Why doesn't the viscosity of water change much with temperature like it does for other substances?

This is something I spent a lot of time thinking about during my PhD, so let me see if I can explain it.

What you are talking about is the relationship between macroscopically observable properties (like viscosity), the relaxation time spectrum (which is a way of looking at the time scales involved in molecular movements), and the structure of the molecules themselves. It's kind of complicated! But, I think I can explain it in way that will make sense.

You know that as we increase the temperature, molecules have more kinetic energy and can more easily break free of the forces holding them together. I have some other answers here and here that explain this in more detail. With something like viscosity, however, there is more to the picture.

The first thing you need to know is:

  Viscosity is a measure of momentum transfer in a fluid

In other words, it is a way of quantifying the effect that movement in one part of a fluid has on the parts immediately adjacent to it. If the movements are highly coupled (meaning that if one layer moves, the layers next to it move almost as much), the viscosity is high. If not, the visocosity is low. Wikipedia has a pretty good explanation of this on the macroscopic scale.

So what causes a fluid to have a viscosity? Since you know about intermolecular attractive forces, this is an obvious candidate. And indeed, the stronger the intermolecular forces are in a fluid, the higher the viscosity tends to be. However, there is another factor involved - intermolecular repulsive forces (sometimes called stearic effects or excluded volume interactions). The idea is that if one molecule wants to move in a dense fluid, the other molecules have to get out of the way first.

As an analogy, imagine moving through the crowd at a packed concert or club. In order for you to move, someone else needs to move, and for them to move, someone else has to move, and so on. You can be polite and wait patiently for them to move, and eventually someone will. Or you could be assertive and ask, and move faster. If you are really impatient, you could barrel through people, knocking them out of the way (and hope you don't get beat up). Your impatience in a crowd is analogous to shear stress in a fluid. At lower temperatures, the molecules are closer together (since their kinetic energy isn't high enough to move very far out of the potential energy wells of their neighbors), and they move more slowly in general. This means that the time it takes for neighbors to move goes up, and as a result, viscosity goes up too - if you force them to move, they will knock others out of the way, transferring momentum through the fluid. At higher temperatures molecules move much more frequently, and so it is relatively easy to "drag" some through the fluid - they don't have to knock other molecules out of the way, because by the time they get there, they have moved on their own. The viscosity is lower.

Now, for simple molecules, the line between "can't move" and "moves easily" is pretty distinct. Once the molecules are mobile enough that excluded volume interactions are overcome, the material melts, and further changes to viscosity are small and mostly due to intermolecular attractive forces. For example, here is a plot of the viscosity of water vs. temperature

As you can see, it decreases smoothly as temperature increases. Below the freezing point, the viscosity is harder to define (although large masses of ice do flow), but the curve is discontinuous and it "jumps up" to some higher value.

But what about complex molecules? Let's link the monomers together:

Now it becomes much harder for the molecules to move. At a given temperature, you have to wait about the same amount of time for a single monomer to move, but it takes much longer for a pair to move together. Going back to the crowded room analogy - imagine trying to move through a crowd in which everybody - including you - was handcuffed in pairs to someone else. It would be a lot more difficult! In general, it turns out that the relaxation time for linear polymers increases as (approximately) the square of the chain length (for short lengths below the entanglement length). This means that if you double the number of monomers in a polymer chain, the average time it takes for them to move out of the way goes up 4 times. This is true up to a certain length of chain called the entanglement length (which depends on the polymer), at which point the likelihood of coordinated movement of the whole chain becomes so low that each polymer is sort of "trapped" in a tube made of other polymers. The analogy often used to explain this is a nest of snakes - they have to slither around each other. As a result, this type of motion was named reptation - after reptiles.

The interesting thing is that for shorter lengths (below the entanglement length) each length scale of the chain moves essentially independently of the rest. This means that in a longer chain, on short time scales the shorter lengths act as though they were in a low-viscosity fluid - they move easily and motion is not correlated with their neighbors. For longer length scales, the chain is essentially "frozen" - it can't move until its neighbors move - and so it acts more like a solid.

The result is that short lengths of chain can "wiggle" on short time scales, but longer lengths move more slowly. This gives us a relaxation time spectrum - it means that if you apply very fast changes (high shear rates) the molecules don't have time to get out of the way, and so the viscosity is high. If you increase the temperature, things move faster, and so at the same shear rate, the viscosity will be lower (more of the chain can move out of the way). The same is true at lower shear rates for the same temperature - the shorter length scales can move out of the way, and so it becomes more liquid - the viscosity decreases. On the longest time scales (lowest shear rates, coldest temperatures) the entire length of the chain can move, and so you see the lowest viscosity.

This is called shear-thickening, and is commonly seen in linear polymer melts or concentrated solutions. It is an example of viscoelastic behavior, which means that depending on the conditions (temperature or time scale of interaction) the material will exhibit a range of liquid-like (viscous) or solid-like (elastic) behavior. An example commonly used to demonstrate this is silly putty (although silly putty is not a pure linear polymer.) On short time scales, the polymers don't have time to relax completely and so it behaves like an elastic solid. This is why you can bounce a ball of silly putty on a table. At longer time scales, the polymers can fully relax, and so the silly putty flows. This is why if you let a lump of it sit out for a while, it will form a puddle. Silly putty is unique in that the relaxation time spectrum ranges from elastic to viscous over time scales that are commonly observed at room temperature, so you can see these effects in high contrast very easily.

There are many other types of non-Newtonian behavior (which just means viscosity is not constant), but they are all fundamentally connected to the relaxation time spectrum of the fluid, which is directly related to the molecular structure.

To sum up:

Water has a narrow viscosity spectrum (as a function of temperature) because it has a narrow stress relaxation time spectrum.
It has a narrow stress relaxation time spectrum because it is a relatively simple, small molecule.


Dissociation constant of molecular dinitrogen as a function of temperature?

Update: If you plot the $\log(K_p)$ vs temperature it does not give a straight line over such a large temperature range so you are unlikely to be successful.

However, you can estimate $K_p$ as the molecular properties of dinitrogen are known so you can calculate the partition functions Z and so the equilibrium constant at any temperature. You need the vibrational frequency (and possibly anharmonicity), rotational constant and dissociation energy. Work out the partition functions for translation, rotation & vibration for dinitrogen and translation for N atoms and that is all you need.

The equations needed are given in most phys. chem. textbooks and specialist books on statistical mechanics. At a standard state of $1$ bar pressure
$$k_P= \frac{Z_{trans}(N)^2}{NZ_{trans}(N_2)Z_{rot}(N_2)Z_{vib}(N_2)}\exp(-E_0/(kT))$$

The experimental values at low temperature are likely to have the biggest error since dinitrogen is such a stable molecule, and it probably does not matter much how small the $K_p$ is its always tiny at low temperatures. Using the method above I calculated values as $2.4 10^{-160}, ~ 8.4 10^{-19}$ and $ 62400$ at the three temperatures you quote (using the dissociation energy as 9.756 eV).


How to calculate the theoretical yield of a Diels-Alder reaction?

You are missing a  couple of zeros in the number of moles of your 9-anthracenemethanol.  I calculate 0.00033 mol of that reagent, which therefore becomes your limiting reagant, and I calculate a total yield of 0.105 g of product, with about 0.73 g of  N-Methylemaleimide left over.


Why active mass of solids and pure liquids taken as unity

Maybe the following reasoning may help Param 1729. The thermal decomposition of $\ce{NH_4Cl}$ may be pedagogically seen as occurring in two steps. First, solid $\ce{NH_4Cl}$ is simply vaporized producing the hypothetical gaseous species $\ce{NH_4Cl(g)}$
$$\ce{NH_4Cl(s) <=> NH_4Cl(g)}\label{rxn:nh4cl}\tag{1}$$
And this transition is described by an equilibrium constant $\ce{K_1}$ which is similar to Henry's law
$$\ce{K_1 = \frac{p(NH_4Cl)(g)}{[NH_4Cl(s)]}}$$
Immediately after this vaporisation phenomena, a gas phase decomposition happens with the following equation :
$$\ce{NH_4Cl(g) <=> NH_3(g) + HCl(g)}\label{run:nh3+hcl}\tag{2}$$
with an equilibrium constant $\ce{K_2}$ defined by :
$$\ce{K_2 = \frac{p(NH_3)·p(HCl)}{p{(NH_4Cl)(g)}}}$$
So what is really observed is the sum $(1) + (2)= (3)$ $$\ce{NH_4Cl(s) <=>  NH_3(g) + HCl(g)} \label{run:dnh4cl}\tag{3}$$ with a constant $\ce{K_3}$ equal to
$$\ce{K_3 =  \frac{p(NH_3)·p(HCl)}{[NH_4Cl](s)} = K_1·K_2}$$
This last expression shows that the pressure of the gases $\ce{NH_3}$ and $\ce{HCl}$ does not depend on the amount of $\ce{NH_4Cl}$ in the solid state. And, as $\ce{[NH_4Cl](s)}$ is a constant, it may be included in $\ce{K_3}$ giving a new constant $\ce{K_3'}$
$$\ce{K_3' = p(NH_3)·p(HCl) = K_1·K_2·[NH_4Cl](s)}$$ This is "equivalent" to stating that the concentration of $\ce{NH_4Cl(s)}$ is equal to $1$.


What type of analytical chemistry (such as spectroscopy) is suitable for high resolution air analysis

Headspace GC-MS is the way to go. It's used to detect vapors coming off a solution. For example, it is used in the fragrance industry to test the time-dependent development of a perfume, since the lighter molecules will evaporate first and form a higher concentration in the headspace initially.

Unfortunately, for detection of noble gases in an extremely low concentration this is difficult. The technology has just been created and involves concentrating the rare gases via adsorption over high surface area carbon before analysis. Each gas adsorbs in a specific temperature range that is different.

From an article on the International Noble Gas Experiment, a modern method to detect trace amounts of Xenon from nuclear explosions:

  “Xenon is isolated from the air by a charcoal-containing purification device resulting in higher concentrations of the gas in its stable and unstable, i.e. radioactive, forms.  The isolated xenon is then measured for its level of radioactivity which helps to quantify the radioactive xenon contained in the sample.” Source

I know that a company called Air Products liquifies air before distilling Nitrogen and Oxygen, and then purifies Xenon for medical use. An in-depth process is described here.


Does a reaction have to have a rate determining step?

Any kind of chemical reaction can be modelled as a (complex) network of elementary reaction steps. For instance, if we model the conversion of R to P via the intermediates $\ce{I^1}$, $\ce{I^2}$, and $\ce{I^3}$, we have the following kinetic network:

$$\ce{R <=> I^1 <=> I^2 <=> I^3 <=> P}$$

Note that on the basis of microscopic reversibility, all these reactions can go both ways and the boundary conditions of the system (pressure, temperature and chemical composition) determine in which way and at what rate the reactions occur.

To answer your question: there exists a formal way of identifying the rate-determining step. If we look at the overall reaction rate $r$; then we can define a concept called degree of rate control:

$$
\chi_{i} = \left( \frac{\partial \ln r}{\partial \ln k} \right)_{k_{i \ne j}, K_{i}}
$$

Here, $\chi_{i}$ is the degree of rate control coefficient that indicates the extend to which an elementary reaction steps controls the overall rate and $k$ is the rate constant of the particular elementary reaction step you want to calculate the DRC coefficient for. Importantly, in this calculation you keep the rate constants for all other reaction steps and for the equilibrium constant of the elementary reaction step you are looking at constant. (in other words, you only change the height of the reaction barrier). When you find that this coefficient is 1 and all other elementary reaction steps have a coefficient of 0; in that case this elementary reaction step is the rate-determining step.

Interestingly, when investigating a lot of chemical systems, it turns out that only roughly half of these systems have a rate-determining step. A lot of the systems (especially in catalysis) have multiple rate-controlling steps and even rate-inhibiting steps (where the DRC coefficient is less than zero).

I will not go too much in-depth, but there exists a complete mathematical treatment and derivation of this theory that shows that the sum of all coefficients has to be unity. This means that a chemical system should always have either a single rate controlling step (the RDS) or multiple rate-controlling steps and possible one or more rate-inhibiting steps.

To conclude: Not every overall chemical reaction has an RDS, but they should have one or more rate-determining steps.

Notable references:
The above theory was proposed by Campbell and co-authors in this research paper: J. Am. Chem. Soc. 2009, 131 (23), 8077–8082. DOI:10.1021/ja9000097.


What is the need of hyper conjugation in this structure when there is
  charge separation resulting in lesser stable structure.

I think we'd all agree that the charge separated structure you've drawn is a resonance structure.  Charge has been separated so it may well be higher in energy than the first (neutral) resonance structure.  Nonetheless, to whatever extent the charge separated resonance structure contributes to the description of propene - even if it is just a small amount - it will stabilize propene.

Since there are 3 hydrogens on the methyl group you can draw three identical resonance structures, this will increase the importance of the structure.  Note too the charge separated resonance structures involve an allyl carbanion (resonance stabilized) and a proton; two very stable structures - this will further increase the weight, or importance, of the hyperconjugated structures.

These hyperconjugated resonance structures are really showing us that an $\ce{sp^3}$ carbon can donate electron density to an $\ce{sp^2}$ hybridized carbon.  Electron movement in this direction lowers the energy of the electron because it moves the electron from an orbital that is 25% s-character to an orbital that is 33% s-character.  The more s-character in an orbital, the lower the energy of the orbital (an s orbital is lower in energy than a p orbital).  As we stabilize the electron, we stabilize the molecule.

This hyperconjugation argument comes in handy.  It can be used to explain

why double bonds with more alkyl groups are more stable than double bonds with fewer alkyl groups (ref1, ref2,)
why carbocation stability follows the order: tertiary > secondary > primary (ref1, ref2)
why alkyl groups are ortho-para directing in electrophilic aromatic substitution (ref)


What role does H2O2 have in copper (II) acetate formation?

Presumably hydrogen peroxide causes copper metal to donate 2 electrons to form $\ce{Cu(II)}$ -- which is soluble in water -- and we'd say the hydrogen peroxide is acting as an oxidizing agent.


Safely lowering bleach pH without chlorine gas?

The oxidizing effect of chlorine is quite powerful even at concentrations of 1% or less. Mixing bleach and vinegar gives a much more concentrated solution. Although chlorine is soluble to the extent of 1.46 grams per 100 mL cold H2O (CRC Handbook), its vapor pressure above the water may be significant. A solution diluted by a factor of 7 or 8 would be in contact with a surface long enough to disinfect with a much reduced odor. Safer for people, but still deadly to bacteria.


Why is the viscosity of water and ethanol lower than that of a water-ethanol mixture?

This answer just popped up when I searched for viscosity of water-ethanol mixtures, so I just copied the abstract:
Viscosities of Binary and Ternary Mixtures of Water, Alcohol, Acetone, and Hexane
https://www.tandfonline.com/doi/abs/10.1080/01932690802313006?journalCode=ldis20
"This articles studied and determined the viscosities of the binary mixtures of water–methanol, water–ethanol, water–propanol, water–acetone, acetone–ethanol, methanol–ethanol, and acetone–hexane and the ternary mixtures of water–methanol–ethanol and water–ethanol–acetone at 20°C. It is shown that the mixing of water with the alcohols and acetone resulted in a positive deviation of viscosity, which reached the maximum value at the water mole fraction x 1 ∼ 0.7 for water–methanol, x 1 ∼ 0.72 for water–ethanol, x 1 ∼ 0.74 for water–propanol, and x 1 ∼ 0.83 for water–acetone binary mixture. This viscosity deviation can be mainly attributed to the formation of micelles of alcohol or acetone molecules in water because of the hydrophobic attraction between the hydrocarbon chains. The micelle surfaces are surrounded by hydration layers, leading to the positive viscosity deviation in the liquid mixtures because the water in hydration layers has a much higher viscosity than bulk water. Also, the contrary observation was found in the binary mixtures of acetone–ethanol and acetone–hexane, having a negative viscosity deviation."


How can I interpolate between two densities at different temperatures?

You could use tabulated literature data and then interpolate between the data points. For example, the following plot was created using data from NIST Reference Fluid Thermodynamic and Transport Properties Database (REFPROP) – NIST Standard Reference Database 23, Version 9. It shows the liquid saturation line as well as the vapour saturation line.

In this case, interpolation for a liquid density of $780.00\ \mathrm{kg/m^3}$ yields a temperature of $31.643\ \mathrm{^\circ C}$.

By way of comparison, the calculated liquid density is $781.55\ \mathrm{kg/m^3}$ for a temperature of $30.000\ \mathrm{^\circ C}$, and $772.10\ \mathrm{kg/m^3}$ for a temperature of $40.000\ \mathrm{^\circ C}$.


Why does the isoelectric range decrease as an Alanine chain grows longer?

It is because of the inter- and intramolecular associations that occur.

In Ala, the first loss of a proton forms a two point intermolecular association which is very favourable.
Since the equilibrium tends to the right, more $\ce{H^+}$ are liberated and the p$K_a$ is lower.

For the second step, this stable dimeric association must be destroyed, so now the equilibrium tends to the left and the p$K_a$ is higher then it would be without this association.

In the case of Ala-Ala-Ala, this type of dimeric association can not occur because the distance is too long.
There is intermolecular association but with different molecules, not in pairs.
Here the first p$K_a$ is higher than in the case of Ala and the second lower because the equilibrium is not forced so much to the associated forms.

Finally, in the case of Ala-Ala, besides the intermolecular association between different molecules like in Ala-Ala-Ala,
there is also an intramolecular association.
This one is less stable than the association in pairs of the individual amino acids, but more than the random intermolecular ones.
Therefore, the two p$K_a$  are in between of the ones for Ala and Ala-Ala-Ala.


Finding vendor information programmatically

You can try molport, they are an intermediate. Just ask them directly, it will be done in a jiffy.
Or if you like you can do it in knime and use any API (like molport's API), which is much more fun, but for just a one-time thing a bit complicating.


The stability of an ion in an aqueous medium is determined by the tendency of attraction between the ion and the water molecules termed as hydration enthalpy. It, in turn, depends on the charge on the ion and its size.
Smaller the ion greater the hydration power of the ion. It is because when the ion is small the distance between its centre and the molecule that solvates it is less hence providing a stronger attraction, thereby releasing more energy attributing to stability. The example of this fact is provided by Cu2+ which has a smaller cationic radius due to a stronger hold of the nucleus on the remaining 27 electrons after the loss of two as compared to Cu+ where only one electron is lost and the hold of the nucleus is less. The stability gained by energy loss in hydration is more stabilising than the destabilising effect of the removal of 2 electrons.
Also, as mentioned earlier, the greater the charge on the ion greater is its tendency to attract and keep the water molecules solvated to it. Cu2+ having a greater charge has a greater tendency of hydration than Cu+.


Possibility of π–π stacking in non-aromatic systems

There is a paper titled "Amide-pi interactions between formamide and benzene",  doi: 10.1002/jcc.21212. It has been cited by a paper on a crystal structure on a urea transporter (the interaction is between a peptide bond of a ligand and aromatic side chains of the protein).

Technically, the interaction between the arginine side chain and aromatic groups could also fall under your heading, but it is usually described as a cation pi interaction.


Where does the additional hydrogen in the Fischer depictions of glucopyranose come from?

Throw away that book, you have found a very terrible error. Okay the rest of the book may still be good, but at least black out that picture or stick a correct version of the picture over the top.

In the image below, you can see four depictions of α-D-glucose taken from Wikipedia (where a full list of authors is available). From left to right these are: Fischer projection, Haworth projection, chair depiction and ‘flat circle’ (a depiction which follows the guidelines for 2D representations of 3D structures).

As you can see, there is no additional hydrogen on the anomeric carbon and it is not pentavalent.

The mechanism of ring-closure is also mentioned in sufficient detail on the corresponding Wikipedia page to which I hereby point you. Basically, the 5-hydroxy group attacks the $\ce{C=O}$ π system as a nucleophile. Proton transfer will lead to the cyclic product. (In actual aquaeous solution, the carbonyl oxygen will be protonated immediately from a nearby water molecule and one will also take away the hydroxy group’s excess proton, so the two are not the same.)


Why does salt dissolved in water taste like salt? If it's just Na+ and Cl- ions

When you taste salt, you're not pushing crystalline $\ce{NaCl}$ into your taste buds. It dissolves in your saliva and dissociates. When one tastes salt, the saltiness taste receptors respond specifically to the sodium cation. That type of taste receptor is a cation channel. This is why lithium and potassium cations also taste salty (though they also stimulate other receptors which make them taste somewhat different).

There seem to be at least two types of receptors that respond to saltiness. One responds almost specifically to sodium at low concentrations, but at higher concentrations, the other type responds to many cations. See here


Friedel-Crafts acylation of anisole with acetyl chloride and aluminium chloride as catalyst is a standard laboratory procedure to synthesize acetanisole.

The fact that this reaction is possible does not imply that there is no complex formation with $\ce{AlCl3}$. On the contrary, it is quite probable that such complexes do form due to the interaction of the methoxy oxygen with $\ce{AlCl3}$.

Therefore, and to compensate the formation of complexes with the resulting ketone, $\ce{AlCl3}$ is added in more than two-fold excess in the above synthesis.


A full verion of the periodic table (For computer game)

You can find all these information on Wolfram Alpha.  You can use the website directly to obtain these information, and you can call their API to directly fetch the data online from your own program.  Of course, you can also use Mathematica to write your program and query the database with WolframAlpha function.


Photocurable Polymer with no oxygen functional group or whose the structure is well known

I'm not sure if this addresses all your criteria,  but acrylonitrile is polymerisable by light and is oxygen free.


Regarding change in temperature in Bomb Caloriemeter

Basically, a bomb calorimeter consists of a small cup to contain the sample, oxygen, a stainless steel bomb, water, a stirrer, a thermometer, the dewar or insulating container (to prevent heat flow from the calorimeter to the surroundings), and ignition circuit connected to the bomb. (*)

The whole device known as calorimeter consists of a bomb and a water bath, they are not two separate entities. So, $ C_{cal}$ is actually referring to the heat capacity required of the steel vessel and also the water. The construction of the calorimeter is such that the thermal mass of the calorimeter system completely absorbs the heat of reaction (**).

*: This Wikipedia
**: Refer 29:27 of this video


In an closed adiabatic container, if we carry out an exothermic process, what will happen to the temperature of the system

Temperature will increase. If system is adiabatic, it doesn't neccesarily mean that temperature of the system can't change because even though system can't exchange energy with surroundings, processes inside can generate heat. In your example, exothermic reaction. Note that, internal energy in your case stays the same even though temperature increased. This is because internal energy is the sum of kinetic energy of all molecules and atoms and all potential enegy of interaction between atoms chemically bonded and between all molecules (intermolecular bonds). When exothermic reaction happens, potential energy of interaction between atoms is partially converted into heat because if reaction is exothermic, chemical bonds formed (of products)are stronger than of reactants. Potential energy is only converted to heat which is kinetic energy of atomic and molecular motion (energy conservation - 1st law of thermodynamics). Since internal energy is sum of potential and kinetic energy of atoms and molecules it stays the same, only potential enegy is smaller and kinetic is bigger after exothermic reaction. If you have any misunderstandings, fire them to me.


In reaction equations, why are j and k sometimes capitalized to J and K?

The entry in Wikipedia misses to define the entries in the formula.  But note equations 1 and 2 in a paper by Griffin et al. (doi 10.1029/2006JD007604, open access) in the Journal of Geopphysical Research, Atmospheres. To establish context of the paper by Griffin et al., the authors introduce the reader to the topic citing earlier work.

«In the troposphere, a molecule of $\ce{NO2}$ is degraded photochemically to form a molecule of $\ce{NO}$ and an oxygen atom ($\ce{O}$). It should be noted that this process depends on the sufficient flux of photons associated with light of the appropriate wavelength to photolyze $\ce{NO2}$. The $\ce{O}$ atom quickly reacts with molecular oxygen ($\ce{O2}$) in the presence of a third body to form $\ce{O3}$. However, the $\ce{O3}$ formed can react rapidly with the $\ce{NO}$ to reform $\ce{NO2}$ and $\ce{O2}$. The net result of these reactions is a null cycle, and when steady state is assumed, the $\ce{O3}$ concentration can be predicted as [Seinfeld and Pandis, 1998]
$$ [\ce{O3}] = \frac{j_\ce{NO2} [\ce{NO2}]} {k_1 [\ce{NO}]}$$
where the bracket notation represents a concentration or mixing ratio (here, parts per billion by volume, ppbv), $j_\ce{NO2}$ ($\pu{s^{-1}}$) is the photolysis rate of $\ce{NO2}$, and $k_1$ ($\pu{ppbv^{-1} s^{-1}}$) is the temperature-dependent rate constant for the reaction between $\ce{NO}$ and $\ce{O3}$.»

The article continues by

«This is referred to as the photostationary state for $\ce{O3}$. If each side of equation (1) is divided by $[\ce{O3}]$, the right hand side is termed the Leighton ratio, $\Phi$ [Leighton, 1961]:
$$\Phi = \frac{j_\ce{NO2} [\ce{NO2}]} {k_1 [\ce{NO}] [\ce{O3}]} $$
[...]»


Where can I find the standard molar entropy for ammonium bicarbonate?

“Standard Thermodynamic Properties of Chemical Substances”, 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. gives values for the standard molar enthalpy of formation $\Delta_\mathrm fH^\circ$, the standard molar Gibbs energy of formation $\Delta_\mathrm fG^\circ$, and the standard molar entropy $S^\circ$ (all at $T=298.15\ \mathrm K$ and $p=100\ \mathrm{kPa}=1\ \mathrm{bar}$) of crystalline ammonium hydrogen carbonate as follows.
$$\begin{align}
\Delta_\mathrm fH^\circ&=-849.4\ \mathrm{kJ/mol}\\
\Delta_\mathrm fG^\circ&=-665.9\ \mathrm{kJ/mol}\\
S^\circ&=120.9\ \mathrm{J/(mol\ K)}
\end{align}$$


When electrons gain energy they become more excited and move to a higher energy level, which increases the tendency of an atom to form a bond with another atom.

Let me assure you that this statement is incorrect. I’m not fully sure where you extracted it from, but I assume it stems from the way many schools teach hybridisation at the beginning of organic chemistry classes; requiring an $\ce{s\bond{->}p}$ excitation in carbon from $\mathrm{[He]\ 2s^2\,2p^2}$ to $\mathrm{[He]\ 2s^1\,2p^3}$, after which the s- and p-orbitals can form $\mathrm{sp^3}$ hybrid orbitals. This idea is nothing more than a school level simplification used to get around teaching of more complex molecular orbital theory and symmetry.

Nothing prevents you from constructing e.g. a methane molecule without initial hybridisation, i.e. starting from an unhybridised carbon atom and four hydrogen atoms in a tetrahedric arrangement. I refer you to the following scheme posted in a different question and originally taken from Professor Klüfers’ internet scriptum for basic and inorganic chemistry at the university of Munich:

As you can see on the right, carbon enters this scheme in the unhybridised ground state. There is no need to invoke a previous hybridisation before mixing orbitals; rather, it is necessary to determine the symmetry of orbitals and thereafter combine symmetry-equivalent orbitals in a bonding-antibonding fashion. Finally, fill in electrons from bottom to top.

This method will always result in stabilisation of bonding orbitals; the trade-off always being the destabilisation of antibonding orbitals in such a way that the (actual) energy gained is lower than the (virtual) energy lost.

Therefore, assuming a positive bond order, forming a bond typically liberates energy while breaking one will typically require energy. I am not aware of any counterexamples, but the sentence is phrased so that it remains true when the obligatory counterexample is posted as a comment.


What should be the name of the given compound?

The most important simplified criteria for the numbering in such cases are:

lower locants for the principal characteristic group that is expressed as suffix
lower locants for multiple bonds
lower locants for prefixes
lower locants for substituents cited first as a prefix in the name

The corresponding actual wording in the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) reads as follows:

P-14.4 NUMBERING
When several structural features appear in cyclic and acyclic compounds, low locants are assigned to them in the following decreasing order of seniority:
(…)
(c) principal characteristic groups and free valences (suffixes);
(…)
(e) saturation/unsaturation:
  (i) low locants are given to hydro/dehydro prefixes (…) and ‘ene’ and ‘yne’ endings;
  (ii) low locants are given first to multiple bonds as a set and then to double bonds (…);
(f) detachable alphabetized prefixes, all considered together in a series of increasing numerical order;
(g) lowest locants for the substituent cited first as a prefix in the name;
(…)

For the compound that is given in the question, the principal characteristic group is $\ce{-OH}$. It is expressed as suffix “ol”. Thus, low locants are assigned first to the $\ce{-OH}$ groups according to Rule (c), which yields “…1,3-diol”.
Next, a low locant is given to the double bond according to Rule (e), which gives “4-ene-1,3-diol”.
This step also establishes the numbering of the other atoms of the ring so that there is no longer a choice for the last substituent. Nevertheless, strictly following the rules, a low locant can be assigned to the remaining substituent according to Rule (f).
Thus, the complete name is 6-sulfanylcyclohex-4-ene-1,3-diol. Note that the prefix “mercapto” is no longer recommended.


There is a simpler way to get fluorinated polyethylene.

Polyethylene is polymerized ethylene $\ce{CH2=CH2}$. As part of the polymerization, the double bond is no longer there. Polyethylene is essentially a very long saturated hydrocarbon:  $\ce{(-CH2-CH2 -)}_n.$

Another polymer already exists that replaces all the hydrogen atoms with fluorine atoms: poly(tetrafluoroethylene) (PTFE), made from tetrafluoroethylene $\ce{CF2=CF2}$. PTFE is a long saturated ferfluorocarbon: $\ce{(-CF2-CF2 -)}_n.$

The most well known trademarked brand of PTFE is Teflon. Get your coated with or made out of Teflon/PTFE. It will be easier, cheaper, and safer in the long run. If XeF2 will fluorinate polyethylene (which is reasonably inert), imagine what it could do to yyou.


How does Borax Solution act as a buffer?

Borax reacts in hydrochloric acid in the preparation of the buffer to give the boric acid through the following reaction $\ce{Na2B4O7.5H2O + 2HCl -> 4H3BO3 + 2NaCl + 5H2O}$

http://www.novedu.ru/anotes/refer/buffers.htm#bor1

The above link gives you the procedures to making the buffer. The reason why the borax solution is able to keep the system at around pH of 9 because the boric acid that is produced from this reaction has a pKa of 9.46. When the concentrations of the acid and its conjugate base, that is, the borate ions is exactly 1:1, the pH will be most difficult to change since any addition of hydrogen ions will only bond with the borate to form the undissociated boric acid.

So to answer you question, boric acid is the weak acid, and borate ions are the conjugate base.


Phosphorus does not form pπ-pπ bonds but how could it form dπ-pπ bonds?

My doubt is then how does phosphorus form dπ-pπ bonds?

Poorly. Influence of the $d$-orbitals on the chemistry of the $p$-block is minor. The current belief is that 'double' bonds are actually dative bonds in most cases.

See the similar cases of chlorine and sulfur

How does chlorine form more than 1 bond?

What is the chemical structure / hybridization of sulfuric acid?

For phosphorus it isn't that bad, but still bad

http://link.springer.com/article/10.1007%2FPL00010730


Is it correct to say that dehydration isn't a subset of condensation?

Alkanols are indeed a good example. Think in the E2 and E1 reactions of primary and tertiary alkanols  in the presence of sulfuric acid at elevated temperatures, yielding alkenes.


Does changing the stoichiometric coefficients affect the Nernst equation?

No, you won't get a different value. The number of electrons transfered in that case is 4, so it will be $E=E_{0} - \frac{RT}{4F} \ln\left(\frac{[\ce{Ni^{2+}}]^{2}}{[\ce{Ag+}]^{4}} \right)$.


You will probably know that the notation for galvanic cell reads from left to right as anode || cathode, where || typically denotes a salt bridge.

Within each half cell, different phases are separated by |.

In your case, the anode is a platinum rod in an acidic solution under a hydrogen atmosphere. This is a Standard hydrogen electrode (SHE).

The cathode consists of a platinum rod which is contact with a layer of mercury and a layer of solid mercury(I) chloride. It is an aqueous cell with 1M KCl solution as the medium. This electrode is known as Saturated calomel electrode (SCE).

Both electrodes are used as reference electrodes.


Upto (D) you are correct From there it's :


How to determine gerade & ungerade symmetry of a MO orbital?

The better way to do it is to check what happens under inversion ($i$ or $\bar 1$). If the orbital stays the same, it is g, otherwise u.

However, as Orthocresol mentioned in the comments, checking that is only possible if the entire molecule contains inversion symmetry. Not all point groups do, and the vast majority of molecules does not (partly because $C_1$ is likely the most prevalent point group out there).

For example, consider transition metals’ d orbitals. In octahedral complexes, they are labelled $\mathrm{t_{2g}}$ and $\mathrm{e_g}$. In tetrahedral complexes, which do not have inversion symmetry, they are $\mathrm{t_2}$ and $\mathrm{e}$.


How much lactose is there in milk (mechanism)?

Using the basics of organic chemistry, solutions of iodine and sodium hydroxide retain enough free $\ce{I2}$ to do other things. I would disocunt $\ce{NaIO}$ as a significant component of this process.

What if instead, we had exactly what your picture shows (except the last few steps are prevented - we'll see how in a few lines). You are worried abut the cyclic form, but the cyclic form and the acyclic form are in equilibrium in aqeous solution, especially if acid or base is present. Note that in the reactions equations below, I am using the notation $\ce{R}$ to refer to parts of the molecule that are not involved.

  Isomerization to the ene-diol

$$\ce{R-CH(OH)-CHO <=>[\ce{OH-}] R-C(OH)=CHOH}$$

  Oxidation by copper(II)

$$\ce{R-C(OH)=CHOH + 2Cu^2+ +3OH- -> R-CH(OH)-CO2^- + 2Cu+ + 2H2O}$$

Now, if that copper(I) cations stumbles into a hydroxide anion, then it's all over since $\ce{Cu2O}$ is not soluble. However, it's iodine to the rescue! Iodine oxidizes $\ce{Cu+}$ back to $\ce{Cu^2+}$.

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

These last two equations add to give the one you quote (only I left out the sodium spectator cations and simplified the carbohydrate).


Acid-molybdate spray of Hanes & Isherwood

I managed to find a re-print of the paper in question– sadly it seems like Nature are mothballing many old papers (dead/missing links).
There are many acidic-cerium/molybdenum type stains that are used nowadays, many of which are commercially available off the shelf, so you might wish to try these before going after something from 50 years ago. (Cf. CAM stain, Hanlessian's stain)
I think this paper may also have pre-dated silica backed TLC plates, which will undoubtedly give you better separation than paper chromatography.

Taken from Nature (Nature Publishing Group)


Why does attractive forces of nucleus increase more than shielding across groups?

Question (1)

To at least a very good approximation, the shielding effect is merely due to a repulsive force.  Like much of chemistry at the atomic level, electrons experience a balance of counteracting forces due to the nucleus and other electrons.

Question (2)

This is usually explained in the following way.  As one goes across a period from left to right, the charge on the nucleus increases at each step, increasing the attractive potential for electrons (which you already noted).  In addition, an electron is added to the valence shell for each step.  Since the added electrons are in the same shell as the preexisting valence electrons, they do not contribute as strongly to shielding as one might initially expect.  Electrons that are in lower shells contribute to shielding more effectively since they are "more likely" to be between a valence shell electron and the nucleus.

Question (3)

One way to justify the increasing effectiveness of shielding in later periods is an extension of the answer above to question (2).  There it was stated that valence shell electrons do not effectively shield other valence shell electrons since they are at the same level.  An natural extension of this would be that the further apart two electronic shells are, the more the lower shell screens the higher shell.  Compared to the valence electrons in Pb, the 1s electrons are essentially right up against the nucleus and, from the view of Pb's 6p electrons, should almost completely neutralize the charge of two protons in the nucleus (i.e. very effective shielding).

Another way of looking at this is that as the principle quantum number (n) increases, the orbitals become less penetrating.  This means they spend less time near the nucleus (where the shielding is weak) and more time further from the nucleus (where shielding is strong).


What is the largest, noncrystalline, non-polymer inorganic molecule?

I agree with the redditors that the answer is likely a polyoxometalate.
$\ce{Na48[H_xMo368O1032(H2O)240(SO4)48] · \text{ca.}~1000 H2O}$ aka
$\ce{\{Mo368\}}$ (core: $M \approx\pu{60.7 kDa},$ dimensions $\approx \pu{2.5 nm} × \pu{4.0 nm})$ is the primary candidate [1]:

Figure 2. Structure of 1a in crystals of 1 in polyhedral (a) and ball and stick (b) representation perpendicular to the $C_4$ axis and along the $C_4$ axis ((c) and (d); building units $\ce{\{Mo1\}}$ yellow, $\ce{\{Mo2\}}$ red, $\ce{\{Mo(Mo5)\}}$ blue with blue-turquoise pentagonal bipyramids; $\ce{O}$ atoms small red spheres, $\ce{S}$ atoms gray spheres).
Note that one capping part is turned against the other by $360/16^\circ.$
The different building blocks and the arrays of different local symmetry are assigned.

A recent review of mass-spectrometry studies of large polyoxometalates is also available [2].
References

Müller, A.; Beckmann, E.; Bögge, H.; Schmidtmann, M.; Dress, A. Inorganic Chemistry Goes Protein Size: A $\ce{Mo368}$ Nano-Hedgehog Initiating Nanochemistry by Symmetry Breaking. Angewandte Chemie International Edition 2002, 41 (7), 1162–1167. DOI: 10.1002/1521-3773(20020402)41:7<1162::AID-ANIE1162>3.0.CO;2-8.
Surman, A. J.; Robbins, P. J.; Ujma, J.; Zheng, Q.; Barran, Perdita. E.; Cronin, L. Sizing and Discovery of Nanosized Polyoxometalate Clusters by Mass Spectrometry. J. Am. Chem. Soc. 2016, 138 (11), 3824–3830. DOI: 10.1021/jacs.6b00070.


The aromatic properties are evident from the heats of combustion, the fact that all bonds have the same length, the ability of porphyrins and metalloporphyrins to undergo electrophilic substitution reactions, the stability of the porphyrin nucleus in electron impact mass spectrometry and the characteristic chemical shifts observed in proton NMR spectra1.

The 18π electron substructure (shown in blue in your figure) was thought to be responsible for the aromatic characteristics. In support of this view it was noted that the outermost bonds in the pyrrolenine rings(the not blue double bonds in your figure) take part in cycloaddition reactions(Fig.1, taken from ref.2) and therefore have a significant double bond character suggesting that their electrons are more localized.

This 18π model is also called "the diaza[18]annulene model" because it was speculated that the aromatic pathway found in porphyrins resembles [18]annulene(fig.2, ref.1). You can also note the similarities in the spectra of [18]annulene(fig.3, ref.3) and porphyrins: both have a major peak in the UV region and a less intense absorption at longer wavelengths.

Other supportive data to the the diaza[18]annulene model are proton NMR studies all mentioned in ref.1

https://www.worldscientific.com/doi/abs/10.1142/S1088424611004063
http://www.scielo.br/pdf/jbchs/v15n6/22664.pdf
Chem. Commun., 2016, 52, 4710


Molecular Oxygen means that it's bonded to another Oxygen (O2). Atomic oxygen would just give you O.


Origin of the 15% difference limit between solute and solvent atom radii in the Hume-Rothery rules

Possibly, there are two extrema to account:
a) the atomic radii of metal A and metal B do not differ significantly, and thus there are chances both metals blend statistically with each other.  Which would be covered by the $< 15\%$-rule, a situation similar to mixing apples and oranges in one crate.
b) the atomic radii differ a lot.  If we assume atoms as hard spheres, even the closest arrangement of spheres of one sort does not occupy fully the space available (see e.g., fcc and hcp packing).  If the diameter of the other metal(s) is/are much less than the one of the host, it is possible for some of these pockets to be filled more, and for others to be filled less. Overall, the distribution of metal B and metal A in the blend of (A .and. B) no longer is even; the smaller atoms may segregate.  As if you would transport a crate of filled with melons and hazel nuts.  This non-regular demixing is possible because -- contrary e.g., to spinels -- you may assume absence of ions charged oppositely charged.
For the purpose of illustration, both the examples as well as the figure below intentionally exaggerate the contrast between case a) and b).


@Loong's encyclopaedic knowledge of IUPAC nomenclature prompted me to have another look through the (irritatingly unsearchable) 2013 recommendations and the 2004 draft upon which they were built.
The only use of the uppercase Δ listed is when describing a double bond that links two cyclic systems:

P.28.2.2: Ring assemblies with a double bond junction
When two cyclic systems are linked by a double bond [...] the presence of a double bond is indicated by the Greek letter Δ, and the point of attachment of the ring is given by superscript locant numbers. This method is not continued in these recommendations

The example given is 1'1-bi(cyclopentylidene) (PIN) which should be used in preference of Δ1,1'-bicyclopentylidene.

Based on this, the example given in the question, using Δ to describe an unsaturated carbonyl would appear invalid, and used for convenience rather than to convey anything specific.


Empirical formula when given percentages?

You will get the empirical formula straight from this data. In order to convert to a molecular formula, you need more information.

By way of another example:

  A binary compound of tin and chlorine contains 62.89% tin by mass. Determine the empirical formula.

Our first goal is to figure out the ratio of moles of tin to chlorine in this compound. Once we know the ratio of moles, we can determine the formula.

What do we know? We know what percentage of tin is in this compound by mass. What is the percentage of chlorine?

In order to get to moles, we need at least a mass of the sample. We were not given that. Since percent by mass is a ratio of masses, it turns out we can choose any mass of sample we want and get the same formula (Try it yourself!).

If we assume that we started with 100.0 g of this stuff, then how many grams of tin would you have? How many grams of chlorine would you have? 100.0 g is a nice starting mass since you are working with percents.

Now, as Ari's answer says, you need to convert mass of tin and mass of chlorine into moles of tin and moles of chlorine by dividing each mass by the atomic molar mass of each element (look it up on your periodic table).

Now that you are in moles, determine the lowest whole number ratio of the two elements by dividing each number of moles by the smaller number of moles. If your numbers are close to whole numbers, you can round. For example, round 1.97 to 2. If your numbers are not close to whole numbers, you may need to do a little more arithmetic:

If you get a number like 1.48, this is close to 1.5 which is 3/2. Multiply both numbers by 3.
If you get numbers like 1.34 (close to 4/3) or 1.67 (close to 5/3) multiply by 3.

Now you have the empirical formula. As Ari's answer says, you need a molar mass of the species to get a molecular formula.

The answer for my example is $\ce{SnCl2}$. Work through the steps and see if you can get to this answer. Then repeat for your lead and oxygen compound.


You did not specify at which geometry you have convergence problems, so at first I thought it happens when all three $\ce{N}$ atoms are far away from each other. But then I run the calculation by myself and found out that it is not the case: the SCF convergence failure happens at R1=1.2 and R2=1.1. At this geometry SCF starts to oscillate after 20 iterations back and force as shown in a graph below, so simply increasing the number of SCF iteration would not help.

For that reason I quickly checked few other well known tricks to avoid the problems with SCF convergence (see, for instance, here) and I recommend to use a quadratically convergent SCF (QC-SCF) by adding scf=qc to the route section:

# HF/6-31G test scan scf=qc

Note though that while QC-SCF is more reliable than the default DIIS algorithm it is also slower, so expect an increase in the computation time. For that reason, there is also an interesting option scf=xqc which first tries the default algorithm and switches to a quadratically convergent one only if the default has not converged.

# HF/6-31G test scan scf=xqc

However, switching to a quadratically convergent algorithm did not help, since SCF still fails to converge but at some other geometries. So, I investigated the matter more closely, and in addition to scf=xqc I also recommend to improve both radial and angular flexibility of you basis set by adding diffuse and polarization functions into it

# HF/6-31+G(d,p) test scan scf=xqc

After running the calculation for an hour it reached 2656 points (which is about 40% of all points) without SCF convergence problems, but in case if this variant still fails:

Consider lowering down the convergence criteria by changing scf=xqc to scf=(xqc,conver=n) with n being, say, 6 (it is 8 by default).
If failures happen when at least one $\ce{N}$ is far away from two others, consider lowering down the maximum distance for your PES scan. In my opinion 8.5Å is way to big, 5Å would probably be more than enough.

I finally succeeded with the following route section

# HF/6-31+G(d,p) Scan SCF=(XQC,Conver=6)

Here are the results and below is the plot of the PES. :D


Following the advice of @Buck Thorn and @user1271772, I asked this same question over at the Cross Validated Stack Exchange.
There, I received a very good answer and am providing the link for the interested reader: % coefficient of variation (%CV) for log-linear and log-log regression (calibrations)
To summarise: yes, the formula is appropriate where the predictor (concentration) and response (peak area/other metric) have both been log-transformed.


In which cases do you say "oxide" when naming a molecule?

The naming nomenclature for inorganic molecules is confusing, but if you are studying chemistry then you just have to grit your teeth and learn.

First there are household names, common names and chemical names. Vinegar, acetic acid, and ethanoic acid are talking about the same molecule. We say water, but aqueous solutions. $\ce{SiO2}$ is silicon dioxide to a chemist, but quartz to a geologist. The list could go on and on...

In general an oxide is compound which has just some element, X, and oxygen in some combination - ${\text{X}_m\text{O}_n}$

Some anion combinations get a name.

$\ce{CO3^{2-}}$ is carbonate
$\ce{SiO3^{2-}}$ is metasilicate
$\ce{CrO4^{2-}}$ is chromate


How to calculate the depression in freezing point of a solution in which there are multiple solutes being added?

Freezing point depression is a colligative property, well defined by this Wikipedia article (emphasis mine):

In chemistry, colligative properties are properties of solutions that depend on the ratio of the number of solute particles to the number of solvent molecules in a solution, and not on the nature of the chemical species present.

This means the problem is the same regardless of the number of different solvent species. You simply add them up to get the total solute concentration and calculate the freezing point depression in the same way you would if they were all the same species.  As you mentioned, you do need to keep track of dissociating species like $\ce{KCl},$ i.e. $\pu{1 mol}$ $\ce{KCl}$ dissociates into $\pu{2 mol}$ of total solutes.


What is the correct numbering & order for this molecule?

The parent hydrocarbon chain for this molecule will be the 7 membered ring.

2.There are no functional groups with suffixes.
3.A halogen substituent(Bromine) and two side chains are present. ( methyl and t-butyl or (1,1-dimethylethyl) )
Having identified all these we can proceed to numbering the parent hydrocarbon chain.
After numbering the Parent hydrocarbon chain in all possible ways(ideally) we choose that numbering that satisfies the following criteria:
Numbering of the chain. This is done by first numbering the chain in both directions (left to right and right to left), and then choosing the numbering which follows these rules, in order of precedence.
1.Has the lowest-numbered locant (or locants) for the suffix functional group. Locants are the numbers on the carbons to which the substituent is directly attached.
2.Has the lowest-numbered locants for multiple bonds (The locant of a multiple bond is the number of the adjacent carbon with a lower number).
3.Has the lowest-numbered locants for prefixes.
In the above molecule since there are no suffix functional groups or multiple bonds we can apply criterion 3.
After numbering the parent chain all possible sets of locants are written down in the ascending order.(in this case)
eg: If the set of locants are (2,4,3) and (1,4,3) we write them as (2,3,4) and (1,3,4)
Then choose the one with the lowest possible locants. First compare the first locant, choose the ones with the lowest locant, among these sets of locants compare the second locant, choose the ones with the lowest locant and so on...
eg: If the sets of locants are (1,3,8) and (1,4,5) we choose (1,3,8)
It is only when we get two different numbering for the same compound following all the above mentioned rules that we should choose the set of locants which assign the lowest locant to the substituent or side chain on the basis of alphabetical precedence.


Crystallographic terminology associated with the honeycomb structure

the concept of lattice is very (too!) often misinterpreted outside crystallography. By definition, the lattice points are all identical, that is each lattice point has identical surroundings to all others. In a honeycomb pattern you can distinguish 2 different sets of points, the red and the blue ones in the figure.
So honeycomb pattern is actually a 2-dimensional structure (not a lattice), resulting from the convolution of a basis (in this casis constituted of one blu + one red point) with a 2D lattice (the hexagonal one in the figure representing the five nets)


This ism the microwave structure of nitrous acid, 1971.  Your mileage may vary condensed phase.

DOI:10.1039/TF9716702179 (1971)
The structure of cis-nitrous acid is found to be: O—H = 0.982, N—O(H)= 1.392, N—O = 1.185 Å, ∠ NOH = 104.0 and ∠ ONO = 113.6°, and for trans-nitrous acid: O—H = 0.958, N—O(H)= 1.432, N—O = 1.170 Å, ∠ NOH = 102.1 and ∠ ONO = 110.7°.

http://nzc.iap.ac.cn/uploadpdf/Wang_et_al._CSB_2007a.pdf
  2007 pape with more recent citations.


Why is maximum buffer capacity for some diprotic buffers not when pH = pKa?

Any aqueous solution increases its buffer capacity toward $\mathrm{pH} = 0$ or $\mathrm{pH} = 14$, as more and more acid or basis is needed to change $\mathrm{pH}$ by a given interval. It happens even without presence of any buffering substance, aside of strong acids or bases.
If an acid, monoprotic or diprotic, is added, a peak(s) of maximum buffer capacity is added. But if $\mathrm pK_\mathrm a$ is small, the peak of maximum buffer capacity does not lay on a steady baseline of the capacity function. It lays on a sloped baseline. If the maximum absolute peak slope is smaller than the absolute baseline slope, the peak does not manifests itself  as a peak(=maximum capacity), but rather as just as a shoulder on the slope, without a maximum.
This phenomena is not specific for chemistry, but rather for mathematics and mathematical analysis. If you have a sinusoid function on a large enough slope, its peaks would disappear at some slope value, becoming just a shoulders.
The similar phenomena occurs in case of multiple $\mathrm{p}K_\mathrm{a}$ in the solution, of the same or of different compounds. On the background of non steady capacity function, a particular peak may not have a shape of a peak.
Additionally, the presence of close peak neighbors affects position of the peak maximum.
If 2 peaks of the similar height are close enough each other, their maxima shifts towards each other, and with enough closeness they merge into a single peak.


You should contact a specialized company, however you should also provide more infos about the diameter and the number of the holes you want. This is an excerpt from FedTech:

  laser cutting is the method of choice for small perforation, as laser
  cutting has a smaller, tighter cutting tolerance and thus has the
  capability to create tiny, intricate perforations. However, if you
  prefer, abrasive waterjet cutting can be used as well but for larger
  perforation (.030" and up, respectively). If you prefer to use
  non-abrasive waterjet cutting, you can create perforations up to .004"
  in diameter, however this form of waterjet cutting is restricted in
  cutting certain types of material. With non-abrasive waterjet cutting,
  you can cut up to about 12" thick from the following: plastic, rubber,
  foam, composites, silicone, foam, felt, cork, Teflon (PTFE), neoprene,
  and other soft materials


How to know which atom is the central atom in VSEPR theory?

You know that phosphorus has 5 valence electrons and that chlorine has 7 valence electrons ... and that you somehow have to combine them.

As chlorine is more electronegative (Allred-Rochow: Cl 2.83, P 2.06) you know that each chlorine will attract one electron from the phosphorus to get the noble gas configuration with 8 electrons in the valence shell.

Now that in turn means that there will be one single bond between the phosphorus and each chlorine, which automatically makes phosphorus the center.


The reaction as pictured above is incorrect, the nitro group is not reduced.  As you suggest this is (predominantly) an electrophilic aromatic substitution reaction.

Usually $\ce{HF}$ is used as a catalyst in these reactions and the following equilibrium is established.

$$\ce{XeF2 + HF <=> XeF+ + HF2^-}$$

The electrophilic $\ce{XeF^+}$ is the reactive species forming a complex with the aromatic ring.  Xenon is expelled from the complex and the reaction proceeds to yield the meta-fluoro isomer as expected for an electrophilic aromatic substitution.  Some other products, such as substituted biphenyls, are usually formed in small amounts suggesting that some minor pathway (radical-cation?) is also in play.


Diazo transfer to 1,3-dicarbonyl with tosyl azide

This starting material is not actually an enone; it is a 1,3-dicarbonyl, confusingly drawn as the enol tautomer. 1,3-Dicarbonyls are mildly acidic ($\mathrm pK_\mathrm a \sim 10$) and can be deprotonated by weak bases, such as triethylamine ($\mathrm pK_\mathrm a \sim 9$), which was mentioned in the question. So, now, we have an enolate ion which is nucleophilic.

Tosyl azide, $\ce{TsN3}$, is an odd reagent. It's easy to think that it is acting as a nucleophile of some sort, because the azide ion is nucleophilic, but actually it is an electrophile.

The next part is the removal of the formyl group, which is a bit tricky. Here I am not 100% sure, so take it with a pinch of salt, but I think Mith was on the right track with the retro-(3+2) cycloaddition. This is my guess:

In fact, the cyclisation of the terminal nitrogen onto a carbonyl group has been proposed before, in mechanistic studies of electrophilic azidation by D. A. Evans.1 That makes me feel a bit more confident in my guess!

Anyway, the product above is simply a resonance form of the diazo product you want.

Reference

Evans, D. A.; Britton, T. C.; Ellman, J. A.; Dorow, R. L. The asymmetric synthesis of α-amino acids. Electrophilic azidation of chiral imide enolates, a practical approach to the synthesis of (R)- and (S)-α-azido carboxylic acids. J. Am. Chem. Soc. 1990, 112 (10), 4011–4030. DOI: 10.1021/ja00166a045. Non-paywall version available from the Evans group website.


Can high-tin bronze be made sharper than low-tin bronze?

Unfortunately I don't have a solid reference for this, but what I remember from my undergrad courses is that the limit to sharpening is basically determined by the porosity and tensile strength of the metal you are trying to sharpen. At some point you will just run into the microscopic roughness of the metal itself and that is where sharpening ends.

Low- and high-tin bronze are pretty rough classifications so I couldn't tell you which will result in a lower porosity or higher strength, but if I would have to guess than I think low-tin bronze can be made sharper, because copper has a higher strength than tin.

Maybe if you can get your hands on this book chapter, you can find out a more exact answer.


Number of water molecules to contribute one molecule to every milliliter on earth

A very quick Googling tells me that the volume of the oceans is 1.3 billion $\mathrm{km}^3 = 1.3\times10^9\;\mathrm{km}^3 = 1.3\times10^{18}\;\mathrm{m}^3 = 1.3\times10^{24}\;\mathrm{cm}^3 = 1.3\times10^{24}\;\mathrm{mL}$.

To get one molecule into each of those mL, we'd need $1.3\times10^{24}$ molecules of water, or $\frac{1.3\times10^{24}}{N_A} = \frac{1.3\times10^{24}}{6.02\times 10^{23}}$ moles of water, about 2.2 mols.  Since water is 18 g/mol, that comes out to 39 grams of water or about 39 mL.


Do polar compounds travel further or less far than unpolar ones in hexane/ethyl acetate TLCs?

Never have I seen a normal phase (i.e. silica gel or alox) TLC where a more polar compounds travels further than a less polar compound — and I have used some extremely (considering normal phase TLC) polar eluents in the past.

The top-running compounds are always the least polar and the slowest are always the most polar in normal phase TLC.


Can one exclude calibration points using a combination of permutation testing and a non-parametric test statistic?

This is more of a comment, but a typical comment doesn't allow a figure...

You have a very wide range (1-1000) for the independent variable. Using a simple linear least squares analysis isn't appropriate.

I grabbed the image below from webpage https://www.mathworks.com/matlabcentral/answers/392636-confidence-band-around-linear-least-squares-line

The point I want to make is shown by the shaded area. Assuming homoscedasticity, you wouldn't weight the errors. Thus the confidence interval for the line is narrowest at the center of the fitted range and gets much wider at either end of the range. So your low samples ( 1, 2, 5, and 10) will essentially be irrelevant.

So you need to assume hetroscedasticity and weight the sample errors somehow.

  1. Does the proposed test statistic seem reasonable as a non-parametric replacement to the t-statistic?

Just because in general non-parametric tests are less powerful than parametric tests, I would try hard to avoid using a non-parametric test on parametric data.

  2. Is it appropriate to exclude samples from the regression, for which the p-value for the test statistic exceeds the defined threshold?

I'll assume that you mean to exclude some of the standards not samples. It is possible to exclude standards, but don't use three standards and then drop the one that you "know" isn't on the line. :-)

Hope this helps a little. I just don't know enough about S/N for LC/MS or the detector system that you're using to tell you what you should do.

Your statement "Given that, for many compounds, the data are often non-‘normal’ around the lower-end of the calibration curve in such dataset,..." is disconcerting.


It is the measured overall reaction rate that counts and not which reaction finishes the sequence.
If you measure the rate of a reaction that consists of two elemental reactions with rate constants $k_1$ and $k_2$ where $k_1 \gg k_2$ you will get an overall rate constant $k_{\mathrm{tot}} \approx k_2$.
This is so, because if the faster reaction takes place after the slower reaction, the faster reaction uses up its reactants (which are the products of the slower reaction) faster than the slower reaction produces its products, so that the faster reaction will consume every new product molecule produced by the slower reaction nearly instantly compared to the time it takes the slower reaction to produce it.
The time it takes the faster reaction to consume the newly formed molecule is insignificant.


Why the "Diverse Ion Effect" increases Ksp?

To illustrate the "diverse ion effect", we will study how the solubility of $\ce{AgCl}$ changes in absence and presence of $\ce{NaNO3}$, given that the molar solubility product of $\ce{AgCl}$ is $1.76 \times 10^{-10}$.

The exact definition of the solubility constant is: $$K_{sp}= a(\ce{Ag+}).a(\ce{Cl-})=[\ce{Ag+}].\gamma(\ce{Ag+}).[\ce{Cl-}].\gamma(\ce{Cl-})$$
Where $a(\ce{Ag+})$ and  $a(\ce{Cl-})$ are the activities of these ions, and $\gamma(\ce{Ag+})$ and $\gamma(\ce{Cl-})$

In dilute solutions, i.e. absence of $\ce{NaNO3}$:  $$\gamma(\ce{Ag+})=\gamma(\ce{Cl-})=1$$
So, the solubility of $\ce{AgCl}$ is equal to $$[\ce{Ag+}]= \sqrt{K_{sp}}= 1.33 \times 10^{-5}$$
In presence of $\ce{NaNO3}$ of a concentration of $0.1 \mathrm{mol/kg}$, then
$$[\ce{Ag+}]= \sqrt{K_{sp}}/ \sqrt{(\gamma(\ce{Na+}).\gamma(\ce{NO3-})}
= 1.33 \times 10^{-5}/ (0.769)= 1.73 \times 10^{-5} $$
i.e. in the presence of a soluble salt, the molar solubility of a slightly soluble salt increases.


How are there three possible tautomers of 2,2,4-trimethylheptane-3,5-dione?

In your attempt the 4th compound is not reported to have significant percentage in solution (acidic or basic). This is because once the tautomerisation occurs for example let's say the 1st compound, there is a tautomer having a six-membered conjugated ring involving H-bonding. This thereby increases the stability of the compound and consequently increasing the activation energy for the tautomerisation you proposed in compound 4. This is why compound 4 is not formed.

Also keep in mind that stereo-chemistry is also important while finding number of possible compounds.


Relation between moles, molar mass and mass

If you have 50 g of a compound that has a molar mass of 213 $\frac{\rm{g}}{\rm mol}$, then you have $\frac{50~\rm{g}}{213 \frac{\rm{ g}}{\rm mol}}=\frac{50}{213} \rm{mol}=0.234~\rm{mol}$ of the compound.

If you want to prepare a 0.8500 molar solution, that means the concentration you want is $0.85~\rm{M}=0.85\frac{\rm{mol}}{\rm{L}}$.

With those two numbers, the final answer should be apparent.  (Just divide them and make sure the units work out!)


How does steam prevent oxidation in tea leaves?

According to Tony Gebely at Tea Epicure (https://teaepicure.com/tea-leaves-oxidation/), it's not steam per se, but rather heating that slows oxidation, by denaturing two enzymes (polyphenol oxidase and peroxidase) that are responsible for it.  Steaming is simply one way to do this:

Much of the oxidation process revolves around polyphenols, particularly the enzymes polyphenol oxidase and peroxidase. When the cells inside tea leaves are damaged and the components inside mix and are exposed to oxygen, a chemical reaction begins. [Specifically when polyphenols in the cell’s vacuoles and the peroxidase in the cell’s peroxisomes mix with polyphenol oxidase in the cell’s cytoplasm. From Michael Harney’s 2008, “The Harney and Sons Guide to Tea.”] This reaction converts the polyphenols known as catechins into flavanoids called theaflavins and thearubigins (which are also polyphenols).

Fixing works by denaturing polyphenol oxidase and peroxidase – the enzymes primarily responsible for oxidation using heat. Fixing is also commonly referred to as de-enzyming, denaturing or kill-green.

The term kill-green is derived from the Chinese term shaqing (杀青), which translates to killing the green. The tea leaves must be heated to approximately 150 degrees Fahrenheit to “halt” oxidation. Oxidation is further slowed by drying the leaves, but it never completely stops. At temperatures over 150 degrees Fahrenheit, oxidation continues to occur at an extremely slow pace.

The process of fixing requires precise control of the temperature and length of heating; each has to be adjusted depending on the size and thickness of the leaves and the amount being processed.

Note also that, for teas, oxidation is not a bad thing.  Rather it is a parameter that is precisely controlled in order to achieve the desired type of tea.  Green teas are produced by minimizing oxidation. Oolongs are produced by allowing oxidation to proceed part-way (and they are thus known as "semi-oxidized teas").  Black teas are produced by allowing oxidation to proceed completely, or nearly so:

Oxidation refers to a series of chemical reactions that result in the browning of tea leaves and the production of flavor and aroma compounds in finished teas. Depending on the type of tea being made, oxidation is either prevented altogether or deliberately initiated, controlled and then stopped. [emphasis mine]

When oxidation is prevented altogether, the catechins are left largely intact. The tea leaves keep their green color and vegetal characteristics.

When a semi-oxidized tea is being produced, some catechins convert to theaflavins and thearubigins, resulting in a slight browning usually along the edges of the leaves and yellower liquor. Lipids, amino acids, and carotenoids also begin to break down into flavor and aroma compounds.

When oxidation is allowed to run its course, the leaves exhibit an aroma and taste profile completely unrecognizable from a finished tea that was exempted from oxidation.


Nonylphenols are a family of closely related organic compounds composed of phenol bearing a 9 carbon-tail (Wikipedia). Following is an example of nonylphenols:

However, "Nonylphenol $n$ mole" is a commercial name for non-ionic surfactants where $n$ stand for the amount of ethelene oxides (EO) in moles in one surfactant molecule. For example, if the non-ionic surfactant is $\ce{R-O-(CH2CH2-O)n-H}$ where $\ce{R}$ is hydrophobic tail of the surfactant, the commercial name "Nonylphenol 10 mole" means $\ce{R-O = 4-C9H17-C6H4-O}$ and $n = 10$ in the structure. An example for a linear chain non-ionic surfactant is pentaethylene glycol monododecyl ether:

where $\ce{R-O = C12H25-O}$ and $n = 5$ (available in Sigma-Aldrich).


Maybe NFS doesn't work well?  Bear in mind that bonds from a second-period element to fluorine are generally stronger and less labile than the corresponding bonds to heavier halogens. If that's true, getting fluorine off a nitrogen-fluorine bond requires more work than, say, getting bromine off a nitrogen-bromine bond.

In https://en.m.wikipedia.org/wiki/Electrophilic_fluorination#Mechanism_and_stereochemistry several common fluorinating reagents are given, these being the compounds mentioned by the OP.  The listed compounds all withdraw electrons from the nitrogen-fluorine bond more strongly than the carbonyl groups would in NFS (they use sulfonyl groups or a cationic function), so it seems this attribute is necessary to get the fluorination to go.


A copper mirror can be obtained by heating two or three grams of dry copper formate $\ce{Cu(HCOO)2}$ in a test tube with a Bunsen burner. For some reason, the copper atoms are deposited on the inner surface of the test tube. Try it! It is easy to do!


Is that correct to say that osmosis happens with water only?

No, it is not correct because it happens with other liquids (solvents) also.

Taking Wikipedia's definition:

  Osmosis is the spontaneous net movement of solvent molecules through a
  semi-permeable membrane into a region of higher solute concentration,
  in the direction that tends to equalize the solute concentrations on
  the two sides.

Only as extra information: osmosis occurs until chemical potentials of every substance on each side of the membrane are equal. As the substance cannot move through the membrane, the liquid (solvent) do.

According to the definition, the book isn't correct, because it says with water only, but take into account that in biology the solvent is always water. So, as an internal definition of biology (or physiology) maybe is not wrong.


I have wondered about this too. Although I could be wrong it is reasonable to assume (or at least handy to keep as a reminder) that the term "special" refers to the fact that compound is protonated in a very specific state, namely before the actual reaction occurs. General acid catalysis can happen both before the reaction or at any time during the reaction (i.e. in some transition state), whatever is likely in some range around the pH of the acid. Hence it is more general version of the special case mentioned above.


Why does crystalline solids have well defined 3d geometry

Keep two concepts separate: the solid crystalline state is not bound to occur on ionic compounds only.

In a simple view, crystalline describes a regular arrangement of matter, with predictable order both in short, as well in long range.  For the crystalline state, it is unimportant if the constituents of the crystal are atoms (like metal atoms), charged particles (like your example of NaCl), or neutral molecules (like benzoic acid) or even a mixture of neutral and charged molecules (many protein crystals contain water).

Some materials are easier packed in such a regular pattern, than others.  In the instance of metals, their (on first approximation) sphere like atomic geometry and similar radius eases such a settlement.  Inorganic salts tend to be a bit more complicated, as often smaller ions are localized in gaps built by larger ions (octahedral / tetrahedral gaps).

Organic molecules may equally build crystals, yet their shape and conformational flexibility around single bonds (for example in alkyl chains) and multiple intermolecular interactions may keep them over a long time in a glassy state (like supercooled benzoic acid, or some brands of honey).  More than for the aforementioned examples: formation of a nice crystal may be thermodynamically favourable, but it is not enough "just to cool down" below the melting temperature to reach this energetic minimum.  There must be seeds of crystallization, and typical conditions of their formation (temperature, concentration, etc.) are not necessarily the same where nicely shaped crystals growth further.

It is not surprising crystals attract attention, for example in material science and crystallography, and that there are journals with focus on the relationship between molecular properties, and macroscopic / crystallographic ones; like CrystEngComm or Crystal Growth & Design.


Do self-growing hydrogels grow and strengthen infinitely?

"Strength" is a dubious quantity. The E-modulus will rise, and as the material swells with more and more monomer diffusing in, it will at some point rupture when the brittle network cracks. Like a fruit gum that was put in water overnight.

I have no idea if the two networks actually stay separated, or if they interconnect more and more. I suspect the latter, with radicals at work. That would make the brittle part softer -> no further hardening, and the soft network more brittle -> it snaps under increasing load.

But the point is moot imho: "Self-healing" makes sense if you had a smaller damage due to light overload. Then this material here can become twice as hard, and you're fine. I don't think it makes sense to "train" sth to become fifty times as strong.


Differentiating esters from glycosidic linkages

In carbohydrates you can find examples of glycosidic linkages; the following example is α-maltose and the image is taken from Wikipedia, where a full list of authors is available.

This following example shows an ester; I chose one of the most simple esters (ethyl acetate) due to boredom. Again, the image is from Wikipedia, where a full list of authors is available.

One side of the molecule — the right-hand side in both of my examples here — always features a typical hydroxy group, i.e. is an alcohol. The difference is on the other side, the coupling partner. Look closely at maltose, and you may spot that the first carbon of the left-hand fragment also contains a hydrogen; its oxidation state is that of an aldehyde. Compare it with ethyl acetate, where the left-hand fragment linking carbon is in the oxidation state of a carboxylic acid (i.e. double bonded to an oxygen).

Technically speaking, glycosidic linkage is just a special case of acetal or ketal linkages; that is the correct name for the functional group $\ce{R^1O-CR^2R^3-OR^4}$ — acetal if either $\ce{R^2}$ or $\ce{R^3}$ is a hydrogen, otherwise ketal if both are carbons. ($\ce{R^1}$ and $\ce{R^4}$ must be carbon-based or we are dealing with hemiacetals/hemiketals/hydrates.) Examples of ketals for glycosidic linkages exist.

(And technically, ortho-esters exist, too, in which the linking carbon is bound to three $\ce{OR}$ groups. But these are very rare and special cases, even though they even exist in natural products such as tetrodotoxin.)


All credit to Zhang et al. "Real-Space Identification of Intermolecular Bonding with Atomic Force Microscopy" Science Vol. 342  no. 6158  pp. 611-614.

Yes, direct images of bonds, not only covalent bonds but also intermolecular hydrogen bonds have been recorded.

It is the electron density that is being observed, covalent and hydrogen bonds involving high electron density between the atoms.

Scanning Tunneling Microscopy can also be utilized to directly observe the electron density of bonds.


How to use a spectrometer to determine NPK

Your spectrometer detection range is 340-780 nm. That is to say mostly visible, and a bit of UV. If your compounds are colorless, it's a first indication that you won't see anything on your spectra. These compounds seem to be inorganic, no transition metal or anything exotic, so it's not a good method.

Apart from this, yes, you would need to calibrate with solutions of known concentration (unless you know the absorption coefficient, which would give you the concentration with Beer-Lambert law).


You may want to read the Wikipedia article on TMAH.  While it does decompose, it may not produce the methanol you are expecting.  Also, heat is also involved in this process, so you may need to heat and, in the process, vaporize much of the water to get this decomposition to go.

TMAH, in common with many other TMA salts containing simple anions, decomposes on heating into trimethylamine.[1] Dimethyl ether is a major decomposition product rather than methanol.[2] The idealized equation is:

$\ce{2 NMe4+OH− → 2 NMe3 + MeOMe + H2O}$

Such a rearrangement suggests an irreversible reaction.  The Wikipedia article also indicates that TMAH is not obtained by combining trimethylamine with methanol anyway.  Instead, a metathesis between a tetramethylammonium salt such as the chloride and an alkali hydroxide is used.
Cited References:
1.
Lawson, A. T.; Collie, N., "The action of heat on the salts of tetramethylammonium". J. Chem. Soc., Trans., 1888,53, 624-636. doi:10.1039/CT8885300624.
2.
Musker, W. Kenneth. (1964). "A Reinvestigation of the Pyrolysis of Tetramethylammonium Hydroxide". Journal of the American Chemical Society. 86 (5): 960–961. doi:10.1021/ja01059a070.


What is the stereochemistry resulting from hydroboration of 1-methylcyclopentene?

The second step is a syn addition of the boron hydride. As depicted below, it will result in an anti-Markovnikov product with the boron on the less-substituted carbon. Both hydrogen and boron are on the same side of the former double bond and these two are the ones we refer to when we say syn addition.

The final step, the oxidation by hydrogen peroxide, is stereospecific with retention. So the newly-formed $\ce{C-O}$ bond is on the same side as the former $\ce{C-B}$ bond. Thus, an anti relationship is established between the methyl group and the hydroxy group.

Since 1-methylcyclopentene is planar, a racemic mixture of enantionmers is formed; neither re nor si side are favoured. The scheme below shows the attack from the si side of the $\mathrm{C2}$ atom.


Confused about the direction in which the bond will break

The reactions happen on different centres because they follow different mechanisms.

Epoxides are actually surprisingly stable. They don’t undergo spontaneous ring-opening reactions by themselves unless the oxygen atom has been protonated previously. That is what the $\ce{H+}$ in reaction path X is there for. After having protonated the oxygen, the ring-opening reaction is charge-neutral and thus releiving the Baeyer strain to generate a tertiary (more stable) carbocation is favourable. This carbocation can be captured by a nucleophile in an $\mathrm{S_N1}$-type mechanism.

This does not happen under basic conditions as removing a negative charge would not help the epoxide opening. However, under these conditions solvent molecules such as methanol are easily deprotonated creating much better nucleophiles. These nucleophiles are then able to attack the epoxide under $\mathrm{S_N2}$ conditions. Hereby, an anion attacks a neutral molecule to generate an anion, so the ring-opening is again charge-neutral. However, $\mathrm{S_N2}$ attacks are more likely to happen on less hindered, primary carbons.

Taking this into account, path X that follows an $\mathrm{S_N1}$ path, generates the most stable carbocation hence the nucleophilic attack happens on the tertiary carbon. Path Y that follows an $\mathrm{S_N2}$ mechanism includes an attack on the less hindered, i.e. primary carbon.


What advantage is there in reporting the Lennard-Jones well depth in wavenumber units?

I don't think there's any inherent advantage as one can always convert to another unit. It seems natural to me, however, because chemists generally have a good idea of the energy of vibrational and rotatial states in $cm^{-1}$, so being able to compare the depth of the well with some familiar chemical process is quite nice. For instance, if the well-depth were quite small, one could immediately identify that the system may not spend a lot of time there if it is comparable to the zero-point energy or something similar.

It's similar to how physicists report everything in electron-volts ($eV$). They do it because it's familiar and they have many physical processes they can use as a reference. For instance physicists know that the band gap of semiconductors is between $2-5\ eV$, so it's quite natural to report other energies in the same unit.


$K_\mathrm a\times K_\mathrm b = K_\mathrm w$.  Some teachers tell their students that the conjugate base of a weak acid is strong but it's not true.  What is true is that the stronger the acid, the weaker the conjugate base and vice versa.

Yes to your second question.  When calculating the pH of a solution, the hydronium ion concentration is usually controlled by the strongest acid in the solution.  This is the case in the type of problems found in lower level chem classes.  In a higher level class problems with acids close enough in strength so that both acids matter may be encountered.


What does the slash in polymer nomenclature, e.g. Acrylates/ Beheneth-25 Methacrylate Copolymer, mean?

Acrylates/Beheneth-25 Methacrylate Copolymer is an INCI name for a copolymer

  … formed from the ester of 2-methylpropenoic acid and docosanol, ethoxylated, 25 mol EO (average molar ratio) and one or more monomers of acrylic acid, methacrylic acid or one of their simple esters

(Source: cosmetics.specialchem.com)

Also known by the trademark Aculyn™ 28.
Dow Chemical provides cumulative MSDS datasheet (pdf) for the entire Aculyn family, so I assume you can use it for your copolymer as well.


Why does the Born equation give the Gibbs free energy of solvation rather than enthalpy of solvation?

1) Because $-\Delta G$ is the maximum possible nonexpansion work done by a system in a constant temperature and pressure process. You can write $$\Delta G \le w_{non PV}\quad (\text{constant T and P, closed system})$$
where the equality sign holds for reversible changes.

Here's proof:
$$\begin{align*}
H &= U + P V&\text{(definition of enthalpy)}\\
dH &= dq + dw + d(PV) &\text{(because dU = dq + dw)}\\
dG &= dH - T dS - S dT &\text{(because G = H - TS)}\\
&=dq + dw + d(PV) - T dS -S dT &\\
&= dq + dw + d(PV) - T dS & \text{(when the change is isothermal)}\\
&= dw_{rev} + d(PV) &\text{(if reversible,}\ dq_{rev} = T dS\text{)}\\
&= dw_{electrical,rev} + dw_{expansion,rev} + P dV + V dP & \text{(partitioning the work)}\\
&= dw_{electrical,rev} + V dP & (dw_{expansion,rev} = - P dV)\\
&= dw_{electrical,rev} & \text{(if pressure is constant)}\\
\end{align*}$$

2) It isn't the scaling to one mole that makes the result a Gibbs free energy (see 1); that just makes it a molar Gibbs free energy. You can use either extensive or intensive Gibbs free energies in calculations; whatever is most convenient. Just make it clear (either by context or notation) which one you're using.


Why is 3-chloro-5-(1-methylethenyl)cyclopenta-1,3-dien-1-ol named so?

The most important simplified criteria for the numbering are:

lower locants for the group that is expressed as suffix
lower locants for multiple bonds
lower locants for prefixes
lower locants for substituents cited first as a prefix in the name

The corresponding actual wording in the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) reads as follows:

  P-14.4 NUMBERING

  When several structural features appear in cyclic and acyclic compounds, low locants are assigned to them in the following decreasing order of seniority:

  (…)

  (c) principal characteristic groups and free valences (suffixes);

  (…)

  (e) saturation/unsaturation:

    (i) low locants are given to hydro/dehydro prefixes (…) and ‘ene’ and ‘yne’ endings;

    (ii) low locants are given first to multiple bonds as a set and then to double bonds (…);

  (f) detachable alphabetized prefixes, all considered together in a series of increasing numerical order;

  (g) lowest locants for the substituent cited first as a prefix in the name;

  (…)

The principal characteristic group of the compound that is given in the question is the $\ce{-OH}$ group. It is expressed as the suffix “-ol”. According to Rule (c), a low locant is assigned first to this group, i.e. “-1-ol”.

In the next step, low locants are assigned to the double bonds according to Rule (e), i.e. “cyclopenta-1,3-dien-1-ol”.

That leaves only one possible locant set for the remaining substituents, i.e. “3-chloro” and “5-(prop-1-en-2-yl)”. The rules (f) and (g) are not needed to make a decision.

Therefore, the complete name for the compound that is given in the question is 3-chloro-5-(prop-1-en-2-yl)cyclopenta-1,3-dien-1-ol.

Note that the preferred prefix for the “isopropenyl” or “1-methylethenyl” (completely: “1-methyleth-1-en-1-yl”) is now “prop-1-en-2-yl”.


This is not the correct definition of salt. The IUPAC definition of salt is "a chemical compound consisting of an assembly of cations and anions". In fact there are many salts which do not fit the definition you give.


What is the relation between surface tension and initial molecular harmonic?

Surface tension arises from the fewer neighbors that molecules at the surface have, relative to the number of neighbors for molecules in the bulk.

For example, consider a simple model of rigid identical spheres in a hexagonal closest packing.  In the bulk, there are 12 spheres in contact with a given sphere, which could be thought of as 6 spheres at vertices of a regular hexgon surrounding the given sphere, 3 spheres in a triangle below and 3 spheres in a tringle above.

At a surface, at least 3 of the neighboring spheres would be absent.

Considering an interaction energy between neighboring spheres, the loss of interaction energy (due to fewer than 12 neighbors per sphere) will be proportional to the surface area of the system of spheres.  The minimum energy of the system will be when surface area of the arrangement of spheres is minimized.

For the above reason, the units of surface tension are Energy/surface area, which the OP correctly points out is the same as force/distance.

From the above explanation, particularly that even for rigid spheres always in direct contact the Energy/surface area (force/distance) relationship arises, it can be understood that surface tension has no particular relationship with a force constant between molecules (or between atoms of a molecule).  Instead, surface tension is related to the energy difference between a molecule with maximum neighbors and a molecule with fewer neighbors.

Considering a simple harmonic oscillator, energy vs distance is a parabola.  There is no horizonal asymptote.  Regardless of what the force constant is, there is always infinite energy required to remove a neighbor. In reality, intermolecular force will approach zero at infinite distance between molecules. It is the difference between energy at equilibrium seperation and infinite separation that is relavent to surface tension, and force constant does not provide this information.

The terms "initial molecular harmonic" and "initial harmonic" are not standard chemistry terms and only have meaning in the context of a particular computer program.  "Initialharmonic" or "Iharmonic" refers to a force constant that can be for linear stretching, bending about a vertex that is an atom, or dihedral bending, etc.


Here is a visualisation that shows the layered structure of the material.

From the side:


Nobody really knows. Using the naive Bohr model of the atom, we run into trouble around $Z=137$ as the innermost electrons would have to be moving above the speed of light. This result is because the Bohr model doesn't take into account relativity. Solving the Dirac equation, which comes from relativistic quantum mechanics, and taking into account that the nucleus is not a point particle, then there seems to be no real issue with arbitrarily high atomic numbers, although unusual effects start happening above $Z \approx 173$. These results may be overturned by an even deeper analysis with current quantum electrodynamics theory, or a new theory altogether.

As far as we can tell, however, we will never get anywhere close to such atomic numbers. Very heavy elements are extremely unstable with respect to radioactive decay into lighter elements. Our current method of producing superheavy elements is based on accelerating a certain isotope of an relatively light element and hitting a target made of an isotope of a much heavier element. This process is extremely inefficient, and it takes many months to produce significant amounts of material. For the heaviest elements, it takes years to detect even a handful of atoms. The very short lifetime of the heaviest targets and the very low collision efficiency between projectile and target mean that it will be extremely difficult to go much further than the current 118 elements. It is possible that we may find somewhat more stable superheavy isotopes in the islands of stability around $Z=114$ and $Z=126$, but the predicted most stable isotopes (which even then are not expect to last more than a few minutes) have such a huge amount of neutrons in their nuclei that we have no idea how to produce them; we may be condemned to merely skirt the shores of the islands of stability, while never climbing them.

EDIT: Note that the best calculation presented above is based on quantum electrodynamics alone, i.e. only electromagnetic forces are taken into account. Obviously, to predict how nuclei will behave (and therefore how many protons you can stuff into a nucleus before it's impossible to go any further), one needs detailed knowledge of the strong and weak nuclear forces. Unfortunately, the mathematical description of nuclear forces is still an incredibly tough problem in physics today, so no one can hope to provide a rigorous answer from that angle.

There must be some limit, as the residual nuclear forces are very short-ranged. At some point there will be so many protons and neutrons in the nucleus (and the resulting nucleus will have become so large) that the diametrically opposite parts of the nucleus won't be able to "detect" each other, as they are too far away. Each additional proton or neutron produces a weaker stabilization via the strong nuclear force. Meanwhile, the electrical repulsion between protons has infinite range, so every additional proton will contribute repulsively just the same. This is why heavier elements need higher and higher neutron-to-proton ratios to remain stable.

Thus, at some atomic number, possibly not much higher than our current record of $Z=118$, the electrical repulsion of the protons will always win against the strong nuclear attractions of protons and neutrons, no matter the configuration of the nucleus. Hence, all sufficiently heavy atomic nuclei will suffer spontaneously fission almost immediately after coming into existence, or all the valid reaction pathways to reach an element will require events which are so fantastically unlikely that if even all the nucleons in the entire observable Universe were being collided with each other since the Big Bang in an attempt to synthesize the heaviest element possible, we would statistically expect some sufficiently heavy atom to not have been produced even once.


Increase in equilibrium concentration of the reactant via decrease in container volume

You're right that at first glance, this is indeed self-contradictory. However, on a closer look, here's how you can realize that it isn't!

The catch here is that when the reaction shifts forward - to counter-balance the decrease in volume - the reduction in the moles of $\ce{A}$ is much lesser than the decrease in volume, hence, the overall ratio, $\frac{\text{moles}}{\text{volume}}$ actually increases!

You can demonstrate this via your own reaction. Assume $K_\mathrm{c}=\pu{1M-1}$, and all equilibrium concentrations to be $\pu{1M}$. When volume is halved, the concentrations double, so $Q=\pu{0.5M-1}$. The reaction thus shifts forward as $Q<K$.

Let the new concentrations be $\pu{2M}-3x, \pu{2M}+x, \pu{2M}+x$ for $\ce{A,B,C}$ respectively. Solving for $x$, you'll get $x\approx0.117$. Hence, the new equilibrium concentrations are $\pu{1.64M}, \pu{2.117M}, \pu{2.117M}$. In fact, not only the reactant, but all species show an increase in their concentration!

Moreover, the same result is obtained in the reverse direction. If the container volume was increased instead, all species would show a decrease in their volume. (the overall ratio $\frac{\text{moles}}{\text{volume}}$ decreases, as the moles increase far less proportionately as compared to the volume)

I must state that this isn't a rigorous proof but a layman's explanation for why this is happening, and how to understand it in simple terms. I would leave it up to a more experienced physical chemist, to demonstrate the applicability of the rules I've stated in bold above, via proofs or other methods.


Why do we sometimes write reactants above the reaction arrow?

Things written on arrows are chemicals that are part of the reaction in some way, but not one of the reactants or products. In this case, the $\ce{KMnO4}$ is your oxidizing agent, which can be swapped out for any other strong oxidizer, and is not an inherent part of the reaction.

Other things you could see on the arrow would be $\ce{H2O}$ or Diethyl Ether, both of which are solvents. Other things include $h\nu$ and $\Delta T$, both of which are conditions under which the reaction takes place. Lastly the most common thing you'll see are catalysts such as $\ce{Pt}$. As you  can see, all of these things are necessary to the reaction, but aren't really what we care about in the reaction/can be substituted for other, similar chemicals.


Balancing a redox equation with only one product

You actually have different possibilities of setting up half-reactions. One way is to note that formally, gold goes from a $\pm 0$ to a $\mathrm{+III}$ oxidation state, thus it can be written as:

$$\ce{Au -> [Au^3+] + 3 e-}\tag{Ox}$$

$$\ce{Cl2 + 2 e- -> 2 Cl-}\tag{Red}$$

Which gives you:

$$\ce{2 Au + 3 Cl2 -> 2 [Au^3+] + 6 Cl- -> 2 AuCl3}\tag{Redox}$$

And, if additional $\ce{HCl}$ is present:

$$\ce{AuCl3 + HCl -> H[AuCl4]}\tag{Aurate}$$

Since $\ce{HCl}$ does not participate in the redox part of the reaction, it does not occur until after the redox part.

Naturally, $\ce{[Au^3+]}$ is not isolateable under these conditions, hence the square brackets.


Where do I go to get details on setting up different instruments?

Your question is very broad and hard to answer without pointing out multiple books on the topic. Here is a good one to start: Lab Techniques in Electroanalytical Chemistry, Edited By P. Kissinger and W. Heineman

However, your specific experiment will depend on the type of your sample and what information you are looking for. The best place to start is to look for papers in your field and read their experimental sections.

Instrument manufacturers also write application notes specifically for this reason. You may find useful information on their websites as well. Here is an example : Application Notes Regarding EIS from Gamry

Full disclosure: I used to work for Gamry Instruments.