React these iron oxide, ammonium nitrate, and sucrose Two questions: what reaction forms between ammonium nitrate, black iron oxide, and sucrose; and a story related question.

So here's the whole story:

So I was playing around a few months ago and stumbled upon a mixture that generated something similar to Black Snakes (firework).
I recently decided to demo this mixture to someone, but I could not find my notes about the experiment, so I decided to try to replicate the mixture from what I remembered.

I am certain that the mixture contained ammonium nitrate, black iron oxide, and some sort of fuel (probably sucrose, impure in the form of powdered sugar).
Combining these three ingredients (one part by volume, whatever particle sizes I have), and igniting with a lighter, the mixture does form black snaking ash, and turns red and crumbles as it cools.

Q1: So, what kind of reaction has taken place between the black iron oxide, ammonium nitrate, and sucrose?

If I leave out any one of the ingredients, the ash snake does not form, so all three compounds are definitely involved. If we leave out either the ammonium nitrate or sucrose, little happens with the application of heat.

Here's a picture after the fact.

I haven't had a very good batch since I lost my notes, so maybe I am missing an ingredient, or maybe I just haven't got the right ratios.
Q2: Any Suggestions?^

Edit:
I have been having more success with this mixture (by volume):
1 part ammonium nitrate
1 part potassium nitrate
2 parts black iron oxide
2 parts sucrose

Mix contents and melt over medium indirect heat (skillet is fine); once it is all melted, immediately remove from heat and form into pellet as it cools. The mixture will reduce in volume by several times.
I suggest doing this in small batches since it can self ignite if you are not careful, or do not heat it at the right rate.

If the ammonium nitrate is omitted or substituted by potassium nitrate, the ash does not expand or cohere. So the ammonium nitrate is doing something more than just providing oxygen here.
The reduction of iron oxide seems to make the ash less monolithic, and moderate reductions in iron oxide forms snakes that fall apart quickly during production.

Disclaimer: what I offer here is an educated guess.

$\ce{NH4NO3}$ is known to decompose quite easily, especially if proper catalyst is present. Thus, one of the reactions is

$\ce{NH_4NO_3 = N_2 + 2H_2O + [O]}$

generated oxygen may be used to oxidize sucrose or black iron oxide. I doubt that $\ce{N2O}$ is formed, since it is quite strong oxidizer and decomposes quite easily, especially if redox catalyst is present

$\ce{C_12H_22O_11 + 24[O] = 12 CO_2 + 11 H_2O}$

$\ce{2Fe3O4 +[O] = 3 Fe2O3}$

However, as you said, the snake is black, so the second reaction does not occur

the final process is dehydration of sugar, producing carbon (actually, the carbon may contain significant amount of hydrogen and oxygen)

$\ce{C_12H_22O_11 = 12 C + 11 H_2O}$

at the same time numerous other processes, producing gaseous reducing organics, happens, so I advise to use a fume hood.

When cooled, reducing atmosphere from sugar decomposition vanishes, and black iron oxide reacts with air

$\ce{4Fe3O4 +O2 = 6 Fe2O3}$

Since black iron oxide acts as a catalyst and does not melt on reaction conditions, it is good to use as fine one as you can get and maybe it may be used in smaller amounts. Since ammonium nitrate acts as main energy source here, it is likely that it can be used in larger amounts. Sucrose forms the body of the 'snake', so it may be used in smaller amount.

The 'snake' consists of (partially) carbonized/caramelized sucrose binding other ashes together and leftovers from iron oxide. The expansion of volume results from gases emitted in the processes of decomposition of ammonium nitrate and decomposition of sucrose, though decomposition of ammonium nitrate is probably the main source here.

BTW, thanks for recipe, I think it would come handy for me this year.


What is the energy released in chemical reactions? I have heard that chemical reactions create energy.  I wanted to know how this energy was created (specifically light energy) and how this energy came about.  I wanted to know if the energy produced was made of particles (for example, light energy is made of photons).

My answer I already knew about heat, kinetic, potential, and sound energy.  What I was really looking for was light energy, which is caused by a phenomenon known as chemiluminescence.

Let's start off on common footing by making sure we understand the definition of energy:

Energy is the capacity to do work.

An object may have many types of energy, we'll get to those in a second, but the total energy of an object will be the sum of its kinetic energy and its potential energy.  Kinetic energy is the energy associated with the movement of an object and potential energy is the energy associated with an object's position.

To take a physical example, hold out a book (preferably your heavy, expensive general chemistry textbook) and drop it on the floor (preferably, not on your little brother's toes).  The book falls to the ground and a loud noise is heard.  As the book moved, the potential energy stored in its original position was converted to kinetic energy as it fell to the ground.  Upon hitting the ground the individual atoms that make up the book crashed in to the individual atoms that make up the ground, causing them all to move.  This movement of atoms is thermal energy.

Let's focus on thermal energy for a moment, the majority of chemical reactions, including the combustion reactions you refer to, involve the transfer of thermal energy.  Hold a cup of coffee in your hand; what happens to your hand?  Assuming the coffee was hot, and your hand was not, your hand starts to feel warm.  We might say your hand is heating up.  This thing we refer to as *heat is the transfer of thermal energy; the molecules in the cup of coffee were bouncing around, crashing in to the cup, which made the molecules in the cup crash around hitting your hand.

So now let's turn to chemical energy which is the energy associated with the position of nuclei and electrons in atoms and molecules. Since we are talking about (relative) positions of objects, this energy is a form of potential energy.  Some molecules, such as methane, have nuclei and electrons that are positioned in such a way that they store a large amount of potential energy.  When the positions of these nuclei and electrons are moved, potential energy is converted into kinetic energy.

So in summary, the energy of chemical reactions is best described in terms of potential and kinetic energy, or in other words the relative position and movement of nuclei and electrons.  Often, the energy we experience in chemical reactions can be adequately described by the transfer of heat, and therefore thermochemistry and thermodynamics take up a large portion of our instruction in Chemistry.  Other types of energy, such as electromagnetic radiation, electrochemical, and nuclear, can all be described in terms of potential and kinetic energy, namely the position and movement of objects.


Which will be major and minor alkene formed if E1 reaction occurs on 2-chloro-2,4,4-trimethylpentane? More substituted alkene should be major product while the less substituted as minor product as it is E1 mechanism.
But in my answer sheet it is given reverse, i.e. major as less substituted and minor as more substituted by giving the reason that it is happening because of steric hindrance of tert-butyl group. But I am confused now that it should not be the case because there is no role of steric hindrance in E1 mechanism, and answer should be as usual.
Please help me in this question with other reason.

The major product in this case is favoured due to Steric hindrance in the product. Drawing the full structure of the molecule in the reaction makes this clearer

We can compare the degree of steric hindrance in the major product and the minor product by using A-values. The Wikipedia article gives the following A-values for common substituents:
$$\begin{array}{|c|c|}
\hline
\text{Substituent} & \text{A-value} \\
\hline
\ce{-CH3} & 1.7 \\
\ce{-CH2C(CH3)3} & 2 \\
\ce{-C(CH3)3} & >4 \\
\hline
\end{array}$$
Clearly, the major product has a lower set of A-values for it's substituents (2 & 1.7) compared to the minor product (4 & 1.7). Also, the minor product does not have any scope for forming stereoisomers, as both the groups on one end of the double bond are methyl. If one of them was a hydrogen group, then the E isomer would be the major product.


How do I calculate the molarity of the Na2CO3 solution? My teacher isn't very good at helping me solve this problem. I keep getting the answer, but I don't know why I am getting it. Could you please help me understand this problem and why you should do this specific step?

  A $\pu{1.00 g}$ sample of $\ce{Na2CO3.10H2O}$ was dissolved in $\pu{20.0 mL}$ of distilled water. Additional water was added so as to give $\pu{250 mL}$ of solution. What is the molar concentration of $\ce{Na2CO3}$?

Here is what I did and got the answer and don't understand why I need to do it:

\begin{align}
  \frac{\pu{1 g}~\ce{Na2CO3.10H2O}}{\pu{286 g}~\ce{Na2CO3.10H2O}} &= \pu{0.003447 mol}\\
\frac{\pu{0.003447 mol}}{\pu{0.020 L}} &= \pu{0.174825 M}\\
(\pu{0.174825 M})(\pu{0.020 L}) &= (\pu{0.250 L})(x)\\
x &= \pu{0.013986 M}
\end{align}

$\pu{0.013986 M}$ is the correct answer, but I don't know why and don't understand the question.
If you were to do the problem, could you explain why you did what you did to get to the answer.

The first part makes more sense writing moles explicitly. The first step is determining how many moles of sodium carbonate are in the original $1.00\ \mathrm  g$ sample:

$$\frac{1.00\ \mathrm{g}}{286\ \mathrm{g/mol}} = 0.003447\ \mathrm{mol}$$

The second step is determining the concentration of the first $20.0\ \mathrm{mL}$ solution, i.e. what concentration results from dissolving the $0.00345\ \mathrm{mol}$ of sodium carbonate calculated above.
$$\frac{0.003447\ \mathrm{mol}}{0.0200\ \mathrm{L}} = 0.174825\ \mathrm{M}$$

The third part is simply a dilution – what concentration results when we dilute the first $20.0\ \mathrm{mL}$ solution to $250\ \mathrm{mL}$:
$$c_2 = \frac{c_1V_1}{V_2}$$
$$c_2 = \frac{(0.174825\ \mathrm{M})(0.020\ \mathrm{L})}{(0.250\ \mathrm{L})} = 0.0140\ \mathrm{M}$$


Van der Waals interaction and magnetic dipole dipole interaction I have a conceptual question that bothers me.

From what I understand, van der Waals forces are the primary source of inter-molecular interaction. There are three different possible origins for van der Waals forces: permanent dipole-permanent dipole (Keesom/short range) forces, the permanent dipole-induced dipole (Debije/short range) interactions and induced dipole-induced dipole (London/long range) forces. (Here we talk about electric dipole.)

We also have magnetic dipole dipole interaction which is $\propto \frac{m_1m_2}{r^3}(3\cos^3\theta-1)$ and has quantum mechanical origins such as Pauli's exclusion principle and exchange interaction and etc.

My question is: since both magnetic and electric dipole dipole interactions are interpreted based on vector model mathematical formulation and these interactions happen at atomic level and they are of a similar range, in what cases we consider interaction between electric charge or a dipole, and magnetic dipole?

Or in general, can magnetic dipole-dipole interaction happen along with electric dipole-dipole interaction in a systems?
When I read text books on inter-molecular forces (Israelaschvili, Cosgrov, etc) non of them talk about magnetic interactions. I wonder why.

Magnetostatic interactions are much smaller than electrostatic interactions. For atomic and molecular systems, the magnitude of the magnetic potential energy is roughly $\alpha^2 \approx 5\times 10^{-5}$ times the magnitude of the electrostatic potential energy. Therefore magnetic dipole (and higher orders) are usually neglected.

The multipole expansions for the magnetostatic long-range potentials are formally equivalent to the usual electrostatic long-range, with $\mu_0/4\pi$ replacing the $1/4\pi\epsilon_0$ prefactor, where $\mu_0$ is the vacuum permeability.

References:

M. Lepers and O. Dulieu,  arXiv:1703.02833v2 [quant-ph] https://arxiv.org/abs/1703.02833v2
1 Z.C. Yan, J.F. Babb, A. Dalgarno, and G.W.F. Drake, Phys. Rev. A - At. Mol. Opt. Phys. 54, 2824 (1996).


Pattern to determine the maximum ionic charge for transition elements? According to my textbook, main group elements follow a simple pattern when determining their maximum ionic charge. The maximum cationic charge is always equivalent to their main group number (group number minus 10 for the groups 13 to 18 according to the new IUPAC terminology) while the maximum anionic charge is eight minus that number.

For example, the maximum charge of an aluminium cation (third main group) would be $+3$, the maximum charge of a sulfur anion (sixth main group) would be $-2$.

Some elements form multiple ions, but no charge will be larger than determined by the above method

Is there a similar method or pattern to determine maximum charges of transition metals? Why or why not?

The pattern you observe for main group elements is due to the number of valence electrons present in the s and p subshells of the outermost shell. The most you can do is either lose all those s and p electrons (giving a ‘cation’[1] charged exactly according to the old main group number — new group number minus 10 for groups 13 through 18) or you can fill these up to a total of eight.

For transition metals, a d subshell is in the process of being filled. Since the d subshell of a lower shell is between the s and p subshells of a higher shell according to the aufbau principle, we can imagine that the highest number of ‘usable electrons’ would be 18. This is not typically the case, however.

For the first few transition metals, you can remove as many electrons as are present in the s and d valence subshells (e.g. 4s and 3d). In the 3d series, manganese is the last one that can do this to give manganese(VII), in the 4d series ruthenium can reach ruthenium(VIII) and in the 5d series iridium can reach iridium(IX).[2] For the remaining elements of a certain transition metal period, the highest accessable oxidation state again decreases. You can find an overview of the maximum oxidation states known for certain elements on Wikipedia.

The only elements that again follow a somewhat more simple scheme are the zinc group (zinc, cadmium and mercury). Since these have a completely filled d subshell in ground state, they are reluctant to give away any d electrons; hence their highest observed charge is $2+$, as if they were in group 2. (Indeed, they behave like group 2 metals in a lot of other respects, too; except mercury which is slightly different again.)

When it comes to the highest possible negative charge or oxidation state, things are more complicated yet again. Since all transition metals are metals, they are not commonly found as anions or anionic particles although some are more likely to than others. For some of these — typically the iron group and the two following groups — this corresponds to $12 - \text{group number}$. This small trend is more of an exception than of a rule, though. Note, for example, how zinc, cadmium and mercury are all known as in the $\mathrm{-II}$ state which does not make sense immediately just looking at the respective shells and subshells.

Many of the actually observable oxidation states can be rationalised by doing a complex molecular orbital analysis of the corresponding compounds. However, for most of these there is no simple and immediately obvious pattern that one can use — with the notable exception of the small bit of pattern outlined in the third paragraph.

Notes:

[1]: A better term here would be oxidation state. For many elements the highest possible free cation is not known but a compound with a corresponding oxidation state is.

[2]: Again, none of these are observable as free cations in standard chemistry; rather they are in compounds. Hence I have stopped using the term charge and started using oxidation state.


Stability of transition metals in aqueous solution Which is more stable in aqueous solution: $\ce{Cr^3+}$ or $\ce{Mn^3+}$? Why?

My approach: $\ce{Cr^{3+}}$ should be more stable as the $\ce{3d}$ electrons will enter the $\mathrm{t_{2g}}$ orbitals. Due to CFSE (Crystal Field Stabilization Energy) of half filled $\mathrm{t_{2g}}$ orbitals $\ce{Cr^{3+}}$ should be most stable.

Both elements are stable under certain conditions in various oxidation states. In chromium they range from -2 to +6 and in manganese from 0 to +7.

However, some of them are more common than others, especially in aqueous solution. In chromium the most common are +3 and +6. As a "free" ion $\ce{Cr^{3+}}$ is most stable due to a filled s orbital and only one electron in the d shell. If you dissolve chromium(III)chloride in water you will get a complex formation.
$$\ce{CrCl3 + 6H2O <=> [Cr(H2O)6]^{3+} + 3Cl-}$$
Assuming an octahedral ligand alignment in crystal field theory, you are correct in assuming, that the $t_{2g}$ orbitals will be half filled, resulting in a more stable configuration.

For manganese the most common oxidation states are +2, +4 and +7. If you dissolve manganese(III)chloride, then you will also form a aqueous complex, which will slowly transform to manganese(III)hydroxide and precipitate.
$$\ce{MnCl3 + 6H2O <=> [Mn(H2O)6]^{3+} + 3Cl- <=> Mn(OH)3 v + 3H3+O + 3Cl-}$$
The resulting hydroxide may be oxidised by air to manganese(IV)oxide hydroxide.
$$\ce{2\overset{\color{\green}{+III}}{Mn}(OH)3 + \overset{\color{\green}{\pm0}}{O2} -> 2\overset{\color{\green}{+IV}}{Mn}O(OH)2 + 2{}^{-}OH}$$
Over time this will convert into manganese dioxide.
$$\ce{MnO(OH)2 -> MnO2 + H2O}$$


Isobaric entropy relation from isothermal Gibbs free energy relation I have been given a problem to derive a relation for change in entropy from change in Gibbs free energy:

Starting with the isothermal equation
$$G(p_2) = G(p_1) + nRT\ln\frac{p_2}{p_1}\label{eqn:1}\tag{1}$$
derive
$$S(V_2) = S(V_1) + nR\ln\frac{V_2}{V_1}\label{eqn:2}\tag{2}$$
using the relation
$$\mathrm dG = V\,\mathrm dp - S\,\mathrm dT.\label{eqn:3}\tag{3}$$

The derivation wants you to assume that pressure is constant and thus $\mathrm dp = 0,$ and \eqref{eqn:3} can be arranged to
$$\left(\frac{\partial G}{\partial T}\right)_p = -S.\label{eqn:4}\tag{4}$$
Now I understand that then differentiating \eqref{eqn:1} w.r.t $T$ and using the fact $p_1 V_1 = p_2 V_2$ (for constant $T)$ can produce the required equation \eqref{eqn:2}.
However, I don't understand quite how we can justify the derivation of an isobaric equation from an isothermal one as this method appears to. Equation \eqref{eqn:1} describes isothermal processes and is a function of pressure, $p_2.$ How is it valid then to suddenly say now pressure is constant and therefore use \eqref{eqn:4} when in equation \eqref{eqn:1} the pressure is clearly not constant?
Might it have something to do with Gibbs free energy being a state function meaning, once there is a relationship for it, constraints (for different state variables) can be changed to derive new relationships using a different constraint?

For an ideal gas at constant temperature $\Delta H=0$, so $\Delta G=-\Delta (TS)$, and, at constant temperature, $$\Delta G=-T\Delta S$$


Is it possible, in principle, for a second row homonuclear diatomic system to show a bond order of 4 in an excited state? Is it possible, in principle, for a second row homonuclear diatomic system to show a bond order of 4 in an excited state?  If so show the energy level diagram for such a system?

I am confused about this problem, mainly because I don't understand what it would mean for a diatomic system to be in an excited state.  Could I just move an arbitrary number of electrons from the n=2 energy level to the n=3 level?

The normal excited state is one in which a single electron is promoted from a lower orbital into one of the anti-bonding ones. Thus the situation is initially $\cdots g^2$ for 2 electrons in a ground state orbital g (need not be the highest orbital but often is), and, say, after absorbing a photon of the correct energy the situation is $\cdots g^1a^1$ where a is an anti-bonding orbital. This combination of orbitals is called the excited state. (If the electron spin remains paired it is called a singlet excited state, if unpaired ('parallel') it is called a triplet excited state). As one electron is now in an anti-bonding orbital the molecule has overall weaker bonding. (However, the NO excited state may be an exception).

The excited state always decays back to the ground state (it in not a stationary state of the hamiltonian) with a lifetime that depends on the particular molecule or atom and between different molecules the lifetime
can have  a huge range, from less than a picosecond to several seconds.


Why doesn't copper react with hydrochloric acid while the other metals do? 5) Which one of the following metals does not react with hydrochloric acid?

  A. zinc
  B. magnesium
  C. iron
  D. copper
  E. aluminium

The right answer must be copper, but why doesn't copper react with hydrochloric acid while the other metals do?

You might want to look up some terms, such as

noble and less noble metals
reduction potential
galvanic series

Here, reaction means that

hydrogen gas is formed
the metal is dissolved

In order to form hydrogen, protons need to be reduced to hydrogen atoms which then combine to $\ce{H2}$.

$$\ce{2 H+ + 2 e- -> H2}$$

The metal serves as an electron donor and is oxidized, e.g.

$$\ce{Zn -> Zn^{2+} + 2 e- }$$

The more noble a metal is, the more reluctant it is to lose electrons. This is the case for copper, which is therefore not oxidized under these conditions.


How will this reaction proceed? Will there be a hydride shift to form a tertiary carbocation or will it proceed as secondary carbocation?

My text book shows that it would proceed as secondary carbocation
But tertiary carbocation is more stable. I am a bit confused here.
Will someone please clarify?

It will surely proceed as a secondary carbocation and no hydride shift will take place. Because if hydride shift takes place, bridgehead carbon will acquire positive charge i.e. will become unstable.

For detailed explanation check this out:
Why are bridge bonds with negative or positive charge on bridge head unstable?


Is this equation of state original? I created this equation of state as a personal undergraduate summer project (here is the link)  seven years ago:

$$\frac{PV}{nRT}-1 \propto \left(\frac{nRT}{PV}\right)^2 \left[\left(\frac{nRT}{PV}\right)^2 - 1\right]$$

I tried to show it to my professors but none of them replied to my emails. I had since forgotten about it until I recently came across it in the cloud. Here is a table comparing its theoretical predictions for the critical temperatures of various gases to the actual experimental values

Yes, this equation of state is original. While I cannot absolutely guarantee that it has never been used or proposed before, I can say that I have not encountered it in any “classical” model — and I teach thermodynamics and statistical physics, so I have seen many textbooks and classical exercices :)

Extending beyond the question of originality, I read your demonstration and what bothers me to some extent is the link you establish between $Z$ (the compression factor) and $U(r)$, the intermolecular potential energy. First, let's state that linking $Z$ to $U$ is indeed a common way to derive equations of state. However, the way you are doing it is lacking a sound basis. The sentence “From these explanations it can be said that $Z-1=U(r)$” (page 1) is the problem:

Yes, if $U(r)>0$ then $Z>1$, and the other way around… but that does not mean that the two quantities are proportional.
In fact, $Z$ is a scalar (function of $P$ and $T$), while $U(r)$ is a function of distance (and not thermodynamic variables). So they cannot be linked directly.

You intuited between $Z$ and $U$, which exists, but it needs to be formalized a bit more carefully. Standard approaches to do so rely on perturbation theory or mean-field method. You can check for examples the classical derivations of the van der Waals equation of state to see how this can be done. But it requires some fundamentals of statistical physics…


Sodium by electrolysis through glass In his 1926 thesis Robert Cady Burts describes the production of sodium by electrolysis of sodium nitrate through glass.

I can't find much else on this method. Are there any modern descriptions of the experiment? Videos maybe?

See Shakashiri's chemical demonstrations. He carefully outlines the steps. As I recall it requires taking sodium nitrate into a molten state and applying a current. The elemental sodium is precipitated on the inside surface of an incandescent bulb.


How to add hydrogens to a crystal structure? PDB protein crystal structures most of the time lack hydrogen atoms. Usually I use PyMOL (h_add) but it does not always seem to be giving the correct result. What can I use for accurate hydrogen addition?

I've used Reduce before, which is excellent. I'm not sure what you mean by ' does not always seem to be giving the correct result', though? Perhaps you could file a bug report with the PyMOL team if you think there is a bug in h_add.


Why is the dissociation reaction of acetic acid in water initially endothermic and then exothermic? Why is the dissociation reaction of acetic acid in water initially endothermic as temperature is increased and then becomes exothermic as it is increased further?
My understanding from Le Chatelier's principle is that a rise in temperature "adds" heat to the system and a drop in temperature "removes" heat from the system. Correspondingly, a temperature increase would favour the endothermic (heat absorbing) direction and a temperature decrease would favour the exothermic (heat releasing) direction. Why is this contradiction? My thoughts are that maybe the autoionisation of water is endo or exo, which somehow dominates the reaction, but I don't know. Could someone please provide a possible explanation?

I have found a paper which does this same experiment and provides a discussion of this phenomenon[1]. I will begin by just quoting from the discussion section:

  It has been stated in the introductory section that a
  maximum in the formal Cp° for a pair of thermally
  stable free ions can be ascribed to an endothermic
  change in the solvent shell around one of the ions.
  However, the physical description of that change can
  cover a wide range of phenomena. At one extreme,
  the change might be a genuine chemical reaction: for
  instance, an octahydrate becomes a hexahydrate. At
  the other extreme, it might be merely an artifact resulting from a peculiar energy level diagram for a single
  species.

Intuitively, one would expect the reaction to stay either endothermic or exothermic with only small changes due to the relative thermal populations of various vibrational modes in the reactants and products. This picture is more relevant to the gas phase. Thus, the authors are stating that this kind of large change in the behavior of a reaction must be related to the environment which each ion experiences, namely, the solvation environment.

The authors go on to give diagrams for each of the cases they describe above. They stress that in the case where a genuine chemical change occurs, you are essentially moving from populating one set of states to another set of states entirely.

On the other hand, and I suspect this is more commonly the case when water is the solvent, the density of states might increase quite markedly as energy increases. That is the solvent may begin exciting librational modes quite strongly or even move to a regime where the solvents can more or less freely rotate. This behavior of the solvent would be very sensitive to temperature. The authors are very careful to note that each of these situations can result in the same thermodynamic outcome, so one can determine which is happening on the basis of thermodynamic data alone. Nowadays, we could easily run a molecular dynamics simulation or carry out some kind of 2-dimensional IR experiment and extract meaningful kinetic information to supplement the thermodynamics.

So, to summarize, this large change in enthalpy which crosses from exothermic to endothermic is almost certainly due to large changes in the solvation environment and not large changes with respect to the reactants and products.

References:

[1]: Leung, C. S., & Grunwald, E. (1970). Temperature dependence of. DELTA.-Cp. deg. for the self-ionization of water and for the acid dissociation of acetic acid and benzoic acid in water. The Journal of Physical Chemistry, 74(4), 687-696.


What makes banana bonds possible in diborane? Diborane has the interesting property of having two 3-centered bonds that are each held together by only 2 electrons (see the diagram below, from Wikipedia).  These are known as "banana bonds."

I'm assuming there is some sort of bond hybridization transpiring, but the geometry doesn't seem like it is similar to anything I'm familiar with Carbon doing.  What sort of hybridization is it, and why don't we see many (any?) other molecules with this bond structure?

Look carefully, it's (distorted) tetrahedral--four groups at nearly symmetrically positions in 3D space{*}. So the hybridization is $sp^3$.

As you can see, the shape is distorted, but it's tetrahedral. Technically, the banana bonds can be said to be made up of orbitals similar to $sp^3$ but not exactly (like two $sp^{3.1}$ and two $sp^{2.9}$ orbitals--since hybridization is just addition of wavefunctions, we can always change the coefficients to give proper geometry). I'm not too sure of this though.

$\ce{B}$ has an $2s^22p^1$ valence shell, so three covalent bonds gives it an incomplete octet. $\ce{BH3}$ has an empty $2p$ orbital. This orbital overlaps the existing  $\ce{B-H}$ $\sigma$ bond cloud (in a nearby $\ce{BH3}$), and forms a 3c2e bond.

It seems that there are a lot more compounds with 3c2e geometry. I'd completely forgotten that there were entire homologous series' under 'boranes' which all have 3c2e bonds (though not the same structure)

And there are Indium and Gallium compounds as well. Still group IIIA, though these are metals. I guess they, like $\ce{Al}$, still form covalent bonds.

So the basic reason for this happening is due to an incomplete octet wanting to fill itself.

Note that "banana" is not necessarily only for 3c2e bonds. Any bent bond is called a "banana" bond.

Regarding similar structures, $\ce{BeCl2}$ and $\ce{AlCl3}$ come to mind, but both of them have the structure via dative(coordinate) bonds. Additionally, $\ce{BeCl2}$ is planar.

Sneaks off and checks Wikipedia. Wikipedia says $\ce{Al2(CH3)6}$ is similar in structure and bond type.

I guess we have less such compounds because there are comparatively few elements ($\ce{B}$ group pretty much) with $\leq3$ valence electrons which form covalent bonds(criteria for the empty orbital). Additionally, $\ce{Al}$ is an iffy case--it like both covalent and ionic bonds. Also, for this geometry (either by banana bonds or by dative bonds), I suppose the relative sizes matter as well--since $\ce{BCl3}$ is a monomer even though $\ce{Cl}$ has a lone pair and can form a dative bond.

*Maybe you're used to the view of tetrahedral structure with an atom at the top? Mentally tilt the boron atom till a hydrogen is up top. You should realize that this is tetrahedral as well.


Proportion of gases in Ideal Gas Mixture? Suppose methane ($CH_4$) and ethane ($C_2H_6$) in a container. The specific gravity is $r_g=0.6$. I think this means $0.6=\frac{\rho_{CH_4}}{\rho_{C_2H_6}}$. Now I am trying to understand the question.

  Question: what is the proportion of gases?

Solution according to my teacher: $C_{CH_4}=0.1$ and $C_{C_2H_6}=0.9$

My Solution: $C_{CH_4}=24.2\%$ and $C_{C_2H_6}=75.8\%$

  Values (Curtis Phase Behaviour book p.162)

    $M_{CH_4}=16.04 \frac{lbm}{lbm-mol}$
    $M_{C_2H_6}=30.07 \frac{lbm}{lbm-mol}$

  What is $x_{CH_4}=\frac{n_{CH_4}}{n_{CH_4}+n_{C_2C_6}}=\frac{n_{CH_4}}{n}$?

    By Amagat's law and Dalton's law here,
    I know $\frac{n_j}{n}=\frac{P_j}{P}=\frac{V_j}{V}$.
    Specific gravity $0.6=\frac{\rho_{CH_4}}{\rho_{C_2H_6}}$ where $\rho=\frac{m}{V}$ and constant $V$ so
    $0.6=\frac{m_{CH_4}}{m_{C_2H_6}}$ so $m_{CH_4}=0.6m_{C_2H_6}$.
    $PV=nRT\rightarrow n=\frac{PV}{RT}$ where constant $\frac{V}{RT}$ so
    $x_{CH_4}=\frac{P_{CH_4}}{P_{CH_4}+P_{C_2C_6}}=\frac{P_{CH_4}}{P}=_{(1)}\frac{n_{CH_4}}{n}=\frac{V_{CH_4}}{V}$.
    By $M_i=\frac{n_i}{m_i}$ so $n_i=M_i m_i$ where $n_{CH_4}=M_{CH_4}m_{CH_4}$

  so

  $\frac{n_{CH_4}}{n}=\frac{M_{CH_4}m_{CH_4}}{M_{CH_4}m_{CH_4}+M_{C_2H_6}m_{C_2H_6}}=_{(2)}\frac{M_{CH_4}0.6m_{C_2H_6}}{M_{CH_4}0.6m_{C_2H_6}+M_{C_2H_6}m_{C_2H_6}}=\frac{0.6M_{CH_4}}{M_{CH_4}0.6+M_{C_2H_6}}=0.242455$

  $\rightarrow\frac{n_{C_2H_6}}{n}=1-\frac{n_{CH_4}}{n}=0.757545$

  where the results are not 0.1 and 0.9.

Your interpretation of specific gravity is incorrect:

  Specific gravity (SG) refers to the ratio between a given density and that of a reference.  Usually it is defined for liquids with respect to water and gasses with respect to air so in this case

    $0.6=\frac{n_{CH_4}\rho_{CH_4}+n_{C_2H_6}\rho_{C_2H_6}}{\text{air}}$

  where $n_i$ are proportions by mole factor. This is the incorrect assumption

    $0.6=\frac{\rho_{CH_4}}{\rho_{C_2H_6}}$.

  Gasses are always a little dicier since the reference is much more important with changes in temperature and pressure affecting the resulting density.

Example

  The density form of the ideal gas laws goes like:

    $\rho = \text{mass / volume = molecular weight * pressure / constant / temperature}$.

  Therefore for constant temperature and pressure, the specific gravity
  is equal to the ratio of molecular weights.  This follows because
  $\frac{\rho_1}{\rho_2} = \frac{MW_1}{MW_2}$ at constant conditions.

Equations and Solution

    $M_{\text{air}} = 28.97$ (I got $28.84$ using the simple $0.21*32+0.79*28$ and this agrees with  this)

    $\text{SG}=0.6=\frac{n_{CH_4}\rho_{CH_4}+n_{C_2H_6}\rho_{C_2H_6}}{\text{air}}$

    $1=n_{CH_4}+n_{C_2H_6}$

  so by the first equation and the second equation the density of the mixture is $0.6 * 28.97 = 17.38$. The third equation holds only if $0.9*16 + 0.1*30 = 17.4$ so

    methane is 90% and ethane is 10%.


Mechanism of acid-catalyzed Robinson annulation? My professor posed my organic chemistry class a challenge problem for bonus points and I've been racking my brain over the last three days trying to work it out and I want to know if my solution is plausible. The problem asks us to draw the best curved arrow mechanism for the reaction in the bottom left corner of the photo included in this post (a substituted cyclobutane molecule to a substituted cycloalkene). At first, this seemed like a simple intramolecular Robinson annulation, however it appears to be catalyzed using the hydronium ion, rather than the hydroxide ion. No other reactants are stated, including light, heat, nor a double equivalence of the starting compound. Working through every mechanism that I could think of, I eventually ended up with the solution in the picture below. I assumed that reaction rates and percent yields were unimportant, as long as the mechanism is plausible. Does this solution break any rules of organic chemistry? Is there a better solution?

As I mentioned in my Comment the only serious issue was the formation of an allylic anion in acid medium. You seem to know how to form enols from ketones and how to form ketones from enols. I will ignore those steps in the diagram. Protonation of the carbonyl in 1 can cause ring fragmentation to form transient enol 2 that rapidly tautomerizes to diketone 3. Cyclization via enol 4 leads to β-hydroxyketone 5. Although loss of water from 5 as you have described is viable, I prefer formation of the enol 6 because loss of water is facilitated over a tertiary cation given the allylic nature of the species 7. Loss of a proton from carbocation 7 leads to 3-methylcyclohex-2-en-1-one (8).
Addendum: The process is not a Robinson annulation, which involves an initial Michael addition followed by an aldol condensation. The latter reaction includes the transformation 3 $\rightarrow$ 8. The transformation 1 $\rightarrow$ 3 may described as a retro-aldol reaction.
You may want to see the formation of mesitylene (1,3,5-trimethylbenzene) from acetone under acid conditions.


Does the density of a liquid and gas phase of a substance(nitrogen or water) converge when reaching critical point I know that the there is no difference in density in a liquid and a gas phase of a substance after the critical point. But I do want to know if this difference in density is more continuous or if it is more like a big step when reaching the critical point? I.e if we are like 0.01 Kelvin beneath the critical point of water (with the critical pressure). Is the difference in density still big at this point or is it near zero?

Does the density of a liquid and gas phase of a substance converge when reaching critical point?

Yes. This is exactly what happens. Let's take a look at a qualitative $P-T$ (pressure–temperature) phase diagram of an arbitrary substance:

source: Cansell, F. et al. Supercritical fluid processing: a new route for materials synthesis. J. Mater. Chem. 9, 67–75 (1999). 10.1039/A804964E

As you can see, moving directly from gas to liquid involves crossing the phase boundary. However, if you're doing it in the supercritical, the change in density is continuous. Maybe you can get a better, more quantitative view with this example of $\ce{CO2}$:

source: http://what-when-how.com/nanoscience-and-nanotechnology/microweighing-in-supercritical-carbon-dioxide-part-1-nanotechnology/

This one is also a a $P-T$ phase diagram, with equal density $(\rho)$ lines. You can see that the lines radiating from the gas–liquid boundary lines are not equal on each side. For example, 0.1 and 0.9. However, as you get close to the critical point, $\rho$ on each side of the line converges to a single value: 0.2 and 0.7, 0.3 and 0.6 and so on.

YouTube has some terrific examples of this happening. For example: http://youtu.be/GEr3NxsPTOA
What you see is a function of density - you can see the phase boundary between liquid and gas because they have differing densities. Notice in the videos how the phase slowly and gradually disappears as the density of both phases converges, effectively turning the $\ce{CO2}$ into a supercritical fluid.


Ideal Gas equation and its Assumptions The two assumptions are that the gases are points of mass that move, they have no volume and that there is no interaction between other molecules.

But even the points of mass can collide with each other so wouldn't that be considered interaction between molecules?

Part of the assumptions set of an ideal gas is that there are no attractive or repulsive forces between them.

Also, part of the assumption is that the volume of molecules obtain a negligible amount when compared to the overall volume of the gas.


Illustrator to chemdraw I am not sure that this is the right place for this question, but I'll try.

I am working on a scheme that I am developing on illustrator. I have a lot of chemical structure and plus I have some vectorial pictures included.

The thing is that I would like to save/export my illustrator file in a way that I can reopen it and modify it in chemdraw. I am aware that the opposite procedure is possible.

Is there any way of doing that?

From p. 327 of ChemDraw 18 User Guide:

  Importing

  You can import graphics or documents from other applications into ChemDraw.

  Files
  Files that you insert become part of the drawing. You can edit a file only if it is in a ChemDraw format.
  To insert a file:

  Navigate to Edit → Insert File.
  In the Open dialog, select the file type from the drop-down list.
  Select the file and click Open. The file is embedded in the drawing.

Further, there is a table of file formats ChemDraw is capable of importing and exporting, from which it is evident that only the following graphical formats can be imported:

Bitmap (BMP)
Graphic Image Format (GIF)
JPEG (JPG, JPEG)
Portable Network Graphics (PNG)
Windows Metafile (EMF, WMF)
Tagged Image File (TIF, TIFF)

PDF or PostScript files can be attached as objects, but I suppose it's not what you are after.
Either way, it looks like there is no way to add AI files directly, so you need to convert them first to either raster (e.g. TIFF or PNG) or vector (WMF) format.
I agree with @Ivan that crafting structures in ChemDraw in the first place and then post-processing them with Adobe Illustrator is arguably a better workflow.


Nitrocellulose (flash paper) synthesis without Sulfuric Acid? According to the Wikipedia page on Nitrocellulose:

  The process uses nitric acid to convert cellulose into cellulose nitrate and water:
  $$\ce{3HNO3 + C6H10O5 ->[H2SO4] C6H7(NO2)3O5 + 3H2O}$$
  The sulfuric acid is present as a catalyst to produce the nitronium ion, $\ce{NO2+}$. The reaction is first order and proceeds by electrophilic substitution at the C−OH centers of the cellulose.[8]

Since I have been on quarantine, I have lost access to the lab and therefore must carry out reactions from the comforts of my home. I have some sulfuric acid drain cleaner that is sufficient for many at-home applications. However, it consists of only about 93% $\ce{H2SO4}$, and contains a lot of dark dyes. I am trying to synthesize nitrocellulose paper from letter printer paper and envelopes, for the purpose of developing a self-destructing letter that burns away when the letter is returned to the envelope after being removed. However, the dyes in the sulfuric acid severely discolor the paper, and make it stand out considerably in a pile of mail, defeating the purpose of secrecy. So, at least until the quarantine is lifted, I may need an alternative catalyst to Sulfuric acid to synthesize the nitronium ions in vitro. Does anyone have an alternative to $\ce{H2SO4}$, or a way to eliminate the dyes without distillation? (because I broke my condenser column, and can't use the lab's)

Update: I'm still curious about my question, but it might not even be legal to send nitrocellulose envelopes through the mail. One spark and you could end up with flaming mail processing machines. Yikes!

I do not think it will be possible to form nitrate esters of cellulose without sulfuric acid present, I am sure that you need the strong bronsted acid to make the reaction work. But I think that there is another issue here, your safety / health / the law on energetic materials.

I hate to have to say this, but I do not think it is safe to make energetic materials using nitric acid at home. You should also consider for a moment the question of "is it legal" to make the energetic material at home.

Keep in mind years ago some "home chemists" would make explosives / fireworks and it was regarded as "boys will be boys". But nowadays the "boys in blue" (Police) are very keen to chase after people who make explosives / things which go bang etc.

I think that the distillation of sulfuric acid at home does not sound like a good idea, I think that you would need to work in a good fume hood. Most school labs would be unlikely to have the equipment required to do it.

With all due respect I think you should choose another chemical project to pass the time while you are on "lockdown".


The role of salt in cooking vegetables Today my teacher asked an interesting question that "what is the role of salt ($NaCl$) in cooking the vegetables"? As the class was in funny mode, students gave answer like "it makes vegetables tasty" (which also seems reasonable). When I came home I realised the importance of question, so I am here.

I suppose that addition of salt is related to boiling point of water (If I'm thinking right). But still so confused that I don't know a generalised answer.

I shall be thankful if someone of you can tell me the role of salt in cooking vegetables (Please give as much as information you can).

I think that the main effect of salt in cooking vegetables is to act osmotically to draw water out of the plant cells, reducing the volume of the veggies and hence the time required to cook the vegetables. Perhaps this also changes the flavor and texture in beneficial ways as well. Example: If you soak/cook dry beans in fresh water, they tend to burst open, but remain whole in salted water.


Looking for raw interferogram data / raw FID data from FT-IR / FT-NIR / FT-NMR I'm trying to get my hands on some raw interferogram data / raw FID data from an FT-IR / FT-NIR / FT-NMR so I can run some tests using FFT with it (it needs to be real data). Here's a picture below of the data I'm looking for highlighted in yellow.  I'm looking for Alkali metals (Sodium, Potassium, but also Magnesium, Gold, Silver, Platinum and Palladium).  Any ideas where I can get these or if someone can send them to me as a text/data file?  I've searched NIST but everything is already converted

In The Journal of Chemical Education, published by the American Chemical Society, a rather recent article presented a related project, hosted by the University of Minnesota.  Intentionally it is about raw FIDs-as-obtained, out of the NMR spectrometer, aiming to teach correct data processing (including spectra with two orthogonal frequency scales). The article is sponsored by the authors (i.e. freely accessible), the database requires an e-mail to register.  Reference: J. Chem. Educ., 2013, 90 (7), pp 941–943; DOI: 10.1021/ed300787v


Organic compounds that react with carbonate ions I want to remove carbonate ions from dissolved potassium carbonate (potash) by letting an organic compound react with the carbonate ions. I don't have a very good knowledge of organic chemistry, but are there certain types of organic molecules that can react with these carbonate ions to form new compounds?

Additionally, are there reactions in which the carbon atom from the carbonate ion can be inserted into the main carbon chain of another organic compound (making, for example, propane from ethane)?

Cook it up with ethylenediamine or diethylenetriamine- but don't bet on it.  What do you plan to have as the counterion for $\ce{K^+}$?  Check your carbon oxidation states for the second half of your question.

No, wait!  Add citric acid.  Yeah, that's the ticket.  Plop, plop, fizz, fizz.


What should be the name of this substituted cyclopentane? I think its name is 5-amino-3-bromo-4-chloro-2-hydroxycyclopentan-1-carbonitrile. Is that correct?

According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), the preferred suffix for the principal characteristic group $\ce{{}-CN}$ is ‘carbonitrile’. Therefore, the name of the parent structure of the compound given in the question is ‘cyclopentane-1-carbonitrile’ (note that the locant ‘1’ would be omitted in monosubstituted homogeneous monocyclic rings, i.e. ‘cyclopentanecarbonitrile’).

The relevant rules for the numbering of locants for the four substituent prefixes (‘amino’, ‘bromo’, ‘chloro’, and ‘hydroxy’) are:

  P-14.3.5 Lowest set of locants

  The lowest set of locants is defined as the set that, when compared term by term with
  other locant sets, each cited in order of increasing value, has the lowest term at the first
  point of difference; (…)

and

  P-14.4 NUMBERING

  When several structural features appear in cyclic and acyclic compounds, low locants are assigned to them in the following decreasing order of seniority:

  (…)

  (f) detachable alphabetized prefixes, all considered together in a series of increasing numerical order;

  (g) lowest locants for the substituent cited first as a prefix in the name;

  (…)

Note that Rule (f) takes precedence over Rule (g).

In accordance with Rule (f), the compound given in the question could be named as 2-amino-4-bromo-3-chloro-5-hydroxycyclopentane-1-carbonitrile as well as 5-amino-3-bromo-4-chloro-2-hydroxycyclopentane-1-carbonitrile since both names correspond to the locant set ‘2,3,4,5’. However, according to Rule (g), this example is named as 2-amino-4-bromo-3-chloro-5-hydroxycyclopentane-1-carbonitrile rather than 5-amino-3-bromo-4-chloro-2-hydroxycyclopentane-1-carbonitrile since ‘amino’ is cited first as a prefix in the name.


Copper and hydrochloric acid When hydrochloric acid is added to a sample of copper why does it look pink at first then form a pale blue solution ?

If pure hydrochloric acid is added to a sample of pure copper, there is essentially no reaction; the solution should not turn blue.

At a guess, you're seeing the copper turn pink because it has a significant oxide layer that is reacting with the acid, forming copper (II) chloride and leaving bare the pinkish metal.  The solution could be pale blue or pale green, and the time delay of the color would be a result of the amount of time required for the reaction to occur.  Note that if this is the case, you would not expect to see any bubbling on the copper, as copper (II) oxide will react with hydrochloric acid to form copper (II) chloride and water.

As Ivan Neretin pointed out in comments, large pieces of copper will react with hydrochloric acid over a period of several hours to days in the presence of air.  Permeakra added that a fine copper powder will react in minutes with a solution of hydrochloric acid in the presence of air.  In both cases it is the addition of oxygen that causes the copper to react, and in both cases the solution will be a greenish yellow color due to the excess of hydrochloric acid required.

As an example, a student of mine recently asked about this with regard to a penny.  We cut open a modern US penny (95% zinc, 5% copper by mass) to expose the zinc core, and placed it in 200mL of 6M HCl.  The oxidation layer on the penny was gone almost immediatley after it entered the HCl solution.  The zinc reacted quickly and by the next day the penny had been converted into a hollow copper shell.  The shell sat in the remaining HCl solution (still 5.7M) over the weekend with no visible reaction but the solution slowly turned green and the copper dissolved completely by the following Friday.

If your solution is turning pink, then I'm not entirely sure what's going on.


Calculating entropy change of surroundings I have seen many similar questions but have not found my answer. Why do we use the actual heat involved in the process to calculate entropy change of surrounding? The only answer I can think of is that we are treating the process as reversible with respect to the surroundings even if it is irreversible with respect to the system. But, this is an irreversible path for the system, and a reversible path will not involve the same heat, so now, the entropy of surrounding calculated must be different. This seems to contradict the idea that the entropy of surroundings must be a state function. So why is this reasoning incorrect?

When you evaluate the change in entropy of an entity (in this case the surroundings) between an initial and final (equilibrium state), you separate that entity from everything else, and then you only look at how that entity has changed between the two states, disregarding the actual process path.  In the case of the surroundings (treated as an infinite reservoir), it had received a certain amount of heat between the initial and final states, and its internal energy and entropy increased as a result.  In the formalism of thermodynamics, changes in an  ideal infinite reservoir are always regarded as taking place reversibly, irrespective of other irreversibilities in the actual process.  So if the actual process is irreversible, the irreversibility (entropy generation) is all considered to take place in the system, rather than in the surroundings (in the case of an ideal infinite reservoir).

You may also find my response in the following link pertinent to this question:  Thermodynamics entropy and temperature


Are my intuitions about how batteries work right? Being really fascinated by how electricity actually works, I started studying batteries.
I initilally understimated the complexity of this apparently-simple objects, and thorugh studies and researches I have understood how they are the result of complex chemical and physical discoveries, blendend together in a triumph of engineering.
Anyway there are  some things that I still have to figure out completely, I made some pictures that aim to clarify my doubts and the question objects.
This is the picture of a zinc-copper cell with a non-specificated electrolyte:

It represents essentially how I imagine a cell of this kind.
Zinc and Copper have both reached a dynamic equilibirum with the electrolyte, so charges leave electrodes at the same rate as charges enter.
This creates an electrode's electric field, and the elctrode potential at the same time.
The net electric field represents the net force driving the elctrons from the zinc to the copper, resulting from the superoposition of the elctrodes fields.
Given that this analysis is right ( If it's not, please help me understand where is flawed), I don't understand how this electric field ( and so the potential) changes as time passes.
If electrons are leaving zinc to reach copper (and eventually react with some positive ions in the electrolyte,right?), how the electric field is mantained so that the flow can go on?
As electrons leave, how the zinc electrode field is affected?
Are positive ions leaving the zinc at a faster rate in order to compensate for th electrons deficency? (That would explain the zinc corrosion)
Is there an electrode's size limit (corrosion limit) where the electrons  that make up the elctrode aren't enough to keep the same voltage(the chemistry's votlage)?
Is it at the point that we start to see a decrease in the voltage of the cell?
Of course, all the considerations presume the precence of a salt bridge capable of mantaing the elctrolyte charge stable.

I'll assume you are someone less than 16 years of age, since usually it's around 16-17 years (11th or 12th grade) that you'll probably learn how batteries work.

You're assumptions about how electrons are taken from the zinc cell and moved to the copper cell are correct. But considering your setup, as the charge moves, an electric field and a potential difference would build up, won't it? In such case a battery would eventually stop functioning.

While batteries do stop functioning after a while, it's due to the reaction being completed (all of the zinc converted to $\ce{Zn^{2+}}$ and/or all of the $\ce{Cu^{2+}}$ deposited as copper onto the copper electrode. Charge separation does not occur. Then how do the charges remain balanced, you might ask. And that's where the salt bridge comes into the picture.

A Salt Bridge is a tube that connects the two half-cells to maintain electrical neutrality. This tube makes sure that the ionic solutions on both sides are electrically connected, but at the same time prevents diffusion of both solutions to become one (see the link to know exactly how this works).

Feel free to comment if you need any help or clarifications.


Is a metal hydroxide fully or partly oxidised in insufficient base? Firstly, my question only applies to transition metal compounds with more than one hydroxide group, otherwise it would be more obvious.
The reason I'm asking is in an effort to oxidise some nickel dihydroxide into nickel oxyhydroxide for practical use in an electrode. I've only come across somewhat laborious methods (any help in this would be appreciated) so I've considered the oxidation like in a NiMH cell (the second equation).
With nickel in a basic solution:
My question could stem from the following situation; If I stir a 2:1 ratio of NaOH:Ni(OH)2 together I would assume the following reaction would be favoured:
$$\ce {4 Ni(OH)2 + 8OH- -> 1O2 + 2 Ni2O3 + 8H2O}$$
But, if I stir a 1:1 ratio of NaOH:Ni(OH)2 and ground the solution:
$$\ce {Ni(OH)2 + OH- -> NiOOH + H2O + e-}$$
*I may or may not be misssing some electrons...I'm just starting out in electro-chemistry.
Is this assumption true? Why or why not? Am I missing the bigger picture? Cue Gibbs?  I did try it, but with the setup I had, I couldn't check for gaseous oxygen, and unfortunately both Ni2O3 and NiOOH are black and powdery.
Similarly with iron in an acidic solution:
In insufficient acid, which one is what would happen and why?:
$$\ce {Fe(OH)3 + H+ -> [Fe(OH)2]+ + H2O}$$ vs.
$$\ce {Fe(OH)3 + H+ -> Fe^{3+} + 3 H2O + excess \,\,\, Fe(OH)3}$$

I guess the question boils down to; How do reduction / oxidation potentials change with multiple individual reduction / oxidations?
I hope everything makes sense. If not I will gladly try to explain my question further.
Thank you for your help!

What you are worried seems to be in your comment, which I'm explaining[maybe not]

Oxidations:

  $$\ce{NiOOH + H2O + e- -> Ni(OH)2 + OH-} \quad E^{\circ}=+0.49\\
\ce{Ni2O3(s) + 6H+ + 2e- -> 2Ni^2+ + 3H2O} \quad E^{\circ}=+1.753\\$$

I suppose if you know about electrode potential you might be able to understand now,If you wish I might explain it, if you can't on your own.


Need of refining aluminium and Hoope vs Hall–Héroult processes Recently I was taught how aluminium is commercially extracted. The ore is first concentrated by leaching either by Bayer's process for red bauxite (impurity: $\ce{Fe2O3}$) or by Serpeck's process for white bauxite (impurity: $\ce{SiO2}$).

Then alumina is reduced electrolytically via Hall–Héroult's process directly to 99% pure aluminium. But for further purification to remove remaining copper or iron impurities, Hoope's process is executed to get 99.99% pure $\ce{Al}$.

Since both processes involve electrolysis, and the materials added are similar in both (except $\ce{BaF2}$) with same graphite electrodes, shouldn't we get the same purity in both process (thus not requiring a further refinement)?

Also, the arrangement of the electrodes are exactly opposite to each other. Hoope's argues that pure aluminium is lighter and thus floats so cathode is kept at top, but why doesn't that logic work in Hall–Héroult's? The difference in purity varying by 0.99% and the electrolyte used being same.

Any insight into this matter is helpful because I haven't been able to come to a convincing conclusion.

I think the difference is due to the way eddy currents form Magnetohydrodynamics.

As the current flows through the conductive liquid, a magnetic field is produced. This in turn causes the liquid to be pushed (at right angles to the current and field).

In one version of the cell, this current causes stirring of the aluminum, alumina, flux mix. This is more efficient from a time and electrical point of view. The trade-off however is that some of the impurities which would otherwise get captured in the slag (waste), end up mixed into the final product.

Edit: Thinking about it some more, the reason for the mixing may be due to where the $\ce{CO2}$ gas is formed. ($\ce{Al2O3 + C_{electrode} -> Al +CO2}$) Bubbles formed a the bottom of the cell would rise up and cause more stirring than if they formed near the top of the cell.


Suggestions for solvents capable of efficiently performing alpha-bromination and nucleophilic substitutions We performed the one-pot preparation of bupropion in class where the bromination of the carbon adjacent to the carbonyl group was performed using DCM as solvent, and upon completion the solvent was switched to n-methyl-2-pyrrolidone and was then reacted with an amine. I know that the nucleophilic substitution won't occur in DCM, the reason for which was also explained here on this site, so what alternate solvent could be used that allow for both bromination and substitution to occur without a solvent switch, AND still allow this procedure to still be conducted within a single class session? I asked the professor, but he said that he would have to think about it and get back to me... yeah, I have a feeling I am being blown off.

I believe that water would work for the bromination as long as the mixture is kept acidic, but then that complicates things with the amine as it will need to be in its basic form, unless that doesn't matter because by the time you add the amine, the bromination has already taken place and you would just neutralize first. However, aren't these reactions very slow in aqueous conditions? I know that water lowers nucleophilicity, so would that explain the hindered reaction times?

What about acetonitrile? I know this can be used in bromination, as I have read procedures that claim increased selectivity for monobromination.

In the event that you are not familiar with the preparation I am referring to, I am providing a link to the ACS paper that we used as the basis of the experiment.
http://pubs.acs.org/doi/abs/10.1021/ed077p1479

EDIT: There is also a follow up question regarding the usage of more than an equivalent of the amine during the substitution step as an acid scavenger. What is up with that? Why isn't the acid being formed in the substitution step being neutralized with an appropriate base that won't hydrolyze the bromoketone first? This just seem rudimentary to me, but I lack the encompassing understanding at this time. Is there a procedure I could reference as an example of what I just proposed?

As a general comment published procedures have been optimised. I don't have access to the paper you link so I don't know if the procedure is from process chemistry (where you expect a high degree of optimisation) or from med chem (where the priority is to get the compound out fast but there is still some optimisation).

But your questions are still valid. So I agree that doing the bromination - amination in one pot using the same solvent is a reasonable experiment to try and actually acetonitrile is a good choice. Why do you think acetonitrile is nucleophilic? it is not (unless you deprotonate it with a strong base). So it is very possible that the first researcher tried your approach initially and possibly found that the bromination was not very selective or the amination step was too slow, or low yielding and switched to the current procedure.

I also agree that you can use a base instead of excess of amine, for example an amine base like triethylamine or inorganic bases like $\ce{K2CO3}$ are standard in this case and this is what they would have used if they had to do the reaction with a different amine (as the nucleophile) of higher cost.


How does water evaporate completely when at room temperature? If water evaporates at room temperature because a small percentage of the molecules have enough energy to escape into the air, then why does a kitchen counter with a small amount of water eventually evaporate completely when at room temperature?

As your small percentage of molecules with high enough kinetic energy evaporates, the remaining liquid water cools down. But in doing so, it drains heat from its surroundings and thus stays at room temperature (or close to it), so there is still some fraction of molecules that can evaporate, and they do so, and more heat is transferred from the surroundings, and so it continues until all water is gone.


How does Le Chatelier's principle apply to a gas phase reaction at constant volume? This reaction is given in my book and the direction of shift of reaction on addition of $\ce{CO}$ and removal of $\ce{CO2}$ at constant volume is asked.
$$\ce{2CO(g) + O2(g) ⇌ 2CO2(g) + \Delta H}$$
If "at constant volume" was not specified, I understand that the forward reaction would occur since the reactant is added and the product is removed, but I do not know if this will vary with constant volume. What will happen in the case of constant volume? Does it matter that there is a change in the number of particles comparing reactants and products? Which direction will the equilibrium shift?

Let's see your formula (with the difference of enthalpy being negative):

$$\ce{2CO (g) + O2 (g) <=> 2CO2 (g)}\qquad(\Delta H^\circ<0)$$

At constant temperature, for Le Chatêlier purposes,

Adding $\ce{CO}$ produces that it react with $\ce{O2}$ to produce more $\ce{CO2}$ to counteract the effect of the applied change and a new equilibrium is established.
Removing $\ce{CO2}$ produce that $\ce{CO}$ react with $\ce{O2}$ to produce more $\ce{CO2}$.

When you are adding or removing a reagent or product, and all components are in a gaseous state, you are changing the pressure and you can study the reaction by $K_p=\dfrac{P^2_\ce{CO2}}{P^2_\ce{CO}P_\ce{O2}}$.

We also have $K_p=K_c$ $RT^{Δn}$:

$K_p$ is the equilibrium constant written with partial pressure of gases.
$K_c$ is the constant but being written by concentration of substances.
$R$ is universal gas constant.
$T$ is the absolute temperature (Kelvin degrees).
$\Delta n$ are moles $(\Delta n=n_{\text{products}} - n_{\text{reagents}})$.

If you change concentration of $\ce{CO}$ you are changing the pressure it, too.

But you can change also the pressure, by changing the volume. (For Boyle's Law: the absolute pressure exerted by a given mass of an ideal gas is inversely proportional to the volume it occupies if the temperature and amount of gas remain unchanged within a closed system).

For example, if you increase the pressure, volume will decrease and the system will shift towards it will be less molecules to counteract the change. In this case, to products. If you in decrease pressure, volume will increase and the system will shift towards reagents.

Other option is to change the pressure by adding or removing an inert gas, but since it don't intervenes in the reaction, the equilibrium is not affected.


Denatured Ethyl Alcohol Purification I have denatured ethyl alcohol which was obtained this formula;

In 100 L ethyl alcohol there is 0.8 gr denatonium benzoate and 78 gr tert-Butyl alcohol.

Is there any way to obtain absolute ethyl alcohol from that mixture.

Please note that the following is not meant to be used as an instruction!

According to an article on Detection of Denatonium Benzoate (Bitrex) Remnants in Noncommercial Alcoholic Beverages by Raman Spectroscopy, published in J. Forensic Sci. 2014, 59, 1358-1363
(DOI), the manufacturers of illegal alcoholic beverages often use sodium hypochlorite ($\ce{NaOCl}$, household bleach) to remove the bitter taste of tax-free technical ethanol.

In the study above, the authors used legally available legal alcoholic beverages (vodka, cognac, etc.), and denatured/de-denatured them under controlled conditions:

  The separate batch of purchased alcoholic beverages was
  denatured with 20 mg/L of denatonium benzoate, after which 1.33 mL/L sodium hypochlorite was added.

Spectroscopical measurement of the samples thus obtained clearly showed degradation products of the bitterant.

While hypochlorite treatment might be suitable to remove the bitter taste, this doesn't imply that the result is healthy or recommended to consume.


Are bridge bonds in diborane longer than the terminal H bonds? The compound diborane on alkylation loses the terminal $\ce{H}$-bonds rather than the bridge bonds indicating that the bridge bond is stronger but on the other hand it's length is found to be longer than terminal hydrogen bonds. But usually bond length is inversely proportional to bond strength.

What is the reason for this?

As ocresol said, its not wise to conclude the bond strength from one reaction. Rather, considering the structure of the molecule is a more viable option.
Actually the bridged bond is weaker than the terminal $\ce{B-H}$ bonds. This is because of the presence of multielectron density region. Since it is a pair of shared electrons spread between two regions, so more volume and less the electron density.

Image Source: Quora.
Also, the bond order of terminal $\ce{B-H}$ bonds is 1, whereas the bond order of the bridged bond is half of it ie. 0.5. And it is intuitive that bond strength is directly related to bond order.


What is the correct name of 3-isopropyl-4-methylhexane? In the branched alkane shown above, I've been told that 3-isopropyl-4-methylhexane is supposedly the correct name. However, in the alternative numbering scheme on the right, the parent chain is still of the same length (six carbon atoms) and has more substituents, so shouldn't that be the preferred name?

Both the names you have given are not the preferred name for the molecule above. You are correct that the preferred name maximises the number of substituents when there is a choice: see Alkane nomenclature - choosing the parent chain - maximize or minimize substituents? So, a name of the form a-ethyl-(b,c)-dimethylhexane (where a, b, and c are appropriately chosen locant numbers) is preferred over a name of the form a-isopropyl-b-methylhexane, as it has three substituents as opposed to two.

However, the numbering on the right is not correct either. If you number the carbon chain the other way around instead, you would obtain the name 3-ethyl-2,4-dimethylhexane:

This is preferred over your proposed name 4-ethyl-3,5-dimethylhexane, because the locant set (2,3,4) in 3-ethyl-2,4-dimethylhexane is lower than the locant set (3,4,5) in 4-ethyl-3,5-dimethylhexane.

Note also that according to current IUPAC nomenclature recommendations (P-57.1.1.1 of the 2013 Blue Book), the preferred prefix for $\ce{-CH(CH3)2}$ is "propan-2-yl", not "isopropyl".


Does washing soda (sodium carbonate) turn into baking soda (sodium bicarbonate) over time? I have some sodium carbonate. But it's over 20 years old and was not kept in tight bottle. I wonder if this compound absorbs $\ce{H2O}$ and $\ce{CO2}$ from air.

I guess I'd need to burn it a little to turn it back in case the answer was yes.

Industrially, sodium bicarbonate is produced by reacting sodium carbonate with water and carbon dioxide.  However, under normal household conditions sodium carbonate simply reacts with water or moisture in the air (it's hygroscopic) to form a hydrate - it doesn not form sodium bicarbonate. The water of hydration can be removed by mild heating (~100 C).  At high temperatures (>850 C) sodium carbonate can decompose according to the following equation.$$\ce{Na2CO3(s) -> CO2(g) + Na2O(s)}$$


How to calculate the ratio of rate constants of competing reactions? There are two compounds $\ce{A}$ and $\ce{B}$ that react with one common reagent $\ce{C}$ following first order kinetics.
  Suppose 99.9% of $\ce{A}$ must react before 0.1% of $\ce{B}$ has reacted then what is the minimum ratio of their respective rate constants?

My Attempt at the question:

We know that the reactions follow first order kinetics, hence I tried to find the time at which the given percentage of $\ce{A}$ and $\ce{B}$ is left using the formula
$$ t = \frac{2.303}{k} \cdot \log\left(\frac{C_0}{C_1}\right)$$
where $t$ is the time, $k$ is the rate constant for each reaction and $C_0$ and $C_1$ are initial and final concentration respectively.

After this I tried to find the ratio of rate constants directly but couldn't arrive at the answer.
I derived the above formula using the integrated rate law.


Zinc oxide reduction in ethanol I'm trying to prepare a ZnO thin film deposited on a glass substrate. The precursor solution prepared by sol gel process through dissolving 1 M zinc acetate dihydrate $\ce{Zn(CH_3COO)_2.2H_2O}$ in ethanol with ethanolamine as stabilizer in 1:1 mol ratio with $\ce{Zn(CH_3COO)_2.2H_2O}$. I aged the solution for about 1-2 weeks in a closed glass container. But, after I actually deposited and annealed at 600 °C, the film's XRD measurement seems to show ZnO as well as zinc peak, with zinc somewhat showing stronger peak.

However I also have the one with precursor aged for only 1 day but annealed at 800 °C, this one show peak characteristic that of ZnO with no noticeable Zinc peak. Other samples I have are aged for 3-4 days added with 1-5at% Nickel doping and annealed at 600 °C, they also show strong Zinc peak along with ZnO peak.

In the ones with Ni doping, Zn peak relative to ZnO also seems to get stronger as Ni concentration increasing. It seems like nickel (or its derived compound) acts as catalyst in the reduction of ZnO. The ones with 3at% and 5at% nickel concentration also have solution with green-bluish turbid solution with white precipitates (this probably zinc) unlike 0at% and 1at%, which yields clear solution. This might support that nickel act catalyst.

So, I guess there must be more rate of reduction reaction of ZnO during aging process, or maybe some other reaction that produces Zinc. Maybe something happens during annealing as well, but that seems unlikely since ZnO seems to be stable at 800 °C.

The question:

Looking at the precursor solution, what possible reactions that would produce Zinc from ZnO, and how likely they are?
Is it possible that the reaction of producing Zinc happens during annealing? How likely they are? How much is the rate, can they be neglected?
Do Nickel doping contributes to the production of Zn? If yes, what mechanisms that seem most likely to affect the production of Zn?

As for your first question, I have personally had better luck only aging 24-48 hours (actually unsealed, just covered with Kimwipe to prevent contamination, but allowing for some of the solution to evaporate).  We're getting good $\ce{ZnO}$ XRD peaks on this process, and are currently developing an SOP/Optimization for $\ce{ZnO}$ by sol-gel technique. I'll let you know when we publish.

When you say you're getting $\ce{ZnO}$ and Zinc peaks, do you mean you're seeing XRD peaks matching Zinc as crystalline metal?  To your question precisely: "What possible reactions would produce $\ce{Zn}$ from $\ce{ZnO}$, and how likely are they?" Remember that your sol-gel is not necessarily $\ce{ZnO}$ yet.  The reaction that produces $\ce{ZnO}$ begins with the dissociation of $\ce{Zn}^{2+}$ and $\ce2{CH_3COO}^{-}$ ions in ethanol, followed by a hydrolysis. As to how you're getting the extra electron to precipitate metal, I don't know. What we have seen has been the precipitation of $\ce{Zn(CH_3COO)_2}$ after about 96-120 hours after originally mixing.  (We're pretty sure it's acetate because it smells like vinegar when we wash it out of the flasks). There is a very good article going over the chemistry and technique of $\ce{ZnO}$ sol-gel: article


How do we know that hydrogen gas consists of two hydrogen atoms? Of course I know that both 1s orbitals will combine and that the electrons will occupy the bonding MO and that the two atoms will reach a lower energy state – but how do we really know that is the case?

Assuming ideal gas behavior we can determine the molar mass $M$ of the hydrogen gas from its density $\rho$ (which can be measured) by the following formula
$$
    M = {{RT\rho}\over{p}}\ \, .
$$
At STP $\rho \approx 0.09 \, \mathrm{g} \, \mathrm{L}^{−1}$ so the following back-of-the-envelope calculation
$$
    M = {{8.31 \, \mathrm{L} \, \mathrm{kPa} \, \mathrm{K}^{−1} \, \mathrm{mol}^{−1} \cdot 273.15 \, \mathrm{K} \cdot 0.09 \, \mathrm{g} \, \mathrm{L}^{-1}}\over{101.325 \, \mathrm{kPa}}}\ = 2.02 \, \mathrm{g} \, \mathrm{mol}^{−1} \, .
$$

shows that $M$ is indeed close to 2 g/mol which means that hydrogen gas consist of $\ce{H2}$ molecules rather than $\ce{H}$ atoms


How many isotopologues of water are there? If ordinary water molecules can contain any of the three hydrogen isotopes, $\ce{^1H}$, $\ce{^2H}$, $\ce{^3H}$, and any of the three oxygen isotopes, $\ce{^{16}O}$, $\ce{^{17}O}$, $\ce{^{18}O}$,  how many kinds of water molecules $\ce{H2O}$ are there?

I'm a little confused because since there are $\binom{3}{2}$ ways of choosing two hydrogen atoms and 3 ways of choosing oxygen atoms, shouldn't there be 9 different kinds of water molecule? Why would it be 18 (as the answer says)? In other words, why would the order of hydrogen atoms matter here?

In an example as small as this, you could hypothetically construct all the cases, or at least enough of them to see that $9$ is incorrect.

The only thing you need to change to get the right answer is instead of using $3\choose{2}$ which is a combination without replacement, you need to do combinations with replacement so ${2+3-1}\choose{2}$$=6$ ways to choose the hydrogens. We can very quickly verify this result by writing out the pairs of masses and seeing how many unique combinations we can make: 11 22 33 12 13 23. And then of course each of those combinations can have one of the three different types of oxygen, so 18 possibilities total. If you haven't seen combinations with replacement before, go here to learn more about them.


By how many pH units does rain vs distilled water change a soil pH test? Sources state that the concentration of $\ce{H2CO3}$ in rain is of a few µmol; I don't know if that's per ml. The buffering effect of most soils is >1000 µmol $\ce{H+}$/kg.

Is the difference from low acid rain water versus lab distilled water on a pH test for soil less than 0.1 in pH values?

Pure water (rain as well as distilled water) in equilibrium with the atmosphere ($p_{\ce{O2}}=10^{-3.5}\ \mathrm{atm}$) can be calculated to contain about
$$\begin{align}
\mathrm{pH}=-\log[\ce{H+}]&=5.65\\
-\log[\ce{HCO3-}]&=5.65\\
-\log[\ce{CO3^2-}]&=10.3\\
-\log[\ce{H2CO3^*}]&=5.0\\
-\log[\ce{CO2}]&=5.0\\
-\log[\ce{H2CO3}]&=7.8\\
\end{align}$$

(The calculation can be found in: Stumm, W.; Morgan, J. J. Aquatic Chemistry, Third Edition; John Wiley & Sons: New York, NY, 1996; pp 159–161.)

Note that $[\ce{H2CO3^*}]$ is the total analytical activity of dissolved $\ce{CO2}$, i.e. $[\ce{H2CO3^*}]=[\ce{CO2(aq)}]+[\ce{H2CO3}]$. $[\ce{H2CO3}]$ is the concentration of true $\ce{H2CO3}$.

Thus, $[\ce{H2CO3^*}]=10^{-5}$ and $c(\ce{H2CO3^*})=10^{-5}\ \mathrm{mol/l}$; i.e. a few µmol/l.


How is the standard electrode potential measured without allowing the electrolyte concentrations to change? Let's say we need to measure the standard electrode potential of a zinc electrode dipped in an aqueous solution of $\ce{ZnSO4}$. To do this, we use the standard hydrogen electrode(SHE)-the reference electrode-as the anode and the zinc electrode as the cathode, and measure the emf of the cell.

The electrolyte corresponding to the SHE is a solution whose hydronium ion concentration is 1M. To construct the zinc half cell, we dip a strip of zinc in a 1M solution of $\ce{ZnSO4}$(for the STANDARD value). We connect the two electrolytes(of the zinc electrode and the SHE) with a salt bridge. Now, we measure the emf of this cell using a high resistance voltmeter, to find the potential difference between the electrodes without allowing any significant current through the cell.

When the zinc strip is dipped in the solution, it initially has no excess electrons on it. It is just a neutral piece of zinc metal. Now an equilibrium will be established between the solid zinc and zinc ions in the solution, finally leading to an excess of electrons on the electrode and a specific concentration of zinc ions in the solution. This, and a similar process at the SHE would cause an electrostatic potential difference between the two electrodes, since the relative positions of the two equilibria would be different. The voltmeter measures this potential difference to be the emf of the cell.

Here's my question: In setting up the two equilibria, won't the concentrations of the electrolytes change? Initially the zinc electrode had zero charge. But later it acquires a negative charge. This means that a certain number of zinc ions had to join the solution, leaving behind the electrons on the metal. So the concentration of zinc ions in the solution would now have to be more than 1M. So how would the STANDARD electrode potential be measured? How would the electrode potential be measured at ANY arbitrary concentration of zinc ions? It seems to me that the electrode potential can only be measured when the resulting concentration of zinc ions is equal to the equilibrium concentration. Does such a concentration even exist? Similarly, how is the hydronium concentration of the electrolyte corresponding to the SHE maintained at 1M?

According to me, the concentrations have to change for any net charge to develop on either electrode, hence establishing a potential difference. How does a net charge develop on each electrode without the ion concentration in the solution changing?

A voltimeter has very high resistance and consumes a very small amount of current, so if you make the measurement quickly, the concentrations are essentially constant for the duration of the measurement.


Why is N- a better nucleophile than S-? I can't rationalize why S- is a worse nucleophile than N-. I thought nucelophilicity was related to electronegativity, the more electronegative an atom is, the less of a nucleophile it is. Sulfur is less electronegative than Nitrogen. What concept am I not considering or misunderstanding?

I think it would be nice to consider the charge density. The S- is more stable than N- because it has more electron shells, so the negative charge density is better distributed in the atom. It makes that the N- is more likely to give up electrons and act as a nucleophile


Thermal conductivity and enthalphy data for nitrogen gas I am struggling to find the data for thermal conductivity and ethalphy of nitrogen gas at reduced pressure (500 mbar). Most of the information is for nitrogen gas at 1 bar or above. Does anyone know where I can find this information? Specifically, I need to know the thermal conductivity and enthalphy of nitrogen at 295.15 K (22 degrees) and pressure of 0.5 bar (500mbar)

By way of comparison, using NIST Reference Fluid Thermodynamic and Transport Properties Database (REFPROP) – NIST Standard Reference Database 23, Version 9, I have got a specific enthalpy of $h=306.26\ \mathrm{kJ/kg}$ and thermal conductivity of $\lambda=25.601\ \mathrm{mW/(K\ m)}$ for nitrogen at a temperature of $T=295.15\ \mathrm K$ and a pressure of $p=0.5\ \mathrm{bar}$.

Note that the absolute value of enthalpy at a single state point is meaningless; it is only the difference between two different state points that matters.


Does removing solution affect a supersaturated solution Given a mixture of: a saturated solution of a solute in a solvent together with excess of the solute compound as crystalline material.

If one were to change the bulk composition of the mixture by removing some of the solution could this cause dissolution or precipitation of the solid in the remaining mixture, or the concentration of the remaining solution?

Presumably one should expect some crystallisation in the portion of the solution that is removed from the mixture?

When a solution is saturated, the remaining solid is in equilibrium with the solvent:
$$\ce{AB<=>[solv.]A+_{(solv)} + B-_{(solv)}}$$

That means that there is as much of your solid solved. If you take out some of the solution then the concentration remains the same, in your original flask as well as in the portion you withdraw.

If you remove only solvent, i.e. by heating your solution or diffusion over time. In this withdrawn portion crystals will form. The concentration will not change, only the amount of substance $n$ in your solution will change as the volume shrinks ($c=\frac{n}{V}$).


Luminescent materials under IR radiation? is there a database somewhere of materials that are luminescent  when exposed to Infra Red radiation?

There are two types of materials I'm aware of that produce light in the way I think you want: two-photon fluorescence and photon upconversion. Two-photon fluorescence is basically the same as normal fluorescence but two photons are absorbed simultaneously, emitting a single photon when returning to the ground state. Photon upconversion similarly involves the absorption of multiple photons, but sequentially (by various mechanisms), followed by emission of a single photon.

Two-photon fluorescence is mostly seen used for two-photon excitation microscopy and useful dyes tend to be big conjugated organic systems. Materials used for photon upconversion are generally lanthanide-doped nanoparticles and they've found many uses in applications traditionally served by quantum dots and organic fluorophores. They're both particularly useful in medical imaging as IR can penetrate tissue deeply, but upconverting nanoparticles have a big advantage in that two-photon fluorescence requires very high excitation intensity (pretty much just lasers) as both photons must be absorbed simultaneously, but upconversion processes are sequential and can thus occur at lower intensity. Though, for some applications, the intensity requirements for two-photon absorption are useful as one can play tricks with focusing such that only a small region receives enough excitation light to emit, making imaging small areas possible.

As for a database of compounds, Thermo-Fisher has a catalogue of two-photon dyes, but I don't know that there are a lot of commercial manufacturers of upconverting nanoparticles, though Sigma-Aldrich formerly carried up converting nanocrystal under the Sunstone brand.


How can an inexperienced chemist determine the chemical structure of a molecule? I'm a fairly inexperienced, but nonetheless interested chemist, currently planning a small project for the coming summer: I want to isolate an unknown chemical from a plant and determine it's chemical structure.

I have ample access to Carex grayi, a sedge plant, and have chosen this plant for a few reasons.

Of the plants growing near where I live, I can definitively identify Carex grayi. I am relatively inexperienced in taxonomy and plant identification, and I want to know which species I am working with.
As far as I can tell, the plant is relatively unresearched, and may or may not contain alkaloids.

This brings me to my next point. I know it may seem immature or disheartening, but my interest in chemistry stemmed from my interest in neuroscience and neurotransmitters. From countless hours of reading, I have learned much about the pharmacological (and sometimes psychoactive) tendencies of alkaloids. I realize that ingesting unknown alkaloids is extremely dangerous, and I do not intend to ingest anything. However, due to their numerous applications in medicine and recreational use, there is a massive amount of information on the internet regarding their extraction and isolation. I am entirely open to the idea of isolating a chemical from another group, but as I am inexperienced and my interest stems from neuroscience, I only really know about alkaloids. I am only somewhat familiar with terpenoids, and know little about any other group.

So, assuming I am able to extract an unknown chemical of some specified group from a plant, is there any way for me to determine the chemical's structure? The Chemical structure Wikipedia page lists the following methods for determing chemical structure:

X-ray diffraction
Proton NMR
Carbon-13 NMR
Mass spectrometry
Infrared spectroscopy

I don't really have access to these tools as student over the summer. I am willing to spend a small amount of money on this project, but no more than about one hundred dollars. Is it too much or too difficult for a student to determine a chemical's structure over the summer? Should I do anything differently (as in, is there a certain group of chemicals that would make this project possible or easier)?

Also, is there a "classical" chemical extraction similar to this that most fledgling chemistry students would conduct? This is obviously a ridiculous made-up example: extract dandylionine from a dandy lion flower.

So, assuming I am able to extract an unknown chemical of some specified group from a plant, is there any way for me to determine the chemical's structure?

Yes.

I don't really have access to these tools as student over the summer. I am willing to spend a small amount of money on this project, but no more than about one hundred dollars.

Then probably no (at least not with the same accuracy you would if there weren't these constraints).

Mass spectrometry and nuclear magnetic resonance are the two most powerful analytical tools to identify completely unknown substances. This doesn't mean it's as trivial as putting your purified substance in the equipment and finding its structure. For reliable results you need a purified substance, an optimized experimental routine and of course the instrument properly set up.

In fact, before trying to determine the structure of the substance, you must purify it, and that alone is a challenge. As I'm sure you are aware, the plant contains thousands, millions of different molecules. How are you going to extract something you haven't defined? The first step would have to be defining what is it that you're looking for.

Let's consider you're looking for whatever makes plants green:

You could macerate the plant, add some water, filter the mixture, check the colors of the filtrate and of the residue, and you would notice most of the green wasn't extracted. You would find different results with different solvents though, and a better option would be ethanol. Even then, it wouldn't mean you have a pure chlorophyll solution, so several other steps would probably be required to purify the substance*. Generally speaking, you would use some kind of chromatography.

For your project, I suggest you try to first precisely define what is it you're looking for. The substance that gives the plant its color? Its smell? Its pharmacological effects? You must use this property to analyze your experimental results. You can visually inspect the color, you can smell extracts, and perhaps test physiological effects (this is obviously dangerous and shouldn't be done unless you really know what you are dealing with!). Use the target substance properties' to decide what's the best way of extracting and purifying it.

Then, considering you did purify something, without the high-tech analytical tools mentioned, there are some experiments which can indicate what you have, but they require a reference substance and usually can't be used to say that a substance is the same as another, but rather that a substance isn't the same as the reference. For example, you can compare boiling/melting points and/or thin layer chromatography retardation factors, both which can be done at home home (to some extent), and also UV/Visible light absorption/emission, which might not be trivial to do at home but require instruments that are a lot easier to find in chemistry teaching labs.

For precise tips on the extraction and identifications, it would be necessary to know what you're looking for.

It's also important to note that there's a lot you can do besides determining a chemical structure. There are several properties of a substance you can study in order to characterize it. Some examples are solubility (in different solvents, at different conditions (temperature, pH, salinity)), density, reactivity (in face of pretty much anything you find at home) and physical properties such as texture, color, smell, etc.

Finally, do not think I'm saying you should give up, your wish to discover what something is made of is what made chemistry a science, but you need to have a clearer problem to be solved. Decide precisely what it is you're looking for (in terms of macroscopic properties, chemical group or something like that), and then research (and ask!) what could you do about that without a full lab.

*Chlorophyl is actually a group of substances which have a critical role in photosynthesis, and are, of course, green.

In summary:

You already have your sample. Now choose your analyte, the molecule(s) you want to study.
Do research. Look for information on how people usually study this (group of) substances. You'll have better luck looking in education-related sources, which usually have simpler experiments.
Once you've found something that fits your interests and available material, do experiments. Observe the results. Change parameters, do them again, try to understand what's happening.
If you can't explain the results, do research, check your references, and of course, ask here. We'll be glad to help.


Are non-ionic compounds generally as dehydrating as ionic compounds in drinking water? In order to turn sea water into drinkable water, virtually all of the salt must be removed.  By definition, salts are ionic compounds.  I am assuming that there are other non-ionic compounds devolved in the sea water, must these compounds be removes as well before water becomes safe to drink?  In general, do non-ionic compounds present the same dehydration dangers as salts when consumed in drinking water?  Why or why not?

For this question, lets not consider other toxic effects besides dehydration.

Actually, the salt in seawater doesn't kill you directly. It dehydrates you and you die of thirst. If you mixed 1 part of seawater with 2 parts of fresh water, it would not dehydrate you, but you would need to eliminate the excess salt so you would need to drink a lot of this salty water.

There are many living things in seawater that can kill you. Red Tide (excuse me, harmful algal bloom) is an algae common to warm water that produces a toxin. Very little of this toxin is needed to kill a person.


Explanation of steric hindrance and existence of PI5 I have come across a following line in my textbook:

  $\ce{PI5}$ does not exist due to steric hindrance.

I searched for the explanation of steric hindrance in Wikipedia but couldn't understand it. Can anyone explain steric hindrance in simpler language?

Wikipedia says:

  The existence of $\ce{PI5}$ has been claimed intermittently since the
  early 1900s. The claim is disputed: "The pentaiodide does not exist
  (except perhaps as $\ce{PI3\cdot I2}$, but certainly not as
  $\ce{PI4+I-}$)".

What does this mean?

The simplest explanation is that you can not fit five iodine atoms around a phosphorous atom.

Consider the pentachloride equivalent - this is already quite crowded. Since chlorine atoms have a smaller covalent radius than iodine, you can get five of them around the phosphorous. Of course, all this depends on the bond length of P-I and P-Cl and the size of the phosphorous atom.

Interestingly, there seems to be some suggestion of an Uranium Pentaiodide compound, for example this PubChem compound. However, there are no references on that page, so presumably this has never been synthesised either?

In general, steric hindrance means geometric conformations of a molecule that are impossible (or difficult to achieve) due to clashes between groups and atoms within the molecule.


Draw a simplified MO diagram for the pi system of Methyl vinyl ether My attempt at the answers:

i) 2 bond pairs and 2 lone pairs on oxygen atom therefore bond angle of slightly less than 109.5 degrees, around 104 degrees.

ii) 120 degrees bond angle therefore $\mathrm{sp^2}$ hybridization.

iii) From what I understand, the sigma-pi separation means that you can neglect interactions between the $\mathrm{p_z}$ and $\mathrm{p_x,p_y}$ orbitals?

The $\ce{CH2}$ carbon would form $\mathrm{sp^2}$ hybrid orbitals.

The $\ce{CH3}$ carbon would form $\mathrm{sp^3}$ hybrid orbitals and would then form 3 sigma bonds with the hydrogen leaving only one orbital to interact with the oxygen. I assume that the oxygen $\mathrm{s}$ orbitals do not match the energy level of the carbon and will form non-bonding orbitals. Therefore the $\ce{C-O}$ sigma bond would be $\mathrm{sp^3-p}$.

I am unsure how to use this knowledge to actually draw the MO diagram? I have tried an attempt below. I can draw MO diagrams for simple diatomic molecules. And how would the MO diagram explain the observed 120 degree bond angle?

iv) I haven't used the MO diagram to draw these resonance structures but I have come up with two based on the octet rule:

I am unsure how to use this knowledge to actually draw the MO diagram?

First, chemists used Lewis structures to explain molecular properties, but eventually this method was replaced with resonance structures (like you've drawn in part iv above).  Resonance structures focus primarily upon p orbitals in a molecule and usually do not consider the sigma system.  Molecular Orbital (MO) theory was a further improvement on resonance theory.  MO theory still focuses on p orbital connectivity and typically disregards the sigma system, but it lets you view the actual patterns that the molecular orbitals generate.  It is very similar to the orbital diagram you included above, but we leave out the sigma system and focus on the p orbitals and the electrons they contain.

In part ii) above you conclude that the ether oxygen is $\ce{sp2}$ hybridized.  This means that there is one p orbital on oxygen that contains a lone pair of electrons (there are also 3 $\ce{sp2}$ orbitals on the oxygen, 2 are used to form sigma bonds to carbon and the third holds the second lone pair of electrons on the oxygen).  This p orbital can interact with the 2 p orbitals involved in the carbon-carbon double bond.  These 3 p orbitals can interact to produce 3 molecular orbitals (see pictures below).  These 3 molecular orbitals are produced by combining (adding and subtracting) a p orbital and the 2 ethylenic molecular orbitals (the bonding and antibonding pi orbitals for the double bond).  All of the orbitals shown in the MO diagram below are p orbitals.  We will fill the MOs with 4 electrons, 2 from the pi double bond and 2 from the oxygen p orbital that contains one of oxygen's lone pair of electrons.

So, we have determined that we have 3 p orbitals (one on oxygen, 2 on the olefinic carbons) that can overlap and form 3 molecular orbitals.  The system we have created with 3 p orbitals from methyl vinyl ether is similar to the all carbon allyl analogue with 3 p orbitals, one p orbital on each of the 3 carbons.  Specifically your system is analogous to the ally anion with 4 pi electrons (2 from your double bond and 2 from the  oxygen lone pair).

Image Source

For more information on how to construct molecular orbitals in organic molecules see here and here.