Are there any (simple) molecules with very different absorption and emission dipole directions?

In part of the naphthalene molecule you draw you indicate a transition moment. This is the direction for absorption into the lowest excited state $S_1$. As naphthalene is highly symmetric the fluorescence is also emitted from a transition moment aligned in the same direction. The second excited state $S_2$ has a transition dipole set perpendicularly to the long axis and so to the $S_1$ direction.  Again excitation and emission would be from similarly aligned transition dipoles if emission was from $S_2$. However, in solution $S_2$ decays very rapidly ( a few picoseconds) to $S_1$ thus the observed absorption and emission dipoles would be at right angles to one another.

The wavefunction (molecule orbital) in an excited state is different to that in the ground state since an electron is excited to an anti-bonding orbital. This means that bond lengths and angles can change a little. If the molecule has polarisable substituents that are asymmetrical placed then the excited state dipole may differ in direction from that of the ground state. However, in molecules such as rose bengal, rhodamine 6G and many dicarbocyanines, which are not strictly symmetric ($C_1$ point group), the absorption and emission dipoles are parallel as determined by experiment.  (Presumably this is because substituents are not sufficiently close in energy to affect the electronic state in a significant way)

How can we know what the angle between absorption and emission is? The probability of absorption is proportional to the cosine squared of the angle $\theta$ between the electric field direction of the light and the transition moment direction, $\approx (\bar e \cdot \bar\mu)^2= e^2\mu^2\cos^2(\theta)$. So we can describe the molecules are being photo-selected, the probability of absorption being greatest when $\theta=0$, i.e. absorption and excitation directions are parallel.

If we use polarised light to excite the molecules, and assume that it is polarised along the $z$ direction, then the  emission intensity observed through a polariser parallel to the $z$ (or vertical) direction is $I_v\approx (\bar e_z \cdot \bar\mu_z)^2$  and observed at right angles to this ($x$ or horizontal direction)  $I_h\approx (\bar e_z \cdot \bar\mu_x)^2$ where $\bar e_z$ is a unit vector aligned along the electric field and  $\bar \mu_z$ the unit vector component of the fluorescence dipole along the $z$ and $\bar \mu_x$ along the  $x$ direction. The amount of polarisation of the fluorescence is usually measured via the  anisotropy which is defined as

$$r=\frac{I_v-I_h}{I_v+2I_h}$$

where the denominator is the total emission intensity.

In fluid solution molecules undergo rotational diffusion and if the molecule is regarded as a sphere, in fact not such a bad approximation in many cases then, the anisotropy becomes $\displaystyle r(t)=r_0e^{-6Dt}$ where $D$ is the rotational diffusion constant. The anisotropy can be measured by recording the fluorescence decay such as by time-correlated single photon counting. Taking the decay back to time zero produces $r_0$, this is the anisotropy that would be observed if the molecules were in rigid solution and effectively measures the angle between absorption and emission dipoles.

Now this relates directly to your question. The value of $r_0$ can be also be calculated as well as being measured and, again, considering the angle between absorption and emission dipoles $r_0=(3\cos^2(\alpha)-1)/5$ where $\alpha$ is the angle between absorption and emission dipoles. When the angle is zero then $r_0=0.4$ which is easily measured and when at right angles $r_0=-0.2$ so this limits the range of anisotropy.

(references. Text. M. Daune. Molecular Biophysics publ. OUP 1999. Some original literature; T Tao, Biopolymers v8,p609, 1969, Y. Yguerabide Meth. Enzymology, v26, p498, 172, Chuang & Eisenthal, J. Chem. Phys. v57, p5094, 1972)


A bit more complex per a source 'Kinetics and mechanism of the iodate-iodide reaction and other related reactions', to quote from an article in the journal Physical Chemistry Chemical Physics (November 1999):

The reduction of iodate by iodide, known as the Dushman reaction, is one of the components of the oscillating Bray-Liebhafsky reaction. The large body of work devoted to the kinetics of the Dushman reaction gives seemingly contradictory results. This work analyses and summarizes them and shows that they are special cases of a complicated kinetics and that they can be explained by a common mechanism. This mechanism can be generalized to all the XO3- + Y- reactions (X, Y = I, Br or Cl). They all involve an asymmetric intermediate YXO2.

See reaction system depicted in Equations (8) to (13). Here is a partial discussion of the system from the available PDF to quote:

The essential hypothesis of this mechanism is the existence of as a definite intermediate species and not as I2O2 an activated complex. Its existence was already proposed in  1930 by Bray [17] in order to explain the classical rate law. We make no distinction between I2O2 and H2I2O3 (I2O2 + H2O) as these two forms are kinetically equivalent. In reaction (12), B~ represents the anion of the buffer....

As to the question on the role of H+, note that the starting Equations (8) and (9) of the system both involve H+, which results in the cited intermediate $\ce{I2O2}$:

$\ce{IO3- + H+ ⇌ IO3H}$  (8)
$\ce{IO3H + I- + H+ ⇌ I2O2 + H2O}$  (9)


Optimum pH range for formation of Schiff bases

Dehydration requires protonation of the hydroxy intermediate, this clearly is slower at neutral or alkaline pH. Hydroxylamine is basic so at acidic pH it protonates making it non-nucleophilic hence lower pH slows the rate.
Time taken for the reaction to go to completion. If the rate is slowed then other competing reactions (e.g. self-condensation) can and do reduce your yield.
The pH will vary for each substrate and needs to be experimentally determined. This is where a literature search saves you a lot of grief.


It is said: one formula unit of sodium chloride.


I'm not 100% certain on this, but my best guess on this would be referring to 'lagged' as in 'lagging', as in insulation material. So, a lagged bath is an insulated bath.


Is reduction potential and stability related?

Remember that reduction potentials do not have any meaning by themselves. They are measured against a hydrogen cell to which arbitrarily we assign the value $E^0=0$. You can compare between different potentials. For example, in your case, you could say that that $Cu^{2+}$ has a greater tendency to be reduced than $Na^+$  because its reduction potential is more positive. However, I do not think the potential itself tells you anything with regard to the stability (and, in any case, what do you mean by that) of the species involved in a particular half reaction.


Electrolysis of brine without a membrane

So if the membrane is removed, will the chlorine gas react with NaOH to form sodium hypochlorite?

Yes.

  Because I've read that the chlorine will instead just react with the hydroxide ion to produce hypochlorite.

Yes. This is just another way to say the same thing.

As for the sodium, it does not go anywhere. It simply continues to stay there in the solution, like it did before the reaction. It does not even "know" about the reaction, or that it now "belongs" to the hypochlorite; it is just a spectator ion. Since the dissolution, anions and cations go their separate ways and do not "care" much for each other. This is always the case with ionic salts.


Are they metal pipes?  If you want to kill algae as such, flush with a diluted quaternary ammononium disinfectant, let sit a day, then flush everything very well indeed, and again.  If you have rubber or polymer fittings, don't do that, either, for the stuff diffuses in.


Cram's rule for nucleophilic addition to carbonyl groups

Cram's rule, proposed by Donald Cram in 1952,[1] was the earliest model proposed for rationalising the stereoselectivity of nucleophilic additions to carbonyl groups with α-stereocentres. The model involved assigning each α-substituent a relative "size" (small, medium, and large), then placing the carbonyl oxygen antiperiplanar to the largest of these three groups. The nucleophile then attacks the carbonyl group opposite the larger of the two remaining groups (i.e., the medium group). This is best explained with a diagram:

In the reduction of (S)-3-phenylbutan-2-one with L-Selectride (a bulky source of hydride ion), the anti alcohol is formed as the major diastereomer. Cram's rule rationalises this by placing the phenyl group (RL) anti to the carbonyl oxygen. The hydride nucleophile, drawn here as H− for simplicity, then preferentially attacks over the proton (RS) instead of the methyl group (RM).

Although Cram was able to demonstrate that his rule worked for a number of nucleophilic additions, there are some conceptual issues with it. Most obviously, in the reactive conformation, the phenyl group is eclipsing the methyl group on the ketone; this torsional strain is worsened when the nucleophile adds into the carbonyl, as the resultant product is initially in a staggered conformation.

In 1968, Hugh Felkin pointed out two observations which the Cram model could not explain.[2] One is to do with the reduction of cyclohexanones, which we will not go into here, as nowadays a slightly different model is used to explain reduction of cyclic ketones. The other was about how the stereoselectivity of C=O reduction varied when the ketone was made progressively  more bulky.

According to Cram's model, as the size of R increased, the steric repulsions between R and the phenyl group (RL) should increase, leading to a decrease in stereoselectivity of reduction. However, Felkin showed that the opposite was true: the anti product was formed with higher stereoselectivity when R increased in size from Me to t-Bu.

To cut a long story short, Cram's rule was eventually replaced by the Felkin–Anh model. Here, the large α-substituent RL was instead placed perpendicular to the C=O bond. This leads to a choice of two possible conformations. The nucleophile attacks the carbonyl group along the Bürgi–Dunitz angle, and the attack over RS instead of over RM is favoured:

Like Cram's rule, the Felkin–Anh model correctly predicts the diastereoselectivity observed in this reduction. However, it also allows us to rationalise why the stereoselectivity increases as the size of R increases. A larger R makes the gauche interactions between R and Ph more destabilising, which biases the conformational equilibrium towards the reactive conformer (on the right).

For more detail about the Felkin–Anh rule and other modifications which are necessary (depending on the exact substrate and conditions), see: Effect of protecting group on diastereoselectivity of LiAlH4 reduction of α-hydroxyketones and the answers therein.

References

Cram, D. J.; Elhafez, F. A. A. Studies in Stereochemistry. X. The Rule of “Steric Control of Asymmetric Induction” in the Syntheses of Acyclic Systems. J. Am. Chem. Soc. 1952, 74 (23), 5828–5835. DOI: 10.1021/ja01143a007.
Chérest, M.; Felkin, H.; Prudent, N. Torsional strain involving partial bonds. The stereochemistry of the lithium aluminium hydride reduction of some simple open-chain ketones. Tetrahedron Lett. 1968, 9 (18), 2199–2204. DOI: 10.1016/S0040-4039(00)89719-1.


You should use the concentration dilution equation

$$C_1V_1 = C_2\cdot V_2$$
If you re-arrange it so the volume of 60% ethanol needed is:

$$V_1 = \frac{C_2\cdot V_2}{C_1}$$

You can then insert all the values you do know and there you go.
For example, if you need a $\pu{100 m}$ of 5% solution: $$V_1=\frac{5\% × \pu{ 100 ml}}{60\%}$$

Which means you need $\pu{8.33 ml}$ of 60% ethanol in $\pu{100 ml}$ solution.

As max pointed out, since you only have a scale, use d=m/v. D = density, m = mass and v= volume You re-arrange so mass = d.v


when Al2O3 melts which bonds are broken?

Right, the text you are mentioning is not clear enough. It is not correct  to say that the molecule $\ce{Al_2O_3}$ exists and is separated from the next $\ce{Al_2O_3}$. Such a molecule does not exist. A sample of $\ce{Al_2O_3}$ is made of a superposition of $\ce{Al^{3+}}$ ions and $\ce{O^{2-}}$ ions. It is not made of a superposition of molecules $\ce{Al_2O_3}$. Now the fact that MgO melts at a higher temperature than $\ce{Al_2O_3}$ is due to the stronger attraction between $\ce{Mg^{2+}}$ and $\ce{O^{2-}}$ ions. For some reason, the attraction between $\ce{Al^{3+}}$ ions and $\ce{O^{2-}}$ is weaker.


What makes protons give the property of acidity?

While it is true that oxidizing strong acids like nitric and sulfuric are more corrosive than a non-oxidizing strong acid like hydrochloric, the latter still is able to dissolve materials (e.g, metals, minerals, proteins) and cause burns to the skin. A strong acid like HCl can (in the case of human skin or other biological materials) catalyze the hydrolysis of proteins, fats, and carbohydrates down to their water-soluble components. This is what happens in your stomach, where there is an abundance of hydrochloric acid, as well as enzymes to speed up the process.

The most widely accepted measure of how strong an acid is, its its acid dissociation constant, not the pH of a solution of it.


In guidelines published by the International Union of Pure and Applied Physics or IUPAP$^{[1]}$, upright letters are recommended. Whether these letters are capitalised depends on their use. Lowercase is for quantum numbers of a single particle, and uppercase for describing a system.

A letter symbol indicating a quantum number of a single particle should be printed in lower case upright type.  A letter symbol indicating a quantum number of a system should be printed in capital upright type.$^{[1]}$

Almost identical text is found in the IUPAC Green Book.$^{[2]}$
The letters themselves ignoring capitalisation are $\mathrm{s}$ – sharp, $\mathrm{p}$ – principal, $\mathrm{d}$ – diffuse, $\mathrm{f}$ – fundamental, $\mathrm{g}$, $\mathrm{h}\ \ldots$
Continue according to the English alphabet skipping letter j.$^{[1]}$
In summary, for orbital angular momentum quantum numbers$^{[1]}$

\begin{align}\begin{split} l\ =\ &0\ \ &1\ \ &2 \ \ &3\ \ &4\ \ &5 \ \ &6\ \ &7\ \ &8 \ \ &9\ \ &10\ \ &11\ \ldots\\ \text{symbol}\ \ \ \  &\mathrm{s}\ \ &\mathrm{p}\ \ &\mathrm{d}\ \ &\mathrm{f}\ \ &\mathrm{g}\ \ &\mathrm{h}\ \ &\mathrm{i}\ \ &\mathrm{k}\ \ &\mathrm{l}\ \ &\mathrm{m}\ \ &\mathrm{n}\ \ &\mathrm{o}\end{split}\tag{for a single particle}\end{align}
\begin{align}\begin{split} L\ =\ &0\ \ &1\ \ &2 \ \ &3\ \ &4\ \ &5 \ \ &6\ \ &7\ \ &8 \ \ &9\ \ &10\ \ &11\ \ldots\\ \text{symbol}\ \ \ \  &\mathrm{S}\ \ &\mathrm{P}\ \ &\mathrm{D}\ \ &\mathrm{F}\ \ &\mathrm{G}\ \ &\mathrm{H}\ \ &\mathrm{I}\ \ &\mathrm{K}\ \ &\mathrm{L}\ \ &\mathrm{M}\ \ &\mathrm{N}\ \ &\mathrm{O}\end{split}\tag{for a system}\end{align}

As an example, here are two ways$^{[1]\ [2]\ [3]}$ of writing the predicted ground state electronic structure$^{[4]}$ of oganesson, element $118$:
$$\ce{Og: [Rn] {5f^{14}}{6d^{10}}{7s^2}{7p^6}\\ Og: [Rn] (5f)^14(6d)^10(7s)^2(7p)^6}$$
Typography regarding spacing is not explicitly stated, but is clear from the examples provided. Nevertheless, if you prefer to use
$$\ce{Og: [Rn] 5f^14 6d^10 7s^2 7p^6},$$
it should be fine as well.

$[1]$ E. Richard Cohen, Pierre Giacomo. $(1987)$. Symbols, Units, Nomenclature and Fundamental Constants in Physics. ('IUPAP Red Book'). International Union of Pure and Applied Physics. Commission C2 – Sunamco. ($2010$ reprint) (pp 12$-$13)
$[2]$ E. R. Cohen, T. Cvitas, J. G. Frey, B. Holmström, K. Kuchitsu,
R. Marquardt, I. Mills, F. Pavese, M. Quack, J. Stohner, H. L. Strauss, M. Takami, A. J. Thor. ($2007$). Quantities, Units, and Symbols in Physical Chemistry. ('IUPAC Green Book'). Third Edition. International Union of Pure and Applied Chemistry. ($2008$ IUPAC & RSC reprint). ISBN: 978-0-85404-433-7. (p 32)
$[3]$ Anne M. Coghill, Lorrin R. Garson. ($2006$). The ACS Style Guide. Effective Communication of Scientific Information. American Chemical Society. DOI: 10.1021/bk-2006-STYG, ISBN: 9780841239999 (print), 9780841228306 (online). (p 256)
$[4]$ 'The Element Oganesson'. $(2016)$. Steve Gagnon. Thomas Jefferson National Accelerator Facility – Office of Science Education. (February 1, 2017)


Are modern newspaper inks a health risk?

Before anybody may wrap some food in a newspaper, it needs to be printed.

Given that workers in print shops are the first that are exposed to potentially dangerous compounds, this is a matter occupational safety their national regulations.

In Germany, the BG ETEM provides an extensive exclusion list for inks and solvents. Compounds covered by this list are not be be used!

In order to avoid an endless recital of compound names or CAS numbers, the exclusion list is based of GHS hazard statements and includes (among others):

materials with CMR properties:
H340, H341, H350, H350i, H351, H360, H360F, H360D, H360FD, H360Fd, H360Df
materials with specific organ toxicity, acute toxicity and other health risks:
H370, H372, H300, H310, H330, H301, H311, H331, EUH 029, EUH 031, EUH 032

Summary

Does this mean that German newspaper are safe for handle in production and daily use? Yes, unless printing houses violate existing regulations.

Is the same true for Indian newspapers? I have no idea, since I am not familiar with the legal situation in India.


Why is the boiling point of sulfuric acid much higher than that of phosphoric acid?

In addition to the factors Howsikan mentioned, my intuition leads me to believe there's a parallel with the boiling points of $\ce{NH3}$ and $\ce{H2O}$. I think the oxos ($\ce{O=P}$ and $\ce{O=S}$) are better acceptors than the hydroxyls, so the acids would prefer $\ce{OH...O=}$ over $\ce{OH...OH}$. $\ce{H3PO3}$ has three oxos to one hydroxyl whereas $\ce{H2SO4}$ has two of each, so $\ce{H2SO4}$  can create a favourable network more easily.


2-methylpropene to tert-butyl alcohol Reaction Mechanism

It is an acid-catalyzed hydration of the double bond. First, the double bond is protonated to yield the tert-butyl cation which is most stabilized by hyperconjugation (protonation at C2 would form a less stable primary cation). The cation is captured by nucleophilic attack of water, which after deprotonation yields tert-butanol and the regenerated proton.

$$\ce{(H3C)2C=CH2 + H3O+ <=> (H3C)3C+ + H2O} \tag 1$$

$$\ce{(H3C)3C+ + H2O <=> (H3C)3C-OH2^{+}} \tag 2$$

$$\ce{(H3C)3C-OH2^{+} + H2O ->~ (H3C)3C-OH + H3O+} \tag 3$$


Some radioactive materials are made by man, others have been parts of earth's rocks since they were formed

Radioactive materials are not all the same. some are man-made and some have been with us since the creation of the planet (so are indeed the product of old exploded stars).

The best way to tell which is which is to look at the half-life of the atoms involved. Half-life is a measure of how fast an isotope decays. It is the length of time it takes for half of the amount of substance to have decayed to something else. So half-lives with the same sort of size as the age of the earth are quite likely to be primordial (which basically means they were present when the earth was first formed). Other compounds with shorter half-lives would have all disappeared by now even if they were present then. They are synthesised by man.

238U (by far the most common isotope of uranium) has a half-life of ~4.5 billion years (roughly the same age as the Earth) so quite a lot of the uranium present at our planet's birth is still around. 239Pu (the isotope of plutonium used in nukes) has a half-life of ~24 thousand years so even if there was some around when the earth was formed there wouldn't be any of that left. Plutonium is created in nuclear reactors which can be designed to generate enough to make an indecent supply for the military. Having said that, a very small amount may be generated naturally in uranium containing rocks as part of the normal decay pathways of uranium or as part of natural nuclear reactors (occasionally in geological history uranium rich rocks have created the conditions for significant nuclear reactions which produce more plutonium and other materials than simple decay of isolated uranium atoms).

Other, more common, elements sometimes have radioactive isotopes. A small proportion of potassium, for example is 40K which has a half-life of ~1.2bn years (so quite a lot of the earth's original stock is still around and is the largest contributor to radiation in living things which all contains a lot of potassium).

There is little danger of running out of primordial radioactive elements as we are not using them fast enough. The man-made ones can be made in nuclear reactors if they are set up the right way (and we have way more plutonium than we know what to do with; but many rare isotopes used in medical research are made in specialty reactors where the risk is that a reactor shutdown might limit the supply).

If you want to know where all the elements originally came from, there is a good Wikipedia article with a periodic table that explains the detailed origins.


Reason behind markovnikov's rule in hydration of alkenes via mercury catalysis

If I could postulate, the reaction is like an epoxide ring opening during acidic conditions, where the oxygen is protonated, making it electron-deficient and have a +1 formal charge. This creates a good leaving group (an electron-deficient O) and in your case, an electron-deficient Hg.

Credits: Clayden Organic Chemistry Vol 2

Since it is a good leaving group, it wants to leave and as predicted by the Sn1 mechanism, the C-Hg bond bonded to the more substituted carbon prefers to break in order to take advantage of the more stable carbocation intermediate. This makes that particular bond weaker. The nucleophile thus prefers to attack the more substituted end in order to take advantage of the weaker bond. As the bond is weaker, the net enthalpy change would be lower, making the attack at the more substituted end more thermodynamically favourable. Additionally, notice that in the transition state, the developing positive charge is more resonance stabilised if the nucleophile attacks the more substituted end, making the reaction more kinetically favourable. From both perspectives, kinetics and thermodynamics, an attack at the more substituted end is favoured.


Datawarrior lets you do it.
Export a csv sheet with two columns: The name of the compound and its SMILES code.
Be sure to use a comma as a delimiter.
Your output file, opened with a text editor, should look like this:

Name,SMILES
Hexazine,n1nnnnn1
Furazan,c1nonc1
Benzisoxazole,c1noc2c1cccc2

From DataWarrior, File>Open and open your file.
Structures are automatically generated from SMILES codes.

Then, File>Save Special>SD-File, and you are done.


Preparation of Barium Hexaaluminate using co-precipitation reaction mechanism

@Maurice, I always thought that barium nitrate and aluminium nitrate will react with ammonium carbonate and barium carbonate and aluminium hydroxide will precipitate down.

$$
\begin{align}
\ce{2 Al(NO3)3 (aq) + 3 (NH4)2CO3 (aq) &-> Al2(CO3)3 (s) + 6 NH4NO3 (aq)}\tag{R1}\\
\ce{Ba(NO3)2 (aq) + (NH4)2CO3 (aq) &-> BaCO3 (s) + 2 NH4NO3 (aq)}\tag{R2}
\end{align}
$$

(As aluminium carbonate is not stable and readily decomposes to form aluminium hydroxide.)

After that, if we keep the ratio of barium and aluminium $1:12$ and raise the temperature above $\pu{1200 °C}$ by calcinating the dried precipitate, barium hexaaluminate forms.

Also, can you clarify what $\ce{BaAl6(OH)20}$ is, because according to my understanding the general formula of barium hexaluminate is $\ce{BaO.6(Al2O3)}.$

Reference

Li, J. Q.; Wang, R. K.; Chen, C. Y. Preparation of Barium Aluminate with $\ce{BaCO3}$ and $\ce{Al(OH)3}$. AMR 2015, 1096, 156–160. DOI: 10.4028/www.scientific.net/AMR.1096.156.


Six independent papers reporting epitaxial formation of silicene in 2012

Since this is a Nature News article, usually it is a way of promoting a scientific article in one of the current Nature publications - that's why it can get away with stating "Six independent reports" without six citations attached.  In this case, it is citing this 2015 article:

  Tao, L. et al. Nature Nanotechnol.
  http://dx.doi.org/10.1038/NNANO.2014.325 (2015)

There are many citations in that article from the 2010-2013 time period reporting silicene sheets and nanoribbons.  Keep in mind, the Fleurence et al. paper you mentioned might not be one of the six mentioned, because it uses a diboride substrate instead of silver.


Basic behaviour of bromide ion with sulfuric and phosphonic acid

A common misconception is that the more acidic component will always protonate the less acidic componend and displace it from its salts. That misconception would be exemplified by equation $(1)$

$$\ce{HBr + KH2PO4 -> KBr + H3PO4}\tag{1}$$

This is incorrect. There is a constant acid-base equilibrium and even bromide ions will be protonated to a non-neglegible extent. The principal difference between $\ce{HBr}$ and $\ce{H3PO4}$ is that the former is a gas at room temperature and standard pressure. Thus, in an open vessel, any $\ce{HBr}$ in solution is again automatically in equilibrium with $\ce{HBr}$ in the gas phase. Thus, the equilibrium should better be written in form $(2)$:

$$\begin{multline}\ce{KBr (aq) + H3PO4 (aq) <<=>\\ <<=> HBr (aq) + KH2PO4 (aq) <=> \\ <=> HBr (g) ^ + KH2PO4}\tag{2}\end{multline}$$

As you can see, the gaseous $\ce{HBr}$ can diffuse away. This reduces the partial pressure of $\ce{HBr}$ above the solution, which affects the right-hand side equilibrium by Le Chatelier’s principle, reducing the concentration of $\ce{HBr}$ in solution, which in turn affects the left-hand side equilibrium, reducing the concentration of $\ce{KBr}$. Thus, if left to equilibrate in a well-vented area, large portions of $\ce{HBr}$ will diffuse away leaving a solution of potassium phosphate or potassium hydrogen phosphate.

The principle alluded to in the first sentence is thus correct, if rather than using acid strength it is quoted as:

  The less volatile acid will displace the more volatile acid from its salts.

It is worth noting that an acid-base equilibrium $(3)$ also is observeable in your first reaction. However, the redox reaction is faster than the removal of hydrogen bromide. It is also irreversible, hence those are the final products obtained.

$$\ce{KBr + H2SO4 <=> HBr + KHSO4}\tag{3}$$


Why is Sodium Transported Across the Cation Exchange Membrane in the Chlor Alkali Process?

There is a rather interesting explanation as to why only sodium ion can pass through a cation exchange membrane and chloride ion is repelled by it.
Let us recall some properties
a) Ion exchangers are very good electrical conductors. This is well studied in the 1940s. The current is carried, not by electrons but by mobile ions in ion exchange resins. In a cation exchanger, current is carried by mobile cations.
b) Imagine in the cation exchanger, it consists of a sulfonated polymer. Let us call that R-SO3H, where H is mobile and R-SO3(-) unit is immobile.
c) Ignore the flow of the chlor-alkali cell for the time being.
d) The punch line is that the bulk solution has to be electrically neutral all the time.
Now consider an electrolytic cell which consists of NaCl solution. There is a partition of a cation exchange membrane in between. On the left there is an anode where chloride ions are being oxidized and the right, water is being reduced to hydrogen.
Right hand side:
When water is being reduced, neutral hydrogen gas is leaving the solution, leaving behind negatively charged hydroxide ions. There is a charge imbalance in the bulk which cannot exist. What has to happen? Na(+) ions must cross the cation exchanger from the left hand side to ensure there is no bulk charge imbalance.
Now chloride cannot pass through the ion exchange membrane because the R-SO3(-) has a negative charge. All it can do is to repel the negatively charged chloride ion. You can read more about Donnan effect and Donnan potential.
Left hand side:
You are losing negatively charged chloride ion as neutral chlorine gas. There is excess positive Na(+) ions there. The positive ions have to cross the membrane in order to maintain electrical neutrality in the bulk and eventually join the free hydroxide ions.
As a combined effect, the right hand side becomes a solution of NaOH (contaminated with NaCl) and the concentration of NaCl on the left hand side depletes.


Assuming you already have the optical setup built around, aiming for the least amount of expensive chemicals, you may try semicrystalline extruded polymers, e.g. polyethylene foils, or PET bottles.

The «trick» is that you do not melt them, but that you reach just the glass transition temperature ($T_g$) which is below melting. Here, crystalline parts in the polymer break up, rendering the material isotropic and in transmittent observation with crossed polarizer / analyzer black.  Depending on material and degree of polymerization, this solid-solid transition already may happen in a range of 60 to 80 Celsius.  Put the material on a glass slide which is heated gently from beneath with an low-powered heating plate.  With a tooth-pick you may test that the material becomes softer.

Upon cooling you will see how spherulites re-appear, e.g.

(source)

Size and shape of them will depend on the rate of cooling, for example if let cool at air (or on the table of your microscope), or if you cool the glass slide by putting it on an ice bag.
If not overheated this may be repeated over and over easily for a dozen of times.  Apart from the heating plate, there is little potential danger.

A low-cost set up including a heating stage, where you could continue the observation of the sample's recrystallization was described in the February issue of the Journal of Chemical Education (J. Chem. Educ., 2019, 96, 823-826, doi 10.1021/acs.jchemed.8b00879).


Tried charging a penny/dime battery. Yellow goo bubbled. What did I create?

Hahaha, I had to translate for myself what a penny and what a dime is. Always the same for me; Fahrenheit, miles, ...

A penny is a copper-plated zinc coin. A dime is a nickel-plated copper coin. Knowing this, I would assume that the yellow color arose from the nickel content. If it would be blue, it would have more likely been copper.
The goo. Such a structure often occurs with proteins, polymers, and similar compounds. It's most likely that the bacterial farm on the coins you used, lead to the formation of this goo. It's impressive how many bacteria and what thick biofilms you can find on a coin. Additionally, your saliva also contains a multitude of living organisms and their remains.

Therefore, you should definitely reproduce this experiment! Clean the coins beforehand using ethanol, acetone, and water while scrubbing. Polish them with a linen cloth and not with sand paper. Otherwise, you would damage the plated metal film and surface the underlying metal (makes things now complicated). And don't spit in it yet! :-)


I no longer am a practicing chemist, so I have no access to a structure drawing program; I'll do my best with words:

Accept the initial premise that the NO2 is strongly deactivating so only the 'end' rings will be attacked.  As we know o,p- substituion have functionalyl equivalent resonance forms, so we only have to consider 4 possible products: p-, m-substitutuion on the 3-ring and similar on the 4-ring.

Drawing all four of these structures and the reasonanc forms indicates that the p-,4-ring susbtitution and the m-,3-ring susbtitution are poor choices because of high energy intermediates.

The final question to resolve is whether or not the m-,4-ring is better than the p-,3-ring.  Based on the intermediates, this is a even, but the nitro group is in conjugation with the 4-ring, whle only in cross-conjugation with the 3-ring  You can see this by drawing a rsonance form with a carbocation at the para position, and rearranging the bonds via resonance all the way to the nitro group.  This can't be done with the with the 3-ring, meaning the nitro group deactivates the 4-ring more than the 3-ring.

Thus the product is likely p-substitution on the 3-ring.

Advanced:
The resonance for arguments require all three structures be coplanar.  This is not the case; biphenyls have some torsion and twisting because o-hydrogens bump.  However, there is still enough 'flatness' and orbital overlap to make some use of the resonance arguments presented above, albeit the magnitude of better and worse are smaller than they would be for more classical EAS problems.


If originally made of silver, the dark layer on your jewlery is silver sulfide, $\ce{Ag2S}$.  The removal suggested by your mother is an electrochemical one, that will reduce the silver back into metallic, shiny $\ce{Ag}$, if in addition of hot water and bicarbonate, salt and an other metal acting as reducing component is in electrical contact with the silver piece are present.  Here, some foil of aluminum will serve well as sacrifical anode.

To quote UMass Amherst, presenting it as a lecture demonstration, the chemistry may be summed up to this equation

$$\ce{3 Ag2S(s) + 2 Al(s) + 3 H2O (l) –> 6 Ag(s) + 3 H2S (aq) + Al2O3(s)   }$$

The addition of salt (which is absent in the equation) serves to increase the electric conductivity of the solution, hence increasing the rate of this process.  Immerse the jewlery in the hot solution (it needn't be boiling), for example wrapped in the creased foil for some minutes; then rinse and wipe it with a cloth.


Just found something:

So the most dominant thermal change, and we are talking here about annealing single crystals for several days to around 150°C seems to be a polytypic change from the 2H-Type to the 12R-Type. Or, as there are several hexagonal and rhomboedric phases let's just assume a change from H to R. You may know polymorphs, like allotropes for example, diamond vs graphite or different phosphorus allotropes. And polytypes are structures that usually appear in layers where the overall composition and connections are pretty much the same but the stacking order of the layer is different. I found a sketch how the layers have to change:

1 E.Salje, B.Palosz, B.Wruck, Structural and Magnetic Phase Transitions in Minerals, Advances in Physical Geochemistry, vol 7, Springer, New York, p.231, 1988.

It was often reported that the change from H to R seems to be thermally while the change from R to H takes a long time. As the crystals were grown from sodium based gels traces of sodium could be found and the higher the sodium content the easier and faster they turned back to H. I didn't grow any crystals but precipitated it using NaI, so I probably have a high sodium content which acts as an impurity to fastly change the structure back, which appeared to me as being reversibly thermochromic.

I don't know whether this also causes the color change but many other color changes, like in TlI depend on minor distortions in the crystal structure as well so it could be plausible.


ICP-OES data treatment: correct for dilution and inweight, the units

I think you have a typo in your calculation (also, using both dots and commas as the decimal point is rather confusing), but you're on the right track. The unfortunate thing about elemental analysis is that %, ppm, ppb, ppt, etc. get misused in a confusing way. These units are supposed to be used for only dimensionless ratios, the units you start with, ppm is actually mg/L which represents mass concentration, not mass ratio. Your calculations are correct:

$$\left(186.49 \frac{\mathrm{mg}}{\mathrm{L}}\right)\left(0.0142\mathrm{\ L}\right) = 2.648 \mathrm{\ mg}$$
gives the mass of the Mn in your sample, so when you divide by the sample mass:

$$\frac{2.648 \mathrm{\ mg}}{10.17\mathrm{\ mg}}=0.260$$
you get the mass fraction of Mn in your sample, and you know it's a mass fraction because you have mass units in both the numerator and the denominator, leaving you dimensionless. You can express it just like that, or you can convert it to 26.0% (and this is a dimensionless ratio, so it actually makes sense to use % here), or else express it as the mass of Mn in an arbitrary sample mass: 26.0 g/100 g.


Does concentration or pKa define acid strength?

The term ‘strong acid’ is sometimes used in a rather fuzzy way and you ran into problems doing so. I prefer to use the term ‘strong acid’ only with respect to an acid’s $\mathrm pK_\mathrm{a}$ value and disregard all other influences. This gives a clearly defined measure of acid strength and we can easily sort various acids by their strength into stronger or weaker acids.

However, this is looking at the acid as a molecule. In real-life applications you are typically more interested in the property of a solution. To give a real-world example, imagine a $\pu{1M}\ \ce{HBr}$ solution and a $\pu{12M}\ \ce{HCl}$. solution. Obviously, $\ce{HBr}$ is the stronger acid, but the concentration of $\ce{HCl}$—also a strong acid and thus fully deprotonated—is higher. Therefore, the $\ce{HCl}$ solution is more concentrated or, as some would say, more acidic. It can do greater harm and it is able to protonate more Brønsted base molecules than its $\ce{HBr}$ counterpart.

If instead of examining a $\pu{1M}$ and a $\pu{12M}$ solution I had examined a $\pu{e-3M}$ and a $\pu{1M}$ solution, we could even base the discussion around the solution’s resulting pH: the stronger acid is much more diluted and will result in a solution of pH 3 while the weaker acid results in a solution of pH 0.


How much does oxygen enrichment improve some combustion engines?

What does "energy efficienct" mean?
It depends on what you mean by "efficient".
For combustion reactions that involve engines, if "efficiency" means how much work (the "good" type of energy) that comes out of the engine per unit input of fuel, then yes oxygen-enriched mixtures would be more efficient than regular old air.
If instead "efficiency" means how much of the fuel is combusted all the way to $\ce{CO2}$, the answer is also probably yes, but for different reasons.
The efficiency of engines
To understand why, keep in mind that no engine can ever be more efficient than an ideal Carnot engine.  The efficiency of a Carnot engine is $\frac{T_H - T_C}{T_H} = 1 - \frac{T_C}{T_H}$, where $T_C$ is the temperature of a cold reservoir (e.g. our everyday environment)   in Kelvins and $T_H$ is the temperature in Kelvins of a hot reservoir.
Thus, the hotter you can get the "hot" side of Carnot engine, the better.  In a car engine or other Otto engines, this would correspond to the maximum temperature of the combusted air (or oxygen)-fuel mixture just before the expansion stroke.
Adiabatic flame temperature
With air instead of pure oxygen as the oxidizer, some of the energy of combustion goes into heating up inert components of the air, mostly nitrogen.  This lowers the maximum temperature of the air-fuel mixture and decreases theoretically attainable efficiency.
The adiabatic flame temperature is a quantitative number behind this idea.  Wikipedia lists the adiabatic flame temperature, the maximum temperature attainable in the combustion of certain fuels, and the values for fuels burning in air are a bit less than the same fuels burning in oxygen.
Limits of this analysis
Of course, there are many factors that make real engine performance different than ideal engine performance.  In fact, designing engines that can operate with at the extremely high temperatures that would maximize efficiency is tough because of materials problems.  (The adiabatic flame temperature of gasoline in pure oxygen is somewhere around 3800 K, far higher than the melting temperature of steel, for example.). So whether a given real engine designed to work under particular sets of conditions will necessarily do better with pure oxygen instead of air is less clear.  Probably yes?  And for sure an engine designed appropriately for use with pure oxygen would be more efficient than an equivalent one designed for us with air.


Why is there a greater degree of separation in a dipole-dipole interaction?

Like I answered my other question, this is in some ways very similar. Since electrostatic attractions are so strong, it is decreases less at a longer distances. The $+$ & $-$ are attracted together so strongly.

Dipole-dipole attractions aren't strong attractions, electrostatic attractions are stronger, since they are created by ionic bonding. Dipole-dipole occur in covalent bonding. Thus dipole-dipole attractions decrease with increasing distance faster.

This is just a piece of the whole picture.

Check the refs: 1, 2, 3. They provide more in depth detail to the entire picture.


Not only is there no reason to think anomerization does not occur in the disaccharide you mention, it most definitely does occur.

As long as water has access to carbon-1 of a glucose unit, there will be an open-chain form at that site, and then when it re-closes, the possibility of alpha- and beta-anomers.  This will be the case for di-, tri- tetra-, oligosaccharides, polysaccharides, etc... and applies to monosaccharides of any carbon length or type (aldoses or ketoses).

The reason is that hemi-acetals or hemi-ketals are intrinsically unstable in water. By contrast, when that carbon is a full acetal or full ketal (a so-called non-reducing sugar) then it is stable in water, and anomerization does not occur.  Examples include trehalose and sucrose.


Epsom salts - heavy metal contamination myth?

Here in the US, they sell different grades of Epsom salt, from USP down to plant food.  I'd presume that depending upon what the product will ultimately be used for, the manufacturer may use MgO with different purity levels - the lower the purity, the cheaper the product.  Even the "bath salts" grade contain trace amounts of arsenic and selenium (see the first link below).  Among the "medicinal" offerings, some are labeled "for external use" and others are labeled "for external and internal use."  I suspect that anything with the "for external and internal use" label would be more than safe enough to use in handwash.

Here are some links you might find helpful:

http://k.b5z.net/i/u/2182313/f/Tech%20Data%20Sheets/Tech_Data_Sheet-bath_salt-epsom_USP-3.pdf

http://www.drugs.com/otc/104235/pepsom-flying-p-epsom-salt.html


Solubility of succinic acid in water vs ethyl acetate?

Hydrogen bonding interactions are also much stronger between succinic acid and water as opposed to EA since EA doesn't have any H-atoms that can participate in the hydrogen bonding.

Ionization is also a significant point to consider, but without pH data we can't say for sure.


Are salt substitutes good alternatives to sodium chloride?

Small quantities of potassium are fine. Windsor Salt of Canada makes a 50% blend of sodium chloride and potassium chloride called Half Salt. You could use it like regular salt, although it would be good to decrease the amount you use.

Nerve impairment and muscle function are affected from too much potassium.


How can you deduce from the structure that probenecid is an acid and ephedrine is a base?

The nitrogen atom in Probenecid is not part of a basic functional group (eg: amine), but of another kind of functional group: sulfonamide.

Sulfonamides are, just like amides, quite acidic: the presence of an electron-withdrawing group next to the nitrogen atom makes the electron pair less available, and makes the functional group quite acidic (see Lewis' definition of acids and bases).

For comparison, ammonium has a pKa of around 9, amides have a pKa above 15.

Sharing with amides the presence of an electron withdrawing group, sulfonamides can be expected to share the same acidic behaviour of amides.

You should check pKa list for the functional groups you have studied.

An example of such chart:

http://evans.rc.fas.harvard.edu/pdf/evans_pKa_table.pdf
(Be careful: values between parentheses are referred to the acidity in DMSO).

To better understand the delocalization of the electron pair in amides, see:
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Amides/Properties_of_Amides/Structure_of_Amides


Effect of relative strength on nature of solution

The conjugate base of a weak acid is basic, whereas the conjugate acid of a strong base is not acidic. When a weak acid and strong base neutralize each other, the basicity of the conjugate base formed makes the solution basic.


Which would be more acidic, benzene or cyclohexane?

See: Relative acidities of alkanes, alkenes, and alkynes. Essentially the same explanation holds here, as benzene carbons are sp2-hybridised.

The newly formed lone pair is in a σ-type orbital (blue) which does not overlap with the π system (red), hence no delocalisation is possible.


Gas chromatography/mass spectrometry: Can I use peak ratios and molecular mass to derive quantity?

Is this even how GC/MS results work?

As cbeleites said the method you described is a proper technique but not likely to be appropriate given the information you cited.

In GC/MS you should have two sets of information. The first is the GC Total Ion Chromatograph (TIC) which will have time as the x-axis and response (abundance) as the y-axis. For each retention time on the TIC there will be a corresponding Mass Spectra (MS). In the MS the x-axis is m/z (ion) and y-axis is also response (abundance).

Different compounds have different responses so if you inject the exact same amount of two different compounds you could get a much larger response from one compared to the other. For example tramadol gives much higher response compared to hydrocodone. This is why you want your internal standard to be structurally similar to your analyte.

Quantification of a compound is often done by running 3 - 5 calibrators at known concentrations in order to make a calibration curve. Once an acceptable calibration curve is made the sample along with controls can be fit onto the curve to calculate the amount of the compound of interest in the sample and controls. If the controls are correct then the sample value can be used.

To make a calibration curve you need an internal standard in each calibrator, control and sample. You can then make the curve either with ion ratios between paired internal standard and analyte ions OR if your internal standard response is consistent in all the calibrators, controls and sample you can use the GC peak response.

In conclusion if you only have one GC/MS run with the two dissimilar compounds in it you will not be able to easily calculate the quantity of the second compound based on the GC response of the first unless you have additional information that was not listed in this question.


How to calculate the heat capacity of a calorimeter?

This is impossible to answer. Usually you have to assume that when no calorimeter heat capacity is given, then it negligible (i.e. you only use the heat capacity of the 70g $\ce{H_2O}$). You know the temperature drop of the metal and the energy increase of the water, combine both to obtain the heat capacity of the metal.


Reasons for existence of eutectic point for alloy systems

First, there are several other answers that you might be interested in looking at first (full disclosure - they include some by me).

A question on Physics SE on eutectics

Ideal vs regular solutions and eutectics on Chemistry SE

These together go in to some detail on why the eutectic forms (when the liquid free energy drops down and touches the tie-line between the two solid phases - the Physics one), and the interactions one needs to get the liquid phase to drop below the tie line between the solids (the Chemistry one). To get a eutectic you need an interaction term in the liquid Gibbs free energy.

As for why it happens at such a definite composition, the glib answer is the Gibbs phase rule. In a system at fixed pressure, three phases can coexist at only one point.

Given that this question is, in general, a duplicate of the ones listed above, particularly of the Chemistry SE one, I have no particular need of the bounty. Go read those and see if they help.


How does the Mosher ester method allow determination of absolute configuration of secondary alcohols

The NMR spectra of two enantiomeric molecules are necessarily identical (just as all properties of enantiomers are identical in the absence of an external chiral environment). In order to tell the enantiomers apart, some additional chirality must be introduced. Two methods exist for this in NMR spectroscopy:

Chiral solvating agents, such as lanthanide shift reagents
Chiral derivatising agents, such as Mosher esters

In modern organic synthesis, the use of chiral derivatising agents is far more common, and the process can be summarised as follows:

Outline of the Mosher ester analysis. Taken from Organic Spectroscopy and Structure Elucidation lecture notes (not available online)

a. The alcohol of unknown configuration is derivatised by forming both the (R) and (S) Mosher esters. This can either be done under Steiglic conditions (DCC,DMAP) or via initial formation of the acid chloride.

b. Proton NMR spectra are ran of the two diasteromeric esters. The analysis is often ran on the crude esters, therefore the acquisition of COSY/HSQC is also useful to aid in assignment.

c. The difference between the chemical shifts of the (R) and (S) esters are tabulated (by convention we consider R-S, it makes no fundamental difference to the outcome however).  Note that the method does not consider the shift of the underivatised alcohol at all, however the peaks for the (R) and (S) ester must be unambiguously assigned. Note also that we do not take into account the oxymethine proton (the shifts of this are erratic and don't contribute to the analysis).

d. The configuration is assigned by considering which side of the molecule is most effected by the presence of the Mosher ester.

Step d is often the hardest to grasp conceptually. When we make the Mosher esters, what we're counting on is that they adopt a specific conformation (which they do). In this conformation, the phenyl group of the Mosher ester will shield part of the molecule (which is different depending on which enantiomer of the Mosher ester we're looking at). Consider the following example, in which the side shielded by the phenyl ring has a lower delta overall:

Shielding by the Mosher ester. Taken from Harvard CHEM2016 lecture notes, Eugene Kwan

The following conformational diagram is also (possibly more) instructive:

Shielding by the Mosher ester. Source unknown

The only way to actually understand the Mosher ester method is to carry it out (ideally with something of unknown stereochemistry so you can't convince yourself of the answer simply because you know its the case.

The next best thing is a worked example:

Outline of the Mosher ester analysis. Taken from Organic Spectroscopy and Structure Elucidation lecture notes (not available online)

Looking at the two spectra, it becomes apparent that (R)-(S) is positive on the LHS methyl group, and negative on the protons on the RHS propyl group. Fitting this to the model, we can assign the sterocente of the secondary alcohol as (R).

The reason both Mosher esters should be formed is due to the fact that change in chemical shift between one Mosher ester and the free alcohol is often not huge. Other derivitising agents (such as AMA) can be carried out with formation of a single ester.


What fraction of hydrogen in air results in an explosive mixture?

Hydrogen is explosive in mixtures in air at room temperature and ambient pressure when it is between 4% and 75% by volume.  These limits are respectively called the "lower explosive limit" (LEL) and the upper explosive limit (UEL).

Values for hydrogen (and many other gases) are provided in many locations on the internet.  Engineering toolbox is one such site.

More info on LEL and UEL is available from Wikipedia.


How to assess sunscreen effective via blueprinting?

From Wikipedia page on the blueprint process:

  The blueprint process is based on a photosensitive ferric compound. The best known is a process using ammonium ferric citrate and potassium ferricyanide. The paper is impregnated with a solution of ammonium ferric citrate and dried. When the paper is illuminated, a photoreaction turns the trivalent ferric iron into divalent ferrous iron. The image is then developed using a solution of potassium ferricyanide forming insoluble ferroferricyanide (Prussian blue or Turnbell's blue) with the divalent iron. Excess ammonium ferric citrate and potassium ferricyanide are then washed away. The process is also known as cyanotype.

Exposing blue print paper to UV light will cause it to develop blue. Sunscreen is meant to block UV light so applying sunscreen over blueprint paper then exposing to UV light and developing will give you a semi-quantitative way to evaluate the sunscreen based on the intensity of the blue.

See: Colorimetry


Wikipedia gives a structure with a central nitrogen atom:

The nitrogen atom (blue, using the standard color scheme) can form all three covalent bonds without building up an excess positive or negative charge, and so is favored in the central position through which all three bonds pass.


Answer to your question, "could the density of an aqueous solution be less than $\pu{1g/mL}$," is yes. See following examples:

The density of the commercially available 28% ammonia solution ($\ce{NH3}$ 1n $\ce{H2O}$) is listed as $\pu{0.9 g/mL}$ at 25 °C (Sigma-Aldrich Catalog).

MERCK Tabellen für das Labor has listed the density of 35% ammonia solution ($\pu{18 mol/L}$ concentration) as $\pu{0.88 g/mL}$ at 20 °C (Steffen's Chemistry Pages)

As obvious, the densities of various aqueous ethanol solutions are also equal to less than unity since the density of absolute ethanol is $\pu{0.789 g/mL}$ at 20 °C (Wikipedia). By a quick google search, I found the densities of various aqueous ethanol solutions listed in MERCK Tabellen für das Labor  in Steffen's Chemistry Pages, which is pretty interesting.


Finding electron configuration of lanthanide ions

I agree with bobthechemist's comment.

When you are determining the electronic configuration of anions/cations, you will first need to determine electronic configuration of neutral atom. Like in the case above:

$$\ce{Gd} : \mathrm{[Xe] (6s)^2 (4f)^7 (5d)^1}$$

Now arrange this configuration according to ascending value of $n$:

$$\mathrm{[Xe] (4f)^7 (5d)^1 (6s)^2}$$

Now, according to your need start subtracting electrons from highest value of $n$, like for $\ce{Gd^3+}$, the electronic configuration will be:

$$\ce{Gd^3+} : \mathrm{[Xe] (4f)^7}$$

See this to know more on calculating electronic configuration of cations and anions.


Yes, pH can be negative.

The official definition of pH is: negative log of H+ activity, not negative log of H+ concentration, however.  See pH Paradoxes: Demonstrating That It Is Not True That pH ≡ -log[H+] J. Chem. Educ., 2006, 83 (5), p 752.

By either definition pH can be negative in concentrated solutions of strong acids.  For example 5M HCl would have a negative pH.


Ground state oxidation potential and Ionization potential from density functional theory (DFT)

Short answer - compute the relative energies of the neutral and cation species

Typically, when people discuss the "oxidation potential," they're considering electrochemistry, so there's solvent around. Keep this in mind if you're trying to compare a calculation done in gas phase - the best comparison would be with an appropriate solvent calculation, although calibrating between a calculation and an experimental electrochemical potential requires work.

On the other hand, ionization potential is typically used for a gas phase measurement - for example, with XPS or UPS spectroscopy, where an electron is removed from a species.

Finally, there's a difference between the vertical ionization potential when an electron is ripped out of a species, but the cation hasn't had time to relax (change geometry) and the adiabatic ionization potential, which is the difference of energy between the neutral and the relaxed cation.

That's a long way of saying that if you want to compute an accurate ionization potential, you need to:

1) Optimize the geometry of the neutral species using an accurate method.

2) Take the optimized geometry, and run a geometry optimization with one fewer electron. (You will get both the vertical ionization energy as the first SCF energy, and the adiabatic ionization energy of the optimized cation.)

Subtract the total energies and convert to your unit of choice.


Is there a T-V or P-V chart relating LiBr temperatures for recovery?

Without some fancy equipment to measure the salt concentration in the solution,  a simple saturated solution of lithium bromide would have some relative humidity which would be a function of temperature.  I looked quickly for some relative humidity table of lithium bromide and found the following reference which has an equation that can be used for calculating the relative humidity ($RH_{sat}$) over a saturated solution of lithium bromide as a function of the temperature in Celsius ($T_C$).

  "Humidity Fixed Points of Binary Saturated Aqueous Solutions" JOURNAL OF RESEARCH of the National Bureau of Standards, Vol. 81A, No.1, January-February 1977, Lewis Greenspan (pdf link)

$RH_{sat} = 7.75437 - 0.0654994T_C + 0.420737\times10^{-3}T_C^2 $

Some values for $RH_{sat}$ are:

7.75 % at 0 C
5.41 % at 100 C

So as long as some solid lithium bromide remained then the solution would be saturated. If all the lithium bromide dissolves, then you'd simply boil the solution to regenerate it. Lithium bromide is about twice as soluble in a boiling solution (somewhat over 100 C) as at room temp (20 C). So boil down to some starting volume of solution or until you see solid lithium bromide in the boiling solution. When the solution cools to room temperature about half the lithium bromide will crystallize out regenerating the saturated desiccant solution.

If you do want to boil this to regenerate then be sure and use some boiling stones. The solution would seem likely to bump.


We have 2 gases in the container, oxygen ($O_2$) and nitrogen ($N_2$). Now we are given that $N_2$ has a pressure of $17$ atm out of $30$ atm, the total volume of the container is $5.00$ L and that the temperature is $25^o$C or $298$ K. So somehow, we need to find out the volume that oxygen occupies and from there we will be able to find out the grams of oxygen present in the container.

First let's find out the mole fraction $N_2$.
You can find mole fraction by the following formula,
 $$Mole\, Fraction = \frac{Partial\,Pressure\,of\,Gas}{Total\,Pressure}$$

So,
Mole fraction of $N_2$= $\frac{17}{30}$=$0.6$

Therefore mole fraction of $O_2$=$1-0.6$=$0.4$

Now we need to find out the total no. of moles by the formula,
$$PV={nRT}$$

Putting the values given for the whole system,
$$30\times5=n\times0.0821\times298$$
$$\Rightarrow\,n=6.1$$

So moles of oxygen in container = $0.4\times6.1$ = $2.5$
Consequently, grams of $O_2$ present in container = $2.5\times32$ = $78.4$

So that is how you get the grams of oxygen. Hope you understood everything. If not, feel free to ask by commenting. Good Luck!


Polarity and elementary electrostatics have nothing to do with acid-base definitions. Sometimes we try to find a reason, using our current understanding, and that ensues the danger of over-rationalization. At a truly fundamental level perhaps no chemist can tell you why an amine can accept another proton, may be someone can do quantum mechanical calculations and say why. So avoid over-rationalization... accept some observations as facts (temporarily until we develop a full understanding).

One can titrate an amine with an acid, by this experimentally observed fact, exactly just like we titrate NaOH with HCl. So an amine group is indeed a base or say it behaves like base. The acidic proton cannot annihilate itself during a titration, it has to go somewhere. The amine group gets protonated, in return.


How can I get a hand sanitizer solution to form a gel?

According to the wikipedia article on hand sanitizers, polyacrylic acid and polyethelyne glycol are commonly used thickeners.

You could try any hydrophilic polymer if those aren't available. Cornstarch might even work if you dissolved it in the alcohol before adding the other ingredients.


I see as I write the answer, that many comments are spot-on.

Whether or not $\ce{F2}$ has lone pairs depends on your view of molecular orbitals and valence bond (or Lewis) descriptions of bonding.

Covalent radii for fluorine compounds are usually around 0.64 Å for F, but yet the F-F bond distance is 1.43 Å.

If we go for the Lewis or valence bond view, we'd argue that the large number of lone pairs on a small atom will be highly repulsive.

If I go for the molecular orbital view, I'd point out that in reality, the anti bonding orbitals $\pi^*$ and $\sigma^*$ are higher in energy than their bonding counterparts, particularly in the 2nd row, so the F-F molecule will be weaker than a simple single bond.

There's a really nice article by Peter Politzer in JACS that I like to use when discussing this "anomaly": "Anomalous properties of fluorine" J. Am. Chem. Soc., 1969, 91 (23), pp 6235–6237

  It appears rather that the anomaly involves the fluorine atom, in whatever circumstances, and its interaction with an external electron which enters its outer shell. [...] In any case, it is suggested that because of the exceptionally small size of the fluorine atom (see Table I), the entering electron feels unusually large forces of attraction from the nucleus and repulsion from the electrons already associated with the atom. It appears that the magnitude of the anomaly in the repulsive forces is greater than in the attractive forces.

In other words, regardless of the MO or VB view, the issue is one of electrostatic repulsion.


Can I make a buffer with a weak acid without adding it's conjugated base, when the desired pH of the solution is exactly the pKa of the weak acid?

So your solution is already at a pH equal to the pKa of the weak acid you are about to add. Your hunch is that the pH remains the same after adding acid to a solution.

  [...] we will end up with what we wanted: The same concentration of the weak acid and it's conjugated base.

It depends on the concentration of what is already in solution and of what you are adding. If you add a weak acid at very low concentration (e.g. an indicator) to a buffered solution, the indicator's protonation state will be governed by the pH of the solution. On the other hand, if you add a weak acid at very high concentration (e.g. a substance that is supposed to act as buffer) to an unbuffered or hardly buffered solution, the buffer will govern the pH.

In your case, the pH will be off (too acidic), and most of the buffer substance will be in the acid (protonated) state. You would have to adjust the pH by adding a strong base to the amount that deprotonates half of the just added acid to get the pH to match the pKa again.

  So, in principle, it's possible to make a buffer without adding the salt?

Yes, but not in the scenario you suggested. Most buffers are made by starting with a weak base (pH is much higher than pKa of conjugate weak acid) and adjusting the pH with strong acid, or by starting with a weak acid (pH is much lower than pKa of the weak acid) and adjusting the pH with a strong base. In both cases, you effectively get a mixture with the "salt", i.e. you introduce some spectator ions (often chloride or sodium ions).


Is it possible to have a dispersion interaction with no inductive interactions?

The first-order correction in the interaction operator is an integral over the frozen reference densities, so in a product basis $|i_A,j_B\rangle$ that would read
$$ E^{(1)} = \langle 00|V|00 \rangle $$
and as such is the electrostatic interaction. This is the same as the Coulomb interaction of the two non-interacting reference densities.

Induction is defined as the  polarization of one monomer due to the reference density of the other monomer, so must involve integrals over the virtual orbitals of the monomer being polarized in order to capture its polarization. So induction is a second-order correction and has the form
$$ E^{(2)}_{ind} = \sum_{j\neq 0} \frac{|\langle 00|V|0j \rangle|^2}{E_{00}^{(0)}-E_{0j}^{(0)}} + \sum_{i\neq 0}\frac{|\langle 00|V|i0 \rangle|^2}{E_{00}^{(0)}-E_{i0}^{(0)}}$$
where the first term represents the polarization of monomer $B$ due to the reference density of monomer $A$ and the second term is the reverse process.

What sets the dispersion energy apart from the induction energy is that it is the simultaneous polarization of both monomers, and so the integrals run over the virtual orbitals of both monomers as in
$$
E^{(2)}_{disp} = \sum_{i\neq 0} \sum_{j\neq 0}\frac{|\langle 00|V|ij \rangle|^2}{E_{00}^{(0)}-E_{ij}^{(0)}}
$$
Mathematically, the dispersion and induction terms are both second-order correction terms in the energy expansion, but they have been grouped in a physically-intuitive way to give two different types of interactions. There are systems where dispersion is the dominant interaction, such as rare gas dimers.

EDIT:

An example of a dispersion-dominated bound system is the neon dimer. Below is the SAPT decomposition of the binding curve at the HF-SAPT(2)/aug-cc-pVDZ level.

At the equilibrium distance, dispersion dominates the binding interaction but there is a non-zero induction contribution. I am inclined to say that although the ratio of dispersion and induction may become large, you can never truly have zero induction. Some naive mathematical reasoning follows.

We need $ \langle 00|\frac{1}{r_{12}}|0i\rangle, i\neq 0$ to be zero in a Gaussian product basis. At a long-enough distance we can expand the Coulomb operator in a multipole expansion to give something like
$$
\frac{1}{r_{12}} = \alpha x_1x_2+\beta(x_1x_2^2-x_1^2x_2)+...
$$
in a one-dimensional model picture. I've also folded all the coefficients into the constants $\alpha,\beta,...,etc.$ The important thing here is the symmetry of the integrals. So, lets take $i=1$ and see what happens...
$$
\langle 00|\alpha x_1x_2+\beta(x_1x_2^2-x_1^2x_2)|01\rangle = \langle 00|\alpha x_1x_2|01\rangle + \langle 00|\beta x_1x_2^2|01\rangle-\langle 00|\beta x_1^2x_2|01\rangle+ ...
$$
The first two integrals are zero because $\langle 0|x_1|0\rangle=0$. The third integral is non-zero and survives in the sum. This can happen any time the operator in monomer $A$ is an even function. As $i$ gets larger, higher-order terms in the expansion contribute to the induction energy. So mathematically, there is a tiny but non-zero induction contribution.


U- or RO-method for Singlet-Triplet Gap?

To expand on user1420303's answer a bit:

When I want to compare the energy between two different spin states of the same molecule, i.e. singlet-triplet-gap, do I better use unrestricted or restricted open formalism to compare the energies between both?

It depends.
The unrestricted formalism will almost always give lower absolute electronic energies (i.e., closer to the "real" value) than restricted-open will, due to the greater variational flexibility introduced by the separate spin-up and spin-down orbital sets.  So, from a purely energetic perspective, yes, unrestricted is likely preferred.
If you're interested in other properties, though, the spin contamination introduced by the unrestricted formalism may be problematic.  Of course, in systems with sufficient static correlation (viz., a degenerate or near-degenerate ground state) to result in spin contamination (radicals, bond-breaking states, transition metals, etc.), single-reference methods like Hartree-Fock and DFT may not give reliable results anyways.

Further on, do I need to calculate the singlet state also with restricted open/unrestricted method?

It depends.
For most 'non-exotic' organic systems, there is no need to calculate the singlet state using an unrestricted method since, as you note, the results will often be indistinguishable.  For systems with static correlation, though, if you still want to try to use a single-reference method, you will need to explore the unrestricted orbital space, as there may be a lower-energy solution where the alpha and beta orbital compositions appreciably differ.  The primary method I'm aware of for such exploration is the "broken-symmetry" method.  Two papers I know of that discuss the method, albeit not in extensive detail, are the following reviews by Frank Neese:

F. Neese. "Prediction of molecular properties and molecular spectroscopy with density functional theory: From fundamental theory to exchange-coupling." Coord Chem Rev 253: 526, 2009. doi:10.1016/j.ccr.2008.05.014

F. Neese. "A critical evaluation of DFT, including time-dependent DFT,
applied to bioinorganic chemistry." J Biol Inorg Chem 11: 702, 2006. doi:10.1007/s00775-006-0138-1

The method is integrated into the current version of ORCA (see Section 5.9.10 of the v3.0.3 ORCA manual), and tips for running such calculations are available at the ORCA Input Library.  Other software packages likely include this capability; search their manuals for more information.


With which 2 of these 5 chemicals would Chlorine not react? N, NaOH, O2, NaBr, Fe

$\ce{NaOH}$

$$\ce{2NaOH + Cl2 → NaCl + NaClO + H2O}$$

  Reaction between chlorine and sodium hydroxide to form sodium
  chloride, sodium hypochlorite and water. The reaction takes place in a
  cold dilute solution. (Chemiday 1)

$$\ce{3Cl2 + 6NaOH → NaClO3 + 5NaCl + 3H2O}$$

  Reaction interaction chlorine and sodium hydroxide to form sodium
  chlorate, sodium chloride and water. The reaction proceeds in a
  concentrated solution with heating.(chemiday 2)

$\ce{NaBr}$

$$\ce{2NaBr + Cl2 → 2NaCl + Br2}$$

  Reaction between sodium bromide and chlorine to form bromine chloride
  and sodium. Hot - sodium bromide. The industrial method of producing
  bromine.(chemiday 3)

$\ce{Fe}$

$$\ce{2Fe + 3Cl2 ->[\Delta] 2FeCl3}$$

  The reaction of chlorine and ferric chloride reaction with the
  formation of iron (III). The reaction proceeds at a temperature above
  250 C °.(chemiday 4)

But I couldn't find any reaction of chlorine with neither nitrogen or oxygen. Guess what, chlorine doesn't have sufficient energy to break the triple bond of nitrogen or double bond of oxygen.


Proton removal during benzyne formation in substituted haloarenes

I appreciate Waylander's answer, but it did not address the OP's curiocity of why the base abstracts the proton from only ortho-position. Yet, Waylander correctly pointed out that since the aromatic $\pi$-system is at right angle to the triple bond (the second $\pi$-bond of the triple bond), the ability to donate a lone pair to the ring by a substituent (like $\ce{OCH3}$ here), or accept a lone pair from the aromatic nucleus (like $\ce{NO2}$ would do) is not relevant in this case. However, inductive effects of the substituents in aromatic nucleus are still relevant (i.e. electron-withdrawing effects that occur through single $\sigma$-bonds).
However, $\ce{t-BuO-}$ is not as strong a base as $\ce{H2N-}$ in usual cases of benzyne reactions. Thus, it would abstract relatively most acidic available proton from to aromatic nucleus. Compared to para-position, ortho-position is more acidic due to proximity (ortho is one $\ce{C-C}$ bond away compared to para which is three $\ce{C-C}$ bonds away). Following is an example you can find in literature (Ref.1):

References:

José-Antonio García-López and  Michael F. Greaney, "Synthesis of biaryls using aryne intermediates," Chem. Soc. Rev. 2016, 45(24), 6766-6798 (DOI: https://doi.org/10.1039/C6CS00220J).


Why is the boiling point of stibane higher than that of ammonia?

Though it does go against your intuition, you've actually mentioned the answer in your question. Stibane has a higher boiling point than ammonia/azane on account of van der Waals interactions (owing to the larger size of the antimony atom).

Our teacher had actually posed this question to us during my first year of high-school. All of us were incredulous when we were told that stibane had a higher boiling point than ammonia, and like you, we were fixated on H-bonding and chose to dismiss van der Waals' interactions as insignificant in this regard.

First, a look at the H-bonding argument:

At school we're taught that H-bonding only occurs in molecules where you find hydrogen atom(s) bound to either nitrogen, oxygen or fluorine ($\ce{N,O,F}$). This is because $\ce{N, O}$ and $\ce{F}$ are sufficiently electronegative (about 3.0, 3.5 and 4.0 on the Pauling scale respectively) to polarize the $\ce{N/O/F-H}$ bond. Do note, that nitrogen is only barely able to establish H-bonding, while it's much easier for oxygen and fluorine to polarize the $\ce{O/F-H}$ bond on account of 1) their higher electronegativites and 2) their smaller sizes.

As antimony is much larger than nitrogen, and since the hydrogen bonding due to the nitrogen in ammonia is pretty weak, the van der Waals interactions in stibane 'outdoes' the H-bonding in ammonia.

However the same argument does not follow for water and hydrogen fluoride, which have higher boiling points than tellane and hydrogen iodide respectively.

$$\textbf{Melting } (\vartheta_\mathrm{m})\textbf{ and boiling } (\vartheta_\mathrm{b}) \textbf{ points of}\\\textbf{Group 15, 16, and 17 hydrides  in K}$$
\begin{array}{lrrlrrlrr}\hline
&\hspace{-2em}\text{Group 15}\hspace{-1em} &&& \hspace{-2em}\text{Group 16}\hspace{-1em} &&& \hspace{-2em}\text{Group 17}\hspace{-1em}&\\ \hline
\text{Hydride} & \vartheta_\mathrm{m} & \vartheta_\mathrm{b} & \text{Hydride} & \vartheta_\mathrm{m} & \vartheta_\mathrm{b} & \text{Hydride} & \vartheta_\mathrm{m} & \vartheta_\mathrm{b}\\ \hline
\ce{NH3} & 195.5 & 239.6 & \ce{H2O} & 273.0 & 373.0 & \ce{HF} & 180.7 & 292.4\\
\ce{PH3} & 138.0 & 185.0 & \ce{H2S} & 190.3 & 211.0 & \ce{HCl} & 161.0 & 189.4\\
\ce{AsH3} & 159.0 & 218.0 & \ce{H2Se} & 209.0 & 231.0 & \ce{HBr} & 184.5 & 206.0\\
\ce{SbH3} & 184.0 & 256.0 & \ce{H2Te} & 222.0 & 271.0 & \ce{HI} & 222.2 & 237.0 \\ \hline\end{array}

Now here's where your logic comes into play; the H-bonding's pretty strong in case of water and hydrogen fluoride (after all, oxygen and fluorine are the two most electronegative elements we know) and this time 'round the H-bonding outdoes the van der Waals interactions.

So you see, you should've made a comparision between the higher and lower hydrides (for groups 15, 16 and 17, to really be able to spot the exception).

As for your spin-off question:

  Why does $\ce{NH3}$ not show hydrogen bonding in gaseous phase?

Well, the H-bonding here is strong enough to keep the ammonia molecules together at normal temperatures. But you must never forget that H-bonding is not a true chemical bond, but an intermolecular interaction. If you provide sufficient energy to a sample of ammonia (i.e. by heating it) the ammonia molecules wiggle and wiggle and wiggle, until at some point they wiggle so rapidly, that they overcome the attractive forces brought about by hydrogen bonding and become a gas. This is why you don't observe H-bonding in gaseous ammonia: because the particles are too energetic to be held down.


How many moles of aspartame are present in 4.00mg of aspartame? (3 sig figs)

Try putting in 0.0000136. Remember that zeroes between the decimal point and the last numbers do not contribute to significant figures. Good luck.


What makes R-MgX a more reactive Grignard reagent than R-Cu?

The difference in electronegativity between copper (1.9) and magnesium (1.3) is the key difference.

Since copper is more electronegative than magnesium its electronegativity is much closer to that of carbon (2.55).  This results in carbon-copper bonds being less polarized and more covalent than carbon-magnesium bonds.  The electrons in a carbon-copper bond are better shared between the two atoms, whereas the electron distribution in a carbon-magnesium bond places more electron density on carbon.

As a result of this difference in electron density on the carbon atoms the carbon portion of a cuprate is a softer nucleophile, while the carbon portion of a Grignard is a harder nucleophile.  These difference in nucleophilicity produce different reactivity patterns for the two reagents.  For example, the softer cuprate will undergo 1,4 addition to an $\alpha,\beta$-unsaturated carbonyl compound (a softer electrophile), while the Grignard adds directly to the carbonyl (which is a harder electrophile).


How to control the optimization step size in Gaussian?

Gaussian by default uses a dynamic trust radius for optimizations to a minimum (you also ensured this by including the trustupdate option). When you specify a maxstep this is only guaranteed to be the initial trust radius. In fact, the second step will ensure the trust radius is at least 0.05.
You can see the trust radius for a given step (if you have #p in the route section) by searching for "Trust test". In that same line, it will say "DXMaxT set to 5.00D-2", with the value printed being the trust radius used for this step.
To prevent these updates, simply replace TrustUpdate with NoTrustUpdate. Keep in mind that they likely force the radius up to 0.05 for a reason and your optimizations may fail or take much longer without these updates. You should only use a step this small if you think you have a geometry very close to the minimum or a very messy potential energy surface that you want to step through cautiously.


Book recommendations for protein structure refinement

I found an article here about PM6/7 which might be useful.


What is the physical state of the gas adsorbed on a solid adsorbent?

Well, it usually depends on the temperature: if temperature is too low, molecules cannot move so freely so they are localized on top of the solid without much movement.

However, in general cases the adsorbed gas behaves like a liquid in two dimensions. Near the surface, the gas molecules are 'trapped' in the force field of the solid, so they don't escape so easily. Then the freedom of movement occurs only in two dimensions (the molecules travel on the surface).

I made some time ago a movie calculating (using molecular dynamics) movement nitrogen molecules adsorbed on a graphene surface at 77 K. This can give you a clear image of what happens on the surface. As you can see, the molecules can travel on top of the graphene in two dimensions (we can call it a 2D phase). This movie is the case when still a monolayer of gas does not complete. There can be cases with less molecules, say, less pressure of nitrogen in the gas phase; or more molecules.

The physical states can be observed more clearly in phase diagrams of the adsorbed surfaces. Typical examples shows a plot of the adsorbed phase transitions in a P vs. 1/T plot. This topic is extensively covered in the nice monograph on Physical Adsorption by Bruch, Cole and Zaremba, for example.


Metallic and magnetic nature of transition metal oxides

This question should be cross-posted to Physics.SE, it is not a simple chemistry question. I'll give short rationalization of the fact, but beware: it is a rationalisation, not a deduction. It is also given in layman's terms.

We know the crystal structure of $\ce{TiO2}$, it is stereotypical and quite common. It contains octahedrons $\ce{TiO6}$ with oxygens shared between three octahedrons. Since the four compounds have similar composition and the cations have similar size, we may expect them to have similar structures. This is, however, a mere guess, but lets' go with it.

For a compound to have metallic conductivity, it must have mobile electrons, i.e. partially occupied group of orbitals. Thus, $\ce{TiO2}$ is obviously out.

It is known, that $d$-orbitals of a metal in octahedral neighbourhood split: three are lower in energy then two other.

In $\ce{MnO2}$ there are three electrons, so all three lower orbitals are occupied. This makes electrons immobile: for an electron to move to a neighboring atom, it has to pair with one of the already present electrons, which requires energy. I do not expect, however, that $\ce{MnO2}$ shall be a great insulator: the pairing energy is quite low, and furthermore, any impurity shall create charge carriers. So, $\ce{MnO2}$ is probably a semiconductor.

To choose from remaining two, we should recall that ferromagnetism is common among compounds with a lot of half-filled d-orbitals. This suggests that $\ce{CrO2}$ has higher chances, than $\ce{VO2}$ Still, the latter should be paramagnetic and conducting under right conditions.

Again, you should cross-post the question on Physics.SE and probably read a good solid state physiscs book.


In a redox reaction, why does more than one oxidation state of an element form?

They all appear because they are all finite amount of energy apart, but note that some concentrations are many orders of magnitude greater than others, and the concentration of $\ce{MnO4-}$ is so low that the table shows a zero.

Predicting the concentration requires being able to read tables of potentials. In standard tables like http://bilbo.chm.uri.edu/CHM112/tables/redpottable.htm  you will find equations such as $$\ce{MnO2{}+4H+ +4e-  \rightleftharpoons Mn + 2H2O}\quad (1.23\mathrm{V})$$ and $$\ce{Mn^{2+}{}+2e^-  \rightleftharpoons Mn}\quad (-1.185\mathrm{V})$$ (the numbers in parentheses are the standard reduction potentials, $E^\circ$, always measured in volts).

With this and the Nernst equation you can relate the concentrations of any set of species that form a balanced reaction.  In a complicated mix like this you don't know a priori the concentration of any species, only the certain sums (for instance, for all sulfur-containing ions) but you still can, perhaps iteratively, solve the corresponding set of algebraic equations and find the individual concentrations.  All the equilibrium equations have to be satisfied simultaneously; if they aren't the system is not in equilibrium.

(In this answer I've used the word "concentration" but the equations written with concentrations are only approximately satisfied; to account for second-order effects the notion of "activity" was introduced, and while it is an essentially empirical notion, it's still useful if the dependence of activity on concentration for a particular problem has been tabulated.)


Why does calcium carbonate decompose into calcium oxide?

Some quick googling did not turn up a definitive reaction mechanism, and most of the papers I did find seemed to instead be focused on the overall reaction rate and controlling factors - probably because this is such an industrially important reaction and (at least according to this article) gas transport is the rate-limiting factor - this is common for heterogeneous reactions.

However, it is not too hard to propose a plausible mechanism. I would suggest the same mechanism you did - what is likely happening is that the oxygen near one calcium ion can pull some electron density away from the carbon, leaving $\ce{CO2}$ which diffuses away as a gas. The calcium ion stabilizes the new oxygen anion and forms a new, different lattice structure.

As @Martin points out, the reason that this process is thermodynamically favorable is that the products have a higher entropy. The standard Gibbs free energy of reaction (using data from Tro's Chemistry) is:

$$
\begin{align}
\Delta G &= \Delta H - T\Delta S \\
&= 179.2\space kJ/mol - T(0.1602\space J/mol*K)
\end{align}
$$
Note that $\Delta H$ is positive, but so is $\Delta S$.
At 298 K (room temp) $\Delta G = 131.3 \space kJ/mol$ - the reaction is not favorable.

The equilibrium temperature (ignoring temperature corrections) should be around 1120 K.


Why does treatment of a geminal dihalide with sodium amide lead to an alkyne and not an allene?

Sodamide is bloody dangerous.  If your bottle is too old, it likely has begun to oxidize some and you have potential azides or other polynitrogen compounds that may lead to an explosion.

Even with good stuff, generally the reagent is not heated to avoid similar complications.  This means normally this reagent works at low temperatures (often times dry ice/acetone).  Under these conditions kinetic products are favored over thermodynamic ones.

The exomethyl group is much more accessible for deprotonation, and so kinetically, I would expect the amide ion to abstract the terminal hydrogen.

After that, the resulting bromoalkene still has the possibility of forming the allene product, but the transition state resulting in elimination to the alkyne is favoured as the vinylic hydrogen and the bromine are always locked in an anti conformation due to alkene geometry, as indicated in the above diagram. On the other hand, the allylic hydrogen may not necessarily be anti to the bromine due to C–C bond rotation.


As Jan mentions in the other answer, in general, a Lewis structure is preferred if it retains the octet rule. There can be more complicated bonding, and sulfur and phosphorous are two common examples (e.g., $\ce{SF6}$).

I find that many students can come up with "non-traditional" Lewis structures that, like yours, satisfy the number of valence electrons and minimize the formal charge. I usually consider these correct responses on an exam.

The "missing piece" is ring strain, which is not typically discussed until organic chemistry courses.

Note that your diagram as two O-S-O three-membered rings. These are extremely high in energy, because the O-O-S angle won't be anywhere near the expected 109.5°.

I tried a few quick calculations using the PM7 method and MOPAC. The best ring structure I could find looks like this (a distorted octahedral shape) and is estimated to have a $\Delta H_f^0$ = +33.05 kcal/mol. The O-O-S ring angles are ~65°.

The lowest energy geometry is the traditional $\ce{H2SO4}$ Lewis structure, estimated at $\Delta H_f^0$ = -177.88 kcal/mol. While PM7 isn't a highly accurate method, the difference in stability isn't close.


Yes, it is.
In both the cases, the number of ligands remains same, which is according to the definition in Wikipedia.

For any type of structural isomerism, the premise is that the primary valence and secondary valence (as defined by Werner) of central metal/ ion remains same for all the isomers.

In this case, the primary valence is 3 and secondary valence is 6, which remains same for all isomers.

Other possible coordination isomers are:

$\ce{[Co(NH3)4(CN)2]+[Cr(NH3)2(CN)4]-}$

$\ce{[Cr(NH3)4(CN)2]+[Co(NH3)2(CN)4]-}$

$\ce{[Cr(NH3)5(CN)]^2+[Co(NH3)(CN)5]^2-}$

(cation is always written first followed by anion)

I don't think $\ce{[Co(NH3)3(CN)3][Cr(NH3)3(CN)3]}$ can exist because, now both parts become neutral.


Do bare protons exist, even transiently, in aqueous solution?

According to Myths about the proton. The nature of H+ in condensed media Accounts of Chemical Research, vol. 46, pp 2567–2575:

  But to put this into
  perspective, an electron-free proton has an estimated acidity $10^{56}$(!) times greater than 100%
  H2SO4, the threshold of superacidity.$^2$ Adding a bare proton to a molecule is a real event in the
  gas phase but only a solvated hydrogen ion can be transferred in condensed phases.


Why do some salt ions in solution conduct electricity better than others?

As you add $\rm AgNO_3$ to $\rm KCl$, solid $\rm AgCl$ forms. The silver ion you add will at first be completely consumed by the chloride. Once all of the chloride is locked up in the precipitate, the silver ion concentration will rapidly increase. The Nernst equation shows how the potential is expected to change when this happens. If you're using a silver/silver chloride electrode with a reference electrode, the indicator potential is

$$\rm E = E^\circ_{\rm AgCl,Ag} - \frac{R T}{\cal F}\ln\frac{K_{sp}}{a_{\rm Ag^+}}$$

To see whether you'll get a drop or a rise in $E$, ask yourself what happens when silver concentration rises. If you've got the leads hooked up to your multimeter properly (negative lead to the silver cathode, positive lead to your anode), shouldn't you be seeing an increase in potential? What exactly are your $x$ and $y$ axes on that graph?

You'll get a bigger change when you form $\rm AgI$, because $\rm AgCl$ is about a thousand times more soluble than $\rm AgI$. The $\rm AgI$ will lock up relatively more ions when it forms (the $K_{sp}$'s of $\rm AgCl$ and $\rm AgI$ are $1.8\times 10^{-10}$ and $1.5\times 10^{-16}$, respectively).

Addendum 1: You say you're just using a silver electrode; the indicator potential is

$$\rm E = E^\circ_{\rm Ag^+,Ag} - \frac{R T}{\cal F}\ln\frac{1}{a_{\rm Ag^+}}$$

(but if you dip a silver wire into a chloride solution you'll get a very thin layer of AgCl and it'll begin to be a silver/silver chloride electrode (Source: G. D. Christian, Analytical Chemistry, 4th ed., Wiley 1986, p. 282)).

You also say you don't have a true reference electrode, just a graphite electrode... the fact that its potential can change as well complicates things. Your cell potential will depend on what the oxidation process is at the anode.

Addendum 2: You asked for a source for the $K_{sp}$'s. CRC Handbook 58th edition gives $K_{sp}$'s at $25^\circ$C for AgCl and AgI as $1.56\times 10^{-10}$ and $1.5\times 10^{-16}$; the 84th edition gives $1.77\times 10^{-10}$ and $8.52\times 10^{-17}$; Keller's Basic Tables in Chemistry gives them as $1.0\times 10^{-10}$ and $1.0\times 10^{-16}$ for "temperatures near 298 K". They're very temperature dependent and they depend on conditions like ionic strength as well, so it's not surprising that the values you're finding on the web are all over the place.  I took the values above from a genchem text that happened to be lying on my desk.


Are glycosides more soluble in dichloromethane than in water?

Anybody with glycoside isolation from bio-matrices would tell you that your plan is not going to work. Amygdalin is classified as a cyanogenic glycoside because it contains nitrile moiety. It is contained in the seeds of bitter almond, apricot, apple, black cherry, peach, plum, etc. Amygdalin was first isolated in 1830 from bitter almond seeds (Prunus dulcis) by French chemists, Pierre-Jean Robiquet and Antoine Boutron-Charlard. Traditionally, the extraction method of the chemical ingredients in the medicinal materials is decoction in boiling water, which was the method French chemists has used to isolate amygdalin. Presently, it has been isolated using several other methods and its structure has been characterized by several groups around the world.

Amygdalin can be dissolved in water and ethanol easily. Solubility of amygdalin in water and ethanol is $\pu{83 g/L}$ and $\pu{1 g/L}$, respectively (Ref.1). Ref.1 described the isolation of amygdalin from plum seeds for the first time, has used ethanol to extract soluble compounds, and finally, used diethyl ether to isolate amygdalin from the complex ethanol extraction mixture. Before extract, the plant material was dried to the moisture content of 6% in a dark place and then ground in a blender to obtain a fine powder with average particle size of $\pu{0.3 mm}$. The ethanol extraction and diethyl ether isolation procedure used in Ref.1 have been described as follows:

  The powdered solid plant material ($\pu{2.0 g}$) was transferred into the Plant material (2 g) was transferred into the $100$-$\pu{mL}$ round bottom flask and treated with exactly defined volume of ethanol. The extractions at higher temperatures were carried out under reflux, while the extractions at lower temperatures were carried out in closed vessels. The system temperature was maintained using a water bath. After extraction, the extract was separated from the solid plant material by filtering process. Plant extracts were evaporated in rotary evaporator under reduced pressure to remove the solvent, and then stored in desiccator until it reached the constant mass. After addition of diethyl ether ($\pu{10 mL}$) to the complex extraction mixture, amygdalin was precipitated, while fatty compounds were dissolved. The dissolved fatty compounds were removed by decanting process. The solid phase was dried at $\pu{30 °C}$ in order to remove the residual solvent. In this way, the sample was prepared for determination of its purity using HPLC analysis and structural characterization by FT-IR, UV, and MS methods.

The effect of extraction time ($\pu{10–120 min}$), ethanol concentration ($20–100\%,\: v/v$), solid-to-liquid ratio ($1: 5\: – 1: 25,\: m/v$) and extraction temperature ($\pu{22–78°C}$) on the content of amygdalin in the dried extract has been studied. As a result, the optimal conditions were achieved: The extraction time of $\pu{120 min}$; $100\% \:(v/v)$ ethanol; solid-to-liquid ratio of $1:25\: (m/v)$; and extraction temperature of $\pu{34.4°C}$. The optimal yield of $\pu{25.30 g}$ of amygdalin content in the extracts per $\pu{100 g}$ of dried extract was reported.

Reference:

Ivan M. Savic, Vesna D. Nikolic, Ivana M. Savic-Gajic, Ljubisa B. Nikolic, Svetlana R. Ibric, Dragoljub G. Gajic, “Optimization of technological procedure for amygdalin isolation from plum seeds (Pruni domesticae semen),” Frontiers in Plant Science 2015, 6, Article 276 (11 pages) (DOI: 10.3389/fpls.2015.00276).


Flame test color for non-ionic compounds

Flame emission (or better atomic emission) does not care whether your starting compound was ionic or not. Everything decomposes into flame either into atoms or very simple molecules. The blue color you see from the flame is also from very small molecules which exist in the flame.

https://en.wikipedia.org/wiki/File:Spectrum_of_blue_flame.png

Ordinary flame like the Bunsen burner is way too cold for non-metals. However if we were to heat HCl or ammonium hydroxide in a confined "container" at very high temperatures say, 5000 - 10,000 $^oC$, eventually you would start to see atomic emission from hydrogen, oxygen, nitrogen and chlorine. Most of the non-metal emission (p-block) lies in the UV or deep UV. You won't be able to see anything with bare eyes.


How to balance the reaction equation of potassium permanganate, calcium oxalate, and sulfuric acid step by step?

This is a redox reaction. While it can be balanced by the method you describe, it is much more reliable to balance it by the half-reaction method.

Step 1 - Identify your spectator ions (so we do not spend time trying to balance them)

Three ions appear unchanged on both sides of the reaction: $\ce{K+},\ \ce{Ca^2+},\ \& \ \ce{SO4^2-}$

Step 2 - Identify your redox pairs.

You have one manganese species on each side. They must constitute one half-reaction:
$$\ce{MnO4- -> Mn^2+}$$
Likewise, you have on carbon species on each side. They must constitute the other half reaction:
$$\ce{C2O4^2- -> CO2}$$
For each half reaction:

Step 3 - Balance O by adding $\ce{H2O}$
Step 4 - Balance H by adding $\ce{H+}$
Step 5 - Balance charge by adding $\ce{e-}$
Step 6 - Combine the half reactions

Can you take it from here?