Calculating Molar Volume from crystal volume I am trying to calculate molar volume of certain crystal lattices to be used for crystal growth kinetics calculations. The information I have is the crystal volume & crystal lattices. For example the crystal $\ce{Cu41Sn11}$ has the structure cF416 and the information taken from a certain crystal database is as follows:

Atom Formula: Cu41.5Sn10.5

Cell volume:    5.81258 nm3

Cell density (calculated):  8.85 Mg m-3

Now I have tried 2 approaches to it.

First was the wikipedia formula for molar volumes

Vm= NAVcell / Z

molar volume = (Avorgados number x crystal cell volume) / number of formula units in a crystal

This gives the following output for the above mentioned crystal

Molar Volume = 6.022 x 1023 x 5.81258 x 10 -27 / 51

Molar volume = 6.731 x 10 -5

The second one is i figured out the formula

Molar volume = Molar Mass of crystal / Cell Density

So in this case:

Molar mass of crystal $\ce{Cu41Sn11}$ = 41 x 63.546 + 11 x 118.71 = 3911.2 gms (3.9112 Kg)

Molar Volume = 3.9112 / 8850 = 4.41944 x 10 -4

These two approaches yield two different results, so i wanted to make sure where am i making the mistake?

In case you're interested in the crystal database, here it is:

well figured out the answer, I had to take the correct value of Z from the database. The difference in values was because of the incorrect value of Z (which was actually 8) . When you use Z =8 , the two solution yield 'approximately' equal results.


Whats the difference between ionization energy and orbital energy? If you look at the trend in orbital energies as you go across a period the pattern is clear (orbital energy decreases with increasing effective nuclear charge) and, to my knowledge, it has no anomalies like the trend in ionization energies does when a 2p orbital is filled for the first time with two electrons (when this happens the ionization energy decreases). I expected that the graphs for ionization energy would be mirror images in the x-axis given that orbital energies are negative and measured relative to an ionized electron (which equals 0) albeit with different y-scales because ionization energy is the energy to ionize a mole of a substance in a gaseous state. Is my understanding that orbital energy = -(energy required to remove one electron from that shell) incorrect? Fundamentally, the question is, why doesn't the trend in orbital energies mirror the trend in ionization energies?

First off, the idea that the orbital energy is the negative of the ionization energy is actually Koopmans' theorem. It works qualitatively because of cancelation of errors.

But I digress...

Orbitals are single-electron constructions. That is, in Hartree-Fock theory, an orbital is a single-electron wavefunction assuming the average electrostatic potential of the remaining electrons.

Ionization energy trends are experimental observables and you see effects of multi-electron interactions.

So the short answer is that orbital energies are not physical quantities (i.e., being single-electron mathematical results) and ionization energies are based on multi-electron interactions.

Probably the simplest way to understand the trends in ionization energies vs. orbital energies is to look at Slater's rules and also remember Hund's rules. That is, when we add electrons, we are partially shielding the nuclear charge, but we also have to consider the stabilization due to exchange of electrons and destabilization when we pair electrons in an orbital. These are multi-electron effects, and only some of that can be recovered in a simple one-electron picture.

For the $2p$ orbitals, as you mention, there's a decrease in the ionization potential when you put two electrons in the same orbital. This happens because this new electron does not benefit from the exchange effect (Hund's rules) and has an electrostatic repulsion with the electron in the same orbital. Thus it's easier to remove it than might be expected from a one-electron orbital picture.


Volume of sucrose (solute) contributing to volume of solution? (solvent = water) I'm trying to make stock solution of 3M sucrose in water.  In 25ml, this is 25.672g sucrose (FW 342.3).  However, that mass of sucrose, in the powder, occupies a pretty substantial volume, such that if I add 25ml of water to it, I end up with a dissolved solution ~42ml in volume (it is fully dissolved).  Puzzled by this, I've also prepared a second solution in which I am only adding water up to the 25ml line in my 50ml conical (so far only 9ml).  This is obviously struggling to dissolve as the saturation solubility for sucrose is like 5M or something like that.

So what's the deal?  Is my 40ml solution with the correct mass of solute and volume of solvent really 3M?  My head says yes but my gut is uneasy.

The English wikipedia page for sucrose notes a solubility of 2000 g/L at 25 degrees Celsius, corresponding to an approximately 5.8 M solution. 3 M shouldn't be a problem in that respect, but to make dissolving easier you could warm up your solution slightly if that is appropriate in your protocol.

Now, molar is a concentration equivalent to mole per liter and this volume is the total volume including solvent and the dissolved material. So 25 ml of a 3 M solution of anything, say x, should have a final volume of 25 ml, including the solvent and 75 millimole of x.

This means your second approach is correct (being 25.6725 g of sucrose, to which you add water until the total volume is 25 ml).


Equilibrium constant in terms of partial pressures $K_p$ is defined as $\Pi_j(\frac{P_j}{P^{\Theta}})^{\nu j}$ where $P_j$ is the partial pressure of component j and $\nu$ is the stoichiometric coefficient. I have a couple of questions already. Firstly, $K_p$ is meant to be dimensionless so therefore, $\frac{P_j}{P^{\Theta}}$ must be dimensionless. However, surely $P_j$ alone is dimensionless as it's a partial pressure defined by dividing a pressure by a pressure - leaving no units. How does that work? Or am I wrong by assuming that a partial pressure has no units?

Given that, suppose I want to write the expression for $K_p$ in terms of partial pressures for the following reaction:

$$\ce{\frac12 O2_{(g)} + C_{(s)} <=>CO_{(g)}}$$

Do I assume that $\frac{P_{c}}{P^\Theta}=1$ leaving me with:

$$K_p=\frac{P_{\ce{CO}}}{(P_{\ce{O_2}})^{\frac12}}\frac{1}{P^\theta}$$

However, surely $P_j$ alone is dimensionless as it's a partial pressure defined by dividing a pressure by a pressure - leaving no units. How does that work? Or am I wrong by assuming that a partial pressure has no units?

You are wrong to assume that partial pressure has no units.  Partial pressure has units of pressure, such as mmHg, atm, bar, psi, etc.

For ideal gas mixtures, $y_j=\frac{P_j}{P_{tot}}$ or equivalently $y_j P_{tot}=P_j$, i.e. the partial pressure $P_j$ of a gas is equal to its mole fraction $y_j$ times the total pressure $P_{tot}$.

The expression you gave for $K_p$ is dimensionless because $P^\Theta$ has the same dimension as $P_j$.

  Do I assume that $\frac{P_C}{P^Θ}=1$....

That part sounds good.

  leaving me with $K_p=\frac{P_{CO}}{(P_{O_2})^{1/2}}\frac{1}{P^\Theta}$

Not sure about that one.  How is $K_p$ dimensionless in what you wrote?


Can an iodine clock reaction work without using a strong acid? I'm going to do an iodine clock reaction for a project and we had to submit the materials and safety sheets for the experiment a while ago. I thought I could do one by using hydrogen peroxide, sodium thiosulfate, potassium iodide and starch, but online I'm only seeing experiments that use sulfuric acid or another strong acid of some sort which is worrying me. I talked to my teacher and it's too late to change/add materials, so I'm really hoping that it can work? I'd also like to know what the strong acid does in the reaction, and what would happen (or wouldn't happen) without sulfuric acid?

Here's a bit from wikipedia for reference about the reaction:

This reaction starts from a solution of hydrogen peroxide with sulfuric acid. To this is added a solution containing potassium iodide, sodium thiosulfate, and starch. There are two reactions occurring in the solution.

(note: I didn't try formatting anything so the numbers after the ^ are the charges)

In the first, slow reaction, iodine is produced:
H2O2  +  2I^−  +  2H^+  →  I2 + 2H2O

In the second, fast reaction, iodine is reconverted to 2 iodide ions by the thiosulfate:
2S2O3^2−  +  I2  →  S4O6^2−  +  2I^−

After some time the solution always changes color to a very dark blue, almost black.
When the solutions are mixed, the second reaction causes the triiodide ion to be consumed much faster than it is generated, and only a small amount of triiodide is present in the dynamic equilibrium. Once the thiosulfate ion has been exhausted, this reaction stops and the blue colour caused by the triiodide – starch complex appears.

You wrote the answer yourself:  the hydrogen peroxide consumes hydrogen ion according to the reaction

$\ce{H2O2 + 2 I^- + 2\color{blue}{H^+} -> I2 + 2 H2O}$

You need the acid to provide the hydrogen ions, especially since if you allow the solution to become basic then the iodine disproportionates (see "Chemistry and Compounds" section, "Iodine Oxides and Oxoacids" subsection).

In terms of pure chemistry, maybe you could have used a moderately strong acid such as phosphoric acid or a bisulfate salt.  But then the limited dissociation of such species might make the iodine formation reaction slower and thus slow down the "clock".

Looking at the Wikipedia article on this reaction, one alternative is to use a peroxydisulfate salt instead of hydrogen peroxide as your oxidant.  This reacts with iodide ion according to

$\ce{S2O8^{2-} + 2I^- -> 2 SO4^{2-} +I2}$

with thiosulfate again as the reducing agent to regenerate iodide ion.  Unlike hydrogen peroxide or the other oxidants, the above reaction does not consume hydrogen ions and is your best bet to run the reaction without a strong acid addition.  However, peroxydisulfates have their own hazards (one of which is requiring a strong sulfuric acid solution to make it in the first place), and if you make the common choice of potassium peroxydisulfate rather than the ammonium salt you could have solubility issues.


What does the ‘−1’ superscript mean in units? For example, the rate of a chemical reaction can be expressed in $\mathrm{mol}/\mathrm{L}^{-1}/\mathrm{sec}^{-1}$. Why is it ‘−1’ and not, say, ‘−2’? Does it change the meaning if the minus is removed and we simply express the rate in $\mathrm{mol}/\mathrm{L}/\mathrm{sec}$?

The -1 means "per" unit. So your first example mol/L-1/s-1 is not correct - it would actually be written as mol L-1 s-1, OR mol/(L s). It is also sometimes written as mol/L/s, but the double division is ambiguous and should be avoided unless parentheses are used.

If it were mol L-1 s-2, this would mean moles per litre per second per second.

This is really just a question of notation, and is not chemistry-specific at all. Yes, all the minus/plus signs and the value of numbers are important. Good examples of units can include:

area, measured in m2, or metres squared
volume, measured in m3, or metres cubed
pressure, measured in N m-2, or Newtons per metre squared
velocity, measured in m s-1, or metres per second
acceleration, measured in m s-2, or metres per second per second


Stereoselectivity of hydroboration of dihydrofuran The above compound undergoes hydroboration. I originally thought that the $\ce{OH}$ group can attach to any of the 2 carbons plus there are 2 possible enantiomers at each position. So there should be 4 different products formed.

However the answer states that only below compound would be produced:

Why is this the only compound that is produced? Also if the $\ce{OH}$ group had to be attached to one of the carbons wouldn't it be more likely to attach to the carbon next to the oxygen since it will be more electrophilic?

Generally, the regioselectivity (and stereoselectivity, depending on the situation) of hydroboration reactions is determined by steric factors. This is a prime example of when it isn't.

Regioselectivity

Borane, $\ce{BH3}$, is electrophilic on boron because of the vacant p orbital (low-lying LUMO). It will therefore preferentially add to the alkene carbon that is more nucleophilic. The enol ether you have drawn is clearly nucleophilic on the carbon β to oxygen:

and consequently it is this carbon that the boron atom becomes attached to:

Reading this:

  Also if the OH group had to be attached to one of the carbons wouldn't it be more likely to attach to the carbon next to the oxygen since it will be more electrophilic?

I suspect you need to check the mechanism of hydroboration-oxidation again. There is no attack of $\ce{OH-}$, or indeed any oxygen-based nucleophile, on an electrophilic carbon. The C-O bond is formed by migration of carbon from boron to oxygen. Which carbon is more electrophilic is inconsequential.

Stereoselectivity

It is very common practice in organic chemistry to only draw one enantiomer when a racemic mixture is implied. The purpose of this is to save time: if you were discussing the diastereoselectivity of a particular reaction, you don't want to have to keep on drawing two enantiomers at every single step of the reaction.

Since attack from the top and bottom faces is equally likely to happen, an equal mixture of enantiomers is formed. As described above, the single enantiomer drawn therefore refers to a racemic mixture:

There is no diastereomerism present in this compound, but I would guess that the other reason why stereochemistry is indicated is to alert the reader to the fact that the product is chiral. Sometimes, this can be considered to be misleading. The most unambiguous way to indicate "undefined stereochemistry" is therefore using the wavy line as I did above.


Chemical oscillators that behave like a double pendulum Oscillating chemical reactions such as BZ-reaction or chemical clock typically result in sequential systematic changes, most notable of which is arguably coloring. Though underlying mechanism can be extremely complicated, we always expect the same sequenced cycling set of properties. For iodine clock it's always will be

  ... dark-blue > colorless > dark-blue > colorless ...

colors that we observe. Just like ordinary pendulum. But are there oscillating chemical reactions with the non-linear outcome and "chaotic" sequences for each cycle, e.g. behaving like a double pendulum?

And if such systems do exist, can we also just use similar mathematical apparatus like a system of differential equations used for a double pendulum to describe such behavior?

Before considering a double pendulum, let us consider a simple pendulum, which is modeled by

$$\ddot \theta + \frac{g}{\ell} \sin (\theta) = 0$$

Choosing units that make $\frac{g}{\ell} =1$, we have

$$\ddot \theta + \sin (\theta) = 0$$

Let $x_1 := \theta$ and $x_2 := \dot \theta$. Hence, the 2nd order ODE above can be rewritten as follows

$$\begin{array}{rl} \dot x_1 &= x_2\\ \dot x_2 &= - \sin(x_1)\end{array}$$

The 2nd ODE is not merely nonlinear, it is non-polynomial (due to the sine).  However, the ODEs that model elementary chemical reactions should be polynomial due to the law of mass action. Hence, it should be impossible to simulate the simple pendulum using a chemical analog computer. Simulating the double pendulum should also be impossible.

What may be possible is to simulate an approximation. For example, if $x_1$ is sufficiently "close" to zero, then we could use the approximation $\sin(x_1) \approx x_1$, which produces a system of coupled linear (and, thus, polynomial) ODEs

$$\begin{array}{rl} \dot x_1 &= x_2\\ \dot x_2 &= - x_1\end{array}$$

which approximate the original pendulum provided that $x_1$ remains "small" (in absolute value).

Lastly, since concentrations cannot be negative, we have non-negativity constraints $x_1, x_2 \geq 0$.


How does SO2Cl2 react with alcohols? Is the mechanism similar to the reaction of $\ce{SOCl2}$ with alcohols?
If not, then how is it different? We have been taught about the latter at school but my friend quizzed me about $\ce{SO2Cl2}$.

Please give the detailed reaction mechanism.

According to this article a mixture of products is formed, with alkylchlorsulfonates $\ce{Cl-SO2-O-R}$ being dominant products. These molecules are resistant to abstraction of sulfur trioxide, probably because trioxide is a very strong Lewis acid, unlike dioxide.


Transferring suspended substance from isopropyl alcohol to water? I have this UV-reactive invisible ink (a clear liquid.) According to the data sheet, it is suspended in an isopropyl alcohol base. Unfortunately, this is terrible for my particular application for two reasons: (1) it dries too fast (I need it to stay wet on a rubber stamp for ~15 minutes before stamping.) (2) It is too "thin" and thus bleeds into the porous material a little too much.

I've used regular black ink refills that (I believe) are water-based and they are perfect (except for the visible, non-UV-reactive part.)

If I understand this stuff correctly, the UV ink would be UV-reactive particles suspended in isopropyl alcohol base. If I were to mix this with water to get a workable liquid, I would likely dilute the ink too much.

Is there a way to transfer the suspended particles from the isopropyl-alcohol to water? Perhaps by mixing them, letting them separate and then let the alcohol evaporate off the top?

Alternatively, to reduce dilution I've considered finding a clear viscous liquid (think simple syrup without the stickiness) or even a clear gel as a thickening agent. Unfortunately, I don't know of a good viscous liquid or gel that would evaporate without a visible/sticky residue. Possibly one of the inactive ingredients in a hand sanitizer gel?

Slowly drive off the alcohol in a vacuum or by gentle heating. Replace the lost alcohol with distilled water. Experiment with small volumes first to ensure that heat does not degrade the ink.


The effect of baking soda on water electrolysis Ok, so now I have a new question, what type of electrode should I use. I am intending to use a simple stripped wire as my electrode, but am concerned whether the average copper wire would react and create unwanted byproducts. Is it ok if I use copper, or is some other metal better suited to this experiment.

EDIT:

I believe more baking soda would result in more gases because more free floating ions means more conductivity, which would probably allow more redox reactions to occur, and thus allow more gases to be produced.

Definitely do not used stripped copper wire, unless you fancy a nice blue solution and electrodes that need constant replacement. You see, copper oxidizes quite readily, so you will end up with lots of $\ce{CuO}$, and a loss in the yield of $\ce{H2}$ and $\ce{O2}$. The best things would be noble metals like platinum, but stainless steel should do fine for your purposes/likely budget.


How does this reaction occur? I was doing homework for my chemistry class, and I came across a question asking to complete and balance the following acid-equation:

$$\ce{Cu(OH)2 (s) + HClO4 (aq) -> ?}$$

At first I thought it was an acid-base reaction, but clearly $\ce{Cu(OH)2}$ is a precipitate. Thus, it could not be an acid-base reaction, and I was under the impression there was no reaction at all.

However, it turns out that the answer was to the question was (courtesy of the back of the textbook):

$$\ce{Cu(OH)2 (s) + 2 HClO4 (aq) -> Cu(ClO4)2 (aq) + 2 H2O (l)}$$

How is this possible? The answer to the question was exactly the same as an acid-base reaction, despite a precipitate instead of a base in the reactants.

Anyone know why this is the answer, or if it is even correct?

An acid or base can react with a solid. For instance HCl will dissolve $\ce{CaCO3}$ with much fizzing of $\ce{CO2}$ gas.

You also miss the point that $\ce{Cu(OH)2}$ has a solubility product. So there is a small amount of $\ce{Cu^{2+}}$ and $\ce{OH^-}$ in solution.

$\ce{K_{sp} = [Cu^{2+}][OH^-]^2}$

The perchlorate anion is a powerful oxidizing agent, but since $\ce{Cu^{2+}}$ is fully oxidized no redox reaction occurs.


Lead(II) oxide CRT monitor extraction I have a large CRT monitor that, by the weight of the monitor, I estimate to contain over two pounds of lead.  I want to know if the lead oxide is a mixture with the glass, or if it is chemically bonded to the glass.

I also want to know if there is any way (preferably extraction using heat) I can remove the lead in pure form or in oxide form. I know for sure you'd crush the glass.

I bring up heat extraction because chemicals are expensive, and I've heard of it being done that way on a large scale.

Edit: I'm not concerned about making a profit because I don't really have the setup to do so. I'm just curious about it because I've seen it done on an industrial scale using heat, so could it be replicated on home scale.

In what form would the lead be if you smelted it out?  Does heating lead oxide reduce it to lead again? If so, would the lead melt out and flow to the bottom?

A simple Google search will find you a number of informative articles.

I went and looked at one of the links and it was an abstract which I will quote below in full.  It looks like a possible method though I strongly advise against proceeding without a lot of forethought into pollution control measures.  There will be various lead compounds formed and keeping the dust and products out of the air and environment will not be easy.  As a business case this is going to be irrelevant soon as the number of CRT displays being recycled is decreasing continuously as 2nd generation LCD monitors have already reached end of life, only a small percentage of industrial control equipment still operates on CRT monitors and they will slowly be recycled over time, the big gold rush is over.

Here is a video from SWEEEP KUUSAKOSKI showing their industrial processing system, the method is not described but this started operation in 2013. The same time as the paper below was published.

Another video by Green Lyon describes a patented process to extract the lead from glass using an aqueous phase process at moderate temperatures with non-scary reagents.  Not sure if this has been reproduced elsewhere, this video was also published in 2013, a good year for CRT glass.

Extraction of Metallic Lead from Cathode Ray Tube (CRT) Funnel Glass by Thermal Reduction with Metallic Iron
Xingwen Lu†, Kaimin Shih*†, Chengshuai Liu‡, and Fei Wang†
Environ. Sci. Technol., 2013, 47 (17), pp 9972–9978
DOI: 10.1021/es401674d
Publication Date (Web): August 5, 2013

"A novel and effective process of thermal reduction treatment with the addition of metallic iron (Fe(0)) to recover lead from cathode ray tube (CRT) funnel glass is introduced. The key technological breakthrough of this process is the use of a relatively lower temperature and an inexpensive reducing agent to extract the metallic lead. The influences of temperature, the reducing agent content, and the holding time for lead reduction were examined to determine the optimal extraction efficiency. The lead extraction efficiency first increased and then decreased with increasing temperature. The maximum lead extraction efficiency occurred at 700 °C. The growth of crystalline lead first increased significantly with an increase in the Fe content, reaching maximum growth at an Fe addition of 50 wt %. The most effective treatment time was determined to be 30 min, as the vitrification of lead back to the glass matrix occurred under longer treatment times. The experimentally derived results indicate that a 58 wt % lead extraction can be achieved with the optimized operational parameters (50 wt % Fe addition, heating at 700 °C for 30 min) in a single extraction operation."


Reaction of white phosphorus with aqueous sodium hydroxide What are the products obtained in the reaction of white phosphorus with aqueous sodium hydroxide?

I know one will be phosphine ($\ce{PH3}$).

How can we find the oxidation state of phosphorus in the other product?

Thanks for help in advance.

Actually, there are two reaction involving:

$$\ce{P4 + 3NaOH + 3H2O -> PH3 + 3NaH2PO2}$$

  White phosphorus react with sodium hydroxide and water to produce
  phosphine and sodium hypophosphite . Sodium hydroxide - cold,
  concentrated solution. This reaction takes place slowly. (Chemiday 1)

The O.S. of phosphorus in sodium hypophosphite is calculated in @Sayantan.G's answer.

$$\ce{P4 + 8NaOH + 4H2O ->[\Delta]  4Na2HPO3 + 6H2}$$

  White phosphorus react with sodium hydroxide and water to produce
  disodium hydrogen phosphite and hydrogen. Sodium hydroxide -
  concentrated solution. The reaction takes place in a boiling solution.(Chemiday 2)

The O.S. of phosphorus in disodium hydrogen phosphite is:

$$\ce{ +2*1 + 1 + x + 3*(-2) = 0}$$

$$\ce{ x = +3}$$


How coordinate bond is formed? Here nitrogen can donate three electrons as it has p orbitals with three unpaired electrons. Now it is filled by electrons of hydrogen and everything is paired. It has 2 paired electrons in an s orbital. Now $\ce{H+}$ is coming and getting electron from nitrogen. How can it get from the inner s orbital?

Understand that ammonia typically adopts a trigonal pyramidal structure with a bond angle of $107^\circ$, which assumes $\mathrm{sp^3}$ hybridisation, rather than trigonal planar, which would assume $\mathrm{sp^2}$. Therefore, the lone pair is in a similar orbital as the other three $\ce{N-H}$ bonds, allowing for ammonia to act as a Brønsted base (accept $\ce{H+}$) and as a Lewis base (in coordination complexes).

The two states interconvert rapidly at room temperature.

The nitrogen atom is simply too small to allow for a bond angle of $90^\circ$ (unhybridised). The heavier homologue phosphorus, on the other hand, can allow this angle. Thus, phosphane $\ce{PH3}$ is much less basic than ammonia (even though phosphonium $\ce{PH4+}$ exists — but that is much more acidic than ammonium and can be used as a source for phosphane).


What does dl- mean in nomenclature I came across the compound dl-stilbene dibromide, and I am confused by exactly what this means. I thought it might have meant a racemic mixture but there is only one chemical structure given.

That picture is showing relative stereochemistry. Essentially, it is just saying that, in the shown confirmation, the hydrogens will be on one side and the bromines will be on the other, but half of the mixture will have the bromines coming out and the other half will have the bromines going back.

If the S or R labels had been shown in the molecule (or they had used just a d- or l- prefix) the picture would have been showing absolute stereochemistry. By saying dl-stilbene dibromide, they are saying that it is a racemic mixture of the compound shown and its enantiomer, but not any diastereomers.


What is the relationship between pressure and stoichiometry ratios This is the question from the textbook.

Ozone O3 can be prepared in the laboratory by passing an electrical discharge through oxygen gas. Assume that an evacuated steel vessel  with a volume of 10L is filled with 32 ATM of O2 at 25 degrees Celsius and an electrical discharge is passed through the vessel, causing some of the oxygen to be converted into ozone. As a result, the pressure inside the vessel drops to 30.64 ATM at 25 degrees Celsius .

The solutions manual subtracts 3x from the initial pressure of O2 to determine the final pressure of O2 after the electrical discharge(32-3x=final pressure of O2). X is the amount of pressure lost (32-30.63=1.36=x). My question is, how are we justified in using the stiochiometric ratios to determine the final pressure. In other words, how is (32-3x) justified when finding the final pressure of O2.

If my question is not clear. Please let me know and I'll try my best to rephrase it.

The actual reaction being discussed is $$\ce{3O2 -> 2O3}$$

We also know from the gas law that $$\ce{PV=nRT}$$

In the current experiment the only things changing are the pressure and the number of moles; the volume and temperature remaining constant.  In this case the gas law simplifies to $$\ce{\Delta P = \Delta n}$$

Therefor, we know the change in moles for the overall reaction (-3+2=-1) is $\ce{\Delta n}$ and by the above stoichiometry the change in the moles of oxygen will be three times this amount or $\ce{3\Delta n}$ or $\ce{3  \Delta P}$ or $\ce{3x}$.


Pressure increase on rate of reaction For reactions involving gases, an increase in pressure is no different from increasing the concentration of reactant particles, thus increasing the rate of reaction. But for reactions involves different phases, such as both solid and liquid, what will the pressure increase cause? Will it increase the rate of reaction? Or, because the rate also depends on other reactants and in reactions involving different phases, will pressure not change significantly even if gas is present?

There is probably going to be a separate reaction rate law for each phase, assuming that the reaction even occurs in other phases.  Of course, in liquid and solid phases, the pressure is going to have much less of an effect than in a gas phase.  Another consideration is that there may be interphase mass transport between the various phases (at their interfaces), and this may have to be taken into account in the chemical species mass balances for the individual phases.


Is this structure make-able? I was wondering if it would be possible to make this chemical?

Any structure, is, in theory, make able. The real question is whether anyone would want to waste a sufficient amount of time to make it.

The structure you've drawn doesn't seem so unreasonable. With enough time planning a synthesis, and enough hours of graduate student time I see no reason to object


Is it true that iron (Fe 0+) cannot exist when there's water around? So somebody told me that iron metal (zero charge) cannot exist at all when $\ce{H2O}$ is present because it will all turn into rust.
How is this possible? Isn't there some sort of equilibrium?
Does it have to do with the pH/pE? And if iron really can't exist, then what happens if I put an excess of iron into the water?

Iron metal can exist in presence of water. As you stated, important point is the $\ce{Fe^0 -> Fe^3+}$ reaction. Water is not an oxidizing agent, therefore is not capable of inducing this process.

To create rust, you need also oxygen, or as you noted, electrochemical potential.


Balmer and Paschen Series If electron makes transition from, in $H$ atom, $6\rightarrow 2$ (to Balmer), then $\bar{\nu}_{{H}_{6\rightarrow 2}}=x$ (let).

Also if electron makes transition from, in $He^{+}$ ion, so that $\bar{\nu}_{{He^{+}}_{a\rightarrow b}}=x$, gives me $a=12$ to $b=4$, by numerical caulations.

Now, what is observed is if I multiply $Z_{He^{+}}=2$ to transition of $H$ i.e. by $2\times\big({6\rightarrow 2}\big)=12\rightarrow 4$, gives me same result without any nasty calculation, why this works I don't know can you help me with this.

Thanks.

$$E_n = Z^2\mathcal{R}\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)$$

You can easily verify that if you replace $Z$ with $2Z$, then the energy of the corresponding transition will be four times the original energy. Now if you replace both $n_1$ and $n_2$ with $2n_1$ and $2n_2$, then you can take out a factor of $1/4$; so the transition $2n_1 \leftarrow 2n_2$ in an ion with nuclear charge $2Z$ occurs at the same energy/wavelength/wavenumber as the transition $n_1 \leftarrow n_2$ in an ion with nuclear charge $Z$.

Same applies for any other factor $k$: the $1 \leftarrow 2$ transition in hydrogen occurs at the same wavelength as the $4 \leftarrow 8$ transition in $\ce{Be^3+}$. Obviously this discussion ignores the variation in the Rydberg constants for different nuclei.


Toxic levels of baking soda What is the amount of baking soda that would be considered toxic for skin? I ask because there is a lot of misinformation online about using baking soda as a "natural" underarm deodorant, and how it is or is not harmful to use as such. Please clarify.

Thank you for your time in advance.

Baking soda is sodium hydrogencarbonate, $\ce{NaHCO3}$, also known as sodium bicarbonate.

The only toxicological data in the GESTIS database on hazardous substances
 is a LD50 of 4220 mg/kg for the oral uptake by rats. There are no data for dermal uptake.

In water, $\ce{NaHCO3}$ is weakly alkaline. I do not know whether it might cause skin irritations, particularly in more sensitive regions, such as armpits.

In occupational safety, I'd recommend to avoid skin contact and use personal protection equipment when handling it.


Inductive effect in carbanions In $\ce{CH3CH2-}$  what should be direction of inductive effect?

If going by order mentioned in my book, it is to be from carbon containing negative charge to CH3, but is not the carbon containing the negative charge more electronegative? So should not the direction be towards it?

The charge on CH2- doesn't say it's electronegative. The direction of inductive effect is from a negatively charged atom to the rest of the uncharged molecule (if there's only one negatively charged atom in a given (canonical) structure) and the effect decreases as we go away from the negative charge. The electronegativity of carbon atoms in organic molecules is more closely related to their hybridization; sp-hybridized carbons are  the most electronegative, followed by sp2- and sp3- hybridized carbons. Thus, your textbook is right.


What do graphene "armchair(5,5)" and "zigzag(5,0)" refer to? I am reading some articles about the electronic properties of graphene and while reading it I came across graphene armchair and zigzag structures, which have different bond structures.

The two different orientations (armchair and zigzag) are due to the arrangement of carbon atoms in the periphery (wiki), but I am unable to understand what armchair(5,5) and zigzag(5,0) mean.

The third picture, subpic (d) here shows you the geometric armchair(red) and zigzag(blue) structure:

It's derived from  structure description of Carbon nanotubes similar to surface reconstruction/overlayer structures:

  The integers n and m denote the number of unit vectors along two directions in the honeycomb crystal lattice of graphene. If m = 0, the nanotubes are called zigzag nanotubes, and if n = m, the nanotubes are called armchair nanotubes. Otherwise, they are called chiral. The diameter of an ideal nanotube can be calculated from its (n,m) indices

I hope you have some background, otherwise here is a nice visualisation, or look the lecture linked above. It's important to understand what the unit cell of a 3D crystal or 2D surface is.


Rearrangements of α-aminoketones in Clemmensen reduction What is the possible mechanism for this reaction? Activated zinc in presence of concentrated hydrochloric acid is used for Clemmensen reduction, that is conversion of ketone group to methylene group. I'd expect the product to be 1-methyl-2-propylpyrrolidine:

However, ring expansion occurs leading to a piperidine derivative. Could anyone please explain or provide a mechanism for the conversion? Thanks!

Regarding this question, there are two things that should be stated beforehand.

Firstly, the mechanism of the Clemmensen reduction is not well-understood. There are two plausible pathways that have been proposed:1

so, the mechanism for this rearrangement is equally unclear.

Secondly, this compound does not actually undergo rearrangement in the Clemmensen reduction, according to studies by Leonard and Ruyle.2 Curiously, the authors could only isolate the alcohol product. No explanation was offered as to why the alkane was not obtained.

However, α-aminoketones are known to undergo rearrangement when the nitrogen and the carbonyl carbon are in the same ring. Two examples which Leonard and Ruyle studied are offered below:2,3

Their proposed mechanism occurs via migration of the C–N bond, following protonation of the carbonyl oxygen. The carbocation is captured by what appears to be hydride, although the existence of hydride is inconsistent with the proposed mechanisms we have seen above. It might just be their shorthand for sequential electron + proton transfer; I can't really tell.

The second half of the mechanism is not explained, but it is plausible to protonate the hydroxyl group and use the nitrogen lone pair to expel water.

The authors posit that the tendency to undergo rearrangement here is related to a "geometrical directing effect": in the case where the nitrogen and the carbonyl carbon are in the same ring, the nitrogen lone pair has a larger opportunity to approach the carbonyl carbon.2 In modern-day terms, this might be considered to be a stereoelectronic effect.

How can this be reconciled with the proposed mechanisms in Kürti-Czakó? Considering that the mechanism of the reduction itself is unclear, I do not want to make any assertions about what the exact mechanism of this rearrangement is. However, I think it would be plausible that C–N bond migration can occur at some stage in the above mechanism - especially in the intermediates where the carbon bears a positive charge.

References

Kürti, L.; Czakó, B. Strategic Applications of Named Reactions in Organic Synthesis; Elsevier: Amsterdam, 2005; p 92.
Leonard, N. J.; Ruyle, W. V. Rearrangement of α-Aminoketones During Clemmensen Reduction. II. Contraction of a Six-membered Ring in the Monocyclic Series. J. Am. Chem. Soc. 1949, 71 (9), 3094–3098. DOI: 10.1021/ja01177a043.
Leonard, N. J.; Wildman, W. C. Rearrangement of α-Aminoketones During Clemmensen Reduction. I. Bicyclic Compounds Containing a Bridge-head Nitrogen. J. Am. Chem. Soc. 1949, 71 (9), 3089–3093. DOI: 10.1021/ja01177a042.


Are the names for chemical elements the same in both modern Greek and classical Greek? I am curious whether the modern Greek names for chemical elements were adopted from Classical Greek, both Latin and Classical Greek, or some international standard?
I'm unsure if this is the correct place to address this question as there's no Greek SE site and meta seems a bit open ended about it.

For "old"  elements (i.e., known / discovered before the advent of a body like IUPAC), there was no uniform and universal rule.  You see this heritage by comparison of the name of an element assigned in different languages; languages in neighboring countries may but need not to share their approach.  E.g., $\ce{_7N}$: nitrogen (English), nitrógeno (Spanish), nitrogen (Nowegian Norsk bokmal); azoto (Portuguese), azoto (Italian), azote (French), azot (Polish), azotas (Lithuanian), aзо́т (Russian), άζωτο (Greek); stikstof (Dutch), Stickstoff (German); dusík (Czech), etc. for a few examples you may encounter in Europe.
For newly discovered / newly synthesized elements beyond atomic number 100, IUPAC published (back in 1979) a recommendation how to name them systematically.  For this you have the roots

0 = nil, 1 = un, 2 = bi, 3 = tri, 4 = quad, 5 = pent, 6 = hex, 7 = sept, 8 = oct, 9 = enn

(Pure & Appl. Chem. 51, 1979, 381-384, open access)
Thus it was until specific agreement that unnilunium ($\ce{_{101}Unu}$) eventually was named mendelevium ($\ce{_{101}Md}$), and IUPAC continues to publish how to name the new elements (example).
Despite this report, however, the roots are neither pure Latin, nor Greek, but convention.  Because there no tenners (like decem, vīgintī), or hundreds (like centum, ducentī), etc.  Even spelling the numeri only by position, you would expect unus, duo, tres, quattuor, quinque, sex, septem, octo, novem for $1\dots9$ for the former.
For the names eventually adopted, IUPAC set the rules that these

"In keeping with tradition, elements are named after

a mythological concept or character (including an astronomical object);
a mineral, or similar substance;
a place or geographical region;
a property of the element; or
a scientist.

[...] The names of all new elements should have an ending that reflects and maintains historical and chemical consistency. This would be in general “-ium” for elements belonging to groups 1–16, “-ine” for elements of group 17 and “-on” for elements of group 18. N.B. The present recommendation is here more specific than that
written in the 2002 document."
(Pure & Appl. Chem., 88, 2016, 401-505, open access)


According to the Cahn-Ingold-Prelog system assign a priority to each of the following groups. (Highest priority is 1) 1-CH(CH3)CH2F
2-CHCl2
3-CH2Br
4-C(CH3)3

this is the order i did using the system but i am having a hard time understanding it and was wondering if anyone could tell me if my answer is correct, and if it is not, explain to me please

The keys are:

going outwards, only look one bond at a time
look for the first point of difference
to break ties follow the chains of highest priority first

So, going out from the point of attachment, the first atom in each chain is a carbon, so we expand from there to the next three atoms in each case (in red). Draw these out in priority order (ie. highest to lowest atomic number), top to bottom. Now compare the highest priority atom on each chain, ie. C vs Cl vs Br vs C. Bromine has the highest atomic number, so this makes that whole substituent number 1 priority, chlorine is next, so that whole group is number 2 priority, the remaining two groups are a tie, so we go down their list of attached atoms, looking for the first point of difference. The second atom on each is a carbon - ie. a tie. The final atom is a hydrogen on the first group, but a carbon on the final group, so the latter has the higher overall priority. In this particular example, there is no need to go further out along the chains to determine priority order for the substituents.

So, the final priority order is: (1) -CH2Br, (2) -CHCl2, (3) -C(CH3)3, (4) -CH(CH3)CH2F


Different products using different indicators A question states that 0.1 M solution of compound$$\ce{Na2CO3.NaHCO3.2H2O}$$ is titrated against 0.05M HCL, and that a different amount of HCL is used according to the indicators(phenolphthalein or methyl orange) used.

How does the indicator used affect the titration?

An indicator does not affect a particular reaction. But in your solution, you have two successive reactions when adding HCl to this solution : First $$\ce{CO_3^{2-} + H^+ -> HCO_3^-}$$ and a given indicator must be added to determine the end of this reaction. If you don't, all you see is a colorless solution being transformed into another colorless solution, even though its composition has changed. This added indicator may be phenolphthalein, which goes from red to colorless precisely at the end of this first titration. Then, by adding more HCl, you carry out the following reaction : $$\ce{HCO_3^- + H^+ -> H_2O + CO_2}$$ And you need is another indicator, maybe methyl orange, to determine the end of this second reaction. The color of the methyl orange in solution changes from yellow to reddish exactly after the end of this second reaction.

Phenolphtalein and Methyl orange are molecules which reacts with $\ce{H^+}$ exactly like $\ce{CO_3^{2-}}$ and $\ce{HCO_3^-}$. But they react before or after these two ions. To summarize, $\ce{H^+}$ reacts first with  $\ce{CO_3^{2-}}$ and nothing else, and it takes a couple of milliliters to do it. Then it reacts with Phenolphtalein and nothing else, and it takes less than one drop to do it. Then it reacts with $\ce{HCO_3^-}$ and nothing else, and it also takes a couple of milliters to do it. And last but not least, it will react with methyl orange.


Explaining Ba(OH)2 as a base? It is well known fact that Barium Hydroxide is a base. How does the Brønsted-Lowry concept explain $\ce{Ba(OH)2}$  as a base?

Is it that $\ce{Ba(OH)2}$ first accepts a proton then leaves behind $\ce{H2O}$?

Barium hydroxide – or more exactly, the hydroxide ion – is a Brønsted-Lowry base, because in aqueous solution, it is completely dissociated into $\ce{Ba^2+}$ and $\ce{OH-}$ ions:

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

$\ce{OH-}$ is a proton acceptor, forming $\ce{H2O}$. In aqueous solution, the following reaction takes place:

$$\ce{OH- + H3O+ ->~ 2H2O}$$

Barium hydroxide thus reacts with acids in a neutralization reaction, with water and the barium salt of the respective acid as the products. For example, with hydrochloric acid:

$$\ce{Ba(OH)2 + 2HCl ->~ BaCl2 + 2H2O}$$


How do we know the number of s and p orbitals in a shell? How do we know how many s & p orbitals will be in a shell?

There is one 2s orbital and three 2p orbitals in shell #2.
There is one 3s orbital, three 3p orbitals and five 3d orbitals in
shell #3.

Is there a reason for this? Wouldn't four 2s orbitals for shell #2 be a configuration of lower energy?

The number of orbitals in a shell is restricted by the possible values for the quantum numbers. Those come directly from the solution of the Schrödinger Equation for the Hydrogen atom (except for the spin, but let's keep it simple here), and are applied to all other atoms as well.

The quantum numbers and their possible values are:

the principal quantum number $n=1,2,3,4,5 \dots$, which relates to the period
the angular momentum quantum number $l=0,1,\dots, (n-1)$ which defines the shape of the orbitals $s$, $p$, $d$ and so on
the magnetic quantum number $m_l=-l,-l+1,\dots,+l$ further define the shape, e.g. $p_x$ or $d_{z^2}$
the total spin for electrons is always $s=\frac{1}{2}$ (this may be different for other elementary particles)
the relative spin of an electron is $m_s=-\frac{1}{2},+\frac{1}{2}$, and represents $\alpha$ and $\beta$ or up and down spin

Each electron needs to have its own unique set of quantum numbers (within one atom). This is know as Pauli Principle.

Thus for $s$ orbitals ($l=0$) there is only $m_l=0$ possible, hence there can only be one for each shell ($n$). Having four $2s$ orbitals would violate the Pauli Principle.
Additionally having a $2d$ orbital would also not be allowed as $l=2$ would be larger than $(n-1)=1$. So the $n=2$ shell is restricted to one $2s$ orbital ($m_l=0$) and three $2p$ orbitals ($m_l=-1,0,+1$).


What does 'forces' mean in the output of Gaussian calculations? In the output of a Gaussian calculation of frequencies, I can see a table listing all the 'forces' of all the atoms. It looks like this:

Center Number       Atomic Number       Forces (Hartrees/Bohr) X Y Z

  1        6          -0.001094502   -0.000768130   -0.006919164
  2        6           0.000528733   -0.000637636    0.005949765
  3        6           0.000424174   -0.000366750   -0.002904382
  4        6          -0.002619381   -0.000001885    0.005496979

...

What are these 'forces'? Are they related to those vibrations? But during vibrations, the force on each atom by other atoms is always changing, so why is only one constant 'force' value assigned to each atom?

The forces you have listed are the derivatives of the energy with respect to a displacement of a particular atom in each Cartesian direction. These forces are useful for finding stationary points on the potential energy surface, which occur when the force on each atom is zero. Computationally, it's not possible or worthwhile to try to get all the forces to be exactly zero, so stationary points are determined by a combination of criteria involving the forces and the displacement between optimization steps.

The vibrations in the system are related to the second derivatives of the energy with respect to the atomic coordinates, which are contained in the Hessian matrix.


Solubility when multiple salts are involved Calculate the solubility of $\ce{AgCl}$ ($K_{\mathrm{sp}} = 1.6\times10^{-10}$) in $100.0~\mathrm{mL}$ of $4.00\times10^{-3}~\mathrm{M}$ calcium chloride.

The stated answer is: $2.0\times10^{-8}~\mathrm{M}$

My way of solving was:
$$\begin{align}
\frac{1.6\times10^{-10}}{4.00\times10^{-3}} &= 4.00\times10^{-14}\\\\
\sqrt{4.00\times10^{-14}} &= 2.0\times10^{-7}
\end{align}$$

Is this correct?

I'm not sure what your train of thought was on this problem. Maybe annotating the problem would help see where you've gone wrong.

We're given that $K_{\mathrm{sp}}$ = $[\ce{Ag+}][\ce{Cl-}] = 1.6\times10^{-10}~\ce{mol^2/L^2}$. Since the solution already contains a known concentration of the chloride ion, we will solve for the amount of $[\ce{Ag+}]$ that will dissolve from the added $\ce{AgCl}$ to figure out the molar solubility of $\ce{AgCl}$.

$$[\ce{Ag+}] = \frac{K_{\mathrm{sp}}}{[\ce{Cl-}]} = \frac{1.6\times10^{-10}~\mathrm{mol^2/L^2}}{8\times10^{-3}~\mathrm{mol/L}} = 2\times10^{-8}\mathrm{mol/L}$$

Note that $[\ce{Cl-}]$ is double that of $\ce{[CaCl2]}$ since $$\ce{CaCl2<=>Ca^2+ + 2Cl-}$$

Since the dissolution of $\ce{AgCl}$ is in a one-to-one ratio with respect to the silver ion, $\ce{Ag+}$, the above answer is sufficient to show the molar solubility of $\ce{AgCl}$ in the given solution.

Any further chloride ions that dissolve will be negligible to the amount that is already in solution (according to the final solubility, only about $2\times10^{-8}~\mathrm{M}$ would be added, which is negligible to $8\times10^{-3}~\mathrm{M}$ by a factor of $400000:1$).


Does the amount of solvent interaction energy depend on the absolute gas phase energy? I've been performing Gaussian 16 DFT optimization and single-point energy calculations on different conformers of a rather large organic molecule. My first series of calculations was without solvent model, the second one was with dichloromethane solvent effects included via SMD model. When I analyzed the results, I saw that the absolute energy of the conformers became lower in solvent (as expected due to solvent interactions) but also the energy differences between the conformers became smaller. The order of conformers according to energy however did not change a lot. This would mean that conformers having a higher energy in gas phase than other conformers would experience more stabilization by the solvent. Is this possible? How could this be caused?
If more information is needed about my problem, feel free to ask.

Thanks in advance

One way this can happen is if the different conformers have different dipole moments. Those with largest dipoles might have higher energy due to greater charge separation. However those conformations with larger dipoles will also interact most strongly with the surrounding solvent medium (dielectric), and thereby the solvent will reduce their energy by a greater amount.


How thick is a snowflake? The x, y dimensions of snowflakes get all the attention for obvious reasons, but the z dimension must be interesting too, in its own humble way.  In fact, I have lived more than half a century and never seen anyone refer to snowflakes' third dimension at all!

So I'd like to know what a snowflake's cross-section looks like, and what forces/phenomena make it that way.

There are numerous possibilities:

that snowflakes are a single molecule thick
single crystal thick then grown over with hoarfrost
thousands or millions of nearly identical snowflakes glued together like slices in a loaf of bread
something else

No doubt, the ratio between x and y dimensions is extremely similar across all snowflakes, but an interesting part of the phenomena involved would be that which defines the ratio between x and z dimensions.

There are many different types of crystals and their fromation is far too difficult for me to understand. This answer simply deals with the thickness part and that too merely gives a rough approximation due to the lack of data.
 From CalTech

We have two formulae for calculating the width:

For Dendrite
$$\mathrm{h =a_2 D^f} $$
D: crystal diameter (cm) (major dimension)
h: crystal thickness (cm) (minor dimension)
For Needles
$$\mathrm{d = a_2 L^f} $$
D: crystal length (cm) (major dimension)
d: crystal thickness (cm) (minor dimension)

This is a table for $\mathrm{a_2, f}$

For d and l   (right most column )

I calculated the thickness of a $\pu{3 mm}$ Dendrite which is coming around $\pu{24  \mu m}$.

Source

https://web.archive.org/web/20180416083128/https://www.nsstc.uah.edu/data/uahsevere/owles/parsivel/references/Snowflake_Size_Distributions_Lecture.pdf


Ligand substitution reaction in coordination compounds I know that the following substitution reaction can take place:

$$\ce{[Cu(H2O)4]^2+ (aq) + 4NH3 (aq) -> [Cu(NH3)4]^2+ (aq) + 4H2O (l)}$$

This is because ammonia is a stronger ligand than water.

What can we say about the product formed from the following reaction?

$$\ce{[Cu(NH3)4]^2+ (aq) + excess HNO3 (aq) ->}$$

Nitric acid is a strong oxidizing agent but $\ce{Cu}$ can not be further oxidised. Is there a possibility of forming $\ce{[Cu(H2O)4]^2+ (aq)}$?

The reaction you wrote down is wrong on two counts. The reactant is not a hypothetical tetraaqua complex and the product is not a hypothetical tetraammin complex. The correct reaction is as shown below:

$$\ce{[Cu(H2O)6]^2+ (aq) + 4 NH3 (aq) <=> [Cu(NH3)4(H2O)2]^2+ (aq) + 4 H2O (l)}\tag{1}$$

Note that I have used an equilibrium arrow here: the reaction can proceed in both directions depending on the concentration of ammonia in your sample (in reality, there is also the competing precipitation of $\ce{Cu(OH)2}$ at slightly basic pH levels but I will ignore that here).

As an equilibrium, this can be pushed either way. So let’s take a look at what nitric acid can do. It is both an acid and an oxidising agent – but as you noted, copper(II) is already very highly oxidised and it is generally not possible to sustain copper(III) or higher copper oxidation states in aquaeous solution (it does exist as an intermediate in various copper-catalysed organic reactions). On the other hand, $\ce{HNO3}$ can very much act as an acid in your solution as there is a basic compound able to receive protons (see spoiler after figuring out the reaction yourself).

   $$\ce{HNO3 (aq) + NH3 (aq) -> NH4+ (aq) + NO3- (aq)}\tag{2}$$

Note that while $(2)$ is also a reversible reaction, the equilibrium is far to the product side so we can essentially treat it as completed. Obviously, this has an effect on the original reaction – and with a simple application of Le Chatelier’s principle the resulting answer should be obvious.

   Yes, since ammonia will be protonated to give ammonium, equilibrium $(1)$ will be shifted back to the reactant side, regenerating hexaaquacopper(II) $\ce{[Cu(H2O)6]^2+}$.


How does a hydrophobic environment help in bond formation? The glucose-induced structural changes are significant in two respects. First, the environment around the glucose becomes more nonpolar, which favors reaction between the hydrophilic hydroxyl group of glucose and the terminal phosphoryl group of ATP.

I am unable to understand the chemistry behind this. How is a hydrophobic environment helping bond formation between two polar molecules?

Source: Biochemistry 8th edition by Jeremy M. Berg John L. Tymoczko Gregory J. Gatto, Jr. Lubert Stryer

The major reason of the ‘hydrophobic environment’ is to block the enzyme-complex from accessing the solvent (water), which might otherwise enter the active site and hydrolyse ATP.

The reduction of polarity also helps to speed up the subsequent nucleophilic substitution.

Note: Hexokinase is a soluble, cytosolic protein

The reasoning behind this statement comes from the X-ray structural studies of yeast hexokinase (HK) by Thomas Steitz.

  Comparison of the X-ray structures of HK and the glucose–HK complex
  indicates that glucose induces a large conformational change in HK.
  The two lobes that form its active site cleft swing together by up to
  11.5 Å so as to engulf the glucose in a manner that suggests the closing of jaws.

This movement places the ATP in close proximity to the group of $\ce{-C6H2OH}$ glucose such that water is sequestered from the active site.
Water is certainly small enough to fit into the phosphoryl acceptor group’s enzymatic binding site. Moreover, phosphoryl transfer from ATP to water is more exergonic than it is to glucose.  If the catalytic and reacting groups were in the proper position for reaction while the enzyme was in the open position, ATP hydrolysis (to yield ADP + Pi) would almost certainly be the dominant reaction.

However, the large conformational change of the enzyme induced by glucose causing the exclusion of water from the active site making the phosphorus atom more accessible for the nucleophilic attack of the C6-OH group of glucose ensures that HK catalyses the phosphoryl transfer to glucose rather than ATP hydrolysis and also reduces the polarity of the active site, which help accelerate the nucleophilic reaction process.

This substrate-induced conformational change in HK is responsible for the enzyme’s specificity.

References

Lehninger Principles of Biochemistry
Voet and Voet Biochemistry


How to quench unreacted sodium azide from reaction mixture? The reaction involved synthesis of azide formation from alkyl bromide in DMF using sodium azide. After the reaction has taken place, how do I quench the unreacted sodium azide from the reaction mixture?

Excess of sodium azide can be quenched using nitrous acid.

$$\ce{2NaNO2 + H2SO4 → 2HNO2 + Na2SO4}$$

$$\ce{2NaN3 + 2HNO2 → 3N2 + 2NO + 2NaOH}$$

For further reading, Click here.


When is Serpeck's process preferred over Bayer's process? Both these processes are used in concentration of bauxite.
They use different reagents.

Which is more commonly used (wikipedia says Bayer's) and why?
When is one preferred over the other?
Which is more efficient and why?

From this source

Baeyer's process: This method is mainly applied when ferric oxide is present as chief impurity.
Serpeck's process: This process is used when silica is present in considerable amount of bauxide ore.

I couldn't find any information on the efficiency , but I suspect both are efficient in their own way. We cannot compare these two processes as they are used for two different impurities.


Why is there carbon lining on the steel vessel in Hall Heroults process? I read This answer, and completely understand that why graphite anode is used , but I have read that the vessel acts as the steel cathode but with lining of carbon. What is the need of this carbon lining ?
And this in turn raises the question , what actually is the cathode , steel or carbon ?
Thanks !!

An obvious answer is to why the presence of carbon, per a source Chapter 4: The Role of Carbon in Metallurgy on the historical extraction of iron from its ores, for example, to quote:

Carbon plays a crucial role in the chemical preparation of metals, particularly iron. ... This chapter discusses carbon as an intermediary product used in very large quantities to reduce and insulate metals and semi‐conductors from their natural oxides.

In a few words, carbon is a reducing agent, an insulator and a good oxygen scavenger.
Also, given the electrolysis nature of the Aluminum process, it is a cheap and very noble cathode. The employment of other noble metal cathodes would be more expensive and likely add impurities.


Classifying some molecular orbitals after Hartree-Fock calculation I have done a Hartree-Fock (HF) calculation on a dioxetane molecule with one methyl group on one of the C-atoms (Fig 1), with minimal basis (using Molcas). I want to identify the active space that will be used in later CASSCF calculations on the same molecule. The interesting molecular orbitals (MO's) for this active space are the following ones: $\sigma_{CC}$, $\sigma_{CO}$, $\sigma_{CO}^{'}$, $\sigma_{OO}$, $\sigma_{CC}^*$, $\sigma_{CO}^*$, $\sigma_{CO}^{*'}$, $\sigma_{OO}^*$, $n_{OO}$, and $n_{OO}^{'}$. So basically all the ring bonding and anti-bonding MO's, plus the two non-bonding MO's on the oxygen atoms.

The CO MO's are not localized for just one CO pair, but rather they both exist at the same time in an "out-of-phase" and "in-phase" version. These are given in Figs 2-5. The same goes for the non-bonding MO's; they both exist in an out-of-phase and in-phase version, given in Figs 6 and 7. Then there are only the CC and OO MO's left. In the same calculation for 1,2-dioxetane (zero methyl groups, only H atoms on the carbons), I find separate CC and OO MO's. However, for this one-methyl dioxetane, it looks like the CC and OO MO's are combined, giving out-of-phase and in-phase version of both the bonding and anti-bonding. My issue is that the MO's look a bit ambiguous, in that some of the lobes are not where I expected them to be. These MO's are given Figs 8-10.

For it to be a $\sigma$ orbital, the electron density should be in the bond axis between the atoms. However, in Figure 8 this is not the case. It's like the same-phase lobes attract each other and overlap instead inside the ring. Is it correct to interpret this as the CC/OO in-phase bonding MO? Figure 10 looks more like the $\sigma_{OO}^*$ than the in-phase $\sigma_{CC/OO}^*$, as the orientation of the lobes on the C atoms seem strange to me. Especially the MO in Figure 11 is not obviously classified, in my opinion. The OO contribution seems very small, and the lobe orientations on the oxygens are not what I would expect for an anti-bonding $\sigma$ orbital.

However, there are no other candidates for these MO's, so I'm left with the current classification, even if it feels a bit incorrect. Perhaps this is a common issue in analyzing molecular orbitals in computational chemistry.

P.S. I apologize for the bad figure qualities, I don't know how to visualize molden files in any other way.

Figure 1. The molecule

Figure 2. CO bonding in phase

Figure 3. CO bonding out of phase

Figure 4. CO anti-bonding in phase

Figure 5. CO anti-bonding out of phase

Figure 6. Non-bonding in phase

Figure 7. Non-bonding out of phase

Figure 8. CC/OO bonding in phase

Figure 9. CC/OO bonding out of phase

Figure 10. CC/OO anti-bonding in phase

Figure 11. CC/OO anti-bonding out of phase

Natural bond orbital (NBO) methods may help you to localized the orbitals of your system after a post-HF calculation. I'm not familiar with Molcas but I know that several computational chemistry softwares have NBO methods implemented. Take a look at this simple explanation that uses MolPro:

https://sites.lsa.umich.edu/zimmerman-lab/tutorial/electronic-structure-calculation/casscf/


Relative magnitudes of 2- and 3-bond coupling constants Why is geminal coupling weaker than 3-bond coupling in complex coupling? It would make more sense that atoms that are closer to each other influence each other more. Take methyl acrylate for example. The $\mathrm{H_a}$ is causing the bigger split, and the $\mathrm{H_b}$ the smaller ones. Why?

Be careful when generalizing from a single example.  There are many situations where the geminal coupling is much larger than vicinal coupling.  Pick pretty much any molecule that has an aliphatic CH2 where the two H's are diastereotopic and you'll find that the geminal coupling for those two H's can be 12-16 Hz vs. the typical ~7 Hz vicinal couplings.  However, if you're familiar with 6-membered rings in chair conformations, you've also seen vicinal coupling that range from ~2-15 Hz which leads us to the real answer to your question: the Karplus relationship.

Martin Karplus (who got a nobel prize for completely different work) is most well known in the NMR community for his theoretical work relating the values of J coupling to bond angles, bond lengths, electronegativity of substituents, etc.  In other words, there's no reason that 2 bond coupling is necessarily larger (or smaller) than a 3 bond coupling – the exact values will depend on the functional group and local geometry of the molecule.  Most organic chemistry texts describe a highly simplified Karplus relationship (ignoring all effects except for bond angles) for 2-bond and 3-bond J values.  I've attached two common examples.

You can see that for a typical aliphatic CH2, a bond angle of ~109° would predict a 2J ~ 14 Hz.  Whereas, a terminal alkene with a bond angle of ~120° would predict a 2J ~ 3 Hz.  Comparing this with the 3-bond Karplus relationship, the vicinal coupling for alkenes should be 8 Hz or 10 Hz depending on the cis/trans relationship.  For an aliphatic system with free rotation about a C-C bond, you have to average the Karplus relationship over all angles and you end up predicting ~ 7 Hz.  For a cyclohexane derivative in a chair conformation, you can see that an axial-axial coupling (180°) is predicted to be ~10 Hz, while an axial-equatorial coupling (~60°) is predicted to be < 2 Hz!

Qualitatively, this matches what we observe quite well.  Quantitatively, we quickly run into a lot of problems.  Many axial-axial couplings are 12-15 Hz and the J coupling observed in alkenes typically deviates quite a bit from the predictions given by these graphs.  Again, these graphs are grossly oversimplified - the real theory would fold in information about nearby substituants and steric effects, etc.  However, you'd then need run a bunch electron structure calculations and average these over many different conformations, etc.  So most people just live with the qualitative nature of these graphs to rationalize the patterns that they see.


How to use the calculate velocity correlation from VanHove correlation? I am trying to derive an expression for the velocity correlation function $C(t) = \langle \mathbf{V}(t) \mathbf{V}(0)\rangle$ from the VanHove correlation:

$$
G(\mathbf{r},t) = \frac{1}{N}\left( \sum_{i,j} \langle  \delta(\mathbf{r}- \mathbf{r}_i(t) +\mathbf{r}_j(0)   \rangle \right)
$$

However I am not sure how to get the time dependency and the $\dot{\mathbf{r}}(t)$ from the VanHove correlation. The average $\langle\rangle$ is over the ensemble.

I tired to figure out how to derive a term for velocity autocorrelation from the original paper, but without any luck.

Probably by now you have found the answer elsewhere, but for the benefit of others who may come across this question, here is my answer.

The link to the original paper seems to be broken.
However the problem you describe is fairly well known in the field of liquid state physics,
and the lecture notes that you refer to give a good introduction to the relevant correlation functions,
including the van Hove function. Certainly this is explained in
Theory of Simple Liquids by JP Hansen and IR McDonald (Academic Press, 4th edition).
That is a comprehensive graduate-level text which doesn't pull its punches:
the liquids may be simple, but the theory isn't!

There are two problems in making this connection.
Firstly, as you have noted, the van Hove function relates to the positions of the atoms,
not their velocities.
Therefore, we must anticipate that it will need to be time-differentiated (twice, in fact).

The second problem is that the velocity correlation function
involves the correlation of the velocity of a particle $i$ at time $0$
with the velocity of the same particle $i$ at time $t$.
However, the van Hove function involves cross correlations
of the positions of different atoms $i\neq j$
as well as positions of the same atom $i=j$.
These are
respectively the distinct part and the self part of $G(\mathbf{r},t)$,
as defined in the notes that you cite.
Formally, the velocity correlation function may be linked to the self part.
If you are attempting to connect it to the full $G(\mathbf{r},t)$,
you need to develop a justification for neglecting the cross correlations.

The self part of the van Hove function is defined
$$
G_s(\mathbf{r},t) = \frac{1}{N}\left\langle \sum_{i=1}^N
\delta[\mathbf{r}+\mathbf{r}_i(0)-\mathbf{r}_i(t) ]
\right\rangle
=
\left\langle
\delta[\mathbf{r}+\mathbf{r}_i(0)-\mathbf{r}_i(t) ]
\right\rangle
$$
where I have chosen one arbitrary atom $i$ since they are all equivalent.

It is convenient to Fourier transform this in space,
to give the self-part of the intermediate scattering function
(so-called because of its importance in inelastic neutron scattering experiments):
$$
F_s(\mathbf{k},t) = \int G_s(\mathbf{r},t) \exp(-i\mathbf{k}\cdot\mathbf{r})  d\mathbf{r}
=
\left\langle
\exp(i\mathbf{k}\cdot\mathbf{r}_i(t)) \, \exp(-i\mathbf{k}\cdot\mathbf{r}_i(0))
\right\rangle .
$$
For simplicity, we will define $\mathbf{k}$ to lie along the $z$-axis:
$$
F_s(k,t)
=
\left\langle
\exp(ikz_i(t)) \, \exp(-ikz_i(0))
\right\rangle .
$$
Then we differentiate with respect to time, twice.
Let $v_i$ be the $z$-component of the velocity of particle $i$.
$$
\frac{d^2}{dt^2}F_s(k,t)
= -k^2\left\langle
 v_i(t) \exp(i k z_i(t)) \, v_i(0) \exp(-i k z_i(0))
\right\rangle .
$$
A subtle trick has been used on the right hand side: the so-called dot-shifting trick.
Because of stationarity $\langle \dot{A}(t) B(0)\rangle = -\langle A(t) \dot{B}(0)\rangle$
where $A$ and $B$ are any variables, and the dot denotes time differentiation.
This is how we are able to bring down $v_i(t)$ from the first exponential,
and $v_i(0)$ from the second one.

The derivation is completed by dividing both sides by $-k^2$ and taking the limit $k\rightarrow 0$:
$$
-\lim_{k\rightarrow 0} \frac{1}{k^2}\frac{d^2}{dt^2}F_s(k,t)
= \left\langle
 v_i(t) v_i(0)
\right\rangle .
$$
This is the answer.

In some circumstances one can make the Gaussian approximation,
namely that the self part of the van Hove function is a Gaussian function of $\mathbf{r}$ at all times.
This is true at short times (free-particle limit) and at long times (diffusion limit),
but has been shown to be inaccurate (by about 10 percent) in typical liquids at intermediate times.
If we nonetheless assume that it is true,
then the intermediate scattering function is a simple function of the mean-squared displacement
$$
F_s(k,t) = \exp\left[-\frac{1}{6}k^2 \langle |\mathbf{r}_i(t)-\mathbf{r}_i(0)|^2\rangle\right]
$$
where now $k$ represents the magnitude of the $\mathbf{k}$-vector.
This can be related straightforwardly to the velocity correlation function.


Typography of multi-letter variables Chemical elements are typeset with upright font ($\ce{H2O}$ for water) while variables are typeset with an italic font ($V$ for volume).

But is there a convention what to do with multi-letter variables like IC50, EC50 (or others without subscripts)? Is there any official convention for this?

Special cases may be pH and pOH where on could read p as an operator and H and OH as chemical abbreviations, so I would always write them upright. But I am not sure about the others.

In the International System of Quantities, i.e. the system of quantities that corresponds to the International System of Units (SI), as described in ISO 80000, the symbols for quantities are generally single letters from the Latin or Greek alphabet, sometimes with subscripts or other modifying signs. However, symbols for so-called characteristic numbers are written with two letters from the Latin alphabet, the initial of which is always capital. For example:

Reynolds number $Re$
$$Re=\frac{\rho vl}\eta=\frac{vl}\nu$$
Nusselt number $Nu$
$$Nu=\frac{Kl}\lambda$$
Prandtl number $Pr$
$$Pr=\frac{\eta c_p}\lambda=\frac\nu a$$

(Note that such two-letter symbols should be separated from other symbols if they occur as factors in a product.)

Quantity symbols are always written in italic type.

Various other widespread multi-letter symbols are not quantity symbols in accordance with the above-mentioned systems; they are actually just abbreviations. For example in toxicology, LD50 is used as abbreviation for “median lethal dose” (or “lethal dose, 50 %“). Such abbreviations are typeset normally, i.e. generally upright.

Also note that abbreviations for quantities shall not be used in equations.


Which ligand names have to be parenthesized while naming the Mononuclear Coordination entities While studying about the Nomenclature of Coordination Compounds I came to know that

While writing formulae of Mononuclear Coordination entities polyatomic ligands are enclosed in parenthesis and also their abbreviations.

But while naming the entities should I follow the same rule?
Like for $\ce{[Ni(H2O)2(en)2]SO4}$ should it be named as

Di(aqua)bis(ethylene-1,2-diamine)nickel(II) suphate

Or

Diaquabis(ethylene-1,2-diamine)nickel(II) sulphate

I Mean is there any rule that says you only need to close certain species of ligands like specifically bidentate (such as en) and polydented ligands (such as EDTA) and can keep the others as they are?
Thanks.

Greek prefixes are used to designate the number of each type of ligand in the complex ion, e.g. di-, tri- and tetra-. If the ligand already contains a Greek prefix (e.g. ethylenediamine) or if it is polydentate ligands (ie. can attach at more than one binding site) the prefixes bis-, tris-, tetrakis-, pentakis-, are used instead.
Bottom line - parenthesize only those ligands which already has a greek prefix in its name so that you can highlight the use of the alternative prefixes such as bis-,tris- etc.
Ps -Diaquabis(ethylene-1,2-diamine)nickel(II) sulphate is the correct naming but you can make it more appropriate by using ethane instead of ethylene


Rule of mutual exclusion: is the converse true? There is a "famous" rule in spectroscopy,1 that goes this way:

  If a compound is centrosymmetric, then its normal vibrational modes cannot be simultaneously IR and Raman active.

and this is simple enough to prove: for a vibrational mode to be IR active, it has to transform under the same irrep as (one of) the cartesian axes $x$, $y$ or $z$, which are all ungerade. For a vibrational mode to be Raman active, it has to transform under the same irrep as one of $(x^2, y^2, z^2, xy, yz, xz)$, and since $u \otimes u = g$, these irreps cannot be the same as the irreps of $(x,y,z)$. Hence, a vibrational mode, which transforms under a single irrep, cannot simultaneously satisfy both conditions at once.

However, I am curious about the converse statement.

  If a compound does not have any vibrational modes which are simultaneously IR and Raman active, is it necessarily centrosymmetric?

If the answer is yes, I would like a proof; if it is no, a counterexample, please.

(Just in case anybody is confused, because I am: A counterexample would be a molecule without a centre of inversion, which has no vibrational mode which is simultaneously IR and Raman active.)

1 Famous enough for me to know, at least.

If there is a molecule of $O$ (as opposed to $O_h$) or $I$ (as opposed to $I_h$) symmetry, it would be such a counter example (no mode that is both IR and raman active, and not centrosymmetric).

http://symmetry.jacobs-university.de/cgi-bin/group.cgi?group=903&option=4

http://symmetry.jacobs-university.de/cgi-bin/group.cgi?group=905&option=4

Apparently a snub cube is an example of the $O$ point group and in Chemistry for the 21st Century it is stated:

  9.5.2.5 Snub Cube (6) We have recently demonstrated the ability of six resorcin[4]arenes and eight water molecules to assemble in apolar media to form a spherical molecular assembly which conforms to a snub cube (Fig. 9.3)


Do undissolved salts in saturated solvents enter equilibrium with the dissolved salts? For instance, if there is cup of water saturated with a salt to the point where any more added salt will just float around in the solution, are those undissolved salt molecules in equilibrium with the dissolved ones? For instance, do the undissolved ones become partially dissolved for a moment, while the dissolved ones swap places with them?

Or does this behave in such a manner that whatever salts were dissolved first to reach saturation are the ones that will stay dissolved, in spite of other added salts?

Yes, they enter into a constant equilibrium. Just as we write equilibrium expressions for insoluble salts as equilibrium expressions, like $\ce{CaF2} \leftrightharpoons \ce{Ca}^{2+} + \ce{2F-}$, when you reach the saturation point, solute particles are in dynamic equilibrium between aqueous and solid phase.


Why is the melting point of p-toluenesulfonic acid monohydrate higher than that of anhydrous p-toluenesulfonic acid? Why is the melting point of p-toluenesulfonic acid monohydrate higher than that of anhydrous p-toluenesulfonic acid?

Compare for example the melting point of benzenesulfonic acid hydrate, which is lower than for anhydrous benzenesulfonic acid.

The hydrated forms might be an azeotrope which means there's a content of water present that is difficult to remove via distillation techniques which may be how the P-toluene sulfonic acid was prepared.

The anhydrous form would require further processing.

The moisture present in the hydrate forms may affect the melting point if it were acting as an azeotrope with the P-toluene sulfonic acid, also it wouldn't be as high purity as the anhydrous form which would display the true melting point.


Is Kp and Kc only effected by temperature? Are Kp and Kc only effected by temperature?

So in exothermic and endothermic equilibrium reactions you are increasing or decreasing temperature, which will effect which side is favourable.

But then so does pressure if one side has fewer molecules?

According to LeChatelier's principle on increasing temperature in an exothermic reaction "which is at it's equlibrium initially" the reaction proceeds in backward direction and in endothermic reaction it proceeds in forward direction, this can be thought as in exothermic reactions the heat should be provided to the products to yield the reactants and in endothermic the heat should be supplied to reactants to yield the products!
And the value of equilibrium constants depend only upon temperature and the pressure only effect the equilibrium conditions and not the value of Kp because partial are so adjusted to provide the same value of Kp at the same temperature.


How to weight picric acid? Picric acid is stored under a layer of water to prevent an explosion.

What is the right way to measure the amount of picric acid that taken out? It is wet and containing water.

If accuracy is not essential, weigh a small amount drained for a fixed time but still wet, and then weigh again dried, to find the ratio of wet-to-dry weight (carefully disposing the dried, sensitized, explosive without destroying the scale or personnel).

Though this could give a rough idea of the actual dried weight, it would vary from batch to batch and with temperature.


What is the oxidation state of oxygen in Na2O2? I am asked to determine what is the O.S. of oxygen in $\ce{Na2O2}$

I was under the impression that oxygen had a charge of -2 and I multiplied that charge by 2 giving me a total of -4.

However, the computer program I am using, states that I am wrong. Can someone clarify why?

Oxygen can take multiple oxidation states. This particular compound is sodium peroxide.

You're right that usually oxygen has a charge of -2, but in this case, there's no way that each $\ce{Na}$ can have an oxidation state of +2.

So you then work backwards, deciding if it's $\ce{Na+}$ then you have +2 from the sodium, and oxygen must have an average oxidation number of -1 per oxygen atom.

Besides peroxides ($\ce{O2^{2-}}$) there are also superoxides ($\ce{O2^{-}}$).

It's not surprising that these compounds are usually very reactive.


Seeking safe, water-soluble chemicals for an IR spectroscopy experiment I am working with a custom-built device which is intended to perform some mid-infrared chemical measurements.  It consists of four thermopile sensors with optical windows.  In front of each window is a band-pass filter.  Each filter admits a distinct range of wavelengths.  We performed an initial experiment which suggests that the manufacturer may have installed the filters in the wrong positions.  I am trying to design a more definitive follow-up experiment.  I can't tell the filters apart by looking at them.
These kinds of sensors are typically used in the automobile industry to characterize engine exhaust gases.  We happen to be measuring aqueous solutions instead, so I am trying to identify useful reference solutions we can prepare.  Ideally, the solutions would absorb well at one of the three wavelengths, and transmit well at the other two.  I don't care whether the reagents are organic or inorganic.  Safe, water soluble, inexpensive, and already present in my lab are all desirable qualities though.
The center wavelengths of the three optical filters are 6.7 μm, 8.3 μm, and 9.6 μm.  In wavenumbers, these are around 1490, 1200, and 1040 cm⁻¹.  The bandpass range on each of the filters is about 0.2 μm wide.  I also have a fourth, broad-band filter which I will identify by process of elimination.
Using my limited knowledge of IR fingerprints and the NIST Webbook database of infrared spectra, I have identified two candidate compounds.  Glycine absorbs preferentially in the 6.7 μm band, and trimethyl phosphate absorbs preferentially in the 9.6 μm band.
I still need a good candidate compound for the middle band: 8.3 μm / 1200 cm⁻¹.  The best I have found so far is n-decanoic acid, but it's not really that selective in absorbing at 8.3 μm.  It also isn't particularly water-soluble.
I would greatly appreciate it if anyone has any suggestions.  Thanks!

Disclaimer: I assume you refer to compounds which may be «soluble in water», not necesarily to constrained to «recording the spectra in aqueous solution» for limitations of IR transmission spectroscopy which are a bit lesser present with IR ATR spectroscopy, and basically not present for Raman (M. Farooq's answer). I assume you mention water soluble because you want to clean your setup after recording the data simply with water.  Note: often, the recordings mentioned below are literally about thin films of theses substances, i.e., an optical pathlength across the sample much less than $\pu{1 mm}$.
Complementary to NIST webbook chemistry offering a search by the vibrational energy here is the Japanese Spectral Database for Organic Compounds, SDBS. (Querries may be either in Japanese, or English.)  It equally offers a «reverse search» starting with spectral information you already have in hand:

(credit)
I'm not aware about the spectral width of the sensors you mention, if these record e.g., at $\pu{(1200 \pm 10) cm^{-1}}$, or say $\pu{(1200 \pm 50) cm^{-1}}$.  Obviously, the larger the allowance, the list of compounds indexed in the database matching this criterion will increase. In a query with the former threshold, browsing across the list of hits however were entries like formic acid (attention, potential danger because it is corrosive), or acetone  (attention, potential danger because it is flammable).
By back-and-forth between the listing (to rule out compounds not water soluble, requiring some intuition / knowledge) and the spectra, you may identify if there are other bands of (dis)interest to you, too, e.g.

(example acetone, loc. cit.)


cupola based hydrocarbons My favorite groups of shapes are cupolas. Are there hydrocarbons in the shape of these similar to how cubane is to a cube? If so how stable are they, and could you please give some general information about each one.

They would be highly unstable due to the inverted tetrahedral geometry of the carbons in the top face. Generally, a polyhedron can form the basis of a stable hydrocarbon if it has at most 3 edges for each corner. A big caveat is that the hydrocarbon will distort the shape if the angle strain is too great. I found that out in my computational research on prismanes.


The thermal decomposition of NH4Cl When you heat $\ce{NH4Cl}$ it decomposes into $\ce{NH3 + HCl}$ gases. My question is why does it decompose into ammonia and hydrogen chloride gases on the molecular level. Is it because in the $\ce{N-(HCl)}$ bond, The $\ce{N-H}$ bonds breaks easily since it requires only $\pu{390 kJ/mol}$ and $\ce{H-Cl}$ requires far more heat energy? Thus the $\ce{N-H}$ bonds breaks first?

In the solid state of ammonium chloride, the nitrogen and hydrogen atoms are connected by covalent bonds, forming the ammonium ion. The bond length is about 98 pm. Each ammonium ion is surrounded by eight chloride ions (and in turn, each chloride ion is surrounded by eight ammonium ions). The distance between a chloride and the closest hydrogen atoms is about 237 pm. In contrast, the bond length in HCl is 127 pm.

In the reaction $$\ce{NH4Cl(s) -> NH3(g) + HCl(g)},$$

you lose the ionic interactions, and ammonium gets deprotonated while chloride gets protonated (lose an N-H bond and gain a H-Cl bond). The change in entropy for the reaction is positive, so it is favored at high temperature.

We might think that high temperature makes the H-Cl bond stronger and the N-H bond weaker. The bond dissociation energy for N-H is 390 kJ/mol while that of H-Cl is 432 kJ/mol. However, bond dissociation energies are not directly relevant because we are not transferring a hydrogen atom (making $\ce{NH3+ + HCl-}$). Instead, we are transferring a proton, so the pKa of ammonium and hydrogen chloride are more relevant but not directly either because this does not happen in aqueous solution.

The key is that the products are in the gas phase. If the products of the decomposition were ions and had opposite charges, they would not remain in the gas phase. In aqueous solution, you get $$\ce{NH4Cl(s) -> NH4+(aq) + Cl-(aq)}$$ because water is good at solvating ions. In the gas phase, the most stable combination of the atoms at hand is $\ce{HCl + NH3}$.