Reverse phase is the standard in HPLC so, if you can use reverse phase do not complicate matters.
Normal phase usually gives more problems than reverse phase and the stationary phases degrade faster.
Reverse phase works fine and in this case you would typically use a C18 column and water and acetonitrile.

In the case of salicylic acid and, in general, all the acids and phenols, the peaks in reverse phase are normally broad and the retention times vary.
That is because there is an equilibrium between the acid and the ionized form, the carboxylate (or phenolate).

To avoid that, you have to force the equilibrium to the acid form, which is also more apolar, so the reverse phase makes even more sense.
For this, it is sufficient to add a small amount of acid to the water used in the mobile phase.
Very often, 0.1% formic acid is enough.
Sometimes it is required to use higher concentrations up to 1%.
If you need a higher $\mathrm pK_\mathrm a$, you can use trifluoroacetic acid instead.


How to determine the order of elution in silica gel column chromatography?

Your conclusion is absolutely correct. Column chromatography can be thought of as three-dimensional version of TLC (and vice-versa). So the most polar compound which interacts with silica gel most elutes slowest and the least polar compound leaves the column first.

I also support your suggested order as an initial starting point. However, As Klaus mentioned in a comment, unless one has performed the same separation multiple times one should never rely on paper-chemical polarity analysis alone and always confirm structures independently — typically that means ‘measure an NMR of all fractions’.


I think your thought process is correct, but you can push it further.
The way I would solve this problem is to think about how you could go from the alkyne to either alkene.
Then, via microscopic reversibility, you would argue that the reverse mechanism--the elimination--is very similar.

 The opposite of conjugate addition + protonation would be most similar to E1cb.


This is an exercise in applying the Hammond postulate (multiple times). To accurately compare reaction rates, we need to have the activation energy. We don't have this, and the actual energy of the transition state is unknowable without experimental data.
Hammond's postulate states that the structure of the transition state is more similar  to the intermediate to which it is closer in energy. For these reactions, the formation of the carbocation is endergonic, so the transition states look more like the carbocation than the alkene. We can then make some assumptions about the relative energies of the transition states from the energy change of the mechanism step if some things remain the same.
For example, alkene 3 can be protonated to form two different carbocations: A and B.

Using PM6 calculations in MOPAC, I calculated the heats of formation of alkene 3 and cations A and B.
$$\begin{array}{l|r}
& \mathrm{\Delta_r H^\circ \ (kJ/mol)} \\ \hline
3 & -42.1 \\
\mathrm{A} & 710.7 \\
\mathrm{B} & 702.6\\
\end{array}$$
Since we are starting from the same place (alkene 3), we can postulate that cation B forms more quickly than cation A because it is slightly lower in energy.
You need to do similar analysis for all your alkenes and the cations they can form. In some cases you have two alkenes that might form the same cation. In that case, the higher energy alkene would likely translate into the more rapid reaction.
You do not need to go and calculate all the energies. You should have learned some rules by now to assess the relative stability of alkenes and carbocations based on substitution pattern.


Is the dissolution of sodium acetate trihydate endothermic?

The process of dissolving sodium acetate trihydrate $(\ce{NaC2H3O2.3H2O})$ in water is endothermic. The molar enthalpy of solution at $T=25\ \mathrm{^\circ C}$ is $\Delta_\text{sol}H^\circ=19.66\ \mathrm{kJ\ mol^{-1}}$.*

Accordingly, since crystallization is the reverse process of dissolution, the crystallization of sodium acetate trihydrate from aqueous solutions is exothermic. For practical purposes, the enthalpy of solution with a reverse sign is taken as enthalpy of crystallization.

The enthalpy of solution mainly depends on two energy contributions: lattice energy and hydration energy. Lattice energy is the energy released when the crystal lattice of an ionic compounds is formed. Conversely, energy equal to the lattice energy has to be supplied to break up the crystal lattice; i.e. this process is endothermic. Hydration energy is released when water molecules hydrate the ions; i.e. when water molecules surround the ions and new attractions form between water molecules and ions. This process is exothermic.

Therefore, whether the process of dissolving a salt in water is exothermic or endothermic depends on the relative sizes of the lattice energy and the hydration energy. If the lattice energy is greater than the hydration energy, the process of dissolving the salt in water is endothermic. Conversely, if the lattice energy is smaller than the hydration energy, the process of dissolving the salt in water is exothermic.

For anhydrous sodium acetate $(\ce{NaC2H3O2})$, the lattice energy is actually smaller than the hydration energy. The molar enthalpy of solution is negative; at $T=25\ \mathrm{^\circ C}$ it is $\Delta_\text{sol}H^\circ=-17.32\ \mathrm{kJ\ mol^{-1}}$.* Thus, the process of dissolving anhydrous sodium acetate in water is exothermic.

However, hydration of ions can also occur in crystalline solids. Sodium acetate trihydrate $(\ce{NaC2H3O2.3H2O})$ contains three water molecules for each $\ce{NaC2H3O2}$ unit in the crystal. The corresponding amount of hydration energy is already released when crystalline anhydrous sodium acetate absorbs water and is converted to crystalline sodium acetate trihydrate. Therefore, the additional amount of hydration energy that is released when sodium acetate trihydrate is dissolved in water and completely hydrated is smaller than the total hydration energy that is released when anhydrous sodium acetate is dissolved in water. Hence, the resulting value for the molar enthalpy of solution is larger for sodium acetate trihydrate than for anhydrous sodium acetate.

* “Enthalpy of Solution of Electrolytes”, in CRC Handbook of Chemistry and Physics, 90th Edition (CD-ROM Version 2010), David R. Lide, ed., CRC Press/Taylor and Francis, Boca Raton, FL.


How does UV affect skin colors in dark-skinned people?

There is a journal article almost exactly about the title question, Mechanisms of Skin Tanning in Different Racial/Ethnic Groups in Response to Ultraviolet Radiation Journal of Investigative Dermatology (2005) 124, 1326–1332.

However, the body of the question assumes that eumelanin changes color.  Instead, the article finds that melanin gets closer to the surface of the skin, and the degree to which this effect occurs varies with race/ethnicity.


Is there a blanket term for all variants (both nuclides and ions) of a chemical element?

Yes, for example consider selenium. You can say "selenium species".


You don't need gas laws, temperature or $K_p.$
You overcomplicated the problem: $K_c$ can be found by following its definition with an RICE table:
$$
\begin{array}{lccccc}
&\text{R}\ce{&2 HI(g) &<=> &H2(g) &+ &I2(g)} \\
&\text{I} & 2c_0 && 0 && 0 \\
&\text{C} & -2\alpha c_0 && \alpha c_0 && \alpha c_0 \\
&\text{E} & 2(1 - \alpha)c_0 && \alpha c_0 && \alpha c_0 \\
\end{array}
$$
For the sake of consistency, I denoted the initial concentration with $c_0$ and dissociation degree as $\alpha.$
By definition, the equilibrium constant on a concentration basis $K_c$ can be found as follows:
$$K_c = \frac{[\ce{H2(g)}][\ce{I2(g)}]}{[\ce{HI(g)}]^2} = \frac{(\alpha c_0)^2}{(2(1 - \alpha)c_0)^2} = \frac{\alpha^2}{4(1 - \alpha)^2} = \frac{0.2198^2}{4(1 - 0.2198)^2} \approx 0.0198$$


Is ln(K2/K1) = (∆h°rxn/R)(1/T1 - 1/T2) even when ∆h°rxn varies strongly with temperature?

For each species, the quantities $g^0$, $s^0$, and $h^0$ are defined as the free energy, the entropy, and the enthalpy of the pure species, respectively, at 1 atm. pressure and temperature T. So, for constant pressure of 1 atm., $d\Delta g^0=\Delta s^0dT$.  But, $\Delta s^0=\frac{\Delta g^0-\Delta h^0}{T}$. So $\frac{d\Delta g^0}{dT}=\frac{\Delta g^0-\Delta h^0}{T}$

So, $\frac{d(\Delta g^0/T)}{d{(1/T)}}=\Delta h^0$


Highly conjugated compounds and the effects of substiuents

Yes, your intuition is correct.

Among others properties, systems with pi-electrons delocalisation are actually chromophore whose UV-Vis absorption spectrum is mainly dictated by the HOMO-LUMO transition.

An extension of the delocalised system results in a narrowing of the gap and thus shifts the spectrum to red. The effect is dramatic when extending, e.g., the length of a polyene chain.

However, although with different impact, the same is true if a parent system is substituted with group leading to additional resonance, both because of electron donating or withdrawing effects.

This is fundamentally no different from considering a C=C double bond, which is itself a chromophore, and functionalize it with the various auxochromes that are commonly listed when introducing UV-Vis electronic absorption spectroscopy.

An auxochrome as listed won't necessarily delocalise over the conjugated system in complicated structure, or it can marginally do so. Thus, don't assume that if there is a methoxy - just to mention an auxochrome - somewhere in the molecule or name, the spectrum of the latter is necessarily red shifted as compared to the parent molecule lacking that auxochrome.

If the auxochrome can indeed partecipate in delocalisation, then its effect is bathochromic (red shift) and generally hyperchromic (higher absorbivity).

Rules for predicting the wavelength in the absorption spectrum of conjugated systems do exist, though they fail quite early with increasing the conjugation length.

Consider that the conjugation length of a system is better described by an effective conjugation length, as for the system is susceptible of various effects, first of all departure from planarity due to hindrance. The same is true for the effect of the substituent. If you think of a very long polyene, an end substitution doesn't necessarily extend its effect all along the formally conjugated system.

Similarly the concomitant presence of electron withdrawing and donating groups can have negligible or no effect, as their influences can somehow cancel each others.

Again, is very important to consider that the effective delocalisation is not only influenced by the size of the system, but by an ensemble of parameters, such as the already mentioned steryc hindrance, structural motifs that might pin electrons (think of a phenylene in the middle of an otherwise polyene chain), solvent, etc.

With a bit of caution, you can certainly take that substitution of a conjugated system with groups having mesomeric effects results in absorption maximum at longer wavelength as well as in a higher absorptivity.


Comparison of rate of reaction in bromination and chlorination

According to the Hammond postulate, a highly exothermic reaction is more likely to have a low transition state barrier than a less exothermic reaction.

Lower barriers lead to faster reactions.

This is a multistep mechanism, so for this to be relevant the rate-determining step.  According to this lecture the rate limiting step is the hydrogen abstraction from the alkane, forming a hydrogen-halide bond.

The bond dissociation energy for HCl is 431.8 kJ/mol and 365.7 kJ/mol for H-Br.  The formation of the H-Cl is this much more exothermic, and the reaction will be faster.


what surface would liquid wax adhere to best?

I think that you need a surface with very low wettability, because as you said in the comments, you want to remove the solidified wax layer afterwards. It is possible to make a very thin layer of wax by spreading it between two hard plates pressed against each other. I would try something like this:

You spread the molten wax between the two sheets made from an appropriate material placed between two plates (glass or metal). To precisely regulate the thickness of the wax layer you use some separators placed near the edges. After setting you remove the plates, then carefully remove the sheets. The main problem is to choose a suitable material for the sheets. You can experiment with various materials. I think that a sheet made from, or covered with a silicone rubber would be perfect. You can try a parchment paper, aluminum foil, glassine and so on. If you need a really thin and fragile wax layer, you can try to use a tin or aluminum foil, and then remove it chemically using diluted hydrochloric acid.
EDIT: Now it hit me. There is another, simpler method you may try. If you want a wax plate with a uniform thickness and smooth surfaces use the procedure above. However, if you don't have those requirements, then simply pour the molten wax on the surface of hot water in a large pot. The wax should spread evenly on the whole surface. Leave it to cool down and collect the wax plate. It may be a bit thicker near the edges and its surface will be matte, but maybe it'll suit your needs.


At what point does decreasing solvent temperature cause a decrease in gas solubility?

Rules such as "solubility of A in B rises with lower temperature" are only meant to be used if there is no phase transition of the participating substances.


What are the specific dangers associated with Grignard reagents?

I have to disagree your position entirely.

Every chemical has associated hazards, with even the most unassuming chemicals having the potential to cause severe harm if not fatality.

However, the actual risk to the chemist is often severely minimised by taking sensible precautions, understanding the hazards, and working under a controlled set of conditions.

The whole purpose of organic chemistry education, especially the purpose of organic chemistry labs is to teach students how they should safely handle chemicals, not to tell them that they're too risky.


A question about the equality of heat and changes in enthalpy

In the absence of non-PV work (at constant pressure), the change in enthalpy of the system is equal to the heat exchanged between system and surrounding. This follows from the first law of thermodynamics and the definition of enthalpy.
The equation you are showing is based on the definition of the Gibbs energy (the chemical potential is defined via the Gibbs energy, and the Gibbs energy and the internal energy are different by TS). Just because you define some new thermodynamic functions and express the change in enthalpy with it does not invalidate the first law of thermodynamics.

How can we just ignore the summation for something like a chemical reaction?

Here is a silly calculation to model what is going on. Let's start with a variable $a$ known to be five:
$$a = 5$$
Now we add $b$ to both sides:
$$a + b = 5 + b$$
Let's define a new variable to simplify the right hand side:
$$c = 5 + b$$
and substitute that into the equation:
$$a + b = c$$
Now we solve for a:
$$a = c - b$$
It looks like $a$ depends on $b$ and $c$. However, we know $a$ is equal to 5, so it is constant. If you substitute the definition of $c$ back into the bottom equation, you get back to $a = 5$.


According to this equation for glycolysis, what is reduced and where do the electrons go?

NAD+ accepts a hydride ion (proton + 2 electrons) and is reduced to NADH.


Why does SYBR green emit more signal when bound to DNA?

In dyes such as this rotation about bonds that can be involved in the conjugated part of the molecule contribute to radiationless transitions causing the excited state to transfer its energy to  a triplet or back to the ground state rather than fluorescing. When intercalated between base pairs as you suggest in your diagram, the molecule is held more rigidly and so radiationless transfer is reduced and fluorescence increased.

In molecules with a dipole it may additionally be the case that in a polar solvent the stabilisation due to the solvent polarity brings the initially formed excited state closer to a crossing to a charge-transfer state and this non-radiative process will lower fluorescence yield. Intercalated in the DNA the polarity will be lower than in water and again this will result in more fluorescence.  If your molecule changes wavelength in polar vs non-polar solvents then this is indicative of this process.


Have you checked the original papers by C. Eckart, R. P. Bell, H. Shin, J. O. Hirschfelder and E. Wigner? These contain a fair degree of mathematical and physical treatment.

There is also a nice paper from S. G. Christov called "Some aspects of the theory of proton-transfer processes" published in J. Res. Inst. Catalysis 1968, 16 (1), 169-194 (pdf).

Other possible references:

Johnston, H. S.; Heicklen, J. Tunnelling Corrections for Unsymmetrical Eckart Potential Energy Barriers. J. Phys. Chem. 1962, 66 (3), 532–533. DOI: 10.1021/j100809a040.

Some papers published by D. G. Truhlar together with T. N. Truong, B. C. Garret, R. T. Skodje.


Carbonic acid-hydrogencarbonate buffer system pH calculation?

Your calculation is correct. However, there needs to be more bicarbonate in the blood because the majority of all metabolic processes tend to produce more acidic than basic products. There are also other ways to help the body maintain a pH of 7.4, not only the carbonic acid/bicarbonate buffer.

So both you and the book are right. If the calculations are done assuming there is just carbonic acid/bicarbonate acting as a way to maintian the pH in the blood, the ratio should be about 10 to one. Since there is a lot more going on though, the ratio is different than that.


Abnormal deviation of CO2 from ideal gas behaviour

When the compressibility factor z for CO2 is expressed in terms of reduced pressure and reduced temperature (i.e., using the law of corresponding states), the PVT behavior of CO2 is not very different from that of other gases.  The reduced temperature and reduced pressure are defined as the actual pressure and temperature normalized in terms of the critical pressure and critical temperature, respectively.  The real question is, "what causes the critical temperature and critical pressure of CO2 to be what they are?"


Why is the radius of the caesium cation smaller than that of the chloride anion?

This is something students often forget when comparing radii: shells are not simply additive.

If you move along eight elements to go from lithium to sodium, you are not only filling the second shell and putting one electron into the third, you are also adding a proton to the nucleus with every additional electron. The larger the nuclear charge is, the greater the electrostatic potential the electrons experience. And the greater the potential the stronger the force that draws them towards the nucleus, thus the more compact an orbital will become. Thus, while each new period will enlargen the atomic radius each added proton will reduce it. Therefore, the atoms do not become infinitely larger in size — although a sodium atom will always be larger than an electronically comparable lithium atom.

A similar discussion can be made for charges. Adding an electron to the mix increases the electron-electron repulsion (more negative charge) while removing one decreases it. Thus, with each positive charge a cation will get smaller while an anion will get larger with each negative charge. All of this results in rather complex calculations; one should always remember that an atomic radius cannot be easily guessed unless it is a straightforward up/down or left/right movement in the periodic table.

As for your second question: an anion may be smaller than a cation if the anion has very few shells and the cation very many. Unfortunately, the Wikipedia article on ionic radii does not quote a size of the hydride anion which would otherwise be a strong contestant. Instead, the smallest anion in the tables is $\ce{F-}$ with a crystal ionic radius of $119~\mathrm{pm}$ in hexacoordinate state. Searching for a cation that is larger quickly gives potassium with $152~\mathrm{pm}$ crystal ionic radius. In effective ionic radius terms (the second table), fluoride is mentioned as having $133~\mathrm{pm}$ while $\ce{K+}$ has $138~\mathrm{pm}$. Thus, by both sets of values the cation in $\ce{KF}$ is larger than the anion.

In crystal ionic radii, chloride is smaller than caesium and fracium while the latter two’s effective ionic radii are larger than chloride’s.

As you can see, there is no hard and fast rule. Fluoride to potassium is a period and two elements difference while chloride to eka-francium is four periods and two elements.


Is this a valid structure for the nitrate ion?

Nitrate actually exists as a superposition of these three resonance forms:

(Source: Wikipedia)

Essentially, a lone pair from an oxygen with a negative formal change forms a double bond to nitrogen. However, as nitrogen is a period II element, and hence must obey the octet rule, it can not have 5 bonds to other atoms. The already-present double bond then turns into a single bond and the pair of electrons moves onto the oxygen. This "movement" can happen with all three of the oxygen atoms and all three of the bonds. This is called "resonance".

Due to resonance, the energy of the system is minimised making it more stable.

Your Lewis structure is not far off. The "true" structure of nitrate is:

(Source: Wikipedia)

Approximately, each oxygen atom contributes one third of a bond with the nitrogen, lowering the formal charge on each atom from what you have written, which does not exist in nature.

The true structure is sort of an average of the properties of each possibility for where the electrons could be located, including the three resonance forms above, your structure, and any other configuration. However, these other forms are of minuscule probability to be formed (perhaps read up on wavefunctions and electron distributions if you're interested), and so the main contributors to the "real" structure are the three resonance forms.

Note that none of these structures ever exist in isolation for more than the most infinitesimal of time periods.


How do energy changes in reactions avoid violating the conservation of momentum?

How do you measure temperature? The thermometer has to be in thermal equilibrium, so you need to send the thermometer with the object. But that means that temperature, for our purposes, is going to be invariant between different reference frames. No matter how you're moving, the reading on the thermometer is the same.

This is clearly not the case with translational kinetic energy of the bulk material. In the rest frame of the object moving with constant velocity, the kinetic energy is zero.

Therefore, you should not mix up the thermal kinetic energy of the system with the translational kinetic energy of the system. They can't be the same thing. In other words, when the exothermic system loses energy, you don't get to subtract that from the translational kinetic energy.

See also these discussions: here and here.

I would also like to point out that you've just described a rocket. And rocket science, while complicated, is still obeying conservation of energy and momentum.


By far the safest solution is to physically abrade the surface until you have worn away everything that isn't copper. This is slow and requires some sort of grinding agent but has the upside that you are not likely to injure yourself.

There is a chemical solution that will work but it requires access to some dangerous chemicals and is extremely hazardous even in a well-equipped lab. It is insane to try anywhere else as you won't have adequate safety equipment.

The best way to remove carbon black from anything is to use a mixture of concentrated sulfuric acid and hydrogen peroxide. This is often used (but also often highly discouraged because of the danger involved) in laboratories to clean sintered glass filters contaminated with things like activated carbon. In the case of a copper bowl, the solution will remove the carbon but will also etch the copper surface (it is sometimes used as an etchant to remove copper from printed circuit boards).

The mixture of sulfuric acid and hydrogen peroxide is sometimes known as Piranha solution of reasons that are obvious when you have seen it work.

I would not encourage anyone without good chemical safety training and access to very good safety equipment to work with the mixture. And even then they should work only with small amounts and in a fume hood behind a shield and while wearing eye protection, strong gloves, and a fireproof coat.


What is the substance with the highest specific heat?

This may not be the absolute highest, but on a mass basis hydrogen gas has more than three times the specific heat as water under normal laboratory conditions.  Diatomic gases under ambient conditions generally have a molar specific heat of about $7\text {cal/(mol K)}$, and one mole of hydrogen has only $2\text {g}$ mass.  Thus $3.5\text {cal/(g K)}$ for hydrogen versus $1\text {cal/(g K)}$ for water.

Helium, contrary to the question, is $5\text {cal/(mol K)}$, not $7\text {cal/(g K)}$.


Etching PCB - dealing with anhydrous ferric chloride

$\ce{FeCl3}$ ( and similarly $\ce{AlCl3}$ ) has very different properties, comparing its anhydrous and hydrate form. They cannot be easily transformed one to the other. Heated hydrate decompose to the oxychloride or oxide eventually. Anhydrate put in water undergoes violant hydrolysis.
I suggest obtaining the hydrate, or preparing it yourself, e.g. reacting of iron, $\ce{HCl}$ and $\ce{H2O2}$ ( this mixture of the 2 reagens was an alternative Cu etching solution, when $\ce{FeCl3}$ hydrate was not easily available in the communist Czechoslovakia, but is more aggressive then $\ce{FeCl3(aq)}$ ).
Another option is to let react the anhydrate with limited water outdoors and that adding concentrated HCl. For the dissolving itself it would be the best to put in directly to the acid:water 1:1, but it would develop dangerous fumes.
Avoid mixing concentrated stuff, it would turn into a chlorine chemical weapon. Start with 10% peroxide and 1:1 acid, calculate the ratio. Tune it later at will. Keep the etching solution rather mild than strong, to simulate etching performance of FeCl3. My experience with etching $\ce{Cu}$ circuit boards was that too strong solution was often etching metal under the protection layer. Be aware of proper after treatment, to prevent future corrosion.
Usual gloves, coat and eventually shield. Well vented place is needed, at least for first experiments, until you know the behaviour ( I mean rather passively - like outdoor - and just temporary for initial experiments. when it is settled, it is not much needed. ).


Why is ammonium chloride rather than an alkylammonium chloride salt the product of nucleophilic substitution of a chloroalkane by ammonia?

You have actually answered the question right in the first sentence of the description.

You are right in saying that $\ce{RNH_3Cl}$ will be formed.

The alkyl halide, $\ce{R-X}$, is the alkylating agent. The reaction medium is basic in nature, so I'd expect the $\ce{S_N2}$ mechanism. So,

$\ce{H3N +RX\to H3NR+X-}$

We know that an excess of ammonia has been used. This simply means that the forward reaction is favoured:

$\ce{H3NR+X- +NH3\rightleftharpoons RNH2 + NH4+X-}$

It's all Le Chatelier's once again - makes extraction simpler, doesn't it?


Determining amounts of product and excess reagent remaining post reaction

Note the following relations from the equation you wrote:
1 mol $\ce{FeS2}$ = (11/4) mol $\ce{O2}$;
1 mol $\ce{FeS2}$ = (1/2) mol $\ce{Fe2O3}$;
1 mol $\ce{FeS2}$ = 2 mol $\ce{SO2}$.

You already determined that $\ce{FeS2}$ is the limiting reactant because we have 600/120 mol of iron sulfide and 800/32 mol of $\ce{O2}$;

This implies that 5 mol $\ce{FeS2}$ will require =5x (11/4) mol $\ce{O2}$.

Remaining mol oxygen = 25-13.75=11.25

Now use the relations established above: 5 mol iron sulfide would produce (5/2) or 2.5 mol iron oxide, and 5 mol iron sulfide would produce 5x2 = 10 mol sulfur dioxide.


It's possible that the gallium is not as pure as advertised. If it has (say) some indium as an impurity, the indium-gallium alloy would have a lower melting point, potentially below room temperature.

I thought that partial freezing might have crystallized out more pure gallium, and left less pure gallium as the melt, but I'm not sure that's how indium-gallium alloys work. I made a few grams of the alloy, and I never saw any such behavior in it -- then again, I never experimented much along those lines.


What (stored-as-solids) acids/liquids can clean corroded copper?

Per a source to generally clean copper metal surface, to quote:

A few hard rainstorms may clean the surfaces sufficiently to start operations. However, it is always advisable, particularly if the coloring is to be done immediately after installation, to go over the surface with a commercial chemical metal cleaner. Cleaners of the trisodium phosphate type should be satisfactory. Avoid cleaners which leave a coat of oxide on the copper surface.

However, assuming one has already a basic copper carbonate, the application of an acid, like HCl or more friendly acetic acid, is likely suitable, especially with the addition say NH4Cl or NaCl to increase the activity coefficient of the acid (so less strong acid may be employed).
However, a review of the cited source notes that these reagents are the very agents suggested to create a patina coating (which can be characterized as electrochemical based corrosion) occurring in the presence of air/oxygen.
So, do limit air/oxygen exposure by immersing the copper surface to be cleaned in say a HCl/NH4Cl mix.
Also, employing the concept of a sacrificial anode, experiment with performing the latter cleaning in the presence of a more anodic metal (say a small piece of Aluminum metal in acetic acid and NaCl), especially when the requirement of limiting air exposure could be difficult to implement. This may prove to be actually beneficial in avoiding surface
oxidation.


Number of equivalents and normality are out-of-date concepts in chemistry. For acids, they are just like number of mols and molar concentration (or molarity), with an extra factor indicating how many hydrogens can be effectively ionized in solution. Because acetic acid ($\ce{CH3COOH}$) is monoprotic, you can just exchange the terms by number of mols and molarity and you are good.

  Why is normality being decreased here?

Some of the acetic acid adsorbs - meaning it bonds - in the charcoal. So the concentration in solution is decreased. Pay attention to the fact that the charcoal remains solid and does not dissolve.

  Why are they taking volume to be same when it generally increases when we add solutes?

As I wrote above, the charcoal stays solid, so it has no effect on the solution volume. The loss of acetid acid in solution might have a slight effect on volume, but that is mostly negligible.


The Van't Hoff equilibrium box is a chamber containing a mixture of reactants and products at equilibrium.  It is connected to an array of cylinders, each containing one of the pure reactants or products.  The chamber itself is operated at constant total pressure and temperature.  The cylinders are each connected to the equilibrium box through a semipermeable membrane that only passes that species into the chamber.  Pure species are introduced in stoichiometric proportions to the chamber through these membranes at the same partial pressures as in the reaction mixture, and pure products are removed in stoichiometric proportions at the same partial pressures as in the reaction mixture.  So the operation of the equilibrium box is reversible, and the delta G in going from pure reactants in to pure products out is zero.


What are the effects of the repulsive forces between electron on the potential energy graph for the dihydrogen molecule?

First of all, you have to understand that the potential energy depicted on the graph in your question already accounts for repulsive forces between the electrons. This potential energy is the sum of two terms, namely, the sum of electronic energy $E_{\mathrm{e}}$ and nuclear repulsion energy $V_{\mathrm{nn}}$, where the first term contains contributions from kinetic energy of electrons motion $T_{\mathrm{e}}$, potential energy of their interactions with the nuclei $V_{\mathrm{en}}$, as well as potential energy of their interactions with each other $V_{\mathrm{ee}}$. This sum of electronic energy $E_{\mathrm{e}}$ and nuclear repulsion energy $V_{\mathrm{nn}}$ plays a role of potential energy for the motion of the nuclei, hence its name.

All these components of the potential energy, $E_{\mathrm{e}} = T_{\mathrm{e}} + V_{\mathrm{en}} + V_{\mathrm{ee}}$ and $V_{\mathrm{nn}}$, behave differently with respect to internuclear separation. For instance, as you have mentioned, the last term $V_{\mathrm{nn}}$ grows when nuclei get closer to each other. The same has to be true for $V_{\mathrm{ee}}$ term since the closer the nuclei are the closer are the electrons. However this effects of increasing the energy could be overcompensated by a decrease in $V_{\mathrm{en}}$ term thanks to attractive nature of the corresponding forces between nuclei and electrons.

When a decrease in $V_{\mathrm{en}}$ overcompensates an increase in $V_{\mathrm{nn}} + V_{\mathrm{ee}}$ the bond is formed. Not that an overall decrease in potential energy is accompanied by an increase in kinetic energy in accordance with the virial theorem. But the increase in kinetic energy is only half of the lowering in the potential energy, so that the total energy goes down when a bond is formed.

Here is the graph of all these terms for $\ce{H2}$ molecule obtained from quantum chemistry calculations1:

Wilson, C. W. & Goddard, W. A. The role of kinetic energy in chemical binding. Theor. Chim. Acta 26, 195–210 (1972). PDF from caltech.edu.


Reduce the problem to one dimension first. The overlap integral between two functions, $f = f(x)$ and $g = g(x)$, is defined by

$$S_{fg} = \int_{-\infty}^\infty [f(x)]^*g(x)\,\mathrm{d}x$$

where $^*$ denotes complex conjugation. $S$ is the common symbol for the overlap integral; the subscript $fg$ indicates that it is the overlap between $f$ and $g$.

So, for example, let's try a simple case: let

$$f(x) = g(x) = \left(\frac{m\omega}{\pi \hbar}\right)^{1/4} \exp\left(-\frac{m\omega x^2}{2\hbar}\right)$$

(the ground state of the quantum harmonic oscillator). This function is real, so the complex conjugation doesn't do anything. We have

$$\begin{align}
S_{fg} &= \int_{-\infty}^\infty \left(\frac{m\omega}{\pi \hbar}\right)^{1/2} \exp\left(-\frac{m\omega x^2}{\hbar}\right)\,\mathrm{d}x \\
&= \sqrt{\frac{m\omega}{\pi\hbar}} \int_{-\infty}^\infty  \exp\left(-\frac{m\omega x^2}{\hbar}\right)\,\mathrm{d}x \\
&= \sqrt{\frac{m\omega}{\pi\hbar}} \sqrt{\frac{\pi\hbar}{m\omega}} \\
&= 1
\end{align}$$

(I used a standard integral, $\int_{-\infty}^\infty \exp{(-ax^2)}\,\mathrm{d}x = \sqrt{\pi/a}$). So this tells you that the overlap of this function with itself is equal to 1. That's not very interesting, because if you substitute $g(x) = f(x)$ into the original definition, you'll find that that is simply the normalisation condition for a wavefunction. I just wanted to use it as an illustration.

In general, the overlap between two normalised states ranges from 0 (when they are orthogonal) to 1 (when they are identical). This applies to three dimensions as well.

Of course, the notion of overlap is usually talked about in a three-dimensional context. So, all you have to do is to generalise the definition to three dimensions. Let's say we have two functions $\psi_1(x,y,z)$ and $\psi_2(x,y,z)$. Then we simply have

$$S_{12} = \int\limits_{-\infty}^\infty \int\limits_{-\infty}^\infty \int\limits_{-\infty}^\infty \psi_1^*\psi_2 \,\mathrm{d}x\,\mathrm{d}y\,\mathrm{d}z$$

or in more concise notation

$$S_{12} = \int \psi_1^*\psi_2 \,\mathrm{d}\tau$$

where $\mathrm{d}\tau$ is the volume element (note that Wikipedia denotes it as $\mathrm{d}V$) and the limits of integration are assumed to be over all space.

This won't get you anywhere with atomic orbitals, though. Orbitals are frequently expressed in terms of spherical coordinates. So, you need to know what the volume element in spherical coordinates is, and that is:

$$\mathrm{d}\tau = r^2\sin\theta\,\mathrm{d}r\,\mathrm{d}\theta\,\mathrm{d}\phi$$

(note again that Wikipedia can be confusing sometimes, because the symbols used for spherical coordinates are different: $\theta$ and $\phi$ in chemistry have exactly the opposite meaning from $\theta$ and $\phi$ in mathematics.)

The limits of the integral are still "over all space", but in spherical coordinates this is expressed differently. $r$ takes values from zero to infinity, $\theta$ from $0$ to $\pi$, and $\phi$ from $0$ to $2\pi$. So:

$$S_{12} = \int\limits_0^{2\pi} \int\limits_0^\pi \int\limits_0^\infty \psi_1^*\psi_2 r^2 \sin\theta \,\mathrm{d}r\,\mathrm{d}\theta\,\mathrm{d}\phi$$

All you need to do now, is to find the form of the function, plug it into the integral, and evaluate the integral. Warning: That's best done by a computer.


Which azo compounds are found in DVD-R and CD-R media?

Each company in the DVD business has their own proprietary twist on the actual dye they use and they don't go out of their way to reveal the structure of the molecule they use.  Also the actual dye used changes fairly rapidly (several years or so) as technology shifts (e.g. higher density, Blu-ray, etc.).  That said, metal-azo dyes are popular today.  A Mitsubishi patent from a few years ago (WO 2006/104196 A1 (10/03/2006)) says "The recording layer contains a pyridone azo organic dye having a 6-hydroxy-2-pyridone structure as a coupler component and an isoxazole, 1,2,4-triazole or pyrazole as a diazo component, coordinated to a metal, such as nickel, cobalt, iron, zinc, copper or manganese." Here is the structure of 2-pyridone (6-hydroxy would be attached on the other side of the nitrogen) and pyrazole.

Here is a link to a Ciba patent that has more detail. And here is a general structure a 2-pyridone-based metal-azo dye


It means that the boundary surface is a closed surface or set of closed surfaces, such that there is a 90% chance that an electron is interior of the closed surface and a 10% chance that the electron is exterior of the closed surface.

A better way of specifying how an orbital is represented is "equal probability boundary surface" which further specifies that each point on the surface must have any equal electron probablity density, and the surface must be a closed surface or set of closed surfaces with a 90% (or other percent) chance of containing the electron at a given time.

Without the "equal probability" statement there are infinite different boundary surfaces for each orbital (infinite different ways to enclose 90% of the probability density).


If in a binary solution of two components A and B, the interaction between unlike components A-B is weaker than the interaction between the like components A-A and B-B, then it means that in such solutions molecules of A (or B) will find it easier to escape in the state A-B, than when they are in their pure states (A-A or B-B). It means that molecules of A (or B) will find it easier to escape from the A-B form.

This will result in having greater vapour pressure of each component of the solution than expected on the Basis of Raoult's law, and hence the total vapor pressure will also be higher than in the case of ideal solution. Thus the deviation will be positive.

In such solution $\Delta_{mix}H$ is positive because energy is required to break A-A and B-B bonds. For such solutions, the dissolution process is endothermic, i.e, the solubility will increase with temperature (Le Chatelier's principle) .

$\Delta_{mix}V$ is also positive because there is decrease in the magnitude of intermolecular forces in the solution, the molecules are loosely held, and therefore the volume of mixing increases.


Hydrogen donor without hydrogen acceptor?

As mentioned in the comments by Philipp, ammonium ions $\ce{R3NH+}$ are a good direct answer to what you are looking for.

The nitrogen in an amide $\ce{RCONHR'}$ is similar - the N lone pair is conjugated into the C=O bond and that makes it a very poor hydrogen bond acceptor, but the N-H lone pair can be a hydrogen bond donor. For more details, see this previous question with an excellent writeup by Jan. Now an amide can of course function as a hydrogen bond acceptor via the lone pairs on oxygen, so it's not strictly what you're looking for, but it was just something related. And it is of course directly relevant to the secondary structure of proteins.


Does fluorescein decompose from long term exposure to light?

Yes, Fluorescein decomposes due to exposure to light.

See The Conservative Behavior of Fluorescein for quantitative data.


How do Electrons Cross Nodes in Orbitals?

Electrons (indeed all matter and energy) behave both as wave and as particle, depending how the observation is made. The most obvious example is the single-electron interference pattern.

"In 1974 the Italian physicists Pier Giorgio Merli, Gian Franco Missiroli, and Giulio Pozzi repeated the [interference] experiment using single electrons, showing that each electron interferes with itself as predicted by quantum theory. In 2002, the single-electron version of the experiment was voted "the most beautiful experiment" by readers of Physics World." Wikipedia.


Need an explanation for how they got this compound name out of this structure

The names that are proposed in the question are in accordance with obsolete IUPAC recommendations:

2-ethyl-5,5-dimethyl-1-hexene (1979 recommendations)
2-ethyl-5,5-dimethylhex-1-ene (1993 recommendations)

The corresponding rule in the 1979 recommendations is Rule A-3.4:

  3.4 – Unsaturated branched acyclic hydrocarbons are named as derivatives of the unbranched hydrocarbons which contain the maximum number of double and triple bonds. If there are two or more chains competing for selection as the chain with the maximum number of unsaturated bonds, then the choice goes to (1) that one with the greatest number of carbon atoms; (2) the number of carbon atoms being equal, that one containing the maximum number of double bonds. In other respects, the same principles apply as for naming saturated branched acyclic hydrocarbons. The chain is so numbered as to give the lowest possible numbers to double and triple bonds in accordance with Rule A-3.3.

Therefore, the parent structure is the unbranched chain that contains the maximum number of double and triple bonds (here: one double bond). Since there is still a choice, the parent structure is the longest chain that contains the double bond, i.e. hexene. Numbering starts at the end of the chain that gives the lowest numbers to the double bond, i.e. 1-hexene. The 1-hexene is substituted using the usual principles of substitutive nomenclature, which yields 2-ethyl-5,5-dimethyl-1-hexene.

According to the 1993 recommendations, locants are placed immediately before the part of the name to which they relate. Therefore, the name 2-ethyl-5,5-dimethyl-1-hexene is changed to 2-ethyl-5,5-dimethylhex-1-ene.

The current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) incorporates a major change from earlier recommendations given in 1979 and 1993.

  In acyclic parent structures the order of seniority between unsaturation and length of chain given in earlier recommendations is reversed. Thus, the first criterion to be considered in choosing a preferred parent acyclic chain is the length of the chain; unsaturation is now the second.

Therefore, the parent structure of the compound that is given in the question is now the longest chain, i.e. heptane. The heptane is substituted using the usual principles of substitutive nomenclature; i.e. the prefixes (here: ‘methyl’ and ‘methylidene’) are considered together in a series of increasing numerical order. This yields the preferred IUPAC name 2,2-dimethyl-5-methylideneheptane rather than 6,6-dimethyl-3-methyleneheptane since the locant set ‘2,2,5’ is lower than ‘3,6,6’.


Ring closure via intramolecular nucleophilic substitution or intermolecular dimerization: which is favoured?

I'd also like to add that the intramolecular cyclisation that you proposed is a 5-exo-tet reaction according to Baldwin's classification.

The ring being formed is 5-membered.
The bond being broken (the $\ce{C-Br}$ bond) is outside the ring, or exocyclic.
The carbon being attacked is a $\mathrm{sp^3}$-hybridised carbon with tetrahedal geometry.

If you look up Baldwin's rules, you'll see that 5-exo-tet reactions are perfectly fine and allowed. Evidently the compound you drew can twist itself a little bit to accommodate the linear transition state. (I am using "linear" loosely to refer to the arrangement of the incoming nucleophile, the carbon, and the leaving group.) Therefore, in this case, the intramolecular reaction will predominate, as is usual.

The disfavoured reactions that you cited from those textbooks are 6-endo-tet reactions. In such compounds, having a linear transition state is an issue as it would lead to too much ring strain.

As an ending note: Baldwin's rules are nothing more than an empirical observation. You might ask why 5-exo-tet reactions are fine, but not 6-endo-tet: where is the threshold or boundary which makes something favoured or disfavoured? The answer is, they did the experiments and found out that that's just how it is. No more, no less.


While it's true that there are polyiodides, the $\ce{I4^2-}$ ion has a completely different geometry from $\ce{IF3}$ molecule. The ion is most easily regarded as two iodine molecules stuck end on and the ensemble is bound because two additional electrons are popped into the in-phase combination of each side's $\ce{I(p_z)-I(p_z)}$ antibonding orbital. This is bonding between the fragments and antibonding within the fragments. Since the individual molecules can't do anything with the extra two electrons, it's best just to stay together and stabilize them together.

As to why there isn't a $\ce{I4}$ that is isostructural to $\ce{IF3}$, this is because the stability of interhalogens is heavily influenced by the ionic character of bonds. Recent work[1,2] has shown that the bonding in halogens is dominated by "charge-shift" interactions in which the ground covalent state is stabilized by ionic states in which both electrons of the bond have been transfered to either atom (e.g. $\ce{F-F}$ is stabilized by $\ce{F+-F-}$ and $\ce{F^--F+}$ states). This stabilization is due to highly electronegative atoms. The stability of higher interhalogens relies on this "charge-shift" stabilization. Since I is a lot less electronegative than F, this charge-shift stabilization is much weaker (and this is also the reason why nearly all of the higher interhalogens have fluorine as a bonding partner). Since this charge-shift stabilization is absent, the 4 iodines are much happier to break apart into 2 $\ce{I2}$ molecules.

Sason Shaik, David Danovich, Wei Wu, and Philippe C. Hiberty, Nature Chemistry 2009, 1, 443 - 449. DOI: 10.1038/nchem.327
Daniel S. Levine, Paul R. Horn, Yuezhi Mao, and Martin Head-Gordon, J. Chem. Theory Comput. 2016, 12 (10), 4812–4820. DOI: 10.1021/acs.jctc.6b00571


Finding the Freundlich adsorption isotherm from the graph of log(x/m) v/s log(p)

One version of Freundlich adsorption isotherm equation is:
$$ \frac xm = Kp^{\frac12}, $$
which can also be written as:
$$ \log \frac xm = \log K + \frac12\log p $$
This is a straight line equation of type $( y = c + mx)$ as given in the question. However, I think the question has made a mistake saying the slope is $2$, but instead it should be $\frac12$. Accordingly, the equation with given numeric values should be:
$$ \log \frac xm = \log 3 + \frac12\log p $$
Or without logarithm:
$$ \frac xm = 3p^\frac12 $$
On plugging $p = 4$, You'll get:
$$ \frac xm = 3 \times 4^\frac12 = 3 \times 2 = 6 $$
Hence, you get the given answer.


Why would electrode reaction in galvanic cell stop when there's no neutrality in each cell?

It would work vanishingly briefly. The separation of charge that occurs when you connect two separated half cells without a salt bridge produces an electrostatic potential difference that opposes the electrochemical potential difference. The cell reaction would proceed until the two voltages balance, and it doesn't take much charge separation to hit the 1–2 volt electrochemical potential of typical cells, so this process is very fast.


Calculate the volume of ammonium hydroxide needed to get desired pH

Technically, you calculated the concentration correctly (math-wise). However, your initial hypothesis is wrong.

First, you cannot get ammonia hydroxide solution of $\mathrm{pH}$ $12.8$ at the room temperature to work with. At $\pu{20^\circ C}$ concentrated solution would be approx. $35\%$ with $\mathrm{pH} \approx 12.0$. The higher the temperature, the lower the solubility and the $\mathrm{pH}$ will be. Having $29\%$ stock ammonia solution, there is no way to reach $\mathrm{pH}$ $12.8$.

Second, when you work with borderline/concentrated solutions, you are no longer allowed to use concentrations when describing the equilibrium, you need to use activities instead. The formula you derived would give a reasonable result for the moderately concentrated solutions only. You can adapt it to establish the relationship between the mass fraction $\omega$ and $\mathrm{pH}$, knowing the density of the solution $\rho$ ($M$ is molar mass of ammonia)

$$\mathrm{pH} = 14 + 0.5\lg{\frac{\omega\rho K_\mathrm{b}}{M}}\, ,$$

but that's pretty much all to it.


Syntax and Typography of Atomic Orbitals

Question 1. The Green Book p. 32 defines the letters s, p, d, f, g, h, i, k ... (omitting j). They have to be typeset upright.

  How far does this "alphabet" reasonably go? (Wikipedia goes up to i).

Section 2.6.3 of that version of the Green Book shows the term symbols (Wikipedia) which are capital letters S,P,D,F,G,H,I,K,... and used to indicate the electronic configuration of the valence electrons and for excited states.
The lower case letters are used for the azimuthal quantum number (ℓ) which is one of the four quantum numbers for an electron in an atom. You can typeset $99.9~\%$ of chemistry of even more using only the letters s, p, d and f, corresponding to the left-hand main group elements, the right-hand main group elements, the transition metals and the lanthanides/actinides, respectively. Everything above f is and will be for quite some time not relevant to chemistry outside a very small circle that would know how to deal with stuff.

  Question 2. On the same Greek Book page, electron configurations are explained, e.g. $\mathrm{(1s)^2 (2s)^2 (2p)^1}$.

  Many other sources write it without parenthesis: $\mathrm{[Kr]\mskip2mu4d^{10}\mskip2mu5s^1\mskip2mu5p^3\mskip2mu5d^4}$

  Question a: What values can the left-side number take? (At Wikipedia, I see 1 to 7.) Can it be 2-digit?

  Question b: What values can the right-side exponent take? (At Wikipedia, I see 1 to 10.)

This has two sub-questions.

By a similar reasoning as presented in the answer to 1, I think 8 will be the limit until the not-so-close future. But there is nothing intrinsically preventing it from becoming two-digit at some distant point.
The superscript can take values of up to $4l+2$, where $l$ can be considered ‘computer science counting’ of the letter sequence s, p, d, …; so for s $l = 0$ and thus $4l+2=2$; likewise for p ($l=2$) the value is maximum $4l+2=6$ etc. Note that $4l+2$ is actually $2(2l+1)$: each subshell gives up to $2l+1$ orientation possibilities that each can be populated with two electrons.

  Question 3. I think I did not come across the configuration with one-electron orbitals (in Greek). $\mathrm{(1\sigma)^2 (2\sigma)^2 (3\sigma)^2 (1\pi)^1}$. (The Greek letters should be upright, but that's a drawback of MathJax.)

  Would this be useful? Or not needed?

Yes, this is useful, especially for discussing molecular orbitals. The Greek letters you want to be aware of are σ, π and δ. Maybe a fourth will be added soon corresponding to f orbitals (υ?) but I don’t see that coming anytime soon.

  Question 4. The examples introduce another notation, "electronic states"  $\mathrm{\cdot\cdot\cdot (2a_1)^2 (1b_2)^2 (3a_1)^2}$

  a: What values could the letter take? b: The left-side number? c: The subscripts? d: The right-side exponent?

I’m going to answer that as one.

This notation uses irreducible representations to define orbitals. The bit between the initial number and the superscript can be any of the symbols that appear in Little Bobby Tables (courtesy of Orthocresol). They each represent a type of orbital with a certain symmetry.

Orbitals of the same irreducible representation (i.e. the same centre) are numbered from lowest to highest energy — so in principle the sky is the limit. Practically, however, you will be hard-pressed finding three-digit examples.

The exponent can at maximum be $2n$, where $n$ is the number in the first column of the respective character table (the E column). This number corresponds to a certain symmetry’s degeneracy (i.e. so many orbitals of the same energy) and each orbital can be populated by two electrons.

  Question 5. From Wikipedia, I learned that there are notations IUPAC does not mention: $\mathrm{p_x}$, $\mathrm{p_y}$ and $\mathrm{p_z}$. (All subscripts should be upright, because they are labels, no variables.)

  But it does not end there. Here and here I saw:
  $\mathrm{d_{xy}, d_{xz}, d_{yx}, d_{x^2{-}y^2}, d_{z^2}}$
  This was confirmed by this book. There are alternative styles
  here ($\mathrm{{p_x}^1}$).
  and
  here ($\mathrm{p_{x1}}$).

  a: I guess the $\mathrm{d_{z^2}}$ notation (with index z-square) is correct?

  Oh, I just found this book that also includes $\mathrm{4f_{x(x^2{-}3y^2)}}$ and $\mathrm{4f_{y(3x^2{-}y^2)}}$.

  b: Are there other indexes possible than x, y, z?

  c: Can it be just squares? Or are other index-superscripts possible?

  d: What about the 3 in the last examples? Could this also be any other numer?

  e: Could there be any other forms of indexes?

  f: Typographically, what does this dash stand for? Is it a hyphen or a "transition between a higher energy state and a lower energy state"?

(Somebody please add and correct this.)

It was my assumption that the axis directions should be italicised but I don’t know. To discuss this, please go to the break-out discussion for this question.

p orbitals will always have $x$, $y$ or $z$. d orbitals will have one of the five you noted ($\mathrm{d}_{z^2}$ is correct — as I said, I am not sure on the upright versus italic question but I would have assumed italics). The f orbitals themselves can be labelled by various different polynomials in $x$, $y$, $z$, and $r$. See these two web pages: 1, 2. In general the subscript is just a polynomial.

Which symbols and exponents can appear here is fixed already by the results of the hydrogen-like orbitals — again, for most of chemistry d orbitals are sufficient and you can cover $99.9~\%$ by including f orbitals.

The ‘dash’ is a mathematical minus-sign operator. Yes, it truly is $x^2$ minus $y^2$.

  Question 6. Finally, there is orbital hybridization, which will lead to notations like $\mathrm{sp, 2sp, sp^2, 2sp^3, s^{0.5}p^3, s{-}s, sp^3{-}sp^3}$.

  a: Can the superscript be any other value than 2 and 3? Can the "base" be any other thing than s, p and sp?

  b: Is $\mathrm{2sp^3}$, is it $\mathrm{2s}$ and $\mathrm{p^3}$ or is sp one ... thing with a 2 before and a 3 in superscript.

  c: Is it a hyphen or a "transition between a higher energy state and a lower energy state"?

Sub questions.

Yes. You have $0.5$ as a superscript in your own examples. But, as posts by Martin and Ron somewhere point out, in $\mathrm{sp}^x$ the superscript $x$ can, in principle, be any number.
$\mathrm{2\,sp^3}$ is $\mathrm{\{2\}\{sp^3\}}$ — and there could potentially be another superscript making it $\mathrm{(2sp^3)^2}$.
I don’t know what that hyphen is supposed to be.


What is the first organic compound to be discovered ? Urea or alloxan?

The problem herein lies in the different possible definitions of organic chemistry.

The IUPAC defines an organic compound as a structure containing at least one carbon atom that can be named using the principles of organic nomenclature.

A less concise definition is

All compounds containing carbon and at least one other element that are not carbon monoxide, carbon dioxide or the latter’s inorganic salts.

Another one:

All compounds that contain at least one $\ce{C-H}$ bond.

Originally, the definition meant compounds isolated from organic materials i.e. organisms, living beings.

You see that the exact definition of organic compounds is difficult. But it was once very clear, historically: Anything not isolated from organisms or a derivative thereof was inorganic.
Note that isolation of organic compounds can be traced extremely far back in history: The ancient Egyptians isolated indigo as a colouring agent. However, they didn’t synthesise it and most attempts pre-19th century failed. Nobody questioned the possibility of isolating pure organic compounds from living matter, it was merely questioned whether they could be synthesised without involvement of organisms.
Alloxan was synthesised by oxidising uric acid with nitric acid as Wikipedia states. Thus, it was synthesised from an organic compound and would have been considered to contain enough vis vitalis even for further synthetic steps.
The compound your link states, 5-amino-3H-1,2,4-dithiazole-5-thione, is not described in any prime source as having been isolated from organisms. In fact, it is unknown to Wikipedia, so with all due skepticism, one can assume it to be of little to no relevance in living beings.
Compounds we today would call organic have been synthesised even earlier. Ethylene was first discovered in 1669 and it was successfully chlorinated to give 1,2-dichloroethane in 1794 — way before 1818, 1821 or 1828. Due to its name Dutch oil it is possible that it was considered an inorganic chemical. If it was synthesised from ethene that was, in turn, synthesised from ethanol, that would have just meant it carried over the vis vitalis from the original ethanol.
Wöhler’s discovery was that significant, because he used compounds that were (and are) unanimously declared inorganic (silver cyanate and ammonium chloride) and arrived directly at a compound that had previously been isolated from living material namely urea.


Which coordination complex does not absorb visible light?

Absorption of visual light is associated with an energy difference between two orbitals — one occupied, one unoccupied — and electrons must be able to be excited from one to the other.

In coordination complexes, these excitations typically happen within the metal’s d subshell, so it is usually sufficient to examine that and approximately determine which excitations are possible. The main selection rules are:

the spin rule. The electron must be excitable without a spin flip
the Laporte rule. Basically, d to d transitions are forbidden in octahedral complexes

The spin rule is very strongly observed. The colour of manganese(II) whose transition is spin-forbidden is extremely faint. The Laporte rule only holds true as long as the complex is inversion-symmetric, so any asymmetric vibration is enough to make it void; thus, Laporte-forbidden transitions are typically still visible but somewhat faint.

Let’s examine the complexes:

$\ce{[Ni(CN)4]^2-}$ This is expected to be square planar and $\mathrm{d^8}$. The energy difference between the two highest orbitals — $\mathrm d_{xy}$ and $\mathrm d_{x^2 - y^2}$ — is expected to be high. The former is expected to be fully populated, the latter to be unpopulated.
$\ce{[Cr(NH3)6]^3+}$ This is a $\mathrm{d^3}$ system. It is expected to be octahedral with a standard difference between the lower and higher energy levels.
$\ce{[Fe(H2O)6]^2+}$ This is a $\mathrm d^6$ octahedral system. There is no reason to assume a low spin state. The energy difference is expected to be slightly less than in the previous case.
$\ce{[Ni(H2O)6]^2+}$ this is expected to be a high-spin $\mathrm d^8$ system and octahedral. The same expectation regarding energy levels as on the previous case applies.

We realise that the of our complexes are average high spin octahedral complexes. For these, visible light absorption is always expected. Only one case is different. In that different case, the HOMO-LUMO difference is large. While we can still expect absorption, it seems most reasonable to assign this absorption band an ultraviolet wavelength.

Thus, $\ce{[Ni(CN)4]^2-}$ is the answer.


How to extract water from a solution in formamide

I'd like to determine the water content in a sample without buying an expensive machine (like [a] Karl Fischer titrator).

Rather than try to extract the water and measure it volumetrically, I think you could try to measure its activity, or else perform your own Karl-Fischer titration, which does not necessarily require any equipment.

Water activity measurement

Practically this means hygrometers.  That may sounds like another expensive machine, but E-bay tells me that capacitive hygrometers are available for under $20.  If your sample consists of sucrose, glucose, and glycerol, those substances are unlikely to interfere with capacitive hygrometry measurements of water activity.

You might even get very crude accuracy by incubating a mirror at a known (chilled) temperature, and then waving it above your water/sucrose/glucose/glycerol sample, seeing if dew forms, and then repeating the process at several different temperatures.

Karl Fischer titration on your own

As an alternative, couldn't you just set up your own Karl Fischer titration?  Commercial titrators often use coulometric detection for better accuracy, but setting up a volumetric titration is pretty easy.  You need:

Methanol (or some other alcohol)
Imidazole (or some other amine base)
Sulfur dioxide
Iodine

Imidazole and sulfur dioxide should be combined in a 1:1 molar ratio in an excess of methanol.  Iodine should be dissolved in methanol.

A 2012 paper has a nice description of how to do KF with visual end-point determination.  You just need to buy a KF titration solution, or make your own:

  Commercially available [...] Test Kit from [...] was used for reference moisture determination. All the equipments needed were provided with the kit. It consisted of two-component ethanol-based reagents: [a] working medium containing [a mixture of] imidazole, sulphur dioxide and diethanolamine, and [separately a titrant solution] containing iodine. According to manufacturer's instructions for use, 20 mL of [working medium] was added to the titration vessel, which was then tightly sealed with a septum. The titrant was delivered through the septum with a syringe to react with the present water, until a color change from colorless to yellow occurred. A known amount of sample was added by a syringe to the titration vessel via septum, and a newly filled titration syringe was used for a new titration until the color change from colorless to yellow occurred. Consumption volume of the titrant was read off the titration syringe. The water content, in percent by volume, was calculated from the consumption of the titrant and the sample volume.

If you've diluted with formamide

  I've dissolved a sample (approx. 20 g; composed of sucrose, glycerol and glucose syrup, water) in 50 ml formamide and like now to separate the water fraction again.

The nice part about both water activity measurement and Karl Fischer titration is that they are both sensitive and accurate enough to still work on your sample even after dilution or dissolution in formamide.


First, why other options are not really the options:

A: vinegar, being a weak acid, doesn't neutralize sulfuric acid and only dilutes it;

B: solid sodium hydroxide, a strong base, does neutralize sulfuric acid, but it does so vigorously releasing substantial amount of heat per unit of time:
$$\ce{2 NaOH(s) + H2SO4(aq) -> Na2SO4(aq) + 2 H2O(l)}$$
Using solid NaOH it is also tricky to guarantee it will cover the spill and there won't be any unreacted acid or excessive hydroxide (which is equally unwanted) left behind as there is little to no visual clue whether the neutralization is complete, unless you test various spots with, let's say, pH paper.

D: Magnesium sulfate solution doesn't react with sulfuric acid and only dilutes it.

Second, sodium bicarbonate solution not only neutralizes the acid

$$\ce{2 NaHCO3(aq) + H2SO4(aq) -> Na2SO4(aq) + 2 H2O(l) + 2 CO2(g)},$$

but also can be (and should be!) used in excess to assure complete acid neutralization.
Once the acid is neutralized, you are left with solution of sodium bicarbonate and sodium sulfate which is only slightly basic due to cation hydrolysis.
Another important advantage of using $\ce{NaHCO3}$ is visual control: once the process is complete, gas evolution stops.


How do you connect a battery to a galvanic cell in order to reverse the flow of electrons?

The electrochemical signs and labels have confused several generations of students. You have to remember a couple of things about the labels for the electrolytic and galvanic cells.
Anode: The electrode at which oxidation takes place.
Cathode: The electrode at which reduction takes places.
The signs of the electrodes of a galvanic cell are electrostatic signs (positive and negative) of the electrodes. This is an important point to keep in mind, which is often not emphasized during teaching. So (-) sign on the battery terminal indicates excess of electrons and (+) sign indicates deficiency in electrons.
In a galvanic cell, the cathode is positively charged and the anode is negatively charged. Work this out.
In order to reverse the flow of electronic current in the 2V galvanic cell, you need to connect the electron rich electrode of the 9V battery to the electron rich electrode of the galvanic cell i.e., you will connect the negative terminal of the 9 V battery with the negative terminal of the 2V galvanic cell and connect the positive terminal of the battery with the positive end of the galvanic cell.
Kaplan's wordings is quite bad!
Consider the analogy of water supply in your own house: When you open a tap, water spontaneously flows out because the actual source of water is at a higher gravitational potential than your tap. What do we need to do to reverse the direction of water flow in the tap. We need to connect the tap to a powerful water pump which overcomes the gravitational potential energy of the water source and pumps the water against the gravitational potential. This battery connection is exactly the same because the battery generates 9V and the galvanic cell produces 2V.


Here is my opinion, which may be subject to errors and corrections are very welcome:

Using the definition of Brønsted, acidity is the tendency to give off protons. This, unlike the definition of Arrhenius, is not limited to aqueous solutions. However, if you do have an aqueous solution of an acid, something interesting happens: any acid HAc (or base B) stronger than H$_3$O$^+$ (or OH$^-$) completely dissociates via:

$$\ce{H2O + HAc <=> H3O+ + Ac-}$$ or
$$\ce{H2O + B <=> OH- + BH+}$$

So the maximum acidic compound in water is the hydronium ion H$_3$O$^+$; you couldn't, for example, differentiate aqueous hydrochloric and perchloric acid solutions by pH. To rank very strong acids by their acid strength, non-aqueous solutions are used to determine the dissociation constants and the data obtained is then transferred approximately onto water as a solvent. Also note that we are talking about solutions of acids.

Specifically for HCl, we can denote the pairs of conjugated acids and bases:

$$\ce{H2O ~(~base~1) + HCl ~(~acid~2) <=> H3O+ ~(~acid~1) + Cl- ~(~base~2)}$$

So we can compare the strengths of the two acids: $\ce{HCl}$ and $\ce{H3O+}$. We know that this equilibrium is shifted to the right, so hydrochloric acid is stronger.

Very late edit to try and answer comment by MukulGoyal (unfortunately, he didn't elaborate on his questions), by number:

Is the definition of acidity the tendency to give H+ ions, or
H3O+ ions? That seems to matter here because in first case H3O+ will
not even be acidic in water as the equilibrium H2O + H+ ---» H3O+
has a very high equilibrium constant.
Similarly even if the solvent is NH3, the equilibrium constant of
NH3 + H3O+ ---» NH4+ + H2O would be very low due to the same reason,
and hence it would actually be a very weak acid.
Does your answer imply that any acid which has a Ka greater than
1 will be a stronger acid than H3O+?

First of all, please remember that equilibria are dynamic.

1) As I've mentioned, acidity, according to Brønsted, is the tendency to give off $\ce{H+}$. In $\ce{H2O}$, $\ce{H+}$ are solvated, on average, by 6 water molecules, giving rise to species like $\ce{H3O+(H2O)6}$ (for more information, check out Wikipedia and this).

1) and 2) you seem to be talking about the dissociation of $\ce{H3O+}$. However, you're using terms like "very high" and "very low", so that, in my opinion, becomes kind of a misleading discussion.

When you mention

$$\ce{H2O + H+ <=> H3O+}$$

you need to remember that the proton has to come from somewhere. In fact, what you are talking about there is

$$\ce{H2O + H2O <=> H3O+ + OH-}$$

which is the autoprotonation of water. Let's just calculate that equilibrium constant you talk about all the time:

$$K = \cfrac{c(\ce{H3O+})~ \cdot~ c(\ce{OH-}) }{c^2~(\ce{H2O})}$$

Well, guess what? The concentration of water (c = n/V = m / (V * M) = $\rho$ / M = 55,51 mol/L) remains about constant for pure water and diluted solutions (because not many water molecules form ions), so it is brought into the constant, yielding

$$K_w = c(\ce{H3O+})~ \cdot~ c(\ce{OH-})$$

$K_w$ depends on temperature and equals $10^{-14} $ mol$^2$ / L$^{2}$ at 25 °C. It remains constant in pure water and also in diluted solutions.

In pure water where $c(\ce{H3O+}) = c(\ce{OH-})$, we then get

$$K_w = c^2(\ce{H3O+})$$

and

$$c(\ce{H3O+}) = 1.0 * 10^{-7} ~mol~ / ~L~ $$

which essentialy quantifies what you, as I suspect, were talking about, in that it tells you the concentration of $\ce{H3O+}$ in the autoprotonation equilibrium for pure water. By the way, because pH is defined like so:

$$pH = -lg ~c~(\ce{H3O+})$$

at pH = 7 (neutral), $\ce{H3O+}$ and $\ce{OH-}$ are equal. In diluted solutions, when they are not equal, you can calculate one using the other.

Also, remember that in the Brønsted–Lowry theory acids and bases are defined by the way they react with each other, as mentioned in Wikipedia. So far, we have only looked at water and its autoprotonation, no solvents and other molecules, which have their own acidies and basicities.

2) As for the reaction you mentioned in ammonia as a solvent, well, first of all, $\ce{H3O+}$ needs to come from somewhere (let's imagine we drop some aqueous solution of hydrochloric acid in there). Second, I am not sure that constant would actually be as you describe it - I think that here, ammonia (which behaves ampholytic) would act as a base and pretty happily accept a proton. If anyone can provide insight into this, that would be great.

3) Acid strength is defined using $pK_a$, which is given by $pK_a$ = -lg $K_a$. From my experience, if $pK_a$ is defined via concentration, then $pK_a$ of hydronium ions is 0, and if it is defined via activity, it is -1.7 (-1.74, to be more exact; similarly, for $\ce{OH-}$, the analogous $pK_b$ is 14 via concentration and 15.7 via activity).

Coming back to your question about "Ka greater than 1", that seems about right, because -lg (1) = 0.

I refer you to an acid strength table, for instance, here. Also, I'd like to refer you to this excellent discussion on the matter, where, by the way, orthocresol lent a hand.

Also, just to add this, strong acids have $pK_a$ < 0, while weak acids have $pK_a$ > 0 (again, this is via concentration - for example, here in Wikipedia they use -1.74).

Everyone, corrections are very welcome!


What does it mean to calculate hardness in terms of ppm of calcium carbonate?

It is meant as molar equivalent. $\ce{CaCO3}$ has the molar mass about  $\pu{100 g / mol}$, $\ce{Ca(HCO3)2}$ has the molar mass cca $\pu{ 162 g / mol}$. Therefore $\pu{100 g}$ of the former is equivalent to $\pu{ 162 g }$ of the latter. The similar for the concentration expressed in ppm. E.g. if water contains $\pu{1.62 ppm}$ of $\ce{Ca(HCO3)2}$, it  is equivalent to $\pu{1 ppm}$ of $\ce{CaCO3}$


Do the groups involved in geometrical isomerism have to be in similar planes to show geometrical isomerism?

To have geometrical isomers you need a structurally rigid bond. This bond can be double and this is a case where groups are in the same plane:

Or can be a cyclic compound where the  carbon-carbon single bond is restricted(2 examples below):

1st example:

2nd example:

Your textbook definition is incomplete and I wouldn't use the word "must" there.


Which species can't appear in the rate expression for this reaction scheme?

Answering this question requires assuming that all steps are "elementary", e.g. that $\ce{A + B -> C}$ is an irreversible bimolecular reaction that is first-order in $[\ce{A}]$ and first-order in $[\ce{B}]$.  (Obviously we are also assuming that activities can approximated by concentrations.)

With that assumption, the fact that $\ce{C}$ conversion is "fast" means that we can replace the existence of $\ce{C}$ in the scheme entirely:

\begin{align}
\ce{A + B~(slow) &\rightarrow D + B} \\
\ce{D + E~(fast) &\rightarrow F}\\
\ce{F~(fast) &\rightarrow G}\\
\end{align}

Likewise if $\ce{F}$ conversion is "fast", then it can be replaced:

\begin{align}
\ce{A + B~(slow) &\rightarrow D + B} \\
\ce{D + E~(fast) &\rightarrow G}\\
\end{align}

Hopefully this makes it clear why $\ce{E}$ must appear in any rate law for formation of $\ce{G}$: there is no source of it in the two reactions we are left with.

For $\ce{D}$, I agree that the situation is a bit more complex.  It might appear in the rate equation, or it might not.  Necessary reactants at the beginning of the reaction if we are ever to see product $\ce{G}$ are $\ce{A}$, $\ce{B}$, and $\ce{E}$, so it would make sense to eliminate $\ce{D}$ from the rate law if possible.  However, that could be algebraically difficult, or perhaps $\ce{D}$ is easier to measure, etc.  There is no reason $\ce{D}$ can't appear in the rate law.

Huge caveat: "fast" and "slow" are relative terms and as such without further context they are ambiguous.  The problem as constructed sounds a bit sloppy.  For example, is the "fastness" of $\ce{F -> G}$ faster or less fast than the fastness of $\ce{D + E -> F}$?  The problem doesn't say but the simplifications I did above actually depend on these sorts of details.


Effect of an inert gas on steam vapor pressure at equilibrium

If the gas phase can be treated as an ideal gas mixture and the Poynting correction for the liquid can be neglected, then the partial pressure of component 1 in the gas phase will be the equilibrium vapor pressure of component 1 at the temperature of the system.  The partial pressure of the non-condensible component 2 will be determined by the ideal gas law, based on the number of moles of this component, the temperature of the system, and the volume of head space.

If the gas phase cannot be treated as an ideal gas mixture but can still be regarded as an ideal solution, then the Lewis Randall law can be used to determine the fugacity of component 1 in the vapor, and this can be matched to that of component 1 for the liquid, with the Poynting correction included.

Does this come anything close to answering what you are asking?


Acid chlorides are normally formed from carboxylic acids by the action of reagents such as thionyl chloride (SOCl2). This is not a SN2 reaction well not at carbon.

Normally a carboxylic oxygen attacks the sulfur of the thionyl chloride displacing an chloride from the thionyl chloride. After loss of a hydrogen ion from the carboxylic acid we will have

R-C(=O)-O-S(=O)-Cl

A chloride anion then attacks the carbonyl carbon as a nucelophile, the sulfur group acts as a leaving group. It will form sulfur dioxide and another chloride anion. In terms of the reaction at the organic part of the molecule it should be classified as a nucelophilic attack on a carbonyl compound which results in the substitution of one group for another. In terms of mechanism it is very similar to the formation of an ester from an alcohol and a carboxylic acid.

Here is the mechanism


Is cyclohexene an alicyclic or aromatic compound?

Setting aside the severe steric  strain that cyclohexyne would require, an aromatic all-carbon ring must have a full cycle of unsaturated atoms.  Cyclohexene and (if you can make it) cyclohexyne have saturated carbons in the ring and thus must be alicyclic instead of aromatic.


Why do atomic radii decrease across a period?

Atomic radius decreases across a period because  electrons are being added to the same energy level and at the same time the nucleus is increasing in protons. The increase in nuclear charge $$Z_{eff}$$ attracts the electrons more strongly, (addition of one proton has a greater effect than addition of 1 electron) thus pulling  electron closer to the nucleus and making atomic radius decrease.


Agreed: unless air is pushed through the activated charcoal bed, few impurities would be adsorbed.
It would depend on how tightly the substance clings to the carbon, i.e. it would vary with impurity, but Trox Technik KS Filters recommends, "saturated activated carbon is reprocessed... with a gradual increase of temperature from 20 to 830 °C and a delay period of approximately 30 minutes."
Finally, how long does the air-stream remain in contact with the filter? It would seem that most air bypasses the pores of the carbon on inexpensive commercial air filters. Activated charcoal might be of more use in reducing foul odors in a closed container, such as bottles of medicines.


photometric titration happens only in one condition, that one of the following substances will absorb the electromagnetic radiation.

Almost ;-)

Regardless whether you monitor a complexometric titration by eye or with the help of a spectrometer, it is a good idea to use an indicator, such as Eriochrome Black T. Then, it doesn't matter, if or how  strong the EDTA complex of the metal absorbs!

Instead, you monitor the absorption of the indicator, that is liberated from the metal-indicator complex upon addition of EDTA.


Identifying a chloride of the unknown element from the masses of chlorine and compound as well as vapor density

I agree with the starting point of the Matthew's solution, namely the definition of vapor density, but I fail to understand the follow-up math, so I just post my approach.

$$M(\ce{ECl_x}) = D\cdot M(\ce{H2}) = 85\cdot\pu{2 g mol-1} = \pu{170 g mol-1}\tag{1}$$

By definition, molar mass $M(\ce{ECl_x})$ is also

$$M(\ce{ECl_x}) = M(\ce{E}) + x\cdot M(\ce{Cl})$$

$$\implies M(\ce{E}) = M(\ce{ECl_x}) - x\cdot M(\ce{Cl}) = (170 - x\cdot 35.5)~\pu{g mol-1}\label{eqn:2}\tag{2}$$

The unknown variable $x$ is defined by the molar ratio between the elements:

$$x = \frac{n(\ce{Cl})}{n(\ce{E})} = \frac{m(\ce{Cl})\cdot M(\ce{E})}{m(\ce{E})\cdot M(\ce{Cl})} = \frac{\pu{0.835 g}\cdot (170 - x\cdot 35.5)~\pu{g mol-1}}{(1 - 0.835)~\pu{g} \cdot \pu{35.5 g mol-1}} = 24.14 - 5.04\cdot x\label{eqn:3}\tag{3}$$

Solving \eqref{eqn:3}:

$$6.04\cdot x= 24.14 \quad\implies\quad x = 4.0 \tag{4}$$

Using $x$, we can now determine the element from \eqref{eqn:2}:

$$M(\ce{E}) = (170 - 4.0\cdot 35.5)~\pu{g mol-1} = \pu{28.0 g mol-1}\tag{5}$$

which appears to be silicon, valency 4.
The unknown compound is silicon tetrachloride $\ce{SiCl4}$, a fuming volatile liquid at NTP.


What is the Difference Between the Conjugate Effect and the Inductive Effect?

Conjugate effect (or delocalization) is an effect in which molecular orbitals (MOs) are conjugated to new molecular orbitals that are more delocalized and therefore generally lower in energy (the amount of MOs stays the same of course). The electrons can move freely in these new extended orbitals. As an example consider this MO diagram of butadiene.

Note that the lowest energy MO is the MO with the least amount of nodes.

Inductive effect on the contrary does not increase delocalization but just pulls electrons towards the inductive (polarized) atom, this, for example, makes lone pairs less available decreasing the nucleophility of a molecule and increase its acidity. A lot of the time this inductive effect will be achieved trough a $\ce{C-X}$ (fluorine and chlorine) bond and you can observe its inductive effect simply by looking at the acidity of a series of more and less substituted molecules. Compare for example trifluoroacetic acid (pKa of 0.23) with regular acetic acid (pKa of 4.76).

I'm not exactly sure though what effect induction has on the size of a MO (but it won't match the effects of conjugation for sure). It is clear however, that the electron distribution will shift toward the spatial direction of the inductive atom.


What is the molecular formula of 2-ethynyloctahydro-1H-indene?

I suspect you forgot one hydrogen at the terminal carbon in alkyne group $\ce{R-C#C-H}$. With this in mind the formula would be $\ce{C11H16}$:


Why do collisions in elementary reactions of higher-orders appear to be more likely?

I second J M’s answer: the rate does not depend only on concentrations, but the rate constant is key. However, I will add two more comments:

First, you cannot compare values of rate constants for reactions of different orders… they don't have the same dimension (unit). So, you cannot say things like “rate constants of third order reactions are typically smaller than those of first-order reactions”.
However, you can say (and it is true) that reactions rates of high-order elementary reactions tend to be small, because of the low probability of the molecules crossing path.


Birch reduction of 1,2-Cyclopentanedione and Diketones

The course of the reaction depends on whether or not a proton source other than ammonia is present. In the absence of an alcohol (ethanol, tert-butanol) the α-dione 1 adds an electron from sodium to give a resonance stabilized radical anion 2. A second electron is added to afford enediolate 3. [This is the same species formed with sodium in the acyloin condensation on diethyl glutarate.]  Water workup provides the α-hydroxyketone 4. In the presence of alcohol, enolate 3 is protonated to form 5. [The reason I am reluctant to protonate the alkoxide of 5 is to avoid reductive elimination of hydroxide from 4 to leave the enolate of cyclopentanone. Further protonation/reduction may lead to the alkoxide of cyclopentanol.]  One-electron reduction of 5 yields ketyl 6, which via protonation and a second reduction leads to dialkoxide 8. Workup affords the diol 9.
The bottom line is if you want to form 4 by metal/ammonia reduction, avoid alcohol.


Calculating rate constant of intersystem crossing from Jablonski diagram

The lifetime of the excited state is the reciprocal of the sum of rate constants which you are trying to find:

$$\begin{align}
\tau(T_1) = \frac{1}{\sum_i k_i(T_1)} &= \pu{6 \times 10^-3 s} \\
\sum_i k_i(T_1) &= \frac{1}{\pu{6 \times 10^-3 s}} \\
&= \pu{167 s^-1}
\end{align}$$

The two lines going from the $T_1 \to S_0$ state represent intersystem crossing (ISC) and phosphorescence. Squiggly lines represent nonradiative transitions, of which ISC is one, so the quantum yield for ISC is 10%. (Straight lines represent radiative transitions, in this case phosphorescence.)

Therefore:

$$\begin{align}
k_\text{ISC} &= \Phi_\text{ISC} \cdot \sum_i k_i(T_1) \\
&= 0.1 \cdot \pu{167 s^-1} \\
&= \pu{16.7 s^-1}
\end{align}$$


Confusion in finding electrophilic centre

In all of the compounds illustrated, the starred carbon is connected to a more electronegative element.

Let's examine this one by one- I shall begin with methyl iodide. Here, the starred carbon is connected to a Iodine which is more electronegative, thus naturally, we can guess that the electron density shifts towards the halogen. Thus, the carbon in question is slightly electron deficient (slightly positive or electrophilic)

Now, let's look at the examples with the multiple bonds.
Consider ethanal: the carbon is again connected to a more electronegative atom i.e oxygen. So, if you were to cleave the pi bond, the electron density would shift to the oxygen, and again you are left with an electrophilic carbon right next to the oxygen.

The same can be said for the acetonitrile: the electron density shifts to the more electronegative element i.e nitrogen and the carbon right next to is left "slightly positive" or electrophilic.

The electron density would always be slightly greater in the vicinity of a more electronegative element across a bond between two atoms differing in electronegativity (essentially, what we mean when we say a bond is polar)


Why does stannous chloride occur despite the octet rule?

According to Theoretical and experimental studies of the electronic structure
of SnCl2 by photoionization with synchrotron radiation AIP Conf. Proc. 258, 60 (1992)

$\ce{SnCl2}$ has a molecular orbital structure as below (HOMO at the top)

Sigma-antibonding: 36% Sn s, 22% Sn p, 41% Cl s

Nonbonding 99% Cl p

Nonbonding 99% Cl p

Nonbonding 7% Sn p , 92% Cl p

Nonbonding 3% Sn s, 6% Sn p , 90% Cl p

Sigma bonding 14% Sn p, 83% Cl p

Sigma bonding 48% Sn s, 6% Cl s 45% Cl p

Nonbonding 3% Sn p, 96% Cl s

Nonbonding 5% Sn s, 2% Sn p, 92% Cl s

So overall it seems:

There is a total bond order of only 1.

Much of the Sn p and some Sn s electron density has been donated to Cl p orbitals.

There is little mixing of Sn s and p orbitals expect in the antibonding orbital.

additional references:

A Symmetry Rule for Predicting Molecular Structures (This reference explains how, based upon second order Jahn Teller effect, the symmetry of molecular orbitals predicts geometry, for example bent vs linear, and says that the symmetry predicts bent for SnCl2)

and Metal oxidation state effect in photoionization of gas‐phase metal halides (which further discusses the molecular orbitals of SnCl2 discussed above)


Obtaining calcium iodide from calcium and iodine

Looks like the Bunsen burner is not providing enough energy for the reaction to occur. AFAIK, you need around 200-400°C of heat for the reaction to occur. Yes, you can always powder the calcium metal to increase surface area and see if the reaction is occurring or adopt simple(chemical) methods. You can react any calcium salt with hydroiodic acid to form calium iodide. I am using calcium carbonate as example[1]:

$$\ce{CaCO3 + 2 HI → CaI2 + H2O + CO2}$$

Or alternatively make a iodide of iron and make it react with a calcium salt[2]. There are many more ways to make calcium iodide in [2]. You can look at them but these two methods is by far the most simple methods.

References:

https://en.wikipedia.org/wiki/Calcium_iodide
Caty J. Braford;  H. A. Langenhan; A preliminary report on the composition of syrup of calcium iodide N. F. V. DOI:https://onlinelibrary.wiley.com/doi/pdf/10.1002/jps.3080180218
http://albumen.conservation-us.org/library/monographs/sunbeam/chap10.html


Why can't terminal alkynes take part in Birch reduction?

The alkyne proton is acidic enough to react with the dissolving metal to give the anion, giving the metal alkynide as it shown by these procedures sodium acetylide prep,  sodium phenylacetylide prep; this does not reduce.


Why is the chemical formula for vinegar CH3COOH what it is?

CH3COOH is an accepted and common form of writing the structure of acetic acid (commonly known as vinegar when diluted in water). You are right in saying the formal notation should be used - which is C2H4O2 - since it gives the reader the types of atoms and their quantities, and these two characteristics define the exact composition of the molecule, and there's nothing wrong with that.

However, the notation CH3COOH takes you one step further and tells you something about the actual structure of the molecule, since it groups together the atoms that comprise a functional group, and a very important one at that. Grouping together COOH is an accepted and convenient form of communicating to the reader: "there is a carboxylic group here", or "this molecule is actually a carboxylic acid", which is something you'd like to know when dealing with a certain substance.

To summarize, C2H4O2 can be the molecular structure for acetic acid or formate (formic ester, as mentioned in another answer). In order to avoid confusion and, more importantly, indicate that there is a carboxylic group in the molecule, it is more convenient and straightforward to note it in the formula than letting readers figure it out themselves.


I want to make NaOH from NaHCO3, what is the easiest way to do it without electrolysis?

You could heat it to turn bicarbonate into carbonate, and let it stand with Ca(OH2) and water mix. That's essentialy how potash lye was made for millenia


A solvent such as water would be required and carbonate or bicarbonate salts will release CO2(g) when reacted with most acids.

Dry ice will turn to CO2(g) when exposed to room temperature.

These are only a couple of examples as others have stated there could be many other methods.


Do acetals and hemiacetals give 2,4 DNP test?

The standard mixture for a 2,4-DNP test contains 2,4-DNP and conc. Sulfuric Acid in EtOH/water here. This is sufficient to hydrolyse most acetals and hemi-acetals so they will give a positive test. Cyclic acetals such as dioxolanes are more stable and may not.


Why P2O5 does not show oxidising properties while N2O5 is the strongest oxidising agent among pnictogen oxides?

A very easy half-answer would base this on electronegativity.

Nitrogen’s electronegativity is $\approx 3$ while phosphorus’ is $2.19$. Oxygen’s is, of course, $3.5$. So the electronegativity difference between nitrogen and oxygen is much lower, meaning that oxygen is much less able to stabilise nitrogen’s high oxidation state.

Contrarily, phosphorus’ low electronegativity means oxygen can stabilise its high oxidation state rather well and the compound is much less likely to act as an oxidating agent by itself.


Mass transfert in process chemistry engineering, two film theory

You already derived this relation (eq 1),
\begin{align} (y_1-y_1^*)&=(y_1-y_1^I)+(y_1^I-y_1^*)\\ &=(y_1-y_1^I)+ K_1^{eq}\left(x_1^I-x_1\right)\end{align}

Now, to keep the molar flux density between the gas phase concentration $y_1$ and fictitious conc. on the other side $y_1^*$ we assumed a fictitious overall transfer coefficient $K_{cg}$. So, now we have another relation for molar flux density
$$N_{1n}=K_{cg}c_g\times(y_1-y_1^*)$$
that follows,
$$y_1-y_1^*=\frac{N_{1n}}{K_{cg}c_g}$$
Similarly,
$$y_1-y_1^I=\frac{N_{1n}}{k_{cI}c_I}$$
and
$$x_1^I-x_1=\frac{N_{1n}}{k_{cg}c_g}$$
Plug these above three relations into eq 1 and you will get your last equation.


Which way does this reaction between NaCl and acetic acid naturally proceed?

Mostly the reaction will proceed from left to right, unless you have ridiculous amount of acetic acid. Hydrochloric acid is much stronger, so it suppresses dissociation of acetic acid.

However, this is only part of the truth. In fact, in water solutions many compounds (partially) dissociates:

$HCl = H^+ + Cl^-$

The process is strongly endothermic in vacuum, but subsequent interaction of ions with water makes it mostly energy neutral or even highly exothermic for some compounds. This means, that in you case the solution would actually contain $H^+$, $OH^-$ (very little), $CH_3COO^-$, $CH_3COOH$, $HCl$ (almost none, may be ignored in most cases), $Na^+$. The actual reaction would be

$CH_3COO^- + H^+ = CH_3COOH$

The actual direction of the reaction would depend on initial concentraion and guarded by equilibrium constant.

The keywords for googling/textbook reading are 'electrolitic dissociation', 'ionic reaction', 'chemical equilibrium', 'equilibrium constant'. Generally, the procedure of determining which concentration of which ions the system will have requires solution of system of equations, including equilibrium constants, electroneutrality equation and material balance equations. The system often may have no analytic solutions, in which case it is usually approximated by simpler system, dropping some elements of sums in the system.


Is there an equilibruim in an unsaturated calcium hydroxide solution?

Equilibria in solutions are given as solubility products, $K_{\text{sp}}$. Your reaction is:

$$\ce{Ca(OH)_2 <=> Ca^2+ + 2OH-}$$
for this, $K_{\text{sp}}=5.5\times 10^{-6}$ (from Wikipedia). What this means, is that in a saturated solution, at equilibrium, you will have:
$$K_{\text{sp}}=[\ce{Ca^2+}]  [\ce{OH-}]^2=5.5\times10^{-6}$$

However, for an unsaturated solution, you will have:

$$[\ce{Ca^2+}]  [\ce{OH-}]^2 <K_{\text{sp}}$$

As mentioned the solution does get cloudy for different reasons, namely the formation of $\ce{CaCO3}$ by reaction with $\ce{CO2}$.


It doesn't work.  The ring geometry holds the carbonyl oxygen and hydroxyl hydrogen too far apart.  If you have a molecular model kit, construct the molecule with it and see what happens to the oxygen and hydrogen atoms.

Hydrogen bonding between different molecules is not constrained by the ring geometry, and thus works perfectly well.


There are non-enzymatic methods for removing the three glucose moieties of stevioside 1. Unfortunately, dilute mineral acid hydrolysis not only removes the glucose entities but also effects Wagner-Meerwein rearrangement of the aglycone steviol 2 to isosteviol 3.1 While vinegar (5% aqueous acetic acid) is a weaker acid than hydrochloric acid, I am certain that it would have been tried to avoid the rearrangement of steviol 2. Exposure of stevioside 1 with aqueous KOH effects saponification of the ester group leading to steviobioside 4.---continued---

A more fruitful procedure that avoids rearrangement involves periodic acid cleavage of the five vicinal diols of the glucose units (marked in red in structure 1 below).2 This procedure liberates hexa-aldehyde 5 in addition to two equivalents of formic acid.3  Vigorous base treatment is required to saponify the hindered ester while base also liberates tetra-aldehyde entity by β-elimination forming steviol 2 in 75% yield. And NO, you can't do this at home.

1) Avent, a. G., Hanson, J. R., de Oliveira, Bras H., Phytochemistry, 1990, 29, 2712.
2) Ogawa, T., Nozaki, M., Matsui, M., Tetrahedron, 1980, 36, 2641.
3) On periodic acid cleavage of vicinal diols, see: Periodic cleavage of carbohydrates


How to select for 1-4 carbon sugar bond in Chemical Glycosylation?

You have to protect the OH groups you don't want to react with protecting groups.

In this case you could install a benzylidene acetale between the 4 and 6 OH, then protect all the other OH and then do selective reductive opening of the benzylidene so your benzyl group will be on position 6 and you got a free OH.
That said: it's quite expensive, labor intensive and has very bad atom economy. And you might run into problems with deprotection once you got polymer structures, so that's not a feasible option for synthesis of cellulose.

I don't think it's possible to optimize a synthesis without protecting groups, so the best way would be doing it enzymatically, which could be very expensive too.


Different types of solder can give a very high operating temperature, are cheap and offer excellent conduction and heat contact when molten. We have used lead and cadmium free solder to 500 degrees C with no problem. Your issue may be the melting point (ours was 230C), but the right mix may bring this down for you. If you have the time and inclination, you could modify a commercial mix yourself. Otherwise,  sand would get my vote.


Optimization and DFT-D3 energy calculation in GamessFireFly version 8.1.1

In original GAMESS-US code there are no analytic gradients for "double hybrids" and it looks like it is true for Firefly as well. You can use numerical gradients by adding NUMDER=.T. in $CONTRL section. See this for details.


Measuring turbidity using a spectrophotometer

You can use a spectrophotometer to measure turbidity since the suspended particles will scatter light and so a reduced amount of light is detected dependent upon how much scatter occurs. The relationship is $I_{trans}=I_0\exp(-k\cdot C)$ where k is a proportionality constant depending on the particle size and shape, particle refractive index, sample path-length and C is the mass concentration in what ever units you choose. Usually in a spectrophotometer  the log is produced by the machine as optical density ( OD ). Thus $OD = -\log_{10}(I_{trans}/I_0)= kC$.

It is essential when comparing samples that the particle size is the same otherwise k varies between measurements and as C is also varied this confuses any measurements. Gum arabic or gelatin are often used to stabilise a suspension. The more finely divided is the suspension the greater is the ability to scatter light. Often experiments are made by comparison to a standard suspension say of Ca oxalate, $\ce{BaSO4}$, or Zn ferrocyanide.


Can the Galvanic cell use a different ionic solution?

Can the Volta Battery use a different ionic solution?

Yes you can use any soluble salt as long as it does not react with electrodes, and certainly such a battery would be able to produce current. My chemistry high school teacher had a clock which ran on salt water. See for example, https://science.howstuffworks.com/environmental/green-tech/sustainable/water-powered-clock2.htm. This technology is at least 400 years old. volta used common salt as well for zinc and copper.

Now the problem is that you can generate voltage by using copper/ zinc and salt water however one cannot predict the voltage by using the Nernst equation. It requires that the electrode dip in its own ions. Hence such batteries are theoretically difficult to deal with.


Forming the resonance structures for the nitrite anion

Yes. I think that there is a mistake in your imagination. Nitrogen is evidently $sp^2$-hybridised in this case. Nitrogen has 5 valence electrons. Two electrons are used to make 2 $\sigma$ bonds with two hybrid orbitals. Another two electrons become the lone pair and reside also in a hybrid orbital. This leaves us with one electron in one unhybridised $p$ orbital. This $p$ orbital is then used to make the $\pi$ bond to either oxygen atom, contributing one electron to the $\pi$ bond while the oxygen atom also contributes one electron from its $p$ orbital.