What is the difference between transaldolase and transketolase in the pentose phosphate pathway? What is the difference between transaldolase and transketolase in the pentose phosphate pathway? From what I understand, they both catalyze the transfer of carbon chains from 1 aldose into 1 ketose to produce 1 aldose and a ketose, though transketolase transfers 2 carbons while transaldolase transfers 3 carbons. Is that the only difference? If so, then why they were named "aldolase" and "ketolase"?

An aldol is a $\beta$-hydroxy ketone or aldehyde, and is the product of an aldol addition.

The enzyme that cleaves the bond formed (red) is known as an aldolase. The transaldolase reaction begins by aldol cleavage that is, cleavage of the bond in red.

Transketolase on the other hand simply cleaves off the ketone group from the substrate. It does not perform an aldol cleavage.

Apart form this, the mechanisms of action of the two enzymes vary considerably. For instance transketolase requires thiamine pyrophosphate (TPP) as a cofactor but transaldolase does not.


Influence of the neighbouring molecules in a crystal at the example of XeF4 For reasons of symmetry I (without deeper chemical knowledge) supposed, that $\ce{XeF4}$ has a shape like $\ce{CH4}$. But according to Wikipedia its

  crystalline structure was determined by both NMR spectroscopy and X-ray crystallography in 1963. The structure is square planar, as has been confirmed by neutron diffraction studies.

In a crystal the neighbor atoms have some influence on another and sometimes the chemical formula shows the numerical ratio between the elements but not draw the full picture of all interactions between the neighbor elements. Does the influence of the neighboring molecules in a $\ce{XeF4}$ crystal makes the square planar shap? Is it imaginable that an isolated $\ce{XeF4}$ molecule has a tetrahedral shape?

Additional question. What is the crystalline structure of $\ce{XeF4}$?

There are two single crystal-structure investigations done at room temperature, both published in 1963. Both suggest that $\ce{XeF4}$ crystallizes in monoclinic system ($P 1 2_1/n 1$) and preserves square-planar geometry in solid phase; however, the angle $\ce{F-Xe-F}$ slightly deviates from $90°$.

Figure 1. Packing of $\ce{XeF4}$ molecules. Color code: $\color{#90E050}{\Large\bullet}~\ce{F}$;$\color{#429EB0}{\Large\bullet}~\ce{Xe}$.
Ibers and Hamilton [1] published a structure with an $R$-factor of $11\%$ (ICSD #27467)

The molecule has symmetry $D_\mathrm{4h}$ within the limits of error of this study, although such symmetry is not required by the space group. The angle $\ce{F1-Xe-F2}$ is $\pu{86 ± 30°}$ and the two independent distances $\ce{Xe}$ to $\ce{F}$ do not differ significantly and have an average value of $\pu{1.92 ± 0.03 Å}$ (Fig. 1). The shortest intermolecular $\ce{F...F}$ contact is $\pu{2.95 Å}$. This and all other intermolecular distances are reasonable; the molecule fills space well, providing further confirmation of the formula $\ce{XeF4}$. Since all of the intermolecular $\ce{F...F}$ contacts in this structure exceed slightly the expected van der Waals contact of $\pu{2.7 Å}$ it is not surprising that a second phase of $\ce{XeF4}$, with a density some $10$ percent higher than that of the present structure, exists (2). The structure of this denser phase is at present unknown.

Fig. 1. The structure of the $\ce{XeF4}$ molecule in the solid.

Templeton et al. [2] did a subsequent XRD experiment and presented improved crystallographic data (ICSD #26626) with $R$-factor of $8.6\%$ and stated that

The structure consists of a molecular packing of square planar molecules of $\ce{XeF4}$.

With a comment on [1]:

Ibers and Hamilton have deduced two structures by refinement of data with $h + k + l$ even. These data do not permit determination of the relative signs of the two $y$ coordinates. One of these two structures is in approximate agreement with our result.

References

Ibers, J. A.; Hamilton, W. C. Xenon Tetrafluoride: Crystal Structure. Science 1963, 139 (3550), 106–107. https://doi.org/10.1126/science.139.3550.106.
Templeton, D. H.; Zalkin, A.; Forrester, J. D.; Williamson, S. M. Crystal and Molecular Structure of Xenon Tetrafluoride. Journal of the American Chemical Society 1963, 85 (2), 242–242. https://doi.org/10.1021/ja00885a038.


Unit cell structure of ionic crystal I have a question about the structure of an LiH unit cell, and while this is related to a homework problem, it isn't the problem itself, I'm just looking for conceptual understanding. I've already found that there should be 4 Li+ ions and 4 H- ions in one unit cell. However, I am struggling with how they will be arranged. Because the number of LiH molecules is 4, I am fairly certain that the structure intended for this problem is face-centered cubic lattice. This is where my conflict begins. In my class we have had the face-centered cubic unit cell described as (for one element) an atom at each corner and an atom in each side. From this arrangement we get (1/8)*8 + (1/2)*6 = 4 atoms. Now, I've seen a description online of the face-centered cubic structure for NaCl as "We can think of this as chloride ions forming an FCC cell, with sodium ions located in the octahedral holes in the middle of the cell edges and in the center of the cell. The sodium and chloride ions touch each other along the cell edges. The unit cell contains four sodium ions and four chloride ions, giving the 1:1 stoichiometry required by the formula, NaCl" (Chemistry LibreTexts). For me this makes a certain amount of sense, because this would give 4 cations when you use the same counting principle as before. However, I've seen multiple diagrams online that contradict this, such as: https://images.app.goo.gl/CGAsc4xAJM39tMgi8.
This seems to claim that the center image is the unit cell, and while it does have 4 of each ion "on" it, it does not seem to fulfill the same structure as the aforementioned unit cell suggestion. Is the problem here just that they are showing 1/8th of the full unit cell for simplicity? If not, I am even more confused.

Thank you for reading this extraordinarily long question for what could be a very simple answer! And please don't roast me too hard, I've never posted here before.

I looked and there are a lot of questions on the site about units cells, many of which are pertinent. However I couldn't find one that answers your question directly.

First there is no reason to guess that $\ce{LiH}$ has a FCC structuree. You can look that up, for instance in Wikipedia.

Second $\ce{LiH}$ doesn't really have molecules in the solid. Think of $\ce{LiH}$ as  an empirical formula rather than a molecular formula.

UNIT CRYSTAL

I lifted the following from website https://opentextbc.ca/chemistry/chapter/10-6-lattice-structures-in-crystalline-solids/

Now the text says:

  The unit cell contains four sodium ions and four chloride ions, giving the 1:1 stoichiometry required by the formula NaCl.

That is correct. The unit cell must have the correct stoichiometric ratio. However the section seems to imply that the middle drawing, as show below, is the proper unit cell. It isn't!

First realize that when looking at proper unit cells is that that the lattice points between which the edges are drawn are the nuclei, not points in space that would contain whole atoms. So consider the drawing below. For the cell as shown in the illustration, there are 8 corners with 1/8 atom each. Four corners are Na, and four are Cl. Thus the cell shown only has $\frac{1}{2}$ of a Na cation and $\frac{1}{2}$ of a Cl anion.

Now down the rabbit hole!

The proper unit cell for NaCl is shown below:

From Wikipedia article on Cubic crystal system:

  The face-centered cubic system (cF) has lattice points on the faces of the cube, that each gives exactly one half contribution, in addition to the corner lattice points, giving a total of 4 lattice points per unit cell (1⁄8 × 8 from the corners plus 1⁄2 × 6 from the faces). Each sphere in a cF lattice has coordination number 12. Coordination number is the number of nearest neighbors of a central atom in the structure.

So the unit cell contains $\ce{Na4Cl4}$ !?!

The reason is that the proper unit cell must also contain all the symmetry elements of the crystal. Look at the drawing below:

I won't get into all of how group theory applies to crystals and molecules, but the  red line shows a four-fold rotation axis. In other words all the atoms are identical. You can rotate the crystal 90 degrees about the red axis and the sodium and chlorine atoms are in identical spots. The axis is a four-fold axis because you have to rotate 90 degrees four times to get the identical atoms back to their starting position.


Which phase is more efficient for temperature changes? In the case of attempting to heat and cool down an object of different states, if you in some case managed to achieve the same temperature of a solid, liquid and gas what state would scientifically heat or cool the object faster?

On my view point I suggested a gas to be more efficient simply due to how the particles move. However I was soon objected by someone else stating that a liquid would be far more efficient.

I was not given a reason , therefore I am asking which would be correct to assume and why.

A liquid is better than a solid for cooling an object, because it can make more complete contact with the object being cooled, plus it can undergo convection, so the layer of liquid next to the object being cooled is replaced by cooler liquid as the liquid circulates around the object.

A liquid is denser than a gas (giving better contact with the object being cooled) and also will likely have a higher heat capacity, which translates into a higher "cooling" capacity as well. But don't forget about "wind chill factor" - a moving blast of air is a lot more efficient at cooling an object than stationary air. Convection ovens take advantage of this principle too, but for heating rather than cooling. Liquids will probably be more effective than gases for most cooling applications, but there will probably be some exceptions.


Is my chlorine sample real? I bought a pressurized liquid chlorine ampoule online and I was surprised how thick it was.  It was more like oil than water.  Here's a YouTube video showing someone else with a very similar sample.

I trust my seller but I’m wondering why is it so thick? It doesn’t flow easily.

At $\pu{20 °C}$ dynamic viscosity of liquid chlorine is about $\pu{0.35 mPa·s}$ (see Figure 4 from [1, p. 7]), whereas dynamic viscosity of water is about $\pu{1 mPa·s}$, e.g. water is nearly 3 times more viscous at room temperature than liquid chlorine, which should not appear that thick.

Figure 4. Viscosities of chlorine gas and liquid.

Note that your perception of viscosity and "thickness" may be distorted due to capillary effect if your chlorine sample is sealed in a tiny ampule.
References

Chlorine: Principles and Industrial Practice, 1st ed.; Schmittinger, P., Ed.; Wiley-VCH: Weinheim; New York, 2000. ISBN 978-3-527-29851-8.


Change of ionic packing upon crystals are pressurized When rock salt crystal is pressurized, its FCC structure changes to BCC structure whereby coordination ratio changes from 6:6 to 8:8. At the same time packing efficiency of the crystal decreases from 74% of FCC to 68% of BCC. So will the crystal actually expand or contract during this process?

Your termination of BCC is incorrect about the $\ce{NaCl}$ crystal structure upon pressurization. It is actually simple cubic (SC as $\ce{CsCl}$) where eight $\ce{Cl-}$ ions ate the corners of unit cell and one $\ce{Na+}$ in the center or vise versa. Although either $\ce{Na+}$ or $\ce{Cl-}$ in the center, it is not BCC because ion possessing the center of unit cell is different from the ions at the corners of unit cell:

Thus, lattice constant $a$ can be expressed as:
$$a_\mathrm{SC} = \frac{2(r_\ce{Na+} + r_\ce{Cl-})}{\sqrt{3}} \tag1$$
For FCC $\ce{NaCl}$ crystal structure:
$$a_\mathrm{FCC} = 2(r_\ce{Na+} + r_\ce{Cl-}) \tag2$$
Since $a_\mathrm{SC} \lt a_\mathrm{FCC}$, we can conclude that upon pressurization, $\ce{NaCl}$ crystal structure would be  contracted.


Reactions of Depressants In the froth flotation process, the concentration of a sulphide ore takes place using $\ce{NaCN/KCN}$ as a depressant. It forms a soluble complex with the impurity sulphide.

Taking the particular case of lead extraction from Galena($\ce{PbS}$), $\ce{NaCN}$ reacts with the impurity $\ce{ZnS}$ present in it forming a soluble complex.
$$\ce{ZnS + 4NaCN → [Zn(CN)4]^{2-} + 4Na+ + S^{2-}}$$

We also know $\ce{Pb^2+}$ ions react with $\ce{NaCN/KCN}$ to form a white precipitate of $\ce{Pb(CN)2}$.

My question therefore is why when both $\ce{ZnS}$ and $\ce{PbS}$ are present in the solution only $\ce{ZnS}$ reacts whereas $\ce{PbS}$ doesn't? Or is it simply that it reacts but from an insoluble compound?

Ok, after having poked around a bit I doubt that $\ce{Pb(CN)2}$ forms.

I'm guessing that you're collecting the froth to get a concentrate of the galena $\ce{PbS}$ out of some finely crushed ore.

An aspect of the operation seems to be chemically reducing the concentration of $\ce{ZnS}$ by dissolving via the following reaction:

$$\ce{ZnS(s) + 4NaCN(l) -> Zn(CN)4^{2-}(l) + 2Na2S }$$

Now there are two ways that $\ce{ZnS}$ could be mixed in with the $\ce{PbS}$.

(1) The individual $\ce{Zn^{2+}}$ cations replace individual $\ce{Pb^{2+}}$ cations in the $\ce{PbS}$. In this case the random distribution of the $\ce{Zn^{2+}}$ cations wouldn't allow much extraction of $\ce{ZnS}$ as $\ce{Zn(CN)4^{2-}}$. The $\ce{Zn^{2+}}$ cations just are not going to diffuse through the solid galena particles to any appreciable extent.

(2) There could be also be that the crushing creates a mixture of relatively pure $\ce{PbS}$ and $\ce{ZnS}$ particles which are mixed together. In this case an appreciable amount of the Zn impurity could be removed.

Now for what seems to be the question. $\ce{PbS}$ is very insoluble with a $K_\mathrm{sp} = 9.04\times10^{−29}$. I couldn't find a solubility product for $\ce{Pb(CN)2}$ but it is doubtful that the solubility would be enough to cause the lead sulfide to dissolve and then form lead cyanide.


Methanol mixed with varsol If I have varsol in a tank and methanol is put into the same tank will methanol rise to the top or will it separate? Or will it mix together ?

Note that I have never heard about Varsol until now, living in the Central Europe. By searching Wikipedia ( what you could do as well before asking ), I realized the below:

Varsol alias White spirit alias many other synonyms, used as a painting solvent, is based on C7-C12 oil hydrocarbons, with density 0.765(type T1), 0.780(type T2), 0.795(type T3)
( divided by volatility, T1 the most volatile )

Methanol has density 0.792. So it would rather sink under T1, T2, or being indecisive with T3 ( this may be temperature dependent )

Note that methanol has very limited solubility in non-aromatic hydrocarbons.

Note: with reported partial aromaticity, methanol will be somewhat soluble.


Effect of evaporation on the concentration of a saturated solution When evaporating a saturated solution, does the concentration remain the same?

My thinking is that because the solution is saturated, and the volume is decreased, the excess solute would precipitate out of the solution. Therefore maintaining the original concentration.

On the other hand, would the precipitate be included when measuring the concentration? And therefore evaporation of a saturated solution would lead to an increase in concentration?

For example: if you have a 100mL of 2.5M saturated NaCl solution and evaporate it until the volume is 50mL, what would the final concentration be?

When a solution is saturated, it contains the maximum amount of solute which can be dissolved at that particular temperature. So effectively it is at its "maximum concentration" (for a given temperature).

This means that if your solvent is evaporating and the temperature remains constant, then solute will precipitate out of solution. The concentration of the solution will still be at its maximum as you are concerned with the amount of solute dissolved.

If you are heating up the solution in order to evaporate the solvent then the increase in temperature might mean that your solute solubility increases. This will depend on what solvent/solute you have.

However, a saturated solution which cools down without being disturbed may form a super-saturated solution whereby its concentration has gone beyond the solubility of the solute at that temperature.


Why is helium used in diving apparatus? In my textbook it said that helium is used in oxygen tanks (diving apparatus) by deep sea divers.

When I searched the internet for the reason it said that :

  A mix of helium and oxygen is used in diving apparatus by deep sea divers, due to low solubility of helium in blood even at high pressure.

My doubts:

Why to mix helium when we can just fill the tanks with pure oxygen?
Why can't other inert gas be used in place of helium?

As TAR86 said pure oxygen is problematic.

A little something from the Wikipedia page on oxygen toxicity

  Severe cases can result in cell damage and death, with effects most often seen in the central nervous system, lungs and eyes. Oxygen toxicity is a concern for underwater divers, those on high concentrations of supplemental oxygen (particularly premature babies), and those undergoing hyperbaric oxygen therapy.

Turns out nitrogen is also toxic at high pressure see nitrogen narcosis.

Helium is used primarily because it's cheaper as compared to other alternatives and is not narcotic.

Other gases are too heavy and they diffuse into the bloodstream.

For more information see this PDF on exotic diving gases.

Fun Fact
Argon is used but can you guess why?

   Used to inflate scuba suits, helps in insulating them.


What is the difference between an Aryl Halide and a Vinyl Halide? Are aryl halides a type of vinyl halide? From my knowledge, when a Halogen atom is directly bonded to a doubly bonded carbon atom it is a Vinyl halide. When we look at halogen atom directly bonded to a benzene ring we see a similar thing. So are these terms never correlated? Also is Phenylic Halide a term which exists?

No, aryl halides are not a type of vinyl halides.
A compound is called an aryl halide when it contains a halogen group directly attached to an aromatic system. For example,

Whereas a compound is called a vinyl halide when it contains a halogen group directly attached to a doubly bonded carbon atom. For example,

The double bonds in an aromatic system are not the same as the aliphatic double bonds. They differ quite a bit in terms of electron density and thus there derivatives behave very differently is many circumstances.
The term phenylic halide doesn't exist but a very similar term phenyl halide exists which is a subset of aryl halides family.
An aryl halide can be called a phenyl halide when the aromatic system of the compound is specifically the phenyl system($\ce{-C6H5}$). For example,


Pigmentation of aromates So, they try to color naftalene, nitrobenzeen and benzene. Only one of them colors. They can say with certainty that this is nitrobenzeen. Right/wrong and why?
Sorry if my English isn't that great, not my native language!!
Thanks a lot x

It is indeed nitrobenzene.

You really should look at the compounds, literally, because chemistry is ultimately an experimental science.

That being so, you can still note that the conjugated nitro group with its electronegative atoms gives relatively low-lying vacant orbitals into which electrons can be excited.  These orbitals can be accessed by absorbing short wavelength (violet/blue) visible light leading to a yellow color.


What will be the possible conditions to react bromo alkane with diethyl phenylmalonate? How to react 1-bromopentane with diethyl phenylmalonate?
I used NaH (1.5eq) but the yield is less than 30%. I used KOH but yield is low. I used tBuOK and it did not work, I used K2CO3, but not working.
I tried it, K2CO3 2eq with 18-crown-6 and KI 0.1eq, in acetone but starting material is not consumed, almost only 30% has been finished after more than 72 hours and TLC shows a different spot, it does not look like desired product. Starting material is recovering when I use K2CO3, but not in case of NaH 1.5eq with DMF or THF but yield is less than 30%. Toluene with NaH under reflux condition is also behaving like K2CO3 and my next step is hydrolysis and decarboxylation.
I need guideline about what I am missing here.

You have tried a number of sensible approaches and nothing is working well. I think the problem here is that the anion you are creating is very stable and hindered and no amount of tweaking will significantly improve it. You either accept the yield as it is and scale the reaction accordingly, or change your strategy.

Your next step is hydrolysis and decarboxylation to give a substituted phenylacetic acid so I suggest you start with ethyl phenylacetate. You could alkylate it with NaH (1eq) in DMF with KI then 1-bromopentane. That should alkylate well, but there will be a percentage of di-alkylation so I'm going to suggest you condense ethyl phenylacetate with pentanal and then hydrogenate (or otherwise reduce the double bond). I think this will give you a cleaner reaction mixture and higher yield.


What is the reason for isotropy in amorphous solids? In my book, it's been mentioned that crystalline solids are anisotropic whereas amorphous solids are isotropic in nature. The reason for these has been explained (with a diagram) as:

Crystalline solids are anisotropic in nature, that is, some of their physical properties like electrical resistance or refractive index show different values when measured along different directions in the same crystals. This arises from different arrangement of particles in different directions. Since the arrangement of particles is different along different directions, the value of the same physical property is found to be different along each direction.
Amorphous solids on the other hand are isotropic in nature. It is because there is no long range order in them and arrangement is irregular along all directions. Therefore, value of any physical property would be same along any direction.

My question is: similar arrangement of particles in different directions in crystalline solids gives rise to different physical properties. Then, how can disarrangement in amorphous solids in all directions give rise to same physical properties?
In other words, why do crystalline solids have properties (like electric conductivity, thermal conductivity, mechanical strength, and refractive index) different in different directions, and the amorphous solids have these properties same in all directions?

Assume that the spheres in the drawing are atoms.

In the second picture atoms are not orderly arranged. No matter where you go in crystal of a amorphous solid, the disarrangement will be same. Since in every direction, disarrangement is same, physical properties along every direction is also same. Therefore, amorphous solids are isotropic.

In the first picture, the arrangement is orderly. So, when you go in any direction, the arrangement will be different. It is also clear in the rectangles shown in the picture. Try doing it yourself on a piece of paper. Try finding any four different directions along which the arrangement is same.

Ps: Ignore my poor drawing skills.


Rate of Reaction Graph - Gas Produced I did a reaction and I measured the volume of gas produced against time. I need to find the rate constant of the reaction.

I am trying to plot a graph of ln(vol of gas produced) against time so that the gradient will be equal to -k so i can calculate the rate constant k. I am using the equation ln[A]t = -kt + ln[A]o

The problem is I can't plot ln0. I timed the reaction every 5 mins and gas wasn't produced until 15mins had passed. This means that the volume of gas produced for time values 0, 5 and 10 mins is 0cm^3, but ln0 doesn't exist.

How do I use my data to create a graph to calculate the rate constant?

My results were:

Time 0___5___10___15___20___25___30___35___40___45___50___55___60 ____(in mins)

Vol _0___0___0_____5___10___18____22___28___36___44___51____60____70___(in ml produced)

I think the first bit of gas produced probably dissolved in the water, so you didn't see it for a little while.

You can skip your three initial "points" with zero volume, and you're not going to have $\ln(0)$, which is what's hanging you up. Note that you did this in your report, and why!

Do a linear fit with the remaining points. Your y-intercept will be $\ln \mathrm{[A]}_0$ and your slope will be $-k$. It's not a terribly good fit.


Magnetic moment of trivalent lanthanide cations The effective magnetic moment $\mu_{\mathrm{eff}}$ of tripositive rare earth elements, is calculated by $$\mu_{\mathrm{eff}}=g_J\sqrt{J(J+1)}\mu_\mathrm{B}$$

Why can't we use normal formula as $\sqrt{n(n+2)}\mu_\mathrm{B}$ for them? I calculated $\mu_{\mathrm{eff}}$ for $\ce{Ce^3+}$ ion and got $2.54\mu_\mathrm{B}$ using first formula and $2.83\mu_\mathrm{B}$ using second one, there's a significant difference in two. How is the first formula derived?

Your question, Why can't we use normal formula to find effective magnetic moment of tripositive rare earth elements (REs)? is answered excellently by  porphyrin' comment (vide supra). Thus, I won't attempt to answer it again except for to mention the calculated values from the equation, $\mu_\mathrm{eff} = g_J \sqrt{J(J+1)}\mu_\mathrm{B}$ (where $\mu_\mathrm{B}$ is Bohr magneton and $g_J$ is Landé factor), have good agreement with experimental values.

Table 1: Theoretical magnetic moment of tripositive rare-earth ions
$$
\begin{array}{c|ccc}
\text{Ion} & 4f^n & S^c & L^c & J^c & g^c & \mu_\mathrm{eff} = g_J \sqrt{J(J+1)}\mu_\mathrm{B} & \text{Observed }\mu^a \\
\hline
\ce{La} & 0  & 0 & 0 & 0 & - & 0.00 & -\\
\ce{Ce} & 1 & \frac{1}{2} & 3 & \frac{5}{2} & \frac{6}{7} & 2.54 & 2.4\\
\ce{Pr} & 2 & 1 & 5 & 4 & \frac{4}{5} & 3.58 & 3.5\\
\ce{Nd} & 3 & \frac{3}{2} & 6 & \frac{9}{2} & \frac{8}{11} & 3.62 & 3.5\\
\ce{Pm} & 4 & 2 & 6 & 4 & \frac{3}{5} & 2.68 & -\\
\ce{Sm} & 5 & \frac{5}{2} & 5 & \frac{5}{2} & \frac{2}{7} & 0.85 & 1.5\\
\ce{Eu} & 6 & 3 & 3 & 0 & - & 0.00 & 3.4\\
\ce{Gd} & 7 & \frac{7}{2} & 0 & \frac{7}{2} & 2 & 7.94 & 8.0 \ (7.23^b)\\
\ce{Tb} & 8 & 3 & 3 &6 & \frac{3}{2} & 9.72 & 9.5 \ (9.58^b)\\
\ce{Dy} & 9 & \frac{5}{2} & 5 & \frac{15}{2} & \frac{4}{3} & 10.65 & 10.6\\
\ce{Ho} & 10 & 2 & 6 & 8 & \frac{5}{4} & 10.61 & 10.4\\
\ce{Er} & 11 & \frac{3}{2} & 6 & \frac{15}{2} & \frac{6}{5} & 9.58 & 9.5\\
\ce{Tm} & 12 & 1 & 5 & 6 & \frac{7}{6} & 7.56 & 7.3\\
\ce{Yb} & 13 & \frac{1}{2} & 3 & \frac{7}{2} & \frac{8}{7} & 4.54 & 4.5\\
\ce{Lu} & 14 & 0 & 0 & 0 & - & 0.00 & -\\
\end{array}
$$
$^a$ Rare-earth and metals really show "Spin"?; $^b$ Reference 1; and $^c$ Reference 3.

The two experimental values from Ref.1 are in $\pu{emu mol−1 Oe−1}$ and have been converted appropriately to be able to compare with the values in hand. The reference also point out that crystals of mixed salt have approximately average of the pure monometal salts. The abstract of Ref.1 states that:

  Magnetic moment and susceptibility measurements of single crystals of pure and mixed rare earth fumarates of gadolinium and terbium were carried out at room temperature. The experimental values of molar susceptibilities for $\ce{Gd2(C4H2O4)3.7H2O}$, $\ce{Tb2(C4H2O4)3.7H2O}$, and $\ce{GdTb(C4H2O4)3.7H2O}$ are $2.68 \times 10^{−2}$, $3.89 \times 10^{−2}$, and $3.18 \times 10^{−2}$ (in $\pu{emu mol−1 Oe−1}$), respectively. The calculated effective magnetic moments are in good agreement with the theoretical predictions on rare earth ions.

The magnetic properties of rare earth elements (REs) are due primarily to the electrons in the incomplete shell ($\mathrm{4f}$). In fact the magnetic moment per atom is determined by the number of electrons in the $\mathrm{4f}$ shell according to the Russell-Saunders coupling scheme. The derivation of the equation $(1)$ is partially explained in Ref.2:

$$\mu_\mathrm{eff} = g_{J} \sqrt{J(J+1)}\mu_\mathrm{B} \tag{1}$$

According to Ref.2:

  The Russell-Saunders coupling scheme is normally adopted to describe the $\mathrm{4f}$ magnetism of the REs. The spins of the individual electrons are coupled to form a total spin $S$ by the exchange interaction and the individual orbital angular momenta are coupled to form a total orbital momentum $L$ by the Coulomb interaction. The state with the lowest energy is found from Hund’s rules by maximizing $S$ and thereafter $L$. The total angular momentum $J$ is given by $J = |L ± S|$, where the minus sign is used for less than half filled shells and the plus sign for more than half filled shells. The total magnetic moment due to the spin angular momentum is $µ_S = 2µ_B\sqrt{S(S + 1)}$, where $µ_B$ is the Bohr magneton. The total moment due to the orbital angular momentum is $µ_L = µ_B\sqrt{L(L + 1)}$. These moments process around the direction of $J$, therefore the magnetic moment due to the total angular momentum looks slightly more complicated: $µ_J = g_J µ_B\sqrt{J(J + 1)}$, where $g_J$ is the Landé factor [12].

The sighted Reference ([12]) is actually Ref.3 here. According to Ref.3:
$$
J =
\begin{cases}
L-S,  & \text{if $4F^n \le 7$} \\
L+S, & \text{if $4F^n \gt 7$}
\end{cases}
$$

Note: The equation $(1)$ does not obey experimental values of $\mu$ found for $\mathrm{3d}$ elements (Rare-earth and metals really show "Spin"?).

References:

B. Want, M. D. Shah, “Magnetic susceptibility measurements of pure and mixed gadolinium–terbium fumarate heptahydrate crystals,” Journal of Magnetism and Magnetic Materials 2016, 401, 391-393 (https://doi.org/10.1016/j.jmmm.2015.10.065).
I. L. M. Locht, Y. O. Kvashnin, D. C. M. Rodrigues, M. Pereiro, A. Bergman, L. Bergqvist, A. I. Lichtenstein, M. I. Katsnelson, A. Delin, A. B. Klautau, B. Johansson, I. Di Marco, O. Eriksson, “Standard model of the rare earths, analyzed from the Hubbard I approximation,” Phys. Rev. B 2016, 94, 085137 (https://doi.org/10.1103/PhysRevB.94.085137)(PDF).
Jens Jensen, Allan R. Mackintosh, In Rare earth magnetism: Structures and Excitations; The International Series of Monographs on Physics, J. Birman S. F. Edwards, C. H. Llewellyn, & Smith M. Rees, General Eds.; Clarendon Press: Oxford, United Kingdom, 1991.


Does a neutral dimercury molecule exist? Is a neutral $\ce{Hg2}$ molecule possible, as a gas under extremely low (partial) pressures? What is the enthalpy of formation or similar?

It is my impression that usual mercury vapors are monatomic? I think it is well-known that a dimer ion $\ce{Hg2^{2+}}$ exists under some conditions.

Somewhat related threads: Is it possible to have a diatomic molecule of sodium in gaseous state? and Which elements can be diatomic?

There has been recent research on mercury dimers in its ground state. Bond energies, transition energies, band spectrum and other spectroscopic parameters has been calculated. Here are the abstracts of two papers on mercury dimer:

  The potential energy curve of the ground electronic state of the $\ce{Hg}$ dimer has been calculated using the CCSD(T) procedure and relativistic effective core potentials. The calculated binding energy ($\pu{0.047 eV}$) and equilibrium separation ($\pu{3.72 Å}$) are in excellent agreement with experiment. A variety of properties, including the second virial coefficient, rotational and vibrational spectroscopic constants, and vibrational energy levels, have been calculated using this interatomic potential and agreement with experiment is good overall.(Source)

  The mercury dimer is among the most weakly bound metal dimers and has been extensively studied. The ground state $\ce{O^+_g}$ dissociation energy has been considered to lie between $\pu{0.55 eV}$ ($\pu{440 cm−1}$) and $\pu{0.091 eV}$ ($\pu{730 cm−1}$). We report here a spectroscopic study of $\ce{Hg2}$ in a supersonic jet. The first optical transition, $\ce{1_u←O^+_g}$, was characterized by its fluorescence excitation spectrum and the binding energy of the ground state has been measured precisely through the threshold of collision induced dissociation of $\ce{Hg2 1_u}$ to $\ce{Hg(^1S_0)+Hg(^3P_0)}$.(Source)


Does purified water corrode copper pipes? In other words would a reverse osmosis system corrode copper pipes due to it's purification?  Does purified water speed up copper pipe corrosion leading to pin holes, etc.?

Although it seems pretty clear that removing potentially corrosive compounds from drinking water will result in a net of no effect or possibly decreased corrosion, there is at least one consideration worthy of mention.

While metallic copper has a solubility that is well below most regulatory standards for drinking water (i.e. 2 mg/L in the UK), enough copper can still potentially dissolve to result in a significant loss of copper mass, such that pinholes and destructive oxidative processes can take place.

However, according to the UK's National Physics Laboratory, Good Practice Guide No. 120, "Avoidance of corrosion in plumbing systems", September 2011:

  "...the rate of corrosion and the release of copper ions slow rapidly in the first few weeks of operation as protective films form on the bore of the copper pipes."

Further:

  The surface films (often referred to as a patina) are composed of thin brown or black coloured copper oxide ($\ce{CuO or Cu2O}$) overlaid with a thicker green or turquoise basic copper carbonate layer. These layers, although protective, are not true passive layers (like those which form on stainless steels or aluminium and which are much thinner and invisible to the naked eye). It is when these films on copper do not form properly in the first place or become disrupted later that problems of pitting or blue water can occur.

According to this related document from the UK's "Foundation for Water Research", the low alkalinity and buffering capacity of purified water can result in conditions of slight acidity that can prevent the formation of the desired patina, and recommends the addition of 50 mg/L bicarbonate alkalinity in order to promote the development and maintenance of the protective copper oxide and carbonate layers.

To reiterate, my answer to the question is a net "No" with an interesting caveat.


How could the ideal gas law be discovered from experiments on real gases? The gas laws, namely Boyle's Law, Charles' Law, Avogadro's Law and Gay-Lussac's Law, are all experimental laws. Combining these laws, we get the ideal gas law $pV=nRT$. Also, "real life" gases do not exactly follow this law, so there are more laws for "real life" gases: van der Waals' law, Dieterici equation, etc., which approximately describe these laws within certain boundaries of the gas parameters: pressure $p$, volume $V$ and temperature $T$.

But there seems to be an apparent logical flaw: the ideal gas law $pV=nRT$  was found by experimenting on "real life" gases, but these "real life" gases do not follow the ideal gas law. How could this be the case?

The ideal gas law is a very good approximation of how gases behave most of the time

There is no logical flaw in the laws. Most gases most of the time behave in a way that is close to the ideal gas equation. And, as long as you recognise the times they don't, the equation is good description of the way they behave.

The ideal gas equations assume that the molecules making up the gas occupy no volume; they have no attractive forces between them and their interactions consists entirely of elastic collisions.

These rules can't explain, for example, why gases ever liquefy (this requires attractive forces). But most of the common gases that were used to develop the laws in the first place (normal atmospheric gases like oxygen or nitrogen) are usually observed far from the point where they do liquefy.

As for the volume taken up by the molecules of the gas, consider this. A mole of liquid nitrogen occupies about 35mL and this is a fair approximation of the volume occupied by the molecules. at STP that same mole of gas occupies a volume of about 22,701mL or about 650 times as much. So, at least to a couple of decimal places, the volume occupied by the nitrogen molecules is negligible.

The point is that for gases not close to the point where they liquefy (and few components of the atmosphere are), the ideal gas laws are a very good approximation for how gases behave and that is what we observe in experiments on them. The fancy and more sophisticated equations describing them are only really required when the gas gets close to liquefaction.


Why is the atomic radius of krypton smaller than that of potassium? A potassium atom comprises 19 protons and 18 core electrons. Those electrons screen 18 protons worth of Coulomb force. Therefore, there is one proton worth of Coulomb force attracting the valence electron of potassium. Whereas, krypton comprises 36 protons and 28 core electrons. Its electrons screen 28 protons wroth of Coulomb force. Therefore, there are eight protons worth of Coulomb force attracting the valence electrons of krypton.

Consequently, there is more Coulomb force attracting the valence electrons of krypton than there is attracting the valence electron of potassium; accordingly, the atomic radius of krypton is smaller than the atomic radius of potassium.

Is that correct?

Your concept is close to the real answer. You should think it as no. of protons per electron over all. This is the way things have been done in literature and if you are not satisfied then consider the fact that not all electrons are shielding the nucleus to same extent. There are certain rules [slaters rule] made and which works in most cases by which you can actually predict how much of nuclear charge is acting on the electron. Even with that you will see valence electron for K is more shielded than for Kr. But the numbers you came up with don't make that much sense. It does a bit but better if you go by the way it has been discussed.


Phosphorus standard state Why is white phosphorus considered to be the standard state of phosphorus although it is only metastable? Red phosphorus is thermodynamically more stable than white phosphorus, yet it is not the standard state.

It is correct that white phosphorus is usually used as the standard state for tabulating thermodynamic data, in particular by the Committee on Data for Science and Technology (CODATA).

A comparison of thermodynamic data for phosphorus and some of its compounds can be found in Rard, J. A.; Wolery; T. J. The Standard Chemical-Thermodynamic Properties of Phosphorus and Some of its Key Compounds and Aqueous Species: An Evaluation of Differences between the Previous Recommendations of NBS/NIST and CODATA. J. Solution Chem. 2007, 36 (11), 1585–1599.. This publication also includes an explanation for the use of white phosphorus as the standard state:

  Although white phosphorus is not the thermodynamically stable allotrope, the red and black forms are difficult to prepare in pure form, which makes them less suitable for quantitative thermodynamic measurements.

Similarly, from the other side of the Atlantic, A.T. Dinsdales' SGTE Data for Pure Elements, originally published as CALPHAD 15(4) 317-425 (1991), says:

  [The] reference phase is normally the phase stable at 10$^{5}$ Pa and 298.15 K. The exception to this rule is phosphorus for which, by convention, the white form is chosen as the reference phase because the more stable red form is difficult to characterise.


According to MO theory, why is the dilithium cation more stable than the dilithium anion? According to molecular orbital theory, $\ce{Li2}$ fills up two electrons in the $\sigma_\mathrm{g2s}$ molecular orbital;
$\ce{Li2+}$ fills up one electron in the $\sigma_\mathrm{g2s}$ MO, and
$\ce{Li2-}$ fills up two electrons in the $\sigma_\mathrm{g2s}$ MO and one electron in $\sigma^*_\mathrm{u2s}$, resulting in the same bond order for the ions.

I first thought that $\ce{Li2+}$ is stable, because they don't have paired electrons in $\sigma_\mathrm{g2s}$, therefore there is no pairing energy.
I soon found that $\ce{Li2+}$ is stable, but someone said it is due to the unfilled antibonding orbital. Is this a right explanation?

Because $\ce{Li2}$ is homonuclear, the weight that imposed on each $\mathrm{2s}$ $\ce{Li}$ atomic orbital would be almost the same. Even though $\ce{Li2-}$ is unstable by filling an antibonding orbital, it will be offset by filling two electrons into the bonding orbital and it has the same bond order as $\ce{Li2+}$ does.

What am I missing now?

First of all some visual aid: an MO Diagram for $\ce{Li2}$.

The concept of bond order allows a very rough estimate of stability. Here we get $\mathrm{BO}(\ce{Li2})=1$, $\mathrm{BO}(\ce{Li2+})=0.5$, and $\mathrm{BO}(\ce{Li2-})=0.5$. So in first approximation, $\ce{Li2-}$ and $\ce{Li2+}$ are equally stable, while $\ce{Li2}$ more stable. But we can consider some more details:

  I first thought that $\ce{Li2+}$ is stable, because they don't have paired electrons in $\sigma_{g2s}$, therefore there is no pairing energy.

Adding a second electron to some orbital, would at first double the energy of that bond. But it also introduces some electron-electron repulsion between the two. This however is usually smaller than the gained bonding energy. So the contribution of the $\sigma_\mathrm{g2s}$ contribution in $\ce{Li2}$ is a little less than double its contribution in $\ce{Li2+}$. Please take this last statement with care, because things are more complex than that.

  I soon found that $\ce{Li2+}$ is stable, but someone said it is due to the unfilled antibonding orbital. Is this a right explanation?

Basically yes. The thing is, that the energy split of bonding and anti-bonding orbitals is not symmetric with respect to the originating atomic orbitals. The anti-bonding orbital is a little bit higher, than what the bonding orbital is lowered:

$$
E(\sigma^*_\mathrm{u2s}) - E(\mathrm{2s}) >
E(\mathrm{2s}) - E(\sigma^*_\mathrm{g2s})
$$

This is because if you add more electrons, you add more electron-electron repulsion. In case of $\ce{Li2-}$ this means the electron in the anti-bonding orbital is more destabilizing (with respect to $\ce{Li2}$), than what the additional electron in the bonding orbital is stabilizing (when going from $\ce{Li2+}$ to $\ce{Li2}$).

  Because $\ce{Li2}$ is homonuclear, the weight that imposed on each $\mathrm{2s}$ $\ce{Li}$ atomic orbital would be almost the same.

Not just almost. The electronic wave function must follow the symmetry of the molecular structure, therefore the weights of the orbitals have to be exactly the same. But this argument is not really helpful for this question.


What mixing ratio of ethanol and acetone has the lowest freezing point? I am in the lab trying to make a sample for 1H NMR. It has to remain liquid at the lowest possible temperature.

I was recommended to use a mixture of ethanol and acetone because when they're mixed in the right proportions, the mixture has a very low freezing point.

I can not find this proportion anywhere. Does anyone know where to look for these things?

Here is a picture of what I'm looking for, but with ethanol and acetone instead of water - solvent:

PD:it's very likely that its methanol instead of ethanol, please let me know... I have access to both, I just want the one that will go to the lowest temp.

its me again, I  want to post the answer I found because I think it might be useful for somebody in the future; it appears a 2:3 molar acetone:methanol proportion gives the lowest freezing temperature ~158 K (see attached picture below).

It appears the difficulty finding these curves was because they are usually called "solid-liquid equilibrium" curves instead of "freezing point curves", as its the same physical situation (Explanation Here)

There is a large database of these in the following link.

EDIT: As commented by "Ivan Neretin", ethanol-acetone mixtures are also good for low freezing temperatures:


Mechanism for Diels–Alder of alkene with substituted thiophene sulfone with loss of sulfur dioxide I came across a synthesis of centrohexaindane that has the following steps:

This is a Diels–Alder reaction that gives an aromatic ring. I know that alkynes will give this reaction as follows:

But in the scheme, after the initial Diels-Alder to form the product with bridging $\ce{SO2},$ it cannot just lose $\ce{SO2}$ to become aromatic. Does the molecule with the bridging $\ce{SO2}$ get oxidized to have two double bonds like in the second image and then lose $\ce{SO2}?$
The sulfone is the only possible oxidizing agent here (toluene solvent doesn't seem like it would oxidize) and Wikipedia says on its page on sulfones that:

Sulfone is a relatively inert functional group, being weakly basic(compared to sulfoxides). They are non-oxidizing.

A few references into the rabbit hole and we have an answer. The major reference was given in the link mentioned by OP which was Chem. Ber. 1994, 127 (2), 409–425[1], in which the reaction mechanism of the synthesis is given as follows:

[...] Reaction of 33 with tetrathiophene-S,S-dioxide is performed by heating the reactants in highly concentrated toluene solution to $\pu{120 °C}$ for a total of
24 h. The cycloadduct 34 is obtained in high yield.

Now the question that has arisen in the step from 33 to 34 is the following, under normal conditions of Diels-Alder, there would only be one double bond in the final product. Here there are two. This can be answered going one down into the rabbit hole and using J. Org. Chem. 1980, 45 (5), 856–867.[2]
This paper deals with the specific reaction that causes confusion - the annulation using tetrachlorothiophene 1,1-dioxide. According to this paper:

The compound has a combination of stability and reactivity that makes it a useful synthetic reagent. It undergoes Diels-Alder addition with double bonds to form an adduct which spontaneously loses sulfur dioxide to leave the double bond annelated with a tetrachlorobutadienediyl group.

Furthermore, it talks about the reactivity to olefinic bonds and states the following:

Example 18, from l-hexen-5-yne, reveals the greater reactivity of alkene vs. alkyne. Reaction is exclusively with the alkene bond as far as NMR reveals. Thus, the kinetically favored product is formed rather than the thermodynamically favored one.

There is a minor error in the reaction mechanism as given in the link of OP as the first paper shows that the tetra-chloro derivative needs to be oxidised in order to be converted into cyclohexaindane
References:

Kuck, D. Benzoanellated Centropolyquinanes, 15. Benzoanellated Fenestranes with [5.5.5], [5.5.5.6], and [5.5.5.5] Frameworks: The Route from 1,3-Indandione to Fenestrindan. Chem. Ber. 1994, 127 (2), 409–425. DOI: 10.1002/cber.19941270218.
Raasch, M. S. Annulations with tetrachlorothiophene 1,1-dioxide. J. Org. Chem. 1980, 45 (5), 856–867. DOI: 10.1021/jo01293a019.


Spectroscopy of excited state lithium I have used data from a question regarding the reciprocal wavelengths observed when lithium transitions from excited states to a common ground state. The lowest energy transition is from n=3 to n=2. The question goes on to say that this lowest energy transition, in high resolution, has 7 peaks, and asks for an explanation as to why there are not 8 lines. Can I have some guidance on this? I think spin-orbit coupling is relevant. I am aware that the dipole moment of an orbiting electron interacts with a magnetic field (from the orbital angular momentum of the electron?) and so some orbitals are raised/lowered slightly, those with an "l" greater than 0.

The energy levels are in increasing energy  3S, 3P, 3D and the final state is 2P. The P-P transition is not allowed by $\Delta L =\pm1$ selection rule. The two transitions are 3S-2P at approx 812 nm and 3D-2P at approx 610 nm. As you are asked about the lowest energy transition this must be the 3S-2P. (Li also has non-zero nuclear spin depending  on its isotope which lead to hyperfine transitions and are ignored )

With some spin-orbit coupling the (Russell-Saunders) terms symbols are for the upper level $^2S_{1/2}, $ and for the lower level $^2P_{3/2}, \, ^2P_{1/2}$. The $M_J$ levels are assumed to be split so the degeneracy is 2 for the S and 2 or 4 for the P states as given by $2J+1$.  The selection rules on $M_J$ levels are $M_J = 0,\, \pm 1$.

Thus there are four transitions possible from $^2S_{1/2}$ to $^2P_{1/2}$ starting in  $  M_J = 1/2 $ in $^2S$ to $(1/2, -1/2)$ in $^2P$ and similarly $  M_J = -1/2 \rightarrow (1/2,\, -1/2)$, making 4 transitions in all as the splitting of the S and P levels will be different. (This is the same case as the splitting of sodium $D_1$ line but 'upside down' in energy to that case).

The other transitions are from $^2S_{1/2}$ to $^2P_{3/2}$. In this case there are 6 transitions as the P splits into 4 levels. The transition between the $\pm 1/2$ make four of these (as just outlined above) and in addition there are transitions between $M_J = 3/2 \rightarrow 1/2$ and $M_J = -3/2 \rightarrow -1/2$.

This makes ten transitions in all, thus it seems I have calculated the wrong transitions (or misunderstood completely),  however, you should be able to follow the method. Note that the observed spectral lines may not correspond to transition as distinct transitions may have the same energy gaps.

I've added a sketch below, it is hand drawn but should be clear enough. The $S_{1/2} - P_{1/2}$  transitions are a subset of these as you can see.


Identifying covalent character in ionic compounds when varying multiple ions By Fajans' rules, we can easily find out which compound shows covalent character.

Example: Among $\ce{NaCl}$, $\ce{MgCl2}$, $\ce{AlCl3}$ which one is more covalent?

Answer: There is a point in Fajans' rules that the compound which has more charge number that compound will show covalent character. In $\ce{AlCl3}$, Al has the highest oxidation state (+3) so it will show more covalent character than $\ce{NaCl}$ and $\ce{MgCl2}$.

The point is that since all three compounds I have mentioned have chlorine in common, it was easy to catch the solution but if the compounds were given as $\ce{NaF}$ and $\ce{LiCl}$, which one shows more covalent character, and how can we apply Fajans' rules?

You will have to consider two factors here: polarising power of the cation and polarisability of the anion.
There is more than one part of Fajan’s rule(Source-J. D. Lee, Concise Inorganic Chemistry):

  A small positive ion favours covalency.
  A large negative ion favours covalency.
  Large charges on either ion, or both ions, favours covalency.
  Polarisation and hence covalency is favoured if the positive ion does not have noble has configuration.

According to these, the $\ce{Li+}$ ion is more polarising than $\ce{Na+}$ ion, and $\ce{Cl-}$ ion is more polarisable than $\ce{F-}$ ion.

Since both factors support it, $\ce{LiCl}$ is more covalent than $\ce{NaF}$.

Somtimes, like in your question, both the factors support each other. Sometimes, when they are opposite, like in the case of $\ce{LiF}$ and $\ce{NaCl}$, it is tough to determine relative covalency using Fajan’s rule.


Order of halide nucleophilicity in DMF The order of nucleophilicity for halide family in DMF is:  $\ce{Cl- > Br- > I-}$. I understand the reason for this must be that DMF, being a polar aprotic solvent, is unable to solvate the halide ions.

I was told by my teacher that this is true only for DMF and DMSO, while in other aprotic polar solvents, the order is as usual: $\ce{F- < Cl- < Br- < I-}$.

I'd like to know what is special about DMF and DMSO (and which is not possessed by other aprotic solvents like acetone) that leads to this order.

I have verified the order in DMF from March's 7th edition (page 427). The book mentions the usual order ($\ce{Cl- < Br- < I-}$) for acetone (another aprotic polar solvent).

It depends on the ability of the solvent to solvate anions. If you draw structures of $\ce{DMF}$ with its resonance structure and check the side of the molecule that should interact with anions, you will see, that methyl groups prevent the interaction pretty much. Same for $\ce{DMSO}$. Thus these two solvents are commonly used to enhance the reactivity of nucleophiles. They solvate cations well but anions poorly. Maybe a bit strange, but it works well and it's pretty reliable.


Carbon monoxide detection without a detector I have just been waken up by an alarm and found out that my neighbour did something terrible with his lighter. Luckily, there was no fire, only a plenty of smoke, and our flats were hermetically isolated, so that smoke didn't go to my flat. But, most probably because of my suspiciousness, I noticed some smell of burnt organic, which usually accompanies CO. I switched on ventilation and went out - just in case. There were no CO detectors in the house - only smoke detectors - my bad.
Now the adventure is over (I hope so, at least) and I have got an (im)practical question:

Are there any ways to detect CO using only "kitchen chemistry"?

Surf a civilian aircraft supply.  They sell a little absorbant white tab (chemical spot detector) containing $\ce{PdCl2}$.  When exposed to CO it goes black.

http://www.aircraftspruce.com/menus/ps/carbonmonoxidedetectors.html

Canary.  Goldfish won't work - they can survive on anaerobic respiration,

http://aquaticcommons.org/3663/1/14P089.pdf
   who needs powdered alcohol?


Is it true that under certain conditions, Mg can reduce SiO2? Is it true that under certain conditions, Mg can reduce $\ce{SiO2}$ and the latter the former? What are those conditions?

Yes it is true.

Conditions are temperature in the 650 to 850 degrees C range.

There is a cool looking youtube video of the reaction, not that that makes it true.

For a more serious discussion see Production and Purification of Silicon by Magnesiothermic Reduction of Silica Fume

and Ordered Mesoporous Silicon through Magnesium Reduction of Polymer Templated Silica Thin Films


Can you model intermolecular interactions by using the Born-Oppenheimer approximation to get a potential function for nuclei? Here is a claim about the modeling of chemical systems. I can’t remember if it’s true or not:

Suppose we have a bunch of molecules and we want to know how they interact. Eg we want to look at the physics of two $\ce{CO_2}$ molecules bumping into each other. We can do this by treating the nuclei as classical particles moving in a potential function determined by the ground state electronic wavefunction for a given configuration of nuclei.

Eg, we want to know how hydrogen molecules bounce off each other. We do this by modeling it as two protons with momentums and positions, and we iterate forward through time using a Hamiltonian where the potential energy operator uses the Hartree-Fock method or something to calculate the electronic potential.

This approximation obviously ignores some important things, for example the momentum of electrons. But that’s small compared to the momentum of the nuclei and plausibly doesn’t matter.

Is this approximation good? Is it commonly used?

The Born-Oppenheimer approximation is very used in ab initio molecular dynamics (AIMD). When the Born-Oppenheimer approximation is used, it is more specifically called Born-Oppenheimer molecular dynamics (BOMD). Various software packages have implemented this method, here is a list of a few of them:

CP2K
CPMD
Gaussian
Orca

Other types of AIMD is also used, including Car-Parrinello molecular dymanics (CPMD) and Ehrenfest molecular dynamics.

As always it is important to remember that the Born-Oppenheimer approximation is only a good approximation if the relaxation time of the electronic structure is much faster than that of the movement of the nucleus.


Resultant temperature of mixture **Question: ** There are 2 flasks of equal volume, connected by a narrow tube of negligible volume, filled with nitrogen gas. When both are immersed in boiling water, the gas pressure inside the system is 0.5 atm. Calculate the pressure of the system when one of the flasks is immersed in ice water mixture, keeping the other in boiling water.

Problem: There are two ways that I am going to describe that I can think of to solve the question, but they give different answers.

Way 1: Net Temperature of the system should be 50 degree Celsius because there is no heat loss and there is symmetry. As both the flasks are identical and at 100 and 0 so it should be 50.
Calculating the answer with this assumption and the formula
P1/V1=P2/V2
The answer comes out to be about 0.432 atm

Way 2: The sum of the moles of gas in both the flasks would remain same.
And using the ideal gas equation to get n = number of moles = PV/RT
After equating the values. The pressure come out to be about 0.422 atm

I was bothered with the 0.1 change. So I took the liberty to calculate the temperature from way 2, and it came out to be about 41.82 degree Celsius.

Note : calculations were done in Kelvin itself.

So we assumed that temperature would be 50 but way 2 contradicts that. I could not find a fault in either of the ways so could not understand what temperature would it really be.

Attempt to solve: is it something about Gibbs Paradox? I was Googleing and that came, but I couldn't understand...

Please help with proper explanation on why the other way is wrong.

In the first condition when both flasks are boiling, the temperature of the gases in both flasks are the same (100 °C).  Using the ideal gas law on the system that comprises both flasks, where $n$ is the number of moles initially in each flask and $V$ is the volume of one flask.

$$ P_{init} = \frac{(2 n) R\;T_{boil}}{2 V} = \mathrm{0.5~atm}$$

$$ 2n = \frac{P_{init}(2 V)}{R\;T_{boil}} $$

Now if Flask #2 is immersed in ice water, we should assume that the pressure in the system will be constant but that the flasks will have different temperature.  For the pressure to be constant at different temperatures, material will have to flow from the hot flask to the cold flask; now, $n_1$ and $n_2$, the moles of gas in flask #1 and flask #2, will not be the same, but since no gas leaves the system comprising both flasks, then we know $n_1 + n_2 = 2n$

$$ n_1 = \frac{P_{final}V}{R\;T_{boil}} $$
$$ n_2 = \frac{P_{final}V}{R\;T_{ice}} $$
$$ n_1 + n_2 = 2n$$

Substituting:

$$ \frac{P_{final}V}{R\;T_{boil}} + \frac{P_{final}V}{R\;T_{ice}} = \frac{P_{init}(2 V)}{R\;T_{boil}}$$

$$ P_{final}\frac{V}{R}\left(\frac{1}{T_{boil}} + \frac{1}{T_{ice}} \right)= 2 \frac{P_{init}V}{R\;T_{boil}}$$

$$ P_{final}\left(\frac{1}{T_{boil}} + \frac{1}{T_{ice}} \right)= 2 \frac{P_{init}}{T_{boil}}$$

$$ P_{final} = 2 P_{init}\frac{\frac{1}{T_{boil}}}{\left(\frac{1}{T_{boil}} + \frac{1}{T_{ice}} \right)} = 2 P_{init} \frac{\frac{1}{T_{boil}}}{\frac{T_{ice} + T_{boil}}{T_{ice}{T_{boil}}}}$$

$$ P_{final} = 2 P_{init} \frac{T_{ice}}{T_{ice} + T_{boil}}  $$

This answer makes sense in that (i) if the temperature $T_{ice}$ is actually the same as $T_{boil}$, then the final pressure would be the same as the initial pressure, and (ii) the answer doesn't depend on the volume or moles of gas in the flasks, just the temperatures and the initial pressure.

  Way 1: Net Temperature of the system should be 50 degree Celsius because there is no heat loss and there is symmetry.

This way is incorrect because a flask immersed in boiling water ($T=$100°C) is unlikely to have a temperature of 50 °C.  Similarly for a flask immersed in ice.

  Way 2: The sum of the moles of gas in both the flasks would remain same.

This is the correct method.  It does require assuming that the rate of heat transfer from the boiling / ice baths to the flasks is much much greater than the rate of heat transfer between flasks via diffusion through the tube.  However since the tube is small, this assumption is probably OK.


Fock operator invariance under unitary transformation I know how to show that the Coulomb operator of the Fock operator is invariant under a unitary transformation of the orbitals, as on page 121 of Szabo and Ostlund, but the indices in my proof for the exchange operator are just not working.  What might I be missing?
I have

\begin{align}
\sum_i\hat{K}_i'(1) &=\sum_i\int dr_2\chi_i'^*(2)\chi_j'(2)/r_{12} \\
 &=\sum_i\int dr_2\sum_kU_{ki}^*\chi_k^*(2)\sum_lU_{lj}\chi_l(2)/r_{12}\\
 &=\sum_{ikl}\int dr_2U_{ki}^*U_{lj}\chi_k^*(2)\chi_l(2)/r_{12}\\
 &=\sum_{i}\int dr_2\chi_i^*(2)\chi_i(2)/r_{12}\\
 &\neq\sum_{i}\int dr_2\chi_i^*(2)\chi_j(2)/r_{12}.
\end{align}

The exchange operator is not a operator by itself, it is only defined with the orbital it is working on:
$%
\newcommand{\ll}{\left\langle}\newcommand{\rr}{\right\rangle}
\newcommand{\lb}{\left|}\newcommand{\rb}{\right|}
\newcommand{\op}[1]{\mathbf{#1}}$
$$\begin{align}
  && \op{F}_i &= \op{H}^\mathrm{c} + \sum_j (\op{J}_j - \op{K}_j),\\
\text{with}&&
     \op{J}_j\lb \phi_i\rr &=
       \ll \phi_j(\op{x}_1) \rb r_{12}^{-1}
       \lb \phi_j(\op{x}_1) \rr \lb \phi_i(\op{x}_2) \rr,\\
\text{and}&&
     \op{K}_j\lb \phi_i\rr &=
       \ll \phi_j(\op{x}_1) \rb r_{12}^{-1}
       \lb \phi_i(\op{x}_1) \rr \lb \phi_j(\op{x}_2) \rr.
\end{align}$$

While in the Coulomb operator case the $\lb \phi_i\rr$ doesn't do anything, so you can take it through the proof as a constant:*
\begin{align}
  \sum_j\op{J}_j\lb \phi_i\rr
  &= \sum_j
     \ll \phi_j(\op{x}_1) \rb r_{12}^{-1}
     \lb \phi_j(\op{x}_1) \rr \lb \phi_i(\op{x}_2) \rr\\
  &= \sum_k \sum_l \sum_j U_{kj}^*U_{lj}
     \ll \phi_k(\op{x}_1) \rb r_{12}^{-1}
     \lb \phi_l(\op{x}_1) \rr \lb \phi_i(\op{x}_2) \rr\\
  &= \sum_k \sum_l \delta_{kl}
     \ll \phi_k(\op{x}_1) \rb r_{12}^{-1}
     \lb \phi_l(\op{x}_1) \rr \lb \phi_i(\op{x}_2) \rr\\
  &= \sum_k
     \ll \phi_k(\op{x}_1) \rb r_{12}^{-1}
     \lb \phi_k(\op{x}_1) \rr \lb \phi_i(\op{x}_2) \rr\\
  &= \sum_k \op{J}_k \lb \phi_i\rr
\end{align}

That is not the case in the Exchange operator, as it switches the orbitals:
\begin{align}
  \sum_j\op{K}_j\lb \phi_i\rr
  &= \sum_j
     \ll \phi_j(\op{x}_1) \rb r_{12}^{-1}
     \lb \phi_i(\op{x}_1) \rr \lb \phi_j(\op{x}_2) \rr\\
  &= \sum_k \sum_l \sum_j U_{kj}^*U_{lj}
     \ll \phi_k(\op{x}_1) \rb r_{12}^{-1}
     \lb \phi_i(\op{x}_1) \rr \lb \phi_l(\op{x}_2) \rr\\
  &= \sum_k \sum_l \delta_{kl}
     \ll \phi_k(\op{x}_1) \rb r_{12}^{-1}
     \lb \phi_i(\op{x}_1) \rr \lb \phi_l(\op{x}_2) \rr\\
  &= \sum_k
     \ll \phi_k(\op{x}_1) \rb r_{12}^{-1}
     \lb \phi_i(\op{x}_1) \rr \lb \phi_k(\op{x}_2) \rr\\
  &= \sum_k \op{K}_k \lb \phi_i\rr
\end{align}

* Let
$$
\mathbb{U}=\left(\begin{matrix}
  U_{11} & U_{12} & \dots \\
  U_{21} & U_{22} & \dots \\
  \vdots & \vdots & \ddots\\
  \end{matrix}\right);
\mathbb{U}^\dagger\mathbb{U}=\mathbb{E}
\implies U_{ij}^* U_{kl}= \delta_{ij} = \begin{cases}
   1; & i = j = k = l \\ 0; & i \neq j \dots\\ \end{cases}
.$$


Difference between Superheated steam and Superheated water? I'm working on a project (school level) and in it I'll be dealing with the pressure generated by water when superheated.

I could understand the concept of superheated water but superheated steam is hard to understand. Are both inter-related, i mean that when water is superheated then it changes into superheated steam?

Or is it something different? And since in my project I require high-pressure-energy to be generated from water, so which one will be better, superheated steam or superheated water?

Boilers and similar steam generators usually produce wet steam, which is a mixture of saturated steam and liquid water droplets. Saturated steam is steam that is in equilibrium with liquid water. The temperature of the steam is equal to the temperature of the liquid water; it corresponds to the boiling point at the given pressure. At normal pressure ($p=101\,325\ \mathrm{Pa}=1\ \mathrm{atm}$), the temperature is $T=99.974\ \mathrm{^\circ C}$. The corresponding specific enthalpy of the steam is $h = 2676\ \mathrm{kJ/kg}$; the specific enthalpy of the liquid water is $h = 419\ \mathrm{kJ/kg}$.

If heat is supplied to the wet steam, liquid water vaporizes. At constant pressure, the temperature remains constant until all liquid water has been vaporized, i.e. until the steam is dry.

If further heat is supplied to the dry steam, the temperature and the specific enthalpy of the steam increase. Such steam at a temperature higher than its boiling point is called superheated steam. For example, superheated steam at normal pressure ($p=101\,325\ \mathrm{Pa}=1\ \mathrm{atm}$) and a temperature of $T=150\ \mathrm{^\circ C}$ has a specific enthalpy of $h = 2777\ \mathrm{kJ/kg}$.

Superheated water is liquid water under increased pressure at temperatures above the boiling point at normal pressure. For example, water at a temperature of $T=150\ \mathrm{^\circ C}$ and a pressure of $p=506\,625\ \mathrm{Pa}=5\ \mathrm{atm}$ is liquid. The specific enthalpy of the liquid water is $h = 632\ \mathrm{kJ/kg}$.

The energy content of superheated steam (e.g. $h = 2777\ \mathrm{kJ/kg}$ at $T=150\ \mathrm{^\circ C}$ and $p=101\,325\ \mathrm{Pa}=1\ \mathrm{atm}$) is much higher than the energy content of superheated water (e.g. $h = 632\ \mathrm{kJ/kg}$ at $T=150\ \mathrm{^\circ C}$ and $p=506\,625\ \mathrm{Pa}=5\ \mathrm{atm}$) because of the latent heat, i.e. the enthalpy of vaporization $\Delta H_\text{vap}$.


The proper explanation for the general formula of the DU? I found a formula for the degree of unsaturation (DU) in Clayden’s Organic Chemistry. It explains, simply, the unsaturation with the difference in H atoms. Then I stumbled upon the general formula:

$$DU=1+\frac12\sum n_i\left(v_i-2\right)$$

$n_i$ is the number of atoms with valence $v_i$.

Can someone explain from where this equation was derived? And how come the valence term was introduced?

My answer is based on V. Pellegrin, J. Chem. Educ. 1983, 60 (8), 626 which might be inaccessible for you by a pay wall.

Starting with Eulers characteristic for planar graphs wich states that adding the faces and vertices of a planar graph and substracting all the edges must be 1 $$F+V-E=1$$ which can be rewritten as $$F=1+E-V$$
Now in more chemical words we can redefine this by saying that edges are bonds and vertices are atoms and the $F$ will be our $DU$ from now on. If we know the structural formula we could stop now, count all atoms and bonds and calculate what we want to know. Those are two examples from the paper:

But what if all we have is a sum formula from let's say mass spectrometry?

Then we have to calculate how many bonds this molecule will have based on the knowledge of how many atoms the molecule contains and how many valence electrons each atom has.

Each molecule consists of $$N=\sum_i n_i$$ atoms of element $i$ whereby $n_i$ denotes the amount of each element. Each atom has to be assumed in it's lowest valency state and has $v_i$ valence electrons which would make in total $$\sum_{i=1}^N n_i~v_i$$ valence electrons. As every single bond contains two electrons, the amount of all possible bonds is half the amount of all valence electrons, namely $$\sum_{i=1}^N \frac{n_i~v_i}{2}$$

Let's try it with the mass formula $\ce{C9H7NO}$:

The molecule consists of $N=18$ atoms.
The atoms are grouped in $4$ groups containing the elements $i=\{C,H,N,O\}$
Each element is contained $n=\{9,7,1,1\}$ times.
The lowest valency of each contained atom is $v=\{4,1,3,2\}$
The sum of all valence electrons is $\sum_{i=1}^N n_i v_i=9\cdot4+7\cdot1+1\cdot3+1\cdot2=48$
The sum of all possible bonds is $\sum_{i=1}^N \frac{n_i~v_i}{2}=48/2=24$

Now let's insert those numbers into the equation $$DU=1+E-V=1+24-18=7$$ Because I knew the moleule that I have chosen we can compare

and see that the calculation was correct ($5$ for the double bonds + $2$ for the ring closures).

As it was correct, we can combine it and derive your equation.
$$\begin{align}
F=&~1+E-V\\
DU=&~1+\text{number of bonds}-\text{number of atoms}\\
DU=&~1+\sum_{i=1}^N \frac{n_i~v_i}{2}-\sum_{i=1}^N n_i\\
DU=&~1+\sum_{i=1}^N \left(\frac{n_i~v_i}{2}-n_i\right)\\
DU=&~1+\frac{1}{2}\sum_{i=1}^N (n_i~v_i-2~n_i)\\
DU=&~1+\frac{1}{2}\sum_{i=1}^N n_i~(v_i-2)\\
\end{align}$$


How can I permanently write on HDPE plastic? So I just spent the better part of an hour graduating a food grade HDPE 5 gallon bucket with permanent marker so I can easily fill it with water of a certain amount for homebrewing.

I'm all done. The only problem is that permanent marker wipes off of HDPE quite easily. I guess I should have tested that first...

Anyways... what CAN I use to write on HDPE plastic? Or will I be forced to etch the markings in with a nail or something?

I tried a Google search for "How to write on HDPE plastic?" and came up empty.

From experience, one good way to permanently mark on HDPE is a paint marker. The following is just the first one I found online; no endorsement intended:

   Promotional image from uline.com

They tend to be stinky and a tad messy, but they usually leave clean, visible lines and are basically indelible, requiring a lot of scrubbing with alcohol, acetone, etc. to remove. No idea how well it'll survive power washing, though.


Are there substances people can put on themselves to prevent catching fire? This can be very useful for people who are afraid of spontaneous human combustion! But seriously, I heard there are non-harmful chemicals used in some fire extinguishers, so can they maybe be put on our skin, clothes, etc. in advance to prevent ever catching fire?

It depends how long one is exposed to heat, and how hot the source is, but yes, there are "fire retardants" for skin:

A mix of alcohol (ethanol or isopropanol) and water or butane in soap sud can be set afire on your palm without burning you, if done very carefully. The water soaks up heat energy both because of its high specific heat and through evaporation.
Insulation applied to the skin or between skin and flame can protect (momentarily) from a torch.
A wet hand is temporarily protected from molten lead by the leidenfrost effect. As water on the skin boils, it acts, briefly, barrier to the heat.

NB: Do not try these experiments yourself (or on others). You get no points for being a crispy critter.


How to get the first step of the derivation of the Maxwell-Boltzmann distribution To derive the Maxwell-Boltzmann distribution the first step requires the knowledge that the equation will take the form $$f(v)=Ke^{-\epsilon/ kT}$$ where $\epsilon$ is the sum of the kinetic energies in the directions of the x,y and z axes. I understand that there needs to be exponential decay and why there will need to a constant. Also, I understand that the eponent needs to be dimensionless but why is KT used? This seems to pop up everywhere without much of an explaination of why (as in this case).  Can you help me understand how to get this first step? Thanks.

KT comes in from requiring agreement with the ideal gas law.

See "section 1.5.3 $\langle v^2\rangle$ should agree with the ideal gas law" and section 1.3 "The functional form of the dependence of T on $\langle\varepsilon\rangle$ cannot be determined solely from kinetic theory, since the temperature scale may be chosen in many possible ways. In fact, one way to define the temperature is through the ideal gas law: T = pV/(nR)" in Kinetic Theory of Gasses


Can a stoichiometric mixture of oxygen and methane exist as a liquid at standard pressure and some (low) temperature? This answer to the question Pre-mixing cryogenic fuels and using only one fuel tank written by a non-chemist (me) begins with:

At STP:

LOX's boiling point is 90.19 K
Methane's freezing point is  90.7 K

This does not a priori prove that a solution of the two can not exist. However it does mean that they can not be handled as liquids at the same temperature, making mixing the two more difficult.
We know that liquid air exists which shows that LOX and LN2 can mix together. But methane is an organic molecules and we know that heavier $\text{C}_n \text{H}_{2n+2}$ hydrocarbons include oils and waxes don't like to dissolve in non-organic solvents.

A stoichiometric mixture of oxygen and methane would be 2:1 molar:
$$\ce{ 2O2 + CH4 -> CO2 + 2H2O }$$
Though the two can not be conveniently maintained as liquids at the same temperature, can a stoichiometric mixture of the two exist as a liquid at some (low) temperature and standard pressure?

There's a NASA report that looks into this: "ON THE SOLUBILITIES AND RATES OF SOLUTION OF GASES IN LIQUID METHANE", Hibbard and Evans, 1968 and concludes that such mixtures are possible.

Starting on page 8:

  Figure 5(a) presents the curves for oxygen, argon, carbon monoxide,
  and nitrogen. Also shown are the two experimental values for nitrogen.
  Agreement is excellent at 99.83K and good at 110.9K. The curves for
  these gases show that solubility should decrease with increasing
  temperature and the nitrogen data confirm this. This figure shows the
  mole fraction solubility of oxygen to be 1.0 at 90K. This means that
  oxygen, which has a normal boiling temperature of 90.1K would
  continuously condense in, and be miscible in all proportions, with
  liquid methane at 90K. This is confirmed by reference 11 where, in a
  study of the solubility of methane in liquid oxygen, it was concluded
  that these formed a near-ideal solution at -297 F (90K)

(emphasis added) Reference 11 mentioned in there is "Hydrocarbon-Oxygen Systems Solubility", McKinley and Wang, 1960 (unfortunately paywalled) which also has interesting discussion of the stability (i.e. presence or absence of a tendency to explode) of various mixtures.

Figure 5 is reproduced below. Note how the solubility of oxygen rises rapidly as temperature drops.


Is silicon dioxide more acidic than the oxides of nitrogen? Is silicon dioxide more acidic than the oxides of nitrogen? How to find acidic strength of compounds which don't have hydrogen atoms?

I think your question boils down to:

What is the meaning of acidic strength of oxides? How do we determine the acidic strength of oxides?

In general, the electropositive character of the oxide's central atom will determine whether the oxide is acidic or basic. The more electropositive the central atom, the more basic its oxide. The more electronegative the central atom, the more acidic its oxide.

Electropositive character decreases from left to right across the periodic table and increases down a group.
The trend of acid-base behaviour is from strongly basic oxides on the left-hand side to strongly acidic ones on the right, via an amphoteric oxide (for example, aluminium oxide) in the middle.

This trend applies only to the oxides of the individual elements  in the highest oxidation states for those elements. The pattern is somewhat less clear for other oxides.

A non-metal oxide's acidity is defined in terms of the acidic solutions formed as a result of reaction(s) with water. For example, sulfur trioxide reacts with water to forms sulfuric acid.

In general, acidic oxides are oxides of nonmetals and basic oxides are oxides of nonmetals.

Note: There are three nonmetal oxides from the upper right portion of the periodic table, $\ce{CO}$, $\ce{NO}$, and $\ce{N2O}$, which have low (not the highest possible) oxidation numbers for the central atom that they give neutral aqueous solutions.

Since the acidity of a cation rises rapidly with charge, d-block elements which exhibit a wide variety of oxidation numbers may have one or more oxides that exhibit only basic properties and one or more oxides that exhibit only acidic properties.

In general, the higher the oxidation number, the more acidic the corresponding oxide (similarly, reverse trend for basic character of oxides). Chromium makes a decent example here, as it exhibits a variety of oxidation states. $\ce{CrO}$ is basic, $\ce{Cr2O3}$ is amphoteric and $\ce{CrO3}$ is acidic.

P.S.
Let me introduce you to a theory developed by Hermann Lux and Håkon Flood, named Lux-Flood theory to explain the acidic or basic character of oxides. Here's what they had to say:

  An acid of Lux-Flood is an acceptor of $\ce{O^2-}$
  A base of Lux-Flood is a donor of $\ce{O^2-}$

For example:

$\ce{CaO}$ is a base of Lux-Flood because $\ce{CaO}=\ce{Ca^2+}+\ce{O^2-}$

$\ce{SiO2}$ is an acid of Lux Flood because $\ce{Si}$ has vacant d orbitals, allowing it to accept more $\ce{O^2-}$ ions.

The two compounds may react to give you $\ce{CaSiO3}$.


Is there a compound that contains water as a subcomponent? Is there a molecule that contains water chemically bonded to another molecule/element? Could someone provide an example or two?

Off the top of my head: Coordination complexes that have water ligated to a (transition) metal atom.

See corresponding Wikipedia article
A familiar example would be (hydrated) copper sulfate $\ce{CuSO4.5H2O}$, which has 4 water molecules ligated to the copper atom. (The fifth water molecule mentioned in the chemical formula is associated copper sulfate via hydrogen bonds)

 The picture is sources from the Wikipedia article for copper sulfate, and the diagram is sourced from this Quora post

If you're new to the idea of complex ions, this chemguide article offers a good intro to the topic.


Spectra from Dipole Moments If there is a change in dipole moments such as absorption spectroscopy,  there are infrared absorptions (and/or transmittance) corresponding to them. Is the reason change in dipole moments are in infrared frequency simply because of the size of the molecules or scale of wavelength?   Or are the dipole moments producing particular frequencies simply because of their sizes just like in antenna? If not, why are they mostly in IR?

(Note: image from https://en.wikipedia.org/wiki/Electromagnetic_spectrum)

The transition energy, or wavelength if you prefer, are governed by the nature of the molecules involved. Whole body free rotational motion (in the gas phase) gives rise to microwave spectra ( fractions of wavenumbers cm$^{-1}$  or THz frequencies). The frequency is determined by the molecules moment of inertia. At higher energy comes vibrational motion, stretching and bending of chemical bonds, a few hundred to a few thousand cm$^{-1}$. These bond frequencies are approximately given by $\sqrt{k/\mu}$ where $k$ is force constant which relates force to bond extension as in Hook's law, and $\mu$ is the reduced mass. There are 3N-6 types of vibration for N atoms, (3N-5 for linear molecules). At higher frequencies are electronic transitions where and electron gets promoted from a bonding to anti-bonding orbital. Typically these transitions are in the visible and uv regions.

The exact number of transitions is difficult to estimate because each vibrational level has rotational levels superimposed on them and there are also vibrations that are combinations of levels. There are, however, fewer electronic transitions because there are generally few molecular orbitals than types of vibrational modes, and each excited state has its own set of vibrational and rotational levels.

One of a chemist's essential tools, arguably more important that any other spectroscopy, is NMR spectroscopy which is caused by transitions between nuclear spin states when in a magnetic field and occurs at very low frequency, 100-800 MHz range.


How are the atomic orbitals for multi electron atoms obtained? The problem with Schrödinger's equation is that it isn't exactly solvable for multi-electron species. In the past, atomic orbitals were used to construct a solution for molecules (LCAO). My question is, how were the multi electron atomic orbitals (the ones used for the LCAO) discovered? . Were the atomic orbitals solutions for a one electron system? In other words, were the atomic orbitals calculated the orbital for a $\ce{He+}$, $\ce{Li^2+}$, $\ce{Be^3+}$, etc? What were some approximations that are used to obtain the solution for multi-electron species?

Already in year 1928 Hartree proposed an approximate method of solving the Schrödinger equation for a many-electron atom that became known as the Hartree method (1, 2, 3). The Hartree method is based on representing the many-electron wave function as a product of one-electron ones with the subsequent application of the variational principle. Two years later, Slater (4) and Fock (5) independently corrected the Hartree method for that did not respect the antisymmetry of a many-electron wave function by representing the wave function by a Slater determinant, rather that a simple product of orbitals. The resulting physically more accurate method became know as the Hartree-Fock method.

In the Hartree–Fock method one solve a set of one-electron equations, called the Hartree–Fock equations, of the form
$$
  \newcommand{\op}{\hat}
  \newcommand{\core}{^{\mathrm{core}}}
  \op{F} \psi_i(\vec{q}_{1}) = \varepsilon_i \psi_i(\vec{q}_{1}) \, ,
$$
where $\op{F} = \op{H}\core + \sum\nolimits_{j=1}^{n} \big(\op{J}_{j} - \op{K}_{j} \big)$ is the Fock operator and $\psi_i$ are the spin-orbitals of an $n$-electron system. It is customary to suppose that spin orbitals come in pairs: for each pair of electrons two spin orbitals corresponding to two different pure spin states are constructed out of the same spatial orbital,
$$
    \psi_{2i-1}(\vec{q}_{1}) = \phi_{i}(\vec{r}_{1}) \alpha(m_{s1}) \, ,
    \quad
    \psi_{2i}(\vec{q}_{1})   = \phi_{i}(\vec{r}_{1}) \beta(m_{s1}) \, .
$$
Substitution of such spin orbitals into the Hartree-Fock equations results in a similarly looking system of equations for the corresponding spatial orbitals,
$$
    \op{F} \phi_{i}(\vec{r}_{1}) = \epsilon_{i} \phi_{i}(\vec{r}_{1}) \, ,
    \quad
    i = 1, 2, \dotsc, n/2 \, ,
$$
although, the expression for the Fock operator is different $\op{F} = \op{H}\core + \sum_{j=1}^{n/2} (2 \op{J}_{j} - \op{K}_{j})$. Note here, that for the case of the original Hartree method there was no exchange terms $\op{K}_{j}$.

So, essentially, atomic orbitals of many-electron atoms were obtained by solving the Hartree and Hartree-Fock equations "by hands". Of course, the spherical symmetry of atomic systems greatly simplified the problem, as it can be seen already in the Hartree's original work (1), plus some additional simplification, such as the central field approximation, were often used. But still, in general, a numerical integration was required and it was usually done on some calculating machines (6). Below is the photo (courtesy of AIP) of Douglas Hartree (left) and Arthur Porter (right) viewing one such machine, the meccano differential analyzer.

And the results of numerical integrations were simply tabulated as the values of the radial part of an orbital for different values of distance from the nucleus (7).

1) Hartree, D. R. The Wave Mechanics of an Atom with a Non-Coulomb Central Field. Part I. Theory and Methods. Math. Proc. Cambridge Philos. Soc. 1928, 24 (1), 89–110. DOI: 10.1017/S0305004100011919.

2) Hartree, D. R. The Wave Mechanics of an Atom with a Non-Coulomb Central Field. Part II. Some Results and Discussion. Math. Proc. Cambridge Philos. Soc. 1928, 24 (1), 111–132. DOI: 10.1017/S0305004100011920.

3) Hartree, D. R. The Wave Mechanics of an Atom with a non-Coulomb Central Field. Part III. Term Values and Intensities in Series in Optical Spectra. Mathematical Proceedings of the Cambridge Philosophical Society, 1928, 24 (3), 426–437. DOI: 10.1017/S0305004100015954.

4) Slater, J. C. The Theory of Complex Spectra. Phys. Rev. 1929, 34, 1293. DOI: 10.1103/PhysRev.34.1293.

5) Fock V. Näherungsmethode zur Lösung des quantenmechanischen Mehrkörperproblems. Z. Physik. 1930, 61 (1), 126–148. DOI: 10.1007/BF01340294.

6) Hartree, D. R. The Differential Analyser. Nature. 1935, 135, 940-943. DOI: 10.1038/135940a0.

7) Hartree, D. R. and Hartree, W. Results of Calculations of Atomic Wave Functions. III. Results for Be, Ca and Hg. Proc. R. Soc. London, Ser. A. 1935, 149 (867), 210-231. DOI: 10.1098/rspa.1935.0058.


What happens when HCl is added to acetylene in presence of mercuric ions? Normal addition of Hydrochloric Acid to acetylene would give me a 1,1 Dichloro product.
But I read that in presence of mercuric ions only one molecule of HCl is added to give Vinyl Chloride.
Why does this happen?

I'm not entirely sure but I'll have a go.

As a strong acid, HCl will be dissociated into Chloride and protons in water. I'd suggest that the mercury will first associate with the acetylene forming an Hg- Pi complex which will weaken the triple bond. Then the chloride ion will attack one side of the acetylene and the electron density one of the triple bonds will form a new single bond, first to the mercury ion and then this bond will hydrolyse and pick up a proton. This is similar to an oxymercuration reaction but with an attack from chloride rather than water.

The resulting vinyl chloride is more electron poor than the starting species so should should undergo further 'chloromercuration?' slower than the starting species. I would suggest that the selectivity is then derived from the milder conditions this reaction must be run at. Probably you have to force acetylene to react with HCl without a catalyst by heating the hell out of it (if it happens at all) therefore the addition happens twice.


Producing hydrogen sulfide at home - how dangerous is it? When I was a kid, I used to go to the pharmacy to buy sulfur and iron powder, then mix it together and heat it until it starts to react into iron sulfide. Put a chunk of iron sulfide into some vinegar, and you get a room smelling badly of rotten eggs. This experiment was also described in some "chemistry for kids" boxes from the 80's.

Now I'd like to do this experiment with my son, but in the meantime I learned that hydrogen sulfide is a very potent poison, even more poisonous than hydrocyanic acid. For this reason, I'd like to set up some precautions.

Of course I won't do it in a closed room, and I won't let my son breathe in much of the gas.

The gas can be smelled long before it starts to get dangerous, so I think it should be pretty easy to use an amount of sulfur and iron that will cause the characteristic smell, but at the same time is guaranteed to not cause any danger. So, I'm looking for some quantitative estimation:

How much iron sulfide would be enough to create a dangerous amount of hydrogen sulfide? To make the calculation easier, let's assume we grind the iron sulfide and use a stronger acid than vinegar, so that all of the gas will be created almost instantly.

If possible, it would be great to have two estimations: one of the beginning of the "danger zone", and one of the beginning of the "lethal zone". According to Wikipedia, the gas starts to smell at $\pu{0.00047ppm}$, the danger zone starts at about $\pu{10 ppm}$, and at about $\pu{320 ppm}$ it can be lethal.

On the question as to how dangerous is H2S let me detail some facts that I find problematic. Here is a good paper.

I summarize some of my findings:

First, H2S after initial exposure can deaden your sense of smell (10-minute exposure at 50-100 ppm impairs the sense of smell). So, one may not really have an accurate conception of the magnitude of one's time exposure to the gas.

Second, skin exposure may be a path to ingestion, especially if moist I suspect. Note, some sources actually disagreed on skin exposure poisoning of Hydrogen sulfide with one study alluring to skin toxicosis from high concentrations over a long period.

Third, following a significant exposure event, the very fact that you are still alive should not be comforting. People having incurred a lethal dose may be a de facto walking dead with a slow hourly progression to death.

Fourth, the ppm considered fatal for H2S is not that far removed from some known very toxic gases (like HCN). However, my opening reference doubts if one can even assign accurate figures on ingestion via the lungs. Here is an extract addressing dosing level ranges:

  Findings in the canary, cat, dog, goat, guinea pig, rabbit and rat (2) are consistent as to the effects of hydrogen sulfide: at 150-225 mg/m3 signs of local irritation of eyes and throat after many hours of exposure; at 300-400 mg/m3 eye and mucous membrane irritation in 1 hour and slight general effects with longer exposure; at 750-1000 mg/m3 slight systemic symptoms in less than 1 hour and possible death after several hours; at 1350 mg/m3 grave systemic effects within 30 minutes and death in less than 1 hour; at 2250 mg/m3 collapse and death within 15-30 minutes, and at 2700 mg/m3 immediate collapse, respiratory paralysis and death.

Viewing all these together, if one has a choice to avoid working or experimenting with the gas, I would do so.


Why is the dipole moment of ethyl chloride larger than that of vinyl chloride? The dipole moment of ethyl chloride is 2.10 D (J. Mol. Struct. 1990, 216, 9–26), and that of vinyl chloride is 1.42 D (J. Chem. Phys. 1960, 32 (1), 205–209).

Why is this the case? I would have thought that hyperconjugation would lead to vinyl chloride possessing a larger dipole moment instead.

I see two explanations for why the dipole in vinyl chloride is smaller than ethyl chloride.

Consider a resonance structure where the chlorine donates electrons to the alkene. This puts a formal positive charge on chlorine:

This would be a very minor contributor given the electronegativities and the mismatch in orbital size. However, this is not possible in ethyl chloride at all.
The more likely explanation is that an sp2-hybridized carbon is more electronegative than an sp3-hybridized carbon. Since the electronegativity of chlorine will be closer to the electronegativity of the carbons in vinyl chloride than the carbons of ethyl chloride, the C–Cl bond will be less polarized in vinyl chloride, giving a smaller dipole.


Computing pH of a buffer, and concentrations You pour $50~\mathrm{mL}$ of a $0.0200~\mathrm{M}$ $\ce{HCOOH}$-solution to $150~\mathrm{mL}$ of a $0.0500~\mathrm{M}$ $\ce{HCOONa}$-solution. The $K_\mathrm{a}$ of $\ce{HCOOH}$ is $1.8 \times 10^{-4}$ and $\mathrm{p}K_\mathrm{a}$ is $3.74$.

a) Compute the concentrations of $\ce{HCOOH}$ and its salt after dilution to this $200~\mathrm{mL}$ buffer solution.

b) Calculate the $\mathrm{pH}$ of this buffer solution with the buffer formula.

Attempt at solution: Not sure how to do part a). For part b) I used the following method. We have $50 \times 10^{-3}~\mathrm{L}  \cdot \frac{0.0200~\mathrm{mol}}{1~\mathrm{L}} = 0.001~\mathrm{mol}$ of $\ce{HCOOH}$. Doing this again for $\ce{HCOONa}$ gives $0.0075~\mathrm{mol}$. Hence: \begin{align*} \mathrm{pH} = 3.74 + \log(\frac{0.0075}{0.001}) = 4.61 \end{align*}

But how can I solve a)? I'm having trouble with writing the reaction down. I'm not sure what's happening. Which chemical is neutralizing which? And what are the ions?

We have to compute the new concentrations after dilution:
$$\ce{[HCOOH]}=\frac{0.02\times 50}{200}=0.005\,\, \mathrm{mol/L}$$
$$\ce{[HCOO^- ]}=\frac{0.05\times 150}{200}=0.0375\,\, \mathrm{mol/L}$$

Now, we can compute the $\mathrm{pH}$ of the buffer solution:$$\mathrm{pH}=\mathrm{p}K_a +\log \frac{\ce{[HCOO^- ]}}{\ce{[HCOOH]}}$$

$$\mathrm{pH}=3.74+\log \frac{0.0375}{0.005}=4.62$$


Why does the isoelectric range decrease as an Alanine chain grows longer? The pK1 and 2 values for alanine are 2.3 and 9.7. In the dipeptide Ala-Ala, they are 3.1 and 8.3 and in tri-peptide Ala-Ala-Ala, they are 3.3 and 8.0. Why does this trend in pKa values exist? Why does the isoelectric range decrease?
The reason I can think of is the extra Alanine dumps electrons towards the C terminal and so makes the O-H bond less polar and thus less acidic while it also pulls more electrons from the N terminus to make that more polar and thus more acidic.
Thanks!

It is because of the inter- and intramolecular associations that occur.

In Ala, the first loss of a proton forms a two point intermolecular association which is very favourable.
Since the equilibrium tends to the right, more $\ce{H^+}$ are liberated and the p$K_a$ is lower.

For the second step, this stable dimeric association must be destroyed, so now the equilibrium tends to the left and the p$K_a$ is higher then it would be without this association.

In the case of Ala-Ala-Ala, this type of dimeric association can not occur because the distance is too long.
There is intermolecular association but with different molecules, not in pairs.
Here the first p$K_a$ is higher than in the case of Ala and the second lower because the equilibrium is not forced so much to the associated forms.

Finally, in the case of Ala-Ala, besides the intermolecular association between different molecules like in Ala-Ala-Ala,
there is also an intramolecular association.
This one is less stable than the association in pairs of the individual amino acids, but more than the random intermolecular ones.
Therefore, the two p$K_a$  are in between of the ones for Ala and Ala-Ala-Ala.


What substance consumes oxygen from atmospheric air I'd like to find an economical substance that will consume oxygen from a container. Steel wool comes to mind but I don't have an intuition how feasible it would be.

The chamber is 5 gallons in size and I'd like the oxygen to be consumed in 48 hours or less.

As mcole's comment suggests, it may be easier to do this by physical rather than chemical means.

There are two common approaches:

  Purge

In the purge method, the vessel is filled from a high pressure source of another gas. The air needs some way to escape. If the new gas is more dense, all the better, but even repetitive purges of less dense gases will eventually push all the air out.

  Pump and backfill

In this method, the vessel is connected to a vacuum pump, and the air inside is pumped out. The vessel is then refilled with an inert gas. Depending on your needs, you can repeat this process several times.


Abundance of sulfur containing compounds in the air I noticed a piece of jewelry has tarnished, so I looked up information about silver tarnishing. I read that this is due to the formation of silver sulfide on the surface of my necklace and that the sulfur is pulled from the air.

My question is, how abundant are sulfur containing compounds in the air? I know sulfur is by no means rare, but I guess I'm just confused as to why pretty much everything silver ends up tarnishing -- a box of silverware unopened for years can be tarnished.

How abundant are sulfur containing compounds in the air? Is the tendency of silverware to tarnish often/easily due to reactivity? If so, why doesn't carbon react to silver in the same manner and thus give way to a carbon-based tarnish?

The tarnishing of silver by sulfur is primarily due to hydrogen sulfide. Hydrogen sulfide is found throughout the Earth's lower atmosphere, and has many sources including the human body$\pu{^1}$.

The amount of hydrogen sulfide naturally found in the air has been estimated as $\pu{0.11-0.33 ppb}$ ($\pu{0.15-0.46 ug/m3}$). Lower levels ($\pu{0.02-0.07 ppb}$; $\pu{0.03-0.1ug/m3}$) have been observed in some remote areas$\pu{^2}$.

Although these are relatively low concentrations compared to many other trace gases found in our atmosphere, silver is highly reactive towards hydrogen sulfide and thus tarnishing of silver is a ubiquitous issue. Additionally, some tarnish results from the oxidation of silver by oxygen.

Regarding your question about silver tarnishing from carbon, it's a matter of reactivity. The overwhelming source of carbon in the air is carbon dioxide, and methane to a much lesser extent. Neither of these gases are reactive towards silver nearly to the degree of hydrogen sulfide and oxygen.

1) Toxilogical Profile for Hydrogen Sulfide and Carbonyl Sulfide, U.S. Dept. of Health and Human Services, November, 2016
2) Ambient Air Guidelines for Hydrogen Sulfide, CAS Registry Number: 7783-06-4, March 27, 2006


Are the tars in cannabis considerably more dangerous than those in tobacco? The claim that "the tars in cannabis are considerably more dangerous than those in tobacco" is made in Patricks An Introduction to Medicinal Chemistry (4th edition, p. 169). It's just a simple one line statement and not much attention is given to it, but from what I already knew this is to say the least a very controversial statement (see for example the independent website about drugs Erowid) and I expect better for a frequently used University level book.

The question is to what degree (from a chemical point of view) there is truth (or falsehood) to this statement given current more or less objective (not like the Heath monkey studies) knowledge? Is the author in his right to boldly state this?

This is not a complete answer (merely a step in a particular direction, namely that of the falsehood of the claim), but I contacted the author of the book (Introduction to Medicinal Chemistry) Graham Patrick and he wrote the following:

  Dear Jori,

  Many thanks for your e-mail and observations. Unfortunately, that
  statement was made in a very early edition of the textbook published
  over ten years ago and so I doubt if I will be able to track down the
  original reference on which it was based. As you say there is
  controversy and the statement is too sweeping. Work on the 6th edition
  is starting this year and so it will be possible to modify that
  particular passage. Many thanks for your input.

  Best wishes Graham


What is the highest relative abundance for diatomic bromine? I was introduced to this thing today, and saw a graph showing the detection of ions with different relative abundances.

Just as an example here, $\ce{Br2}$ is shown a detection chart. And at $160~\mathrm{m/z}$ the ion has the highest relative abundance, but why? Since the $\mathrm{m/z}$ ratio is just the mass, for $\mathrm{z}$ is always $+1$, the one which is supposed to have the highest peak should be at $162$?

In $\ce{Br2}$, the possible combinations of isotopes are:

$\ce{^{79}Br-^{79}Br} : \mathrm{m/z}~158$
$\ce{^{79}Br-^{81}Br} : \mathrm{m/z}~160$
$\ce{^{81}Br-^{79}Br} : \mathrm{m/z}~160$
$\ce{^{81}Br-^{81}Br} : \mathrm{m/z}~162$

Bromine's isotopic distribution is essentially a $50:50$ ratio of mass $79$ and mass $81$. That means that there's a $25\%$ chance of $\mathrm{m/z}~158$ $(0.5 \cdot 0.5 = 0.25)$ arising from $\ce{^{79}Br-^{79}Br}$. There's a $50\%$ chance of $\mathrm{m/z}~160$ since there are two combinations ($\ce{^{79}Br-^{81}Br}$ and $\ce{^{81}Br-^{79}Br}$) that give that mass $(0.5\cdot0.5 + 0.5\cdot0.5 = 0.25)$. There's a $25\%$ chance of $\mathrm{m/z}~162$  $(0.5\cdot 0.5 = 0.25)$ arising from $\ce{^{81}Br-^{81}Br}$.