How is fluorine gas contained and transported, why aren't the metal cylinder and tube in this video attacked? I included the GIF below as part of an answer to Can paper burn without oxygen or air?. It was made from the video Fluorine - Periodic Table of Videos.

I noticed that the cylinder containing the fluorine gas "filled up on the roof" in the video looks like it is made out of metal, as is the tube feeding the gas to the combustion sample.

The liquification experiment also shown in the video uses a flexible tube that appears to be plastic.

I know hydrofluoric acid will dissolve gas, but can dry fluorine gas be contained in glass? Since metals, even precious metals are generally attacked by fluorine, how is it stored, and how are the metal cylinder and tube in the video not also consumed?

Incidentally you can look up individual elements on Periodic Table of Videos at http://www.periodicvideos.com/

I think it is mentioned somewhat around 4:35 or so in the link you added. The metal cylinder that has to store elemental fluorine will have to be made from a special metal like nickel or monel. Those form metal fluoride layers on the inside but they remain on the metal and allow no further fluorine to attack the rest of the metal which is called passivation.

About the plastic tube, it could be Teflon but I don't know if that works or not. Basically, the problem with fluorine especially about demonstrations was always the container. You could have easily done experiments in metal containers but then no one would be able to see through. Then there were, of course, fluorinated plastics but they tend to be quite rough and it all becomes blurred. Which is probably also the reason why there is no real fluorine ampoule for element collectors. That was at least how I understood it until quite recently when I was listening to a talk by Florian Kraus, who is a fluorine chemist. And they are using a special polymer that can hold fluorine and other highly fluorinating interhalogens but is much more transparent. So I guess by now there is quite a variety of polymers available that will hold fluorine.

Elemental fluorine could be stored in glass at least theoretically. It is not the fluorine that attacks the glass but hydrogen fluoride or I think even more precise something like $\ce{H2F^+}$. You could avoid that to some degree by drying the fluorine but the problem is the glass itself. As glass forms a polymer structure of $\ce{[SiO4]^4-}$-tetrahedra that join all vertices there has to be a sudden end once there is no more repeating unit added. But that would mean that the outside of a glass surface is all negatively charged (ignoring metal cation additives). Therefore there are many $\ce{Si-OH}$ groups on the outside which tend to be acidic. When fluorine reacts with them it will form hydrogen fluoride that will attack the glass.


Solubility of oxygen in liquid hydrogen; is anything near 1:2 (molar ratio) possible? In a question on another Stack Exchange site a proposal was forwarded to premix oxygen and hydrogen in the stoichiometric molar ratio of 1:2 to be used as rocket fuel, as opposed to the standard method of storing each liquified gas (LOX, LH2) separately.

According to Wikipedia (here and here) the freezing and boiling points of oxygen (presumably at standard pressure) are around 54 and 90 Kelvin, while those of hydrogen are around 14 and 20 Kelvin, respectively, suggesting that oxygen would freeze well before gaseous hydrogen would condense into a liquid.

However, one might (naively) wonder if oxygen might still exist as a solution in liquid hydrogen at some useful fraction, and at a pressure of roughly one atmosphere or less for practicality of storage on a weight-conscious rocket.

Even the wording of this abstract makes this sound fairly unlikely, but without some confirmation from someone more knowledgable it's hard to address the other linked question.

It would be the other way round.
Since the oxygen is the first to liquify, it would be hydrogen dissolved into liquid oxygen.

In this article they show the following calculated vapour liquid equilibrium diagram.

The 2:1 mixture, $x_{\ce{H2}}=0.67$, would exist at around 85 K and 50 bar.


Help with naming molecule I am trying to name the molecule pictured below. I thought it would be 5-bromo-4-isopropyl-2-methylhex-2-ene, but googling that name does not show any results, which makes me think it is not right. Can anyone suggest what I might be doing wrong?

The suggested name 5-bromo-4-isopropyl-2-methylhex-2-ene is not wrong; however, it is not the preferred name according to current IUPAC recommendations.

Since the compound has a double bond but no principal characteristic group that would be expressed as a suffix, it is clear that the parent structure is a 2-methylhex-2-ene. However, there is a choice between two different principal chains, which lead to different names:

5-bromo-2-methyl-4-(propan-2-yl)hex-2-ene and
4-(1-bromoethyl)-2,5-dimethylhex-2-ene

Both possible chains have two additional (but different) substituent groups in the same positions; therefore, the usual nomenclature criteria related to number and location of substituent groups do not help. Finally alphabetical order is used to identify the preferred name as 4-(1-bromoethyl)-2,5-dimethylhex-2-ene.


Is it possible for already charged metal ions to form a metallic bond? I would like to know if it is possible for already charged metal ions to become part of a metallic bond.
I was thinking of a possible metallic bond formation with other already charged metal ions or a metallic bond formation between elemental uncharged metal atoms and already charged charged metal ions. By formation a mean the moment before the metallic bond has formed because my understanding is that in a metallic bond there are only charged metal ions and no elemental uncharged metal atoms.
If one of the two scenarios or even both could occur I would assume that the already charged metal ions would form or join a metallic bond by either giving up or donating additional electrons to the so called sea of electrons of a metallic bond or the already charged metal ions would form or join the metallic bond by not giving up additional electrons.
It would be very helpful if someone could provide a link for citation about whether or not charged metal ions can become part of metallic bond since I could not find anything explicitly ruling out that possibility.

Your reasoning is correct. Such things do exist, and they came to be known as clusters. The link above does not quite do them justice; in fact, they are vastly more numerous and diverse.
Come to think of it, the clusters go beyond the scope of your question, in that many of them are not charged. To pick a few examples, see the osmium carbonyl (which, arguably, is too small to consider the bond really metallic, but its bigger brothers are known), or gold clusters which range all the way up to the immense five-layer icosahedron $\ce{Au561}$ with some ligands on top.
To pick something charged and at the same time big enough to be called metallic, have a look at this: $\ce{Ta6Cl18^4-}$.

[source]

Upd. I thought I'd better stress this again. Yes, metal ions that still have some valence electrons can use those to form bonds, given the right circumstances. Whether those bonds qualify as metallic is a different question altogether, and depends on the number of atoms (or ions) in the cluster. Two is clearly too few. You can't possibly have a metallic bond between two atoms. A bond with electrons delocalized between two atoms is called covalent. That's the case of $\ce{Hg2^2+}$. Then you have clusters with 3 atoms, then 4, and so on, all the way up to 100 and beyond.
So it goes.


Why is Pyridinium Chlorochromate soluble in organic solvents? I would presume that Pyridinium Chlorochromate is not organic soluble due to the relatively strong ionic bonding between the anion and cation on account of their similarly large size. Although the charge on the pyridine is not stabilised by resonance, the charge on the chromium molecule is. Does this weaken the ionic bond enough so that the weak London forces between organic solvent molecules and the ions is enough to pull them apart? Are there other factors at play here?

Apart from the fact noted by Mithoron that it's soluble in polar organic solvents, we may consider what lattice binding energy the solvation forces must overcome.  In a compound like sodium chloride with compact, monatomic ions the lattice forces are too strong for all but the most aggressive solvents (but lithium salts have a way around this problem).  With pyridinium chlorochromate (PCC), however, the ions are multiatomic, bulky, and differently shaped as well as having only one charge, so the crystal lattice is only weakly bound compared with sodium chloride.  PCC is far from unique; to cite another example, the triangular cyclopropenyl cation was identified through its nmr specta when salts were dissolved in polar solvents.


Best oxidizing and reducing agents: Na, Zn^2+, Ba, Ba^2+, and Ag? Which is the best oxidizing agent and the best reducing agent from the species listed below?

$\ce{Na,~Zn^2+,~Ba,~Ba^2+,~Ag}$

I determined that oxidizing agent as $\ce{Zn^2+}$ because it appears on the table with the most positive reduction potential and determined that $\ce{Ba^2+}$ as the best reducing agent because it has the most negative reduction potential. I am not sure of my answer.

It is a good idea to recap the the definitions of oxidation and reduction when approaching this kind of question, and also what it means to be an oxidising or a reducing agent:

Oxidising is defined as the loss of electrons, reduction as the gain.
An oxidising agent is good at oxidising other compounds while being reduced itself; and vice-versa.

Take another quick look at your list; all you have are neutral elements or the most common cations of said elements. If you want to oxidise or reduce, then we would always be going back and forth between either. Therefore, only sodium, barium and silver are potential reducing agents on your list (being oxidiseable themselves) while only zinc(II) and barium(II) can be oxidising agents. Here are the relevant equations:

$$\begin{align}\textbf{Reducing agents:}\\
&&\ce{Na &-> Na+ + e-}\\
&&\ce{Ag &-> Ag+ + e-}\\
&&\ce{Ba &-> Ba^2+ + 2 e-}\\
\textbf{Oxidising agents:}\\
&&\ce{Ba^2+ + 2 e- &-> Ba}\\
&&\ce{Zn^2+ + 2 e- &-> Zn}\end{align}$$

Using your table or our chemical intuition, we can then decide which reduction or oxidation is the easiest. The best oxidising agent without doubt is zinc(II) as it is much more noble than barium (although still not a noble metal). The best reducing agent is a slightly more tricky call without data, but barium(0) being further down in the periodic table allows us to assume it to be stronger. Luckily, this is supported by the data.


Calculate the mole fraction of component B in this solution? (partial pressures) The vapour pressure of a pure liquid solvent A is $\pu{0.80atm}$. When a non volatile substance B is dissolved in the solvent, its vapour pressure drops to $\pu{0.60atm}$. Calculate the mole fraction of component B in this solution.

All I can think of is that $P_A=x_A * P_A^*$, which is Raoult's Law. But I don't know how to apply it.

Raoult's law is the correct approach to solving this problem. The definition of Raoult's law is explained well in this Wikipedia page as:

  a law of thermodynamics established by French chemist François-Marie Raoult in 1887. It states that the partial vapor pressure of each component of an ideal mixture of liquids is equal to the vapor pressure of the pure component multiplied by its mole fraction in the mixture.  Mathematically, Raoult's law for a single component in an ideal solution is stated as:
  $$P_i = P_i^*x_i$$
  where $P_i$ is the partial vapor pressure of the component $i$ in the gaseous mixture (above the solution), $P_i^*$ is the vapor pressure of the pure component $i$, and $x_i$ is the mole fraction of the component $i$ in the mixture (in the solution).

Because you will be solving for mole fraction, you can rearrange the equation to:

$$x_i = P_i/P_i^*$$
$$x_i=\pu{0.60atm}/\pu{0.80atm} = 0.75$$

Of course the value calculated above is for the volatile solvent A. Since the total mole fraction of both components combined is equal to unity, the mole fraction of the non-volatile solute B must be $1 - 0.75 = 0.25$.


Order of ionization enthalpy The order of successive ionization enthalpies of group 15 are as follows :

$$\Delta H_1<\Delta H_2<\Delta H_3$$

But as all the fifteen group elements have a very stable electronic configuration with half filled p orbital so it would require more energy to remove an electron from an extraordinarily stable configuration so the order according to me should be $\Delta H_1>\Delta H_2>\Delta H_3$.

please guide me if I am wrong.

While you are correct that the first ionization energy should have a larger value for all the group 15 elements, this has a minuscule effect on the ionization energy compared to the energy that is required to separate charges. Second ionization energies are almost always higher than first ionization energies because the nucleus already has a positive charge, creating a larger coulomb force that needs to be overcome as part of the ionization process. In addition, additional ionization increases the charge on the nucleus of the ion, which progressively makes the ion less and less stable.


Solublity values interpretation "Solubility:    Soluble in water (>25 mg/ml), and DMSO (50 mM)."

It happens to be talking about a drug but that's irrelevant to my question.  Does this solubility statement reasonably indicate that I can dissolve 10mg/ml in water and expect it to stay in solution?  if possible, what does the dmso bit mean?  (maybe mM is a typo of mL?)

Yes.

The compound is soluble in water to at least 25 mg/ml - which is (obviously) the same as g/l. Presumably they didn't test at higher concentration.

Alternatively, it is soluble to 50 mM, i.e. mmol/l, in pure DMSO.

Why they would use different units is beyond me,  but I would guess this has to do with the units conventionally used in the two experiments they've run.


What is dust made of? I was cleaning my blinds today, and wanted to know what the primary components of dust are. I know that it is made of microorganisms and other particles, but I do not want to guess that it will be $\ce{N}$ or $\ce{Si}$.

I can guess that other materials outside our house might have more than one component for dust. But what about inside the house?

The dust in the house mostly contains sand, dirt and dead skin cells.

First, let's start with the easiest, sand. The chemical formulae of sand is $\ce{SiO2}$, so it contains silicon and oxygen.

Next, dirt has rock, sand, and clay. Sand is made of silicon and oxygen. Rock is mostly made up of graphite (carbon) and some minerals like phosphorus and silicon. Clay is made up of silicon and oxygen and some minor minerals like aluminium.

Lastly, dead skin, like other organic compounds, contains carbon, hydrogen and oxygen, maybe some nitrogen as it may have protein in it.

So basically, dust is made up of carbon, hydrogen, oxygen, nitrogen, silicon, phosphorus and some minor minerals.


How would I calculate the Normality of a 0.00167 molar solution of KIO₃? I need a 0.01 N solution of $\ce{KIO3}$ to use as a standard in a Winkler titration.  The protocol that I am using specifies that this is 0.3567 g L-1, which I have calculated to be 0.00167 M based on 214 g mol-1 for $\ce{KIO3}$.

I like to understand the calculations in the protocols that I am using, so how would I calculate that a 0.00167 M solution of $\ce{KIO3}$ is 0.01 N?

EDIT (More detail on the reactions)

The $\ce{KIO3}$ solution is added to an $\ce{NaOH} \cdot \ce{NaI}$ solution under acidic conditions for:  $\ce{IO3- + 5I- + 6H+ \rightarrow 3I2 + 3H2O}$

The $\ce{I2}$ is then titrated to standardize a thiosulfate solution.

Hmm.. I think that your data is wrong, I get that the normality:molarity ratio must be 5 in this case.

This ratio (I call it the $\eta$-factor) for a compound participating in a reaction is the net change of its oxidation state.

In this case, one iodine in $\ce{IO3-}$ (+5) becomes one iodine in $\ce{I2}$ (0). The net change in oxidation number is 5, so $\rm normality=5\times molarity$. Normality is like the reducing/oxidising "power" of a solution--or the concentration of electrons that it can furbish/absorb.

Your density, your $\mathrm{g~L^{-1}}$, or your value for normality is wrong here.

Note that one needs to be careful with this reaction. It is comproportionation, where the same element is oxidising itself (Basically, two forms of the element in different oxidation states are reacting with each other). We can have some trouble when dealing with the products of such a reaction, though dealing with the reactants is fairly simple, as shown above. I'm writing a detailed self-answered post on this later.


How do I calculate the humidity inside sealed containers containing mixing salts? At $\pu{25 ^\circ C}$:

saturated Magnesium chloride $\ce{MgCl2}$ keeps humidity 33% inside sealed containers.
saturated Sodium chloride $\ce{NaCl}$ keeps humidity 75% inside sealed containers.

I want to keep humidity at 60%. Which ratio should I mix this two chemicals?
How we can calculate humidity for mixing salts? Is there a formula or table list for this?
At $\pu{25 ^\circ C}$ temperature, $\ce{NaCl}$ sets 75% humidity and $\ce{KCl}$ sets 85% humidity. NaCl mixed 1:1 with $\ce{KCl}$ will give an RH% around 70–71 between $20$ and $\pu{30 ^\circ C}$.
What is the formula to calculate 70% with using 75% and 85%?

As Ivan Neretin pointed out in his comment, there is no simple formula on this matter. Howevr, if you have access to ammonium nitrate ($\ce{NH4NO3}$) or sodium nitrite ($\ce{NaNO2}$), saturated solution of both give 64% relative humidity (RH) at $\pu{25 ^\circ C}$ (Engineering Toolbox).
Interestingly, magnesium nitrate ($\ce{Mg(NO3)2}$) gives exactly 60% RH at $\pu{0 ^\circ C}$, but it was only 54% at $\pu{25 ^\circ C}$. A few salts and their measured RH values in different temperatures are given in the following Table for readers' convenience:
$$
\bf{\text{Relative Humidity over the Saturation Salt Solution (%)}}\\
\begin{array}{c|ccc}
\hline
\text{Salt} & \pu{0 ^\circ C} & \pu{10 ^\circ C} & \pu{20 ^\circ C} & \pu{25 ^\circ C} & \pu{30 ^\circ C} & \pu{40 ^\circ C} & \pu{50 ^\circ C} & \pu{60 ^\circ C}\\
\hline
\ce{NH4NO3} & - & 75 & 67 & 60 & - & - & - & - \\
\ce{(NH4)2SO4} & 82 & 82 & 81 & 81 & 81 & 80 & 79 & - \\
\ce{MgCl2} & 34 & 33 & 33 & 33 & 32 & 32 & 31 & 29 \\
\ce{Mg(NO3)2} &99 & 98 & 98 & 97 & 97 & 96 & 96 & - \\
\ce{KNO3} & 96 & 96 & 95 & 94 & 92 & 89 & 85 & - \\
\ce{KCl} & 89 & 87 & 85 & 84 & 84 & 82 & 81 & 80 \\
\ce{NaCl} & 76 & 76 & 75 & 75 & 75 & 75 & 74 & 75 \\
\ce{NaNO2} & - & - & 65 & 64 & 63 & 61 & - & - \\
\hline
\end{array}
$$


What is the sense of electronic configuration of an atom if an electron can be in any superposition of energy eigenfunctions? I recently started learning abou the quantum mechanical explanation of hydrogen atom. In the derivation of the wave functions of hydrogen atom we find its energy eigenfunctions. But according to the Schrödinger equation, the electron can exist in any linear combination of the energy eigenfunctions. So what I don’t understand is that why do we say that the electrons in an atom belong to specific orbitals? Why do we always treat electrons in an atom as if their energy is fixed?

The answer that will satisfy you depend on how familiar you are with the mathematics and principles of quantum chemistry.

So, let's start small. Hydrogen atom. Exact expressions for eigenfunctions for the Hydrogen are known, they are the orbitals we know and love(?). Although the electron can be in any combination of eigenfunctions, it can only stay that way when you are not looking. When you do look, that is, take a measurement, it will collapse to a specific eigenfunction. That eigenfunction could be 1s or it could be 6h, but it turns out that 99.999% of the time, when you measure the energy of a Hydrogen atom, it will display the familiar 1s. That is the sense in which we can say that the electron in a Hydrogen atom is is in orbital 1s, what we really mean is that, typically, for the vast majority of time, the electron of the Hydrogen will be occupying the ground eigenstate.

Now that is for Hydrogen. For heavier elements, you no longer have orbitals, strictly speaking, you have more complicated eigenstates, obtained by solving the Schrodinger Equation to each atom separately. Currently, there is no analytic solution to the equation when you include a second electron. We approximate the ground state of each atom as though the electron were occupying different hydrogen-like orbitals (hydrogen-like because they differ from the pure hydrogen orbitals in a few parameters that are added).

Because electrons cannot occupy the same quantum state (considering spin), the way to do that is to assume that each electron is in a spin orbital. The way to distribute the electrons through the infinitely many hydrogen-like orbitals is to distribute them in the way you have learned, following the Pauli diagram. The resulting electron distribution is an approximation to the true ground state, using hydrogen-like orbitals, because the true ground state is not known.

O hope I made myself understood, but if you need clarification, please ask and I shall edit the answer as much as possible. I have fond memories of the time when I was asking the same questions :)

EDIT 1

You asked why is the probability of finding the electron in 1s orbital highest when we measure its energy?

The doubt comes because from your understanding these probability comes from writing the initial state as a linear combination of eigenbasis of our choice (in this case the energy eigenfunctions) and the initial state can be any function (normalisable and continuous)

Now, while in principle this is correct, in practice that is not so. When you have a vast sample of hydrogen atoms and you measure the energy of each one of them, the vast majority of them will return the ground state for most experimental conditions. That is because experimental setups that actually manage a high degree of superposition with excited states are very hard to prepare.

This means that the experimentally expected state for an electron in a Hydrogen atom is 1s for the vast majority of experimental setups. This is the relevant information. The preparation of an arbitrary superimposed state is not trivial, and they do not occur easily in nature.

This leads us to the approximation using hydrogen-like orbitals and the aufbau principle, according to which the ground state (the most experimentally relevant eigenstate) of an atom can be approximated as though each electron occupied one spin orbital around the nucleus, following the Pauli diagram of energies.

It is not so much a theoretical result as much as a way to make theory and practice agree. There is a way to reach the aufbau principle from first principles, but it involves doing statistical thermodynamics using Fermi Statistics. These calculations would tell you how electrons usually distribute through their available energy levels, and it would become clear that the aufbau principle is pretty sensible, from an experimental standpoint.


Is there a pH range for "neutral" detergents? There exist many home made pH scales like these

But do any of you know if there is an "official" scale that defines if a neutral detergent is 7 sharp or within a range _______?

Neutral detergents are not neccesarily at pH 7, and I know of no legal restriction (at least in the EU) on labelling detergents as neutral based on pH. For example, this patent claims protection for a set of neutral detergent formulae with pH of "from about 5.5 to about 8.5".

An example of a commercial product is Proctor and Gamble's "Tide", which has a pH of 8.1-8.5 according to their safety data. This is advertised as "100%" neutral !


Why does ClO₄⁻ only have 3 double bonds? It seems that most people draw $\ce{ClO4^{-}}$ with three double bonds and the negative charge on the oxygen:
http://www.chemspider.com/Chemical-Structure.109953.html

However, it would seem that it is more stable for the negative charge to be on the chlorine as it is more electronegative? (thus giving 4 double bonds to each oxygen)

I think the best argument comes from resonance.

Drawing perchlorate with three $\ce{Cl-O}$ double bonds and the negative charge on the oxygen allows for three resonance structures to be drawn (four total). This resonance stabilization  dominates the Lewis structure over simply drawing four double bonds to chlorine.

In other words, energetic benefit of charge spread out over four oxygens > charge localized on one chlorine.


Helium gas expand isobaric isothermal If $4\ \mathrm{mol}$ helium gas expand against constant pressure in isothermal process, how can calculate $W$?

Is it $W=-nRT\ln\frac{V_2}{V_1}$ as defined by isothermal or is it $W=-p\left(V_2-V_1\right)$ as define from isobaric?

Also, is enthalpy equal to $0$ in such process?

If the helium is expanding isothermally against a constant external pressure $P_{ext}$, the work is calculated as
$$w = -P_{ext}(V_2 - V_1)$$
This process is irreversible. The first expression you gave for work is (sort of) the expression for work for a reversible isothermal ideal gas expansion.

If helium is assumed to be an ideal gas, then $\Delta U = 0$ because the process is isothermal. You have $q = -w$ (and not zero) for this process. $\Delta H$ will be zero too, since $$\Delta H = \Delta U + \Delta(PV) = \Delta U + nR\Delta T = 0$$.


Toxicity of metallic lead (Pb) I have some pieces of lead ($\ce{Pb}$) at home. I keep them in a box in the closet and never touch them.

Can it be harmful?
Should I throw them away?

I would like to keep them, but I am not certain about their toxicity.

Metallic lead is very low risk.

Lead compounds are fairly poisonous: they slowly build up in the body and cause many harmful effects. But lead metal is very inert and you would need to do something fairly risky with it to create much likleihood of generating dangerous lead compounds.

Don't eat it, for example. Don't put it in contact with food (especially acidic foods which can solubilise lead). But normal handling should not be a problem.

Lead is so inert that it has been used to pipe clean water supplies to people for most of history (or at least the parts of history where somebody had access to clean water). This is safe unless the water supply is particularly acidic and leaches lead from the pipes.

In short: don't worry about lumps of lead.


Will the solubility of a solute increase over the time a solute dissolve? would like to ask about a question that I came across in one of my assessment books. Appreciate the help that I can get.

Qns: In an investigation, two substances X and Y of the same mass are dissolved in the same volume of water. It is found that X dissolves faster than Y. Which of the following graphs correctly shows the solubility of the substances against time in this investigation?

The answer given is A.

I would like to ask why is that so. Intuitively, I would think that the solubility would be a constant since solubility is defined as "the maximum amount of solute that will dissolve in a given volume of solvent at a given temperature and pressure", therefore it will not change with time as the solute is dissolving.

Even if we're talking about rate of dissolving here, shouldn't the rate be decreasing, as when a higher concentration of particles are dissociated, there would be more of the dissociated particles in the solvent which will shift the equilibrium to the left?

Thanks for all your help!

Ivan's comment really nailed the problem, but I'll throw in a bit more.

In the question the term solubility would be ok in an English class, but the term is incorrectly used within chemistry. The more correct tense would be solvation. Likewise the diagrams should have solvation rather than solubility.

  In an investigation, two substances X and Y of the same mass are dissolved in the same volume of water. It is found that X dissolves faster than Y. Which of the following graphs correctly shows the solubility solvation of the substances against time in this investigation?


CIP Rules - How to deal with a double bonded nitrogen? If a substituent were to exist, how will one deal with the doubly bonded nitrogen? In the tree will the nitrogen with the double bond have (C,C) or (C,C,C) or (C,C,lone pair)?
I could not find this case on the CIP Wikipedia.

Also if it were (C,C), then how would one compare the priority of this with say (C,C,C)?

It would be helpful if you could draw the complete hierarchical digraph.

Here is my shot at it:

I have taken your functionality and I have incorporated it into structure 1 that has stereochemistry. The two double bonds in 1 are shown in the CIP shorthand of structure 2. C3 and C4 are non-duplicate atoms and they are attached to duplicate carbon atoms (4) and (3), respectively. Similarly, in the C=N double bond C1 is attached to the duplicate red nitrogen (N) while the non-duplicate red nitrogen is attached to duplicate carbon (1). Both non-duplicate and duplicate atoms of the same element have the same atomic number, however, duplicate atoms are attached to three phantom atoms of atomic number zero. In addition, all atoms (C,N,O,Cl,etc.) are treated as though they are quadrivalent.
In the digraph the blue numbers are the successive spheres emanating from the stereogenic center C2. In sphere 1 hydrogen has the lowest priority while C5 has the next lowest priority owing to the 3 H's in sphere 2. C1 is attached to N,(N),N in sphere 2, having the highest priority, while C3 is connected to 4,(4),H in sphere 2. The overall priorities are C1>C3>C5>H.
Now to the gist of your question. While N=N=(N), in sphere 3, (N) is attached to 0,0,0. (N) has the lowest priority of the three nitrogens. Of the remaining two nitrogens, C,C,0 = C,(1),0 in sphere 3 but this tie is broken in sphere 4 where H,H,H/H,H,H has priority over H,H,H/0,0,0. The nitrogen priorities are black N>red N>(N).


3-index integral schemes for density fitting Density fitting allows us to approximate the 4-index 2-electron integrals using 3-index integrals:

$$(ij|kl) \approx \sum_{Q}^{N_{aux}} (ij|Q)(Q|kl)$$

My question is, what schemes are used for the evaluation of the 3-index integrals? Do we simply modify existing schemes such as Obara-Saika and McMurchie-Davidson or are completely separate schemes given for this purpose?

I think in the early days one really used just the normal integral schemes Obara-Saika, McMurchie-Davidson, Rys where for one exponent just a Gaussian with exponent zero was used. Later the people examined the schemes for this special purpose and modified them for calculating 3 index coulomb integrals. A quite recent paper on this topic can be found in Gyula Samu and Mihály Kállay, J. Chem. Phys. 2017, 146, DOI: 10.1063/1.4983393.

One minor thing I want to point out is that you presented RI/DF in a slightly unusual way (notation). Using RI/DF the integrals are evaluated as

$$
(ij|kl) \approx \sum_{PQ}^{N_{aux}} (ij|P)[V^{-1}]_{PQ} (Q|kl)
$$

where

$$
V_{PQ} = \left(P|Q\right) = \int \int {{\phi_{P}({r_1}) \frac{1}{r_{12}}  \phi_{Q}({r_2})}}d{r_1}d{r_2}
$$

Of course you can rewrite it a bit with forming $V_{PQ}^{-1/2}$ to arrive at a similar expression as given by you.


Chemistry of cereal in milk In a nearly finished bowl of oat cereal, why do the pieces of cereal stick together? I figure that since they are composed of starches it could be hydrogen bonding. Is this true or is there some other reason?

Surface tension, not hydrogen bonding, is the primary phenomenon which causes cereal rings such as Cheerios to adhere together while floating in milk.

The total linear perimeter of the 3-phase cereal-air-milk interface is minimized when pieces cling together.

Here are some links:

http://www.livescience.com/9350-cereal-science-floating-objects-stick.html
http://phys.org/news/2010-09-well-known-effect-breakfast-cereal-physicists.html
https://en.wikipedia.org/wiki/Cheerios_effect


What is the greenish substance formed after the electrolysis of two graphite electrodes in a sodium sulphate solution? Today at school we tried the electrolysis with two graphite electrodes in a sodium sulphate solution. We put the electrodes in two test tubes and filled them with some more sodium sulphate so that the solution could react with the electrodes and after that, we quickly turned them over and put them in a beaker filled with sodium sulphate. After that, we added some bromothymol blue in it to see where the hydrogen and oxygen would have gone. Afterwards, we turned the power on and saw that it was producing gas and everything was normal at first (1st image).

Some time after that we didn't see the volumes between the two test tubes changing so we waited some more. Later, we saw that the solution started assuming a green color and something started forming on the surface (2nd image).

It became more and more green so we decided to turn the power off. Then we noticed that on the bottom of the beaker was something yellowish and brownish (3rd image). We checked if the wires were consumed by the solution or if they had reacted with it but nothing was wrong with them. I have no idea what it is and my teacher told me and my lab group to find an explanation to that. What is the problem here?

List of the materials used:

1 beaker
2 test tubes
two electrodes connected to wires (graphite)
$\ce{Na2SO4}$ (sodium sulphate)
bromothymol blue

I think that what possibly happened here is that the copper wire that connects the cable with the graphite electrodes is exposed in the solition (if you look carefully in Fig.1 there seem to be a lot of bubbles forming near the red cable insulation at the bottom of the tube which escape out of the tube). So copper is oxidising to give copper cations which give the coloured material in the solution.

This is just a theory though. To confirm, repeat the experiment first with copper electrodes then with your graphite electrodes but ensure that only the graphite is in the solution (so get rid of the tubes). If using the copper elecrodes gives the same result as above (blue green stuff) while using the graphite electrodes just gives O2 and H2 it means that the above theory is likely. Finaly you could isolate the greenish solids by filtration, dry them and do a flame test. If the flame becomes green it suggests presence of copper further confirming the above.

The only thing that doesnt fit with the above is that Cu2+ ions are blue and soluble in solution not green, so maybe the green stuff also (or only) result from some reaction with the indicator. Hence, it would be good to repeat the experiment with and without the indicator to see what happens (that makes 4 experiments in total).


3 component (ternary) system in equilibrium If the components of a system were Cl2, Br2, and I2, how could I describe this system under equilibrium conditions when the number of phases is at a maximum?

If there are 3 components, using the formula, F=M+2-P and solving for P under equilibrium conditions there would be a total of 5 phases.

Since these above components are all gases, and we know there are 5 phases, can I assume that 3 of the 5 phases will be Cl2 (gas), Br2 (gas), and I2 (gas).

For the other two phases, can I assume they consist of liquid iodine as it's commonly in the liquid phase, and cl2 (solid) as it's commonly in the liquid phase.

I'm not sure if this is correct.

If there are 3 components, using the formula, F=M+2-P and solving for P under equilibrium conditions there would be a total of 5 phases.

I understand from that you set $F=0$. Equilibrium does not necessarily mean that you have no degrees of freedom. In this case, you actually have two degrees of freedom: the molar fractions of two components, for example $X(\rm{Cl}_2)$ and $X(\rm{I}_2)$.

$X(\rm{F}_2) = 1 - X(\rm{Cl}_2) -X(\rm{I}_2)$ and cannot vary independently, so it is not a degree of freedom. Once you consider $F=2$, you only need three phases.

What about liquids and solid? Let's say that you vary temperature, and you reach the melting/condensation/etc curve in $T$ space. On the curve itself you will have four phases, three gases and (for instance) liquid chlorine. This causes you to lose a degree of freedom, which in this case is temperature. These four phases can exist only in that specific temperature.

EDIT

Following the comments on the original question, I would like to note that I was addressing a simplified case where the only phases possible are the ones listed in the original question. Once you have intermediate compounds (ICl for example) it's a whole different story.


If Mendelev hadn't put the periodic table in its current order, would it still have been made popular? What I mean is, does the current model make the most sense? The QWERTY keyboard was made popular because someone won a typing contest with it, and is widely used today, even though other keyboard designs might have made more sense. Is our current periodic table also just popular for an arbitrary reason, or is it actually the best design possible? If things had played out differently, is there a high chance we would use another table?
Thank u

The periodic table is very different to things like the QWERTY keyboard layout which is essentially arbitrary.

The structure is determined by the fundamental chemical properties of the elements. When the table was first developed Mendeleev was wise enough to arrange the table with both gaps and with some elements in chemical property order rather than the order indicated by know physical properties such as atomic mass (which usually gives the same order as chemical properties but gets it wrong a small number of times).

These properties are fundamental to the elements and not a matter where human choice can have any impact at all. Sure, there are alternative ways to get the graphic design right, but none of these work if the violate the original principles of grouping similar elements together. Therefore, no alternative choices by Mendeleev could work since they would fail to achieve this.


Lewis structure of chlorite ion When solving the lewis structure for the $\ce{ClO2-}$ ion, taking into consideration formal charges, the structure is represented as:

A lewis model with 2 double bonds also fits the formal charge and lewis model requirements, however this time the negative formal charge would be on the chlorine atom. Is there a reason this form is not favoured?

When drawing structures and taking into consideration formal charge, negative charges are assigned to the most electronegative atoms. Since the electronegativity of chlorine is less than the electronegativity of oxygen, it follows that the model with 2 double bounds is less favorable.


A first-order reaction at infinite concentration I have a first-order reaction, for the binding of two molecules, of the form: $\text{k}_\text{f} = \ce{k_1 (M*s)^{-1}}$.  What happens at infinite concentration?  How is this upperbound typically handled for approximations in chemistry?  Do we need to measure additional kinetics parameters in this regime?

If anyone has any specific knowledge, I'm particularly concerned with DNA hybridization, where we have $\text{k}_1 \approx 10^6$.  Will we simply converge on the elementary step time for the formation of a base $\text{k}_+ \approx 10^6$ to $10^7$ (Craig et al., 1971; Porschke & Eigen, 1971)?  It seems reasonable to me that as $\text{C} \to \infty$ we'd have $\text{k}_\text{f} \to \text{k}_+$, which, I'd guess is the point where the rate limiting step is the time for the molecule to sample different rotamer/etc. configurations.  Perhaps it isn't an accident that these orders of magnitude match up?

Also, should we think about the entropy of complex formation in the same way in the limit of infinite concentration?  My guess for this would be "yes" since the entropic penalty of holding two molecules together, derived from studies on monoatomic gases, should be concentration independent as long as diffusion/mixing is allowed.  In practice, however, we'll probably have a gel at these higher concentrations.

When you go to the limit of infinite concentration (obviously you can never get there), a reaction cannot stay first order, because you will always run into some other rate-limiting mechanism thus making the reaction zeroth order in the highly concentrated species.

To give an example, for the binding of 2 molecules if both molecular species are 'infinitely' concentrated, the rate will not be limited by molecules 'finding' each other (as is normally the case for first order reactions), but it will be limited by the finite time binding process itself takes.


Imine as substituent nomenclature I am trying to name the following compound:

I thought the name would be: (E)-3-[(4-methylphenyl)methylideneamino]benzoic acid

I stayed away from benzylidene since P-57.1.2 indicates no substitution on this preferred prefix. However, Chemdraw shows the name as: (E)-3-((4-methylbenzylidene)amino)benzoic acid

Is my Chemdraw outdated? If so, is my name correct or is there a better way to name this compound? I am also a bit confused as to when to use parenthesis, brackets, or braces with substituent names, am I using them correctly? If not, why? Also should it be "amino" or "aza", I am confused as to when to use one or the other (similar to "oxa" vs "oxo", I think; unrelated but I also don't get the difference between those).

Your assessment is correct. The rules concerning the prefixes ‘benzyl’ and ‘benzylidene’ have been changed. In the 1993 IUPAC Recommandations, these prefixes could be substituted, but only on the ring. According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), these prefixes cannot be substituted for preferred IUPAC names, although for general nomenclature restricted substitution is permitted (see P-29.6.2).

Except for the use of enclosing marks to set off parts of the name dealing with specific structural features, the suggested name is correct. The preferred IUPAC name is 3-{(E)-[(4-methylphenyl)methylidene]amino}benzoic acid.

Note that enclosing marks are used around compound (see Rule P-29.1.2) and complex (see Rule P-29.1.3) prefixes. When multiple types of enclosing marks are required, the nesting order is {[({[( )]})]}, etc.

Concerning the use of ‘amino’ in this case instead of ‘aza’, Rule P-62.2.3 explains

  (…)

  Preferred IUPAC names for prefixes corresponding to $\ce{-NHR}$, $\ce{-NRR'}$, or $\ce{-NR2}$ are formed by prefixing the names of the groups $\ce{R}$ and $\ce{R'}$ to the prefix ‘amino’, for example ‘methylamino’ for $\ce{-NH-CH3}$. Prefixes such as azanyl and azanylidene may be used in general nomenclature.


What is the least dense liquid under normal conditions? What is the least dense liquid under normal conditions, room temperature, one atmosphere of pressure, doesn't combust upon contact with air, also wouldn't kill a human just for being in the same room as it?

Isopentane $\ce{C5H12}$ has the density of $0.6201~\mathrm{g\,cm^{-3}}$ at $20~\mathrm{^\circ C}$ [1, p. 3-330].

References

Haynes, W. M.; Lide, D. R.; Bruno, T. J. CRC Handbook of Chemistry and Physics: A Ready-Reference Book of Chemical and Physical Data.; CRC Press, 2017; Vol. 97. ISBN 978-1-4987-5429-3.


Those calculations are tricky because some functionals work well, others don't, some basis sets work well while even a better basis might give bad results and so on.

We did some studies in the past using the GIAO Method in G09 with B3LYP and 6-311+G(2d,p) als basis for the NMR calculations and M06-2X/6-311++G(d,p)for geometry optimization. And that worked quite well. However, we always used solvated systems (PCM) which seems to be the common way to do it. Try to reoptimize in solvent and run the NMR calculation in solvent if that's possible in NWChem.


How do I name a molecule with a compound substituent? I added a branch to the example shown in the textbook. What do you do when your longest carbon chain has a $\ce{CH2Cl}$ attached to a certain carbon?

Previously the book treated it as a methyl group and a halogen but my modification changes the length of the longest carbon chain so how do I name this molecule?

I think it is named 5-chloromethylheptan-3-ol.

The given compound includes the characteristic group $\ce{-OH}$, which is expressed as suffix ‘ol’. Since there is only one suffix, the order of seniority of suffixes is not relevant in this case. Functional suffixes are always attached to the name of the parent structure. Note that alcohols are named by attaching the suffix ‘ol’ to the name of the parent structure, with elision of the final letter ‘e’ in the parent structure, if present.

When there is a choice among acyclic parent structures, a principal chain must be chosen. The current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) stipulates various criteria for the selection of the principal chain. The first relevant criterion in this case provides that the principal chain has the greater number of skeletal atoms; i.e., the principal chain is the longest chain. The longest chain in the given structure has seven skeletal atoms; i.e., it is a ‘heptane’.

In addition to the characteristic group $\ce{-OH}$, the parent structure has another substituent ($\ce{-CH2-Cl}$), which is expressed as the substitutive prefix ‘chloromethyl’. Note that parentheses are used around such compound prefixes.

Finally, the current IUPAC recommendations stipulate various criteria for the assignment of locants when several structural features appear in a compound. In particular, the order of seniority provides that low locants are given first to the principal characteristic group (here: ol) before low locants are given to prefixes (here: chloromethyl).

Therefore, the resulting name is 5-(chloromethyl)heptan-3-ol.


"No variables in Z-matrix" error in scan calculation of Gaussian I asked a similar question earlier this year but I'm still stuck with a similar problem.

Basically I'm trying to implement a scan calculation using both a z-matrix and Cartesian coordinates. I have the optimized Cartesian coordinates for a nanotube and z-matrix for benzene; I would like to set up the calculation to calculate the potential energy of the system when the benzene molecule is a distance R away from the NT along the z-axis. I'm trying to scan from 2.5 to 5.5 angstroms in 0.1 A intervals.

I've read the Gaussian documentation and it seems like mixing Cartesian and z-matrix coordinates in a scan function is possible but I've tried many different inputs and I still get a "No variables in Z-matrix error."

Here is a my current input file

%mem=40GB
%nproc=4
#P BP86/6-31G scan

Scan using 1 unit nanotube with benzene at 2.5 A

0 1
[Nanotube Cartesian coordinates]

X
X 1 R1
C 2 XC 1 A1
C 2 XC 1 A1 3 60.0
C 2 XC 1 A1 4 60.0
C 2 XC 1 A1 5 60.0
C 2 XC 1 A1 6 60.0
C 2 XC 1 A1 7 60.0
X 3 R1 2 A1 1 0.0
H 3 HC 9 A1 2 180.0
H 4 HC 3 A2 2 180.0
H 5 HC 4 A2 2 180.0
H 6 HC 5 A2 2 180.0
H 7 HC 6 A2 2 180.0
H 8 HC 7 A2 2 180.0

Variables:
R1 2.5 25 0.1

Constants:
XC=1.39893
HC=1.09100
A1=90.0
A2=120.0

System in consideration

Visualization of the z-matrix

Some questions:

Is there an issue with the route section? I tried replacing "scan" with "opt=modredundant" but that introduces its own errors and its clear I don't know what I'm doing.
I suspect this is a syntax issue since the documentation for "z-matrix" and "scan" suggests it is possible to keep the nanotube coordinates frozen and just scan the benzene molecule. Is there an issue with my variables and constants section? I've looked at many sample calculations but I didn't find any that consider moving one molecule toward another. If someone can link me to relevant literature regarding (reaction coordinate calculations?) I'm more than willing to look into it; as for now this is the method my professor suggested.

Ok, so I guess I'm on the right way now. Trying to fix such a complex system is always very difficult so what I did was using a similar system, I got one sodium atom instead of your NT and I'm moving the benzene away.

First of all: you got two different dummy atoms in there, atom 1 and atom 9, which I guess makes some problems. Also the extra lines between the sections are a problem.

So here's the input for this, for a first try just from 2.5 in 5 0.5 A steps which at least runs without any errors:

# opt=z-matrix hf/3-21g

Title Card Required

0 2
 Na                 0.00000000    0.00000000    0.00000000
X 1 R1
C 2 XC 1 A1
C 2 XC 1 A1 3 60.0
C 2 XC 1 A1 4 60.0
C 2 XC 1 A1 5 60.0
C 2 XC 1 A1 6 60.0
C 2 XC 1 A1 7 60.0
H 3 HC 8 A2 2 180.0
H 4 HC 3 A2 2 180.0
H 5 HC 4 A2 2 180.0
H 6 HC 5 A2 2 180.0
H 7 HC 6 A2 2 180.0
H 8 HC 7 A2 2 180.0

XC=1.39893
HC=1.09100
A1=90.0
A2=120.0
R1 2.5 S 5 0.5

So here's the point. The format basically needs to be exact like this. Everything else will confuse Gaussian. You can substitute the "Na" line with your cartesian coordinates, but then the numbering in the z-matrix section changes! you have to adopt those numbers. Oh and of course if you got a more complex system than one atom you want to not only define the distance of that dummy atom but also the angles, so it's in the right position. You might also need to adept the whole matrix so the benzene has the right orientation in relation to the NT, so at least three atoms in there need to be somehow defined by atoms in the NT.

If you want to have fixed coordinates (what you labeled as constants) in the z-matrix, you can add them after a blank line just below the last section.

  However, after a test run I'm not sure if it actually scanned that one
  coordinate, I just get one structure right now, I have to check how to
  do the scan correctly.

It works after adding that one 'S' in the coordinate which needs to be scanned.


Is carbon dioxide organic or inorganic? Today in chemistry class we were discussing Organic Chemistry. We discussed what organic compounds basically are and then I asked the teacher whether $\ce{CO_2}$ is organic or not. She told that it is as it contains carbon and oxygen with a covalent bond. I told her it can't be as it is not found in animals (naturally). I am very confused about it.

I need some good reasons to agree with either explanation. (I have searched the internet already but found no great insights as of now).

It is entirely arbitrary whether you call it an organic compound or not, though most would not.

The distinction you make that organic compounds should be found in living things is not a useful criterion. Moreover you are wrong that carbon dioxide isn't: it is both made and used by living things. Animals make it when they metabolise sugars to release energy; plants consume it when they build more complex organic molecules through photosynthesis. In fact most organic molecules are, ultimately, derived from $\ce{CO2}$.

Even more importantly most molecules considered organic are neither made by nor are found in living things. Chemists make new carbon compounds all the time (tens of millions in the history of chemistry) and most have never been made by animals or plants.

The organic/inorganic terminology is mostly very simple: covalent compounds containing carbon are organic. The only fuzzy area is around very simple molecules like $\ce{CO2}$ where the distinction doesn't matter much. So we would not normally think of diamond or silicon carbide as organic. But we might (though many would not) call calcium carbide organic because it contains a $\ce{C2}$ unit with a carbon-carbon triple bond.

However since the terminology is mostly very obvious and also somewhat arbitrary, it isn't worth much argument to sort out those very simple but awkward edge cases.


Which stabilizes a carbocation more : Back Bonding or Conjugation? Suppose we take these two compounds:

The first carbocation can rearrange to produce the second one by hydride shift. Which is more stable ? Why ?

I am inclined to consider the second one as a "bona-fide" carbocation, if that makes sense. The first one resembles a protonated aldehyde rather than a carbocation, so relatively speaking, the first one should be more "stable".


Is there any substance that's a 4-4-4 on the NFPA diamond? A lot of the organometallics are rather... interesting compounds to work with. The most famous (among those who care, anyway) is tert-butyllithium or t-BuLi. It is the textbook example of a pyrophoric substance, demonstrated to pretty much every chemistry major as an air-sensitive chemical requiring special handling (Syringe and cannula transfers, gas-tight septa, argon/nitrogen blankets, that kind of thing). Despite the inherent hazards of the stuff, it's widely used in the industry (hence widespread notoriety) to add butyl groups to an organic molecule (most organometallics are useful for this kind of carbon-carbon bond formation), and you can buy it by the - airtight - gallon canister.
There are, surprisingly (except to most organic chemists), even more dangerous compounds in the average organic chemistry lab. Along the pyrophoric line, many of the multi-methyl-metallics are violently pyrophoric (even compared to BuLi), including trimethylaluminum, dimethylzinc, and dimethylmagnesium. All of these are also extremely poisonous (anything these compounds can do to your wonder-drug-in-progress, they can also do to various key structures in your own body), with dubious honors reserved for dimethylmercury. Because many of these, especially the organoalkaline compounds, react violently with water, they also automatically rate at least a 3 on the "reactivity" scale.
When leads me to wonder, because I wonder about these things from the safety of my office chair in a completely unrelated field; just how bad can it get? Specifically, is there any compound with an accepted use in laboratory or industrial application that is nasty enough to max out the entire NFPA-704 diamond? The closest I can find is trimethylaluminum, at a 3-4-3 (heath-fire-reactivity). T-BuLi is a 3-3-4. I can't find NFPA data on straight diazomethane (which may be because nobody in their right mind ever works with the stuff in its pure gaseous form; it's always used in a dilute diethyl ether solution, and even then is never sold or shipped that way) but it would probably be a finalist, as the gas is acutely toxic, autoignites at room temperature, and detonates on standing (something for everyone!). I'm thinking that these two inorganic families - light alkyl-metallics and organo-polyazides - would be the most likely candidates to produce a compound so toxic, so flammable, and so readily reactive, yet so interesting to chemistry, that the NFPA would see fit to rate it, and would give it highest honors.

Answering my own question based on the comments, tert-butyl-hydroperoxide is at least one such chemical. As stated on this MSDS from a government website, it's a 4-4-4, with additional special warning of being a strong oxidizer. The only thing that it does not do that could make the 704 diamond any worse is react strongly with water. It is in fact water soluble, though marginally, preferring to float on top (and therefore traditional water-based fire suppression is ineffective, but foam/CO2 will work).

If anyone else can find a chemical that, in a form that is used in the lab or industrially, is a 4-4-4 that is a strong oxidizer and reacts strongly with water, that's pretty much "as bad as it gets" and they'll get the check.


Probability of the single molecule reaction to happen I am trying to simulate the single particle second order reactions with known rate constants $k_1$ and $k_{-1}$ in Python. The point of this simulation is to fit the real experimental data to the model and see which theory is working the best.

I have found many different equations for the probability of reactants $\ce{A + B}$ going to $\ce{C}$ and vice versa. The good explanations I found here and here, but the equations are not the same.

So far, I have used those equations:

$$
\begin{align}
\ce{A &->[$k_1$] B} &\quad &p(\ce{A}) = p_a = k_1[\ce{A}]\mathrm{d}t \\
\ce{A + B &->[$k_1$] C} &\quad &p(\ce{A}) = k_1[\ce{A}][\ce{B}]\mathrm{d}t / (V ([\ce{A}_0] + [\ce{B}_0]))\\
\ce{A + A &->[$k_1$] B} &\quad &p(\ce{A}) = 2k_1[\ce{A}]\mathrm{d}t/(2V[\ce{A}_0])([\ce{A}_0] - 1)
\end{align}
$$

I don't expect those equations to be correct, but they work. I would be very grateful for some hints or suggestions!

EDIT: I will post the sample code I was using:

 #A - list with 1 for particle in A or 0 for particle in B, starts with all as 1,
 #e.g:
 A = [1,1,1,1,1,1,1,1]

    for iteration, element in enumerate(A):
        A_conc = len(A[A > 0]) # number of particles in state A

        p(A) = k1 * A_conc * dt      # probability of changing to A->B

        if random.random() > pa:     # checking if random event is favouring transition
            A[n] = 0                 #A reacts to B

This is the way I want to use those equations, but still, the question remains if the equations are correct.

When two species $A$ and $B$ react, $A + B \to product$, the rate of reaction is $k\mathrm{[A][B]}$ where $k$ is the rate constant and $\mathrm{[A]}$ and $k\mathrm{[B]}$ the concentrations at any time during the reaction, which clearly change with time. To work out what happens in each small time interval means calculating the change in the number of reactant molecules in a given volume. Doing this means calculating two things. The first is the time at which any of possibly several reactions will occur, and the second is to decide which reaction this will be.
In the reaction $A + B \to product$, there is only one type of reaction but there are very many schemes where there are two or more reactions; for example, the photo-dissociation of halocarbons with UV light and subsequent reactions to destroy ozone.

$$\mathrm{CFCl_3} \overset{h\nu}\to \mathrm{CHCl_2  +Cl}\qquad \mathrm{Cl+O_3} \to \mathrm{ClO+O} \qquad \mathrm{ClO+O_2} \to \mathrm{Cl+O_2} $$

Given such a reaction scheme, the first step is to decide what reaction will happen next given the current state of the reaction at time $t$. At the level of considering individual molecules, once the reaction has started, it does not follow sequentially as shown in the scheme. This only happens when a whole ensemble of molecules, typically as large as Avogadro's number is examined, and then normal kinetic equations apply because of the averaging over many events that occurs. In any given time interval when considering molecules one at a time, it is not possible to predict exactly which reaction will happen next. It is quite possible, for instance, that several $\mathrm{Cl+O_3}$ reactions may occur before either of the other two. The best that can be done is to guess the outcome, and repeat the calculation many times and this approach is now described.

Suppose that the reactions are labelled with a number $R$ that represents reaction 1 or 2 or 3 etc. The chance of a reaction of type $R$ occurring will be defined as

$$prob(T + dT, R) \equiv a_RdT$$

Specifically, this is the probability that a reaction of type $R$ occurs in time interval $T$ to $T + dT$ in the reaction volume, given the current state of all reactions at an earlier time $t$. The values of $a_R$ can be easily worked out, and this is shown later. What is now needed is the probability $P_0(T)$ that no reaction of any type occurs up to time $T$ past the present time $t$, multiplied by the probability that only a reaction of type $R$ occurs in the small time interval $T \to T + dT$, which is defined as $a_RdT$. The probability of reaction type $R$ happening is therefore

$$ P(R,T)= P_0(T)a_RdT$$

and shown schematically in the figure.

Calculation of reaction probability.

To calculate the probability of no reaction occurring, consider any small interval $d\tau$ after some arbitrary time $\tau$, making an interval $\tau \to \tau + d\tau$. The probability of no reaction is then

$$P_0(\tau+d\tau) =P_0(\tau)\left(1-\sum_{R=1}^n a_Rd\tau  \right)$$

where the summation is over all $n$ reaction types $R$. The summation is made because this gives the chance that any one of the $R$ reactions will occur, and therefore one minus this number is the chance that no reaction occurs. Rearranged, this equation gives

$$ \frac{P_0(\tau+d\tau)-P_0(\tau)}{d\tau}=-\sum_{R=1}^n a_R$$

and in the limit of a small time interval $d\tau$, it becomes the true derivative $\displaystyle \frac{dP_0}{d\tau}=-\sum_{R=1}^na_R$.

For clarity only, a notational change is now made so that

$$a_0=\sum_{R=1}^n a_R \tag{8}$$

and therefore $\displaystyle \frac{dP_0}{d\tau}=-a_0$.The chance of no reaction up to time $T$ is found by integrating, and is $\displaystyle P_0(T)=e^{-a_0T}$.

The final result for the probability of reaction R occurring at time $T$ is the chance of no reaction up to $T $ multiplied by the chance of reacting at $T$, which is $a_R$, giving

$$P(T,R) = a_Re^{-a_0T} \tag{9}$$

an expression that shows that these times are Poisson distributed. Equation 9 is a joint probability density function, as it is the chance that reaction occurs in the time interval and that it will be reaction $R$.

Two terms now remain undefined. The first is the individual $a_R$ and their sum $a_0$, and the second determines which reaction happens next. Calculating $a_R$ is very easy, as each $a_R$ is the product of the number of molecules present in a reaction of type $R$ multiplied with the rate constant per unit volume.

If there is just a single reaction $\mathrm{A} \overset{k_1}\to \mathrm{B}$ then $a_1=k_1N_A$ where $N_A$ is the number of $A$ molecules present at time $T$. If the reaction is $\mathrm{A+B} \overset{k_2} \to \mathrm{C}$ then $a_2=k_2N_aN_B$ and if both reactions are present in the overall reaction scheme which would then be

$$\mathrm{A} \overset{k_1}\to \mathrm{B} \qquad \mathrm{A+B} \overset{k_2} \to \mathrm{C} \qquad a_0=\sum a_R=a_1+a_2 \tag{10}$$

In the special case where a reaction combines two identical molecules $\mathrm{A+A} \overset{k_2} \to $ the number of indistinguishable combinations, i.e the number of distinct A-A pairs of the two species must be calculated and this makes $\displaystyle a = k_2\frac{N_A(N_A - 1)}{2!}$. Normally, in dealing with chemical kinetics, $N_A$ is so vast that $\displaystyle a = k_2\frac{N_A^2}{2}$, but in the Monte Carlo simulations, $N_A$ is small because individual molecules are dealt with, and the exact formula must be used. Thus, macroscopic, (kinetic), chemical equations are only valid when deviations from the Poisson distribution are negligible.

An aside

The term $\displaystyle a = k_2\frac{N_A(N_A - 1)}{2!}$ implies that these numbers are averages in the reaction volume at time $t$ and should be written as the average of the product, i.e. as $\displaystyle \frac{\langle N_a(t)(N_A(t)-1)\rangle }{2!}$. The question then arises as how to deal with this. Two approximations can be made; the first is to assume that $N_A$ is large compared to 1 and so $\langle N_A(t)(N_A(t)-1)\rangle \rightarrow \langle N_A^2(t)\rangle $, which seems to be a reasonably sensible approximation, the next step is more problematic and is $\langle N_A^2(t)\rangle \sim \langle N_A(t)\rangle^2$. This is a drastic step and assumes that the variance in the number of molecules is zero, i.e. that $N_A(t)=\langle N_A(t)\rangle$ which means that $N_A$ is a deterministic process, so that the random nature of the numbers of $N_A$ entering into the reaction volume is always the same, which is not what was assumed at the outset. (To see that the variance is zero use the definition $\sigma^2_{N_A} = \langle N_A^2 \rangle -\langle N_A \rangle^2$ where $\sigma$ is the standard deviation)

In any normal kinetic experiment the number of molecules is vast, $\langle N_A(t)\rangle \sim 10^{18}$ would not be untypical, so that the fluctuations in the  number of molecules is tiny and the approximation made above is valid.

$$ \frac{\sigma_{N_A}}{\langle N_A \rangle} \sim \frac{\sqrt{\langle N_A \rangle}}{\langle N_A \rangle} \sim \frac{1}{\sqrt{\langle N_A \rangle}}\sim 10^{-9}$$

Even if there are as few as 1000 molecules the relative fluctuation  is not too drastic but in experiments using small numbers of molecules, as in the case of many 'single molecule' experiments then clearly the rate equation approach needs to be modified.

The final part of the calculation determines which particular reaction step is going to occur and at what time this is done by choosing pairs of numbers that have the distribution $P(T, R)$ equation 9. Gillespie (1997, 2007) argues that $P(T, R)$ is equivalent to two distributions, $P_1$ and $P_2$;

$$P(T,R)= P_1\cdot P_2 = \left(a_0e^{-a_0 T} \right)\left( \frac{a_R}{a_0}\right)$$

and which can be determined by choosing two uniformly distributed random numbers, $r$, each in the range 0 to 1. The equation

$$T=-\frac{1}{a_0}\ln(r) \tag{11}$$

generates a random number $T$ according to the probability density function $\displaystyle P_1=a_0e^{-a_0 T}$.

Equation 11 produces times that are Poisson distributed, and the time $T$ is added to the current time of the reaction. The second distribution, $P_2$, is calculated by choosing a uniform random number according to the probability density function $a_R/a_0$. For example, the chance that reaction 1, (equation 10), occurs is $a_1/(a_1 + a_2)$ and for reaction 2 this chance is $1 - a_1/(a_1 + a_2) = a_2/(a_1 + a_2)$. Which reaction occurs is determined by choosing a random number between 0 and 1 and seeing which of the two regions it falls into. If it falls between 0 and $a_1/(a_1 + a_2)$, reaction 1 occurs; otherwise, reaction 2 occurs. If reaction 1 occurs, $\mathrm{A} \to \mathrm{B}$, then 1 is subtracted from the original number of $A$ molecules in each step of the reaction and 1 is added to the number of $B$ molecules.

If reaction 2 happens, $\mathrm{A+B} \overset{k_2} \to \mathrm{C}$ then 1 molecule is removed from each of the numbers for $A$ and $B$. How many molecules are added and subtracted, of course, depends on the stoicheiometry of each particular reaction.

In the case of a termolecular reaction $\mathrm{ A + B + C} \rightarrow $ when the probability of three molecules simultaneously colliding is calculated the probability the chance that this occurs in  an infinitesimal time interval $d t$ is proportional to $dt^2$ so becomes vanishingly small and thus does not occur with any physical reality. It is often convenient, however, to consider a termolecular scheme as a simplified approximation to a scheme such as $N_A+N_B \leftrightharpoons X^*; \; X^* + N_C \rightarrow$

The outline plan for the Gillespie algorithm is

(1) Define the initial amounts of each species.

  Define the arrays of data points to hold results.   Set the maximum time. Set the time to zero; t = 0.

(2) Start a loop around the calculation until the time is up or no molecules are left.

  Calculate $a_1,\;a_2,\;a_3$ etc. and $a_0 =a_1 +a_2 +a_3 +\cdots$ Calculate time $\displaystyle T = -\frac{1}{a_0}\ln(r)$
  Increment time $t = t + T$
  Calculate which reaction occurs: if a random number falls between 0 and $a_1/a_0$, reaction 1 occurs, and so forth for other reactions.
  Add or subtract numbers of molecules from the totals, depending on which reaction has occurred and its stoicheiometry.
  Store the results.

(3) Continue the loop until finished.

Refs

'Markov Processes. An Introduction for Physical Scientists' D. T. Gillespie, publ Academic Press 1992.

D. Gillespie  J. Phys. Chem. 81, p2340, 1977 eqn 29

The python code for a Lotka-Volterra predator-prey reaction is given below. I'm assuming that matplotlib and numpy, as np.blabla() etc are used.

def lotka():                # Lotka reaction Gillespie method
                            # x0 + Y = 2Y  (k1) ,  Y +  Z = 2Z (k2) , Z = Q(k3)

    #k1 =10.0   k2 = 0.01   k3 = 10.0
    #y0 =2000.0  z0 =2000.0
    ycount = np.zeros(bins,dtype=int)
    zcount = np.zeros(bins,dtype=int)
    dtime =  np.zeros(bins,dtype=float)
    for i in range(bins):
        dtime[i] = maxt*(i+1/2)/bins
        pass
    ny = y0
    nz = z0
    ycount[0] = y0
    zcount[0] = z0
    t = 0.0
    indx = 0
    while indx < bins :
        r1 = k1*ny
        r2 = k2*nz*ny
        r3 = k3*nz
        a0 = r1 + r2 + r3+1e-20
        t0 = -np.log( np.random.rand() )/a0
        t = t + t0
        indx = int(np.trunc( t*bins/maxt ))
        if indx < bins :
            r = np.random.rand()
            if  r > ( r1 + r2 )/a0  :
                nz = nz - 1
                zcount[indx] = nz          # react 3
            elif r > r1/a0 :
                nz = nz + 1
                ny = ny - 1
                zcount[indx] = nz
                ycount[indx] = ny          # react  2
            else:
                ny = ny + 1
                ycount[indx] = ny          # react 1
                pass
            pass
        pass         # events

    return zcount,ycount,dtime      # end lotka

    bins = 1000
    maxt = 200.0
    k1 = 0.5
    k2 = 0.001
    k3 = 0.5
    y0 = 400.0
    z0 = 100.0

    zcount,ycount,dtime = lotka()

    fig1= plt.figure(figsize=(10, 6))
    plt.rcParams.update({'font.size' : 16})  # set font size for plots

    plt.plot(dtime,zcount,color='red')
    plt.plot(dtime,ycount,color='blue')#'ro',markersize=2)

    plt.xlabel('time')
    plt.ylabel('population')
    plt.xlim([0,maxt])

    plt.tight_layout()

    plt.show()


Is F2 two fluoride ions or just one? I'm attempting to write the balanced equation of the combination of lithium metal and fluorine gas. I understand that this will produce an ionic solid because it is a reaction of a metal and a nonmetal. However, I'm confused with the fluoride ion having a charge of -1 and fluorine only occurring as a diatomic molecule ($\ce{F2}$). Is $\ce{F2}$ two fluoride ions or just one?

$\ce{F2}$ will produce  $2\space \ce{F-}$ ions.

You can write the balanced half cell reactions as follows:

$\ce{F2 + 2e- ->2F-}$

$\ce{Li -> Li+ + e-}$

Multiply the second equation by 2 to get $\ce{2Li -> 2Li+ + 2e-}$. This is to ensure that the final equation does not have any electrons on either side.

Now add this equation with the fluorine one.

You'll get $\ce{2Li + F2 -> 2Li+ + 2F-}$ which you can write as $\ce{2Li + F2 -> 2LiF}$


Determining electronic configuration One writes the electronic configuration in terms of shells, spdf subshells, orbitals, and spin of electrons. There are exceptions to principles we know. How does one confirm that the way of writing is correct? How did scientists understand the exceptions in case of nobium and platinum in electronic configuration? Is it possible to take an atom and find which shell its electrons are in? How did this spdf electronic configuration (along with exceptions) be first understood.

I have discussed this problem once in the past with an Italian colleague, teaching in Palermo University, and who was in charge of training the Italian team at the International Chemistry Olympiad. I remember he told me that this order can be established by plotting the square root of the successive ionization potentials of any atom versus its atomic number. Of course there is one point for H, two for He, three for Li, etc. If you report these √I values versus the atomic number Z, you will find that all highest values (relative to last $1$s electron) are steadily increasing, of course. But these points remain alined across the page. Likewise, all next to highest values (also relative to $1$s) are alined, and surprisingly the slopes of both lines are equal. The first line starts at H, and the second at He. But they are parallel, after my Italian colleague.
Now if you consider the next but two highest √I values, the curve starts at Li, and the corresponding points ($2$s) are also alined through the diagram, with a slope which is nearly the same as the first two curves. Likewise the points corresponding to the next but three values (starting at Be) are alined and the line is parallel to the preceding line.
Something new happens with the next points. If you consider the fifth √I values by decreasing numerical order, these points start with Boron. And they are alined across the diagram. They correspond to $2$p electrons. New surprise : the points are still alined. But the slope of this line is much smaller than before. The slope seems to be typical of p electrons.
If you join the sixth √I values, stqrting with Carbon, the curve is still a line, with the same slope typical for p electrons. Same thing for the $7$th, $8$th, $9$th and $10$th electron. All lines are parallel.
First difficulty occurs with the $11$th √I values, starting with Na. The curve joining these points is much steeper than for the preceding electrons. It has the same slope, typical of the $1$s and $2$s electrons. So, this line cuts preceeding lines. As a consequence, when studying the points through Mg, Al, Si, etc., it is not easy to decide which √I value belongs to what.
I have not done it myself, but my colleague told me that this job may be carried out through all the periodic table, and that it is possible to recognize a bundle of steep lines for all s electrons, another bundle of not so steep lines for p electrons, another bundle of still less inclined lines for d electrons, etc. and the same with f electrons.
Apparently, this job explains why there are irregularities in the electronic configurations through the periodic table.
If some reader will try to check this statement, I would be pleased to hear of his job.


Are colligative properties applicable for volatile liquids and gases in liquid solvent? Why/Why not? My teacher told me that the expressions for calculation of colligative properties like lowering in freezing point are only applicable if the solute is non volatile liquid or a solid. We cannot apply the formula for gases or volatile liquids. Is this true? If so then why/ why not?
I have not been able to come up with a satisfying reasoning explaining this statement. Also I have not been able to find evidence regarding this statement online. Although one of my books does say this without an explanation but I am not convinced.

Yes, they are, but consider that colligative properties are often described in terms of equations whose derivation require making various assumptions about the substances involved, most importantly that the solvent and its mixtures behave close to ideally (follow Raoult's law), and that the solute does not affect the chemical potential of the solvent in the second phase (solid in freezing point depression or vapor in boiling point elevation). Textbook descriptions of freezing point depression in general assume that the solute does not partition into the solid, for instance, since partitioning would imply formation of a solid solution with a depressed (altered) chemical potential for the solid solvent, a complication too formidable to tackle in an introductory text (and even more advanced ones).


In NaCl electrolysis, why does chlorine ion get oxidized but not oxygen ion? In $\ce{NaCl}$ electrolysis, on the positive electrode (aka. on the electrode where oxidation happens), $\ce{Cl-}$ ions are oxidized to $\ce{Cl2}$. But why don't we see any $\ce{O^2-}$ (naturally present in water's $\ce{OH-}$ ions) oxidized to $\ce{O^2}$ at all?
My pre-existing knowledge, may or may not be correct, that led to this question:
[1] I am under the impression that chemical reaction is a messy business in the real-world with many reactions reversible, many intermediary steps happening and many small reactions happening in the sideways. So the real-world scenario is never as neat as one clean equation.
[2] I am aware of the concept of electronegativity, and I also have feelings that
$\ce{O^2-}$ and $\ce{Cl-}$ should require a similar amount of energy input to get them oxidized. So we should really expect both reactions to happen.
With my pre-existing knowledge laid out above, I find it remarkable that not a trace of $\ce{O^2-}$ is oxidized at all. This led me to guess that the $\ce{OH-}$ ion itself may have imparted some degree of protection to the $\ce{O^2-}$ inside of it?

I am afraid that as stated in another answer there is no simultaneous evolution of oxygen or chlorine during controlled electrolysis. It all depends on the concentration of ions in the solution. Keep in mind that the concentration of water is 55 M, so the ion concentration is nowhere near it no matter what you do!

The key reason is that the electrode potential of chlorine/chloride couple is quite different from that of oxidation of water, so if you have a high concentration of chloride ion, chlorine is oxidized first, once chlorine falls below a certain threshold, water begins to oxidize. Nernst equation can help in predicting that.


What is the meaning of a decimal in a subscript in a chemical formulas? What is the meaning of a decimal in a subscript in a chemical formulas?

I came across the fractional subscripts in a publication,[1] for example:

  We introduce oxygen into a ferromagnetic metallic glass to form a $\ce{Co_{28.6}Fe_{12.4}Ta_{4.3}B_{8.7}O46}$
   magnetic semiconductor with a Curie temperature above $\pu{600 K}$. "

How do I interpret these numbers?

References:

Liu, W.; Zhang, H.; Shi, J.; Wang, Z.; Song, C.; Wang, X.; Lu, S.; Zhou, X.; Gu, L.; Louzguine-luzgin, D. V.; Chen, M.; Yao, K.; Chen, N. A room-temperature magnetic semiconductor from a ferromagnetic metallic glass. Nat. Commun. 2016, 7, 13497 DOI: 10.1038/ncomms13497.

In the case of complex glasses such as $\ce{Co_{28.6}Fe_{12.4}Ta_{4.3}B_{8.7}O46}$, Instead of normalizing the subscripts to have integer subscripts they normalized them by molar percent such that the sum of subscripts is $100\%$  ($28.6+12.4+4.3+8.7+46=100$)

This is fairly common practice in materials science. For many complicated materials such as this, it's hard to gain useful meaning from the chemical formula if the subscripts are normalized to integers so they are normalized to 100. People reading this formula now can compare it to similar materials they have seen and future materials are also more comparable (e.g. if they substituted the tantalum with niobium).


Ketene cycloaddition reactivity Why do ketenes participate in [2+2] cycloadditions with their $\ce{C=C}$ bond and not the $\ce{C=O}$ bond? Can the reactivity be reversed somehow?

Ketenes can participate in [2+2] cycloadditions with their $\ce{C=O}$ double bond.  Ketene dimerization is the best known example of such a cycloaddition.

(image source)

As you add substituents to the ketene $\ce{C=C}$ double bond, you begin to favor cycloadditon across the carbon-carbon double bond at the expense of cycloaddition across the carbonyl double bond.

Some Diels-Alder cycloadditions also involve adding across the ketene carbonyl.

(image source)

With both the [2+2] and [4+2] cycloadditions often the kinetic product involves cycloaddition across the carbonyl, but on mild heating this cycloadduct can rearrange to the product from cycloaddition across the carbon-carbon double bond, the thermodynamic product.

A very nice review of the synthetic and mechanistic aspects of ketene cycloadditions can be found here.


Does a messier room actually have higher entropy? When my chemistry teacher was explaining entropy, she used the analogy of a clean vs. messy room. What I'm wondering is whether a messy room really does have higher entropy. Does this still work on a larger scale? If not, can I trust the analogy?

A room can be a complicated system, (at least mine can be) so I shall assume a simplified version: an enclosed space bound by 4 walls, a ceiling and a floor and the only piece of furniture inside the room is a desk, and  two books on it -- book 1 and book 2.

In the this scenario, I would define a clean room as one with both the books neatly stacked on the desk, and in a messy room you can find one of the books on the floor.

Now, if I were to try enumerate the number of ways of organising the books in a clean and a messy room, I would find that there is exactly one way in which I can achieve a "clean room configuration", and that is by having both books on the desk. On the other hand if you wish to achieve a "messy room configuration", you can do so in two ways-- either have book 1 on the floor and book 2 on the desk, or have book 2 on the floor and book 1 on the desk.

The no. of ways in which you can achieve a configuration is called its weight and is defined as:

$$W = \frac{N!}{N_0!N_1!N_2!....}$$

In the first case, $W=1$ (clean) and in the second $W=2$ (messy)

The boltzmann formula for entropy is: $ S = k_b\ln W$

Plugging in $W$, we get $$S_\text{clean} = 0$$ and $$S_\text{messy} = k_b\ln2$$

Thus, the messier room has a higher entropy. This somewhat "silly" argument can be reproduced for a room with more furniture, and more items in it. Feel free to toss in some clothes, papers, chairs, book cases, toys, shoes, pets...go nuts.

So, I believe the analogy holds good. But again, what I did here is a silly exercise to illustrate a point, and it isn't meaningful as such.

If you wish to learn more, watch this lecture by Leonard Susskind


Why do we scratch the walls of test tube in test for magnesium or calcium basic radicals? In test for calcium ion, we add ammonium carbonate in presence of ammonium hydroxide and then white precipitate (of $\ce{CaCO3}$) is formed which is dissolved in acetic acid and then reacted with ammonium oxalate to form $\ce{CaC2O4}$. But it says you have to scratch the walls of the test tube to get the precipitate. Why is it so?

Similarly, for magnesium ion we add ammonium hydrogen phosphate to get white precipitate. For this reaction too, the walls of the test tube are to be scratched to get the precipitate.

I read somewhere that scratching creates a surface for recrystallization of the precipitate. I just don't really understand what does it mean. Can someone explain it?

Scratching causes microscopic shards of glass to chip off the scratched surface which can break the surface tension of the liquid, allowing the solute to precipitate/crystallize from of the supersaturated solution.


What is the reaction between Magnesium chloride and Sodium bicarbonate? For some reason it has been very difficult to find the right answer for this question as there's so much conflicting information to be found online. Some mention a reaction where carbonic acid is made, others mention carbonates and hydroxide end product but nothing seems to be precise.

But can the reaction be as simple as this (a double displacement reaction) where magnesium bicarbonate and sodium chloride are the end products:
$$
\ce{2NaHCO3 + MgCl2 ->  2NaCl + Mg(HCO3)2}
$$

Having actually perform this reaction, I can say with certainty that when you mix aqueous sodium bicarbonate with aqueous magnesium chloride, there is no apparent reaction. This is consistent with an equilibrium reaction (actually, just ions):

$$\ce{2 NaHCO3 + MgCl2 <=> 2 NaCl + Mg(HCO3)2}$$

However, upon brief heating in a microwave (for example), a white precipitate of magnesium carbonate develops indicating the reaction has moved to the right.

$$\ce{2 NaHCO3 + MgCl2 ->[Heat] 2 NaCl + MgCO3(s) + CO2 (g) + H2O}$$


What are the alternatives to the Periodic Table of the Elements? What are the different diagrams/tables used to organize the elements other than the Periodic Table? What are the advantages/disadvantages of each?

One periodic table that apparently physicists find useful is ADOMAH periodic table, for its usefulness is finding electron configuration of a particular element. More information at this guide. A disadvantage to this table, along with any new table configuration you have not experienced, will be the lack of knowledge in the trends on your new table. There are many spiral looking tables, but the trends such as the change in atomic radii, or electronegative characteristics, might not be as obvious, and you might have to find a guide or relearn old tricks to methods you would learn in your first year of chemistry. At least the ADOMAH Table is built in such a way that these trends should still be recognizable, although clearly not the same as the normal table you find in any high-school science classroom.

Of particular interest to me are dynamic periodic tables. These are tables that are basically computer applications which will change according to what you are looking for. One outstanding example is the widely popular ptable.com which will maintain all the common trends you probably know already, and dynamically show you various states of the elements when changing the temperature. I learned from this table that rhenium has the highest known boiling temperature, while I thought it was tungsten (although tungsten has the highest melting temperature of the transition metals). And better yet, carbon has the highest melting temperature of all elements, very cool table! The disadvantage is obviously that you cannot carry it around in your notebook without some flavor of a mobile computer and/or internet connection.

I am sure there are many other great tables still to be found, but I hope you have learned something new from my post.


Naming double bond sidechain In IUPAC naming double bonds take preference over an alkane chain of any length, right? So the above should be named as 2-ethylnon-1-ene.

However, when I verify with chemdraw it gives the answer as 3-methylenedecane. I cannot seem to find any mention of double bond side chains as alkylene anywhere on the internet... Is this common nomenclature, and if not why would chemdraw name it as this?

It looks like you're correct.

This molecule has been given 2-ethyl-1-hexene or 2-ethylhex-1-ene

Another name for your molecule would be 3-methylidenedecane.

Added from Wikipedia:

The trivial name carbene is the preferred IUPAC name.

The systematic names methylidene and dihydridocarbon are constructed according to the substitutive and additive nomenclatures, respectively.


Gibbs free energy can be expressed as a function of P,T and n but are enthalpy and internal energy also (partially) functions of n? $$G=G(P,T,n)$$
$$\mathrm dG=V\,\mathrm dP-S\,\mathrm dT+\mu\,\mathrm  dn=\left(\frac{\partial G}{\partial P}\right)_{T,n}\,\mathrm dP+\left(\frac{\partial G}{\partial T}\right)_{P,n}\,\mathrm dT+\left(\frac{\partial G}{\partial n}\right)_{T,P}\,\mathrm dn$$

This allows open systems to be considered (where $\mathrm dn$ does not equal zero). However, can enthalpy, internal energy and Helmholtz free energy also be treated in this way to allow for open systems?

Yes, you can define chemical potentials as partial molar enthalpies, or partial molar internal energies, or partial molar Helmholz free energies.  (Sorry for my initial confusion on this point ... see also my answer on your other question on this topic, here).

The definitions as partial molar enthalpies and partial molar internal energies are not particularly useful, since they involve working at constant entropy, which is practically impossible, but they are still mathematically valid.

The definition of chemical potential as partial molar Helmholz free energy at constant temperature and volume may actually have some practical applications, (e,g. perhaps for reactions that evolve gases in closed containers? although I have never come across any.


Ratio of conjugate bases of a tricarboxylic acid A tricarboxylic acid $\ce{H3A}$ has three different types of $\text{p}K_\text{a}$ $(\text{p}K_\text{a, 1} =2.0,$ $\text{p}K_\text{a, 2}=5.0,$ $\text{p}K_\text{a, 3}=8.0)$.

Identify the ratio of the three charged components($\ce{H2A-}$,$\ce{HA^{2-}}$ and $\ce{A^{3-}}$) in $0.2 ~\text{M}$ solution of tricarboxylic acid $\ce{H3A}$.

$\pu{0.55 M}$ of the strong base $\ce{NaOH}$ is added into a solution of $\pu{0.20 M}$ tricarboxylic acid $(\ce{H3A})$. Calculate the final pH. (Assume that the addition of $\ce{NaOH}$ will not influence the volume of the solution.)

I managed to get $\pu{0.04 M/L}$ for the $\ce{H2A-}$ species but I do not know how to find the concentration of the acids.

Let's use citric acid as an example. Although the $pK_a$ values of citric acid might not necessarily be the same as those above, it will be useful for demonstration.
$$\ce{C6H8O7 + H2O <=> C6H7O7- + H3O+ <=> C6H6O7^2- + H3O+ <=> C6H5O7^3- + H3O+}$$
Throughout the polyprotic acid titration, two protons are donated. The way this problem is solved is you have to use successive $\mathrm pK_\mathrm a$ values and concentrations.
Primary proton transfer equilibrium:
$$\ce{C6H8O7 + H2O <=> C6H7O7- + H3O+} $$
\begin{array} {|c|c|c|c|c|}
\hline
C_\mathrm{initial} & 0.20 & - & 0 & 0 \\
\hline
\Delta C & -x & - & +x & +x\\
\hline
C_\mathrm{final}& 0.20 - x & - & x & x \\
\hline
\end{array}
$$K_\mathrm {a_1} = 0.01 = \frac{[\ce{C6H7O7-}][\ce{H3O+}]}{[\ce{C6H8O7}]} $$
$$ K_\mathrm{a1} = 0.01 = \frac{x^2}{0.20-x} $$
Unfortunately, we can't disregard the x in the denominator so we have to solve the quadratic:
$$ x = 0.04 \ \mathrm{or} \ x = -0.05$$
We disregard our negative root. From $[\ce{C6H7O7}]^- = [\ce{H3O+}]$:
$$[\ce{C6H7O7}-] = \pu{0.04 mol L-1}$$
Now we use our second $\mathrm pK_\mathrm a$ to find the concentration of $\ce{C6H6O7^{2-}}$. Since $K_\mathrm {a_2} << K_\mathrm {a_1}$, we can assume that the hydronium ion concentration is unchanged by the second deportonation.
$$\ce{C6H7O7- + H2O <=> C6H6O7^2- + H3O+} $$
\begin{array} {|c|c|c|c|c|}
\hline
C_\mathrm{initial} & 0.04 & - & 0 & 0.04 \\
\hline
\Delta C & -x & - & +x & +x\\
\hline
C_\mathrm{final} & 0.04 - x & - & x & 0.04 + x \\
\hline
\end{array}
$$ K_\mathrm {a_2} = .00001 = \frac{x(0.04 + x)}{0.04-x} $$
$$ x = -0.04002 \ \mathrm{or} \ x = 9.995 \times 10^{-6} $$
Ignoring the negative root:
$$ [\ce{C6H6O7^2-}] = \pu{9.995 \times 10^{-6} mol L-1}$$
Finally, assuming that the concentration of hydronium ions and the conjugate base of citric acid after the 2nd deprotonation is unaffected by additional deprotonation, we can write up our equilibrium table again:
\begin{array} {|c|c|c|c|c|}
\hline\
C_\mathrm{initial}& 9.995*10^{-6} & - & 0 & 0.04 \\
\hline
\Delta C & -x & - & +x & +x\\
\hline
C_\mathrm{final} & 9.995*10^{-6} - x & - & x & 0.04 + x \\
\hline
\end{array}
$$ K_\mathrm {a_3} = .00000001 = \frac{x(0.04 + x)}{9.995 \cdot 10^{-6} -x} $$
$$ x = -0.04002 \ \mathrm{or} \ x = 2.49875 \times 10^{-12}$$
$$ [\ce{C6H5O7^3-}] = \pu{2.49875 \times 10^{-12} mol L-1}$$


Which atoms form which elements? Is there a diagram/flow chart available that shows which atoms fused together to make each of the elements?  I can find examples for some elements, but nothing which shows all the combinations of atoms which make up each of the naturally occurring elements.

The general consensus is that each element is made up of unique atoms of that element's type. I address nucleosynthesis below, but this is not how elements become elements.
For example, oxygen is made of oxygen atoms. Titanium is made of titanium atoms. Neon is made of neon atoms.
The first good atomic theory, formulated by John Dalton and refined by Amadeo Avogadro and others has a the following principles (some of which are invalidated by modern experiments):

All matter consists of atoms as fundamental building blocks (we now know that subatomic particles are the real fundamental building blocks).
Atoms cannot be created or destroyed (we now have the ability to do this by nuclear fusion and fission).
Each element has its own characteristic type of atom (still valid).
All atoms of the same element element are chemically and physically indistinguishable (the discovery of isotopes invalidated this one).
Chemical reactions occur by the rearrangement of atoms (this one is still valid).

More modern atomic theory takes into account the discoveries of the subatomic particles and the development of quantum mechanics.
Atoms are now defined by their composition of the three principle subatomic particles: protons, neutrons, and electrons. Any atom can be characterized by three numbers derived from this composition:

Atomic Number ($Z$) - equal to the number of protons.
Mass Number ($A$) - equal to the sum of the numbers of protons and neutrons ($A=Z+N$).
Charge ($Q$) - equal to the difference between the number of protons and the number of electrons ($Q=Z-E$).

The atomic number, $Z$, defines to which element the atom belongs. For example, all atoms with $Z=1$ are hydrogen atoms. All atoms with $Z=6$ are carbon atoms. The periodic table is arranged by increasing atomic number. The atomic number controls the number of electrons an atom prefers (see below), and therefor controls the majority of the chemistry of an atom.
The number of neutrons (mass number) defines the isotope. For example, there are three isotopes of hydrogen, with $A=1$, $A=2$, and $A=3$: protium, deuterium, and tritium. These three isotopes have 0, 1, and 2 neutrons respectively, but they each have only one proton, so each of them is a hydrogen atom. The mass differences influence the properties of these isotopes subtly, but not in such a way that is usually noticeable to the average human.
The number of electrons (or more precisely the charge) determines whether the atom is neutral or an ion(http://en.wikipedia.org/wiki/Monatomic_ion). The number of electrons controls the behavior of an atom chemically (for example the number of covalent bonds that can be formed), and the number of electrons of a neutral atom is the same as the atomic number $Z$. For example, a carbon atom has 6 electrons (and you can go further into the quantum mechanics of electron configuration to learn that two of those electrons are less accessible than the other four, which accounts for most of carbon's chemical behavior). Carbon can have a number of electrons not equal to $Z$. If carbon were to have 7 electrons, the charge on carbon would be $Q=Z-E=6-7=-1$, making carbon an anion. Carbon behaves differently as an anion than it does as a neutral atom.
Atoms of some elements can be made by nuclear reactions. For example, in the sun hydrogen is converted into helium by the following pathway (and others). The notation below follows the generic example $^A_Z \ce{X}^Q$, where $X$ is the atomic symbol (and here $n$ means a neutron with $A=1$ and $Z=1$ and $Q=0$ and $e$ means an electron with $A=0$ and $Z=0$ and $Q=\pm1$). A proton is represented as $^1_1 \ce{H}^+$.
$$\ce{^1_1H + ^1_1H -> ^2_2He}$$
$$\ce{^2_2He -> ^2_1H + e+}$$
$$\ce{^2_1H + ^1_1H -> ^3_2He}$$
$$\ce{^3_2He + ^3_2He -> ^4_2He + ^1_1H + ^1_1H}$$
In addition to the nuclear transformations, gamma rays, neutrinos, and heat and being generated and flung around like it was nothing.
The number of possible nuclear reactions is humongous. I would go check out a book on nuclear reactions or nuclear chemistry from a library to learn more.


Why can't rust form without water? Shouldn't iron oxide be able to form without water? It is just iron and oxygen. I don't really understand what the dot followed by the $\ce{H2O}$ means either. I was reading on wikipedia, but I have a rather terrible understanding of chemistry.
What purpose does water serve in rust formation, and what does it mean for a molecule to be hydrated-how is that different then just being surrounded by water molecules?

We could explain it from an electrolytic point of view.

Iron has a certain tendency to dissolve in water according to the following equation:

$\ce{Fe (s) -> Fe^2+ (aq) + 2e^-}$

The release of electrons causes a small current to flow in the metal (Remember that metals conduct electricity). This turns the point where iron dissolves into an anode, and the region around this area becomes rich in electrons, a cathode. Now, we have to remember that water dissociates to small extents according to:

$\ce{H2O (l) <=> H^+ (aq) + OH^- (aq)}$

If it was an acid, the $\ce{H^+}$ would have quickly taken all the electrons liberated from the dissolution of iron earlier, but the concentration of hydrogen ions is not large enough in water, so that we get another reaction taking place at the cathode [The hydrogen ions actually 'gather up' near to the surface but cannot do anything, creating a thin protective layer of hydrogen ions around the cathode, but not strong enough to prevent further reaction]. This reaction involves oxygen and other water molecules:

$\ce{2H2O (l) + O2 (g) + 4e^- -> 4OH^- (aq)}$

Now this is the $\ce{OH^-}$ that reacts with the earlier liberated iron (or ferrous) ions:

$\ce{Fe^2+ (aq) + 2OH^- (aq)->Fe(OH)2 (aq)}$

Now more oxygen will react to oxidise the iron (II) hydroxide...

$\ce{Fe(OH)2 (aq) \xrightarrow[O] Fe(OH)3 (aq) \xrightarrow[O] Fe2O3 (s)}$

You probably can recognise the last product.

As the $\ce{Fe^2+}$ are consumed, more iron will get dissolved (Le Chatelier's Principle) and keep the whole process going.

Maybe a little picture to go along with this :)