What is advantage of creating elements that disappear in a flash? More than dozen of heavy elements of periodic table do not occur naturally but are prepared artificially at nuclear reactors.. These elements are said to disappear in a flash..

So, why invest lot of money for the discovery of such elements? What are the advantages of such elements which disappear in a flash?

Researchers are looking for the Island of Stability, a predicted group of heavy nuclei with halflives in the minutes or even days.

While we have some understanding that certain nuclear configurations and compositions are more stable, we are currently unable to reliably to a priori predict the stability, the half-life, or even the decay mechanisms of a nuclide. Thus, when a new element or isotope is produced, no one knows how long it will be around. Every time a new nuclide is produced, observations of its decay help refine the models of nuclear stability and decay.

The Island of Stability is believed to be centered around 120 protons and 180 neutrons. Element 118 $\ce{^194Uuo}$ has been produced, and while its most stable isotope has a half-life of only 0.89 ms, it is very close to the island of stability and there may be a heavier, more stable isotope of $\ce{Uuo}$ out there.

As in many areas of pure science, the benefits to society (or even other fields) may not be evident except for those few who are actually doing the work. The outcome of pure science research is knowledge and information. Consider, for example, that research on cathode rays led to the development of the television decades later.
The scientists studying cathode rays likely could not even envision their work being applied to the projection of moving pictures (especially those transferred over the airwaves as radio signals), as such technology did not exist or was very new at the time.


Why NOx will break the ozone sphere I've found that a reason for people to oppose supersonic aeroplanes was the $\ce{NO_{$x$}}$ gases it released would do harm to the ozone layer. So what is the mechanism of this?

According to Introduction to Atmospheric Chemistry, by Daniel J. Jacob, Princeton University Press, 1999,Chapter 10,

An important component of aircraft exhaust is nitric oxide ($\ce{NO}$) formed by oxidation of atmospheric $\ce{N2}$ at the high temperatures of the aircraft engine. In the stratosphere $\ce{NO}$ reacts rapidly with $\ce{O3}$ to produce $\ce{NO2}$, which then photolyzes:
$$\ce{NO + O3 -> NO2 + O2}$$
$$\ce{NO2 + $h\nu$ -> NO + O}$$

Note that there is no net consumption of $\ce{NO_x}$, thus it is able to act as a catalyst such that a single $\ce{NO_x}$ molecule can destroy many molecules of ozone by these reactions.
Please don't hesitate to ask for any clarifications in the comments below, particularly if I've failed to address any aspect of your question.


Heisenberg's uncertainty principle Can anyone explain Heisenberg's uncertainty principle and it's derivation.
Im a 12th standard student.Please make it simple enough for me to understand.

It is a basic postulate of quantum mechanics. It states that it is impossible simultaneously to measure the position of a particle, say an electron, an atom or a molecule, and its momentum. It places limits on these measurements such that uncertainty in position $\Delta x$ and momentum $\Delta p$ is given by $\Delta x\Delta p \ge \hbar/2$ where (hbar) $\hbar =h/2\pi$ and $h$ is the Planck constant $6.626 \times 10^{-34}~\mathrm{J~s}$. In a classical world no such restriction occurs, but this is not what is observed experimentally.

The important part is simultaneous measurement. We can measure where an atom is to whatever precision an instrument will allow, (e.g. by xray crystallography) but we cannot in the same measurement (i.e. at same time) know what its momentum is. We can measure an atom's speed v as accurately as we like and hence momentum mv. We note its position in two places a and b, say 10 m apart, and divide distance by time to cover this distance.

The uncertainty principle places limits on the measurement of momentum mv and position that allows us to know both the position and momentum an atom (particle) has at any given instance. Ask again if this is not clear


What does the Sm5Ge4 orthorhombic structure look like? I see tons of papers referencing this structure but none show what it looks like.

You can look at the $\text{Sm}_5 \text{Ge}_4$ crystal structure at Materials Project here.


What does it mean by degree of aromaticity of a molecule? This is a fundamental question, though I didn't really get the meaning/idea what does it mean by aromaticity or degree of aromaticity. Does it mean the ability of a molecule to be aromatic?

Aromaticity might be viewed as resonance stabilzation energy, NMR shift, or altered reactivity.  Here is a way of measuring its change among related systems,

Chem. Commun.  48, 8144 (2012)
J. Am. Chem. Soc. 131, 189 (2009)
J. Am. Chem. Soc. 127, 16251 (2005)
J. Org. Chem. 67, 8798 (2002)
J. Am. Chem. Soc., 124, 13495 (2002)
Chem. Rev. 101, 1301 (2001)
Tetrahedron 57, 3689 (2001)

http://nptel.ac.in/courses/IIT-MADRAS/Engineering_Chemistry_III/pdf/1_Aromaticity.pdf
A more practical approach derives from MO theory and Hund's rule of maximum multiplicity.

http://www.youtube.com/watch?v=U6FX5RfLjd0
http://www.youtube.com/watch?v=H0so_NJ2aVw
Frost circle


What is the difference between X-ray photoelectron spectroscopy and X-ray absorption spectroscopy? XPS is used for surface chemical state identification.
But other than that, is XAS just an inverse of XPS?

You could easily google this...

X-ray photoelectron spectroscopy uses low energy monochromatic x-rays to excite electrons from atoms near the surface of a sample. The kinetic energy of the electrons is then measured and then the binding energy of the electrons can be calculated.

X-ray absorption spectroscopy measure the attenuation of a x-ray beam as a function of x-ray wavelength (or energy). So what you are detecting is the absorption edges of the atoms in the sample.


Why can't carbon form an ionic bond? My textbook says that a $\ce{C^{4+}}$ cation cannot be formed because it requires a lot of energy to remove 4 electrons. Formation of  ionic bonds involve "removing" electrons and there seems to be enough energy there. So what's different for carbon?
The textbook also mentions that a $\ce{C^{4-}}$ anion cannot be formed because 6 protons cannot hold on to 10 electrons. Elements like chlorine form ionic bonds and end up with 18 electrons and 17 protons. I know there's just one more electron. So is $\ce{NaCl}$ possible because it's not very hard to hold on to 1 extra electron? In that case at what exact point does it become hard to hold onto the extra electrons?

$\ce{C^{4+}}$ ions:

Single ions are really only observed in the gas phase.  There is absolutely nothing that prevents a C4+ ion from being generated in the gas phase, given sufficient energy.  Each successive electron removal requires additional energy though, so by the time you get to 4 electrons removed, you're looking at quite a lot of energy required.  Note that this does not mean that $\ce{C^{4+}}$ ions are seen in compounds, just that they can be made in the gas phase.  Any cation you can imagine can be produced in gas phase.

The difficulty arises when we want to do something with that carbon cation.  If I want to make a compound with it, it can bond with other substances and make new bonds.  Once that is done we can evaluate those compounds to tell whether the bond is truly ionic or not.  In condensed phases or in solution, we look at how the parts of a compound interact with the particles surrounding them to decide whether to treat them as ions or not.

The best chance at making a C4+ ion would be to bond it to fluorine, which is an extremely electron-hungry element.  When I actually try this, I find that the C-F bond is very polar, but the $\ce{CF_4}$ molecule that forms behaves as a covalent compound normally would.  For example, it does not dissociate into ions when dissolved or melted and it doesn't form an ionic crystal in its solid form.  As much as fluorine wants electrons, it doesn't want them enough to completely steal 4 of them from carbon, because the energy required to remove each electron becomes greater than the energy required to remove the one before it.  It never gets to the point of $\ce{C^{4+}}$ and $\ce{F^{-}}$ ions being bonded.

Lead can form a +4 ion, $\ce{Pb^{4+}}$, in compounds such as lead fluoride.  The reason that this can happen but carbon cations cannot is that the outermost electrons in a lead atom are much farther from the nucleus than those of a carbon atom and are easier to remove as a result.

$\ce{C^{4-}}$ ions:

In gas form this does not happen.  Carbon's first electron affinity is positive, so we would expect to see it form $\ce{C-}$ ions in the gas phase if given free electrons to attract.  Adding a second electron to this substance would require it to attract an electron against its already negative overall charge, and that is not energetically favorable.

In order to make the $\ce{C^{4-}}$ion in a compound, a species would have to donate 4 electrons to carbon.  The attraction of carbon's nucleus to other atoms' electrons is just too low for that to happen.  The best candidates for a cation in an ionic compound of this type would be either lithium or cesium, however, when we try to actually make this compound we get another class of ions, the carbides, which feature $\ce{C2^{2-}}$ ions.  In short, it doesn't happen.

Phillip's comment regarding the carbides is a good one.  There are a few metal carbides that feature carbon atoms bonding to a metal in the ratios we would expect if it were purely ionic, like $\ce{Mg2C}$.  In these compounds the carbon does have a formal charge of -4, but the properties of the substance are such that it behaves like a covalent network compound rather than an ionic compound.

What is the limit of this process?  Anions with -3 charge, like nitride or phosphide, do exist, though they require a very weakly electronegative species, like an alkali metal or alkaline earth metal, to form an ionic compound.

Carbon atoms can carry a charge as part of an ion, but only when they are part of a larger species (again, like $\ce{C2^{2-}}$) and they do not carry 4 additional electrons per atom.  Check out cyanide as another good example of this type of polyatomic ion.


Computing the bond spring constant from first principles Given a set of atoms and their positions in space, is there a way to compute the bond spring constant for each pair of atoms that are considered to be bonded (I can already determine the bonds with OpenBabel)? I don't want to rely on empirical data, and it's okay if the spring constant is wrong by up to about 20%. I already know the bond length and bond energy, if that helps.

For background: my ultimate goal is to compute the normal modes of a molecule, by finding the generalized eigenvalues of an undamped mass-spring system. I already know the mass matrix, but I don't know the stiffness matrix.

tldr, you want the force constant from the Hessian matrix .. ideally in internal coordinates

First off, you're saying 'first principles,' which most computational chemists would take to be some level of quantum chemical method, whether density functional or wavefunction-based. None of that is available in Open Babel - it only implements a very small number of classical force fields.

Even if you have optimal bond lengths and energies, that doesn't directly give you the force constants. For that, you want the second derivative.

In principle, you could start perturbing atoms and evaluating the energy. In that way you could build up a stochastic approximation of the PES and evaluate numerical second derivatives. I don't recommend that.

A more accurate way would be at least to get the forces / gradients at each of those points, which you can use to build up the Hessian matrix. There are some notes about that as part of the psi4numpy notebooks which can be used to drive the Psi4 quantum program.

The problem is that for any molecule over a few atoms, the Cartesian Hessian matrix won't directly give you a bond force constant. Most vibrations include multiple atoms, and are rarely along any particular Cartesian axis.

The psi4numpy notebooks give some indication of how you can evaluate the Hessian in internal coordinates, which should be closer to what you want.


Valence electrons in helium? I am having trouble understanding how valence electrons are counted in Helium.

My references are:

Here where it appears that Helium has 0 valence electrons because the shells are finished
Here where it appears that Helium has two valence electrons (which is what I thought)

How many valence electrons does helium have?

Helium has two electrons in total, and according to the aufbau principle, it adopts the electronic configuration $\mathrm{1s^2}$. This means it has two electrons in s orbitals with a principal quantum number of 1.

The last (and only) level of helium's electronic configuration is $\mathrm{1s^2}$, and therefore $\ce{He}$ has 2 valence electrons.


Classical bioisosteres Classical bioisosteres are functional groups that satisfy with the Grimm's hydride displacement law and Langmuir's definition of isosteres.

My question is why chlorine can be replaced by trifluoromethyl or cyano group and the replacement is categorized in univalent atoms replacement?

Because as seen from the criteria of classical bioisosteres aforementioned, the replacement atoms do not have any hydrogen atom therefore I think Grimm's law is devoid. However both trifluoromethyl and cyano group do not have the same number of atom nor electron as well as electron arrangement as in chlorine.

So what explains that these functional groups are classical bioisosteres?

I think you are right. Grimm's law in its original version (1925) should not apply to the $\ce{-CF3}$ , $\ce{-CN}$ pair. This "law" is very restrictive and has only historical significance. To be honest I could not recall its statement and I have never come across it in the last 10 years or so that I work in the field. The modern definition of classical bioisosteres (as opposed to non-classical bioisosteres like the $\ce{-CO2H}$, tetrazole pair) has been extended exactly for this reason, to include functional group pairs with similar physiochemical properties that do not follow Grimm's law or any other classification of less general scope.


Extract ammonium sulfate from water What is the best method to extract ammoniumsulfate from water?

The assignment is as follows:

  Manure is being processed and the product left is a mixture between water and ammonium sulfate, transporting the mixture costs a lot of money. So we only want the ammonium sulfate.

What is the best method to do this on a big scale? Evaporation isn't a good option since it costs too much energy to heat the big amounts of water.

Evaporation isn't a good option since it costs too much energy to heat the big amounts of water.

Is it? I've seen prices around USD 130-150 per ton (minimum order of 25 tons) for ammonium sulfate with a purity around 99%. You might want to calculate the costs to operate a production plant using natural gas (or other fuels) for evaporation yourself.

There is however a different energy source that you might want to take into consideration: the sun.

Evaporation of water from solutions of sodium chloride in open ponds by solar heating alone has a long tradition in the production of sodium chloride in the form of sea salt.

Update

Ammonium sulfate will not evaporate from the open ponds. At temperatures between 235-280 °C, it will start to decompose.

However, if surface temperatures reach that level under irradiation with unconcentrated sunlight, the fate of the ammonium sulfate is the least of your worries ;)


Compare concentrations using LC/MS Suppose I run a reaction and then pass the results through LC/MS to find the conversion. One idea could be to compare the peaks of the reactants to the peak of the product. But:

MS peak will depend on the levels of ionization of each of the compounds
UV spectra depends on the molar extinction coefficient of the compounds

So it looks like I can't compare the peaks. Is this right and if so what are the possible ways of finding the conversion?

You can indeed find the conversion (not the absolute concentrations), even without calibrating the system.

At time t = 0, take an exactly measured volume of reaction mixture (e.g. 1 mL) and dilute it in a volumetric flask with water and/or acetonitrile to an exact volume (e.g. 100 mL).

Inject an exact volume (e.g. 50 uL) of the diluted solution in the LCMS and note down the (UV-based) peak area of the limiting reagent. This is your ${A_0 = \epsilon \cdot C_0 \cdot k}$, where ${C_0}$ is the concentration of your limiting reagent at this time, and ${k}$ is a constant related to the sample size and dilutions.

At any time t, do exactly the same as above, and you get an area ${A_t = \epsilon \cdot C_t \cdot k}$. Under these conditions, ${k}$ is the same, and of course so is ${\epsilon}$.

As you know, the conversion is ${y = 1 - \frac {C_t}{C_0}  = 1 - \frac {A_t}{A_0} }$.


Does higher [OH] not entail higher pH? A student has equal volumes of $\pu{1.0 mol dm-3}$ sodium hydroxide and ammonia solutions. Which statement about the solutions is correct?
A. Sodium hydroxide has a lower electrical conductivity than ammonia.
B. Sodium hydroxide has a higher hydrogen ion concentration than ammonia.
C. Sodium hydroxide has a higher pH than ammonia.
D. Sodium hydroxide has a higher hydroxide ion concentration than ammonia.

The answer is C, and I understand why. What I do not understand is why D isn't correct also? Could someone clarify this for me?

See this website. Essentially it's a misprint in the IB Chem testbank.

https://www.thinkib.net/chemistry/blog/8944/multiple-choice-answers

There's a correct rewording for D that makes C the only correct answer, according to the link I just posted.


Quantum numbers required to specify the position of an electron In a test today, there was this question:

  How many quantum numbers are required to specify the position of an electron?
  $$(A)\ 1 \quad (B)\ 2 \quad (C)\ 3 \quad (D)\ 4$$

The answer key says that the answer is $(D)$ but I am feel that the answer should be $(C)$ as the spin quantum number describes the angular momentum of the electron.

Please let me know what should be the correct answer keeping in mind that I am a sixteen year old.

Thanks!

P.S. let me know if I should include more or change tags.

Edit: Since it was not clear previously (or so I think), my point is that spin quantum number is not necessary for specifying the position of an electron.

First of all, the question is badly formulated. It asks about "position" but clearly means "orbital", which are two different things. (An orbital is a wave function, which depends on the electrons position: $\psi(\vec r)$.) Although the difference may be difficult to explain at high school level, this choice of words is imprecise and often leads to wrong ideas about the basic concepts of quantum mechanics.

The question is obviously about the 4 quantum number ($n$, $l$, $m_l$ and $m_s$), thus including spin. Whether spin is really required to describe the spatial part of an orbital is arguable. But if we stick with high school level chemistry and consider what is typically called restricted orbitals, then both electrons ($\alpha$ and $\beta$) share the same spatial part of the orbital, and the spin quantum number is unrelated to the "position" of the electron. But the spin quantum number is still required to "distinguish" the two electrons, what seams to be what the question has in mind. (I used quotation marks, because the electrons are actually indistinguishable.)

If we go beyond high school level and into actual quantum chemistry, then it depends on what type of orbitals are considered. I already mentioned restricted orbitals. But one may also use unrestricted orbitals, where $\alpha$ and $\beta$ electrons do not share the same spatial part. Here it is obviously necessary to include the spin quantum number.

One final remark: There are actually 5 quantum numbers that specify an orbital. Analogous to $l$ and $m_l$ of spatial angular momentum, in spin angular momentum there is $s$ and $m_s$. However for electrons we always have $s=\frac{1}{2}$, therefore explicit specification of $s$ is not necessary.


Naming of bicyclo[2.2.1]hept-5-ene-1,4,5,6-tetramethyl-3-bromo-2-ethyl carboxylate I've synthesized a molecule (depicted above) using a Diels-Alder reaction, and I can't find it in any database (I used scifinder for my search), so I need advice about the name.

The name I came up with, based on similar structures is bicyclo[2.2.1]hept-5-ene-1,4,5,6-tetramethyl-3-bromo-2-ethyl carboxylate. Is that correct?

The individual parts of the proposed name are correct; however, their order of citation in the name is wrong.

The name of the parent hydride is ‘bicyclo[2.2.1]hept-2-ene’. Note that the ending ‘ene’ receives the lowest locant possible (here: ‘2’ since the bicyclic ring system is numbered starting with one of the bridgeheads).

The principal characteristic group is expressed by means of a suffix, which yields the name ‘bicyclo[2.2.1]hept-5-ene-2-carboxylate’. Note that now the characteristic group receives the lowest locant possible (here: ‘2’), which
is cited immediately in front of the suffix. Thus, the locant for the ending ‘ene’ in the functionalized parent hydride is changed to ‘5’.

The other substituents are cited as prefixes in substitutive nomenclature: ‘3-bromo-1,4,5,6-tetramethylbicyclo[2.2.1]hept-5-ene-2-carboxylate’

Finally, the name of the ester is formed according to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), by placing the ‘hydroxylic’ component in front of the name.

  P-65.6.3.3.1 Monoesters

  Monoesters formed from a monobasic acid and a ‘monohydroxylic’ component are named systematically by placing the ‘hydroxylic’ component denoted by an organyl group (alkyl, aryl, etc.) in front of the name of the acid component expressed as an anion derived from the appropriate acid (…).

Therefore, the preferred IUPAC name (PIN) of the compound given in the question is
‘ethyl 3-bromo-1,4,5,6-tetramethylbicyclo[2.2.1]hept-5-ene-2-carboxylate’.


Is it wrong to write a coefficient of 1 in a chemical equation? Take $$\ce{2H2 + O2 -> 2H2O}$$
Is it wrong to write $$\ce{2H2 + 1O2 -> 2 H2O}$$
By wrong I mean that it is against a widely accepted guideline (those of the IUPAC?), not just "your teacher said it's wrong, therefore it's wrong". The only information I could find, is that it is assumed there is a coefficient of 1 when there is none written.

It is not wrong, per se, but it goes against conventions.
These rules are in place to simplify how we write formulas and equations, etc.
If we did write the $1$ all the time, then perhaps we would write it for a chemical formula, and water would be $\ce{H2O1}$, for instance.
If you want to write it as a reminder and placeholder that is totally fine.
A teacher might say that they have a rule, or that they want you to follow standard convention. You should do this if requested. When I taught, I would offer to students that it is OK if they want to do it to get in good habit of always writing coefficient.
All that being said, IUPAC guidelines say to write the species with no coefficient if the coefficient is one. So, technically it is not to the standard, but does not make it less clear, so use it in your work, but if you are submitting as classwork follow the protocol of the class.
Below is cited a brief note to this from the The ACS Style Guide[1].

➤ Specify the number of each species (molecules, atoms, ions, etc.) of reactants and products by a numeral written on the line and closed up to the symbol.
$$\ce{2Al + 6NaOH -> 2Na3AlO3 + 3H2 }$$
➤ To indicate the aqueous, solid, liquid, or gas state, use the appropriate abbreviations on the line, in parentheses, and with no space preceding them.
$$\ce{Ag(s) + H+(aq) + Cl–(aq) -> AgCl(s) + ½H2(g)}\\
  \ce{4FeS(s) + 7O2(g) -> 2Fe2O3(s) + 4SO2(g)}$$

Within these few lines, it is stated to write a numeral, but then it is omitted when it is $1$.
House rules prevail, as this is commonly accepted practice.
Reference:
[1]: The ACS Style Guide: Effective Communication of Scientific Information; Coghill, A. M., Garson, L. R., Eds.; Oxford University Press: New York, NY, 2006(pdf).


Chemical Equilibrium - finding equilibrium constant given degree of dissociation $\ce{N2O4}$ is dissociated $20\ \%$ at 27 degrees Celsius and 1 bar. Find the equilibrium constant of that reaction. What percentage of $\ce{N2O4}$ dissociates at $13.15\ \mathrm{kPa}$?

An explanation of the solution is below, but there are two things I do not understand.

  Why do we have $1 + \alpha$ in the denominator in equations ${\text(6)}$–${\text(9)}$?

  How do we get the fraction $\dfrac{(2\alpha)^2}{1-\alpha}$ in equation $\text{(11)}$?

If the degree of dissociation is $\alpha$, then the amounts of substances present $n_0$ dissociated is $n_0 \alpha$, and the remaining amount of undissociated substance $n_0 (1-\alpha)$. Since each molecule $\ce{AB}$ dissociates into molecules of $\ce{A}$ and $\ce{B}$, the amounts of each substance is
$$\begin{align}
n_{\ce{A}} &= n_{\ce{B}} = n_0 \cdot \alpha \tag{1}\\
n_{\ce{AB}} &= n_0 \cdot (1-\alpha) \tag{2}
\end{align}$$
In the case of the dissociation of dinitrogen tetroxide, $\ce{N2O4}$, it will be
$$\begin{align}
n_{\ce{A}} &= 2n_0 \alpha \tag{3}\\
n_{\ce{AB}} &= n_0 \cdot (1-\alpha) \tag{4}\\
\sum{n} &=n_0 (1+\alpha) \tag{5}\\
x_{\ce{A}} &= \dfrac{2\alpha}{(1+\alpha)} \tag{6}\\
x_{\ce{AB}} &= \dfrac{1-\alpha}{1+\alpha} \tag{7}\\
x_{\ce{NO2}} &= \dfrac{2\alpha}{(1+\alpha)} =\dfrac{0.40}{1.20}=0.3333 \tag{8}\\
x_{\ce{N2O4}} &= \dfrac{1-\alpha}{1+\alpha}=\frac{0.80}{1.20}=0.6666 \tag{9}\\
K_p &= \dfrac{(P_{\ce{NO2}})^2}{P_{\ce{N2O4}}}=P\cdot \dfrac{( x_{\ce{NO2}})^2}{ x_{\ce{N2O4}}} = \mathrm{ 1\ bar \cdot \dfrac{(0.3333)^2}{0.6666} = 0.1667\ bar} \tag{10}
\end{align}$$
We obtain the degree of dissociation of $\ce{NO2}$ at a pressure of $13.15\ \mathrm{kPa}$:
$$\dfrac{K_p}{P} = \dfrac{(2\alpha)^2}{(1-\alpha)}=\mathrm{\dfrac{0.1667\ bar}{0.1315\ bar}=1.268} \tag{11}$$
From here, we get:
$$
4\alpha^2 = 1.268 (1-\alpha) = 1.268 -1.268\alpha  \tag{12}$$
Which becomes
$$
4\alpha^2 =1.268\alpha – 1.268 = 0  \tag{13}$$
For the degree of dissociation, we get:
$$\alpha =\dfrac{-b + \sqrt{b^2} -4ac}{2a} = \dfrac{-1.268 +\sqrt{1.268^2}+20.288}{8} \\= \dfrac{-1.268 +4679}{8} = 0.426  \tag{14}$$
Check your results:
$$K_p = \dfrac{(P_{\ce{NO2}})^2}{P_{\ce{N2O4}}}=P\cdot \dfrac{( x_{\ce{NO2}})^2}{ x_{\ce{N2O4}}} = \mathrm{0.1315 \cdot \dfrac{(2\alpha)^2}{1-\alpha} =0.1315 \cdot \dfrac{0.852^2}{0.574}= 0.1663\ bar}  \tag{15}$$

It looks like alpha $\alpha$ is the degree of dissociation. The way it is explained in your image is a little odd, so let me see if I can do a better job. What seems like it might be problematic for you is that the explanation you have been given skips some algebra along the way. Below, I am showing all of the algebra up until your points of confusion.

For your reaction: $\ce{N2O4 <=> 2NO2}$

In the first part of your question, we need to determine the equilibrium constant

$$K_p=\frac{P^2_{\ce{NO2}}}{P_{\ce{N2O4}}}$$

We know the total pressure, but we do not know the partial pressures of each species. Fortunately, thanks to Raoult's Law, we can substitute in mole fractions since $P_i =P \chi_i $.

$$K_p=\frac{P^2 \chi^2_{\ce{NO2}}}{P\chi_{\ce{N2O4}}}=P\frac{\chi^2_{\ce{NO2}}}{\chi_{\ce{N2O4}}}$$

We aren't given any number of moles, but let's say that initially there was some number of moles of $\ce{N2O4}$ before any dissociation occurred. We'll call that number $n_0$. We can write expressions for the number of moles of $\ce{N2O4}$ and $\ce{NO2}$ after dissociation using $n_0$ and $\alpha$. Since $\alpha\%$ of $\ce{N2O4}$ is dissociated, $(1-\alpha\%)$ remains undissociated:

$$n_{\ce{N2O4}}=n_0 (1-\alpha)$$

The moles of $\ce{NO2}$ are $n_0$ times $2\alpha$, since $\alpha\%$ of $\ce{N2O4}$ was converted into $\ce{NO2}$ and there are two $\ce{NO2}$ molecules for every $\ce{N2O4}$ molecule:

$$n_{\ce{NO2}} = 2n_0 \alpha$$

The total moles (need this for mole fraction) is

$$\sum{n}=n_{\ce{N2O4}} + n_{\ce{NO2}} = n_0 (1-\alpha) + 2n_0\alpha = n_0 (1-\alpha +2 \alpha) = n_0 (1+\alpha)$$

When we calculate the mole fractions of each species, the $n_0$ term cancels out, which is good since we do not know what its value is:

$$\chi_{\ce{N2O4}}=\frac{n_{\ce{N2O4}}}{\sum{n}}=\frac{n_0 (1-\alpha)}{n_0 (1+\alpha)}=\frac{1-\alpha}{1+\alpha}$$
$$\chi_{\ce{NO2}}=\frac{n_{\ce{NO2}}}{\sum{n}}=\frac{2n_0\alpha}{n_0 (1+\alpha)}=\frac{2\alpha}{1+\alpha}$$

  Now you know where $1+\alpha$ came from. These values are plugged into the law of mass action for $K_p$ to determine the equilibrium constant.

In the second part of the question, you use the value of the equilibrium constant and the new pressure to determine a new value of $\alpha$. It all starts with the law of mass action and we manipulate some algebra from there. First the law of mass action using mole fractions:

$$K_p = P\frac{\chi^2_{\ce{NO2}}}{\chi_{\ce{N2O4}}}$$

Next, substitute in the expressions for mole fraction in terms of $\alpha$.

$$K_p = P\frac{\chi^2_{\ce{NO2}}}{\chi_{\ce{N2O4}}}=P\frac{\frac{(2\alpha)^2}{(1+\alpha)^2}}{\frac{1-\alpha}{1+\alpha}}$$

The term $1+\alpha$ appears in both the numerator and the denominator and can be canceled.

$$\require{cancel}
K_p = P\frac{\frac{(2\alpha)^2}{\cancel{(1+\alpha)}(1+\alpha)}}{\frac{1-\alpha}{\cancel{(1+\alpha)}}}=P\frac{(2\alpha)^2}{(1-\alpha)1+\alpha}=P\frac{(2\alpha)^2}{1-\alpha^2}$$

Now we move $P$ over to the left side of the equation to separate the values we know ($K_p$ and $P$) from those we do not know ($\alpha$).

$$\frac{K_p}{P}=\frac{(2\alpha^2)}{1-\alpha^2}$$

  Now you have something like your second equation that you are confused about. Perhaps there is a mistake in the original source?


19F NMR line spectra I am trying to figure out the NMR spectra of F3P-BH3
I am solving this out using 2nI+1 rule.  IP = 1/2 and IB = 3/2
Hence for this compound,

(2nIP+1)(2nIB+1) = (2x1X1/2+1)(2X1X3/2+1) =
(2)(4) = 8 lines (quadraplet of doublet)

This is what I came up with, it would be great if you could correct me if I'm wrong. Trying to solve few problems before the exam =)

I think you have got the calculation right. However, I would also consider the coupling between F and H because coupling through 3 bonds is also usually visible. So, each of your 8 lines will be split into four by the H-atoms.
That will be the major part of the spectrum. You would also get weak signals with a different splitting pattern due to the other isotope of boron ($\ce{^10B, I=3}$, abundance=20%).


Van't Hoff Equation with changing Enthalpy Is there a way to derive a modified van 't Hoff equation if we can't make the assumption that $\Delta H_\mathrm{r}$ doesn't change?

We know that $\mathrm{d} \ln K/\mathrm{d}T = \Delta H_\mathrm{r}/(RT^2)$ and $\Delta H_\mathrm{r} = \Delta H_\mathrm{r}^\circ + \int C_p\,\mathrm dT$

Is there a way to integrate the $C_p$ term along with the $T^2$?

There is. Suppose that at 298 K $$\Delta G^○_{298} = -RT\ln{K_{p\,298}} = \Delta H^○_{298} - T\Delta S^○_{298},$$ and if both $\Delta H^○$ and $\Delta S^○$ depend on $T$, $$-RT\ln{K_p} = \Delta H^○_{298} + \int_{298}^T \Delta C_pdT - T\Delta S^○_{298} - T\int_{298}^T \frac{\Delta C_p}{T}dT.$$ Then both sides of it can be divided by $T$ and knowing that $\int udv = uv - \int vdu$, where in our case $v = 1/T$ and $u = \int_{298}^T \Delta C_pdT$, we can "simplify" the last equation: $$-R\ln{K_p} = \frac{\Delta H^○_{298}}{T} - \Delta S^○_{298} - \int_{298}^T \frac{\int_{298}^T \Delta C_pdT}{T^2} dT.$$ I'll not endeavor to differentiate that with respect to $T$, but now you can probably figure out the rest. Since $C_p$ is usually a not-that-complex polynomial of $T$, the math should be rather straightforward.


Can the colour change in litmus paper be explained by conjugated systems? I have been trying to understand the colour change in litmus paper on a more detailed level than "the chemical identity changes, and therefore also physical properties". I have knowledge of the particle in a box model of quantum mechanics, and i would like to apply this to the conjugated system in the active indicator-component of litmus dye.
In other indicators I have looked at, you can clearly see that when the dye molecule dissociates or is protonated, there is change to the length of the conjugated system (number of alternating double bonds in a row), which explains the colour change. An example is given below for BTB (bromo thymol blue)

However, i can't find any explanation like this as to why the indicator dye in litmus paper (7-hydroxyphenoxazone) changes colour when protonated/deprotonated. From what I can see, the length of the conjugated system is the same in both versions of the molecule.

Edit: If someone knows of a model that more easily explains this colour change than conjugation, feel free to share that as well.

You have shown the structural change in bromothymol blue upon deprotonation.

For the anion of 7-hydroxyphenoxazone, the following resonance structures involving a p-quinone imine are conceivable:

The assumption that the $\pi$-system of the neutral species and the corresponding anion are identical is not justified.

A smaller system, in which a similar effect can be observed, is 4-nitrophenol. A solution of the phenol is colourless, whereas the solution of the phenolate is yellow ($\lambda_{\mathrm{max}}$ = 405 nm).

Again, loss of a proton leads to tautomerism with contribution of a quinoid structure.


Are complex ions thermochromic? Does raising the temperature of a complex ion (formed from a transition metal ion and ligand(s)) affect its color? For example, if the temperature of $\ce{[Cr(H2O)6]^3+}$ is raised, will its color change?

Indeed, this effect is observed in certain select cases. The effect of spin-spin cross over has been mentioned in this regard. There can be some other effects at play, namely: phase transitions, change in ligand geometry, equilibria between different molecular structure, and change in the number of solvent molecules in the coordination sphere.
Thermochromic behavior can be divided into two main classes: continuous and discontinuous. Some color changes are gradual; as the temperature increases over time, the color slowly changes. This can be due to the breaking or rearrangement of the lattice. This type of rearrangement is also known as continuous thermochromism.
On the other hand, a dramatic change in color can occur at a specific temperature or over a very small temperature range. This is known as discontinuous thermochromism.
I shall attempt to supplement what others have said by providing some illustrative examples.
I. $\ce{[Fe(NH2trz)3]Br2·H2O}$
This compound exhibits a spectacular reversible purple-white thermochromic transition upon heating−cooling with a large hysteresis centered at room temperature. The thermochromic transition is related to the spin crossover (SCO) from the low-spin state (LS, purple, $S = 0$, diamagnetic) to the high-spin state (HS, white, $S = 2$, paramagnetic).

(A) after cooling at $77~\mathrm{K}$ and (B) after heating at $353~\mathrm{K}$. The white form (B) is attracted to the magnet, whereas the purple form (A) is slightly repelled by the magnet.
Spin crossover (SCO) complexes are essentially $\mathrm{d}^4$−$\mathrm{d}^7$ first-row transition-metal octahedral complexes.
The occurrence of ST in coordination compounds of transition-metal ions is governed by the relationship between the strength of the ligand field (the electrostatic field acting at the central metal ion) and the mean spin-pairing energy. Octahedral complexes of $\mathrm{d}^4$–$\mathrm{d}^7$ ions may be either HS or LS, depending on whether the ligand field strength is weaker or stronger, respectively, than the spin pairing energy.
In order for thermally induced ST to occur the difference in energy must be of the order of thermal energy, i.e $k_\mathrm{B}T$. An increase in temperature favours the HS state, while lowering the temperature favours the LS state.
Thermal spin transition occurs nearly exclusively with coordination complexes of $\mathrm{3d}$ metal ions, because for $\mathrm{4d}$ and $\mathrm{5d}$ transition element compounds on the basis of ligand field theory, because the strength of the ligand field increases considerably, relative to analogous $\mathrm{3d}$ compounds and is generally much greater than the spin pairing energy (hence, these mostly show LS behaviour).
II. $\ce{Cu2HgI4}$
$\ce{Cu2HgI4}$, (and the related $\ce{Ag2HgI4}$) undergo a temperature-dependent order-disorder phase change at ca. $67~^\circ\mathrm{C}$, reversibly changing from a bright red solid material at low temperature to a dark brown solid at high temperature, with intermediate red-purple states. the colours are probably due to $\ce{Cu(I)}$-$\ce{Hg(II)}$ charge-transfer complexes.

All the metals involved are $\mathrm{d}^{10}$ so one can rule out $\mathrm{d}$-$\mathrm{d}$ transitions, and spin-spin crossover. An electron transfer of some kind might be occurring, especially since the colours observed are so intense. Based on the pi-donating characteristics of iodide, one can also rule out metal to ligand charge transfer. The colour could be ligand to metal or metal to metal charge transfer, although the electron would have to enter something other than the d orbitals.
As the temperature increases and the lattice is rearranged, the $\ce{Cu(I)}$ and $\ce{Hg(II)}$ ions remain in the same environment relative to the iodide ions, but their environment in relation to the other metal ions is changing, and this supports a metal-metal charge transfer explanation for the colour.
III. Bis(diethylammonium)tetrachlorocuprate(II)
At room temperature, this compound is a bright green colour and changes
discontinuously to bright yellow at the phase transition temperature.
The thermochromic phase transition in this compound is quite complex Structurally, the salt transforms between two monoclinic unit cells, with three independent $\ce{CuCl4^2-}$ anions in the low-temperature phase and two independent $\ce{CuCl4^2-}$ ions in the high-temperature phase. The phase transition is first order due to the lack of any crystallographic
relationship between the two phases.
From reference 5:

No simple structural transformation pathway can be envisioned for converting one phase to the other. This accounts for the stability of the supercooled high-temperature phase as crystals of that phase, grown from the melt, can frequently be kept at room temperature for periods of
several weeks.The crystallographic study of the high-temperature phase shows the presence of disorder for both the cationic and anionic species
in the structure.

At temperatures below ca. $50~^\circ\mathrm{C}$, the four coordinate compound has a square planar geometry even though it is not a $\mathrm{d}^8$ compound, owing to the short hydrogen bonds between the hydrogens of the counter ion, diethylammonium, and the chlorines on the metal compound (this favours the square planar geometry).  When the temperature passes the phase transition temperature, the colour changes to yellow because the hydrogen bonds lengthen to (because of the phase transition between the two crystalline phases) and the geometry changes to a distorted tetrahedral  The change in geometry changes the arrangement of the d orbitals.

Square planar is pictured on the left, and tetrahedral is pictured on the right.
There is no stable intermediate between the square planar and the distorted tetrahedral geometry, so the change is an example of discontinuous thermochromism. Individual clumps of crystals may appear as either green or yellow depending on their exact temperature.
References

http://www.ncbi.nlm.nih.gov/pmc/articles/PMC3596041/

http://pubs.acs.org/doi/abs/10.1021/ed4000487

http://people.umass.edu/mjknapp/Chem242/2004%20Experiments/CuHg2I4Lab.pdf

http://www3.amherst.edu/~thoughts/contents/amberger-thermochromism.html

http://pubs.acs.org/doi/abs/10.1021/ja00230a020


Does baked bread have the same amount of calories as its ingredients? I bake a loaf of bread containing:

500 g strong flour (1700 kcal)
40 g butter (300 kcal)
24 g yeast (80 kcal)

For a total of 2080 kcal.

How many calories does my finished loaf contain?

My assumption is that at most, some of the energy in the yeast will be used during the proving process but I realised I know very little about the actual processes that take place during proving and baking.

Yes, the yeast does consume sugar to produce the bubbles in the dough, but I don't think it would be more than a few grams. During the baking of the bread, some water and alcohol evaporates and proteins are denatured.

Denaturing proteins does not change the calories. Since the alcohol that was produced is lost, then the calories are reduced. Water doesn't have any calories.

So, to answer to the question in the title: there are fewer calories in the baked bread than in the ingredients. Unfortunately, it cannot be predicted, the bread needs to be tested.


Difference between continuously overlapping p orbitals and conjugated pi bonds Is there any difference between the phrase "continuously overlapping p-orbitals" and "conjugated pi bonds" when referring to aromatic compounds? I've heard both used and I wasn't sure if they mean the same thing.

Interesting question!

To have continuously overlapping p-orbitals, the p-orbitals must be arranged in a cyclic loop.  The p-orbitals in benzene are continuously overlapping.  The p-orbitals in benzene are also conjugated pi bonds.

image source

In buta-1,3-diene the p-orbitals form conjugated pi bonds, however, they are not continuously overlapping - there are two "end" carbons that are not connected with each other, there is no continuous loop.

image source

Let's take it a step further:

Continuously overlapping p-orbitals are not always aromatic.

Example 1: Cyclobuta-1,3-diene (the square planar structure, $\ce{D_{4h}}$) has 4 continuously overlapping p-orbitals but it is not aromatic, it is antiaromatic.  For a continuously overlapping system of p-orbitals to be aromatic, it must contain 4n+2 pi electrons.

image source

Example 2: Have you heard of the Möbius strip? The mathematical (topological) concept has application in chemistry.  Imagine a large cyclic loop of p-orbitals where each p-orbital is twisted a few degrees from the preceeding p-orbital.  If the loop is large enough, then by the time we come back to the first p-orbital we have twisted the last p-orbital by 180°; there is a phase inversion (node) between the first and last p-orbital.  In a Möbius system the p-orbitals are continuous, but not continuously overlapping!  Möbius systems are aromatic if they contain 4n pi electrons.

image source

Continuous pi systems with an even number (0, 2, 4...) of nodes (cyclobutadiene, benzene, etc) are called Hückel systems in order to differentiate then from Möbius systems where there are an odd number of nodes (1, 3, 5...).

Lest you think the Möbius concept is purely theoretical take a look at these two molecules

image source

Isn't this what makes chemistry fascinating!


Am I using these spectra to identify the compound correctly? What is the structure of the molecule whose data is shown below? Explain your answer [...] EI mass spectrum: The highest peak observed is $m/z$ 210 and its abundance is low. The only other significant peak is observed at 167 Th (base peak). [...]

Full question text: https://imgur.com/a/awLve4z

Based on the IR value, I reasoned that it is a ketone. Nothing else in the IR spectrum really stood out to me, so I stuck to the idea that it must be a ketone. I assumed the mass of the compound is 210 g/mol so I subtracted the mass of oxygen and used the rule of 13:

Oxygen's mass = 16 g/mol

(210-16)/13 = 14 + 12/13

So the molecular formula is $\ce{C14H_{(14+12)}O = C14H26O}$

I noted that there were 7 signals recorded in the carbon NMR spectrum. I assumed that meant there were equivalent carbons in the molecule. The molecule I ended up drawing to match this is 7-tetradecanone.

Do you think this is the correct molecule?

Structure elucidation from spectrocopic analysis needs some experience. I have feeling that you are novice to this field by your explanation of the structure. You cannot determine the molecular formula by the data given and, hence you cannot determine the degree of unsaturation of the molecule. Yet, based on $\ce{^1H}$-$\mathrm{NMR}$, you can propose there are at least two aromatic nuclei (probably mono-substituted) in the molecule, be case there is a 10-proton multiplet in the aromatic region ($\pu{7-8 ppm}$). This assignment is supported by the 4 resonances in the region of $\pu{110-160 ppm}$ in $\ce{^{13}C}$-$\mathrm{NMR}$ spectrum. That also suggest these two aromatic nuclei are identical. Two aromatic rings are accounted for 8 degree of unsaturation.

The given FT-IR spectrum suggest there is a carbonyl group in the molecule, which is confirmed by the resonance at about $\pu{205 ppm}$ in $\ce{^{13}C}$-$\mathrm{NMR}$ spectrum. The peak at about $\pu{205 ppm}$ would also suggest that the carbonyl is not an aldehyde, but a ketone by its chemical shift ($\gt \pu{200 ppm}$), which would be confirmed by the lack of aldehydic peak at about $\pu{10 ppm}$ in $\ce{^1H}$-$\mathrm{NMR}$. The carbonyl group  is also accounted for another 1 degree of unsaturation.

Now, let's analyze $\ce{^1H}$-$\mathrm{NMR}$ spectrum. There is a 3-proton singlet at around $\pu{2.2 ppm}$ for isolated methyl group (without a neighboring hydrogen), indicating it is a benzylic methyl or next to a carbonyl group. Also, there is another one-proton singlet at about $\pu{5.1 ppm}$ for a methine group (-$\ce{CH}$-), indicating its highly deshielding nature. That deshielding must come from two aromatic ring and carbonyl group together. Thus, based on these data, I can deduct the structure as 1,2-diphenylpropn-2-one as depicted in the diagram below (blue box):

The highest peak observed in mass spectrum of the compound at $\pu{210 m/z}$ with law abundance and the base peak ($100\%$ intensity) observed at $\pu{167 m/z}$ confirmed the structure. As depicted in the red box (vide supra), the fragmented cationic radical at $\pu{167 m/z}$ is very stable, which explained the absence of any other fragmentations (as stated by OP, it is the only other significant peak in MS).


What is Nano Zinc Oxide? First of all, I have zero chemical background, but I seek for an easy explanation about Nano Zinc Oxide.

I was told that the name of "Nano Zinc Oxide" couldn't possibly exist. Instead, it should be coined as "Nano Zinc Chelate", since an Oxide particle is quite big and Nano is so small, relatively. It sounds very contradictory.

I know there is, of course, Nano Zinc Oxide, but I don't understand what he really means and  I feel enthusiastic to understand it.

Nano zinc oxide is a real material used in sunscreens and looks like this.  It is just little crystals of zinc oxide which are about 30 nm across.

Chelate refers to something which is produced by a chemical process of chelation.  Chelation means that a metal atom has more than one chemical bond to surrounding organics.  (That is kind of a bad definition, but I'm trying to keep the lingo simple.)  Chelation can be used to produce zinc oxide, but I don't think this is the process used for manufacturing most industrial zinc oxide.


What is the purpose of a boiling stick when heating a reaction mixture? What is the purpose of a boiling stick when heating a reaction mixture?

I read some stuff online and still don't understand how it works. Does it just ensure that the mixture will boil evenly? If so, then what's wrong when a reaction doesn't boil evenly?

This is the LibreTexts page I was reading information from. I don't understand the terms "bump" and "superheated". If the liquid is above the boiling point, why hasn't it turned into a gas then?

Boiling chips and boiling sticks are practical solutions to problems chemists face in the lab.

When boiling liquids they often observe the the boiling does not happen evenly. Instead of a gentle warming giving gentle bubbling we find that nothing happens for a while until, suddenly, a great deal of bubbling happens and this causes the liquid to splash overwhelming the capacity of the vessel and throwing the liquid into places we don't want it to be (during distillations, for example, the splash may throw liquid to the top of the distillation column, contaminating the product with raw liquid; during reactions the product can be expelled from the vessel entirely which is rarely what you want).

Boiling chips or sticks ensure the boiling happens evenly,  keeping things under control.

The problem they solve is that, in a smooth vessel, there are often not many rough patches to initiate bubbles. Bubbles are much easier to initiate if there is a small rough patch on the vessel wall. Chips or sticks provide such a rough patch. In the absence of a rough patch some liquids will superheat and may spontaneously vaporise but suddenly and violently. Roughness ensures bubbles start to form evenly and in many places avoiding superheating and making the vaporisation much more even.

The idea that things won't happen this way as "If the liquid is above the boiling point, why hasn't it turned into a gas then?" is a mistake from applying the ideas of thermodynamic equilibrium to a situation that is clearly not in equilibrium. If you are heating something it is clearly not in equilibrium and you have to worry about the practical effects not the ideal thermodynamic view. In practice, vaporisation is encouraged by nucleation sites (rough patches on the glass, for example). If there are not enough of those, it may not happen evenly.


What are some other gases that liquify at around the same temperature as ammonia? I am looking for gases that have a liquefication that falls within or on the edges of the liquefication range of Ammonia. I am mainly looking for gases that are comprised mainly of Carbon, Nitrogen, Oxygen, Hydrogen, Chlorine and Fluorine and liquify in a similar temperature and pressure range. This is for a worldbuilding project, but since the question is real, not theoretical, chemistry, I thought it would be better to ask about it here.

Wikipedia has a large list of gasses in ambient conditions here as well as a list of refrigerants here. You can sort them in order of boiling point by clicking on the small black triangles just underneath the respective column. Then just look for things in the range of ammonia, which boils at -33 °C under 1 bar. Note that a large number of entries in the table for refrigerants are for mixtures of gasses.

I don't think these lists are complete, but there are comparatively few substances with such low boiling points (probably no more than a few thousand that could ever be made), simply because having more than a handful of atoms will already cause the molecules to attract enough to remain as liquids or solids at non-cryogenic temperatures.


Difference in the functioning of alkaline and acidified KMnO4 What is the difference in the functioning of alkaline and acidified $\ce{KMnO4}$ as a reagent in organic chemistry?

E.g., in the oxidation of a primary alcohol to carboxylic acid, we use acidified $\ce{KMnO4}$ and not alkaline $\ce{KMnO4}$. Why?

Permanganate as oxidizing agent works most efficiently in acidic solution, because it is reduced to the greatest extent in this medium, from oxidation state +VII in $\ce{MnO4-}$ to +II in $\ce{Mn^2+}$.

$$\ce{8H+ + MnO4- + 5e- ->~ Mn^2+ + 4H2O}$$

Therefore, the number of electrons transferred from oxidized species per mole $\ce{KMnO4}$ (5 electrons) is the largest, and the least amount of $\ce{KMnO4}$ is needed for the reaction with a certain amount of alcohol. This makes oxidation in acidic medium the best choice for economic reasons.

The oxidation in neutral medium is less efficient, as manganese(VII) is only reduced to manganese(IV), yielding insoluble $\ce{MnO2}$ as a byproduct which needs to be removed from the reaction mixture during purification of the desired product (carboxylic acid).

$$\ce{2H2O + MnO4- + 3e- ->~ MnO2\downarrow + 4OH-}$$

Oxidation in strongly alkaline medium is least efficient in terms of transferred electrons per mole permanganate, as the latter is only reduced to manganate(VI).

$$\ce{MnO4- +e- ->~ MnO4^2-}$$

However, when the reaction is done under these conditions, a higher reaction speed and selectivity can be achieved, presumably because the alcohol will react more readily with $\ce{MnO4-}$ when deprotonated. Nevertheless, oxidation in acidic medium can be preferred for reasons stated above and substrates which are sensitive to strong bases.


What does it mean when "a gas is dilute in CO₂"? I'm reading an article about carbon sequestration, and I came across a strange phrase:  "The flue gas is dilute in $\ce{CO2}$".  I can't quite ascertain its meaning from the context.  I think it means that the $\ce{CO2}$ is in low concentration in the flue gas, but I'm no chemistry expert.

Does this have a technical meaning in chemstry when a gas is dilute in another substance?

In this case, it's comparing post-air-combustion $\ce{CO2}$ capture with pre-combustion capture, and with post oxy-combustion capture.

In the two latter cases, the  $\ce{CO2}$ stream is very highly concentrated.  However, with air-combustion, the flue gases contain a lot of nitrogen and NOx; hence the $\ce{CO2}$ is diluted in the flue gas, compared to the other two methods, where it's concentrated.

And as to Leonardo's supplementary question in the comments, about why the pre-combustion  stream has higher concentrations of $\ce{CO2}$ than the post-air combustion stream:

the comparison here isn't between two points in the same process. It's between analagous points in two completely different processes.

With post-combustion capture, the $\ce{CO2}$ is stripped out after the hydrocarbons (coal, gas, oil) have been burnt in air.

Whereas with pre-combustion capture, the hydrocarbon goes through a gasifier/shift process, resulting in a stream of $\ce{H2}$ mixed with $\ce{CO2}$. And at this point, the $\ce{CO2}$ is stripped out, and the $\ce{H2}$ is burnt.

And that's why the $\ce{CO2}$ concentration is higher in the pre-combustion capture process, than in the post air-combustion capture process.


Eutectic Phase Diagram and Lever Rule A solution having composition $p$ (left side of graph) is cooled to just above the eutectic temperature (point $s$ is at $0.18$ $x_{\ce{Si}}$ and the corresponding tie line intersects the liquidus curve at $0.31$ $x_{\ce{Si}} $); calculate the composition of the solid that separates and that of the liquid that remains.

The liquid phase is obviously 31 % silicon and 69 % gold as given in the question. The solutions manual also gives the weight percentage of the solid phase. Using the lever rule,

$$ m_\text{solid}(0.18)=m_\text{liquid}(0.31-0.18)$$
$$ \frac{m_\text{solid}}{m_\text{liquid}}=\frac{0.13}{0.18} $$
$$ \frac{m_\text{solid}}{m_\text{solid}+m_\text{liquid}} = \frac{0.13}{0.13+0.18}=0.42$$

However, the solution manual states that the wt% of solid phase is 58 %, and that of the liquid phase is 42 %. Is the solutions manual wrong, or am I losing my mind?

You are correct. Another way to calculate it is: $$X_{\mathrm{L}} = \frac{s-b}{e-b} = \frac{0.18-0}{0.31-0}=0.58$$
You can also inspect the diagram visually and see that the $bs$ line is longer than the $se$ line. The lever rule dictates that the line length is proportional to the percentage of the phase opposite from the line. Thus, it is not possible that there is more solid than liquid at this point.

There is definitely a mistake in your textbook's solutions.


SMARTS for cis amide Is there a SMARTS pattern for cis (and trans) amide bonds?

This seems to be the SMARTS for any amide:

[NX3][CX3](=[OX1])[#6]

But I am not sure whether or not the SMARTS format has cis/trans definitions for “single” bonds such as amides.

No, not really.

For a double bond, you'd use the same '\' and '/' characters to indicate cis- and trans- stereochemistry that you use in SMILES. But as you indicate, there isn't a double bond for such stereochemistry.

My personal preference in scripts, code, etc. is to match the double bond, and then use my own code to check the relative positions of substituents around the central double bond. In this case, you could look at a dihedral angle around the C-C-N plane.

For future reference, an excellent resource for testing SMARTS patterns is SMARTSView.


Why is only the carbonate anion used to precipitate Barium? Q: Why is the $\ce{CO3^2-}$ anion used to precipitate Barium from $\ce{BaBr2}$ and not another anion, such as $\ce{PO4^3-}$, or $\ce{SO4^2-}$?

I am studying high school Chemistry, and the current unit is precipitations. The worked example states that $\ce{Na2CO3}$ can be used to precipitate $\ce{Ba^2+}$ from a solution of $\ce{BaBr2}$, which makes sense.

However, it does not explain why the $\ce{CO3^2-}$ anion should be selected, when according to the solubility table $\ce{SO4^2-}$ and $\ce{PO4^3-}$, would both be able to precipitate $\ce{Ba^2+}$ from $\ce{BaBr2}$. Am I missing something obvious? Why is it that the $\ce{CO3^2-}$ anion is favoured over the other two anions?
`

Actually, the textbook should remove this question. You should discuss this with your teacher. It will be a service to future students. The heavier elements of group II elements form insoluble sulfates, phosphates and carbonates.

Now carbonates display another interesting phenomenon. If you keep bubbling carbon dioxide in into a group II carbonate solution (or better say suspension), the group II ions form water soluble bicarbonates. It was a classical experiment of "old school" chemistry. Bubble carbon dioxide into a clear solution of lime, it would become cloudy, if you continue that, it becomes clear again.

Another reason, which may be valid is that you may wish to re-dissolve precipitated barium carbonate after filtration. That would be easy to do so by adding a small amount of strong acid. However re-dissolving barium sulfate or phosphate is very difficult.


Equilibrium constant of water If $\ce{H2O}$ dissociates as follows: $$\ce{2H2O <=> H3O+ + OH-}$$ and not in this way: $$\ce{H2O <=> H+ + OH-}$$ why is the equilibrium constant: $$K_{eq}=\frac {\ce{[H+][OH-]}} {\ce{[H2O]}}$$ and not: $$K_{eq}=\frac {\ce{[H3O+][OH-]}} {\ce{[H2O]^2}}$$

In a strict sense, all quantities in the equilibrium constant are not molar concentrations but activities. Activity is a thermodynamic quantity which is proportional to concentration. Now by IUPAC definitions of standard states, the activity of pure liquids and solids is taken is unity (by convention and for reference purposes. Therefore in the following expressions, $\{\ce{H2O}\}$ or $\{\ce{H2O}\}^2$ are all unity, no matter what the power is (real number). Elementary textbooks tend to say a lot of wrong or oversimplified stuff.

$$K_\mathrm{eq} = \frac{\{\ce{H+}\}\{\ce{OH−}\}}{\{\ce{H2O}\}}$$

and

$$K_\mathrm{eq} = \frac{\{\ce{H3O+}\}\{\ce{OH−}\}}{\{\ce{H2O}\}^2}$$

Here the curly bracket indicate activities. The IUPAC notation is $a$ for activity. The species of interest is written as a subscript.
A more detailed answer as to why activity is taken as unity is already provided:
Why are activities of solids and liquids taken to be unity?


Electrode polarity in fuel cells I am studying fuel cells right now and I'm confused about the polarity of the two electrodes.

In the diagram below, the anode is shown as negative. However, why do $\ce{H+}$ ions move away from the anode, when they should be attracted to it because of its negative charge?

On the other hand, if I swapped the polarity around, movement of electrons from the positive to the negative terminal wouldn't work out.

Please explain which polarity is correct, and why.

Please explain which polarity is correct, and why.

In a fuel cell, the anode does sit at a more negative potential than the cathode. This polarization is opposite that of an electrolytic cell, where the anode is positive relative to the cathode.

The chemical composition of the fuel $(\ce{H2})$ and oxidant $(\ce{O2})$ feeds at the anode and cathode, respectively, cause this.  Thermodynamically, in this system the product of the fuel cell reaction, $\ce{H2O}$, is much more stable than the reactants.  Thus, the system biases itself to move in favor of the forward reaction:

$$
\ce{2H2 + O2 <=> 2H2O}
$$

As the $\ce{H2}$ is oxidized, the electrons from this half-reaction are passed into the anode:

$$
\ce{2H2 -> 4H+ + 4e-}
$$

These electrons have to move through the circuit in order to be able to supply the electrons needed for the oxygen reduction reaction:

$$
\ce{O2 + 4H+ + 4e- -> 2H2O}
$$

Here's the key: In order for the electrons to move through the circuit in the correct direction, the potential has to be more-positive at the cathode: the negatively charged electrons are attracted toward the positively charged cathode, along the conductive pathway of the circuit.

  However, why do $\ce{H+}$ ions move away from the anode, when they should by attracting because of negative charges?

For the most part, the circuit is going to be composed of (semi)conducting materials. Conductive materials in a circuit cannot sustain a very substantial charge polarization at all--as soon as they become very slightly polarized, they allow current to pass that equalizes the charge imbalance. (This is why they are used to carry current!)

As a result, the actual charging of the anode is negligible, even though it develops a finite electrical potential.  Thus, the electrostatic attraction of the $\ce{H+}$ ions to the anode can in general be ignored, as it is orders of magnitude too small to keep the protons from diffusing away, toward the cathode where they are being consumed to produce water.


Can you give examples for a halide ion reacting with interhalogen compound to give polyhalides? Interhalogen compounds are formed when two halogens $(\ce{F},$ $\ce{Cl},$ $\ce{Br},$ $\ce{I})$ react together. They are represented by $\ce{XX'_n},$ where $\ce{X}$ is halogen of larger size and $\ce{X'}$ of smaller size. The total number of halogens in $\ce{XX'_n}$ depends on the radii of $\ce{X}$ and $\ce{X'}.$ The correct statement(s) regarding $\ce{XX'_n}$ is/are

  ❌ 1. $\ce{FCl3}$ is one of the stable interhalogen compounds.

  ✅ 2. Iodine can form an interhalogen compound containing maximum numbers
  of halogen atoms.

  ✅ 3. Interhalogen compounds are covalent in nature and more reactive than
  halogens, in general.

  ✅ 4. Halide ions reacts with interhalogen compounds to give polyhalides.

Regarding the 4th option: will it be like the following:

$$\ce{X'^- + XX'_n -> XX'_{n + 1}}?$$

Something like this? ($\ce{X}$ and $\ce{X'}$ are halogen compounds), so the polyhalide ion here would be $\ce{X'_{n + 1}}.$

What would be you views on this?

TLDR: There is no exact generalized reaction. The reaction depends on type of interhalogen compound(value of n) and the nature of metal halides(ionic or covalent)

$\ce{XX^{'}}$ form

Let us take iodine monochloride($\ce{ICl}$). If it is reacted with ionic chloride like $\ce{KCl}$, it will form $\ce{ICl2-}$ ion.

$$\ce{MCl + ICl -> M+ICl2-, (M=K, Rb)}$$

If a different counter-ion is taken like bromide, it will form $\ce{ClBrI-}$ ion

$$\ce{ICl + KBr -> K+[ClBrI]-}$$

But if we add pure diatomic halogen, it will form a additive product

$$\ce{Cl2(excess) + ICl -> ICl3}$$

Covalent halides like $\ce{AlCl3}$ however gives different product:

$$\ce{ICl + AlCl3 -> I+ + AlCl4-}$$

$\ce{XX^{'}3}$ form

Let us take iodine trichloride. It will form $\ce{ICl4-}$ ions on reacting with ionic chloride:

$$\ce{MCl + ICl3  -> M+ICl4-}$$

If we take any other counter ion like fluorine, it will form $\ce{MICl3F}$ which is again kind of an additive product.

$\ce{XX^{'}5}$ form

Taking chlorine pentafluoride as examples, reacting with covalent fluorides like $\ce{AsF5}$ and $\ce{SbF5}$ will give adducts: $\ce{ClF5.AsF5}$ and $\ce{ClF5.SbF5}$

Iodine pentafluoride gives iodine heptafluoride on reacting with fluorine at 250 C

So, there is exact no generalized reaction. It all depend upon type of inter-halogen compound, reaction conditions and nature of metal halide. The close generalization I can make is that interhalogen compound of form $\ce{XX^{'}_n}$ if reacted with ionic metal halide will form metal polyhalide of form $\ce{MXX^{'}_{n+1}}$ or $\ce{MXX^{'}_{n}X^{''}}$(if a different counter-ion is taken). Additive product is formed when inter-halogen compound is reacted with pure diatomic halogens provided the second halogen is big enough to occupy electrons like bromine or iodine. In some case, covalent halides forms adducts with interhalogen compounds.

Thus the generalized reactions are(of course, there are certain exceptions):

$$\ce{X^{'-} + XX^{'}_n -> XX^{'-}_{n+1} (ionic)}$$

$$\ce{X^{'}_2 + XX^{'}_n -> XX^{'}_{n+2}}$$

Reference

http://www.vpscience.org/materials/Study%20Material%20on%20d%20Inter-halogen%20%20Compounds%20By%20Dr.%20D.%20M.%20Patel.pdf (reaction sources)


Are photolysis reactions considered exothermic or endothermic? I know that photolysis is when a chemical reaction (usually decomposition) is forced by photons. I also note that there is a general warming in the stratosphere, due to the photolysis of ozone $$\ce{O_3 + $h\nu$ -> O + O_2}$$ and the subsequent restorative mechanism $$\ce{O + O_2 + M ->O_3 + M}$$ where $\ce{M}$ is a third body used to carry off excess energy. To me, that says that the photolysis of ozone is an endothermic reaction, and the reformation of ozone is an exothermic reaction, as the free body takes away excess energy (assumably thermal energy).

Am I correct in generalizing that photolysis is generally an endothermic reaction? If so, could the enthalpy released by the second equation be quantified by the frequency of the photon?

In general it could be either. In your example you will have to work out the heat of formation of ozone and of O atoms (presumably it is an endothermic reaction). In general what the photon is doing is providing a way to overcome an activation barrier in the ground state by opening up a pathway from an excited state to the products. This last step is generally exothermic.

The enthalpy released in the second reaction is less than the photon energy by the difference in energy of the photon and that of products, an oxygen atom in your case as the heat of formation of O$_2$ is zero.

The sketch gives the general idea, I hope, for an endothermic reaction :)

Notes: The barriers may also be called transition states. There may be no barrier in the excited state. The crossing from the excited state potential to the product may be controlled by Landau-Zener behaviour or be a conical intersection. The far right upper line (not labelled) is the potential of the excited state of the product


Can all water soluble ionic compounds conduct electricity? I'm taking a first year university chemistry course, and I am reading a lot of contradicting things. Some people say that ionic compounds that are soluble in water are always electrically conductive, others say that it maybe conductive.  Which one is it?

I have this problem to think about:

  Suppose that an unknown chemical compound exhibits the following properties:

  it is crystalline but shows no electric conductivity in the solid state
  it melts at $\mathrm{300^\circ C}$ and decomposes at $\mathrm{400^\circ C}$
  the compound is soluble in water, the solution shows no electrical conductivity.

  What kind of chemical bond would you expect in the
  given compound? Try to describe this type of bond in detail.

Everything screams "ionic" (high melting point, solid at room temperature, crystalline structure, non-conducting in solid form) besides the fact that it doesn't conduct in water.

DavePhD is right! Your material is an organic compound.

In order to be soluble in water, it should have some polar substituents. I'd however exclude carboxylic acids or phenolic $\ce{OH}$ (except maybe in the proximity of a carbonyl group) since these will partly dissociate and yield to minor conductivity.

If the melting point wouldn't be that high, inositol would be a candidate.

The nucleobases thymine and uracil do show melting points in the 300 °C range and are both soluble in water.


in which way calcium promotes cell division I have read in bio chemistry that calcium plays an important role in the cell division but the in depth information is'nt available in the textbook and I cannot memorize any fact without a supporting fact I just want to know what is its role in the process of cell division

see its simple all you need to know is that calcium is responsible for the formation of middle lamellum which is a key structure in the cell wall formation if only karyokinesis occurs and cytokinesis fails to happen then the condition of resultant cell would become multinucleated which most plants do not have


Are titration, recrystallisation, functional group analysis, etc. outdated? In modern chemistry,
- pH meter, ion electrode, AES, HPLC, etc. replacing titration
- HPLC replacing recrystallisation
- IR and NMR replacing melting point analysis, derivative analysis, functional group analysis

In other words, are the techniques that being replacing only being used inside school and teaching labs?

It is true that modern techniques make it faster to elucidate the identity of compounds and isolate them from the reaction mixture, however to call the traditional methods like titration, recrystallization, or column chromatography (basis of HPLC) outdated is far too broad. You could probably get away with calling them outdated if all you wanted to do was to characterize compounds/final product. But when you're in synthesis, you still need the 'traditional' methods to prepare larger amounts of intermediate.

For example, HPLC is only able to separate small amounts of compounds (possibly enough for characterization) and if you wanted enough to go to the next step of your reaction (or if you wanted to make a lot), HPLC would be a highly inefficient way of doing this. Recrystallization, Distillation, and/or Column Chromatography (depending on the compound) would be much better for this purpose.


Why should one perform an extraction with multiple small portions of solvent as opposed to one large portion? I'm performing a liquid-liquid extraction on a 1 l water sample for GC analysis. The procedure calls for three extractions with 50 ml of dichloromethane. Would one extraction with 150 ml produce a different result?

The important thing is the distribution ratio (for the extraction of a metal) and the partition coefficent for the extraction of a given chemical species. This can be used for a first approximation calculation.

The distribution ratio is the total metal concentration in the organic phase (grams of metal per litre) divided by the total metal concentration in the aqueous phase.

If we consider a batchwise extraction where a litre of water bearing x mg of a metal is extracted with an organic solvent. Assume that the distribution ratio is 2. We can do the calculation for the extraction.

The phase ratio (ratio of the volumes of the two phases) is 1, call this theta.

Theta D / (1 + (Theta D)) = the fraction extracted in each stage.

If D is 2, and theta is 1 then we will extract 66 % of the metal.

As an alternative, if we use one litre of water and extract this with 0.5 L of organic phase, then with D = 2 and Theta = 0.5 then we will extract 50 % of the metal.

If we repeat the extraction with another 0.5 L of solvent then we will recover 50 % of the remaining 50 % of the metal. Thus we will have recovered in total 75 % of the metal.

If instead we were to extract the 1 L of water four times with 0.25 L of water then in the first extraction we would leave 66 % of the metal in the lower phase. The easy way of working out the amount left in the aqueous later after n extraction stages with the same D and theta valve is to use

{1 - ( D Theta / {1 + (D Theta)})}^n = fraction left in raffinate. In solvent extraction raffinate is the term for the aqueous layer after an extraction has occured.

This gives me about 20 % left in the aqueous phase after four extractions with 0.25 L of solvent.

It is important to understand that the solubility of the substance is not what you should consider it is the distribution ratio or the partition coefficent which is needed.

If you were to take a dilute solution of DEHPA (150 mM) in aliphatic kerosene and to shake this with 0.1 nitric acid containing a small amount of Fe-55 with no stable iron as iron(III) then you might get a D value of 100 for arguments sake.

Now the organic phase can only extract 50 mM of iron per litre as to extract one iron atom three DEHPA molecules are needed. If we were to repeat the extraction experiment with a solution of 56 grams of iron per litre (roughly 1 moles per litre of iron) spiked with a little Fe-55 then we would measure a far lower distribution ratio.

This is due to loading of the organic phase.

Equally when you extract a large amount of an organic substance into an organic solvent the solvent properties of the organic layer at the end will be different to those of the pure solvent.

For example if I have 1 litre of water saturated with hexanol with 500 ml of water saturated hexanol floating on the top. Then if I was to add 500 ml of ether and then shake it. I would have a different Kd for hexanol to that which I would have if I was to shake a dilute solution of hexanol in water with ether.

I hold the view that the formula used above for the system with a D value of 2 will only hold perfectly true when the concentration of the metal is zero. This is why radioactive tracers are so useful for academic solvent extraction experiments. This is becuase they allow the measurement of distribution ratios with a zero metal concentration. The concentration in moles per litre of a 1 GBq per litre solution of Fe-55 will be very small.

The half life of Fe-55 is about 2.7 years, thus I have calculated that this very radioactive stock solution will contain 205 nanomoles of iron if the only iron atoms present are Fe-55. The best way to make the iron-55 would be to bombard managense with protons with a cyclotron and then to separate the iron from the manganese by solvent extraction or some other separation method.

I would in a lab expect to be able to run a solvent extraction experiment with only 700 Bq of Fe-55. If we assume that I was to shake 700 microlitres of aqueous phase with 700 microlitres of the organic phase. I would then be able to sample 200 microlitres of the two layers with a pipette.

If we assume that the highest distribution ratio I was to measure was 100 then I would expect that at the end of the shaking that the aqueous layer would contain 10 Bq per ml. If I was to sample 200 microlitres and to put this into a LSC vial and then to count it. Assuming that 100 % of radioactive decays will result in a count in the LSC machine then I would in one minute get 120 counts. If I was to count until I had 10000 counts (1 % error due to count number) I would have to wait about 83 minutes.

For a D value of 10 I would be able get 10000 counts in less than ten minutes for the least radioactive liquid phase. Thus it should be clear that with a radiotracer it is possible to measure the distribution ratio with less than 1 nanomole per litre of metal present.


Mechanism of Acidity - Boric Acid How does boric acid behave as an acid? Is it a Lewis acid or a Brønsted-Lowry acid?

Boric acid is a Lewis acid, therefore when in water it accepts the lone pair from oxygen and becomes $\ce{(OH)3B^{-}-OH2^{+}}$.
From this compound a proton can be donated and this is how boric acid behaves as a Brønsted acid.


Why does the solubility of LiCl increase with temperature if the dissolution is exothermic? In an answer to another question, the following data on the solubility of $\ce{LiCl}$ at different temperatures is presented.

This suggests that dissolving LiCl is endothermic (according to the van't Hoff relationship).  However, the enthalpy of dissolution has been measured and extrapolated to infinite dilution, and it is negative (the value given in one of the comments is $\Delta H = \pu{-37 kJ/mol}$, reference).

What is the reason for this discrepancy? Is it because $\ce{LiCl}$ is so soluble that there is some non-ideality effect? Maybe dissolving the first couple of molecules of $\ce{LiCl}$ is endothermic, but then when you dissolve more it becomes exothermic? Or is the activity of $\ce{LiCl}$ at increasing temperatures actually decreasing even though the concentration is increasing? Or is the van't Hoff relationship not a general one?

There is another question ($\ce{NaOH}$ solubility with temperature) which claims that Le Chatelier's principle sometimes does not apply, but it does not mention van't Hoff (and the answer might be incomplete).

The key to the answer lies in this comment in the answer to the linked OP:

  the dissolution of NaOH is exothermic, yet it's solubility increases with temperature.

So in short it's the opposite of #2: dissolving the first couple of molecules of $\ce{LiCl}$ is exothermic, but then if you dissolve more it become endothermic. The heat of dissolution to form an infinitely dilute solution is not the same as the heat associated with formation of a saturated solution.

Alternately you might say that the heat of dilution of $\ce{LiCl}$ is also exothermic (see e.g. Reference 1).

Reference

YC Wu, TF Young, Journal of Research of the National Bureau of Standards Vol. 85, No.1, January- February 1980, p. 11-17.


What are computed energies in DFT and FF methods relative to? I am not talking about zero-point energy.

I wish to understand how the energy of a system is calculated relative to a zero baseline. This is best explained with an example.

From DFT, the energy could be:

Isolated H = -13.6 eV,
H2 = -31.8 eV

What are these numbers related to? What would be a system at 0 eV? Or what is the zero baseline from which these number are calculated from?

I have also noticed that force-field energies tend to be positive. What is the zero baseline here?

In summary:

What is the DFT baseline?
What is the Forcefield baseline?

Start by looking at the Hamiltonian for a molecular system
\begin{equation} \label{eq:coulomb_hamiltonian}
    \hat{H}
    = - \sum\limits_{α=1}^{ν} \frac{1}{2 m_{α}} \nabla_{α}^{2}
    - \sum\limits_{i=1}^{n} \frac{1}{2} \nabla_{i}^{2}
    - \sum\limits_{α=1}^{ν} \sum\limits_{i=1}^{n} \frac{Z_{α}}{r_{αi}}
    + \sum\limits_{α=1}^{ν} \sum\limits_{β > α} \frac{Z_{α} Z_{β}}{r_{αβ}}
    + \sum\limits_{i=1}^{n} \sum\limits_{j > i}^{n} \frac{1}{r_{ij}} \, ,
\end{equation}
where $m_{α}$ is the rest mass of nucleus $α$ and $Z_{α}$ is its atomic number, $r_{αi} = |\vec{r}_{α} - \vec{r}_{i}|$ is the distance between nucleus $α$ and electron $i$, $r_{αβ} = |\vec{r}_{α} - \vec{r}_{β}|$ is the distance between two nuclei $α$ and $β$, and $r_{ij} = |\vec{r}_{i} - \vec{r}_{j}|$ is the distance between two electrons $i$ and $j$.

Now, thinking classically, the total energy is equal to zero when all the terms in the Hamiltonian are equal to zero, so that the zero of the total energy corresponds to the case when, (again) classically speaking, particles are infinitely far away from each other and not moving. Obviously, the total energy of any stable molecular system should be lower than that, i.e. negative, for otherwise the system will be unstable with respect to dissociation towards particles. That is the way it works for any method of computational chemistry that solves the Schrödinger equation or its relativistic counterpart is some way or another (HF, post-HF, DFF, etc.).

But force field methods and energies are completely different beasts, because the meaning and the zero of energy in this realm depends on the functional form of potential energy and parameter set used in a particular force field. So, in molecular mechanics in general absolute energies have no meaning, only relative energies (energy differences) should be used for any kind of analysis.


Hydrogen peroxide and ThickenUp (potassium chroride) Retr0bright is used to whiten yellow plastic.
The original recipe contains xantham rubber that is a bit hard to find.
I'm thinking about using Nestlé ThickenUp that contains xantham rubber, but it also contains potassium chloride. I know you should be careful with mixing hydrogen peroxide with other chemicals.
Does anyone here know if mixing 12% hydrogen peroxide with ThickenUp should be avoided or is considered safe?
Content of ThickenUp is corn and potato maltodextrin, xanthan gum, potassium chloride.

Hydrogen peroxide is an oxidizer and unstable.  It can react with a variety of common materials to exothermically release oxygen gas, exothermically oxidize the material, or both.  Additionally it can damage skin.  Fortunately these safety concerns are only really a major consideration at higher concentrations, where the amount of oxygen released can be problematic, and the heat is sufficient to pose a burning or fire hazard, made worse by the oxygen.  At lower concentrations (such as the 12% you're talking about), it poses mild hazard to skin, and significant hazard to clothes and eyes, but shouldn't present any major safety concerns.
The potassium chloride shouldn't do anything, but the hydrogen peroxide may react with the maltodextrin, or other unlisted ingredients.  To test for this, you can dilute down your hydrogen peroxide (or buy some medical hydrogen peroxide that is already lower concentration), and mix it with the TickenUp.  If bubbles or foam forms, that's an indicator that the hydrogen peroxide is reacting with the maltodextrin or some other ingredient in the TickenUp, and you shouldn't use it, because it will destroy your peroxide before it can bleach your plastic.
Of note, additional care should be taken to wash the plastic after using this form of Retrobright, because the maltodextrin and potassium chloride may leave an undesirable residue otherwise.  Additionally, the xantham gum is only required if you wish to apply the Retrobright as a gel, if you are planning on submerging the plastic, the xantham gum is unnecessary.


What compressible rubber like gasket material is the least permeable to CO2? I am trying to can a carbonated solution at 120 PSI.  I am trying to figure out the best material to use for the gasket on the valve cap. I am carbonating my fluid with $\ce{CO2}$ and it is loosing pressure over time due to the permeability of $\ce{CO2}$ through my gasket. I have tried about 8 materials (aluminum, copper, graphite,  buna, FMK A, Viton, ECH, PVDF) I was not able to compress the metals enough with my equipment we also dismissed a number of other materials that are known to be highly permeable to $\ce{CO2}$ or are not feasible and have not found a suitable gasket yet. I also tried using $\ce{N2O}$ as my gas but it seems to have the same permeability as $\ce{CO2}$. my gas has to be soluble in water and non toxic. I know that Viton and FMK A are the same but one supplier said theirs was more pure than anyone else.

Viton is a trade name for a fluoroelastomer type. There are different grades which have different properties. I've had a Viton seal be deformed in a vessel when using tetrahydrofuran as a solvent, whereas Kalrez does not. I've seen different Viton grades again not have great compatibility with acetone. So do check out the compatibility of each supplier's product.
We have two seals that we use for $\ce{CO2}$ applications. EPDM and PTFE.  These both cope with 100 bar pressure.


Aufbau principle in filling Ti A few weeks ago, I was given this question to complete:

  The ground state electronic configurations of Ti is $\ce{[Ar] 4s^2 3d^2}$
   State whether it follows the Aufbau principle or not. If not, explain why and state what the configuration would be if the principle was followed.

Now, I know the Aufbau principle tells us that when filling orbitals with electrons, the lowest energy orbital is filled first.

Looking at this diagram and the periodic table, I thought the filling was right - $\ce{4s}$ is before $\ce{3d}$ so I thought it was reasonable.

However, according to the Lecturer this is apparently wrong.

He says:

  The $\ce{3d}$ level is at a lower energy than the $\ce{4s}$ and should be filled first according to the principle. It takes the above configuration to minimise the d-d interelectronic repulsion energies. The alternative configuration of [Ar]3d4 would have a higher energy.

I'm slightly confused here.

How am I wrong?

The energy level diagram which you give and is to be seen in the standard textbooks is assumed to apply to all elements. This is not the case, nor would we expect it to be so. After calcium the 3d drops below the 4s so fills first. Coulombic repulsions tend to push electrons up into the larger 4s orbital where repulsion is less. This means the 4s electrons are the outer electrons which are lost first in chemical reactions and define the atomic radius of the atom.


Meaning of "ad" in a pharmaceutical patent: "Acetic acid ad pH 5.0" Can the abbreviation "ad" mean "up to" in pharmacology? I saw a translator translate the Russian line "до pH 5.0" ("pH level up to 5.0") in a table describing the composition of a drug product as "ad pH 5.0". The meaning of the Russian record was "the content of the excipient in the product may be as high as to bring the product's pH to 5.0, but not higher".

I looked up a list of pharma abbreviations, and "ad" there does not have the meaning "up to".

I googled and found this on Google Patents:

  Ammonium acetate 0.77 mg Sodium chloride 7.21 mg Benzyl alcohol 10.0 mg Polysorbate 80 0.2 mg Acetic acid ad pH 5.0 0.1 q,s, NaOH 0.1 N ad pH 5.0 0.1 q,s, Water for injection ad 1.0 ml

But this is unlikely to mean "up to", yes? I wonder what the meaning is there.

To be on the safe side, I used "up to" in my translation, but now I'm curious.

Most likely yes. In latin,(1) ad means "to, towards, until", like in "ad libitum" (aka ad lib) which means "keep going until I am satisfied", or "ad nauseam".

(1) Pharmaceuticals very often use latin.


Under what circumstances can the stereochemistry at nitrogen atom be faithfully retained and identified? Under what circumstances can the stereochemistry at nitrogen atom be faithfully retained and identified?

Because in most cases nitrogen atoms tend to undergo stereorandomisation by flipping due to its lone pair. I know of one example which has a three membered ring where nitrogen flips very slowly due to the high s character in the lone pair which requires significant activation energy to flip. But I wanted to know of the other cases where this is possible (probably in a stereo locked system inhibiting flipping).

You could try to lock the conformation of structure. For example, derivatives of a  1-azabicyclo[2.2.2]octane system could have a chiral nitrogen, and the fact that the substituents are tied back prevents inversion of the nitrogen. Quinine has this kind of structural moiety.

This example isn't exactly what you're looking for, but it does it by using diastereocontrol to limit the amount of undesired isomer: J. Org. Chem. 2004, 69 (3), 714–718.

As Ivan suggested, you can hold onto the lone pair with something. Here's an example where metal coordination prevents inversion: Inorg. Chem. 2012, 51 (15), 8516–8523.

There are plenty of others. I just did a quick search on the JACS website for stereogenic nitrogen.


How are imaginary frequencies and force constant calculated by DFT? Computational calculations allow us to simulate the frequencies of molecules. They can even tell us if the optimized structure is a minimum, a saddle point or a maximum according to the number of imaginary frenquencies.

I was wondering how do quantum calculations could yield these frequencies. Does it have any link to the force constants of a molecule ?

Within DFT the vicinity of the solution to the Kohn-Sham equations is explored treating small changes of nuclei positions as a perturbation. The restoring forces that act in the system upon such perturbations give us information about the force constant matrix, from which we can get the harmonic vibrational frequencies as explained elsewhere, e.g. in the book of Wilson and Decius. Basically, you need the second derivatives.

A key point is to do it efficiently. Doing it numerically is very costly.
Since the basis functions are known and they have some nice properties one can do it analytically. Thus, researchers have devised a set of coupled equations to compute the effect of perturbations on the density matrix, which is the core of the calculation.

You will find a detailed explanation in this article by F. Neese.


The effect of a voltmeter on the electron flow of a voltaic cell Due to a voltmeter’s strong resistance nature, how could electrons flow from anode to cathod? It is more reasonable to assume that a voltmeter would stop the redox reaction because the electrons could never reach the positive electrode. When I measure the potential energy between two electrodes with a voltmeter, is the reaction happening? If the reaction is not taking place, how come the potential difference exists?

An ideal voltmeter has an infinite resistance, so does not let any current flow in the system. However nothing is ideal in real life, it consumes a very small amount of current to move the magnet and needle arrangement or electronics. An ideal ammeter has zero resistance. Assume that you have a water tank on your ceiling which supplies water to your kitchen tap. If I add a pressure gauge before your closed kitchen tap, it will read a positive pressure, much greater than the atmospheric pressure. Note that there is no water flow. Forget about the textbook picture of measuring galvanic cell voltages using a voltmeter (as shown in chemistry textbooks), as nobody every measures cell potential this way. A reliable approach to measure cell potential is called null-point detection (you can search more about it). As the name indicates, it draws zero current at the point of balance condition. It is one of the most precise and accurate techniques in classical potentiometry.

Just think a little deeper, the two electrodes in a galvanic cell do not "know" each other and they are oblivious to each other's presence. So current flow is not a requirement to measure potential difference. Each single electrode is in contact with its ion in the solution. If you have a zinc and copper galvanic cell, each has its own electrode potential e.g. a Zn rod dips in Zn2+ soln and Cu rod dips in Cu2+ solution. Electrode potential is the thermodynamic tendency of ions (in solution) to reduce themselves without any current flow, in this case Zn2+ --> Zn and Cu2+ --> Cu.  The voltmeter simply reads the difference of potential between zinc and copper, however it does not know the individual electrode potential of zinc and copper.


What is there in an atom between the nucleus and electrons? As you can see the atom consists not only of the nucleus and electron but also of "empty space is the space empty or is their something else?

It is not necessarily empty space.

For example the flux of neutrinos on Earth is about $3 \times 10^{12}/cm^2/s$.  At a given moment, a neutrino may or may not be within the atom or even within the nucleus of the atom. Similarly, other particles may be present within an atom at a given time.

Also, an electron may be inside the nucleus in accordance with the Fermi Contact Interaction.  So if you have a hydrogen atom, having only one electron, and the electron is inside the nucleus at a given moment, there will be no "between" for there to be space, empty or not.


Carbocation stability order I was wondering whether the following order (from here) is correct:

In the first inequality, why did 9 hyperconjugations dominated resonance effect of benzene ring?
In the last inequality doesn't the 2 hyperconjugations of vinylic carbocation overpower the methyl carbocation?

In middle there is:
allyl = secondary

Why is there an equality here when resonance dominates inductive effect?

About the first two, it's an exception to the rule that resonance dominates hyperconjugation.

About the last inequality, the vinyl cation is sp hybridized which means it is highly electronegative, and a positive charge on it will be very unstable.

About the middle inequality, the benzyl carbocation (with five resonance structures) is more stable than allyl carbocation (two resonance structures). You may verify this here.


Can a benzyl ether be removed in the presence of an ester? I'd like to deprotect a benzyl group for a Fráter–Seebach alkylation. Would it be possible to do this with a Birch reduction using Li/NH3 (or Na/NH3), without also reducing the ester carbonyl group?

If not, hydrogenation over Pd/C seems to be a plausible alternative. How else can I remove the benzyl group on oxygen?

In Wutz's book on protecting groups,[1] I found an example of almost instantaneous reductive debenzylation which leaves an ester untouched:[2]

The use of BCl3 is also possible:[3]

References:

Wuts, P. G. M. Greene's Protective Groups in Organic Synthesis, 5th ed.; Wiley: Hoboken, NJ, 2014, p 131.
Sulikowski, M. M.; Davies, G. E. R. E.; Smith, A. B., III. Avermectin–milbemycin synthetic studies. Part 7. An approach to the southern hemisphere of milbemycin α1. J. Chem. Soc., Perkin Trans. 1 1992, 979–989. DOI: 10.1039/P19920000979.
Williams, D. R.; Brown, D. L.; Benbow, J. W. Studies of Stemona alkaloids. Total synthesis of (+)-croomine. J. Am. Chem. Soc. 1989, 111 (5), 1923–1925. DOI: 10.1021/ja00187a081.


Basicity of nitrogen heterocyclic compounds Question :
The order of basicity for the following compounds is :

The logic I've applied is that since 4 has a conjugated system, the lone pair of N will be delocalised and hence would be the least basic. Conjugate systems/Resonance is not seen in the other three compounds
I'm not entirely sure about how to differentiate between the other three. For morpholine 2, If we see the stability of the conjugate acid formed, the O would exert (very less) -I effect, which should destabilize the positive charge on N (on formation of conjugate acid). Now between 1 and 3, 1 has a five other C atoms attached to the N, while 3 has four. Now CH3 is a +I group, hence more CH3 in 1 should stabilize the positive charge on N more than 3.
I don't think Inductive Effect should matter much, considering Inductive Effect becomes very ineffective with increase in chain length. But this is the only reasoning that strikes me. So ultimately, order of basicity becomes 4 < 2 < 3 < 1. Is my reasoning and the final answer correct?

Nitrogen inversion is not relevant to this discussion. A faster rate of inversion does not correspond to a greater delocalisation in the electronic structure, as lone pair inversion is not equivalent to resonance. The lone pair does not exist as a delocalised smear; it exists in two different localised forms.

The PDF on nitrogen inversion in the comments does not even bring up the topic of basicity, and quite rightly so.

Inductive effects explain why morpholine 2 is less basic, and resonance effects explain why pyrrolidone 4 is even less basic. However, since the pKaH's of piperidine 1 and pyrrolidine 3 are so similar (11.22 and 11.27 respectively), I think it is pointless to try and qualitatively rationalise a difference in their basicities. It is like trying to explain why the temperature in Spain one day is 11.22 °C and the temperature the next day is 11.27 °C. The relative basicity could well be different if you switch the solvent, which would invalidate any reasoning based on - for example - s-orbital character of the nitrogen lone pair, which I think is the only plausible qualitative argument here.

Simply looking at the data, though, the order you want is 4 < 2 < 1 < 3.