The Evasive Ether: Existence of diethynyl ether? I tried to find online whether the compound diethynyl ether, $\ce{HC#COC#CH}$, exists in a stable state at STP, mainly out of curiosity, but also because it would be interesting to study the properties of a molecule with such interesting characteristics (eg. the very long bent shape, the highly exposed oxygen atom, the potential for multiple different polymerization reactions). However, I spent about 30 minutes scrolling through page after page of search results with only results about diethyl ether, and came up empty-handed. I'm wondering if anyone here has any info/sources/knowledge about this evasive ether. I've included some images to clarify what molecule I'm talking about, as well as to possibly help jog your memory. Any and all information/data about the molecule is appreciated.

You can find a nice discussion of diethynyl ether and some of its derivatives here:

Ooi, I. H., & Smithers, R. H. (1989). Bis [(trimethylsilyl) ethynyl] ether: a moderately stable C4H2O derivative. The Journal of Organic Chemistry, 54(6), 1479-1480.

https://pubs.acs.org/doi/pdf/10.1021/jo00267a053

They write that "little or nothing is known about diethynyl ether", and they were only able to find theoretical (INDO) calculations in their literature search about the compound, as opposed to actual reports of its synthesis. They do mention that the following authors managed to synthesize a dimethyl derivative, $\ce{Me-C#C-O-C#C-Me}$, but reported that it explodes at room temperature (and I infer, from what Ooi and Smithers wrote, that diethynyl ether would be even more unstable):

Brandsma, L., & Arens, J. F. (1962). Chemistry of acetylenic ethers LVI Di‐(alk‐1‐ynyl) ethers and thioethers. Recueil des Travaux Chimiques des Pays‐Bas, 81(6), 510-516.

https://onlinelibrary.wiley.com/doi/abs/10.1002/recl.19620810610

They themselves (Ooi and Smithers) were able to synthesize a di-trimethylsilyl derivative, $\ce{Me_3Si-C#C-O-C#C-SiMe_3}$ which, while stable when dissolved in $\ce{CH_2Cl_2}$ and stored in a freezer, degrades overnight when stored neat at room temperature.

In summary, for $\ce{R-C#C-O-C#C-R}$, larger R's provide added stability:

R = H (diethynyl ether): If it were ever synthesized, it would probably have a very short half-life, except under specalized conditions.

R = Me: Has been synthesized, but explodes at room temperature

R = $\ce{Me_3Si}$: Low to moderate stability, as described above.   Ooi and Smithers write that the added  stability provided by the trimethylsilyl groups "is probably both steric and electronic."

See also:

Arens, J. F. (1960). Ethynyl ethers and thioethers as synthetic intermediates. Advances in organic chemistry, 2, 117.


How to determine concentration of an unknown solution using Beer's Law? What if the concentration of an unknown is higher than the range of concentration used when you were doing an Absorption v. Concentration graph? How can you figure out the concentration of an unknown solution, basically, if it falls above the range? I have literally no idea how to go about doing it.

This can be an issue in many calibration experiments, and falls into the broader and extremely important field of "design of experiments" within the field of chemometrics. Clearly, your range of calibration data is not quite sufficient for your application, but you have several options available to you, each with pros and cons, some of which have been mentioned in the comments

Go back to the calibration curve, and take some standard measurements at higher/lower concentrations than you did. Pros: this will be the most accurate method, and will help you to avoid this problem with future measurements. Cons: takes time and requires more measurements.
As commented, extrapolate from your model. Pros: extremely quick. Cons: likely to be inaccurate unless you are 100% sure Beer's Law holds for your sample at all concentrations. There is also no way of estimating the errors.
As commented, dilute your sample with a known quantity of solvent and remeasure. Then multiply your answer by the dilution factor. Pros: Fairly accurate, takes less time than performing more measurements for your model. Cons: Depending upon your lab skills and precision of your equipment, errors could start to propagate and become larger.

Good luck!


Ab initio effective potentials I have been reading a journal article on ab-initio pseudo-potentials, and I need some help understanding it. The article is C. F. Melius and W. A. Goddard, III. Phys. Rev. A 1974, 10, 1528.

A summary of the article for those that do not have access follows. The authors present a way to construct a pseudo-potential. Basically, their argument goes like this (according to my understanding):

  Suppose that we are using the Hartree Fock method. We can represent
  the ground-state wavefunction of some atom as a function of the valence
  orbitals, core orbitals, and spin.

  $$\Psi= f (\phi_{core}, \varphi_{valence},X)$$

  Note: in the article they use a weird letter instead of $f$.
  After we obtain the ground-state occupied orbitals, we solve for the valence orbitals.
  We obtain the valence orbital by solving the variational equations
  $$H^{HF}\varphi_{i}=\epsilon_i\varphi_{i}$$ Where $H^{HF}$ is the Hartree Fock Hamiltonian and is given by
  $$H^{HF}=-\frac{\nabla^2}{2}-\frac{Z}{r}+2J_{\alpha}-K_{\alpha}$$
  For the valence orbital, $J_{\alpha}$ is equal to the classical Coloumb potential due to a charge density corresponding to the $\phi_{core}$. The exchange operator $K_{alpha}$ is an integral operator resulting from the antisymmetric form of the wavefunction.

  The authors then state that the valence orbital of the ground-state wavefunction is NOT determined uniquely by the variational principle. This is because the solutions of the variational equations may not yield a wavefunction that is purely a valence orbital. In other words, the mixing of core orbitals with a valence orbitals does not change its energy. Hence the solution $\varphi_i$ may be of the form $$\varphi_i=\varphi_{valence}+\sum_i c_i\phi_{core}$$
  The authors then impose the constraint that $\varphi_{i}$ be orthogonal to other occupied orbitals.

What I do not understand is this. How does orthogonality allow for a unique solution? Also later on, the authors state "Although this orthogonality restriction has the desired consequence of leading to a prescription for a unique valence orbital, there is no reason to consider $\varphi_i$ to have a special significance over any other combination of $\varphi_i$ with the various core orbitals." What do they mean by this statement?

First off, I would like to note that you wrongly interpreted the very first equation. Below I quote the relevant part of the paper:

  In the Hartree-Fock (HF) approximation for, say, the $\ce{Na}$ atom,
  the ground-state wave function has the form

  $$
\Psi = \mathcal{A} (\Phi_{\mathrm{core}} \phi_\nu X) \, , \tag{1}
$$

  where $\Phi_{\mathrm{core}}$ is a product of (ten) spatial orbitals
  very similar to the orbitals of $\ce{Na+}$ and $\phi_\nu$ is the
  valence orbital (the one removed in ionizing to $\ce{Na+}$). ($X$ is an
  appropriate product of spin functions.)

So, $\Psi$ is not just some function of spin orbitals designated by a weird letter as you said, it is, as usual, their antisymmetrized product, where I bet the calligraphic A symbol ($\mathcal{A}$) as usual stands for the antisymmetrizer. The only small difference is that small calligraphic a is used in the text rather then much more common capital one.

Now to you actual question. Both core and valence orbitals are thought in the derivation as the solutions of the Hartree-Fock equations, which have a very well-known feature: they are defined to within a unitary transformation (see, for instance, here). So, orbitals which are solutions of the Hartree-Fock equations are not unique: there are infinite different sets of orbitals that minimize the Hartree-Fock energy and you are free to choose any of these set for any further work. This is, essentially, what is said in the paper:

  It will be important in our later analysis to note that the valence
  orbital in (1) is not determined uniquely by the variational
  principle. One can modify $\phi_\nu$ by mixing in an arbitrary amount
  of any core orbital (doubly occupied. in $\Phi_{\mathrm{core}}$)
  without changing the energy.

Just note that this is true not only for the valence orbital: all orbitals (i.e. core as well) are not uniquely determined by the variational principle, as I mentioned above.

Now, if the goal is to have a unique valence orbital, then it can be achieved by restricting it to be orthogonal to core orbitals. In that case when transforming orbitals you can not mix any amount of any core orbital into the valence one since it will immediately make it non-orthogonal with this particular core orbital. And since mixing in core orbitals is the only way you can change the valence one, orthogonality of the valence orbital implies its uniqueness.

  In order to obtain unique solutions for $\phi_\nu$, one generally
  requires that $\phi_\nu$ be orthogonal to the other occupied orbitals.

And it is important to note (as it is done in the next sentence of the paper) that :

  Although this orthogonality restriction has the desired consequence of
  leading to a prescription for a unique valence orbital, there is no
  reason to consider the orthogonal valence orbital $\phi_\nu$, to have
  a special significance over any other combination of $\phi_\nu$, with
  the various core.

Note finally, that due to the orthogonality restriction you can not also mix any amount of the valence orbital into the core ones, but you can mix core orbitals with themselves anyway you like since it won't do any harm for their orthogonality with the valence orbital. Thus, while the valence orbital is uniquely defined by the orthogonality requirement, the core orbitals are still defined to within a unitary transformation.


Chemistry behind Gale's coffee maker in Breaking Bad Is there a scientific basis for the coffee making equipment which Gale Boetticher describes in Breaking Bad?

He talks about maintaining the right conditions for bringing out the coffee flavor without degrading it.

Yes, there is a scientific basis, but I think you can't do anything with the apparatus shown in the figure. Here you can find much information about it. I will try to summarize.

It seems that extracting coffee at a lower temperature makes it tastes better. Gale thinks that the amount of quinic acid is the key variable that makes good coffee. In fact, some studies (McCamey, D. A.; Thorpe, T. M.; and McCarthy, J. P. Coffee Bitterness. In "Developments in Food Science." Vol 25. 169-182. 1990.) report that bitterness of coffee is due to quinic acid but is not the only substance involved (see this link). I don't have any reference about the dependence between temperature and method of extraction and quantitative of quinic acid extracted.

By the way, the apparatus should extract the coffee in a more effective way. The vacuum pump seems to suggest the use of the vacuum to make the water boil at a lower temperature. But the connections in my opinion make no sense, so I will try to describe it:

All start with an autoclave that is connected with a Gast vacuum pump. This is connected with an Allihn condenser (not the right thing to do because in there you should connect the water used to refrigerate), and then to a heated Erlenmeyer flask that is connected with a strange steel cylinder (I don't know what it is, maybe a filter), then to a Florence flask and then to another steel cylinder (...). The inspiration comes from a Florence Siphon but this is much simpler and is generally something like this:

Here you can read and here you can watch how it works.

Of course this is not so nice and intriguing as the previous apparatus and is not a geeky thing that makes us think we are dealing with a great chemist, so the authors thought to make the apparatus more appealing but completely useless!


What does the negative superscript in the term symbol for ground state Oxygen stand for? I think that I understand most of the parts of the symbol shown for ground state oxygen $\ce{{}^{16}O2}$, which is ${}^{3}\Sigma^{-}_{g}$.

However, I was hoping that someone could tell me what exactly the negative superscript sign means? And how do you identify between a positive and negative state? I think it has something to do with a mirror plane.

You're right that it has to do with a mirror plane. The negative sign means that the overall electronic wavefunction changes sign when reflected in a mirror plane that includes the internuclear axis.

For ground-state oxygen, reflecting in a mirror plane will change the sign of one unpaired $\pi$ electron (the one that sticks out of the mirror plane) but not the other, resulting in an overall change in sign. (This is slightly simplified, a more rigorous explanation would take into account the complex-valued orbitals rather than their real-valued linear combinations.)


Difference between intermediates and transition states Can someone please explain what exactly the difference between an intermediate and a transition state is?

I understand that they are formed as part of the process in converting the reactants to products in a chemical reaction. Other than that, all I know is that usually a transition state exists very briefly (if at all) whereas an intermediate can be isolated. But why?

An intermediate is a short-lived unstable molecule in a reaction which is formed inbetween the reaction when reactants change into products.

Whereas, transition state is just the state before formation of new molecule(involves breaking of bonds of reactants and formation of new ones)

An intermediate differs from a transition state in that the intermediate has a discrete lifetime (be it a few nanoseconds or many days), whereas a transition state lasts for just one bond vibration cycle. Intermediates may be unstable molecules (in which case they are called reactive intermediates) or highly stable molecules.
The difference between them can be better described through the energy profile diagram.

Transition states are local energy maximums and have partial bonds. This might be one of the reasons why they cant be isolated as intermediates.


How can the formula of WD-40 remain a trade secret? Wikipedia says

  WD-40's formula is a trade secret

How can it remain a secret today? Can't mass spectrometers, NMR and other equipment be used to determine the structure of the compounds that go into it?

To put the value of the secret into perspective, the WD-40 company (whose flagship product is its namesake) makes \$76M a year and is currently worth \$2.7B

"How can it remain secret?". Trivial answer is - it can't (it isn't).

WD40 is a Fast Moving Consumer Good (FMCG). It is also a reasonably cheap product - as an industrial chemist this suggests at least two things to me:

It doesn't contain anything rare or expensive
It isn't difficult to manufacture

As an ex-petroleum chemical analyst, with access to a reasonably complete lab, checks to do :

Get the Safety Data Sheet (below is composite information from a number of WD-40 Penetrant SDS)

cas        %
64742-47-8      50-70    isoparaffins petroleum hydrotreated HFP

64742-56-9      30-35    paraffinic distillate, light, solvent-dewaxed

61789-86-4      <5       calcium petroleum sulfonate

111-76-2        1-3      ethylene glycol monobutyl ether

Not Available    1       fragrance
Check the SDS info using physical/chemical testing
If you have the gear available :
Density
Distillation
FTIR
GC (GC-MS if you have it)
AAS/ICP - metals (ICP multielement would be best)
CHOSNP analysis for hetero atoms
Rebuild the product from data. Check to see if it performs the same.
Improve on the product (believe me - WD-40 is NOT the be all and end all of penetrating oils - technology has improved since its formulation).


What makes enantiomers so special? We study enantiomers in detail. But what makes them more important to study than, say, diastereomers? What I'm trying to ask is, enantiomers are simply a type of stereoisomers that can rotate plane polarized light in opposite directions by the same amount. What makes them more important than other stereoisomers?

Enantiomers have different physical arrangement of atoms. This difference plays a crucial role in biology.

  But, why do enantiomers have different biological activities?

  Recognition of chiral drugs by specific drug receptors is explained by
  a three-point interaction of the drug with the receptor site, as
  proposed by Easson and Stedman. The difference between the interaction
  of the two enantiomers of a chiral drug with its receptor is
  illustrated below.

  Source: https://www.khanacademy.org/test-prep/mcat/chemical-processes/stereochemistry/a/chiral-drugs


Reactivity of propellanes Why is [1.1.1]propellane considerably less reactive than either [2.2.1]propellane or [2.1.1]propellane as illustrated in the hydrogenation reactions pictured below?

The heat of hydrogenation reflects the thermodynamic difference in energy between starting material and product. It is a thermodynamic measure of the difference in stability of the products and reactants.  Reactivity, on the other, hand is related to kinetics, a reflection of the activation energy  involved in the rate determining step.

Cyclopropane and cyclobutane have roughly the same strain energy (SE=27.5 and 26.3 kcal/mol respectively). To a first approximation, the strain energy of a polycyclic system is the sum of the strain energies of the individual rings in the system.

Each of the 3 starting materials contains a total 3 (cyclopropane plus cyclobutane) rings.  Since the SE of cyclopropane ~ cyclobutane, we would therefore expect the SE's of the 3 starting materials to be similar and larger than the SE's of the less strained products. In agreement with this expectation the calculated SE's of the 3 starting materials are 109, 106 and 103 kcal/mol respectively (see fig. 2.2 on p. 37 here), indeed very similar.

The SE's of the final products are 17, 41 and 62 kcal/mol respectively (ref, see Table 3.2 on page 72).

If we calculate the strain released during hydrogenation it is in the same order and very close to the actual heats of hydrogenation reported in your figure.

$\mathrm{109-17=92}$ kcal/mol
$\mathrm{106-41=65}$ kcal/mol
$\mathrm{103-62=41}$ kcal/mol

Clearly, in this series of propellanes the heat of hydrogenation is a measure of the strain released in the process.

Because bicyclo[1.1.1] pentane (the right-most product) has a much higher SE than either of the two other hydrogenated products, the heat of hydrogenation given off when it is formed will consequently be much less.

[1.1.1]propellane is an extremely reactive compound (see here for example).  Comparing the thermodynamic heats of hydrogenation in your series of compounds cannot be used to meaningfully predict the kinetic reactivity of these compounds.


How to find the plane of symmetry in biphenyls and substituted cyclohexanes? I was going through my text book, (Solomon&Frhyle)
I came across a couple of of examples which I didn't understand.

1.

How is the molecule symmetrical? Upon folding the molecule over its plane of symmetry  the cooh group will be placed over fluorine.

2.

Similarly in this in structures 3,4,5,6 a chlorine atom out of the plane and below the plane coincide.

3.
It's written that for biphenyls to be chiral a can be the same as a'. (Same for b and b')
           But
If a a' and b b' are the same them won't a plane of symmetry lie in the middle perpendicular to the plane of molecule.

Finding planes of symmetry is simple: try all planes, some (or maybe none) of them will fit.

As for your examples, the first image is just poorly drawn. The plane is supposed to be the one of the ring containing COOH and F, while the other ring is perpendicular to it (which, BTW, I don't believe is the native conformation of the molecule, but whatever).

The second example asserts right things, but for the wrong reasons. (This is not what you would normally expect of a textbook, to put it mildly.) True, each of these structures contains at least one plane of symmetry. Too bad the planes are drawn incorrectly in all structures except for the first two. You may take your time and fiddle around to find out where they really are.

In the third example, you would be right if the molecule were planar, which it is totally not. Steric repulsion between a and a' (also, between b and b') makes that impossible, even if all of the substituents a, a', b, b' are simply hydrogens.


What happens to the lost mass in radioactive decay? We were studying kinetics in class which led me to wonder what happens to the decayed matter. After all, there is a loss in mass and according to mass energy equivalence, there should be some release in energy. So I thought that heat may be released by radioactive decay, but I doubt it.
Can anyone tell me what happens to the lost mass?

On the nuclear physics side, it goes in to kinetic energy of the various particles as well as any gammas. If it is a simple reaction (D+D -> 3He+n or 3H+p) you can determine the energy of each particle in the center of mass by simple kinematics. In something resulting in more particles (most fission events result in 2 nuclei and multiple neutrons) you get a spread of kinetic energies of the various bits. And, if one or more of the nuclei are left in an excited state than you can get emission of gammas, energetic photons.

As these daughter particles interact with the rest of the world, they will be thermalized through interactions with electrons and nuclei surrounding them, transferring kinetic energy, resulting in heat to the surroundings. Any gammas will eventually be absorbed/scattered, also resulting in energy transfer to the surroundings. If this did not happen, fission reactors would have a hard time making steam to turn turbines.


What's the usual agent added to discourage drinking of hazardous substances I've once read somewhere that in materials that are hazardous to swallow, a chemical is used in very small quantities to assure any accidental drinker to retch. This also helps to prevent suicide by drinking. Am I correct about this info or just myth? And what is the name of the substance?

It is denatonium benzoate. This comes from wikipedia page of denatonium:

Denatonium, usually available as denatonium benzoate (under trade
names such as Denatrol, BITTERANT-b, BITTER+PLUS, Bitrex or Aversion)
and as denatonium saccharide (BITTERANT-s), is the most bitter
chemical compound known, with bitterness thresholds of 0.05 ppm for
the benzoate and 0.01 ppm for the saccharide. It was discovered in
1958 during research on local anesthetics by MacFarlan Smith of
Edinburgh, Scotland, and registered under the trademark Bitrex.

With respect to discouraging drinking of hazardous substances it again comes from wiki:

The bitterness of the compound guides most applications of denatonium.
Denatonium benzoate is used to denature ethanol so that it is not
treated as an alcoholic beverage with respect to taxation and sales
restrictions.


What difference does a polar protic solvent make in the addition of X2 to alkenes? After the formation of the cyclic intermediate, $\mathrm{S}_{N}2$ reaction takes place. This reaction would be favored by an aprotic solvent.

How will the products of the reaction be affected if I use a polar protic solvent?

I think that the formation of the cyclic intermediate is independent of the solvent used but I can't figure out what would happen after that.

In the presence of a polar protic solvent, the halogen will react with the solvent to form acids ($\ce{HX}$ and ${HOX}$).

$$\ce{Cl2 + H2O -> HCl + HOCl}$$

The high concentration of the cations formed (the hydrohalic acid and the hypohalous acid will exist as ions in a polar protic solvent) can act as electrophiles and initiate an electrophilic substitution reaction which results in the formation of a range of products.

$$\ce{CH2-CH=CH2 + X2 + H2O -> CH2-CH(Cl)-CH3 + CH2-CH(OH)-CH2(Cl) + CH2-CH(Cl)-CH3(Cl)}$$

This reaction is primarily initiated by the $\ce{H+}$ attacking the electron-rich double bond. This causes an electomeric shift of electrons to a carbon such that the carbocation formed is stable (Markonivkoff's rule). As carbocations are formed, rearrangements are possible. Not only are the mechanisms different, the products will be different too. This reaction is entirely different from a $SN_2$ reaction.


How do I know with which oxidation numbers I can form a certain kind of compound? Take, for istance, phosphorus; it has eight different oxidation numbers, but it can only form two different anhydrides, phosphoric (when the oxidation number is 5) and phosphorous (when it is 3). How do I understand which ones I can use to get anhydrides?
 Or take chlorine, it also has eight oxidation numbers but if I wanted to know the corresponding hydracid, HCl, how'd I know which number to use?

Please forgive me if the question is trivial and written in a poor english, but I'm having trouble with both chemistry and English language classes at present!

You need to be familiar with some of the compounds earlier, otherwise you can't tell if it exists. Remember facts come first before explanation. It also depends usually on the atom/ion to which it is bonded. for example since Fluorine is most electronegative it can form $\ce{XeF6,XeF4}$,etc. but $\ce{XeH3}$ doesn't form also oxygen forms large oxidation states as it is electronegative too and there is also more energy released as many bonds are formed (ususlly pi-bonds.)


Relative energies between reactants and products I've been asked to draw a relative energy sketch of the reactants and the two products. The reaction being examined was adding copper acetate monohydrate to glycine in two different fashions to create the two isomers cis-bis(glycinato)copper(II) monohydrate and trans-bis(glycinato)copper(II) monohydrate. I found that the cis-bis(glycinato)copper(II) monohydrate was  thermodynamically stable and trans-bis(glycinato)copper(II) monohydrate was kinetically favored since when adding heat to trans-bis(glycinato)copper(II) monohydrate it changed to the same color as cis-bis(glycinato)copper(II) monohydrate therefore cis-bis(glycinato)copper(II) monohydrate has a lower energy relative to trans-bis(glycinato)copper(II) monohydrate. The part where I'm stuck is where the reactant energy is relative to these two. I know it must be lower than the intermediate trans-bis(glycinato)copper(II) monohydrate, but where the energy level of the reactants compared to the thermodynamically stable cis-bis(glycinato)copper(II) monohydrate has me stumped, whether it is higher or lower.

Cheers

That's interesting; I just did this experiment (I'm actually writing my lab report up now) and for me the trans-bis(glycinato)copper(II) monohydrate was the thermodynamically favoured product and the cis-bis(glycinato)copper(II) monohydrate the kinetically favoured. As far as the reactant energy levels are concerned, I attached a photo of a rough graph my lab instructor gave us that illustrates the energy levels for the two pathways to the final thermodynamically favoured product. I hope that helps!!


What reactions will their reaction rates increase with time? My textbook states that for most reactions, the reaction rate decreases with time. I wonder what reactions will have increasing reaction rates with time. I know that adding a catalyst will increase the reaction rate but the final reaction rate will finally be zero. So, what reactions will their reaction rates increase with time?

Reactions that are autocatalytic can have their rate increase as a function of conversion.

One example is the decomposition of cellulose acetate by hydrolysis, also known as vinegar syndrome. The reaction is acid catalyzed, and it produces acid. In theory, the reaction gets faster. In practice, diffusion and limiting reagents many prevent a rate acceleration. Acetylsalicylic acid decomposes similarly. In the following equation, $\ce{R}$, could be the cellulose backbone or the salicylyl group. The acetyl group $\ce{-COCH3}$ is represented by $\ce{Ac}$. Thus acetic acid is $\ce{HOAc}$.

$$\ce{R-OAc + H2O ->[\ce{H+ catalyst}] R-OH + HOAc}$$

Another example is the alpha halogenation of ketones or aldehydes, which is acid catalyzed and acid-producing.

$$\ce{CH3-CO-CH3 + Br2->[\ce{H+ catalyst}] CH3-CO-CH2Br + HBr}$$

A historically interesting example is the allotropic decomposition of tin, or "tin pest", in which tin metal transforms to a brittle form in cold weather. The book Napoleon's Buttons, which is about the influence of chemistry (wittingly or otherwise) on history, opens with the conjecture that tin pest is what really stopped Napoleon's invasion of Russia.

In addition, as this answer reminds us, the rate of the reverse reaction increases to a maximum during the establishment of dynamic equilibrium.


Why do nonmetal oxides react with water to form acidic compounds? I am confused about the molecular basis by which nonmetal oxides react with water to form acidic compounds. For example, I understand that sulfur trioxide reacts with water to form sulfuric acid as shown in the following reaction:

SO3 (g) + H2O (l) --> H2SO4 (aq)

I recognize that the polarized nature of the O-H bond in sulfuric acid makes it thermodynamically favorable for H2O to abstract a proton and subsequently form an acidic solution with the formation of H3O+. However, I fail to recognize how reactions like this with a nonmetal oxide proceeds to form an acidic oxide. Namely, I fail to understand the role of the covalent bond in the nonmetal oxide for the mechanism of this reaction. I recognize that metal oxides dissolve in water to form the highly polarizing, high charge density ion O2-, which polarizes the O-H bond in H2O to form hydroxide. Can someone give a similar explanation for nonmetal oxides and their dominant acidic nature in water?

Well, in terms of this reaction, sulfur in $\ce{SO3}$ acts as a Lewis acid for an incoming $\ce{OH-}$ nucleophile. The reaction mechanism seems to be any old nucleophilic addition reaction:

$\ce{H2O <=> H+ + OH-}$ : (Water can disassociate into its ions)
$\ce{SO3 + OH- -> SO2(OH)O-}$ : (I couldn't decide on whether this is the right
way to write this intermediate, but I'll be happy to add a diagram for clarity)
$\ce{SO2(OH)O- + H+ -> SO2(OH)2}$ : (This is the final step of the mechanism to yield $\ce{H2SO4}$)


Why does fluorine act as a leaving group in organophosphates? C-F bond is the strongest single bond in organic chemistry with a bond energy of  $\ce{453 kJ\,mol^{-1}}$. And it is very difficult to break this bond and F does not act as a leaving group.

However, in organophosphates, F acts as a leaving group. For example, take an organophospate such as sarin(below). When this binds with acetylcholinesterase (enzyme), F is leaving as a fluoride ion. How can this be explained w.r.t. the P-F bond energy of $\ce{490 kJ\,mol^{-1}}$, which is even higher than that of C-F. Is this possible because of the catalytic activity of the enzyme?

According to David O'Hagana, Chem. Soc. Rev. 2008, *37 , 308-319, C-F bond is very strong due to the difference of electronegativities of C(2.55) and F(4.0), and the resultant ionic character of the bond. However, the electronegativity of P is 2.19 and according to this, P-F bond must be much difficult to be broken. But, this is not the reality. Sarin can be hydrolyzed (Andrew F.   Kingery  and  Herbert E.   Allen, Toxicological \& Environmental Chemistry 1995, 47, 155-184.) at pH 6 ($\ce{25\,^{\circ}C}$) with a half life of 312 hours in aqueous solution.

I am reluctant to post this as an answer because it is not really backed up by much research, but since nobody has said anything yet, here's my theory:

  It is probably to do with the fact that the covalently bound sarin (which necessarily involves kicking out either $\ce{RO-}$ or $\ce{F-}$) mimics the transition state for acetylcholine hydrolysis too well, such that the expulsion of one of those two leaving groups is much more favoured than in aqueous solution.

Acetylcholinesterase (AChE) is very similar to a typical serine protease, and serine proteases are very well-studied. I am going to postulate that the mechanism of acetylcholine (ACh) hydrolysis in AChE is exactly the same as it is in a typical serine protease. So first, let's look at what the enzyme does normally. It's nothing special, just a nucleophilic acyl substitution:

(There's an additional glutamate residue, analogous to the aspartate in a serine protease, that fixes the conformation of His440, but it's not directly relevant to the mechanism. Also, there's an anionic site which binds to the positively charged quaternary nitrogen, but that's also irrelevant.)

The acyl-enzyme intermediate then goes on to be hydrolysed by water (not relevant here).

The reason why the enzyme can catalyse the reaction is because it stabilises the transition state for the ester hydrolysis, which means that the activation energy for hydrolysis is decreased and the rate is increased. (Strictly, it provides an alternative hydrolysis pathway in which the activation energy is lower - but the point is that since the catalysed hydrolysis is mechanistically similar to the uncatalysed hydrolysis, the enzyme must play a role in stabilising a tetrahedral intermediate or transition state.)

The transition state looks something like this (partial charges omitted):

Sarin is a potent AChE inhibitor because it covalently binds to the active serine residue as in the left-hand side of this diagram:

You can see that structurally, this is very similar to the transition state shown above. If the enzyme stabilises the TS above, this covalent binding of sarin to the enzyme must also be extremely stable. I do not pretend to know exactly how it is stabilised, but it must be related to the specific 3-D conformation of the enzyme. If you do not kick out the fluoride ion (right-hand side), then it does not quite resemble the transition state so well.

What this means is that there is a strong driving force for the fluoride ion to be expelled, in order for it to reach that stable, 4-coordinate phosphorus form.

That leaves one question that I feel is worth addressing, which is why fluoride is expelled instead of isopropoxide. Honestly, I do not know. As pointed out by jerepierre, fluoride is certainly a better leaving group than isopropoxide ($\mathrm{p}K_\mathrm{a}(\ce{HF}) = 3.17$, $\mathrm{p}K_\mathrm{a}(\ce{^{i}PrOH}) = 16.5$ - quite a large difference), and it might very well just be that. On the other hand, as with most biological systems, it is difficult to give a full explanation of why things happen the way they do, and applying (relatively) simple chemistry heuristics might not be the full picture. As to what other factors might influence this, I can only guess that is that there might be some hydrogen bonding arrangement that requires the specific size of the oxygen in the $\ce{-OR}$ group over the size of the fluorine (in the catalysis TS, there is an oxygen there as well). Essentially, I am saying that the LH diagram resembles the TS more than the RH diagram (don't quote me on it though...):


What is the formula of bleaching powder? What is the structure and preparation for bleaching powder?  Why there is a controversy in the formula of bleaching powder?

I have encountered many forms of the formula of bleaching powder, like

$\ce{Ca(OCl)2}$ - wikipedia (it also mentions that bleaching powder is actually a mixture having other calcium compounds in it)
$\ce{Ca(OCl)Cl}$ - my textbook
$\ce{CaCl2 + Ca(OH)2 + H2O -> CaCl2\cdot Ca(OH)2\cdot H2O}$ - here the product is "apparently" bleaching powder.

My question is posed in other sites like:-

http://www.chemicalforums.com/index.php?topic=48844.0 -
here it reads that the composition of bleaching powder has been disputed for decades.
http://chem-guide.blogspot.in/2010/04/preparation-properties-and-uses-of_6609.html - many forms  of the formula are given here in this site although none of it has been proved right! Also can you explain about this Odling concept of bleaching powder mentioned in this site?

The bleaching agent in all commonly available bleach powders is hypochlorite.  Calcium hypochlorite is a pure compound with the formula $\ce{Ca(OCl)2}$ and is the active component in most bleach powders.  (Note: active does not necessarily mean the most abundant component).

You presented three examples of formulations for bleaching powder that you have encountered:

  $\ce{Ca(OCl)2}$
  $\ce{Ca(OCl)Cl}$
  $\ce{CaCl2 + Ca(OH)2 + H2O -> CaCl2\cdot Ca(OH)2\cdot H2O}$

The third species, $\ce{CaCl2\cdot Ca(OH)2\cdot H2O}$, cannot be a bleaching agent of any type because it does not contain hypochlorite.  So I view that as plainly incorrect.

According to wikipedia,

  Bleaching powder is made with slightly moist slaked lime. It is not a simple mixture of calcium hypochlorite, calcium chloride, and calcium hydroxide. Instead, it is a mixture consisting principally of calcium hypochlorite $\ce{Ca(OCl)2}$, dibasic calcium hypochlorite, $\ce{Ca3(OCl)2(OH)4}$, and dibasic calcium chloride, $\ce{Ca3Cl2(OH)4}$ [7].

I was able to find the source cited by Wikipedia online; it seems quite authoritative.  It says:

  Calcium hypochlorite solutions are made by suspending lime or hydrated lime in water and adding
  chlorine. However, reacting lime with water generates lots of heat, and this heat must be allowed to
  dissipate before chlorination. The maximum concentration is ∼85 g/L
   of available chlorine, above
  which dibasic calcium hypochlorite begins to precipitate.
  Chlorination can be done in a batch or continuous reactor similar to those used to make sodium
  hypochlorite, except that agitation is needed to maintain the suspension. In a batch reactor, agitation
  can be achieved using a mechanical stirrer, or, more effi ciently, by circulation. A centrifugal
  pump is used to pull the solution from the bottom of a conical tank and then reinject it tangentially
  to create a strong vortex. Alternatively, the solution passes through a packed tower before it returns
  to the tank. The packing in the tower creates shear that breaks the particles apart before the chlorine
  is injected at the bottom of the tower.57 The reaction is as follows:
  $\ce{2Cl2 + 2Ca(OH)2 → Ca(OCl)2 + CaCl2 + 2H2O}$

  [...]

  Hemibasic calcium hypochlorite, $\ce{Ca(OCl)2\cdot 0.5Ca(OH)2}$, can be precipitated from a solution of calcium hypochlorite at 40–80°C.143 These solutions are commercially made by chlorinating a
  slurry of hydrated lime.5
   The precipitate is removed by filtration and dried. Although it may still be
  made in Asia, it has not been made in the United States or Germany since 1955.

  [...]

  Dibasic calcium hypochlorite, $\ce{Ca(OCl)2 · 2Ca(OH)2}$, is commonly used as an intermediate in making calcium hypochlorite, $\ce{Ca(OCl)2}$.   It has never been successfully marketed as a product because it dissolves much too slowly.

[...]

  Although calcium hypochlorite dihydrate, $\ce{Ca(OCl)2 · 2H2O}$, can be made by chlorinating a suspension
  of lime, it is difficult to filter, partly due to the presence of calcium chloride. Calcium chloride also reduces the stability of the dry product. [...] Sodium hypochlorite is commonly added to convert calcium chloride to calcium hypochlorite
  as follows: $\ce{CaCl2 + 2NaOCl → Ca(OCl)2 + 2NaCl}$

Pages and pages follow with detailed process descriptions for the manufacture of various formulations of bleaching powder.  Suffice it to say, an ideal powder would be something like pure $\ce{Ca(OCl)2 · 2H2O}$, but this is very difficult to prepare economically, because of contamination with various amounts of chloride salts, hydroxide salts, or (depending on the route of manufacture) sodium salts.


how to identify reaction medium by looking at the equation ( for h2o2)? $\ce{Ag2O + H2O2 -> 2Ag + H2O + O2}$
My book says the medium is alkaline as $\ce{H2O2}$ gives $\ce{O2}$ on reduction only in alkaline medium only but $\ce{O2}$ is produced in acidic medium too. Is there really a way of recognizing the medium by looking at reaction equations??

Is there really a way of recognizing the medium by looking at reaction equations??

Yes and no...

For the particular reaction, you have to know some chemistry. In acid solutions hydrogen peroxide acts as an oxidizing agent: $$\ce{H2O2 + 2H^+ +2e^- -> 2H2O}$$

In neutral and basic solutions hydrogen peroxide acts as an reducing agent: $$\ce{H2O2 -> 2H^+ +2e^- + O2(g)}$$

The silver oxide is being reduced so the hydrogen peroxide must be acting as a reducing agent.

The real answer is that the balanced reaction isn't really enough information to duplicate the reaction. What temperature? How long for the reaction to take place? Since reaction would use excess hydrogen peroxide, how much was in excess and so on...

PS - Sometimes you have to consult the oracle to know the laboratory conditions. I asked Carnac the magnificent for more information.

This reaction was done by a red-headed girl. She was wearing a white lab coat over a red blouse and a blue skirt. She did have safety glasses which is a really sexy look over her hazel eyes in the chem lab. She was also wearing white leather sneakers.


Why there isn't any unimolecular addition reaction? In my textbook, I have read about unimolecular elimination and nucleophilic unimolecular substitution reaction of organic compounds. In those reactions, firstly one bond breaks automatically then as soon as a carbocation forms then a nucleophile attacks the carbocation to complete the reaction.

Now, my question is: Why doesn't the same happen in case of addition reaction of molecules having double bond? One bond breaking may cause a charge separation and then nucleophile/electrophile can attack the positive or negative charged side? I know molecules with charge separation aren't supposed to be much stable. But even in the case of a carbocation the same happens.

I have Googled for unimolecular addition reaction but all the results were about unimolecular substitution/elimination.

The problem I see with your question is that you mix up two completely different concepts. The first concept is the reaction order (i.e. order of the kinetics, for example unimolecular = first order) which is dependent on the number of components that take part in the rate determining step of the reaction. The other concept is the type of the reaction, e.g. addition, substitution, functional group interconversion.

Naturally, most addition reactions will be of the type $\mathrm{A + B \rightarrow C}$ or something like that, but that neither means that they will be of second order kinetically, nor that all addition reactions are of that type. Intramolecular additions would be one example for a unimolecular addition.

Please be careful though not to mix up the order of the reaction and the stoichiometry of the written down reaction.

Imagine a reaction were you have $\mathrm{A+B \rightarrow C}$ and $\mathrm{B}$ is used as the solvent, so that $\mathrm{B}$ is available in large excess. While there are clearly two components needed for the addition to proceed it will appear to be unimolecular (first order reaction).

Strictly speaking, the stoichiometry of the reaction has nothing to do with the reaction order.

However, most addition reactions can be conceptualized to be of the form $\mathrm{A+B \longrightarrow C}$ whether that means that $\mathrm{A}$ and $\mathrm{B}$ are part of the same molecule or different chemical entities or if the rate law will actually appear to be of second order as well then depends on the specific case.

In that regard, I don't really see the point in asking whether an addition reaction might be unimolecular, as the kinetics will depend on the specifics of the reaction while the stoichiometry can be inferred just by looking at the equation. I hope that clearifies it.


What is the active ingredient in this "stimulant?" It is an easily defensible argument that this is a medical question rather than a chemistry one, however, I feel that it leans ever so slightly towards chemistry, and so I posted it here.

There is a "new" OTC stimulant being pushed around that shall remain unnamed so that this isn't flagged as an advertisement or anything.  In looking at the list of ingredients, I don't see anything that would act as a stimulant anywhere in that list.  Having said that, I'm not an expert.

So, I came to the experts.  ;)

Here are the ingredients according to the bottle:

Just to clarify, the heart of this question is "Which ingredient is the stimulant?"

If you google Schizandrol A ( 1,2,3,10,11,12-hexamethoxy-6,7-dimethyl-5,6,7,8-tetrahydrodibenzo[a,c]cycloocten-6-ol, here's a link to a picture of the structure) you find that it is an herb capable of curing just about anything.  If you poke around a little deeper, you find two repetitive themes 1) it increases focus and 2) synergistically enhances the effect of other chemicals ingested along with it.  You also note this herb has caught the attention of the bodybuilding community.  The lifters use terms like "pump and rush", "being on another planet", etc.  Apparently it can take focus to a whole new level. From what I read it looks like it's always ingested with other stuff so it's hard to separate out what's what.  My guess is that all the ingredients in the bottle come into play and the effect they have together, particularly with the Schizandrol A present, is much greater than any ingredient alone.


Flame Test of Alkali Metals and Alkaline Earth Metals The flame test colors for alkali metal and alkaline earth metal salts are well known, but what about the metals themselves? Do they also give rise to the same colors (eg, brick red for Ca, pale green for Ba etc), or does it differ? And if so, why?

Yes, the metals, when sent in a flame produce the same colors as their salts. In fact, the metallic ions do not produce any color when sent in a flame. It is only the neutral atom that produce the famous yellow double line of the sodium. This means that a tiny amount of the $\ce{Na+}$ and $\ce{Cl-}$ ions are first transformed into neutral $\ce{Na}$ and $\ce{Cl}$ atoms. And the color of the flame is due to the deexcitation of these rare atoms being first electronically excited in the flame. So it means that the following processes are occurring successively in the flame:$$\ce{NaCl(s) -> Na(g) + Cl(g)}$$ $$\ce{Na + heat -> Na^*}$$ $$\ce{Na^* -> Na + h\nu }$$ The famous yellow light of the $\ce{Na}$ atom is due to the return of the outer electron from the level $4s$ to $3p$ of the sodium atom, if my memory is good. Nothing to do with the $\ce{Na+}$ ion.


How do I scale up to semi-prep from an analytical HPLC I am looking for an equation that would allow me to change the variables injection volume, column size, and flow rate scaling up from an analytical HPLC to a semi prep instrument.

Best practice is to try and maintain the same column length between scales.  Then the scale up factor is simply the ratio of the square of the radii.

If the lengths are different,  try and keep the aspect ratio in the same ballpark e.g. if the length to diameter ratio is not the same but between 4 to 5 for both, you're in a better place. Then simply scale by the ratio of the two column volumes, using the radii calculated from the internal diameters.

Using the ratio of column volumes should work in other circumstances but scaling from a long, thin column to a pancake may cause some issues.

Make sure to keep your particle size of your stationary phase the same too so you have similar packing.
All things being scaled appropriately,  the back pressure should be the same too.


Do acids stay acid forever? I mean their pH value never change unless we dilute with water etc?

  An acid is a molecule or ion capable of donating a proton (hydrogen
  ion H+) (a Brønsted–Lowry acid), or, alternatively, capable of forming
  a covalent bond with an electron pair.

So acids stay acids forever?

So acids stay acids forever?

You know a company markets that diamonds are forever. Well may be not, a diamond is not thermodynamically stable. It may wish to turn back into graphite if you somehow provide the right activation energy.

So there is nothing special about acids. Acids are like other molecules some may remain stable for a long time and some may not.

An example is that of sulfuric acid. If someone leaves concentrated acid in an open beaker, in period of few months its volume will increase. How? It absorbs moisture from air.

If you leave conc. HCl bottle open, HCl gas will escape into air and the acid concentration will decrease.

Moral of the story: Nothing is permanent but change.


What happens to the H electron during acid reaction? In this generic acid-base reaction:

$\ce{HX + B <=> X- + HB+}$

It's explained that the acid HX donates a proton H+ to its conjugate base HB+.  What happens to the electron that was originally in the acid's H atom?

The electron belonging to elemental hydrogen forms part of a covalent bond with $\ce{X}$. In this acid-base reaction the $\ce{HX}$ bond breaks heterolytically, meaning that both electrons constituting the bond go to $\ce{X-}$, hence the negative charge.

The base then donates electrons to the proton, forming the electron deficient species $\ce{HB+}$.


Sulfur from pyrite roasting process I'm solving a quantitative problem involving the process of making oleum from pyrite.

The problem specifies that, at the end of the reaction, we find sulfur which does not come from the impurities present in pyrite.

Where is that sulfur coming from? Is it from the thermal decomposition of $\ce{FeS2}$?

$$\ce{FeS2 -> Fe + 2S}$$

Guilin Hu, Kim Dam-Johansen, Stig Wedel, Jens Peter Hansen. Decomposition and oxidation of pyrite, Progress in Energy and Combustion Science, Volume 32, Issue 3, 2006, Pages 295-314, ISSN 0360-1285, http://dx.doi.org/10.1016/j.pecs.2005.11.004.

    "In inert atmospheres, the transformation of pyrite proceeds through a multi-step process in the sequence of pyrite→pyrrhotite→troilite→iron. The whole process is determined by the temperature and the total sulfur gas pressure in the system."

Since there isn't any oxygen in an inert atmosphere, the sulfur in the pyrite must be converted to elemental sulfur to maintain charge neutrality and mass balance as you proposed.


Why do I see two peaks for methyl salicylate in my GC–MS spectrum? I ran a neat sample of methyl salicylate (CAS: 119-36-8) on the GC–MS and used $m/z~120$ as the quantifier and $m/z~92$ and $152$ as qualifiers. Below is the SIM spectrum for these ions, with two distinct peaks.

Both peaks are identified by the NIST spectrum database as that of methyl salicylate. Considering this is a neat sample and is not contaminated or diluted, why do I see two peaks in my spectrum?
GC parameters: $\pu{60 °C min^-1}$ ramp to $\pu{230 °C}.$

The short answer: Do not rely on the SIM trace alone.  Amplify the representation of the trace about the TIC (total ion current) until you see a reasonable peak  intensity at the same retention time as in your SIM traces (perhaps a time spectrum, but not a mass spectrum).  Then integrate with the peak-shaped cursor (currently shown in the magenta coloured trace of one SIM) in the trace of the TIC and equally consult the then displayed EI scan at this time interval (in part already seen at the lower part of your photo) to assess the mass spectrum.  Compare this to a trustworthy reference spectrum; e.g. an authentic sample; your spectrometer's library; NIST, SDBS, etc.
The longer answer:
Your experiment coupled a separation technique, either GC, with an analytical, MS.  Often, the mass resolution of these simpler, less-boxy bench-top MS hyphenated to the chromatography (i.e., not a tandem MS-MS detector) is low compared to, e.g., a sector field mass spectrometer.  Thus, observing the SIM traces alone is more like asking «tell me if whatever ion passes the detector with a $m/z = 119.2$», or even worse «tell me if any ion passes the detector in the range of $m/z = 119 - 121$.  You will see this approach often used (sometimes misused) in HPLC-MS if one is not sure about the setup's sensitivity for $\ce{M+}$ vs. an adduct like $\ce{[M + H]+}$.  Or because the software allows you to check for an $\ce{NH+_4}$-adduct after the chromatography run is recorded because your eluent included an ammonium buffer.
Only if you couple the information by the SIM with the integration in the TIC to yield a complete mass spectrum you may collect additional evidence that the peak in the SIM may belong to your target compound.  Otherwise, it could be (but need not to be) caused by an ion which accidentally shares the same ratio of $m/z$ within the setup's resolution in time (think elution of chromatography, scan rate of the MS), and mass (the MS) to your fragment in question; but shows up regardless if it belongs to the defragmentation pattern of your target compound, or not.  Without additional inspection of the spectra it equally may be an impurity, e.g. the sample just before yours was too concentrated and some is eluated (and subsequently characterized by the MS) now altogether with your sample.
Without the separate inspection of these integrated spectra around $t_R = \pu{4.1 min}$ and $\pu{5.1 min}$, the present form of your question does not allow to answer if one of these peaks (and if so, which one) safely may be attributed to methyl salicylate.  A mixture of isomers of compounds, e.g. (cis)/(trans) dienes may elute at different times and -- depending of the $m/z$-ratio you set for the SIM -- equally yield two peaks in the SIM but distinctly different mass spectra.


Diamond filtering - is it real or bogus? I always hear that some big corporations use "diamond dust" to filter water seven times to make "diamond clear" vodka. I tried googling something about using diamonds for filtering, but I could not find anything except corporate sites that tell us only the best things about "diamond dust filtration technology".

Is this just a bogus claim? Because I can not think about diamonds as a viable filtering material.

It appears that there is some validity to this technology.  Not surprisingly, the published journal literature available is basically non-existent, except for a handful of patent applications describing something along those lines.

The title of one journal publication "Preparation of a filtering material using nanomaterials for removal of a heavy metals from wastewater" uses a diamond nanopowder to aid in filtering out heavy metals from waste water.  The citation is "Antonenko, L. P., Energotekhnologii i Resursosberezhenie, 6, 2011, 49-54"

Additionally, there is a patent appplication from France (where at least one article I read cited as the technology originating from), title "Apparatus for activation and mineralization of water for domestic use".  In this, they use a variety of powdered gemstones (diamond, ruby, sapphire) as well as precious metals (gold, silver) to filter the water.

Certainly, nano-particle scale materials are effective filtration agents, as most contaminants will not pass through such a small sized filtration device.  The use of some gemstones (certainly diamond) are beneficial for their relative chemical inertness, and certainly silver is well-known for it's antimicrobial activity, and synthesis of gold-nanoparticles or gold-silver coatings are also actively under research


Dipole moment of CH2Cl2 and CHCl3 Why does $\ce{CH2Cl2}$ have a higher dipole moment than $\ce{CHCl3}$?

My argument being that the $\ce{CCl3-}$ ion being rather stable, can generate a good amount of polarity in the bond, moreover, shouldnt the dipoles of both $\ce{Cl}$ atoms cancel out?

As the dipolar moment is a vectorial property, you have to take into account both the number of chlorine atoms around the carbon atom, as well as their relative arrangement -- among them, as well in respect to the carbon atom.

Lacking access to experimentally determined data about the molecular geometries,you may sketch the molecules in question, for example with Avogadro, and optimize the geometry (here, MMFF94 was deployed). You are literally able "to walk around" the molecule and its dipolar moment.

You easily will recognize a feature in common -- the similar direction of the resulting dipolar moment.  Simultaneously, you will recognize the different angle defined by Cl-C-Cl, too.  However, beside the angle, distances like C-Cl and C-H are affected by the different steric demand and electronic properties, too; equally influencing the overall dipole moment.  Hence the molecule of dichloromethane is more polar than the one of chloroform.

Sidenote: a force field like MMFF94 is a computationally cheap approach to optimize molecular structures.  There are better methods, but more costly; for the level of the question addressed, however, I thought it already provides the insight necessary here.


What is the pH of ice? The pH of pure liquid water depends on temperature. It is about pH = 7.0 at room temperature, pH = 6.1 at 100 °C, and pH = 7.5 at 0 °C. What happens to the pH (or to the ion product) of pure water when it freezes?

I assume that the proton transfer reactions
$$\ce{2H2O <=> H3O+ + OH-}$$
$$\ce{H3O+ + H2O <=> H2O + H3O+}$$
$$\ce{H2O + OH- <=> OH- + H2O}$$
are too fast, so that any present $\ce{H3O+}$ and $\ce{OH-}$ cannot be easily trapped in the solid ice crystal when it grows. Does that mean that pure ice crystals are free of $\ce{H3O+}$ and $\ce{OH-}$ ions?

According to Martin Chaplin's Water Dissociation and pH:

  In ice, where the local hydrogen bonding rarely breaks to separate the constantly forming and re-associating ions, the dissociation constant is much lower (for example at $-4~\mathrm{^\circ C}$, $K_\mathrm{w} = 2 \times 10^{-20}~\mathrm{mol^2~L^{-2}}$).

So $[\ce{H+}] = 1.4 \times 10^{-10}~\mathrm{mol\ L^{-1}} \Longrightarrow \mathrm{p[\ce{H+}]} = 9.9$

For more information see Self-Dissociation and Protonic Charge Transport in Water and Ice Proceedings of the Royal Society of London. Series A, Mathematical and Physical Sciences Vol. 247, No. 1251 (Oct. 21, 1958), pp. 505-533

This is a review article by Nobel Prize winner Manfred Eigen , after whom hydrated $\ce{H3O+}$ is sometimes referred to as the Eigen Ion.


What is the use/function of lithium chloride as an excipient in sofradex eardrops? Certain drugs have lithium chloride as an excipient, eg. Caprilate, NeyDop N (Revitorgan-Dilutionen N Nr 97), NeyDop N (Revitorgan-Lingual Nr 97), Nisita.

An excipient is not supposed to have a pharmacological effect, but it seems like lithium chloride is an inhibitor of various enzymes (http://www.ncbi.nlm.nih.gov/pubmed/15107743).

Probably the main reason $\ce{LiCl}$ is used in liquid to soften ear wax is that $\ce{LiCl}$ is quite hygroscopic and pharmacologicically inactive under the conditions used (small amount used to retain moisture).

As you state, taken internally, in contact with the eyes, or left in contact with skin for extended time, there are some biological effects, but I'd not be concerned about those with a tested commercial product.


Effective nuclear charge of oxygen atom (O) vs oxygen anion (O2-) The size of an anion is greater compared to its parent atom because former's effective nuclear charge is lesser than that of latter.

I found on wikipedia that the effective nuclear charge can be calculated by the formula:

Zeffective = No. of protons in the nucleus - No. of non-valence electrons

For oxygen atom, the electronic config is: 1s2 2s2 2p4

For oxygen anion, the electronic config would be: 1s2 2s2 2p6

Zeffective for oxygen atom = 8 - 2 = 6

Zeffective for oxygen anion = 8 - 2 = 6

If the effective nuclear charge is staying the same, then how it can be said that the size of an anion is greater than that of its parent atom?

The size of an anion is greater compared to its parent atom because former's effective nuclear charge is lesser than that of latter.

This is not the best explanation. The size of the anion is larger because, as Poutnik wrote in the comment, the effective nuclear charge is almost the same (if you account for nuclear charge and inner electrons only), and there is more repulsion among the valence electrons.

The size of fluorine atoms is smaller than the size of sodium atoms because the effective nuclear charge is larger in fluorine  than in sodium.

This is the typical explanation for the biggest jump in atomic sizes. It makes sense that as you add electrons to a given nucleus, the particle gets larger. It makes sense that going down a group, atoms get larger. What needs an explanation is the dip in size going within one period.

So comparing F and Na (or any other group 17 element and the following group 1 element), we go from a high effective nuclear charge (nucleus and inner electrons combined have a +7 charge) to a low effective nuclear charge (nucleus and inner electrons combined have a +1 charge). On top of that, the outer electrons are also in the next higher shell in Na (3rd period) compared to F (2nd period).

Why are we switching explanations?

Depending on which two species you compare in size, some things change and some things stay the same. Comparing $\ce{O, O- and O^2-}$, the nucleus and inner electrons don't change, so you focus on the number of electrons in the outer shell. Comparing isoelectronic species such as $\ce{O^2-, F-, Ne, Na+, Mg^2+}$, the number of electrons stays the same, but the nuclear charge (and the effective charge) change, so you would focus on that. Comparing atoms in a group, e.g. $\ce{Li, Na, K, Rb}$, the effective nuclear charge is similar (corresponding to similar chemistry of the valence electrons) and you would focus on the valence electrons occupying higher and higher shells.

Comparing neutral atoms in order of atomic number, e.g. $\ce{O, F, Ne, Na, Mg}$, everything changes. Within a period the increasing effective nuclear charge dominates (making the atoms smaller and smaller). Jumping from one to the next period, the drastic decrease in effective nuclear is the most pronounced effect (in addition to using the next shell), resulting in a larger radius.


Why are nitrogenous hydrocarbons sometimes colorful? Alanin and dyes contain nitrogen, is being colorful a special feature in nitrogenous substances?

Introduction from Color Introduction

"What causes a compound to appear as being colored? The answer to this question lies in the way that electromagnetic energy interacts with matter. The portion of the electromagnetic spectrum that is of interest to this discussion is the visible portion, or what we call light. This visible light includes all the colors that we perceive: violet, blue, green, yellow, orange, red, and the various combinations of these. Although light can interact with matter in several different ways, this discussion will limit the modes of interaction to absorption and reflection."
In most cases color comes from the excitation of electrons due to an absorption of energy performed by the chemical. But - there is still ... :)
"(...) The vast majority of simple inorganic (e.g. sodium chloride) and organic compounds (e.g. ethanol) are colorless. Transition metal compounds are often colored because of transitions of electrons between d-orbitals of different energy.
Organic compounds tend to be colored when there is extensive conjugation, causing the energy gap between the HOMO and LUMO to decrease, bringing the absorption band from the UV to the visible region. Similarly, color is due to the energy absorbed by the compound, when an electron transitions from the HOMO to the LUMO.(...)" Wiki - color of chemicals

You can also assume that if a compound has double or triple conjugated (2 or more) bonds it can be colorful.
To answer your question: in alanine there is almost no chance to make double bond in amine group (e.g. $\ce{=NH}$), but nitrogen group ($\ce{O-N=O}$) has a capability to conjunct with other double bonds (can be also delocalized electrons or pi electrons - e.g., nitrobenzene)


pKa and pH behaviour of a drug lignocain Lignocain is a weak base with pKa of 7.9. if injected into an acidic ( low pH medium or tissue ) it does not act adequately. Reason probably is that it becomes more dissociated and ionized drugs donot penetrate tissues adequately.
But according to relationship between pH and pKa in this forum it is given to understand that substances with high pKa in low pH solution get protonated or undissociated ? why the difference in interpretation?

If I understand your question correctly, this should answer it:

An "uncharged" base with a high pKa is protonated to form an ionic species:
$$\ce{B + HX <=> HB^+ + X^-}$$
$$Ex:\ \ce{NH3 + HCl <=>> NH4^+ + Cl^-}$$

A negatively-charged base with a high pKa is protonated to form an uncharged species:
$$\ce{B^- + HX <=> HB + X^-}$$
$$Ex:\ \ce{OH^- + HCl <=>> H2O + Cl^-}$$


Why can one product in an equilibrium system not be produced without the other? I performed an experiment where $\ce{CuSO4 (aq)}$ reacted with $\ce{NH3 (aq)}$ in one test tube, and in the other, $\ce{CuSO4 (aq)}$ reacted with $\ce{NH3 (aq)}$ and $\ce{NaOH (aq)}$. I know two products were formed, one being the complex ion $\ce{[Cu(NH3)4]^{2+}}$.

The goal of the experiment was to see if one of the products could be produced without the other, and I know that is not possible due to the solutions being at equilibrium, but why is this?

In the first experiment, the dissolved $\ce{CuSO4}$ reacted with $\ce{NH3(aq)}$ to form the soluble, deep blue/purple complex via equation 1, $\ce{[Cu(NH3)4]^{2+}}$, just as you said.

1) $\ce{Cu^{2+} + 4NH3 <--> [Cu(NH3)4]^{2+}}$

Depending on the concentration of $\ce{NH3}$, you would also have precipitated some amount of insoluble $\ce{Cu(OH)2}$ from the hydroxide formed by the ammonia reacting with water as in equation 2:

2) $\ce{NH3 + H2O <--> NH4+ + OH-}$

The equilibrium of equation 2 lies far to the left, so when you add more ammonia, you aren't adding an equal amount of hydroxide.

The difference in the second experiment is that you explicitly added excess hydroxide.  In this case the high hydroxide concentration would have pushed the dominant product to $\ce{Cu(OH)2}$ via equation 3.

3) $\ce{Cu^{2+} + 2OH- <--> Cu(OH)2}$ (solid)

Regarding the statement "The goal of the experiment was to see if one of the products could be produced without the other":

The products of both experiments were the same, just to different degrees; in other words, under different pH conditions, different products predominated.  In the first experiment, you had far more ammonia than hydroxide, so equation 1 dominated; lots of $\ce{[Cu(NH3)4]^{2+}}$ and a small amount of $\ce{Cu(OH)2}$.  In the second experiment, you added excess hydroxide so lots of $\ce{Cu(OH)2}$ was formed an very little $\ce{[Cu(NH3)4]^{2+}}$.

So the upshot is that these reactions are in equilibrium, and because of equation 2 you always had both $\ce{NH3}$ and $\ce{OH-}$ in your reaction mixture so that both equations 1 and 3 always produced some small amount of their respective products, $\ce{[Cu(NH3)4]^{2+}}$ and $\ce{Cu(OH)2}$, respectively.


Effects of atmospheric gases on colour of aurora Recently I moved to the north of Norway, so I've seen quite a few auroras. Because the colour is mainly green, I was sure this must be due to electron transitions in $\ce{N2}$ (because the atmosphere is like $75\%$ $\ce{N2}$)
Looking into this, however, I found out that the colours of the aurora are mainly due to electron transitions in oxygen gas. Why is it that the electron transitions of oxygen are most prevalent in the colours of the aurora, even though it is far less abundant than nitrogen?

I also read that oxygen gas gives one colour up to $150$ miles altitude(green), and another above $150$ miles (red). Can someone explain the reason for the difference in colour? Does oxygen exist in different oxidation states at different altitudes, leading to different electron configurations => different available electron transitions?

There is good information at Glowing Gases - Aurorae

There are many factors that need to be considered.

Once an atom or molecule is excited, it can lose the energy by collision or by emission of light.

The longer the lifetime of the excited state, the higher the altitude is required to make radiation vs. collision the way energy is lost.

The atomic oxygen transition with green light has a lifetime of about 1 second, while the transition for red light is 110 seconds, so a higher altitude is required for the red transition to become significant.

For molecular nitrogen, radiative lifetime is only $4 \times 10^{-8}$ seconds according to Electronically excited molecular nitrogen and molecular oxygen in
the high-latitude upper atmosphere Ann. Geophys., 26, 1159–1169, 2008

Also, human eyes are more sensitive to green light than red light.


Creating an exothermic reaction through exposure to oxygen Is there a way to create a label that when ripped or torn, it's triggered by oxygen and creates an exothermic reaction that heats the object it's surrounding? Say just like a cold cup of coffee or something?

There are products like this already on the market.  A product called Hand Warmers are gloves that make use of exothermic reaction to warm your hands.  And there are cold packs (that make use of endothermic reactions) to put on sports injuries.


Can Hydronium Act as a Ligand? Can hydronium function as a ligand? For instance could there exist $\ce{[Al(H3O)4]^{+7}}$ or $\ce{[Fe(H3O)6]^{+9}}$ cations? I asked my chemistry teacher and he said that he's never encountered a coordination compound with cationic ligands (however I know that there are because $\ce{NO+}$ can act as a ligand). What are the properties of these cations, if they exist? How are they synthesized? And finally, how would you name these complexes?

This is not the answer OP would expect because I couldn't find metal complexes of type $\ce{[M(H3O)_x]^{y+}}$ (the reason for non-existence(?) is mentioned in the comments) but I did find a complex of an organic compound where hydronium ion act as a ligand i.e a complex of 18-crown-6 and hydronium ion(emphasis mine):

The observation that cyclic crown ethers can bind cations with high
affinity and selectivity has had a tremendous impact in chemistry. The
complex between 18-crown-6 (18c6) and the hydronium ion, $\ce{H3O+}$,
is a prototypical representative of this important host−guest paradigm
and has been the subject of numerous experimental and theoretical
studies. The computational evidence presented so far has favored a
structure with three linear $\ce{O−H···O}$ hydrogen bridges as the
lowest, stable minimum, whereas a bifurcated arrangement has turned
out to be a transition state for $\ce{H3O+}$ rotation 4−5 kcal/mol
higher in energy.

The closest metal complex I could find where hydronium ion is present as an entity is hydronium zinc(II) pyrazine-2,3-dicarboxylate(again not the type of complex OP is looking) where 2,3-PZDC (μ-pyrazine-2,3-dicarboxylato-N,O,O′) act as ligand(emphasis mine):

The structure of dihydronium
[catena-bis(μ-pyrazine-2,3-dicarboxylato-N,O,O′)zinc(II)],
$\ce{(H3O)2[Zn(2,3PZDC)2]}$, is composed of polyanionic ribbons of
zinc(II) ions linked by double bridging 2,3-PZDC ligand molecules.
Each ligand uses an N,O bonding moiety formed by one carboxylic group
[$\ce{Zn–O}$ 2.071(2) Å; $\ce{Zn–N}$ 2.184(2) Å] and a monodentate oxygen atom of
the other carboxylate group [$\ce{Zn–O}$ 2.092(2) Å]. Coordination around the
zinc(II) ion is strongly distorted octahedral. Hydronium cations
$\ce{H3O+}$ link the ribbons by hydrogen bonds.

References

Hydronium Ion Complex of 18-Crown-6:  Theory Confirms Three “Normal” Linear Hydrogen Bonds
Michael Bühl, Ralf Ludwig, Rachel Schurhammer, and, and Georges Wipff,
The Journal of Physical Chemistry A 2004 108 (51), 11463-11468
DOI: 10.1021/jp045879+
Michal Gryz, Wojciech Starosta & Janusz Leciejewicz (2005) "Doubly bridged molecular ribbons in the structure of an ionic complex, hydronium zinc(II) pyrazine-2,3-dicarboxylate", Journal of Coordination Chemistry, 58:11, 931-935, DOI: 10.1080/00958970500055492


Why does the carbonyl group in an acid anhydride have two stretching frequencies? The carbonyl stretching frequencies for an acid anhydride are approximately $1820~\mathrm{cm^{-1}}$ and $1760~\mathrm{cm^{-1}}$. These are both higher frequencies than a simple ketone owing to the electron withdrawing capability of the adjacent group. However, why are there two distinct infrared stretching frequencies?

The two observed C=O frequencies are due to the symmetric and asymmetric stretching modes of the anhydride.

  Source: Introduction to Spectroscopy, Pavia and Lampman

You can see that the lower frequency symmetric stretch occurs where both C=O bonds are lengthening and shortening in tandem, whilst the higher frequency asymmetric stretch occurs when one C=O group is lengthening as the other is shortening.  You can see an example of this kind of spectral pattern in the example below.

  Source: Introduction to Spectroscopy, Pavia and Lampman

This explains why you get two frequencies for C=O bond in acid anhydrides, whilst ketones/aldehydes etc. only give a single IR frequency for the C=O bond.


Why is ΔE described as Q-W and Q+W in different contexts? Depending on the textbooks looked at, the energy change is described as either Q+W or the signage is changed to Q-W. Which is correct? Is it solely context dependent? Could anyone explain the background?

The difference in sign in the two versions of the first law of thermodynamics is to handle the two ways in which work can be defined.

The work done (assuming only pressure-volume work) can be defined as

$$w=P\Delta V$$

This is the definition often used in in scenarios when we care about the fate of the work. This definition is about the surroundings. Thus, when the system expands, it does positive work on the surroundings, which is a good thing. Think about an internal combustion engine. When the reaction occurs in a cylinder, gas is produced causing the volume to expand and move the piston, turning the drive shaft, etc. Since we care about the work after it leaves the cylinder, it is positive because the surroundings are gaining energy.

The work done by a system can also be defined as:

$$w=-P\Delta V$$

We use this definition often in chemistry and in general when we care more about how much energy is left in the system than we care about what the energy that left the system is doing. This definition is about the system. In the case of the engine above, energy has left the system, so the work done is negative to reflect the fact that the system has less energy than it used to.

The sign convention in the definition of the change in internal energy (and I prefer $\Delta U$ to $\Delta E$) reflects that we can define work done with respect to the system or the surroundings. The sign in front of $w$ guarantees that we always get

$$\Delta U=q-P\Delta V$$

With respect to the surroundings:

$$w=P\Delta V$$
$$\Delta U = q-w=q-P\Delta V$$

With respect to the system:

$$w=-P\Delta V$$
$$\Delta U = q+w=q-P\Delta V$$

A better question might be

  Why then do we not change the sign of $q$ when we switch reference frames from system to surroundings?


What are the fumes that form above liquid nitrogen? What is the composition of liquid nitrogen fumes? Is direct contact with these fumes harmful?

The visible fog that forms when liquid nitrogen is poured into an open container is almost entirely water fog: the boiling nitrogen chills the air above it, causing the humidity in the air to condense into fog.

There may be some microdroplets of liquid nitrogen in the fog too, but the air, even after it's been chilled, is still much warmer than the boiling point of nitrogen, so there won't be much.  (You can condense air by running a continuous stream of liquid nitrogen through a heat exchanger and blowing air over it, but mostly you get liquid oxygen when you do that.)

Liquid nitrogen has a relatively small heat of vaporization, as these things go: 199.2 kJ/kg according to Air Liquide.  (Compare 572.2 kJ/kg for dry ice.)  That plus the Leidenfrost effect means it's actually rather difficult to freeze yourself with the stuff. There's a standard demo where you pour liquid nitrogen into an audience member's cupped hands. The Leidenfrost effect prevents it ever coming in contact with their skin, and it only leaves a mild chill behind.  Try this with dry ice and your "volunteer" will need emergency treatment for frostbite.

Gaseous nitrogen is second only to the noble gases in its inertness, and it makes up 78% (by volume) of the air you're breathing as you read this, so obviously the "fumes" aren't harmful in themselves.  However, as was mentioned in an earlier answer (which seems to have been deleted), if you let liquid nitrogen evaporate in an enclosed space, it can "displace" the oxygen that makes up another 21% of the air, at which point anyone in the area may asphyxiate.  The sensation of suffocation is triggered by having too much CO2 in your bloodstream, not by inadequate oxygen supply, so you might not even notice that anything was wrong until you collapsed.  That's the second most dangerous thing about evaporating liquid nitrogen.

The most dangerous thing is, if you let liquid nitrogen evaporate in a sealed container, it's liable to explode on you.


Is chromate a suitable indicator for the titration of Ag⁺ with Cl⁻? To titrate $\ce{Cl-}$ with $\ce{Ag+}$ we use chromate $\ce{CrO4^2-}$ as an indicator.
The titration reaction is:

$$\ce{Ag+ + Cl- <=> AgCl}\tag{R1}$$

$$K_1 = \frac{1}{K_\mathrm{sp}(\ce{AgCl})} = \frac{1}{1.8×10^{-10}} = 5.56×10^9\tag{1}$$

The theory says that after all $\ce{Ag+}$ are reacted with $\ce{Cl-}$ the end point of titration is detected when excess $\ce{Ag+}$ reacts with the indicator chromate to form silver chromate:

$$\ce{2 Ag+ + CrO4^2- <=> Ag2CrO4}\tag{R2}$$

$$K_2 = \frac{1}{K_\mathrm{sp}(\ce{Ag2CrO4})} = \frac{1}{1.1×10^{-12}} = 9.1×10^{11}\tag{2}$$

However, as you see, $K_1 < 100K_2,$ so when both $\ce{Cl-}$ and $\ce{CrO4^2-}$ are present, $\ce{Ag+}$ will react with $\ce{CrO4^2-}$ and not with $\ce{Cl-}$.

But our teacher and everywhere on Google they say $\ce{AgCl}$ precipitates before $\ce{AgCrO4}$. And that should be true since this method of titration (Mohr's method) has been used long ago.

But, how can that be true? I don't understand why. Where have I mistaken?

You got the solubility part reversed.
The solubility of $\ce{AgCl}$ is lower than the solubility of $\ce{Ag2CrO4}:$

$$s(\ce{AgCl}) = \sqrt{K_\mathrm{sp}(\ce{AgCl})} = \sqrt{\pu{1.8E-10 mol2 L-2}} = \pu{1.34E-5 mol L-1}$$

$$s(\ce{Ag2CrO4}) = \sqrt[3]{\frac{K_\mathrm{sp}(\ce{Ag2CrO4})}{4}} = \sqrt[3]{\frac{\pu{1.1E-12 mol3 L-3}}{4}} = \pu{6.50E-5 mol L-1}$$

Therefore, if the $\ce{AgNO3}$ solution is gradually added to the solution containing the both $\ce{Cl-}$ and $\ce{CrO4^2-}$ ions, then initially the formation of a sparingly soluble $\ce{AgCl}$ salt occurs.
After the $\ce{Cl-}$ ions are almost completely isolated in the form of $\ce{AgCl},$ the $\ce{Ag2CrO4}$ precipitation starts to occur, signifying the equivalence point is reached.

The same reasoning can also be applied to the titration of even less soluble silver bromide $\ce{AgBr}$ with $K_\mathrm{sp}(\ce{AgBr}) = \pu{5.3E-13}$ (try it yourself).

Note, however, that while using Mohr's method it's imperative to titrate halide salts solutions with $\ce{AgNO3}$ and not vice versa.
Otherwise the precipitation condition

$$c(\ce{Ag+}) · c(\ce{Cl-}) > K_\mathrm{sp}(\ce{AgCl})$$

will be overridden by

$$c(\ce{Ag+})^2 · c(\ce{CrO4^2-}) > K_\mathrm{sp}(\ce{Ag2CrO4})$$

due to high concentration of silver ions in solution, favoring silver chromate precipitation (note squared term $c(\ce{Ag+})^2$) and thus shifting the equivalence point.


How to calculate the relationship between the pressures of zinc, carbon monoxide, and carbon dioxide based on mole balance? Zinc oxide is reduced at a constant temperature in a closed reactor using $\ce{ZnO(s)}$ and $\ce{C(s)}$ as the only starting materials. The following reactions are assumed to be at thermodynamic equilibrium.
  \begin{align}
\ce{ZnO(s) + C(s) &-> Zn(g) + CO(g)}\\
\ce{2CO(g)   &-> CO2(g) + C(s)}
\end{align}

  Assume ideal gas behaviour. Based on mole balance, the relationship applicable to the system at equilibrium is:
  \begin{align}
\tag{A} P_{\ce{Zn}} &= P_{\ce{CO}} + 2P_{\ce{CO2}}\\
\tag{B} P_{\ce{Zn}} &= 2P_{\ce{CO}} + P_{\ce{CO2}}\\
\tag{C} P_{\ce{Zn}} &= P_{\ce{CO}} + P_{\ce{CO2}}\\
\tag{D} P_{\ce{Zn}} &= 0.5P_{\ce{CO}} + 2P_{\ce{CO2}}
\end{align}

In this problem seeing that none of the given options consisted of $\ce{C(s)}$, so in order to get rid of that I just added the two equations and arrived at the equation $$\ce{ZnO(s) + CO(g) = Zn(g) + CO2(g)}.$$
Now just sort of comparing with the options I find that $\ce{CO(g)}$ and $\ce{CO2(g)}$ are on the opposite sides. It sort of gives me the intuition that may be the pressure term should have had a minus. Anyway I know that I have gone horribly wrong somewhere with regard to my concepts. Please tell me how to solve the problem henceforth.

First of all, let me explain why your approach is a bit wrong...
Addition of steps will only give you the net reaction, which is what you got. But in the process, the mechanism becomes blurred. For example, the net reaction which you have expressed, doesn't tell us the full story. One completely overlooks the fact the CO is also recombining to give CO2, and that is exactly where the fallacy arises.

My solution to the problem. Hope it helps.


Calculation of VOCs concentration (ppm) in the headspace of a bottle What is the right way to calculate the concentration of volatile compounds in the air in a closed container?

Suppose we have a $\pu{50 mL}$ bottle with $\pu{25 mL}$ of $100\%$ ethanol $(\ce{EtOH})$ in it.

In this case, will it be okay to use $PV = nRT?$

Doing so, it will be:

$$
\begin{align}
p(\ce{EtOH}) &= \pu{0.05872 atm}\\
V &= \pu{0.025 L}\\
R &= \pu{0.08206 L atm mol-1 K-1}\\
T &= \pu{298 K}\\
\end{align}
$$

giving us $n = \pu{6.00314E-5 mol}.$

The ppm: $\pu{1 ppm} = \pu{1 mg l-1}$

$$M(\ce{EtOH}) = \pu{46.07 g mol-1}$$

$$\pu{6.00314E-5 mol} × \pu{46.07 g mol-1} × 1000 =  \pu{2.766 mg}$$

This lead to $$[\pu{ppm}] = \frac{\pu{2.766 mg}}{\pu{0.025 l}} = \pu{110.6 ppm}$$

Usage of ppm/ppb etc units is discouraged, as they are ambiguous without the explicit context. As generally, 1 ppm (w/w) <> 1 ppm (V/V) <> 1 ppm (n/n) <> 1 ppm (w/V).

It is recommended to use explicit units, like e.g mg/L.

Expecting ideal gas behaviour $pV = nRT$ is not reasonable.   Ideal gases are not in equilibrium with their liquid phase.

There should be rather used a real gas equation.The easiest to use is the van Der Waals equation, but even that has significant error near condensation state.

Additionally, the sample(!) temperature  has to be strictly controlled, otherwise $n$ would be variable.

Why not rather to use small volume of ethanol to be known to fully evaporate in given volume ? Or perhaps other solvent, unless ethanol is waterless.
Note that ethanol has significant deviation from the Raoult's law.

I prefer $R = \pu{8.31446 J K-1 mol-1}$

Note that by May 2019 redefinition,
 It's exact value is $\pu{8.31446261815324 J K-1 mol-1}$


Is planarity really necessary for conjugation? If we look at the $\ce{ClO4-}$ anion, the chlorine atom in it is sp3 hybridised. Hence, it has a tetrahedral structure in which conjugation shouldn't be possible, but while explaining the acidic strength of $\ce{HClO4}$, I've seen many people claim that it is very acidic because the $\ce{ClO4-}$ anion is stable due to the conjugation of charge between the four oxygen atoms.

So, all of this boils down to: Is planarity really necessary for conjugation?

this is a rather good question several students have actually asked me about it. it is not any exeption or contradiction .i have written the explanation below


Is 1,2-dihydronaphthalene "partially aromatic"? I've learned that if any one ring of a structure like the below is aromatic, the whole compound is aromatic.

But one of my friends suggested that such compounds cannot be aromatic and gave the following resonance structures as the reason:

I feel that the non-aromatic/anti-aromatic resonance structures wouldn't make significant contribution. Is my reasoning correct?

Aromaticity is a quite difficult concept but in this case we can answer the question by simply looking at similar compounds and experimental data.

What you got here is a styrol derivative, the pi-system (which we need to look at if we are talking about aromaticity) is the same. So if this compound is not aromatic then styrol shouldn't be either.

But styrol is in fact aromatic (or better: the benzene ring in styrol is). It reacts like an aromatic compound should, it got the right shift in NMR, the bond lengths are in accordance with aromaticity and so is the UV/VIS spectrum.

So a conjugated double bond clearly doesn't make a benzene (significantly) less aromatic.


What is the exothermic reaction of superglue with cardboard? I glued a sheet of paper to a cardboard package using superglue (labelled with ethyl 2-cyanoacrylate).
To my surprise the paper became very warm.
What kind of reaction is it?

I believe this is due to the fact that cyanoacrylates polymerize by anionic (or sufficiently strong neutral nucleophile) addition. In particular, they react with hydroxyl ($\ce{-OH}$) groups; paper, being primarily cellulose, is abundant in hydroxyls. The bond formation involved in the polymerization reaction is inherently exothermic (generating a $\ce{C-O}$ bond on initiation and many subsequent $\ce{C-C}$ bonds as the polymerization proceeds), and with a large surface area and many reactive groups, I could certainly imagine the reaction being surprisingly violent. The Wikipedia page on cyanoacrylates notes its reactivity with cotton fibers (again, being chiefly composed of cellulose), and the MSDSs for various similar compounds warn of exothermic reaction with paper and similar materials (q.v., 1, 2, 3).


Calculating the physical properties of isotopes I am trying to determine the theoretical physical properties (namely melting and boiling temperatures) of isotopic substances.  Harold Urey used Debye thermodynamics (which is quite dense) in his 1946 paper to calculate the physical properties of isotopic substances. Most of the equations appear to be for exchange rates of the isotopes within a given system.  I would like to know which equations are available to determine isotope effects on physical properties of isotopologues and the assumptions that go with them.

The Debye frequency of a crystal is a theoretical maximum frequency of vibration for the atoms that make up the crystal.

It seems safe to assume melting will be occurring at the Debye frequency.

It is quite possible that the authors of your paper used the Lindemann model, which was published in 1910, in conjunction with Debye frequency, which is unique to each topoisomer, to predict the melting point using the following equation of Lindemann:

$$ T_m = \cfrac{4\pi^2 m \nu^2 c^2 a^2}{k_B} $$

$$m = \mathrm{atomic\ mass}$$ $$ v = \mathrm{frequency} $$   $$ k_B=\mathrm{Boltzmann\ constant}$$ $$T=\mathrm{temperature}\ \left(^\circ \mathrm K\right)$$ $$c = \mathrm{Lindemann\ constant}$$   $$a = \mathrm{atomic\ spacing}$$

Different diastereomers have different resonant frequencies which can be found through either Debye calculations or IR and Raman spectroscopy.

Substituting in the Debye expression for frequency (I copied this from Wikipedia; I didn't actually derive it myself) gives:

$$ T_m = \cfrac{2\pi m (\theta^2_D k_B)c^2 a^2}{\hbar^2} $$

Where $\theta^2_D$ is the Debye temperature and $\hbar$ is Planck's constant.

Another way of doing it is by using a hard sphere model in combination with Debye/Lindemann (PDF link).


Finding the enthalpy of formation of magnesium oxide using Hess' Law I have the following reactions:

$$\begin{align}
\ce{Mg(s) + 2HCl(aq) &-> MgCl2(aq) + H2(g)}\\
\ce{MgO(s) + 2HCl(aq) &-> MgCl2(aq) + H2O(aq)}\\
\ce{H2(g) + 1/2O2(g) &-> H2O(l)}\\
\ce{Mg(s) + 1/2O2(g) &-> MgO(s)}
\end{align}$$

I also have $\Delta H$ in $\mathrm{kJ/mol}$ for the first three reactions.

Out of this information, I need to find $\Delta H^\circ_\mathrm{f}{\left(\ce{MgO(s)}\right)}$.

I’m pretty sure I have to use Hess’ law but I don’t know how to apply it here. I’ve tried using the second reaction as my “main” reaction and then I can use the third and fourth reaction as formation reactions for two of the compounds of the main reaction but after that I’m not sure how to proceed.

You are trying to find out the enthalpy of formation.
The enthalpy of formation is the energy change when one mole of a substance is formed from its constituent elements in their standard state.

The last equation matches this, so that should be your 'main' or 'target' equation. From there you should be able to use Hess' law to calculate the enthalpy change of the formation of magnesium oxide.


Best bags to contain chemical odours? I have noticed that if you put household or other chemicals in bin liners or freezer bags the plastic doesn't keep the chemicals in. Rather they travel through the bag material and out into the air. However I have noticed that some bin liners have stronger material and are better than others.

Perhaps it will depend on which chemical is in the bag. However I guess I am asking about pollen, pesticides,washing detergent and general cleaning chemicals.

Is there a specific type/grade of plastic bin liners that I can get that will, generally speaking. keep odors in? Actually it doesn't have to be pvc any bag material which does the job will do however I guess PVC bags are most easily accessible so would like to know what options there are. See through bags are preferable.

As you state, the ubiquitous polyethylene (or polythene) bag allows many chemicals to slowly seep through. Pollen and dust particles themselves are far to large to get through, but water vapor, for example does.

For that reason, many foods are sold in plastic (e.g., mylar and polypropylene) sacks layered with aluminum (ah, aluminium for you polythene-sayers) to block evaporation, oxygen (which leads to fats and oils becoming rancid) and loss of flavor.

For storing small amounts of food, reuse these aluminized bags -- which also are effective shields to block RF tags in passports and licenses. They're available commercially, e.g. this list.

These bags are comparatively expensive, though, and last millenia in landfills, and impractical for general use waste bin bags.

Non-chemical work-around: store food wastes and other potentially odorous items in paper or plastic bags that would have been discarded anyway, and that bag inside the bin liner, to delay odor migration.


Composition of desiccant Most types of desiccation packages use silica gel, but are there any that use calcium oxide instead? Here's my reasoning:

Calcium oxide absorbs water, forming calcium hydroxide.
Adding a (used) package to water produces a slight amount of heat (I may be imagining this).
The solution stays cloudy even after sitting for a while.
The solution almost immediately turns clear upon addition of vinegar.

At this point, I'm pretty sure it is calcium oxide and hydroxide in this package, but I just wanted to make sure first. For point 2, I should note that it is very humid in the place that I live, so it would make sense that most of the calcium oxide is converted into calcium hydroxide, and therefore not much heat is produced.

Yes, calcium oxide can be used as an desiccant.

The following quote is an excerpt from this site:-

  Calcium oxide is calcinated or recalcinated lime having a moisture adsorptive   capacity of not less than 28.5% by weight. The
  distinguishing feature of calcium oxide is
  that it will adsorb a much greater amount of water vapor at a very low
  relative humidity than other materials. It is most effective where a
  low critical relative humidity is necessary, and were there is a high
  concentration of water vapor present.

  Calcium oxide removes water from a package very slowly, often taking
  days to reach its maximum capacity. As calcium oxide adsorbs moisture,
  it swells. Proper desiccant packaging is required for effective use.
  For these reasons, its use has been limited to primarily the packaging
  of dehydrated foods.

This following is a table which compares the properties of various dessicants taken from this site:-

And the following graph depicts the adsorption capacity of various dessicants:-

For more information click on the following sites:-

http://www.ohe-chem.co.jp/e_qa.html
http://www.agmcontainer.com/support/selecting-desiccant
https://www.usaemergencysupply.com/information-center/self-reliance/food-storage-frequently-asked-questions/types-of-desiccants


Why is the second acid dissociation constant for sulphuric acid much lower than the first in water? In aqueous solution, $\ce{H2SO4}$ ionises in two steps :

$$\ce{H2SO4(aq) -> H+(aq) + HSO4-(aq)}; \quad K_\mathrm{a_1}=1\times10^3$$
$$\ce{HSO4-(aq) -> H+(aq) + SO4^2-(aq)}; \quad K_\mathrm{a_2}=1.2\times10^{-2}$$

The value of $K_{\mathrm{a}_2}$ caught my attention. I thought that $\ce{H2SO4}$ was a strong dibasic acid, but a very low value of $K_{\mathrm{a}_2}$ implies that the second proton is not that easily donated. Why is the value of $K_{\mathrm{a}_2}$ so less?

For the second deprotonation you are trying to remove a proton from something that is already negatively charged, a process far more difficult than removing a proton from something that is neutral.
Alternatively you can think that the $-2$ charged $\ce{SO4^2-}$ anion attracts really strongly the $\ce{H+}$ resulting in the equilibrium being towards $\ce{HSO4^-}$, making it a weak acid compared to $\ce{H2SO4}$.
It is similar thinking to why the second ionization energies are always much higher than the first.

There can always be more than one ways to explain an observed phenomenon. With this in mind, one might be tempted to believe that there is suppression of dissociation of the $\ce{HSO4^-}$ ion due to the common ion effect.  The reasoning goes along the lines of:

Note that even before $\ce{HSO4^-}$ dissociates, there is an equal concentration of $\ce{H+}$ ions already present in the solution. We know, by Le Chatelier's principle, that the increased concentration of the products favors the reactants. Hence for the second dissociation, since $\ce{H+}$ is one of the products, and is in large concentration, the second dissociation is actually favored in the reverse direction. Thus, deprotonation of $\ce{HSO4-}$ is not favorable.
Hence, the $K_\mathrm {a_2}$ of $\ce{H2SO4}$ is significantly lower than its $K_\mathrm {a_1}$.

However, this is a wrong reasoning. Note that $K_\mathrm{a}$ values are always a constant. The $K_\mathrm{a}$ value for the $\ce{HSO4-}$ ion would be the same whether we took $\ce{H2SO4}$ or $\ce{NaHSO4}$. However, in the case of $\ce{NaHSO4}$, there are no pre-existing $\ce{H+}$ ions in solution when $\ce{HSO4-}$ deprotonates. Hence, it would be wrong to invoke the common ion effect here.
Therefore, the common ion effect should not be invoked when explaining the low $K_\mathrm a$ of $\ce{HSO4-}$ ion as compared to $\ce{H2SO4}$.


How to determine steady state of multi-step reaction with both reversible and irreversible steps? I am trying to determine the relative concentrations of $\ce{U1}$, $\ce{U2}$, and $\ce{E}$ when the following reaction is at steady state. Note that $k_t$ is the zero order rate constant of new $\ce{U1}$ entering the system and $k_d$ is the first order rate constant of substances leaving the system.

$k_1$ and $k_2$ are the first order rate constant for the forward and reverse reaction, respectively, of $\ce{U1}$ turning into $\ce{U2}$, and $k_e$ is the first order rate constant for $\ce{U2}$ reacting to $\ce{E}$. First order reactions are all first order in the respective reactant.

All the individual rate constants are known but the concentrations are not.

Can anyone point me in the right direction for how to determine the relative concentrations at steady state? I'm imagining an equation or set of equations such that when values for each rate constant are plugged in, the concentrations of the products are determined.

Can anyone point me in the right direction for how to determine the relative concentrations at equilibrium?

This system will not reach equilibrium as long as there are irreversible reactions removing U2. If someone claims to have found an equilibrium state, the rates associated with k1 and k2 would be equal, but the [U2] would decrease because of those irreversible reactions.

  I know that kt=3kd must be true for the whole system to be in balance, but I have been struggling to figure out how to relate the other rate constants.

Usually, lowercase k is used for rate constants, with the rate of reaction given by a rate law that involves k, but also concentrations of the reactants and perhaps other species. The way you are talking about it, all reactions would be zero order. For zero order reactions, there is no equilibrium (a melting process has no equilibrium - if the temperature is above the melting point, everything melts, and if it is below the melting point, everything freezes).

  I have read that the Keq for multi-step reactions is the product of each single equilibrium constant, but I'm not sure how to apply this with the irreversible reactions in the system.

You can't. There is the concept of steady state, though, where concentrations of intermediates remain constant while the forward and reverse rate of a given reaction are different. In this case, the rates increasing the concentration of an intermediate are equal to the rates decreasing the concentration of that same intermediate, but there is a net reaction. This is typical for living organisms (they can, however, reach equilibrium after they die).