Molecular orbital theory & predicting the stability of a molecule? A question in my general chemistry textbook gives me a list of formulas for molecules and ions. It then asks me to determine, using molecular orbital theory, whether or not each would exist in a relatively stable form.

How can I tell this? The chapter covers molecular shapes, valence bond theory, and molecular orbital theory which all builds on what was taught in the previous chapter (mainly on Lewis structures.) I can't tell what concepts to use in telling what will be stable and what won't.

You need to mix the orbitals, populate them with the electrons and see if you have net bonding.

Eg:  H + H two 1s orbitals mix to form sigma and sigma*.  Two electrons total, both occupy the sigma orbital, two more electrons in bonding than antibonding orbitals, the compound is stable.

Eg:  He + He; same mixing as above.  Four electrons, two in the sigma, two in the sigma*.  Since there are as many bonding electrons as as antibonding, there is no net bond.  He2 is not possible.

Eg:  He + H; same mixing as above.  Three electrons, two in sigma, one in sigma*.  One more electron in bonding than antibonding.  He-H forms a very weak bond.  Please note the diagram is for He2+ but the He-H is very similar

Eg: Li + H; Li has 1s + 2s, while H has 1s.  This mix to form a sigma orbital from H1s+Li2s, a sigma* orbital and H1s-Li2s, and a non bonding orbital from Li1s (lower in energy than the sigma).  Four total electrons.  Nonbonding sigma is occupied, and then the sigma orbital is occupied.  Net effect: Li-H forms a stable bond.

You might have noticed that the first three diagrams had identical structure, even if their energy levels were different.  This is a property of quantum mechanical symmetry, and solving the problem only requires knowing a few diagram types and filling the electrons in.  Consider the second row.  The diagram is as :

You simply count the valence electrons and fill in the diagram.  Then you subtract the number of bonding orbital electrons from antibonding, and if you have more, you will have a bond.

A couple of webpages are very good for further explanation:

Source of the diagrams:
http://www.chem1.com/acad/webtext/chembond/cb08.html

and
http://www.meta-synthesis.com/webbook/39_diatomics/diatomics.html


What non-toxic non-water substances have a freezing point very close to water's? I was linked to a Kickstarter for a cooling ball to be placed in drinks which claims with constant marketing hyperbole how much it relies on "Phase Change" ! which is a neat concept and seems not unreasonable, but in thinking about it I'm failing to imagine (as I'm no chemist, and of ill education) what refrigerant could be inside that meets the named requirements of non-toxic food safe, as well as passing from solid to liquid (or liquid to gas) such that a typical freezer would freeze it (−2 °C to −5 °C typically?), and it would liquefy around a temperature you'd want to keep your drink (say 1–3 °C).

Obviously water meets these requirements, however due to the expansion pressures water commits I have a very hard time imagining a small stainless steel ball containing that expansion through many freeze-thaw cycles without causing eventual deformities which would be continually weakened.

So the question is in the title, and to note the container is a sealed solid stainless steel ball of unknown thickness 2" in diameter.

Let me know if this question is off-topic, more than anything I'm just curious so I realize sometimes questions not based on a problem are poor fits for SE.

Do note: I am not interested in actually trying to do something with this, it is simply a curiosity, so don't be concerned if the things that come to mind aren't FDA tested, I'll settle for answers that simply show there are such things that are likely non-toxic regardless of thorough safety testing. Also the phase change from liquid to gas is a possible one too if you know of any safe chemicals here that wouldn't cause uncontrollable internal pressure and would have that phase change at the temperature range specified.

Just trying to come up with an idea for a compound that could meet their marketing hyperbole and finding I come up short.

I tried to use the curated dataset from Mathematica to answer this question.  You can ignore the following if you have no interest in Mathematica and just jump down to the end for my ideas.
The following code gets all of chemicals currently in the curated dataset and grabs the melting points and NFPA health rating.  The second line finds the chemicals with melting points between -7 and 0 degrees and the last line looks for chemicals with a health rating of 0, which is low risk.
out = {#,  ChemicalData[#,"Name"], ChemicalData[#,"MeltingPoint"],
   ChemicalData[#,"NFPAHealthRating"])& ChemicalData[];
cull1 = Cases[out, x_ /; -7 < x[[3]] < 0];
cull2 = Cases[cull1, x_ /; x[[4]] == 0];

We go from about 43000 chemicals in the first list to 220 in the second and finally 6 in the last.  I'm not sure I agree with the health ratings of these compounds but here you go (melting points in parentheses):

2-methyl-2-propanethiol (-0.5)
prehnitene (-6)
trans-1,2-dichlorocyclohexane (-6)
hexadecene (-6)
trans-1,2-dibromocyclohexane (-6)
olive oil (-6)

The only chemical I would deem "food safe" would be the last one.  Perhaps some type of food-grade oil is the mystery substance.


Why is ethyne and not ethene used for welding? Ethyne (Acetylene) is used with $\ce{O2}$ for welding. My question is that why is ethene not used? What properties of ethyne make it suitable for welding?

An acetylene molecule is composed of two carbon atoms and two hydrogen atoms. The two carbon atoms are held together by what is known as a triple carbon bond. This bond is useful in that it stores substantial energy that can be released as heat during combustion.The triple bond which makes the oxy-acetylene flame the hottest of all gas flames is also responsible for two rather exceptional properties of acetylene gas.

The first is: If free gaseous acetylene is subjected to severe shock, or a source of ignition, some of the triple bonds may break, releasing enough energy to cause all the other molecules in the enclosed volume to decompose into carbon and hydrogen with explosive force. The force of such an explosion is not so great as that released by the explosion of most mixtures of acetylene and oxygen, or acetylene and air, but it is substantial, and can be withstood only by extra-heavy-wall steel tubing.

The other property of acetylene is that the flammability range of mixtures of air and acetylene is broader than that of any other fuel gas/air mixture. Acetylene/air mixtures can be ignited when they contain anywhere from 2.5 percent acetylene to 80 percent acetylene. Mixtures of methane (the principal component of natural gas) and air are flammable when they contain as little as 5 percent methane and not more than 15 percent methane.

To be suitable for welding operations, a fuel gas, when burned with oxygen, must have the following:

High flame temperature.
High rate of flame propagation.
Adequate heat content.
Minimum chemical reaction of the flame with base and filler metals.

Among the commercially available Oxy Fuel welding fuel gases, acetylene (combination of hydrogen and carbon) most closely meets all these requirements. Other gases, fuel such as MAPP gas, propylene, propane, natural gas, and proprietary gases based on these, have sufficiently high flame temperatures but exhibit low flame propagation rates.


What if solute and solvent exist in equal quantities? A solution is made of solute and solvent, like salt in water, however, what if the solute and solvent exist in equal quantities?
Which one is the solute and which is the solvent? If one of them is water I think it will definitely be the solvent, but what if neither of the two parts are water?

As far as I can recollect, I think in such a situation you can arbitrarily select one of them as solute and one as solvent. However, according to convention, it is better to assign the solvent as a substance which is usually used as a solvent in other situations.


Why does DMF (dimethylformamide) have 3 peaks in its 1-H spectrum? I get that the carbons can't rotate around the nitrogen due to its partial double bond character, but how does this result in the hydrogens being in different environments? As such would this mean all non rotatable bonds would result in hydrogens all with different peaks?

If you understand the partial double bond character induced by the N-CO moeity, then you can draw out the molecule as shown below.

The three signals you get are from the CH and the two CH3 groups. When the energy barrier for rotation is high to prevent averaging (fast rotation), you can see that the two methyl groups are non-equivalent: one is trans to the C-H, and the other is cis to it. This is a very common characteristic of many amide-containing molecules. The energy barrier to rotation is highly dependent on steric interactions. The number of possible conformers arising from even small peptide molecules increases very rapidly, when you consider the combination of cis/trans type orientations of the amide bonds.

Now, as you start to heat this system (or observe at a lower magnetic field - that may be for another question), you will start to see signal averaging, as the two methyl groups begin to exchange on the NMR timescale. As this rate of exchange becomes faster, you will get signal coalescing and finally complete signal averaging to the point where the two CH3 groups are equivalent. This is seen below, with the bottom spectrum being at low temperature (and/or high magnetic field) and the top spectrum being at high temperature (and/or low magnetic field).

Signal averaging via rotation about bonds happens even in single bonds. If you cool a CH3 group down far enough, you will be able to slow down the rotation of the C-C bond enough to be able to distinguish all protons as non-equivalent.


What is the definition of thermodynamic stability? The term thermodynamic stability is used on this site, but I can't find a good definition. Is is a quantitative or a qualitative concept? Does it apply to a single compound, or a pair of compounds? Does it a relative or an absolute value?
An answer to another question, Thermodynamic stability of benzene derivatives , states

Thermodynamic stability of compounds can be determined by obviously enthalpy of formation ($\Delta H_{_\mathrm f}$) of individual compounds. The enthalpy of formation will be lesser if the compound is formed from its constituent elements enjoys some greater stability.

The enthalpy of formation considers the synthesis of a given compound from the elements. If the enthalpy of formation of two substances like bromomethane and chloromethane are different, it might be due to differences in bond strengths (or electronic states) in the two compounds, or differences in bond strengths (or electronic states) comparing elemental bromine and elemental chlorine.
I do understand the concept of thermodynamically favored product. In this case, there are competing reactions leading to different products. Because they start with the same reactants, there is no issue about the reference state. However,  if I compare two substance with different composition (i.e. not isomers), I don't know what it means to compare their thermodynamic stability. I also understand local and global minima, as discussed in this answer which references a single set of atoms that undergoes transformation. The problem arises when comparing compounds with different sets of atoms.
Thermodynamics usually is pretty well defined. Thermodynamic stability, however, is a bit of a mystery to me.

The full phrase should be thermodynamic stability with respect to ____, where the dash indicates a process, or a chemical reaction.
A mixture of hydrogen and oxygen is thermodynamically unstable with respect to water formation.
Similarly, a diamond is not forever (which may not please De Beers and ladies). It is thermodynamically unstable with respect to conversion to graphite.
Also, thermodynamic stability is a relative term which is often contrasted with reactivity or kinetic stability.
Diamond is kinetically stable at room temperature for the same process (lucky ladies can smile again).


Why is mer-AlQ3 more stable than fac-Alq3? Several research papers and books say that the meridional isomer of tris-(8-hydroxyquinoline) aluminium is thermodynamically more stable than the facial isomer.

What is the reason behind this stability?

The stability of meridional isomer over facial isomer can be deduced by considering the interactions between atoms of the ligand 8-hydroxyquinoline.
 In considering the interactions if we take the repulsion between the electron rich atoms dominated in 90 degrees and at 180 degrees these interactions become negligible, we will find that in meridional isomer, there are dominating
i) 2 N-N interactions
ii) 2 O-O interactions
iii) 8 N-O interactions.
 But if we look into the facial isomer of tris(8-hydroxyquinolinato)aluminium we will find there are dominating
i) 3 N-N interactions
ii) 3 O-O interactions
iii) 6 N-O interactions.
 So in facial isomer two N-O interactions of mer isomer has turned into 1 N-N interaction and one O-O interaction. Due to high charge density on small sized O atoms O-O repulsions are much much greater than N-O interactions and thus the electronic repulsions are much more greater in fac isomer which decreases its stability than meridional isomer


When are we justified in making assumptions in questions The density at $20^\circ\mathrm{C}$ of a $\pu{0.500 M}$ solution of acetic acid in water is $\pu{1.0042 g/mL}$. What is the molality of the solution? The molar mass of acetic acid is $M(\ce{CH3CO2H})=\pu{60.05 g/mol}$.
  The density at $20^\circ\mathrm{C}$ of a $\pu{0.258 M}$ solution of glucose in water is
  $\pu{1.0173 g/mL}$, and the molar mass of glucose is $M(\ce{Glc}) = \pu{180.2 g/mol}$. What is the molarity of the solution?

In question one we can assume a 1 liter solution and in question two we can assume a 100 gram sample. I understand these assumptions are needed to find other information, but when are we justified in making these assumptions?

You can make certain assumptions in these and similar problems because concentration of uniform substances (whether it's expressed as molarity, molality, mass density, or something else entirely) is an intensive property, i.e., it has no dependence on the total amount of material in the object or sample being measured. For example, if you have a homogeneous solution, the solute particles are distributed uniformly among the particles of solvent, and the two form a constant ratio. If I split the sample in half, the solute particles are halved, but so are the solvent particles, so the ratio remains the same.

As it happens, there's no need to assume any particular quantities to solve this problem; instead, you can use algebraic substitutions using the known values given and common definitions to solve for either molarity or molality directly. Let $M = \frac{m_{solute}}{n_{solute}}$ be the molar mass of the solute, $\rho = \frac{m_{solution}}{V_{solution}}$ be the mass density of the solution, $c = \frac{n_{solute}}{V_{solution}}$ be the molarity, and $b = \frac{n_{solute}}{m_{solvent}}$ be the molality, and note that $m_{solvent} = m_{solution} - m_{solute}$. Then, we have:

$$
\begin{align}
b\ & = \frac{n_{solute}}{m_{solution} - m_{solute}}\\
 & = \frac{n_{solute}}{\rho \cdot V_{solution} - M \cdot n_{solute}}\\
 & = \frac{n_{solute}}{\rho \cdot \frac{n_{solute}}{c} - M \cdot n_{solute}}\\
 & = \frac{c}{\rho - M \cdot c}\\
\end{align}
$$

You can rearrange this equation to solve for $c$ (molarity). Note, of course, that units need to match, but it's trivial to convert between grams and kilograms, mililiters and liters, etc.


How is the degree of unsaturation derived? The formula for degrees of unsaturation is:

$$1+ \frac{1}{2} \sum ( n_i (v_i-2))$$

$v_i$ = valency of atom
$n_i$ = number of atoms with that valency

Could someone explain how the formula works, and why divalent atoms don't contribute to unsaturation (as evident from the formula).

This is the correct formula:

$$ DU = 1 + \frac{1}{2}\sum _{v=1}^{j} n_{i}(v-2),$$
where

$j$ = maximum valence of any atom in molecule
$v$ = valence of atom(s) 1 to whatever
$n_{v}$ = number of atoms with that valence

Wikipedia gives a good explanation. In simple terms the formula will give the number of double bonds in an alkene. For all alkanes the number would be 0 of course. Rings and ethynes would complicate the use of the formula. Think of using the equation on chemical formulas like $\ce{C_{x}}\ce{H_{y}}$ or $\ce{C_{x}}\ce{H_{y}}\ce{O_{z}}$.


N-O bonds in Nitric acid The wikipedia article on nitric acid mentions the following structure of $\ce{HNO3}$ in which the two $\ce{N-O}$ bonds (other than that of $\ce{N-OH}$) are of different length i.e.,  $119.9\,\pu{pm}$ and $121.1\,\pu{pm}$.

But it also mentions that $\ce{HNO3}$  molecule is a resonance hybrid of - (only major contributing ones)

I can't understand how can both be simultaneously correct since the resonance hydrid should have same two $\ce{N-O}$ bond lengths (other than $\ce{N-OH}$ one). How is it correct or is the source itself wrong?

In addition to permeakra's comment, I want to add the following thoughts:

  1) How were these bond lengths determined?

Were these values determined computationally? If so, what is the standard error in the computational method uses? In other words, is the difference between the two values within the error of the method, and thus the difference is not significant. Additionally, a computational method operates on single molecules "in the gas phase", and there might be just enough interaction between the hydrogen atom and one of the other oxygen atoms. This interaction would also explain the disparity in bond angles:

Were they measured experimentally, for example from x-ray crystallography? If so, then dissymmetry arising from a preferential crystal packing arrangement can lead to differences. In particular, in the solid state, the position of the hydrogen atom can make a big difference in preferential hydrogen bonding, either intermolecularly or intramolecularly. For units like the following hyrogen-bond dimers, the hydrogen bonding could lead to dissymmetry between the N-O bonds.

  2) Bonds vibrate

They do, even in the solid state. Thus, an x-ray crystal structure has error its measurements as well. Again, is this difference in bond lengths within the experimental error?


Is there a difference between Ca++ and Ca(2+)? In this EKG tutorial it says

  Depolarization conducts slowly through the AV node, since it is carried by slow moving $\ce{Ca++}$ ions...

Is there a difference between $\ce{Ca++}$ and $\ce{Ca}^{2+}$?

The answer to this question is: No. There is no difference between the two, they are merely two different notations.


Polar rearrangements: What is this reaction called? I have been trying to find more information about the given reaction but cannot seem to find it? I want to know the name so I can read up more information on it. Can anyone help?

It is a Wagner–Meerwein rearrangement. Your particular example has the following mechanism:

Edit: Regarding your comment, the best way to see what is happening is to draw the product so that it resembles the reactant (like in the image below) and then in another step draw it in a more reasonable conformation:


Properties of inconel I am doing a project on the physics of the spaceX raptor engine. I believe it is made out of a certain type of inconel alloy. I want to understand why this alloy has the properties that it does, but I don't know anything about materials science. I am a senior physics major.
I am looking for a good introductory text/texts that will give me insights into the properties of inconel so I can explain why it doesn't flare at high temperature+pressure in my paper.
Any other advice would be welcome as well. Thanks for the help!

There are several Inconels , the basic one is 600. Because the Inco and Huntington divisions are now gone or acquired ,it would be difficult to get literature . But they produced copious amounts , you may be able to find some ; I have about a 6" stack of brochures I have been too lazy to discard. American Society for Metals Handbooks should be easy to find ( there are several issues). ASM is a common and comprehensive source. What do you mean "Flare" ?

The 600 Inconels do not age harden; The 700 Inconels do age harden. All have good strength and oxidation resistance at high temperatures . Some useful above 2000 F. If you are talking about a rocket ,I guess you do not need blade and vane alloys ( for turbines). 800 and 900 are Incoloys  ; 400 and 500 are Monels
Addenda
Looking at SpaceX info they are using something like Inconel 750 ( formerly X , and X750 ), so it is age hardened for strength due to pressure in the engine. Supposedly they have a modification named SX 500 . Apparently by "flare" they means rapid oxidation because of the oxygen in the engine , but they do not explain.


Minimal basis for Li Currently, I'm trying to get into the field of quantum chemistry and was reading about minimal basis sets. As I understood this is the minimal number of orbital/spatial wavefunctions needed for a specific number of electrons.

For example, H has one electron, so a minimal basis set would be $\mathrm{1s}$ (STO, STO-nG, or something else).

Li has 3 electrons, so my guess would be that we need 1s and 2s. But in the literature, also the $\mathrm{2p_x}$, $\mathrm{2p_y}$ and $\mathrm{2p_z}$ orbitals are included in the minimal basis. Why is this the case? Together with the spin degrees of freedom, the three electrons already fit in the 1s and $\mathrm{2s}$ orbitals?

A minimal basis set is a set that employs just one function to describe each orbital. For $\ce{H}$ you take into account one orbital ($\mathrm{1s}$), so you use one function, and for $\ce{Li}$ you consider five orbitals ($\mathrm{1s}$, $\mathrm{2s}$, $\mathrm{2p_x}$, $\mathrm{2p_y}$, $\mathrm{2p_z}$) and you use five functions, one for each.

What you described ("the minimal number of orbital/spatial wavefunctions needed for a specific number of electrons") would use a function for each occupied orbital - leaving unoccupied orbitals, even in a partially occupied subshell, out of the picture. This is a very bad idea - shells are a set of degenerate solutions for the hydrogen atom, so reducing the number of functions to describe them will seriously distort the depiction of the atom.

The usual approach is, instead, to always provide functions at least for the entire valence shell of the atom (and its inner shells). So for all 2nd period elements, you would use all 2nd shell orbitals ($\mathrm{2s}$, $\mathrm{2p_x}$, $\mathrm{2p_y}$ and $\mathrm{2p_z}$), even in a minimal basis set, no matter how many electrons actually occupy that shell. What still makes that basis set a minimal one is that you are only using one function for each orbital.

Similarly, a minimal basis set calculation for $\ce{Na}$ would include one function for each orbital in the 1st, 2nd and 3rd shells, and so on.


How to interpret a MALDI mass spectrum? I have some questions regarding how to read a MALDI mass spectrum in a paper. I am a biologist by training, so I would like some analytical chemists to help me out.

The researchers ran a western blot, and immunoblotted for a protein of interest. They cut out this band from the western blot, subjected the protein to trypsin digest, and used these peptide fragments for MALDI-TOF analysis. I know for a fact there will be other proteins at the same apparent molecular mass which would have also been included in the analysis, which should be taken as contaminants in the resulting spectrum.

They produced a graph with the absorbance intensity x 1,000 on the dependent axis, and mass:charge (m/z) ratio on the independent axis.

The authors point to two peaks, which they say correspond to two peptide fragments from the tryptic digest of their protein of interest. They provide the supposed peptide fragment sequences and m/z ratio data for these two peaks, and I checked to make sure they are the same as reported. The authors expect to see two peaks: protein of interest is digested to two fragments, A + B.

I have two questions:

Peak 'A' and peak 'B' are of different intensities. Peak 'A' is 20-fold greater than peak 'B'. However, the tryptic digest must yield equimolar amounts of peptide fragments A and B. Is it possible that these intensity peaks can be 20-fold different? Wouldn't the absorbance intensity correlate to total analyzed peptide mass?
Is it normal to multiple absorbance intensity by a (seemingly arbitrary) factor, in this case, 1000? Since the absorbance intensity of peak 'B' is < 0.5 (after scaling up), could this represent noise rather than signal in the analysis?

Please see the wiki articles for the general method:

http://en.wikipedia.org/wiki/Mass_spectrum
http://en.wikipedia.org/wiki/Matrix-assisted_laser_desorption/ionization
http://en.wikipedia.org/wiki/Time-of-flight_mass_spectrometry

Then let's start with question 2: yes, you can and you should. The spectrometer measures in arbitrary units, therefore you can speak only about relative abundances.

And question 1: it depends how easily is the given molecule ionized. So if the fragment A is rich in polar aminoacids, whereas fragment B is mostly hydrophobic and difficult to ionize, than you should expect this big difference in signal intensities.


Determination of structure by 13C and 1H NMR spectroscopy I'm trying to interpret a $\ce{^{13}C}$-NMR and $\ce{^1H}$-NMR spectrum. So far, I've decided that $\ce{CH3-C}$, $\ce{CH2-CH3}$ and $\ce{CH-CH}$ exist in the structure. I'm not sure but I guess there should be different seven carbons too.

I've come up with the structure below but I feel like there is not enough information to be sure. What should I need to consider to go further at this point?

This is a tricky little problem, given there is an element of symmetry to consider. As mentioned in the comments, once you come up with a proposed structure, and your one is not a bad first attempt, you should go through and test the spectrum against what you would predict from your structure. Unfortunately, in your suggestion, there are 4 non-equivalent $\ce{-CH3}$ environments, and the spectrum only shows two, so we can eliminate that structure as a possibility. But don't be disheartened — that is the process of structure elucidation! Propose a stucture, test it, refine or eliminate it. And then start again.

Looking at the $\ce{^1H}$ spectrum, we can see:

$\ce{-CH3}$
$\ce{-CH2}$
$\ce{-CH3}$
$\ce{-CH}$
$\ce{-CH}$

We can construct the $\ce{-CH2CH3}$ spin system based on coupling, and the chemical shifts of the two downfield protons indicate a double bond (and, using some old tricks of the trade, looks to be cis — the splitting here is much less than about $14~\mathrm{Hz}$, which would be the expected splitting between the outside lines of the methyl triplet). The $\ce{CH3}$ at about $3.3$ is in the general ball park for a methyl ether, I'd say.

That gives us 10 protons, but the molecular formula says $\ce{H20}$ … so we must have symmetry in there somewhere which gives us:

2 $\ce{-CH2CH3}$
2 $\ce{-CH3}$ (probably $\ce{-OCH3}$)
2 $\ce{-CH=CH-{}}$

So, looking at the carbon spectrum:
I count 7 carbon peaks.
From the proton information, we have 5 $\ce{C}$ environments, each appearing twice, which gives a total of 10 carbons. The other 2 carbon environments occur once only in the molecule.

From DEPT experiments (which nobody in the real world should ever run when they can instead run a phase-edited HSQC instead) we can see we have

$\ce{CH3}$ at $9.9$ and $51.4$
$\ce{CH2}$ at $38.0$
$\ce{CH}$ $132.8$ and $133.2$
$\ce{C}$ at $47$ and $110$.

We know (or should know) what the non-quaternary carbon cantres all are. We worked that out from the $\ce{^1H}$ spectrum. So let's concentrate on the quaternaries. The peak at $110$ is interesting, and is typical of a carbon centre attached to 2 oxygens. Many sugars have carbons that come around this area. The other quaternary at about $47$ is typical of an aliphatic quaternary; not overly exciting.

So somehow we need to construct a molecule that has 2 non-equivalent quaternary centres that have a plane of symmetry through them, with the $\ce{-CH2CH3, -OCH3}$ and $\ce{-CH=CH-{}}$ attached. Each of these three spin systems must be isolated, otherwise we would see coupling to the vinylic protons.

As this is clearly a homework question, I'll leave it to you to piece together, and ask further questions if you need, but my answer gives the exact same predicted spectra in ChemDraw as your spectra.


Does ICP-OES detect molecules in suspension? I'm curious whether ICP-OES has the ability to detect molecules in suspension?

It will depend on the size of the particles which are in suspension. If the particles are very large compared with the capillary feeding the solution into the spray chamber then the capillary will block up and you will detect nothing and end up very unhappy with a broken machine.

If the particles are much smaller than the capillary then they will be able to enter the spray chamber and then be present in the small droplets which enter the plasma torch. As a result it will be possible to observe the particles by means of the ICPOES machine.

With a normal solution a droplet enters the torch.
The droplet dries out to make a solid particle
The solid particle melts
The molten droplet then boils in the heat of the plasma
The molecules formed are broken down to atoms in the heat of the plasma
The atoms are then ionized.

The inclusion of solid particles in the liquid fed into the ICPOES will give as the droplets boil to dryness additional solids particles. These solid particles may modify the behaviour of the other elements in the torch. I suspect that it may be similar in some ways to the processes which occur in the fireball as it cools after a low altitude nuclear detonation where things can adsorb onto existing particles. If you look in the literature at studies of fallout particles, the particles from the nevada tests have radioactivity deposited on minerals which are similar to those found at the test site.

I have found that if you increase the salt content of what you are feeding into either a ICPOES or ICPMS that the intensity recorded for a given concentration of an element (not found in the added salt) will decrease. This is a possible sign that some of the element is being lost onto the particles formed by the dehydration of the salt solution.

It is best if the ICPOES machine is only ever fed with samples which have been filtered through a very fine filter. I tend to use a syringe filter with holes which are about 0.5 micrometers. This helps to keep the fine particles away from the machine.


In TOF Mass Spectrometry how does the detector 'know' the difference between ions of different elements? After the positive ions have been accelerated, I know that they travel through the drift zone at different speeds due to their masses.

But surely, a heavier ion/element that enters the drift zone first may still reach the detector before a lighter ion if that lighter ion is one of the last ions to enter the drift zone. If so, when these ions reach the detector, how does it know that the heavier ion IS heavier than the lighter ion that does not reach the detector until afterwards.

*I'm aware my question is a tad confusing, let me know if I should try and reword it!

I suppose that you are referring to a Wiley–McLaren time‐of‐flight mass spectrometer [Rev. Sci. Instr. 65, 3344 (1994)] which is basically a tube containing an electrode stack for accelerating the ions toward a detector. As mentioned in the comments, one can discriminate different masses if one uses a pulsed experiment, for instance using pulsed lasers or pulsed electric fields to extract the ions.

Another option is to add a magnetic field to the setup in order to bend the trajectory of the ions. The radius of the trajectory depends on the mass-to-charge ratio of the ion and one can record spectra as a functoin of the magnetic field to get a certain mass selected or use a position-sensitive detector to measure several masses simultaneously.

Yet another solution is to use quadrupole–time-of-flight mass
spectrometry, where radio frequencies are applied to four electrode rods placed on the circumference of a circle. Depending on the frequency that is applied to the rods, only ions of a certain mass-to-charge ratio make it out of the quadrupole and the qaudrupole thus serves as a mass filter.

Note that all these techniques are sensitive to the mass-to-charge ratio only and one cannot distinguish between, e.g., O$^+$ and S$^{2+}$.


An introduction to "oxidation state"/"oxidation number" I just can't get my head around oxidation states. They're really confusing! Why do I need to assign a number to some element?

What are oxidation states?
How and why should we assign oxidation states?
Are oxidation states real?

#TL;DR;

Oxidation States are useful tools used in nomenclature, redox chemistry etc.
There are a few simplistic rules to determine the oxidation state of elements in an atom; but they're not applicable to all the cases.
There are instances where the assignment of oxidation state isn't useful, or where the oxidation state is ambiguous.
We're easier off assigning, rounding off or estimating oxidation states ourselves in some cases, such as intermetallic compounds.

##What is an oxidation state?
Oxidation state (OS), or oxidation number, is an integer or decimal number assigned to an element in a chemical species. This assignment follows certain rules and guidelines, to make it a useful tool whenever we need it. Oxidation states can be negative, positive, or zero.
According to Comprehensive definition of oxidation state
(IUPAC Recommendations 2016), p 833:

The oxidation state of an atom is the charge of this atom after ionic approximation of its heteronuclear
bonds.

All those fancy words up there mean that, when you're assigning OS, you take all of the bonds in the species to be completely ionic. No purely ionic compound exists, and even, say, the $\ce{Cs-F}$ bond in $\ce{CsF}$ has some covalent character. So you have to understand that OS does not happen. It doesn't refer to any real-life phenomenon.
##Why do we assign OS's, then?
Oxidation states are a powerful tool for chemists, and save us a lot of headache. You might question "why do we need to give an element a number anyway?" which arises because you usually only learn how to give elements those numbers, but no real-life application becomes necessary or evident.
OS is a tool, and is not the only tool, wherever it's being used. It's like the shovel that digs the ground. You can also do the digging with a spoon. Why do you choose a shovel?
#Assigning OS
##The Basis
The basis of assigning OS, in simpler cases, is just good old electronegativity. There are exceptions where this would not work, but you don't need to worry about them at this stage.
You're usually given the values of electronegativities, and those aren't what you should worry yourself about. There are different scales for measuring electronegativity, and IUPAC recommends the Allen scale because it's the only scale independent of OS.
##OS Assignment Using the (Molecular) Formula
You're given a formula of a compound you've never met, and you're asked to determine the OS of some, or all of the elements present. This is a classic scare if you don't understand what you're doing.

Determine the OS of all the elements in $\ce{H2C2O4}$.

Like most of the problems you face, there are some "unknown" and some "known" values. You need to identify a starting point, go on from there until everything sorts itself out.
Here's what we know: We need to use the electronegativities to find out what the OS's are. O is the most electronegative, C is next, and H is the least electronegative.
Here's what we want to know: We have three OS's here that we need to determine. C, H and O.
As it happens, the elements in the rightmost and leftmost parts of the periodic table are the most and least electronegative elements of all, respectively. So we expect their stance to be more clear: The ones on the left, the alkaline metals, tend to lose electrons and the ones on the right, tend to gain. As a general rule of thumb, the closer an element is to the middle of periodic table, the fuzzier and harder determining its OS is.

Always try to remember the location of the element in the periodic table, and start from one that's the rightmost and/or the leftmost. Don't try to quantify this rule, since it's merely a guideline. Sometimes, deciding the OS of oxygen becomes difficult!
Thus, we would expect oxygen or hydrogen to be easier to determine here. That is the case: the electronegativities tell us that oxygen must have a negative OS, and hydrogen a positive one.
Time to quantify this. You remember from the intro to electronic configurations that O's electronic configuration ends in $\mathrm{2p^4}$, and that if it ended in $\mathrm{2p^6}$, it would've given oxygen a relative amount of stability. It would've done the same here if it could grab the two electrons from the carbon, but in reality, that hasn't happened. But anyway, when assigning OS, you assume that the bonds are ionic, and the oxygen has already stolen carbon's two electrons.
If a neutral oxygen with an electronic configuration ending in $\mathrm{2p^4}$ gains electrons to become one ending $\mathrm{2p^6}$, how many electrons has it gained? Two. Thus, the OS of oxygen in $\ce{H2C2O4}$ is −2.
You can apply the same process for hydrogen. It has only one electron in its neutral form, and it's about to lose it. How many electrons can it lose?
Determining carbon's OS is more tricky. Carbon is in group 14, and it's not obvious what its OS should be. It's the closest element to the middle. Here's a workaround: Instead of trying to find out which OS's carbon can have, we calculate how much is left as carbon's share.
$$\color{green}{\text{The sum of the OS's in the species is equal to the charge} \\ \text{of the species, be it a molecule, an ion etc.}}$$
This is a rule you have probably also met in your textbook. The charge of the species is an indication of excess or lack of electrons, and should definitely affect what OS's we assign to the elements. Why? Because OS is itself some form of counting electrons.
($\Rightarrow$ if they're just counting electrons, why do we see fractional oxidation states? See OS Assignment Using the Lewis Structure below)
$\ce{H2C2O4}$, oxalic acid, is a neutral molecule. So the sum of OS's must equal zero. Note that we count hydrogen twice, because two hydrogens have two electrons in neutral form, and they will lose two electrons. Likewise, four oxygens will, in a hypothetical purely ionic oxalic acid, gain eight electrons overall.
$$\begin{equation}
2(+1) + 2(\mathrm{OS}_\ce{C}) + 4(-2) = 0 \\
\mathrm{OS}_\ce{C} = +3
\end{equation}$$
##Most Common Patterns in the OS of elements
If you consider all the possible species that can be synthesized, there are great many bizarre trends and anomalies in OS assignment. Refer to the Further Reading section to learn more about them.
However, for most educational purposes, OS's for a number of elements follow simple guidelines worth mentioning:

Fluorine's OS is always −1 (except in $\ce{F2}$ itself).
Halogens below fluorine typically adapt a −1 OS as well, but always check when there are more electronegative elements such as fluorine and oxygen present.
Alkaline metals adopt a +1 OS.
Alkaline earth metals adopt a +2 OS.
You're almost always better off determining the OS of transition metals using the sum rule.
If oxygen is not the leftmost element present, or there are alkaline and alkaline Earth metals present, don't start by assigning an OS to oxygen.
If there are any of the elements in guidelines 1–4; don't start by assigning an OS to hydrogen.

Determine the OS of all the elements in $\ce{NaO3}$.

There are only two possibilities: Start from Na, or start from O. Both are, as said, in left and right parts of the periodic table. So which?
If we take oxygen, and assign the "normal" OS, −2, we see that Na has to be +6 because the sum should equal zero. Something doesn't sound right. As guideline 6 says, we shouldn't start from O when there's Na present in the compound. So we take the other route.
$$\begin{equation}
+1 + 3(\mathrm{OS}_\ce{O}) = 0 \\
(\mathrm{OS}_\ce{O})= \frac{-1}3
\end{equation}$$
###Check your understanding
Find the OS of all the elements in these compounds:

1.$\ce{NaAlF6}$  2. $\ce{C2O2F2}$
3. $\ce{C8O8}$ 4. $\ce{Au2O3}$

##OS's That Don't Exist
The upper bound for OS's that you will meet is +7 for some manganese $(\ce{MnO4-})$, or chlorine $(\ce{ClO4-})$ compounds. OS's more than that are only found in exotic compounds. Thus, if you get a, say, +9 OS, you should start doubting the route you took.

Determine the OS of W in $\ce{H4W2O13}$.

Three possible starting points. We rule out tungsten because it's the nearest to the middle. Of the other two, hydrogen would lose 1 electron, and oxygen would tend to gain two. Right?
Well, if you solve it for tungsten, considering the neutral overall nature of the compound, you'll get an OS of +11! That's definitely not right, since the highest observed OS is +10 (See Further Reading). This could only mean one thing, since hydrogen can't adapt another OS or tungsten would have an even higher OS. This means that some of the oxygens are in the form of peroxides, superoxides etc.
Had we known that the structure of the compound, more elaborately written, is $\ce{(H2O)(O)(O2)2WOW(O2)2(O)(H2O)}$, we wouldn't have made the mistake of thinking all the oxygens have the normal OS of −2.
Hence, it becomes necessary to be able to deduce OS from structures.
##OS Assignment Using the Lewis Structure
At times you need to determine OS in an electron dot diagram, or Lewis structure. This is sometimes useful/needed over determining from the molecular formula because

either the molecular formula doesn't give us the whole picture (see below), or
it gives us impossible answers or no way to go further, like the tungsten compound's case, or
it's already what the examiners gave us!

Determining OS from Lewis structure is even easier than deducing it from the molecular formula, because you already see the bonds. It's the same process in the IUPAC definition above: Ionic approximation

The electrons in a bond go count as the more electronegative atom's. (There are exceptions to this, see Further Reading)
The electrons in a bond between the same atoms are shared equally. So in a $\ce{C-C}$ bond, each C atom gets one electron from the bond.
The OS of each element is the number of valence electrons minus the sum of lone pairs and electrons it gains from bonds.

A good example used to illustrate this is acetone.

Determine the OS of all the elements in acetone, $\ce{C3H6O}$.

The answer is in the spoiler below (no peeking ಠ_ಠ):

 $\mathrm{OS}_\ce{O} = -2, \mathrm{OS}_\ce{C} = \frac {-4}3, \mathrm{OS}_\ce{H} = +1$

There's one oxygen in acetone, and it's bonded to the carbon:

Assigning the rules, we get

$\ce{C1}$ has four valence electrons, but there are "seven electrons around it". Both electrons from the $\ce{C-H}$ bond belong to carbon. Thus, $4 - 7 = -3$.
$\ce{C2}$ has four valence electrons, but only two electrons are around it. One from of each of the two $\ce{C-C}$ bonds. That makes $4 - 2 = +2$.
$\ce{C3}$ is just like $\ce{C1}$, $\mathrm{OS} = -3$.

##Fractional OS's
Didn't we say earlier that OS assignment is some kind of counting electrons? Then how come we encounter fractional OS?
Now you know the answer. Some elements in some compounds show two different oxidation states, and when computing the OS from the molecular formula, you actually compute the average OS. None of the carbons in acetone have an OS of $\frac {-4}3$.
Familiar examples of this 'average' OS are carbon in acetic acid, which is computed to be zero but is in fact −3 and +3, and iron magnetite, which is $\frac 83$.
##Further Reading
The school lessons are over, but the fun is actually just about to start. This is the part where, you know, you see what you've been taught isn't accurate. Read on if you're interested. It might get confusing for first-timers.
###Be Liberal in Thinking
OS is a useful concept. Unfortunately, it's hypothetical. There's this mindset that there should only be one OS for each element in a compound out there. That's simply not the case.
$\ce{S2O3^2-}$, thiosulfate anion, is one such compound where the two approaches we just introduced produce conflicting results. The average $\mathrm{OS}_\ce{S}$ is +2 for sure, but is it +5 and −1, or +6 and −2?
Turns out, studies are conflicting on this matter. The final say here is provided by IUPAC recommendations linked above:

###5.3 Nominal OS
In systematic descriptive chemistry, OS is used to sort out compounds of an element; in electrochemistry, it represents the electrochemically relevant compound or ion in Latimer and Frost diagrams of standard (reduction) potentials. Such purpose-oriented OS may differ from the OS per definition and are termed nominal, or more specifically “systematic” or “electrochemical”.

Consider thiosulfate, the $\ce{S2O3^2−}$ anion: The sulfur–sulfur
bond is practically a single bond, which yields OS = +5 for the central S and −1 for the terminal S atom by both algorithms. The electrochemical OS of sulfur in thiosulfate is their average, +2. On the other hand, a systematic OS of +6 may be chosen for the central S atom, as if the sulfur–sulfur bond were approximated ionic, to emphasize the similarity of the peripheral O and S atoms obtaining OS = −2 as in sulfate.

##Exceptions
These are the cases where the methods above don't reliably work in determining OS:

The most glaring example is when the more electronegative atom is bonded reversibly as a Lewis acid.
We originally introduced electronegativity as the basis to decide OS. Truth is, the real basis is the energy level of molecular orbitals, and this, in most of the cases, corresponds to the more electronegative atom getting all the fish. It's difficult and/or impractical to use complicated MO schemes to determine OS, so we stick to electronegativities.
In the case of Lewis-acid ligands, however, this backfires and gives us the opposite of the correct answer.
Non-innocent ligands are ligands whose OS is ambiguous because they're very redox-active. This article does a nice job of explaining $\ce{H2}$ and nitrosyl non-innocent ligand briefly (§ 10 & 11).
Nominal OS's; like thiosulfate above.
Metallic compounds; there are some obvious cases such as TiO, but it gets more complicated – Source
Intermetallic compounds, such as Zintl phases, where it's safe to assume an OS of 0 for redox purposes or compounds with delocalization over different atoms  – Source (§ 13)
Compounds with a close electronegativity such as Se and C in $\ce{Se-C}$ bonds or H and P in $\ce{H3PO3}$. – Source

##Exotic OS
These are compounds whose certain elements demonstrate unusual OS's:

Alkalides, which are salts containing alkali metal anions with an OS of −1. Lithides are currently not known, but some kalides and natrides are stable in room temperature.
The highest known OS belongs to Ir in $\ce{[IrO4]+}$ which is +9, and potentially +10 for Pt in $\ce{[PtO4]^2+}$. The lowest known OS is −5 for boron in Zintl phases. – Wikipedia
Platinum hexafluoride is a strong oxidizing agent that can oxidize $\ce{O2}$, giving dioxygenyl, with an unusual OS of $+\frac 12$ for O.
Alkaline earth metals are observed to show unusual +1 OS in some complexes. – This article reports the synthesis of complexes with Mg–Mg bonds.

##Useful Links
[1]: List of Oxidation States of Elements
[2]: High-valent Iron
[3]: Using OS to analyze redox reactions; a tutorial
[4]: ChemGuide OS tutorial
<Under construction . . . Probably>
<......>


Hexagonal unit cells I know 3 types of unit cells, the simple cubic, the body centered and the face centered.

But what is a hexagonal unit cell, and why does it have a hexagon base while others are all cut into a cubic? And how can I calculate the number of atoms in every primitive hexagonal unit cells?

The cubic primitive, cubic body-centered and the cubic face-centered are just the three possible Bravais lattices of the cubic crystal system.

In total, we know seven crystal systems, namely (in order of increasing symmetry) triclinic, monoclinic, orthorhombic, tetragonal, trigonal, hexagonal and cubic.

Among these, the triclinic crystal system has the lowest symmetry ($a \ne b \ne c;\ \alpha, \beta, \gamma \ne 90°$), the cubic crystal system has the highest symmetry ($a = b = c;\ \alpha = \beta = \gamma = 90°$).

A careful look at illustrations of the hexagonal closest packing ($7+3+7$ atoms in $ABA$ layers) in a hexagonal prism and a bit of thinking gives the number of atoms:

$3$ atoms in the $B$ layer are not shared by any other prism. They fully count
as $\mathbf{3}$.
$2$ atoms in the centre of the top and bottom $A$ layer are each shared by any other prism (above and below). Each of them counts $\frac{1}{2}$. So there's another $2 \times\frac{1}{2}= \mathbf{1}$ atom.
Top and bottom $A$ layers each have $6$ atoms on corner positions. Each atoms is shared by three neighbors in the the same layer and with another hexagonal prism on top or below. So they count as $2\times6\times\frac{1}{3}\times\frac{1}{2}=\mathbf{2}$ atoms.

The total count thus is $\mathbf{6}$.

These diagrams of the hexagonal closest packing, do not represent the the unit cell as the most simple unit from which the crystal is build by symmetry operations.

Cutting the hexagonal prism (along the axis through the atoms centering the top and bottom faces) into three pieces gives identical rhombohedra. These can be transferred into each other by rotations along the axis described above.


Is there a compound that binds with salt (NaCl) in solution? Is there a non-toxic compound that binds with $\ce{NaCl}$ in a water medium?

$\ce{NaCl}$ itself does not exist as a compound in water. The dissolution of salt in water means that both the sodium cation and the chloride anion are separated from each other and are each surrounded by water molecules. Therefore, no compound can bind to NaCl in water because $\ce{NaCl}$ does not exist in water.

It is, however, possible that a compound "binds" with either sodium or chlorine. For example, by adding copper to the solution and performing a bit of more electrochemical magic (out of the scope of this answer), you can get the solid compound copper chloride ($\ce{CuCl}$).


Is the water in an aqueous solution undergoing a reaction considered part of the system or surroundings? System: In chemistry, a system is a chemical reaction. A system operates within its surroundings. Energy can move between the two.

Surroundings: The environment around a particular chemical reaction. The chemical reaction is the system.

Above are my textbook's definitions for system and surroundings.

For example, mixing two aqueous solutions to produce a precipitate. Although the water doesn't undergo any permanent chemical changes, it is still enabling the reaction (much like a catalyst). So is water then considered part of the system?

The boundary between system and surroundings is defined by you the scientist. Still, some choices make more sense than others, depending on the question you want to answer. When you perform a calorimetric experiment, for instance, you monitor the heat generated or absorbed during a reaction (such as formation of a precipitate). The heat originates or is consumed in the local combination of reagents to form the bonds between precipitate atoms and during segregation of solvent molecules. That heat then dissipates into the rest of the vessel and is registered, for instance as a temperature change. In the final analysis, what you call the system here would probably include only what has changed and you are specifically interested in, namely the molecules involved in the actual reaction. This would include solvent molecules associated with the solubilized reagents (the involvement of solvent is made explicit when reagents are labelled as aqueous, for instance). The behavior of the bulk solvent, which serves an "auxiliary" role, for instance by transferring heat across the vessel, is important during calibrations, but is not the question of interest, and would usefully be regarded as the surroundings.

Water can be regarded as a catalyst in many aqueous reactions, but when a reaction occurs in a solvent that associates strongly "catalyst" is not generally a proper description for the solvent, as the energy of both reagents and products may be stabilized/destabilized by the solvent, and to different degrees (and not just the energy of an intermediate in the reaction).


Synthesis of 1-hexene from 1-octyne This is my synthetic pathway. For one, is it plausible? For another, is it efficient?

1) Reduction of 1-octyne with hydrogen gas and Lindlar's catalyst to 1-octene.

2) Oxymercuration to install a hydroxyl group at the most substituted carbon. This creates 2-octanol.

3) Reaction with tosyl chloride and pyridine to create a sulfonate ester - a much better leaving group than an alcohol.

4) E2 elimination with sodium hydroxide and strong heating to create the internal, Zaitsev alkene.

5) Ozonolysis to create two aldehydes.

6) Reduction of longer aldehyde with hydrogen gas over transition metal to primary alcohol.

7) Dehydration of primary alcohol to alkene using a Lewis acid (aluminium trichloride in this case).

On the plus side, you've clearly thought about this problem and have worked out a detailed synthetic route.  I think most of the steps would work as desired.  The two steps that concern me the most are 4 and 7.  I suspect your step 4 will give a mixture of products.  Perhaps with further experimentation you could find reaction conditions that would give you the desired product without  other products from undesired side reaction (elimination in the other direction and perhaps some rearrangement products too).  I see your step 7 as the most problematical.  Aluminum chloride, heat and the alcohol will produce a carbocation and it will undergo significant rearrangement creating a variety of elimination and addition (polymer formation) products.  So while your reaction scheme is plausible, in my opinion it is not efficient.


Electrolysis of diluted sulfuric acid My teacher told me that electrolysis of dil. $\ce{H2SO4}$ (concentration below $50\%$) using inert electrodes results in gradual increase of the concentration of $\ce{H2SO4}$.

I can't seem to figure out why. I know that the sulfate ions will remain in the solution, but aren't the $\ce{H+}$ ions of $\ce{H2SO4}$ reduced at the cathode as well?

Another doubt I have is that I've seen different equations for the oxidation at the anode for this electrolysis in different books. Some of them show that $\ce{OH-}$ will be reduced, and some show that $\ce{H2O}$ will be reduced! I'm really confused about all this.

In a dilute solution of sulfuric acid, there are the following species present: $\ce {H_2O}$, $\ce {H^+}$, $\ce {OH^-}$, $ \ce {HSO_4^-}$, $\ce {SO_4^2-}$. Of course, the water molecules are present in the highest concentration, much higher than the other species, since it is a dilute solution. For simplicity, we would also represent the proton and hydroxide ion as unsolvated. However, do take note that they are in fact heavily solvated in aqueous solution.

In some textbooks, it may be said that for the electrolysis of dilute $ \ce H_{2}\ce S\ce O_{4}$, the oxidation half-equation is written as $\ce {2H^+ + 2e^- -> H_2}$ and the reduction half-equation is written as $\ce {4OH^- -> 2H_2O + O_2 + 4e^-}$. However, the concentrations of these ions in solution are rather small compared to the concentration of the water molecule in solution. Albeit these reactions may be taking place, they should not be the main reactions and are unlikely to be representative of what is actually happening in solution.

In fact, since water molecules are the dominant species in solution, it is more likely that they undergo the redox reactions. The following half-equations represent the reactions happening in the solution more accurately:

Reduction: $\ce {2H_2O + 2e^- -> H_2 + 2OH^-}$

Oxidation: $\ce {2H_2O -> 4H^+ + O_2 + 4e^-}$

Since water molecules are being used up in the electrolysis process, the concentration of the remaining ions increase as the solution is electrolysed. Thus, the concentration of the sulfuric acid solution should increase.


Help with determining the rate constant of a reaction The question asked was,

The initial concentration of $\ce{N2O5}$ in the first order reaction $$\ce{N2O5 -> 2NO2 + 1/2O2}$$ is $1.24\cdot 10^{-2}~\mathrm{mol\,L^{-1}}$ at 318K.
The concentration of $\ce{N2O5}$ decreases to $0.20\cdot 10^{-2}~\mathrm{mol\,L^{-1}}$ after 1 hour.
Calculate the rate constant at this temperature.

How i attempted to solve this is as follows:-
The answer given, however, is $4.82 \times 10^{-4}\ \mathrm{s{-1}}$. What have I understood incorrectly?

This Khan Academy Video explains it very nicely.

So since this is a first order reaction
$$\ce{ -\Delta N2O5/\Delta t = k[N2O5]}$$

This gives you the rate of reaction at the very start. A second later however,  $\ce{[N2 O5]}$ decreases which will in turn change the rate of reaction. So lets apply calculus. We can write this as

$$\ce{ \frac{-d [N2O5]}{d t} = k[N2O5]}$$

Then we can integrate to find out the total change over a period of time.

$$\int_{[N2O5]_0}^{[N2O5]t} \frac{d[N2O5]}{[N2O5]}= \int_0^t-k dt$$

since $-k$ is constant, we can take it outside of the integral

$$\int_{[N2O5]_0}^{[N2O5]t} \frac{d[N2O5]}{[N2O5]}= -k\int_0^tdt$$

So then we get

$$ \ln([N2O5]_t) - \ln([N2O5]_0) =-kt$$

So putting in your values

$$\ln(0.20\cdot 10^{-2}) - \ln(1.24\cdot 10^{-2}) = -k\cdot 60\cdot 60$$

So I'm pretty sure the answer is $$5.068\cdot10^{-4}~\pu{s^{-1}}$$


Why do tungsten and chromium have different electronic configurations? Why is the outer structure of tungsten $\mathrm{5d^46s^2}$ instead of $\mathrm{5d^56s^1}$?

I try to explain this in comparison  with outer structure  of chromium $\mathrm{4d^55s^1}$  taking into consideration  5d orbitals have lesser energy than 6s orbitals in tungsten by saying that the size of the energy gap between the 3d and 4s orbitals lesser than size of the energy gap between the 5d and 6s orbitals, and the repulsion between electrons in the structure$\mathrm{5d^46s^2}$  more than  the repulsion  in the structure $\mathrm{5d^56s^1}$ which can push the electron from 5d to 6s to produce the lowest energy structure.

I need qualitative answer?

Firstly note that the energetic differences between $\mathrm d^m\mathrm s^n$ and $\mathrm d^{m-1}\mathrm s^{n+1}$ configurations are generally not large. Sometimes these are not amenable to simple qualitative rationalisation.

However in this case it is possible to argue along two lines.

Relativistic effects lead to the stabilisation of the 6s subshell and destabilisation of the 5d subshell, which directly favours population of 6s over 5d.
Electron-electron repulsions in the 3d/4s orbitals are generally larger than repulsions in the 5d/6s orbitals, because the latter are more diffuse. Hence chromium chooses to adopt a configuration in which exchange energy is maximised (i.e. electron-electron repulsions are minimised).


How Does One Create A Cellulose Ingredient For Use In The Creation Of Glucose? I am attempting to begin synthesizing glucose, however I will require a form of ball milled cellulose as a main ingredient. Having been said, what methods would be recommended to create the Cellulose? If we were to use grass clippings or cotton as our source for example, what steps would we take in order to gain the desired material from the aforementioned source?

Cotton wool, also called "purified cotton", is inexpensive and available at pharmacies ("chemists" in the UK). I could not find the percentage of impurities (probably lignin and minerals), but it is considered one of the purest forms of cellulose commercially available, and should not need pretreatment to make glucose.


Solid salts of hydronium ion Are there any solids with hydronium ion in its crystal structure?

If so, what are they?

If not, why not? I understand there are hydroxide-containing solids and gels such as aluminium hydroxide but why not hydronium ions? Does this have to do with the positive charge of hydronium ion? I could see the positive charge of hydronium ion excluding the possibility of forming any salts/solids with the group 1 and 2 metals.

There is a section in this Wikipedia link entitled, "Solid hydronium salts."  It says that any acid with an ionization constant of 10^9 or higher may form stable hydronium ion salts.  Examples they give include hydronium perchlorate and carborane-based hydronium ion salts.


Stability of Gauche and Anti form in conformational Isomerism I came through this question: Which form is stable Gauche or Anti in F-CH2-CH2-F?
My answer: Due to lone pair repulsion Anti form should be more stable but the answer is given Gauche form. Please tell me how is gauche form more stable in this case?

This problem should be referred to concept called stereoelectronic effect. It is stabilized by the donation of $\sigma_{C-H}$ bond to $\sigma_{C-F}$ bond, leading to decrease energy of the whole system.


Why refine tar sands oil? Why refine tar sands oil into petroleum and the rest of the oil processing byproducts rather than burning it directly in the furnaces of private residencies? At least in the snowbelt areas where gas is not readily available.

Tar sand oil is extremely heavy and would produce a lot of soot. It also contains a lot of sulfur, so the smoke would contain a lot of sulfur dioxide. This means, that burning it would require exhaust purifications. This in turn would require costly equipment that is not practical for individual household.

Furthermore, heavier fractions of oils have hard time burning. They have to be heated significantly before a flame can form.

Consequently, tar oils must be refined to be of practical use. This includes three key processes:

hydrocracking, which breaks down larger molecules into smaller ones, that produce less soot and burn easier.
hydrothreatment, to remove sulfur, metals (yes, heavier oil fractions contain measurable amount of metals, vanadium and nickel in particular, that are known cancerogenes)
distillation, separating lighter fractions from the heavier ones. Typically fractions C5-C20 are used for internal combustion engines, some С21+ are usable in industrial-grade furnaces, and the heaviest and most troublesome fractions are used as asphalt.

So, propane or propane-butane mixture is a lot more practical for small furnaces. It is very easy to refine, produce no smoke and contains no sulfur, so burning it is pretty easy. Furthermore, in can be liquefied under pressure, but it is still a gas at most common conditions. So equipment to burn it is extremely simple and exhausts are virtually harmless in most cases. Propane is reasonably cheap in most setups to use.


Clear definition of a 'non-oxidizing acid'? From the Beryllium article in Wikipedia:

  Beryllium dissolves readily in non-oxidizing acids, such as $\ce{HCl}$ and diluted $\ce{H2SO4}$, but not in nitric acid or water as this forms the oxide.

I think it would be helpful to have a definition of 'non-oxidizing acid'. Is it an established term in chemistry? Am I right to understand that it is an acid that can only act through its $\ce{H+}$ moiety, and not by the reduction of its conjugate base part (as in the reaction between $\ce{HNO3}$ and $\ce{Cu}$)?

There's a little obscure page titled oxidizing acid on Wikipedia.

The trouble is, until I came across this term, I imagined all acids to be oxidizing by definition. It's not very intuitive.

P.S.

Related: Weird Wikipedia Section on Oxidizing Acids

A non-oxidising acid is an acid in which there is no stronger oxidant than $\ce{H+}$ at $\mathrm{pH}\ 0$.

Or, in a slightly different wording, a non-oxidising acid would be one that cannot dissolve any noble metal at $1~\mathrm{M}$ concentration, but should (in the absence of effects like overpotential) dissolve any non-noble metal at that concentration.

Nitric acid has the strong oxidation agent $\ce{NO3-}$ present, which will reduce to $\ce{NO}$ or $\ce{NO2}$ depending on the concentration. It is able to oxidise noble metals such as silver.

Hydrochloric acid, on the other hand, has no other oxidising agent. The chloride ion is already fully reduced. Therefore, it is not able to oxidise noble metals such as copper.

Some acids are more strongly oxidising if they are concentrated. Sulphuric acid is one. Only in higher concentrations will reduction of sulphate to $\ce{SO2}$ take place and make the acid oxidising. In lower concentrations, only $\ce{H+}$ takes part in oxidations.


What is the minimum number of given sets of information required to complete an ICE chart for an equilibrium reaction? I'm considering creating a program that automatically completes an ICE chart/table when the minimum given information is provided. The reason I'm posting my question here is because I need to know the possible cases of minimum information required before I can even start the program.
I know you can complete the table if you're given these following sets of information:

K and initial concentrations
K and final concentrations
K and initial concentrations of reactants and final concentrations of products

I also believe there are a ton more, but I'm not sure how to complete the list.

Alright I think you have the basics down, first of all I have a couple pointers for you since I have done many (many) RICE charts. What I would do if I were you? Since I have some experience in programming I would just use molarities of the initial and final.
You must make the user input the data, say if the user was given moles then make him/her convert it to molarity.
I think you have it pretty much down, you need either the initial concentrations and K or find the K through using the initial concentrations of the products and reactants or find the initial using the product and K.
I mean then again I would work through making this first part down because those are the essentials (atleast what I've learned) and then work on making updates to that program.
Since I do have programming experience I would just use if statements to do the entire thing. If this is available then solve for this.
Don't worry about the entire thing yet. Otherwise I would have to say you have pretty much everything down.
NOTE: This is not about Ka, Ksp, Kb, just for Keq. Because then you would have to work with the pH and pOH more equations.


Are stronger bonds always shorter? This question relates to a few earlier questions I have seen on the site (particularly this one about the bond strength of heavier isotopes). Does a bond necessarily have to be shorter to be stronger?

To my understanding, length and strength are definitely correlated, but I don't think it is necessarily a one to one correspondence. The examples I brought up in the comments of ron's answer in the linked question related to data tables that seemed to show no difference in bond length for deuterium substituted bonds despite significant changes in energy. (HCl NIST, DCl, data table)

To expand on my question, if length and strength are always directly related:

How do we explain experimental and theoretical cases where the length doesn't differ with bond energy?
If there is some direct relationship between length and strength, what is that relationship?

For the sake of not being too general, I'm looking for an answer focused on comparing bonds of the form $\ce{A-B}$ and $\ce{A-C}$ (similar to the examples I asked about in the linked question). So for a more specific case related to the linked question, does substituting a protium for a deuterium always make the bond shorter in addition to being stronger?

This is an interesting question and depending on how you define bond strength the answer is different. Let us for simplicity consider only diatomic molecules and let us assume that the electronic potential between the two atoms is well described by a Morse potential

\begin{align}
V(r) &=D_e \left( 1 - \mathrm{e}^{-a(r-r_e)} \right)^2,& \text{with } a &=\sqrt{\frac{k_e}{2D_e}}.
\end{align}

Here $D_e$ is the depth of the potential (at the minimum position $r=r_e$) and $k_e$ is the (harmonic) force constant. The potential depth is related to the dissociation energy $D_0$ by

\begin{align}
D_0 &= D_e - \frac{1}{2}\omega_ehc,& \text{with } \omega_e &=\sqrt{\frac{k_e}{\mu}},
\end{align}

where $\omega_e$ is the harmonic wavenumber, $\mu$ is the reduced mass and $h$ and $c$ have their usual meaning.

We can define the strength of the bond by the magnitude of the dissociation energy $D_0$ or by the "spring constant" $k_e$ of the bond. In your example with deuterium you implicitly assumed the former definition. In the Born-Oppenheimer Approximation the potential (i.e. the Morse potential) does not depend on the mass of the atoms (they are assumed to be infinitely heavy) and the potential is the same for $\ce{H2}$ and $\ce{D2}$. However, because deuterium is heavier than hydrogen, the harmonic frequency is lower and therefore the dissociation energy is larger ($\ce{D2}$ has a smaller zero point energy). Using this definition, the bond length is not easily related to the bond strength (as the reduced mass of the system plays a role as well).

As stated earlier, we can also use $k_e$ as a measure for the bond strength. The larger the value of $k_e$, the steeper the harmonic part of the potential well and the more localized the nuclear wavefunctions will be. In other words, if we take $k_e$ as a measure of bond strength, then stronger bonds are indeed shorter.

As stated by @porphyrin, the separation between $k_e$ and $D_e$ is not very strict (see also the formula's above) and the explanation above implicitly assumed a constant $D_e$, just as we assumed constant $k_e$ for the different isotopologues.


Proton attached to N or O in HNMR spectroscopy I was wondering why a hydrogen attached to a nitrogen or oxygen is not able to couple with p+ on adjacent carbon atoms and hence why it always appeats as a singlet in HNMR spectroscopy?

It depends on the condition of your experiment if a coupling of a proton of an amino or hydroxyl group with a proton attached to the $\alpha{}$-carbon is observable, or not.  As pointed out by the indicator set by @Mithoron, intermolecular exchange processes need to be taken into account.

If your alcohol (or amine) to be analyzed is very pure, and the deuterated solvent you deploy is both very pure, and dry, and free of acids or bases catalysing proton exchange, than you may observe $^3J$-couplings you expect to see.  For example a freshly opened sample of deuterated DMSO (used as solvent) may contribute to this observation.  See the peak data of the 1H NMR spectrum of ethyl alcohol in DMSO here.
If you aim to analyze an alcohol, not well dried, analyzed in deuterated chloroform or methanol that was frequently exposed to ambient (humid) air, then indeed instead of a triplet of an hydroxyl group  ($\ce{-OH}$)  attached to a methylene group ($\ce{-CH2-{}}$) all what may be spot is a broad singlet signal.  The same reference, 90 MHz 1H NMR in deutero chloroform provides one example.  (Note the chemical displacement of the hydroxyl protons altered, too.)


Reduction of terephthalic acid to an alcohol (or phenol?) I want to convert terephthalic acid (TPA) to an alcohol (however I think the correct term would be "a phenol").

I've read that lithium tetrahydridoaluminate ($\ce{LiAlH4}$) can produce an alcohol from a carboxylic acid. And, according to Wikipedia, dicarboxylic acids (such as TPA) react around the same. So does $\ce{LiAlH4}$ reduce TPA to a phenol (or an other product)?

I want to make sure that you understand the different between two commonly confused molecules (their names don't help).

Phenol is an $\ce{-OH}$ group directly attached to an aromatic ring. Because of this connectivity, phenols exhibit different chemical behavior than other alcohols. In particular they are more acidic. Phenol can be abbreviated $\ce{PhOH}$, where the $\ce{Ph}$ represents the $\ce{C6H5}$ group of the aromatic ring. As a class of molecules, phenols can be represented generically as $\ce{ArOH}$, where $\ce{Ar}$ is used not for argon (there are no compounds of argon), but for generic arenes with any substitution.

Benzyl alcohol is a different molecule. In benzyl alcohol, there is a $\ce{CH2}$ group between the benzene ring and the $\ce{OH}$. Benzyl alcohol behaves like other alcohols. Benzyl alcohol is abbreviated $\ce{PhCH2OH}$.

Conversion of benzoic acid to phenol requires loss of a carbon atom.

$$\ce{Ph{\bf C}O2H -> PhOH + {\bf CO}}$$

Lithium aluminum hydride converts carboxylic acids to alcohols, but it does not cause the loss of the carbon atom. Lithium aluminum hydride will reduce benzoic acid to benzyl alcohol. If you want the benzyl alcohol derivative, stop here. If you want the phenol, read on.

$$\ce{PhCO2H} \overset{1. \ \ce{LiAlH4}}{\underset{2. \ \ce{ H2O}}{\ce{->}}}\ce {PhCH2OH}$$

To convert benzoic acid to phenol, we need a way to loose that carbon atom. One way is to oxidize benzyl alcohol to benzaldehyde using perhaps a Swern oxidation. Then benzaldehyde can be converted to phenyl formate using a Baeyer-Villiger Oxidation. The formate could be hydrolyzed in aqueous base.

$$\ce{PhCH2OH \overset{Swern}{->} PhCHO}$$
$$\ce{PhCHO \overset{Baeyer-Villiger}{->} PhOCHO}$$
$$\ce{PhOCHO}\overset{1.\ \ce{NaOH, \  H2O}}{\underset{2. \ \ce{H3O+}}{\ce{->}}} \ce{PhOH}$$

In practice, it would be easier to buy the phenol. In your case, you want hydroquinone.


Thought process behind Rutherford's gold foil experiment In Rutherford's gold foil experiment, if he at first thought the atom looked like the plum pudding (as Thomson had said), why did he think the alpha particles would not be repelled by the positive "pudding" and that they would go straight through?  Is it because of the high energy of the emission of the alpha particles?

Rutherford conducted his gold foil in order to prove Thomson's plum pudding model but the result was contradicting.
He expected that most of the Alpha particle will show very minor deflections  because the positive charge was evenly spread throughout the atom according to the plum pudding model.


is it possible to make ethanol from ethane through a chlorination / hydrolysis two-step procedure? $\ce{C2H6}$ is chlorinated in presence of sunlight to form $\ce{C2H5Cl}$ and $\ce{HCl}$.
$\ce{C2H5Cl}$ is then reacted with water to form $\ce{C2H5OH}$ + $\ce{HCl}$

Yes, it is possible.

Although ,commercially(and ironically!) alkyl halides are produced from the corresponding alcohols.
The main reason why this is so because free-radical chlorination requires high temperatures and catalysts to promote the reaction.

Theoretically, the first reaction will give good yields(and not a mixture unlike most free radical halogenations because of the simple reactant involved)

Edit:As pointed out by bon, I forgot to take into account that because its a free radical reaction, the number of possible products is a lot.So no, the first reaction will not give you good yields.


Schrödinger equation and degeneracy of atomic orbitals How does the application of the Schrödinger equation to model a system, such as a particle in a box, help us understand the origin of the degeneracy of atomic orbitals?

Degeneracy refers to states having the same energy.
If you have two or more coordinates that can be related by a symmetry operation then the system will contain degenerate states because swapping the coordinates associated with those degrees will result in a Hamiltonian with the same solutions.
This is the case in both the hydrogen atom and in the particle in a cube.


For the solubility of AB2, Ksp=[A][B]^2=x*(2x)^2. Why is (2x)^2 not considered double dipping? For $$\ce{AB_2<=> A + 2B} $$

$$K_{\text{sp}}=\ce{[A]*[B]^2} $$

$$K_{\text{sp}}=\ce{[x]*[2x]^2} $$

For the term $$\ce{[2x]^2} $$

Why isn't this considered double dipping on B's behalf?

I.e. why isn't it just $$K_{\text{sp}}=\ce{[x]*[x]^2} $$ or $$K_{\text{sp}}=\ce{[x]*[2x]} $$

Why is B both multiplied by 2 and squared?

It's easy, if you consider:

$$\ce{AB2 <=> A + 2B}$$

It's solubility equations would be:

$$K_{\text{sp}}=\ce{[A][B]^2}$$

That equation follows from the law of mass action.

If you consider the amount of A as $x$, wouldn't the amount of B be $2x$ obviously?

If your issue is with why it's squared, it's just the equilibrium law, it follows from law of mass action. You can't really argue why the square is there.


Heat and water absorption by calcium chloride Can calcium chloride be effectively used as a desiccant to absorb both heat and moisture?

Anhydrous calcium chloride is an effective desiccant, but the absorption of water by anhydrous calcium chloride is exothermic; i.e. the reaction releases heat.

$$\ce{CaCl2 + 6 H2O -> CaCl2 . 6 H2O}\quad\Delta H = -97\ \mathrm{kJ/mol}$$


What elastic polymeric material look like at molecular level? I think there are few crosslinks in elastic material. So probably, the material would look like not so dense, maybe even transparent.

I received this question from my collegue and have difficulty in answering it.
I think this question is about know how the structure of an elastic polymer looks on the molecular level.

What does the elastic polymeric material look like at molecular level?

Broadly, the elastic polymeric materials are structural with the mixture of networks of molecular chains and liquid. There may be a distinction between the terms "elastic polymeric materials" and "elastic polymers". Key features contain repetition of some parts so open to bioengineering such as bioprinting. Each specific elastic polymeric material differs at molecular level. For example some materials can have long repetitive carbon chains while some materials can have short repetitive carbon chains where each structure is attached together by some forces. There are different kinds of forces where their degree is relative to the molecule:

Brainstorming on the question "What look like?" What does it mean?

what is meant by "look like"?
intermolecular forces? Repulsive forces? Attraction forces?
EM fields between each charged particles? How are quantum phenomena in materials such as doping considered?
What does the electron current look like over the material?
What is the distribution of static charge?

Perhaps relevant

The book Edited by Xiang Yang Liu and Jing-Liang Li "Soft Fibrillar Materials Fabrication and Applications".

  "As one of the most important classes of soft materials, supramolecular materials are of a mixture of networks of molecular
  chains/fibrils and a liquid. These self-assembled fibrous/ molecular
  architectures exhibit various functionalities, ie. superhydrophobicity
  or superior mechanical strength, etc. and consist of the controllable
  structures.

  --

  The book covers the most important soft functional materials, including small molecule physical gels, silkworm silk and spider silk
  fibers and functional fibers, with respect both to the fundamentals
  and to development and engineering methods. It provides the reader
  with the necessary knowledge on the chemical and physical formation
  mechanisms of these materials and demonstrates that one can rationally
  design and tune the fibrillar networks so that the resulting materials
  exhibit the desired functionalities.

  This work is a must-have for all Materials Scientists, Polymer Chemists, Condensed Matter Physicists, and Biotechnologists working in
  this interdisciplinary field."


Chemical Kinetics - change to system conditions This is a question from a very old AP Chem exam (2006) prep book which I am amusing myself with in my spare time which is kind of humbling. There is something I am missing with this question, one of several similar questions in the kinetics section.

Given the following reaction (which doesn't establish an equilibrium) :
$\ce{A (g) + B (g) -> C (g)}$
which of the following would increase the concentration of $\ce{A}$?

addition of $\ce{A}$ at constant $T$ and $V$
addition of $\ce{A}$ at constant $T$ and $P$
addition of $\ce{A}$ at constant $V$ and $P$

Now my first thoughts are to reason as follows :
If you add $\ce{A}$ and the volume is constant obviously the concentration of $\ce{A}$ increases. If you add $\ce{A}$ and keep $T$ and $P$ constant the $V$ will just expand and the concentration will be the same. But if I ignore the reaction this is a gas law / concentration question so it is out of place. But it isn't an equilibrium so none of that reasoning applies either.
Now if the reaction happens so slow it takes a year for $1\,\%$ of $\ce{A}$ to react I could just consider the reaction not even happening? If the reaction happens so fast that it is essentially over in 1 second then it seems like any change to A would be undone as fast as you could increase it. But it seems like I am missing something basic as there is nothing given regarding the rate.
There are several questions almost identical to this in the same section which just have other than $1:1$ coefficients in the equation.

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

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

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


Stereochemistry of epoxide hydrolysis under acidic conditions The epoxide on the top of the picture, in aqueous acid hydrolyzes to the compound below. What is the mechanism of this reaction? Standard epoxide opening in acidic conditions would happen at the more substituted end, therefore the secondary -OH group would end up below the plane of the paper, and the tertiary center would racemize to a significant extent. What is happening here?

Standard epoxide opening in acidic conditions would happen at the more
  substituted end

This premise is false. Usually $\ce{S_{N}2}$ is faster than the cation-stabilized pathway (most textbooks don't even bother to explain the subtleties involved), yielding mainly the less substituted (less sterically hindered) product. It is not clear cut which of the two mechanisms will be predominant in an acid catalyzed epoxide cleavage given the reactants and reaction conditions.

See also Regioselectivity of acid-catalyzed ring-opening of epoxides (especially the second answer and then the excerpt from Clayden et al.).


Potassium Alum Crystals This is a question regarding the crystallization of $\ce{KAl(SO4)2 . 12 H2O}$. I have an idea of what to do (I'll explain later in the explanation), but I would like to hear other suggestions encase the following ideas do not work out.

I placed a specific amount of the salt into hot distilled water and began stirring, later setting it out to grow my seed crystals. I have been waiting for the past 3 days to receive any potential seed crystals, however, I have not received even one. I noticed that over the 4 days a significant amount of water has evaporated, and at the bottom of the beaker has accumulated a white powder I'm assuming is the $\ce{KAl(SO4)2 . 12 H2O}$. I fail to understand why it has not crystallized into a small singular crystal?

My idea is to scratch the bottom of the beaker (it is new) to provide a location for the crystal to begin it's growth. Another idea pertains to adding more solute ($\ce{KAl(SO4)2}$) into the solution. Honestly, I don't know why I would add more solute if I am seeing the powder fall out of the solution. This should mean it is already saturated. Finally, I am deciding about adding some of the undissolved salt from the packet, into the solution. Hopefully this would provide a nice "infant crystal" (I am aware that there is no such thing an "infant crystal", I simply want to refrain from referring to "seed crystal" since these are not my seed crystals.) for my ions (dissolved solute) to latch onto and form my seed crystals.

Any suggestions would be appreciated.

Just to mention, my final technique drives through a more mathematical approach. I found the molar solubility from the $\mathrm{p}K_\mathrm{sp}$ of the crystal ($\mathrm{p}K_\mathrm{sp}$ of salt from aqion.de) I did my calculations to see a total of 0.3 grams of $\ce{KAl(SO4)2}$ per $1~\mathrm{mL}$ of water to obtain a saturated solution. Aesthetically, this doesn't make any sense to me. I feel the website I used may not have provided accurate experiments to reveal the $\mathrm{p}K_\mathrm{sp}$. If you can link me to any other websites or locations I would be more than grateful.

Typically, you will want to make sure that your solution is saturated. Don’t add a specific, certain amount; rather, add the solid to hot distilled water until it actually won’t dissolve any more. Then quickly filtrate the hot solution through a hot funnel (don’t burn yourself though, please). Let this filtered solution cool down and you should obtain acceptable seeding crystals in your beaker.

By using this supersaturation technique, you can be sure that

you have a saturated solution so crystallisation can occur immediately and
since you cool down the solution rather quickly and since solubility of aluminium potassium sulphate increases with increasing temperature, it becomes very supersaturated very quickly.

Once you have a seeding crystal, e.g. by this method, you can use the hanging technique mentioned by Carlos with the remaining solution (which is still supersaturated). If you do so, you can use an almost infinitely thin string; a human hair typically works perfectly and is even less visible. Another option is to filtrate the solution again so that the bottom of the flask is clean and place your crystals there to grow. You will need to turn them around every now and again to allow for even growth.

And you can make your aluminium potassium sulphate crystals even nicer by cocrystallising it with chromium potassium sulphate. They crystallise in exactly the same structure but the chromium(III) ion is coloured (aluminium(III) is not).


Help understanding external magnetic field sensitivity of radical pair reaction rate Two recent papers have been published on magnetoreception in birds. In this case it is the ability of the birds to sense the direction of the Earth's magnetic field potentially for navigation.

Expression patterns of cryptochrome genes in avian retina suggest involvement of Cry4 in light-dependent magnetoreception Atticus Pinzon-Rodriguez, Staffan Bensch, Rachel Muheim J. R. Soc. Interface 2018 15 20180058; DOI: 10.1098/rsif.2018.0058. Published 28 March 2018
Double-Cone Localization and Seasonal Expression Pattern Suggest a Role in Magnetoreception for European Robin Cryptochrome 4 Günther, Anja et al.
Current Biology , Volume 28 , Issue 2 , 211 - 223.e4

The idea has been around a long time; cf A Biomagnetic Sensory Mechanism Based on Magnetic Field Modulated Coherent Electron Spin Motions Zeitschrift fur Physikalishe Chemie Neue Folge, Bd. 111, S. 1-5, (1978)

The two new papers address the role of something called "radical pairs", two molecules, each with an unpaired electron. The large molecules are in the retina and tend to have a preferred orientation with respect to the curved retina, and they also appear to affect the perceived intensity of blue light. This causes a modulation of light sensitivity across the retina dependent on the direction of the magnetic field with respect to the direction of the local normal vector of the retina.

The modulation of the reaction rate by the magnetic field direction has to do with singlet vs triplet states and quantum coherence, but I only say that because I've tried to read Cryptochrome and Magnetic Sensing by the Theoretical and Computational Biophysics Group at U. Illinois, Urbana-Champagne, as well as this Science Alert.

Are there simpler examples of the sensitivity of the reaction rate of "radical pairs" to the orientation of a magnetic field? Perhaps two atoms or two small molecules? Or does this phenomenon generally occur only in large, complex molecules?

below: "This is how a bird might see magnetic fields. (Theoretical and Computational Biophysics/UofI)" From ScienceAlert.

(This is really a comment but  has become too long)
There are many examples of  triplet states in magnetic fields as studied by electron paramagnetic resonance (ESR or EPR) experiments. See chapter 8 'Introduction to Magnetic Resonance' by Carrington & McLachlan. It is an old book but does contain all you need to know. (A paper I happen to know showing magnetic field effects is Hyde et al. Chemical Physics v151,p239 1991 but there must be many others if you search)

As to the bird experiments, eqn 1 of the paper by Schulten et al. as described in your question describes the situation: scheme copied below.

$$\begin{align}
(D^*&+A)\leftrightharpoons \;^1(^2A^-+\;^2D^+) \stackrel{B}\leftrightharpoons \; ^3(^2A^-+\;^2D^+)\to \;^3T\\ &\downarrow\\ &\;^1X \end{align}$$

What it seems is necessary is (a) The reaction starts with a light induced electron transfer from excited donor $D^*$ to a nearby molecule that is the acceptor $A$. (b) The singlet state $^1(^2A^-+\;^2D^+) $ species produced has energy levels that cross those of one of the  of $m_I=\pm 1$ levels of the triplet states $^3(^2A^-+\;^2D^+) $ three levels at some value of the magnetic field $B$ and so singlet to triplet crossing occurs($^1$). (c) This enables the yield of products $^3T$ to $^1X$ to vary as the value of the magnetic field varies as the animal moves. (d) The protein has to be fixed relative to the bird otherwise this will not work. (This is the 'anisotropy' mentioned)  Being fixed could simply mean being in a cell membrane in one part of the retina. (e) Somehow the ratio of products has to be able to be distinguished.  Clearly this mechanism clearly will not work at night.

So all in all a complicated mechanism, involving subtle processes, and this makes it very interesting.

($^1$) In a molecule as the triplet often differs significantly in energy from the singlet a huge magnetic field is needed to reach the S-T crossing point, but in pairs of molecule the interaction is very small and so the energy difference is v small also. This is advantageous as then small magnetic fields can cause S-T crossing. (Normally the singlet is above the triplet in energy, unlike the in figure, but the crossing is clear nonetheless)

also see Can magnetic fields affect a chemical reaction?


When is phosphorus a polyatomic molecule? Phosphorus is listed as a single element in the periodic table; however, I see that it is also a polyatomic molecule as $\ce{P4}$. I am trying to determine the number of atoms of phosphorus in an chemical equation, so I can have a balance equation. So, when would I use the polyatomic version, or the single atomic version if the chemical equation is a word equation, and the subscripts are not listed? For example the word equation is $\ce{P + Cl -> PCl5}$. Is $\ce{P -> P4}$ or just $\ce{P}$? Also, would $\ce{Cl}$ be diatomic too ($\ce{Cl2}$) in this equation?

Phosphorus is most common as white phosphorus (but more stable as red phosphorus) and in this form, it exists as a tetrahedron, or $\ce{P4}$. Chlorine is most stable as a diatomic molecule, as you may have learnt it with the other common gases that exists as diatomic molecules


Looking for a mechanism for the demercuration of organomercurial alcohol I am teaching myself organic chemistry and am very frustrated that I cannot get a complete mechanism for the oxymercuration-reduction of an alkene to an alcohol. The reagents are an alkene, mercury(II)acetate, and water for the first step... the step I understand. It ends by forming organomercurial alcohol. This is then reacted in a second step with sodium borohydride in a solution of sodium hydroxide. I have searched for a mechanism for this step (specifically the reduction portion at the very end) but I cannot find one anywhere. The closest I have come to an answer is a source that says it probably had to do with radicals and inorganic chemistry. The products are an alcohol with the C-Hg(OAc) bond transplanted with a C-H bond, tetrahydroxyborate, reduced mercury (standard conditions), and OAc anion. Thanks any help is appreciated

Update:
I am attaching my attempt at a mechanism and would like someone to verify if it is the correct mechanism or tell me why it would not work. Thanks

The direct displacement of the mercury acetate by borohydride, as prooposed by @xavier_fakerat, looks attractive because it (1) makes some chemical sense and (2) gives the correct products. However, that mechanism can be disproven by looking at the stereochemical outcome of the reduction step.

Pasto and Gontarz did just that (Pasto JACS 1969). When cyclopentene was subjected to oxymercuration, followed by reduction with sodium borodeuteride, the product observed was the trans-configuration of the hydroxyl and deutero groups. The replacement of mercury for deuterium went with retention. This rules out direct displacement (SN2 like mechanism) which would lead to inversion of configuration.

When cis- and trans-butene were subjected to the same process in independent experiments, in both cases the result was a 50:50 mixture of syn- and anti-stereochemistries. This indicates that there is cleavage of the carbon-mercury bond and sufficient lifetime for the resulting intermediate to undergo bond rotation before carbon-hydrogen bond formation. For other substrates, C-C bond rearrangements were observed.

The authors propose that hydride displaces acetate at mercury. This is followed by homolytic C-Hg bond cleavage, leaving a carbon centered radical. Finally, hydrogen atom abstraction from the mercury-hydride gives the reduced product and elemental mercury.


Acetic(vinegar) and citric acid as the rust remover I tested, separately, acetic and citric acids for cleaning steel parts. After 24 hours I have moderate results for acetic acid, and good results for citric acid. But in both cases, rust doesn't disappear fully.
I have two question:

If I mix two these acid, do I get the synergy?
Is there another acid which easy to buy?

P.S. I'd like to get clean iron without gray protective coat, like rust remover from auto parts shop.

To answer your first question, no, you would mainly just get dilution of the stronger acid (citric) with the weaker (acetic). If you used a citrus fruit as the source of citric acid, it's important to note that there is a fair bit of variation in acidity from one citrus juice to another. The best bet would be lemon or lime juice, which have among the lowest pKa values (measurement of acidity, lower is more acidic) of the common citrus juices.  If you just used something like orange juice, you might notice significant improvement by switching to lemon or lime juice.

Your second question probably is the right way to go. Rust is fairly soluble in dilute solutions of strong acids like hydrochloric acid (the acids you have used so far are weak acids).  You can get consumer-grade hydrochloric acid from a number of sources, the best probably being an aquarium or pond shop. These products usually have names like "pH down" or something indicating that they are acidic. What you want to do is look at the ingredients for a product that is just dilute hydrochloric acid, without a bunch of other stuff. Be sure to read and follow all of the precautions on the container, particularly with respect to wearing eye protection.

Another alternative source for dilute hydrochloric acid would be rust-cleaning products from a hardware store. These typically have a detergent of some sort mixed in with the acid, which is OK, and this may actually be your best solution depending on availability where you live.