong to me are A4 and B5. Of course, I'm not a chemist.

Any reader will understand A4, but it is very awkward, Do not use A4. I prefer A1. A2 is OK. A3 has two unnecessary words ("a" and "of"), but is OK. As for the Bs -- do you mean nanomoles per liter, or is "per liter" understood? : I prefer B1. B2 is OK. B3, B4, and B5 all have the word "concentration" which is unnecessary if "per liter" is understood.   Of these three, B5 is like A4 -- very awkward: don't use it. B3 is better than B4 -- it has fewer unnecessary words

@ab2: the units nM are for molarity, not moles. See Wikipedia. The unit mole is abbreviated mol. Again, see Wikipedia. So the "per liter" isn't just understood, it's included. And just because "concentration" is unnecessary, it doesn't mean that it's ungrammatical to include it.

I don’t see why this is not a valid question of usage so as a professional biochemist I am going to answer the question as a comment. If the question is released from hold I shall answer it as an answer. A-1 should be used. (A-2 is unusual, A-3 is ok in speech but too verbose for a paper, A-4 with two ofs not English.) B-1 is best. There is no need to expand it as in B-2, B-3, or B-4. B-5 has the ugly double ofs again. If you don't like B-1 you can name the buffer in a phrase such as "3 nM Tris Buffer".

@David : Thank you so much!  If the concentration part is complex, I sometimes need to try B2, but would it be OK? "a solution of 0.325 to 3.265 nM"  Also, I think this questions is not a chemistry question but an English question which can be generally used in Middle/High school English.

I would not use A1, but rather A2': ... a solution (of a chemical) with pH = 3. B1 is okay. B4 is also okay, but better would be using: with a concentration of c = 3 nmol/L.

This is more a question of English usage than Chemistry, but the English Language and Usage people regarded it as “proof reading” and migrated it here. So I will move my original comment here to make it an answer (until the Chemists throw the question out).
First, you really should try to answer this yourself by looking at the style used in scientific journals relevant to your work in papers from English-speaking countries with authors whose surnames suggest they may be native speakers. Several of your alternatives are grammatically correct but unneccessarily long-winded and never used.
Having said that, my opinion (as a native English-speaking biochemist with many publications) is:
A: A-1 (“a solution of pH 3”) should be used.
The use of ‘with’ alone in A-2 is unusual, A-3 is ok in speech but too verbose for a paper, A-4 with two ‘of’s is clumsy, ugly and unnecesary.
B: B-1 (“a 3 nM buffer solution”) seems best to me — brief and correct.
There is no need to expand it as in B-2, B-3 (both of which sound clumsy and I advise you not to use), or B-4. B-5 has the ugly double ‘of’ again.
If for some reason you don’t like B-1 you can name the buffer in a phrase such as “3 nM Tris-HCl Buffer”.

I find A1 at least inaccurate. A solution is not made out of pH 3, it has a property, which is pH = 3.

@Martin-マーチン You'll just have to live with it because that's the scientific English usage. Believe me, I'm a native speaker and have published dozens of scientific papers. Admittedly one would normally write something like "Tris-HCl buffer, pH 7.5", but A1 is the most natural of the alternatives. In any language it is pointless to try to rationalize pronoun use.

I admit, I am not a native English speaker. I am also not a chemist who deals with aqueous solutions a lot. It just sounds incredibly odd and inaccurate to me. I agree that I should have disagreed to migrating it here.


text:
Why does branching leads to lower surface area?
I know that lower surface area means lower interactive forces between molecules and hence lower boiling point. But the statement that branching leads to more compact molecules with lower surface area, I just cannot understand it. Can someone help me with a visual or any otehr kind of explanation?

If you have a lump of, I don't know, plasticine, and you want to minimise its surface area, what do you do? You roll it into a sphere, right? Now which looks more like a sphere: neopentane, or n-pentane?

@orthocresol Cannot I assume it as a square like surface area, where as n pentane is linear. Why does it roll into a sphere?

@ReeshabhRanjan Square is a 2D structure.The compact one is more spherical.Branching causes a molecule to become compact.

@ZOZ  I mean why they tend to get more compact? Why not more open and having more surface area as possible? Obviously square is 2-D structure, by it I mean being more open anyways...

In math the isoperimetric problem is to determine a plane figure of the largest possible area whose boundary has a given length. That is a circle. https://en.wikipedia.org/wiki/Isoperimetric_inequality Likewise the 3D figure that has the greatest volume for a given surface area is the sphere. Look at models for neopentane https://en.wikipedia.org/wiki/Neopentane and n-pentane https://en.wikipedia.org/wiki/Pentane

Say take a string.First extend it completely.Next,cut it into 4 parts and arrange it to make a neopentane like structure.Which is more compact and spherical?

@ZOZ, duh... I completely agree sphere is more compact. I say why a molecule makes itself more compact.

@MaxW, I don't get what you are proving. if surface area is fixed, then the story will end there only? Surface area is decreasing as molecule is becoming more compact.

@ReeshabhRanjan Branching makes a molecule compact.Therby a lower surface area.

@ZOZ sorry for making this so long a discussion, but do molecule have a general tendency to get as compact as possible?

Not necessarily.In small molecules with $\ce{sp3}$ centres continously free rotation occurs.However in very large molecules folding occurs to reduce surface area as you said.

@ZOZ Okay, but I still don't get why a compact structure is preferred.

@ReeshabhRanjan I hope you heard about Van Der Waal's attractive forces?That's what causes folding in large molecules.Also see https://en.wikipedia.org/wiki/London_dispersion_force . You might have read the NCERT textbooks which describe them well...

Also see https://www.youtube.com/watch?v=1iYKajMsYPY ...explains it well...

@ZOZ yeah I am currently reading the NCERT textbook. It has only mentioned Van Der Waal's attractive forces on intermolecular, not intramolecular level. That's why I was confused. Thanks.


text:
Ionic compound gases
Does any molecule containing only ionic bonds exist as a gas in room temperature? In the case of carbon dioxide, it is a covalent molecule which is a gas at rtp. Does the same apply for ionic compounds?

Hydrogen chloride is a covalent compound that reacts with water (and plenty of other compounds) to form ions.

That depends on your definition of an ionic bond...

If anything, being a gas in ambient conditions is a very strong argument against a compound being ionic.

related http://chemistry.stackexchange.com/questions/2511/will-gaseous-ionic-compounds-be-free-moving-ions http://chemistry.stackexchange.com/questions/14174/what-is-sodium-chloride-like-in-gas-state


text:
Where is bifurcation situated on cyclohexane isomerisation PES?
I am wondering what does PES for cyclohexane isomerisation looks like.
I have two possibilities in mind but each has some kind of vagueness.
1) Starting from boat conformation minimum, followed by transition state, then bifurcation, after that two separate paths leading to halfchair transition states and then chair minima.
I am wondering what this first transition state would look like? I can't seem to imagine conformation that would be different from boat but leading to two further conformational possibilities.
2) Starting form boat conformation minimum, followed by bifurcation, after that two separate paths leading to halfchair transition states and then chair minima.
In this case I am wondering how can bifurcation occur without valley ridge inflection point, and I can't seem to imagine valley ridge inflection point coming from boat minimum.
Aditional question: Does bifurcation occur only when moving on PES from global minimum?
I don't know if I'm missing some major point here, but any help is welcome.
Also any suggestions for some book or article about this?

You could model it.  GAMESS (US) is free, as is ORCA.

http://butane.chem.uiuc.edu/jsmoore/chem232/notes_current/Cycloalkanes/NOTES-The_Cyclohexane_Ring_Flip.pdf

http://www.ch.imperial.ac.uk/rzepa/blog/?p=7926

It says nothing about bifurcation on PES of cyclohexane in these articles. Additional help would be welcome.

@Martin 's article implies a bifurcation by noting two paths that pass through the same transition state. This is made more clear in the [precursor article] (http://www.ch.imperial.ac.uk/rzepa/blog/?p=1587) that is linked to in Martin's article and I have included here.

I don't know how helpful this will be, but still https://pubs.acs.org/doi/abs/10.1021/ed047p488


text:
Brown stains on skin after working with Tollens' reagent
I was working in the lab preparing Tollens’ reagent. When I came back home, I observed that my fingertips were brown, and I tried to wash it with soap several times but it didn't work.
It has been a day and they don't seem to go.
What may have been the reason? Is it permanent and how can I get rid of it?

Let that be a massive wake-up call. The three things that matter most in a chem lab? SAFETY, SAFETY, and SAFETY. Have a safety plan and know how you are going to properly dispose of all waste before you start any experiment. We don't about safety nearly enough on this site.

The stain will wear off your fingers in a few days. Just wait...

Here's the canonical cautionary tale: https://en.m.wikipedia.org/wiki/Karen_Wetterhahn

@JimGarrison - I'd take some exception to your example. It is entirely possible to get exposed in some way or by some chemical where the problem is yet unknown. That is a small risk associated with being a chemist. What is STUPID is to ignore the science that we do know. Cyanide is toxic. Rubbing it on your fingers and dying isn't an accident, as in the case of Dr. Wetterhahn, it is plainly just STUPIDITY. 99+% of "accidents" in a chemistry lab could be avoided with a proper consideration for known safety factors.

Of all things that can leave long-lasting marks on your skin, silver is probably the least harmful. In a way, it might even do you a little good because of antimicrobial activity. Your attempts to get rid of it, though, might be REALLY dangerous. Just leave it alone, or you'll end up in a textbook, and not quite in the role of Tollens or Fehling, but rather like Karen Wetterhahn or Sheharbano Sangji.

@MaxW I made a small edit to your comment because I removed the comment you were replying to.

@jonsca - Don't understand what is going on, but ok. The OP read somewhere on the web that cyanide could be rubbed on fingers to remove the silver stains and inquired about that. That is absolutely not safe. Using cyanide like that is of  course is so moronically stupid as to be in Darwin Award territory. https://en.wikipedia.org/wiki/Darwin_Awards // I also tried to make the point that the biggest safety problem is making a small mistake then doing something really stupid to try to fix the mistake.

Turing a small mistake into a stupid - Had a classmate set a beaker of organic solvent on fire. He used a Bunsen burner to heat a flammable solvent.  Very bad planning. Then the numb-numb carried the burning beaker past other students through the lab with tongs to the stone sink. Mistake X 10 . Still unhappy the numb-numb knocked the beaker over in the sink to get the solvent to burn more quickly. Now we are at first mistake X 100. The solvent of course flashed. Thankfully no one was hurt. He should have just put an asbestos mat or cover glass on the beaker in-place and put out the fire.

@MaxW Yeah, I was trying to avoid mentioning the substance in the comments so that someone wouldn't run across this and try it.

Although the four close votes aged away, I still consider “(…) I observed that my fingertips were brown, (…) What may have been the reason? Is it permanent and how can I get rid of it?” to be a personal medical question in accordance with the Help Center: “Personal medical questions are off-topic. We can not safely answer questions for your specific situation and you should always consult a doctor for medical advice.”

I'm voting to close this as a personal medical question.

The stains are due to silver nitrate, the reactive portion of Tollens’ reagent. It is very prone to reduction to metallic silver.
If you got it on your hands (a clear signal you weren’t wearing gloves when you should have been) it will create small silver particles in your skin — this is why it won’t wash off. Since the particles are small, they don’t exhibit the classical metallic characteristics (looking silvery, reflecting etc).
Don’t worry, they will grow out within a week’s time or so, due to your skin renewing itself.
And wear gloves next time.


text:
How do you change atomic valence in Avogadro?
I have a bridging hydroxyl group between two metals, so the oxygen is triply coordinated in this structure.
How do I change the valency to three in Avogadro so that these bonds to the metal are displayed?
Perhaps it might be necessary to also play around with the van der Waals radius.
Or is there a direct way to enforce a drawn bond between two atoms?

I'll point out that there's a discussion forum particularly for Avogadro questions...

While there are some automatic valence features in Avogadro, you can always turn them off. Bonding is mostly a matter of appearance.
For example, let's import $\ce{IF7}$.

Notice that the central iodine doesn't show enough bonds. So we'll draw some in. Switch to the draw tool and turn "Adjust Hydrogens" off.

Now just draw connections between the atoms. Voila!

After using it now several times, I have to say it is a very intutive way of doing it. Thank you very much!


text:
Why is the cis isomer of 1,3-dimethylcyclobutane more stable than the trans isomer?

I expect there to be more steric repulsions between the two methyl groups if they are on the same face as each other. Why is this not the case?

Hint: the four carbon atoms are not coplanar. It is not a flat square.

You should find a realistic stick drawing of the cyclobutane ring, you should then be able to answer your own question.

The four carbon atoms in cyclobutane are not coplanar, as that would lead to a large degree of unfavourable eclipsing interactions. So, it is not a flat square; it adopts a puckered conformation. One of the carbon atoms makes a $25^\circ$ angle with the plane formed by the other three carbons, and this mitigates some of the eclipsing interactions, at the cost of a small increase in angle strain. The conformation is commonly called the "butterfly", and two equivalent puckered conformations interconvert rapidly. Here's a visual reference to help you

Pictured above is an unsubstituted cyclobutane.
Take a closer look at the figure on the left, and notice how there would be some transannular interaction between the two $\ce{H'}$ hydrogens at $\ce{C-1}$ and $\ce{C-3}$, marked in $\color{red}{\text{red}}$. This is bad. In the figure on the right, the $\ce{H'}$ hydrogens are no longer axial, but equatorial, and this is better.
In short, placing substituents in an equatorial position is better than placing them in an axial position.
Now, imagine what the cis and trans isomers would look like for 1,3-dimethylcyclobutane.
In the cis isomer both $\ce{H'}$ on $\ce{C-1}$ and $\ce{C-3}$ would be replaced by $\ce{-CH3}$ groups (since they are on the same side). Of course there will be considerable repulsions if they are arranged in the conformation on the left, which is why ring flip takes place and we will predominantly get the conformation on the right with both the methyl groups in equatorial positions. This is ideal.

The trans isomer would be like replacing one $\color{red}{\text{red}}$ $\ce{H'}$ and one $\color{blue}{\text{blue}}$ $\ce{H}$ at $\ce{C-1}$ and $\ce{C-3}$ with methyl groups. No matter what you do, you can only get one of the methyl groups at the equatorial position, but never both.

[SOURCE]

It is because cyclobutane looks like a square piece of paper with a
  fold along the diagonal. If you add the methyl groups to the corners
  not on the fold, they can be both axial or both equatorial. The axial
  is the less stable conformation than diequatorial. If it is the trans
  compound, then it will have one axial and one equatorial. Just as two
  equatorial groups are more stable, only one equatorial group will be
  less stable.

You need to understand that cyclobutane is a bent structure to reduce angle strain and torsional strain.Read more here http://www.masterorganicchemistry.com/2014/04/03/cycloalkanes-ring-strain-in-cyclopropane-and-cyclobutane/


text:
Is diamond transparent to X-rays?
Why is diamond said to be transparent to X-rays? Does this statement mean that X-rays can pass through diamond? If that is the case, then how does one explain this phenomenon?

You are transparent to X-rays. (Your bones are not, though.) It is all about atomic number, nothing more.

Atoms do absorb x-rays , but this depends on the wavelength and is different for different elements.  So the answer to your question is yes to some x-rays diamonds are transparent.  However, the main interaction of a crystal (which diamond has to be) is to diffract x-rays. See https://en.wikipedia.org/wiki/X-ray_absorption_spectroscopy for absorption

There are x-rays (Li K-$\alpha$ is 50 eV) and there are x-rays (Uranium K-$\alpha$ 98 keV).
In reality all electromagnetic radiation tries to pass through everything. Let $\rm I_0$ be the intensity of the beam at the surface and $\rm I_x$ be the fraction that gets through to depth $x$, then:
$\rm I_x = \rm I_0 e^{-\mu\rho x}$
where:

$\mu$ is the mass absorption coefficient, units = $\rm{cm}^2/\rm g$
$\rho$ is the density of the material, units $\rm g/\rm{cm}^3$
$x$ depth of radiation, units = $\rm{cm}$

As a gross generality as the energy of the radiation increases (wavelength shorter) the transparency of matter increases. So a thin slice of carbon, be it graphite or diamond,  will stop the vast major of Li K-alpha x-rays, but very few U K-alpha x-rays.
Note with some rearrangements and variable substitution this gives the Beer-Lambert Law used in UV-Vis spectroscopy.

Your diamond will absorb x-rays at all x-ray wavelengths, but the extent of absorption changes, and is far less at shorter wavelengths (or higher energy) than at longer ones.  The measured absorption is both due to true absorption and scattering of the x-rays off an atom’s electrons. The 'background' absorption in an atom is due to an electron in an occupied level being excited to a virtual (unoccupied) level. Absorption spectra are typically recoded in the range of approximately 0.01 to 1 nm or 123.9 keV to 1.239 keV respectively.
As pointed out in another answer (@MaxW) the  attenuation due to absorption experienced through any material depends in the Beer-Lambert law $I=I_0 $exp$(-\mu x)$ for path length x . The linear absorption coefficient, $\mu$, however,  depends on wavelength.
As the wavelength is decreased (energy increase) there is a general decrease in absorption until certain critical energies are reached, then there is a sudden and significant increase  in absorption, see figure. (This is a generic sketch and does not refer to any atom) These features are called absorption edges and occur when a core electron is ejected into the continuum as a photo-electron.The minimum energy to eject an electron is the threshold energy, which is equal to the binding energy of the electron.

The probability of ejecting an electron is greatest just when the energy is just enough to eject a core electron from the atom. This corresponds to the absorption edge. As this energy depends on the energy levels in each type of atom these edges occur at different energies in different types of atoms. As the atomic number gets larger these jumps in absorption (edges) shift to higher energy but become smaller. For example,$\ce{C \approx 0.28,  Al \approx 1.5, Cu \approx 8.9, Pb \approx 88 } $ keV. (This difference means that, in practice, one element can act as an absorber to filter and or isolate another element’s x-ray emission  lines. This is particularly true of elements with Z values that are close to one another.)
The highest energy absorption edge is for K shell electron ejection then L and M etc. to lower energies. The K absorption corresponds to ejecting an electron with principal quantum number $n=1$ thus from a 1s orbital ($^2S_{1/2}$) into the continuum. The L absorption ($n=2$) has three closely spaced levels because orbitals with angular momentum (p orbitals in this case) are split in energy due to coupling of electron spin and orbital angular momentum (spin-orbit coupling), the levels are labelled ${^2S_{1/2}, ^2P_{1/2}, ^2S_{3/2}}$ . The M transitions involve $n=3$, involve s, p & d orbitals and are also split.
The absorption here refers to the photoelectric effect when a core electron is ejected and energy is carried away by the photoelectron as kinetic energy. The atom is left in an excited state and the  ‘hole’ left in the atom, is very rapidly ($\le 10^{-15} $ s) filled by an other electron moving down in energy and the release of an photon of x-ray fluorescence or of an electron (Auger process).
The attenuation depends on the amount and type of matter traversed, and it is found that absorption coefficient can be expressed as $\mu \approx \rho Z^4 \lambda ^3/A $, where $\rho$ is the density, A the atomic mass and Z the atomic number; $\lambda $ is the wavelength. Thus as $\mu$  depends  of density and atomic number it is easy to see why metals show up in medical x-ray images more than does bone or soft tissue. (The reason for the $z^4$ dependence is that the matrix element for an electric-dipole transition (bound to unbound / continuum) depends on $Z^2$ and the transition probability depends on the transition matrix squared).
If the excited atom is in a  molecule, when the photoelectron is released it has wavelike properties, and can be diffracted off electrons in neighbouring atoms causing constructive and destructive interference. This behaviour adds a sinusoidal-like feature on, and just past, the absorption edge (in the direction of higher energy, but to small to show in the figure) and is very useful in determining structure of small molecules when diffraction cannot be used. The phenomena are called XANES (x-ray absorption near edge structure) and EXAFS (Extended x-ray absorption fine structure).


text:
What is crystallographic equivalence?
I’ve learnt that in crystallography, there are many crystallographic directions that are “equivalent.” For example, in a cubic crystal system, the [011] and [110] and [101] directions are “crystallographically equivalent.” See the image below:

I cannot begin to fathom how they are equivalent. These are vectors after all, and from what I can tell with my rudimentary knowledge of maths, these vectors are not equivalent in direction. Yet all materials science textbooks and powerpoints insist that, for cubic systems, any permutation of positive and negative indices will yield a same equivalent direction. That is, the [011], [110], and [101] vectors are… the same vector?
I share the same confusion with planes, even if they aren’t vectors. For cubic systems, just as for directions, any arrangement of positive and negative indices produces an equivalent plane. For example, (100) and (010) are the same plane, when to me they clearly look different.
So what exactly is “crystallographic equivalence”?

It's all about symmetry. Cubic system is not just $a=b=c,,\alpha=\beta=\gamma=90^\circ$; it comes with certain symmetry which makes many things equivalent.

They are the same chemically - that is all that is important in this case.

Perhaps it is easier to understand if you assume that the structure is not cubic but extend the lengths so that $a \ne b \ne c$ , (keeping right angles) and plot out some atom positions in 2D then make vectors and planes . Next make the lengths equal and see what happens. You should find that (100) (010) etc are interchangeable.

Ivan is right it is all about symmetry. Think of it this way. You make the cube as a ball and stick model as shown. In your mind you number the balls according to their position. You turn your back and I reorder the cube. Now which ball was your ball 1? You can't tell because all the balls are the same. Now if you make the cube using one red ball and 7 black balls there is no symmetry and you cvould always tell what ball was which.

More specifically, following up on @IvanNeretin - it is about the allowed symmetry operations on the crystal structure. These are operations that take the crystal structure back to itself. In your cubic example, you can rotate 90 degrees about the z-axis, taking x -> y and y->-x, but the structure does not change. Thus, [100] and [010] are equivalent (and, similarly, so is [001]).

Directions & sites are said to be crystallographically equivalent, because there are symmetry operators that relate them.  So

learn about simpler symmetry operations (proper axes, mirror planes, centres of inversion) or advanced ones (like screw axes, glide planes, etc.)
learn how to recognize them in every-day's life.  It may be helpful for crystallography, and in other fields too: IR / Raman spectroscopy, for example, not only for understanding (descriptive), but for forcasting (predictive).  Tutorials like this represent an entry.

All of this in chemistry is an application of an area of mathematics called group theory.

@MaxW I completely agree.  Character tables to recognize the IR active vibrations, too.

you could have a look at https://molecule-viewer.com/  to try out finding symmetry elements in each of the commonly met symmetry species in molecules.

@porphyrin If it only is about molecules' point group: load the structure in Jmol (e.g. load $water), invoke calculate pointgroup; for a determination (e.g., $C_{2v}$) and draw pointgroup for a simultaneous visualization of all elements listed by show pointgroup; (all commands as commands to Jmol's console).

@Buttonwood, i've noticed in the past that sometimes the automatic pointgroup calculation does not always get it correct.

@porphyrin For calculate pointgroup, Jmol documents «pointGroupDistanceTolerance (default 0.2 Angstroms) determines the maximum distance between a rotated atom and another atom in the molecule. pointGroupLinearTolerance (default 8.0 degrees) determines whether a potential axis matches one that has already been discovered. Setting these values to higher numbers allows more flexibility in terms of atom positions, but also may result in molecules being reported with higher symmetry than they really have.» Did you adjust them?

@Buttonwood, I prefer to use https://molecule-viewer.com/ because it allows you to  workout the point group by adding sym elements one by one etc, and there are examples of all point groups found in practice (except diatomics). I found that the automatic estimation is ok if the structure is also v good but sometimes the experimental structure is poor and so automatic estimation  is poor but the point group obvious by eye. However, using this automatic method is a good starting point.

@porphyrin If it is about crystallographic model data (as in the question, then more space group symmetry than point group symmetry of an isolated molecule), Platon's Addsym can be a valuable assistant.  It however requires a model at hand (.cif file), the hkl reflections (well, for a couple of years the .cif obtained with ShelX include them) and works even better with the structure factor .fcf file sometimes still available. Without reading .hkl/.fcf, visualizer Jmol lacks some information of the experimentally measured electron density map

Crystallographic equivalence arises because we can change our coordinate systems. For example, $[1\,1\,1]$ direction and $[\bar{1} \,\bar{1}\,1]$  are equivalent in the cubic class because we can change the coordinate system and end up getting the value of $[1\,1\,1]$ for the same vector:

$$
[1\,1\,1] = \color{red}{[\bar{1}\,\bar{1}\,1]} = \overrightarrow{\mathrm{CE}} \\
\color{red}{[1\,1\,1]} = [\bar{1}\,\bar{1}\,1] = \overrightarrow{\mathrm{AG}}
$$
There are multiple ways to represent the exact same directions. There is a set of red axis and a set of black axis and we can clearly see that $[1\,1\,1]$ in red is $[\bar{1}\,\bar{1}\,1]$ in black although they represent the same direction $\overrightarrow{\mathrm{AG}}.$


text:
IUPAC naming for compounds containing carboxylic acid and anhydride together?
According to IUPAC, the carboxylic acid comes first in priority order for functional groups, while anhydride has no prefix. Then how do you name a compound which contains both a carboxylic acid and an anhydride in it?

Acids are senior to anhydrides; therefore, a compound that contains both functional groups is named as acid.
The acid is substituted using the usual principles of substitutive nomenclature. The prefix ‘oxo’, denoting $\ce{=O}$, is used to indicate a carbonyl group when the group cannot be cited as suffix. The group $\ce{R-CO-O-{}}$ is indicated by an acyloxy prefix (e.g. ‘acetyloxy’, ‘propanoyloxy’, etc.).
For example: 3-(acetyloxy)-3-oxopropanoic acid

(Cyclic anhydrides, however, are named as heterocyclic pseudo ketones.)

But isn't acetyl used when writing common name? Should be 'ethanoyl-oxy-propanoic acid' I think.

@user282856 According to current IUPAC recommendations (2013), acyl groups from the carboxylic acids that have retained names are used as preferred prefixes. Since ‘acetic acid’ is the preferred IUPAC name, ‘acetyl’ is the preferred prefix. The corresponding systematic alternatives ‘ethanoic acid’ and ‘ethanoyl’ may be used in general nomenclature. The name ‘acetyloxy’ is the preferred prefix for the group $\ce{CH3-CO-O-{}}$; the contracted name ‘acetoxy’ may be used in general nomenclature.

Are the parentheses mandatory?

@user34388 Yes, parentheses are used around compound prefixes as in ‘3-(acetyloxy)’ (‘acetyloxy’ = ‘acetyl’ + ‘oxy’) or ‘(chloromethyl)benzene’ (‘chloromethyl’ = ‘chloro’ + ‘methyl’), but not around the contracted prefix ‘acetoxy’, which may be used in general nomenclature.


text:
Are there super(calorie-)dense foods?
It’s commonly reported that there are 9 kilocalories per gram of fat, 7 per gram of alcohol, and 4 per gram of carbohydrates or protein. But these figures (with the exception of that for alcohol, which presumably means specifically ethyl alcohol since it is discussed in the context of a human diet) are approximations given to an assumed "dietary blend" of fats, carbohydrates, or proteins respectively.
My question is this: if you could obtain a pure sample of one specific molecular type of fat, carbohydrate, or protein, is there any molecule you could choose that would provide an energy density significantly greater than 9 kcal/gram?
(I’m asking because my friends and I over at Worldbuilding SE are trying to figure out whether you could make Lembas bread without magic. My answer ("No") has been accepted and so now I’ve got raging impostor syndrome and I’m just wondering if it happens to be factually correct.)

Related: https://what-if.xkcd.com/126/

If you are a T-800 cyborg then your uranium fuel cell provides more than 9 calories of energy per gram.

Indeed, but as far as I'm aware no one classifies uranium as a carbohydrate, fat, or protein.  (Though, I wouldn't put it past the astrophysicists.)

Your answer looks OK. Pure fats have more or less the same energy value, and even gasoline is beating them only a little.

I'd say anything on the menu at McDonald's fits the bill.


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

This is essentially a question about the meaning and significance of the term vapour pressure (or vapor pressure if you're American). From what I understand a liquid in a container will have a certain amount of its vapour above it exerting a certain pressure: the vapour pressure - which is dependant on the identity of the substance and the temperature of the system only. The point at which the ambient pressure equals the vapour pressure the liquid boils - the liquid is in equilibrium with its vapour/gas phase (the line in the diagram separating liquid and gas).
However, this had me slightly confused because in the situation initially described (at a point somewhere within the liquid region - NOT on the phase boundary) some vapour existed above the liquid in the container and thus there must be some kind of equilibrium already existing. So, when the ambient pressure equals the vapour pressure nothing has changed in the sense that equilibrium remains present it's just that the pressure is no longer significant enough to facilitate the liquid.
Is what I've said above correct? Mainly in the definition of vapour pressure being the pressure of the vapour above the liquid AT ANY POINT in the liquid region on the phase diagram and also whether the vapour pressure is just a function of temperature.

Are you speaking of a situation in which there is a second component such as air present in the gas phase, or are you describing a situation in which there is only a single chemical component?

Does that make a difference? I think keep it simple and just consider a single component system and ideal gas behaviour. I'm not too concerned about the details of the system I just want to understand what vapour pressure really means

yes in a closed container vapour is at equilibrium with liquid (and vapour with solid if temperature/pressure low enough) at a given temperature. An open bottle of acetone, say, is not in equilibrium with the atmosphere, there just is not enough of it to become so.

It does make a difference.  It is easier to demonstrate that equilibrium does not necessary exist (and explain why) for a case in which there is a second component such as air in the gas phase.

Very few things in the real world are ever at equilibrium. But if an equilibrium has been established then the answer is yes.

I know this question is very old, but the answers here are unsatisfactory, and so for the benefit for people with same questions in the future, let me clarify important points. I have assumed ideal scenario neglecting inter molecular forces.
Consider an open container with liquid phase of a substance with no molecule in gas phase above it, only air. At the top surface, there would be invariably some molecules with high energy due to their random motion. Therefore, they would begin to escape the liquid since nothing prevents them to do so. This is called evaporation. The liquid would continue to evaporate till any more evaporation would lead to partial pressure of its gas phase get in liquid region. This gas pressure is known as vapor pressure of the liquid, and is equal to maximum pressure in which the substance can remain in gas phase at that temperature. Note that liquid at this point has pressure equal to total gas pressure above it (atmospheric pressure).
Take a control volume inside the liquid. Remember that due to thermal motion, the molecules inside the control volume try to expand into gas phase, but since the gas phase pressure at that temperature is lower than the pressure by liquid, it can't expand. But as soon as the pressure exerted by the liquid becomes equal to the gas pressure, the molecules inside can easily transform into gas phase. (Note that it is not that straightforward--some seeding is required for bubble to form due to surface tension forces while the surrounding liquid is superheated, but I digress.)
And what is the pressure of the liquid equal to? Atmospheric pressure. But only when the container is open. So when container is open, and either atmospheric pressure or the temperature of the liquid is brought such that the pressure of the liquid becomes equal to its vapor pressure, then the thing boils.
If the container is closed, and all the air is removed, the pressure inside the liquid is equal to vapor pressure, and there will always be vapor pressure present, even if in minuscule amount. (You can't have a container filled with only liquid. Think about that for a moment. Similarly you can't have a container filled with only ice. Some water, or water vapour, or even combination of the two will appear alongside.) And in that case, any change in conditions of pressure or temperature will be immediately reflected by boiling or condensation of the system. That is, a closed container with only water inside can boil at any temperature.
So in summary, vapor pressure is the maximum gas pressure possible at that temperature. The liquid always tries to expand and attain vapor pressure, and in case the inside pressure is greater than the vapor pressure, the inside of the liquid can't do so, otherwise it would boil. However, the surface of the liquid, due to imbalanced forces, can do it at any point by evaporating, and thus forming gas above it till vapor pressure is attained.

Liquid (and solid) phases can only form if there are intermolecular forces. I assume you mean you're ignoring intermolecular forces just in the gas phase?

You can have a container filled with only liquid. You can have a container filled with only ice. Fill a zip lock bag under water, you have only liquid. Put it in the freezer, you have only ice after a while. It is best to use degassed pure water, otherwise there might be some bubbles of nitrogen etc. See the question this (in my mind) incorrect statement inspired: https://chemistry.stackexchange.com/questions/117850/is-it-possible-to-have-only-liquid-in-a-closed-container

The definition of vapor pressure means that for any temperature and pressure within the limits of a liquid system, there is a certain amount of vapor present above the liquid. The liquid and vapor at this point are in equilibrium because the vapor pressure does not change over time for that given temperature and pressure. When the liquid starts boiling, the system is undergoing a phase transition and in not at equilibrium because the system volume is changing and energy is continually being added to raise the system's temperature. Of course, we know the temperature does not start to change until vaporization has completed.
Vapor pressure of any given pure liquid is normally a strong function of temperature and a weak function of pressure. If you took a beaker of a pure liquid and put it inside a container that could support a vacuum, then as the pressure external to the beaker starts to decrease, more molecules from the liquid phase have enough energy to escape the liquid phase and exist in the vapor phase. If you continued developing more vacuum, the liquid would boil. This is seen on the phase diagram as moving straight down from somewhere in the liquid region until you reach the gas-liquid equilibrium line.

You swirled around so much that it is hard to follow your logic.
The key here is the concept of equilibrium. If the system is at equilibrium then a liquid will have a particular vapor pressure at a particular temperature.
Think about it. The only reason that gasoline stays in the tank of a car (well at least old cars where the tank isn't sealed) is that the liquid gasoline is not in equilibrium with the atmosphere. If you just pour gasoline on the driveway and looked at the spot the next day then all of the gasoline will have evaporated.
So in general chemistry is about equilibrium conditions before and after some reaction. The major "exception" would of course be chemical kinetics.

May you please expand slightly? Is a liquid in equilibrium with it's vapour at some random value of P and T not on the phase boundary?

The point that I am trying to make is that you can have a system which has some pressure at some temperature which is not on the phase boundary because the system is not at equilibrium. "At equilibrium" is sometimes not true, like the gasoline in the tank of a car.

so a system can't exist in equilibrium at a point not on the phase boundary line? I don't really understand what you're saying

Let's ignore the solid phase for the moment. So we are just concerned with the liquid phase and the gas phase. At equilibrium there can be a gas phase with no liquid, but there cannot be a liquid phase with no gas phase (unless the container is absolutely full). So given the temperature, the curve for the gas-liquid phase boundary, the volume of the container, and the mass of the substance I can figure at equilibrium how much of the substance will be in the liquid form and how much will be in the gas form. Equilibrium is the key.

Look at your question - Is a liquid in a container always in equilibrium with its vapour? No, not always. It is very common to assume equilibrium but it isn't an absolute requirement that a system always be at equilibrium. // You can for example get pure water below  below 0 Celsius in the liquid state because the system (liquid water & ice) is not at equilibrium.

Take a point in the liquid phase area on the diagram, somewhere right in the middle - not on the phase boundary. That is the state of the system (pressure P, temperature, T) It's a container that is not completely filled. Is there vapour above the surface of the liquid in the container that exerts a prescribed pressure on the surface and the walls of the container - this pressure is dictated by the temperature and the identify of the substance? Is it, or at least can it, at this point in the diagram be in equilibrium with its vapour?

A system may or may not be at equilibrium. If the system is at equilibrium then we can use the phase diagram to calculate what is going on. If the system isn't at equilibrium then who knows?


text:
Which is the strongest bond?
Between ionic, covalent and metallic, which is the strongest?
I got really confused looking up on the internet, and found a lot of contradictory informations.
If I have to evaluate the strongest bond, maybe given a list of molecules, what do I do?

There is no definitive answer. For a given molecule, you determine the Gibbs free energy of formation relative to standard state precursors. Uniquely determining that a given bond is ionic, covalent, or metallic is either really hard or impossible.

Strong covalent bond is pretty strong. Weak covalent bond is pretty weak. There is no such thing as bond strength of a covalent bond. Ditto for other types.

@JonCuster but if I need to determinate which one between some given molecules is the strongest, how do I tell?

To determine the relative bond strength difference, between any two atoms pairs. You can simply look up the measured bond strengths in kj/mol, or eV, at STP and directly compare. These values can be found in tables in every chemistry textboom.


text:
Acetone - Can it already be anhydrous when purchased?
If I have 99.9% Reagent ACS, Lab grade Acetone, is it safe to say that it's free of moisture/Anhydrous?

No, that will have some water. I'd guess that you can buy "dry" acetone but it would be expense. Once you open the container and expose it to air then some air will fill the bottle which has at least a small amount of moisture in it and that moisture will migrate into the acetone. You can lower the water content of 99.9% reagent grade acetone by adding  molecular sieves to the bottle. That will help keep the acetone "relatively" dry.

An important thing to keep in mind is that "dry" isn't necessarily a single thing. Your downstream application determines how much residual water is acceptable. Some applications require much "dryer" solvents than others. Your acetone may be fine for some applications but not others. - Most solvents pick up water from the air, though, in amounts depending heavily on storage conditions. That's why most people re-dry solvents according to their needs, unless they purchase special solvent grades which are certified low moisture and are packaged and stored to limit reintroduction.

As correctly pointed out by @MaxW, the technical acetone still contains water.  In addition, it is better to quantify "dryness", so by passing a mark you may draw the line of "dry enough" for the application you aim for; especially as Karl Fischer analysis may provide you a reproducible figure of merit about the water content.
Now, if you need dry acetone, stirr the acetone with potassium carbonate or calcium chloride and subsequently distill it off into a dry container.  In the instance of acetone, however, I refrain from molecular sieves because they catalyze aldol reactions.

Thanks for pointing the problem with molecular sieves and acetone! I've been out of the lab too long. Forgot about that.

@MaxW In the welcoming exchange with each other, we learn by the other; in benefit for both sides.

The acetone you have ($99.9~\%$) may be pretty pure but is certainly not dry (anhydrous). The dry solvents we have in the lab are typically marked as $< 50~\mathrm{ppm}\ \ce{H2O}$, which in percent would be $< 0.00005~\%$. In your acetone, however, there are $0.1~\%$ impurities, likely a large amount of which would be water.
If you buy dry acetone (and note that it is much more expensive than the already expensive $99.9~\%$ stuff), remember that it will come in an argon or nitrogen atmosphere and beneath a septum to prevent moisture coming in. Try and keep a positive pressure of whichever inert gas you have in that bottle so it will actually stay dry. Otherwise you will waste a lot of money.


text:
What is the difference between dynamic and static electronic correlation
I am confused. In the static correlation, we use combination of Slater determinants to account for electronic correlation (known as CI or configuration interaction) and the wave function is represented as a sum of various SD where the HF SD is the base function and higher excited SD contributes some parts. Similarly, in dynamic electronic correlation, we use basis set expansion to account for higher excitation orbitals. What is the fundamental difference between these two correlations? Both are essentially expanding the electronic space to allow more relaxation. What am I missing here? Thanks

Related: http://chemistry.stackexchange.com/questions/50804/why-is-no-dynamical-correlation-recovered-in-the-casscf-method.

$\newcommand{\el}{_\mathrm{e}}$In quantum chemistry, when a nomenclature in which one distinguishes between "static" and "dynamic" correlation is used, "correlation" referrers to all the deficiencies of the Hartree-Fock (HF) single-determinantal approach. For instance, the the correlation energy is defined as the difference between the exact (non-relativistic) energy and the HF energy (calculated with a complete basis),
$$
  E_{\mathrm{corr}} = E_{\mathrm{exact}} - E_{\mathrm{HF}} \, .
$$
Now, what are the deficiencies of a Hartree-Fock approach?

First, electrons in this model do not instantaneously interact with each other, as they do in reality, but rather each and every electron interacts with the average, or mean, field created by all other electrons. Classically speaking, each electron moves in a way so that it avoids locations in a close proximity to the instantaneous positions of all other electrons. And the failure of the HF model to correctly reproduce such motion of electrons is the first source of $E_{\mathrm{corr}}$. This type of correlation is called dynamic correlation since it is directly related to electron dynamics.
Secondly, the wave function in the HF model is a single Slater determinant, which might be a rather poor representation of a many-electron system's state: in certain cases an electronic state can be well described only by a linear combination of more than one (nearly-)degenerate Slater determinants. This is the second reason why $E_{\mathrm{HF}}$ may differ from $E_{\mathrm{exact}}$ and the corresponding type of correlation is called static, or nondynamic, since it ist not related to electron dynamics.

Interestingly, both static and dynamic correlation effects can be taken into account by "mixing in" more Slater determinants $\Phi_i$ to the Hartree-Fock one $\Phi_0$,
$$
  \Psi\el(\vec{r}\el) = c_0 \Phi_0 + \sum_i c_i \Phi_i \, ,
$$
Here, if $c_0$ is assumed to be close to $1$ and a large number of excited determinants $\Phi_i$ are added each of which is assumed to give only a small contribution, then the method primarily treats dynamic correlation. And if, on the other hand, it is assumed that there are just a few excited determinants $\Phi_i$ with weights close to that of the reference determinant $\Phi_0$, then the method primarily treats static correlation.
An example of a method that recovers primarily dynamic correlation is Møller–Plesset perturbation theory (MP$n$), while multi-configurational self-consistent field (MCSCF) method primarily takes account of static correlation. Note the word "primarily" here and above. It is almost impossible in principle to keep dynamic and static correlation effects separated since they both arise from the very same physical interaction. Thus, methods that typically cover dynamical correlation effects include at high-order also some of the non-dynamical correlation effects and vice versa.


text:
Does titration usually always involve neutralization?
I've been doing titration exercises that involve figuring out the molarity of an unknown solution when you know the molarity of another (so like knowing the molarity of solution A but figuring out the molarity of solution B), and it usually always involves neutralization. Is it common that titration exercises involve neutralization?

It is common, but other types of reactions (redox, etc.) are common as well.

A lot of titrations are acid-base in nature, but not all of them are. Some are redox, like permanganate with iron(II). Some are precipitation titrations (e.g. Mohr's method) and some are complexometric titrations (e.g. with EDTA). Some are not even monitored with a colour change, for example you can measure the conductivity of the solution (conductometric titration). It is quite common, but do not think that it is the only thing.

As you can see from the other comments, neutralization strongly implies an acid-base reaction in chemistry but there are other types of reactions which can be used in a quantitative analysis titration. A reaction could be  redox. You could also do a titration that involves creating a precipitate, or a reaction that involves a substance partitioning between two liquid phases. So there are a lot of different kinds of titrations in quantitative chemistry.

To give you some examples of a non-neutralization titration reactions, I have actually titrated $\ce{H2O2}$ using yeast catalase and on another occasion $\ce{H2O2}$ using $\ce{KMnO4}$.

While neutralization of an acid (or a base) is a frequent example of titration, titration is more general. IUPAC's gold book mentions  "Also contains definitions of: acidimetric titration, acid–base titration, alkalimetric titration, chelatometric titration, complexometric titration, coulometric titration, equivalence point, precipitation titration" right in the beginning.  Even the emblematic glass buret is not a must, as publications like My Dear Buret, Your Time Has Indeed Come! (Journal of Chemical Education) showcase.


text:
Deriving the equation for ionic current
High-schooler here reading up on electrochemistry.....
I came across this equation (and its derivation) in the book Physical Chemistry, by Wallwork and Grant:

Now the thing is, the equation is valid for 'uni-univalent' electrolytes (like KCl). But I want a more....'versatile' formula.
So I attempted to modify that equation to suit electrolytes of the type AB (i.e- magnitude of charge on a formula unit cation is equal to that of the charge on a formula unit anion).
I wrote down the whole thing (my formatting skills are terrible, so to type it out would be a HUGE pain in the neck), pardon the handwriting, I hope it's legible enough....

As for that last question.....I was wondering if the equation could be made even more 'versatile' 3:)
( @getafix don't think of me as clingy but...I'm counting on you for this one too! )

I will be using the following equation from my previous answer

$$I = (z \mu \nu c F E)A$$

You are interested in an electrolyte of the type $\ce{A^{2+}}$ $\ce{B^{2-}}$.
So, the "cation" current is
$$I_+ = (z_+ \mu_+ \nu_+ c F E)A$$
and similarly the "anion" current is
$$I_- = (z_- \mu_- \nu_- c F E)A$$
Here, $z_+ = z_- = z$  and $\nu_+ = \nu_- = \nu$. The mobilities, however, are different, so $\mu_+ \neq \mu_-$
So combining the contribution from the cations and anions in the solution, we get a total current
$$\mathrm{I}_{\text{tot}} = (z \nu c F EA)(\mu_+ + \mu_-)$$
After substituting  $\ E = \frac{V}{L}$,  one gets the desired result
$$\mathrm{I}_{\text{tot}} = \left (z \nu c FA \frac{V}{L}\right ) (\mu_+ + \mu_-)$$
In general, if $z_+ \neq z_- $  and $\nu_+ \neq \nu_-$ then simply write down the individual expressions for cationic and anionic contributions to current and add them.