How to test a sample of cigarette tobacco for its nicotine content?

Unfortunately, nicotine analysis is prohibitively complicated and expensive for a home kit. The presence of other alkaloids similar to nicotine pretty much necessitates that somewhat expensive equipment (i.e. a gas chromatograph) is used to first separate out the nicotine prior to analysis (typically by mass spectroscopy).

The only practical means for a layperson to test tobacco products for nicotine content is to out-source the work to a GC-MS equipped laboratory that routinely performs such analysis.


Ortho vs para alkylation of chlorobenzene

In general para isomera are thermodynamically more stable than ortho isomers (except in cases when ther is hydrogen bonding in ortho isomer). There are several reasons:-
1. Steric hindrance makes para isomer more stable.
2.Benzene substituents rotate and change thier plane due to steric hindrance. It stops delecolaization of lone pair of electrons (if present in substituents) causing elevation in energy.
Both proper orientation (steric hindrance) and appropriate energy (temperature) is required for formation of product.
Para isomer satisfies these conditions readily and hence are major product.


Is there a way to determine the orientation of a benzene-ring ethyl group in synthesis reactions?

Unless the temperature is really low, you will not be able to distinguish the two conformation isomers (conformers).

Rotation about single C-C bonds is rarely hindered. It can be, by using bulky groups or a tether, but neither is the case here. I expect the benzyl*-methyl bond and the phenyl*-ethyl bond to rotate freely at reasonable temperatures (say, above 250K).


Why does the dielectric constant of a solid increase due to a Frenkel defect?

When Frenkel defect alone is present, there is no decrease in density, however
the closeness of the charge brought about by Frenkel defect tends to increase the dielectric constant of the crystal.
Compounds having such defect conduct electricity to a small extent.


By how many pKa units will the pKa value of a ketone be minimized by addition of a Lewis acid

I used to work in a group that specialised on the total synthesis of polyketides. For obvious reasons, syntheses of polyketides rely on aldol additions a lot. Yet, I never heard of any simple rule of thumb for the $\mathrm{p}K_\mathrm{a}$ value differences so I suspect there isn’t one.

Both boron reagents enolise ketones practically by themselves, $\ce{NEt3}$ is typically added to capture any excess protons. On the other hand, I have never heard of lithium or sodium enolising a ketone by themselves.

Finally, $\ce{HCl}$ (or any strong proton source for that matter) will protonate the ketone but not to the enolate. Instead, the protonated ketone with $\ce{C=\overset{+}{O}-H}$ will be obtained.


Energies of Virtual Molecular Orbitals (LUMO)

When thinking about orbital energy levels, it is not a good idea to think of them as fixed energy 'levels' that can be filled with electrons. This kind of picture is often found in introductory textbooks, but it is wrong. For example, it is known that even for the Hartree-Fock theory, the energy of a molecule is not a sum of the orbital energies. In this sense, you can never really consider orbital energies as the energies of the given electrons - which is really a meaningless quantity, if you think about it. Basically the only interesting quantity that you can get from an orbital energy is the first ionization energy based on Koopmans' theorem, which also is just approximately true.

So can you trust orbital energies? It really depends what you want to calculate. One thing that people often want is to calculate the UV-VIS spectrum of a molecule, which is, of course, the energy required to move an electron form an occupied orbital to a virtual orbital. But this energy should not be calculated as the difference of the energies of the two orbitals. The reason for this is not that virtual orbital energies can not be trusted, but the fact that when you move an electron, you now have a different electron configuration, that is, a different wavefunction. And with a different wavefunction, of course, you would get different energy levels. You should not use the energy levels of the original wavefunction to describe the energy levels of an excited state.

So in order to calculate UV-VIS spectra, you should rather use a multi-reference method with a large enough active space that includes both the ground and the excited states and also some 'buffering states' to expand the difference in the electron correlation. If you can afford a CCSD(T) calculation, then you might also be able to afford a CAS-SCF calculation with a formidable active space, maybe even add a bit of NEVPT2 to it. I also have some good experience with the SORCI method of Orca, if you want to check that out. And if you can not afford to do such large calculations, you might also fall back to TD-DFT - some good double hybrid methods have been made available this year.


Identification of an unknown organic molecule from only 1H NMR and 13C NMR

The doublet at $4.2\ \mathrm{ppm}$ is almost assuredly an alcohol group or ether group.  Since we only have one carbon that is bonded to an oxygen at $59\ \mathrm{ppm}$, we conclude this is a primary alcohol. The situation of being a doublet says smoething else is lurking around.

$5.4\ \mathrm{ppm}$ is far enough down field, we start thinking alkenes, but the multiplicity is mysterious for the moment.

$1.68\ \mathrm{ppm}$ and $1.75\ \mathrm{ppm}$ are in the range of allyl methyl groups, and the carbon signals are okay with this as well.  Additionally singlets for these groups indicate a geminal dimethyl compound with a third substituent on an alkene justifying different chemical shifts.

Finally the peak at $1.16\ \mathrm{ppm}$ is difficult to explain.  However, we know we have an alcohol, and they can have a fairly wide range of values.  If this peak is broadened at all, I would immediately assign it to the alcohol.  It being a singlet is also worrisome, but if exchange is occurring, this is not out of the question.

The multiplicity of the alkene proton indicates there are some long-range things going on, and these don't always resolve well.   My guess is you are picking up some long-range coupling and is not resolving nicely in the methyl groups, which appear to be singlets.


These indicate "doping" or non-stoichiometric compounds.

In this case, there's some fraction of Fe, and some fraction of Mn in a lithium iron phosphate battery. If I remember correctly, you need to dope a different ion to increase the conductivity of $\ce{LiFePO4}$.

See also here: How are non-Stoichiometric compounds determined?


Methoxide ($\ce{CH3O-}$) is the conjugate base of methanol. Methanol is very weak acid (e.g. its dissociation constant is very small), so its conjugated base is very strong.


How to calculate dew point of air-water system? I want to calculate the maximum water holding capacity of air ... However for most of them the temperature range is not more than 100 C. Are there any relations which cover a higher temperature range, up to 300-400° C? If not, is it not possible to determine saturation point above the boiling point of water?

At very high temperatures I could only find the Mollier enthalpy-entropy chart (for steam) here's one for up to 1931.67 degrees rankine (800 degrees celsius) and another for up to 2200 degrees rankine (949.072 degrees celsius):

Source: "Mollier H-S Diagram"

Source: "Mollier enthalpy-entropy chart for steam, US units"
Source: Wikipedia - Enthalpy–entropy chart

On the diagram, lines of constant pressure, constant temperature and volume are plotted, so in a two-phase region, the lines of constant pressure and temperature coincide. Thus, coordinates on the diagram represent entropy and heat.
The Mollier diagram coordinates are enthalpy $h$ and humidity ratio $x$. The enthalpy coordinate is skewed and the constant enthalpy lines are parallel and evenly spaced.

See the Wikipedia webpage for more information, or assistance reading the diagram.
For restricted ranges (of which you complained, but are useful for comparison and to view the bottom of the chart) see Wikipedia's webpage for "Psychrometric charts":

Source: "Psychrometric Chart for Sea-level pressure using SI units"
Wikipedia's webpage for "dew point temperature" shows some calculations and means to make simple estimates, but it cuts you off at 100°C.


"Synthetic wood" from cellulose and lignin?

The US published application US20110190402 titled “Synthetic wood composite” appears to be very close to answering your question: “Is there research into making wood from its constituents.”  The full text can be found here: https://patentscope.wipo.int/search/en/detail.jsf?docId=US73316434&tab=PCTDESCRIPTION&_cid=P10-KGFN8M-16074-4
It appears to me that the short answer to your question is, Yes!
Below is part of the Summary of the Invention and Independent Claim 1.
SUMMARY OF THE INVENTION
The present invention is directed to a synthetic wood composite comprising biomimetic macromolecules and methods for the preparation thereof.
In one embodiment, the invention is a composite comprising at least two polymeric macromolecules, wherein at least one macromolecule is lignin and the other macromolecule is selected from the group consisting of cellulose and hemicellulose, or a combination thereof.
In some embodiments, the composite comprises lignin, cellulose and hemicellulose. In other embodiments, the hemicellulose is xylan.
Claim 1 – the main claim

A composite comprising lignin, cellulose and hemicellulose, wherein the cellulose content is about 40 to about 60% (w/w), the hemicellulose content is about 20 to about 40% (w/w), and the lignin content is about 10 to about 30% (w/w), wherein the lignin is selected from a kraft lignin or a lignosulfonate, wherein the hemicellulose is xylan, and wherein the composite is transparent.


Is there any difference between medical and cleaning ozone?

“Medical ozone” is a mixture of 0.05 to 5 vol% of ozone produced from medically pure oxygen (usually via silent electrical discharge) and 99.95 to 95 vol% of oxygen.
Using air for medical-grade ozone production is strictly forbidden due to formation of hazardous nitrogen oxides $\ce{NO_x}.$
Note that ozone therapy is still considered an alternative medicine.

Ozone used for cleaning doesn't have such constrains and its concentration and production method vastly depends on application area and regulatory standards therein.
For example, for water treatment Russian federal standards prescribe the ozone concentration in the ozone-air mixture from 15 to 22 mg/l based on the required efficiency and level of pollution of the water source, and average contact time of 5 to 15 min.
Prolonged exposure or increased ozone concentration during water treatment may lead to unwanted effects such as problems with coagulation and extensive phenols formation.


Could observer effect cause electrons in one orbital to be found in another?

Your intuition is exactly right. Let's consider an atom of neon. It has ten electrons, which we describe as being in 1s, 2s and 2p orbitals. But the electrons are indistinguishable from each other, and accurate wave functions allow all of them to be in all orbitals. Only when we make a measurement (or some other sort of "observation") do the electrons behave as if they are localized to individual orbitals.
For example, if we measure the binding energy of the electrons by photoelectron spectroscopy, the result will always be an eigenvalue of the energy operator (aka Hamiltonian). But that does not mean that any one electron had exactly that energy prior to the measurement. As is always the case in quantum mechanics, systems that have multiple eigenstates exist as a superposition of those states until the measurement, the result of which is always one of the possible eigenvalues rather than, for example, a weighted average of them. So we will measure binding energies consistent with 2p orbitals, 2s orbitals and 1s orbitals with an intensity ratio of 3:1:1.


Oxidation number of hydrogen in compounds

You can look at the electronegativity of the atom Hydrogen is bonded too.

For instance, in NaH the electronegativity of H is greater than that of Na (~2.1 as opposed ~.9) so H two would have an oxidation state of -1)

In HCl, we observe that Cl is more electronegative so it will have an oxidation state of -1 and H will have an oxidation state of +1

In borane (even though B is a non-metal) each hydrogen displays an oxidation state of -1 (again electronegativity of H is 2.1 as opposed 2.0 of B)

Hope this makes sense.


Calculating the theoretical Density of a Lithium oxide

Here's the unit cell of $\ce{Li2O}$, which adopts an antifluorite structure. The image is adapted from this webpage; technically it shows the fluorite structure but it doesn't matter.

The cation-anion contacts occur along the diagonal of the cube (black dotted line). By Pythagoras' theorem, the length of the black dotted line is $a\sqrt{3}/2$, as I indicated on the diagram. The sum of the ionic radii is half the length of this line, so

$$r + R = \frac{a\sqrt{3}}{4}$$

Using this I obtained a value of $\pu{1.929 g cm-3}$ for the density. By way of comparison, Wikipedia cites a density of $\pu{2.013 g cm-3}$. By the way, regarding your final equation, it's a good idea to leave the units in. At first I was fairly confused why you were using $10^{-10}$ instead of $10^{-12}$, until I realised it was meant to be in centimetres.


$\ce{Na + ROH}$  reduces:

Aldehydes to primary alcohol
Ketones to secondary alcohol
Esters ($\ce{RCOOR'}$) to mixture of two alcohols ($\ce{RCH2OH + R'OH}$)
Cyanides to primary amine
Isocyanides to secondary amine
Alkyl nitro to primary amine
Aryl nitro to aryl amine, can $\ce{LiAlH4}$ do same?

Couldn't verify for Acid anhydrides and for acid halides, but my instinct says it must reduce acyl halides and anhydrides too. As @Waylander states "anhydrides and acyl halides are likely to be converted to ethyl esters by the EtOH solvent and then reduced."


Chemical reasons for choosing Salt Water or Chlorinated pool?

A salt water pool still has hypochlorous acid $(\ce{HOCl})$ as the active component so essentially it CAN be the same thing as a chlorinated pool. For a chlorinated pool the $\ce{HOCl}$ is directly added, whereas for a salt water pool the salt water is electrolysed to create $\ce{Cl2}$ which then reacts with water to form $\ce{HOCl}$.

However, there is indeed often a difference and the explanation is partly chemistry and partly chemical engineering. In terms of chemistry  the overabundance of chloramines $(\ce{NH2Cl})$ in chlorinated pools over salt water pools makes the difference. When hypochlorous acid, the active component in both, comes into contact with ammonia (from sweat, saliva etc) you get the formation of these chloramines:
$$\ce{NH3 + HOCl → NH2Cl + H2O} $$
Chloramines are the major reason for irritated eyes and skin.

Now the difference between salt water and chlorinated pools has a chemical engineering reason. Salt water pools have a constant generation of chlorine as opposed to a 'batch' delivery in the case of chlorinated pools. Therefore the control over the concentration of hypochlorous acid in the water is better, which also allows better control over the concentration of chloramines.

So to sum up: salt water pools allow higher control over the concentration of chloramines and therefore might be better, however, if this control is not properly applied, it could also even be worse!


Is the pressure of a gas proportional with the concentration of a gas?

Assuming your gas follows the ideal gas law, you have $PV=nRT$. Setting $c=\frac nV$ yields $P(c,T)=cRT$.

If the temperature is fixed, then $P(c)=(RT)c$ indeed tells us that the pressure of a gas is proportional to its concentration.


Does changing the concentration of reactants alter the equilibrium constant

D is an invalid statement because the ''equilibrium constant'' for a certain reaction is constant as long as the temperature of the system is constant.
When the concentration of a species Is changed according to Le Chatelier's principle the equilibrium system will change it's rates of the forward and backward reactions so as to balance the concentrations thus keeping the equillibrium constant the same.


Why do certain signals in proton NMR experience extensive splitting despite only having 1 neighbour?

This question has a few different parts to it.

Part 1 - what's that peak at 3.6ppm?
Well, it's not the -OH peak.  The peak at 3.6 is from the CH(OH).

Part 2 - where is the -OH peak if that isn't it at 3.6ppm?
Peaks from labile protons are frequently not observed, and have variable chemical shifts, unless great care is taken in sample preparation. This sample was run in CDCl3, and the -OH peak exchanges with H2O, and comes at ~1.6ppm.

Part 3 - can we account for the splitting of the peak at 3.6ppm now we know what it is?

The CH(OH) proton should couple to proton on the adjacent carbon (split to a doublet) and also the protons of the methyl group (split to a quartet). It won't couple to the -OH for the same reason we don't readily observe the -OH; it is rapidly exchanging with water. So, we expect that the peak at 3.6ppm should be a doublet of quartets. Looks more complicated than that to me.

Part 4 - It looks  more complicated that a doublet of quartets to me. What's happening there?
Of course, 3-methylpentan-2-ol has 2 stereocentres, and so we have here a mixture of diastereomers in solution in the sample. Both diastereotopic CH(OH) protons have very similar chemical shifts, and so there are in fact 2 sets of doublets or quartets. It's also why the rest of the spectrum looks such a mess.


What is this wave function in Chemical bonding?

$\Psi$ is the wavefunction describing an electron's motion.  $\Psi^2$ is the probability density function for the electron. $\Psi^2$ is usually interpreted as the probability of finding the electron at a given set of coordinates.

In the case that $\Psi$ is a complex function (in that it contains imaginary numbers and not that it is complicated, which it usually is), $\Psi^2$ is defined as the product of $\Psi$ with its complex conjugate: $\Psi \Psi^*=|\Psi|^2$ so that the probability density function contains only real numbers.

Consider a simple case of $\Psi=a+bi$. The complex conjugate would be $\Psi^*=a-bi$.

$$\Psi \Psi^* = (a+bi)(a-bi)=a^2+b^2$$

Often the probability density function is normalized so that its integral over all space is 1 (i.e. there is 100% chance of finding the electron in all of space). In the equation below, $A$ is the normalization constant.

$$1=A \int_o^{2\pi} \int_0^{2 \pi} \int_o^\infty \Psi \Psi^* dr d\theta \phi$$


Indigo and the Environment: is it that bad?

For this process, sodium dithionite (Na2S2O4) has been considered as a best reducing agent in traditional and industrial dying but the reducing agent releases sodium sulphate, sulphite and thiosulphate ions into the media. Those ions not only cause environmental concerns as major pollutants of the modern textile industry waste waters but they present also no negligible risks of toxicity against living organisms.

It is a very convoluted sentence with so many negations. My simple definition of environmentally dangerous stuff is that anything which is not found in nature is bad for the environment in excessive amounts. Sodium thiosulfate and sulfite are strong reducing agents, in excess amounts, they will slowly deplete the oxygen levels in water. Read the concept of "Chemical Oxygen Demand" in natural waters. These ions will heavily affect the COD, which is a measure of water quality. Sodium sulfate should be fine. You have to check the half-life of these ions (thiosulfate and sulfite) in natural waters.

Untreated textile waste is worse in the sense because it is colored. If large amounts of untreated textile waste, such as those of indigo, are discarded into natural streams they might block the light reaching the depths.


It was probably a poor grade of stainless steel. The chromium oxidizes before the iron therefore you obtain a thin layer of protection. In poor grades of stainless steel, I have seen rust form over a period of time or during exposure to non ideal conditions such as high humidity. This is probably because the manufacturer added the minimum chromium required to meet standards, and so you get a smaller ensemble of chromium oxide and more iron oxide.

You can treat the metal with phosphoric acid to remove the rust. You can also treat the stainless steel with concentrated sulfuric acid to produce a phosphate layer for additional protection with minimal loss of material (they do this in industry)


Are the hydrogen atoms of 2-methylindane homotopic, enantiotopic, diastereotopic or unrelated?

The process to determine whether two protons are enantiotopic or diastereotopic is to replace each with deuterium and look at the relationship of the resulting compounds. For the indicated hydrogens in 2-methylindane, replacement of each gives a pair of diastereomeric compounds (Figure, top). The compound also contains enantiotopic protons (Figure, center). The carbon where the replacement happens doesn't necessarily become a stereocenter in this process (Figure, bottom).


Why does LiAlH4 reduce the double bond of cinnamaldehyde, whereas NaBH4 does not?

Cinnamaldehyde (1) does not undergo 1,4-reduction but rather 1,2-reduction with LiAlH4. The product of the reduction is 3-phenylpropan-1-ol (2). If conjugate addition were to occur, enolate 3 would be produced (stable to the reductant; enolates can be prepared to protect ketones from this reductant) and aldehyde 4 would be expected, as long as the aqueous workup were to destroy the reductant before any enolate 3 could be protonated.

All of this aside, the answer was provided experimentally in 1954.1 Reduction occurs 1,2 to afford the aluminate salt of cinnamyl alcohol 5. This species undergoes hydroalumination of the double bond by a presumed intramolecular process to give alkyl aluminate 6. This cyclic aluminate was deuterated (D2O) at the benzylic position to provide 9. In addition, trapping of 6 with benzophenone afforded diol 7 in >80% yield! Reduction of cinnamaldehyde with LiAlD4 and aqueous workup would be expected to give dideuterio 3-phenylpropan-1-ol (8).
Sodium borohydride reduces cinnamaldehyde via 1,2-reduction in hydroxylic solvent to cinnamyl alcohol in 97% yield.2 The intermediate alkoxyborane (Cf.; structure 5) either is incapable of reacting with the double as the aluminate does or the alkoxyborane rapidly exchanges with solvent.

1) Dornow, A., Winter, G., Vissering, W. Chemische Berichte, 1954, 87, 629.
2) Chaikin, S.W., Brown, W.G., J. Am. Chem. Soc., 1949, 71, 122.


Hint:

For 2. Try to apply Le Chatelier's principle. Increasing pressure will make the system try to decrease it and there is one side of the equilibrium with an unequal number of moles.


Is there a way to get hot water without altering plumbing or electrical wiring too much?

Crystallization of a supersaturated solution of sodium acetate will release heat - once. But then it is just a solid. It works nicely for pocket warmers, but since you have to heat the flexible container in boiling water prior to re-use, this and other comparable processes are of little use for a water supply.

At the end of the day, you won't get around plumbing to tap an energy source.

Your options would be:

solarthermal
geothermal
heat released from the fermentation of organic waste


How do I determine the amount of Calcium in a substance using EDTA titration?

Yes, EDTA will chelate to both calcium and magnesium. However, since the complex formation constant for $\ce{Ca^2+}$ is significantly greater than that for $\ce{Mg^2+}$ ($\log \beta_{f,\,\ce{Ca^2+}} = 10.65$; $\log \beta_{f,\,\ce{Mg^2+}} = 8.79$), we are justified in assuming that EDTA preferentially chelates to $\ce{Ca^2+}$ and only chelates to $\ce{Mg^2+}$ when no $\ce{Ca^2+}$ is available.

Remark. How significant is "significant"? In principle we could avoid the approximation and solve for the system analytically, but then we would still have error from (i) imprecision in visual determination of endpoint, (ii) imprecision in determination of formation constants, and so on.


What pulls the electrons from iron in rust formation?

The iron is being oxidized in this process. You just need something to act as an oxidizing agent. Dioxygen will do quite nicely:

$$\ce{O2 + 4e- -> 2O^{2-}}$$


What is the interaction between dissolved ions of opposite valence in a solution at rest?

The answer is: it depends. Dissolution of a salt implies that the entropy gained exceeds the cost of breaking lattice interactions (the solution enthalpy, assuming it is positive). Electrostatic interactions compete with kT (thermal jostling). Under physiological conditions, long range interactions are strongly screened by intervening solvent molecules and other ions. At best ions are subject to an effective potential due to distant ions, in particular between segregated charges as in the cases you present (e.g. on opposite sides of a membrane).

However, ionic solutions are generally regarded as "non-ideal". This means that interactions between the ions and with the solvent cannot be dismissed. At high salt concentration ion pairing will be encouraged. Above the solubility limit salt precipitation will occur, preceded by formation of clusters (seeds). If an ion has multiple charges, then electrostatic interactions will be stronger and pairing will be encouraged. Some metal ions form permanent ligand complexes that can attenuate the overall charge or distribute it over the complex, and the ligands can form a "cage" about the central ion, altering the interaction potential with solvent and other ions. The effect of temperature is complex. The entropic cost of association increases with T, but the dielectric constant of the solvent tends to decrease also.

If you are interested in reading on ion-pairing in NaCl solutions you may want to start with Ref. 1. It explains that association is negligible in dilute solutions:

  For dilute NaCl, we predict a coordination number of 5.3 with only waters in the solvation sphere for Na+ at 25 °C, 1 bar. With increasing NaCl molality, we find increased Cl complexation of Na. At 25 °C and 1 bar the Na+ ions in an 8 m solution of NaCl have 4.3 waters and 1.0 Cl ligands in their first coordination shell. There is some evidence of solvent-separated Na–Cl ion pairs from the peak in the Na–Cl pair distribution function at 5 Å. However, this disappears with increasing temperature.

There is more on formation of NaCl complexes and clusters in that article.

References

Sherman, DM Collings, MD. Geochem. Trans., 2002, 3(11), 102–107


Is it true that an evaporating molecule has the same kinetic energy as a molecule in a pot of boiling water?

Well for one the kinetic energy for a given temperature is an average kinetic energy of a distribution of energis. So for the question:

  "Is it true that an evaporating molecule has the same kinetic energy as a molecule in a pot of boiling water?"

The answer is maybe or there probably are molecules that have the same energy in each phase, but I think the question you want to ask is:

  Is it true that all molecules in steam have the same average kinetic energy as all the molecules in a pot of boiling water?

From a thermodynamic perspective, at the boiling temperature the liquid and vapor phases are in equilibrium and have the same temperature. Since temperature is the measure of the average kinetic energy, if temperature is equal average kinetic energy is equal.

So simplistically speaking yes they are the same. You can stop with reading here or veer with me into super technical semantics and minutia—your choice.

  Reasoning for thinking this: boiling is when the vapor pressure = the pressure of the atmosphere.

Now it is commonly taught as you state: "boiling is when the vapor pressure = the pressure of the atmosphere". The thing is though that this is a restatement of "at the boiling temperature the liquid and vapor phases are in equilibrium and have the same temperature" which is similar but not equivalent. You said:

  It does so because it gained kinetic energy ${x}$, and ${x}$ was great enough to counteract the pressure of the air surrounding it.

Boiling occurs at the bottom of the pot which has a higher hydrostatic pressure due to the height of the water $(P = \rho gh)$. So the molecule must not just over come the pressure of the atmosphere but also the pressure of the water height. This increased pressure mean boiling occurs at a higher temperature than water at the top of the pot (see: water saturation pressure). As the steam bubble rises, the pressure decreases ($h$ decreases) and the bubble expands the steam will cool due to the work done by expansion (see: internal energy) lowering the equilibrium vapor pressure and condense at the interface of the bubble. The thermal transport is not instantaneous though so the interface inside the bubble will absorb heat from the latent heat of condensation, and in turn be hotter than the surrounding water thus the bubble will be hotter in the center than at the interface with water. Since the vapor is (very) slightly hotter, the average kinetic energy will be (very) slightly hotter, thus (very) technically it would not be correct that all the molecules in steam have the same average kinetic energy as all the molecules in a pot of boiling water


Disclaimer: The following information has been extracted from my own Chemistry lecture notes which was built on contributions from many other general chemistry textbooks.  It is a stereotypical example of relating $K_c$ and $K_p$.

$K_c$ and $K_p$ relationship

For a simple reaction involving gas phase substances

$$\ce{aA(g) <=> bB(g)}$$

The equilibrium constant can be written in terms of molar concentrations or partial pressures as

$$K_c = \frac{[\ce{B}]^b}{[\ce{A}]^a} \qquad\text{or}\qquad K_p = \frac{(P_\ce{B})^b}{(P_\ce{A})^a}$$

Assuming that the gas behaves ideally (a common and convenient assumption for general cases), we can write the ideal gas law for each gas such that

$$ PV = nRT \quad\to\quad P_\ce{A}V = n_\ce{A}RT \quad\to\quad P_\ce{A} = \left(\frac{n_\ce{A}}{V}\right)RT$$

$$ PV = nRT \quad\to\quad P_\ce{B}V = n_\ce{B}RT \quad\to\quad P_\ce{B} = \left(\frac{n_\ce{B}}{V}\right)RT$$

We note that
$$ \frac{n_A}{V} \equiv M $$

where $M$ is simply molarity (moles/unit volume).  Since $[\ce{X}] = M_x$, we get

$$P_\ce{A} = [\ce{A}]RT \qquad\text{and}\qquad P_\ce{B} = [\ce{B}]RT$$

Substitute these terms into the $K_p$ equation to give

$$K_p = \frac{(P_\ce{B})^b}{(P_\ce{A})^a} = \frac{([\ce{B}]RT)^b}{([\ce{A}]RT)^a}$$

and recognize that

$$\frac{[\ce{B}]^b}{[\ce{A}]^a} = K_c$$

Plug this relationship into the $K_p$ expression to give

$$K_p = K_c (RT)^{b-a} \quad\to\quad K_p = K_c (RT)^{\Delta n}$$

Where $\Delta n$ is simply the difference of the sum of the stoichiometric coefficients of the products and reactants from a balanced chemical equation like so

$$\Delta n = \sum (\mathrm{products}) - \sum (\mathrm{reactants})$$

It should be clear from this that the only way $K_c = K_p$ within the context of an ideal gas is when $\Delta n = 0$.

$$\begin{align*}
K_p = K_c(RT)^{\Delta n} \quad\to\quad K_p &= K_c(RT)^{0} \\
&= K_c (1) \\
&= K_c
\end{align*}$$

By visual inspection of your image, it is clear why these are the correct answers based upon the derived relationship offered here.


Why does density increases when adding nanoparticles in a mixture of water and some clay?

If your question is why density increases with the addition of nanoparticle then i have to recommend this quora post.

  If the attractive forces between the constituent particles of the
  solid with the water molecules are high enough, the total amount of
  intermolecular space is reduced and the total volume of the solution
  can be smaller that the combined volume of its components, leading to
  an extra increase in density....

Surface of silica nanoparticles have siloxane($Si-O-Si$) and silanol ($Si-OH$) groups. Because of the presence of these functional groups silica nanoparticles tend to form hydrogen bonds between themselves and/or with other polar molecules such as water. Its these interactions which reduces the intermoleculr space which further reduces the volume thus increase in density.

If the silica nanoparticles interect with themselves and form a linked three dimensional networks then it will also increase the viscosity of the solution.


Why is the value of "a" is positive instead of negative in Van der Waals equation?

You understand the chemistry: intermolecular attraction forces cause the pressure to be less than expected. The problem here is with the math. Though unintuitive at first sight, a positive $a$ leads to a negative correction to pressure. Let's look at a numerical example.

  Assume that the pressure inside a certain container is $50$ atm with the ideal gas approximation and $45$ atm with the van der Waals equation of state. Then $P_\text{ideal} = P_\text{vdW} + 5$.

In other words, the corrected pressure is $P_\text{vdW}$, not $P_\text{vdW} + a\frac{n^2}{V^2}$.


How fast is the interconversion triplet → singlet in organic molecules?

It depends on the organic molecule.

When spin-orbit coupling facilitates intersystem crossing, it can be fast: 100 femtoseconds for nitronapthalene (viz. Chem. Eur. J. 2018, 24, 5379 – 5387).

The Wikipedia article on intersystem crossing cites a timescale for radiative decay (phosphorescence) of triplet to singlet of 10−8 to 10−3 s.


Simulation of a particle and chemical molecule

Density Functional Theory is worth investigating if you're looking for a fairly efficient way of investigating these interactions at an electronic level. The term 'nanoparticle' is wonderfully vague as it encompasses two to three orders of size magnitude - the size of the nanoparticle will inform the specific choice of method:

Small (as in, only a couple of atoms) nanoparticles are amenable to computation with atom-centered basis sets in programs such as Gaussian, ADF, ORCA, GAMESS and so on. The latter two may be acquired free for academic use. If your particle/ligand complex is highly symmetric and/or you have supercomputer time you might be able to work with a moderately large particle. I recall a paper abstract where researchers did TDDFT on a 120 atom silver nanoparticle in ADF (here it is!), aided by high symmetry.

Large systems, where it's a) computationally unreasonable to model the whole particle and b) to a first-order approximation the ligand-particle interaction can be treated as occurring on an infinite plane or edge, may be more amenable to periodic DFT using plane-wave basis sets (PW-DFT). There are many codes available, ranging from free to very expensive. Off the top of my head, VASP, SIESTA, Quantum Espresso, Abinit and CASTEP are fairly popular.

There's also an atom-centered periodic franken-version of ADF called ADF-BAND which can model 1D to 3D periodic systems, which is worth mentioning. It is not particularly well documented, however.


What is meant by work done "expanding against atmosphere"?

Imagine someone was pushing you against a wall. If you spend some energy overcoming that push, wouldn't you say that you did some work (expenditure of energy) against that pressure?

With a similar line of reasoning, the atmosphere is exerting a pressure of one atmosphere trying to compress your gas down. That's a huge pressure (nearly $10^5$ pascals!) and the gas in question has expanded against it. The wording makes sense.

In general, whenever gases are produced as a result of some reaction (in this case combustion), they are initially at a higher pressure than atmospheric pressure. This isn't an equilibrium situation. So, they expand, and lose their pressure until it finally becomes equal to that of the atmosphere.


Find oxidation number of a metal in a compound

Oxidation number of Fe is +3. This is because oxygen has a charge of -2. There are 3 oxygen atoms. So net charge becomes -6. Thus to balance it, each of the 2 Fe atoms must have a charge of +3 (-6 + 6 = 0).


How is the yield of a complex organic synthesis defined?

In theory

For simplicity's sake, let's consider a two-step sequence:

The usual terms for these are as follows: A is the "starting material", D is an "intermediate", F is the "final product", and B, C, and E are "reagents".

Why are B and C listed over the same arrow? This means that they have been added together to the starting material A, in the same flask, and together they effect the transformation of A to D.†

Now, depending on the stoichiometry of the reactions, as well as the quantities of A, B, C, and E which are added while performing the synthesis, it will turn out that one of A, B, C, or E will be a limiting reagent for the overall synthesis. Note that D cannot be a limiting reagent, as it is not an "input".

For example, let's say that the exact stoichiometry of the two reactions is as follows:

$$\begin{align}
\ce{A + 2B + 2C &-> D} \\
\ce{D + E &-> F}
\end{align}$$

Now, it is clear that if you add 1 mol of A, and if everything else is added in excess, then the theoretical maximum you can get is 1 mol of D and hence 1 mol of F. In this scenario, A is therefore the limiting reagent, because the quantity of A limits how much of F you can get.

Sadly, reactions don't ever work perfectly. You perform the synthesis, and at the end of the day you get 0.05 mol of F instead of the theoretical maximum of 1 mol. Your yield is therefore:

$$\frac{\pu{0.05 mol}}{\pu{1 mol}} = 0.05 = 5\%.$$

Identifying the limiting reagent

Usually the limiting reagent is chosen to be the most expensive, or most precious reagent you have (or else part of it goes to waste). This is most often an organic compound, and is nearly always the given starting material. So, one usually does not have to calculate the theoretical yield from every single input, as the limiting reagent can usually be identified at a glance.

MacMillan's strychnine synthesis can be used as an example here. The sequence of synthetic steps is given on SynArchive. In this case, it is reasonable to assume that the organic starting material is the limiting reagent, and that all the other "inputs" are added in excess.

In practice

In the lab, organic chemistry is done one step at a time; therefore the intermediate D is usually isolated and stored for a period of time.

If the above theoretical analysis is to hold true, then all of the D that you formed in the first step must be carried forward to the second step. i.e. if you form 0.8 mol of D in the first step, the second step must start with all of the 0.8 mol.

In practice this is never the case, due to some reasons:

Some sample must be sacrificed to obtain spectroscopic data on them; this helps the chemist know that they have made the right thing. In other words, we need to make sure that what we think is D is really D.
Some sample must also be sacrificed to small-scale experiments. Reactions often need to be tested before being performed on a full scale. If a chemist made 0.8 mol of D from the first step, and happily carried all of it forward to the second step, they would be left fuming should the second step fail, because they would have to perform the first step again. The way around this is to take a tiny amount: say 0.01 mol of D, and carry out the reaction on it. That way, if the test run works, you can then take the remaining 0.79 mol and do the reaction again; and if not, you still have plenty of D left to experiment on. Ultimately, this leads to losses in the amount of D which is carried forward.

So, the better way to do it is to measure the yield of each individual step. That is to say, measure the yield of D that you made from A; and then measure the yield of F that you made from D. If you multiply those two numbers together, you will obtain the overall yield.

For example, if you put in 1 mol of A and got 0.8 mol of D, the yield of the first step is 80%. Let's say you then took 0.75 mol of D, treated it with E, and got 0.075 mol of F. The yield of the second step is then 10%.

It looks like the overall yield is 0.075/1 = 7.5%, but in truth the yield is (80%)(10%) = 8%. Why is the true yield larger than expected? It's because you didn't carry forward all the D that you could have. In theory, if you repeated the synthesis from start to end and carried forward the entire 0.8 mol of D, you would then have gotten 0.08 mol of F, which corresponds to the correct yield of 8%.

In a paper, yields for every step are reported. This helps people identify which steps work well, and is also useful for other chemists who wish to only replicate a certain step or steps from the paper (for other purposes). The quoted yield of the overall synthesis is usually simply the product of the yields for the individual steps.

† The exact definition of a "step" has been subject to some recent controversy (1, 2), but it's not worthwhile getting into it here.


There is a bit of a philosophical debate as to whether orbitals exist only when they're populated, or if they're always there. In both oxygen and sulfur, there is no occupied $d$ orbital in the ground state (so both $3d$ orbitals are vacant), but in sulfur the promotion energy of an electron from a $3s$ or $3p$ orbital to a $3d$ orbital is much less than the promotion energy of an electron in the $2s$ or $2p$ orbitals in oxygen to a $3d$ orbital ($2d$ orbitals don't exist, of course). Arguably, this means that sulfur can access its $3d$ orbitals under the right conditions since the promotion energy required is relatively low and could be supplied in chemically relevant situations (so the sulfur $3d$ orbital is accessible).

It is not expected that an oxygen atom could ever populate its $3d$ orbital in a stable substance. It is possible to occupy an oxygen $3d$ orbital for a short while, however, by exciting the atom with a photon of the proper frequency.


Non-existence of time operator in quantum mechanics

On the not Hermitian point.

Let $\hat{F}$ be a non-stationary observable (but does not depend explicitly on time $t$, hence $\partial \hat{F}/\partial t = 0)$. in the Heisenberg picture it satisfies the Heisenberg equation of motion for $\hat{F}$, namely
$$i \frac{d \hat{F}}{dt} = [\hat{F},\hat{H}] \tag{1}$$.

Now assume that whenever $d \hat{F}/dt \neq 0$ then it has an inverse
$$(d \hat{F}/dt)^{-1} = i([\hat{F},\hat{H}])^{-1}\tag{2}$$
Now, if
\begin{align}
[(d \hat{F}/dt)^{-1}, \hat{H}] &= 0 \\
\implies [([\hat{F},\hat{H}])^{-1},\hat{H}] &=0.
\end{align}
Now introduce a Time operator, $\hat{T}$ as
\begin{align}
\hat{T} &= \frac{1}{2}\left(\hat{F}\left(\frac{d \hat{F}}{dt}\right)^{-1} + \left(\frac{d \hat{F}}{dt}\right)^{-1} \hat{F}\right) \\
&=\frac{1}{2}\left(\hat{F} ([\hat{F},\hat{H}])^{-1}+([\hat{F},\hat{H}])^{-1}\hat{F}\right)\tag{3}
\end{align}
Now using equations $(1)$ and $(2)$ into $(3)$ one finds that
$$[\hat{H}, \hat{T}] = i \tag{4}$$

That is, $\hat{T}$ and $\hat{H}$ form a canonically conjugate pair. Owing to the fact that the Hamiltonian spectrum is bounded from below, the time operator $\hat{T}$ is not self adjoint, in other words, not Hermitian.


Classification of oxides,transition elements

'Metallic oxide' has nothing to do with basicity/acidity of the oxide and is  about solid phisics.

Basically it means, that the solid is electrically conductive like a true metal. This involves a half-filled conductivity zone (that can be roughly imagined as a number of half-filled overlapping orbitals of same energy), so electrons can move freely through the crystal lattice.

There is no simple rule to guess metallicity of multi-element compounds, unfortunately; because this property is strongly tied to electronic structure of the solid and crystall lattice. As an example: of $\ce{TiO2}$, $\ce{VO2}$, $\ce{CrO2}$ and $\ce{MnO2}$ only $\ce{CrO2}$ is truly metallic, while $\ce{VO2}$ looses metallic conductivity at lower temperatures (it has two allotropes, one metallic and one insulating). For TiO2 it is, at the very least, expected, as it is a $d^0$ compound, but for other metallic oxides it is unclear, or, at least, aboive average chemist's level.

=================

As for acidic/basic classification of oxides, there are few solid rules. Generally, any oxide with oxidation state 5+ is a strong suspect for acidic nature, with 3-4 being anyware from acidic to basic and +1 usually basic except for halogens.

For an oxide to be basic, corresponding metallic cation must be reasonably stable. This implies relatively large radius, and moderate to low charge density. Consequently, +1 metal oxides are practically exclusively basic, +2 metal oxides are usually basic except for very small cations (Be and few late d-elements like copper and zinc) and +3 are usually amphoteric except for very large cations (like La) Anything of higher oxidation state is usually acidic or at most aphoteric, as corresponding cation has too high charge density to be stable in water solutions. The only true salts of +4 cation I can come with are $\ce{Zr(SO4)2}$ and $\ce{Hf(SO4)2}$ and even for them the corresponding hydroxide is classified as amphoteric.

Still, in some cases such oxides still may act as bases, but corresponding cation has lower charge. Probably, the simpliest examples are $\ce{VO2}$ and $\ce{UO3}$, with corresponding salts containing $\ce{VO^{2+}}$ and $\ce{UO2^{2+}}$ respectively. This, however, is not usually condidered as an excuse to classify these oxides as basic/amphoteric.

Some oxides are not acidic/basic/amphoteric as there is no 'simple' salt corresponding ot the oxide in question. This claim is usually debatable, as such a salt usually may be found (for example, NO, typically classified as not producing salt, may be tied to salts like $\ce{Na2N2O3}$ aka  Angeli's salt), but for simplicity are not considered. There are reasons for it, as basic/acidic oxide classification is very rough, and should not be relied upon in extreme, borderline and simply too complicated cases.


Why are magnesium chloride and calcium chloride more soluble than sodium chloride?

I don't think there is any simple answer.There are some rules of thumb, all of which have exceptions. Most chlorides are soluble. That includes both MgCl2 and CaCl2.

Solubility is the result of the balance between competing interactions: between the ions in the salt and between those ions and the water.

Many factors contribute to the solubility of a compound in water. In this case, the main difference between the compounds is the bonds’ strength.

The longer answer is to do with something called the "enthalpy of solvation".'

you can think it this way that the lattice enthalpy in MgCl2 and CaCl2 will be less and can be easily be overpowered by hydration enthalpy


Regioselectivity of deprotonation of 1,4-dicarbonyl

Here, both the α-$\ce{CH2}$ and α-$\ce{CH3}$ groups have very similar acidities since they are both α to only one carbonyl group. [This is different compared to, for example, a 1,3-dicarbonyl such as acetylacetone, where one $\ce{CH2}$ group is α to two carbonyl groups and is therefore significantly more acidic.] In our substrate, the two protons differ very slightly in acidities, and in general, whether one deprotonates at the more or less substituted carbon can be controlled by reaction conditions.

The mechanism you are drawing is the first step of what is known as an aldol condensation. Because all the steps in this reaction are easily reversible, simple aldol condensations with $\ce{OH-}$ are typically under thermodynamic control, meaning that the most stable product is preferentially formed.

Therefore, let's consider what would happen if we deprotonated the $\ce{-CH2}$ group. The deprotonation itself is definitely a possibility, but what happens after that is quite implausible:

Once you form the enolate, the next step in an aldol condensation is the enolate attacking the other carbonyl group. In this case, it would lead to a three-membered ring, where there is a lot of angle strain and the substituents eclipse each other. The dehydration to form the α,β-unsaturated ketone is even less likely, because the cyclopropene ring thus formed would possess incredible angle strain.

On the other hand, deprotonation of the terminal $\ce{-CH3}$ group eventually leads to the formation of a five-membered ring after nucleophilic attack on the other carbonyl group. This is much more stable, and you can eliminate water via an E1cb-type mechanism to get a cyclopentenone.

So, you could certainly deprotonate the $\ce{-CH2}$ position. It is just that, after it gets deprotonated, nothing very good can come out of it. This deprotonation is said to be unproductive.


Calculating concentration of H+ Ions using Ka Value

Answer:

$K_a = 4.5*10^-4$
Concentration (From ICE Table) of products/reactants:

HNO2 = 0.2 - x
H+ = x
NO2 = x

Therefore:

$$4.5*10^-4 = x^2/(0.2-x)$$
Rearrange:
$$x^2 + x*(4.5*10^-4) - (0.2(4.5*10^-4)) = 0$$
Using quadratic formula:
$x \approx 0.009$

$$pH = -log(10)$$
$$pH \approx 2.05$$.

That should be correct!


Can I boil off chlorine? Does it evaporate quickly at high temperatures?

Yes, solubility of chlorine decreases drastically as the temperature rises, and it's almost insoluble in boiling water.
That's also the reason why in the areas where tap water is chlorinated, it advised to boil it before drinking.
Data from [1, p. 8]:

Figure 5. Solubility of chlorine in water, hydrochloric acid (two concentrations), and sodium chloride solutions (three concentrations) All percentages are weight percents.
In aqueous solutions, chlorine is partially hydrolyzed, and the solubility depends on the pH of the solution. Below 10 °C chlorine forms hydrates, which can be separated as greenish-yellow crystals. Chlorine hydrate is a clathrate, and there is no definite chlorine: water ratio.

References

Chlorine: Principles and Industrial Practice, 1st ed.; Schmittinger, P., Ed.; Wiley-VCH: Weinheim ; New York, 2000. ISBN 978-3-527-29851-8.


According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), the trigonal pyramidal centre of the sulfinate group is converted to a tetrahedral centre using a phantom atom of low priority.

P-93.3.3.2 Trigonal pyramid
The configuration of molecules containing a trigonal pyramidal center (TPY-3) is described in a similar way to that of tetrahedral centers (T-4) described above in P-92 (see IR-9.3.4.3, ref. 12). The tetrahedral configuration is achieved by adding a phantom atom (0) to the central atom perpendicular to the base of the pyramid. (…)
Traditionally, sulfoxides have been considered as a tetrahedral system composed of a central atom, ligands, and a lone pair of electrons (or phantom atom). No polyhedral symbol is used.

P-93.3.4.1 The chirality symbols ‘R/S’.
The stereodescriptors ‘R’ and ‘S’ (as defined in P-92.2) are used to indicate the absolute configuration of a trigonal pyramidal system discussed in P-93.3.3.2 (see Rule IR-9.3.4.3, ref. 12). A phantom atom of low priority, and not a pair of electrons, is used to create the tetrahedral configuration permitting the use of ‘R/S’ stereodescriptors in the manner described for tetrahedral stereogenic centers. As no locants are present, the name following the stereodescriptor is placed in parentheses, or bracket, according to the required nesting order.

The examples given for this rule in the Blue Book include ethyl (R)-(4-nitrobenzene-1-sulfinate), which confirms that this rule is applicable to the similar compound methyl (S)-methanesulfinate, which is given in the question.
Notably, an equivalent rule also exists in the current version of the current version of Nomenclature of Inorganic Chemistry – IUPAC Recommendations 2005 (Red Book).

IR-9.3.4.3 The R/S convention for trigonal pyramidal centres
Molecules containing a trigonal pyramidal centre (TPY-3) may exist as a pair of stereoisomers. The configuration of this centre can be described in a similar way to that of a tetrahedral centre. This is achieved through notional placement of a ‘phantom atom’ of low priority in the coordination site that would create a tetrahedral centre from a trigonal pyramidal centre. The centre can then be identified as R or S by the methods described above.
The use of some bonding theories leads to the placement of a lone pair on a trigonal pyramidal centre. If this is done, the absolute configuration of the centre is also described by the R/S convention, in this case by placing the ‘phantom atom’ in the site that is occupied by the lone pair. Examples of this practice may be found in the description of absolute configurations for sulfoxides in which the alkyl substituents are different.

After placement of the phantom atom, the R/S convention for tetrahedral centres can be used. Rules for the tetrahedral configuration of elements other than carbon are given in Section P-93.2 of the Blue Book. In particular, the rule for compounds containing sulfur refers to the rule for similar phosphorus compounds.

P-93.2.5 Sulfates, sulfonates, and related compounds
Sulfates, sulfonates, and related anions are treated in the same way as phosphate anions (see P-93.2.4). (…) Sulfoxides are discussed in P-93.3.3.2.

The corresponding rule for compounds containing phosphorus reads as follows.

P-93.2.4 Phosphates, phosphonates, and related compounds
The ‘$\ce{P=O}$’ bond, as conventionally written in phosphates, phosphonates, and related compounds, is considered as a single bond, as there are already four atoms or groups in the tetrahedral configuration. Similarly, the formal arrangement of charges is not considered when determining the configuration of a chiral molecule. As the stereodescriptors ‘R’ and ‘S’ describe the entire structure, either a salt or an ester, the full name is placed in parentheses to denote the global configuration.

Therefore, the analogous ‘$\ce{S=O}$’ bond in the sulfinate group is considered only as a single bond since the tetrahedral centre of the sulfinate group with an attached phantom atom already has four groups. (Note that this is a deviation from the usual Sequence Rules for the Cahn-Ingold-Prelog (CIP) System. Using the Sequence Rules, double bonds are normally split into two bonds with duplicate representations of the atoms at the other end of the double bond.)
Thus, for the compound given in the question, the $\ce{-O-CH3}$ group ranks higher than the $\ce{=O}$ group. This leads to the configuration S and the systematic name methyl (S)-methanesulfinate.


Why does nitrous oxide have 300 times the global warming potential of CO2?

There are two really important properties of N2O that make it such an important greenhouse gas, both of which you have raised in your question but need some clarification.

Infrared absorption - it's good to think in terms of gaps here. There's a lot of overlap between CO2, CH4 and (a frequently ignored GHG, but very important) water vapour in the wavelengths which they absorb. At the 4-5 micron and 7-8 micron range, N2O is extremely efficient at absorbing infrared radiation. These are important regions, precisely because there is no overlap with other GHGs in these regions. N2O is absorbing infrared very efficiently, and is doing so without "competition".
Lifetime - you noted that the lifetime of N2O and CO2 are comparable, which is sort of true. N2O, as a stable, inert molecule that is well mixed in the atmosphere, has a very long lifetime of 120 years compared to some other GHGs (CH4, for example is 8 years, based on the fact that its removed by the hydroxyl OH radical). It's possible to determine this fairly accurately as the only sinks for N2O are its stratospheric photo-dissociation and reaction with O(1D) radical. CO2 by comparison has a lifetime that varies anywhere between 5 and 200 years. The comparable part comes from the fact that for calculations for global warming potential, CO2 is given an "averaged" lifetime.

With N2O, therefore, we have a very efficient infrared absorber at two "windows" at which there is no overlap with other GHGs, and a long lifetime. The Global Warming Potential is the time integrated radiative forcing for a 1 kg pulse emission of the compound, where the upper limit of the integration is time horizon (usually set at 100 yr), relative to the same quantity of the reference compound CO2. It becomes clear from this definition why a stable, inert molecule with a long lifetime and which is very efficient at absorbing at particular infrared wavelengths would have a greater global warming potential than CO2 (shorter lifetime, absorbing in "windows" shared by other compounds) on a kg to kg comparison.


How to calculate energy gap when determining electron configuration?

You could calculate the energy of your initial electron configuration, then do the same for your final state and work out the difference. The Los Alamos Flexible Atomic Code (https://github.com/flexible-atomic-code/fac) can do that for you. There should be no state where you fill shells from the ground up that has a lower energy than what we now call the ground state of nobium.


Graphite is odd in its categorisation and does by no means rigorously follow the conventions of either a covalent solid, or a molecular solid (for example it can conduct electricity). Graphite has properties of both molecular solids (it is soft), and covalent solids (it has a very high melting point. In fact it will break down before it melts.).

For this reason, the categorisation of some materials, such as graphite, can be somewhat complicated, and the idea of graphite being a covalent solid really just comes from its similarity with other covalent solids.


Synthesis of ZnO by wet chemical method and reducing particle size by ball milling

I had a parallel experience and noted it in my thesis, 20 years ago.

Was working on an oxide ternary phase diagram (not wet chemistry, traditional shake and bake).  There were some conducting phases (dark) and some insulating ones (white).  I had a ZnO pellet (post firing) and I ground it under acetone (just mortar and pestle).  There was a definite color change to yellow.

I had a lot of other results and never ran this issue to ground.  I did note it and speculated that it had to do with the grain interiors having some conductivity (e.g. from oxygen defects or from impurities) and thus a different color, with the grain exteriors more insulating and white (not having the defects).  This would explain the color change.  There have been papers written (more than the 20 years ago) about differences in defects at grain boundaries and interiors of ZnO.

Very much a speculation at the time and never looked into it more.  But very interesting to see you have a parallel experience.  Maybe you will do more than I did and run it to ground?

P.s.  Consider the appearance of In2O3 and Ga2O3.  Indium oxide out of the jar is yellow and has semiconducting properties, from oxygen defects or the like.  Gallium oxide is much more insulating and is bright white.  Perhaps your ZnO grain interiors are a little In2O3 ish?  After all, both materials are dopable (despite the large band gap) and are TCO materials.  Ga2O3 on the other hand is extremely insulating and undopable, with a wider band gap.


Dehydration of primary alcohol with a cyclobutadiene substituent

You have misinterpreted what Wikipedia has said. I'd put whole quote here:

Cyclobutadiene is an organic compound with the formula $\ce{C4H4}$. It is very reactive owing to its tendency to dimerize. Although the parent compound has not been isolated, some substituted derivatives are robust and a single molecule of cyclobutadiene is quite stable. Since the compound degrades by a bimolecular process, the species can be observed by matrix isolation techniques at temperatures below $\pu{35 K}$. It is thought to adopt a rectangular structure.

These somewhat stable cyclobutadiene derivatives are not just derivatives with a simple alkyl substituent. They are with bulky substituents, e.g., tetrakis(tert-butyl)cyclobutadiene. Therefore it is safe to assume that given compound (2-cyclobutadienylethanol) does not exist. However, as Aniruddha Deb has said, we can treat it like it is existing only for this theoretical question (we don't have other choice):
Since it is a primary alcohol it can't undergo usual $\mathrm{E1}$ elimination with an acid. The $\mathrm{E1}$ pathway through formation of a primary carbocation is not the most likely pathway here, because primary carbocations are extremely unstable. Therefore, more likely, this reaction undergoes through an $\mathrm{E2}$ mechanism where the transition state should be lower in energy. Accordingly the protonation of the alcohol group to give the good leaving group $\ce{-OH2+}$:

And then, a weak base (which could be either $\ce{H2O}$ or $\ce{H2PO4-}$, or even another molecule of the alcohol) could abstract $\beta-\ce{H}$ from nearby $\beta$-$\ce{CH2}$ group, leading to give the conjugated alkene with cyclobutadienyl group (vinylcyclobutadiene):

We really don't know how stable vinylcyclobutadiene is, but it should be more stable than original alcohol since it has additional resonance contribution with external vinyl group. This compound can undergoes either Diels-Alder or Cope rearrangement to give a bicyclo[2.2.0]hex-1,5-diene derivative. This derivative should be unstable due to severe strain, and expected to rearrange to Dewar benzene derivative, which then rearrange to benzene upon heating (Ref.1).
Reference:

L. Watts, J. D. Fitzpatrick, R. Pettit, "Cyclobutadiene," J. Am. Chem. Soc. 1965, 87(14), 3253–3254 (https://doi.org/10.1021/ja01092a049).


how to determine the relative differences in activity between metals

You have a standard pot of liquid electrolyte into which are dipped non-touching same-spaced paired strips of different clean metals.  The external circuit between the strips is a sensitive voltmeter (with extremely high electrical resistance - hard by zero current flow).  You measure the potential (noting direction!) between all combinations of paired strips, (n)(n-1)/2, and then draw certain conclusions.

Why are amalgams saturated with respect to excess solid metal interesting substrates for this experiment (e.g., aluminum)? How do you construct a standard hydrogen electrode?


What is the meaning of the "colon" notation in chemical formulas?

All it means is that there's an intimate physical mixture of the specified components in a solid or liquid phase (no chemical bonding between the components as happens in a copolymer, for example).

In the context of solar cells, LEDs and related devices, they are often used as part of shorthand device architecture notation. Slashes ( / ) are used to separate different layers of the device, while colons ( : ) are used to enunciate all separate components in a single layer. Do note that every so often a slash is used incorrectly in place of a colon. Checking the experimental procedure should clarify whether two consecutively listed materials are in the same or separate layers.

For example, consider the notation ITO/PEDOT:PSS/P3HT:MEH-PPV:PCBM/Ca/Al for an organic polymer solar cell. There are five layers in total, deposited in sequence from left to right:

A layer of ITO (usually on top of a glass substrate which is omitted).
A layer containing a binary mixture of PEDOT and PSS, usually deposited via spin casting of an aqueous solution with both materials dissolved.
A layer containing a ternary mixture of P3HT, MEH-PPV and PCBM, often deposited by spin casting from a solution of all materials in a chlorinated solvent.
A layer of pure calcium metal, likely deposited by thermal evaporation.
A layer of pure aluminium metal, deposited on top of the calcium layer.

Sometimes the thickness, component proportion and concentration of the spinning solution for each layer is reported inside of parentheses, such as in ITO/PEDOT:PSS (40 nm)/P3HT:PCBM (1:0.8, 150 nm)/Ca (80 nm). Be sure to pay attention to small differences in notation resulting in completely different cell architecture. For example, a solar cell with a ITO/PEDOT:PSS/P3HT:$\ce{C_{60}}$/Al represents a bulk heterojunction solar cell, while ITO/PEDOT:PSS/P3HT/$\ce{C_{60}}$/Al represents a bilayer (or planar heterojunction) solar cell.


The response to your concern depends on the application. If it were just an academic exercise, forget about it, since you may have already done the experiments. However, the point to keep in mind is that paper based weighing materials should not be rinsed or washed for standard preparation until and until unless it is specifically allowed to do so. Paper can have leachable materials which can contaminate the standard- this applies very well to trace analysis. If you note, Fisherbrand specifically mentions "low nitrogen" content.
The second reason why the paper based weighing paper should not be rinsed is that the bottom of the paper has been in contact with the balance pan (which are often horribly dirty in academic labs due to misuse). Since that paper is thin and permeable to solvents, rinsing the paper is akin to washing the dirty pan into your standard solution rather indirectly. Of course, balances in a professional or research settings are better maintained.
Sometimes it appears that some solids particles are still sticking to the weighing paper after transfer process. In such cases, one uses the "weighing by difference" method. You know the weight of the paper+solid, after transferring, you would weigh the paper again, and record the actual weight transferred to the flask or other container. No washing is required.
Analytical work, where extreme care is required, paper is not used at all for standard preparation. One should use "weighing funnels" which are commercially available and there is no concern in washing or leaching or soluble materials.


Why is Diels–Alder adduct exo when we leave it for a while?

As you probably know, the Diels–Alder (DA) reaction is reversible — the easiest example is the reaction of cyclopentadiene, which polymerizes at room temperature, which — if needed as monomer — has to be cracked just in advance.

In terms of the selectivity you observed, the endo-oriented product of the DA reaction often is the kinetic one, the exo-oriented product the thermodynamic preferred one.  Which is, if the rate of a DA reaction is to be increased while still obtaining endo-preference, an increase of the reaction temperature is skipped in favour a Lewis-catalysis, addition of a perchlorate, or high (external) pressure.


Are specific heat capacity and boiling point of a substance related or proportional to each other?

According to the Ehrenfest classification, during a first-order transition (such as boiling) the heat capacity $C$ goes to infinity. This makes sense, since at the transition point the temperature does not change with input of heat and $C = dQ/dT $. This and other considerations add a little complexity to the problem, as it requires you to state which heat capacity (at which T) you are talking about.

More importantly, however, the boiling point depends on the magnitude of attractive intermolecular forces compared to kT, whereas the heat capacity depends on all the available degrees of freedom (including intramolecular). Therefore you can envision large and small molecules with similar overall solution composition (for instance hydrocarbons) - and thus similar heat capacities - but very different BPs. See for instance this
table of specific heats, where C for hexane through dodecane are very similar, whereas their BPs differ quite a bit.


Why did the Rio 2016 pool water turn green?

Although not strictly alkalinity, this article quotes a changed pH level and an expert gives two possible explanations

  One: A low or high pH can bring out minerals. "Calcium would show itself as cloudy water, copper would show up as a blue-green," he says.

  Two: A too-high pH would render the chlorine ineffective, which could then allow for the proliferation of algae. (Hernandez says it's doubtful, though, that the green color is mainly from algae. It typically doesn't bloom that fast in such large pools.)


Why does ordinary water have a triple point?

Summarizing the relevant point (pun intended) of my answer to a related post: a mixture does not exhibit a "triple line" if its composition is constant. It is only a "triple line" insofar as you get to vary the proportion (mole fractions) of the components. Therefore different samples of water with different isotopic composition will differ in the value of the TP, but for a given sample the TP is just that, a single constant point (not a line).
This question however does point out a very important problem for the implementation of the standards adopted during establishment of an absolute temperature scale based on the triple point of water. A good discussion of this is provided in white papers (1,2) from Isotech (manufaturers of triple point measurement apparatuses):

There are 3 main sources of error in a water triple point cell[6,
7] (excluding errors associated with measurement):

The isotopic composition of the cell’s water may not be that of SMOW.

There may be air trapped inside the cell.

There will be impurities in the water.

Isotopic composition can be measured to great accuracy from a
small sample of the cell’s contents.

SMOW and V-SMOW are water samples of standard composition (this is amply discussed in Isotech white papers, referring also to the documentation behind the International Temperature Scale established in 1990 (ITS-90)):
\begin{array} {|r|r|}
\hline
\text{Component}  &  \text{Mole Fraction} \\
\hline
\ce{^1H} & 0.999 842 \\
\ce{^2H} (\text{deuterium}) & 0.000 158 \\
\ce{^3H} (\text{tritium}) & 0 \\
\ce{^{16}O}  & 0.997 640 \\
\ce{^{17}O} & 0.000 371 \\
\ce{^{18}O} & 0.001 989 \\
\hline
\end{array}
A second white paper provides guidelines on consideration of these effects (see the section "Isotopic Composition"):

We think of water as $\ce{H2O}$, more precisely we should write $\ce{^1H2^{16}O}$. This has been termed ‘light’ water and is unobtainable. If it were obtainable it’s triple point would be approximately +0.008°C. In practice what we drink from the tap, and what is in the ocean is a soup of $\ce{^1H}$, $\ce{^2H}$ with $\ce{^{16}O}$, $\ce{^{17}O}$ and $\ce{^{18}O}$. This means real water does not have a single triple point temperature. Its ‘combination triple point’ will vary as the isotopic composition varies. This  was  understood  more  than  50  years  ago  and  after  much  analysis  of  water  from  different  sources recent temperature scales state that  the water in a triple point cell shall have the isotopic composition  of ocean water.  Although not explicitly stated, there is a Standard Mean Ocean Water (SMOW) and it is assumed that the isotopic content of SMOW fulfils the ITS-90 requirements.

concluding:

Analysing the cell contents for isotopic composition is a useful and valid addition to the usual techniques of intercomparing cells and gives an accurate indication of the cells ideal triple point value.
[...]
Such cells, complete with isotopic analysis provide the closest association to the ITS-90 value of the water triple point and hence the KTTS Scale. It is now possible to produce cells, based on Stimpson’s original design which can be associated to V-SMOW within ±3mK and this is a smaller uncertainty than can be provided by traditional measurements.

References

The Establishment  of  ITS-90 Water Triple Point References  to  ±2μK,  and  the Assessment of 100 Water Triple Point Cells Made Between 2001 and 2006, John P. Tavener and Nick Davies, Isothermal Technology Ltd. (England).

The Water Triple Point - A Reference Cell Close to the ITS-90 Value. John P. Tavener & Anne Blundell. Northern Temperature Primary Laboratory, Pine Grove, Southport, Merseyside, PR9 9AG, England.

Supplementary Information for the International Temperature Scale of 1990, BIPM. ISBN 92-822-2111-3.


Conversion of oxalic acid to malonic acid

The Arndt-Eistert procedure suggested above seems very applicable to this problem. To use it we need ethyl oxalyl chloride. This can be prepared in 2 steps from oxalic acid by first esterification to form diethyl oxalate  procedure here then reaction with PCl5 to form ethyl oxalyl chloride procedure here. This can undergo the Arndt-Eistert mechanism here with excess diazomethane then either irradiation or treatment with silver oxide to give monoethyl malonate. Simple base hydrolysis then gives the required.


Comparison of acids to dissolve phosphates

N, or Normality is a way of standardizing the # of acid equivalencies a particular acid has across multiple acids.  Acids that have only 1 acidic hydrogen for example $\text{HCl}$, the Normality will equal Molarity.

Normality is defined by:

$$
Normality = \frac{c_i}{f_{eq}}
$$

Where $c_i$ is the molar concentration of the acid, and $f_{eq}$ is the equivalency factor.  In the cases of something like $\text{H}_2\text{SO}_4$, the equivalency factor is 0.5 because it is a diprotic (two proton) acid.

This test also is showing the effect of weak vs strong acids.  $\text{HCl}$ is a strong acid, so it dissociates essentially 100%, while citric is a weak acid.  A weak acid exists in equilibrium between the dissociated acid species, and the neutral species.  This equilibrium is defined by the acid-dissociation constant $K_a$, or it's logarithmic version $pK_a$.

For citric acid though, it is triprotic, meaning there are 3 acidic hydrogens, so it's $f_{eq}$ value is $1/3=0.3\overline{3}$

Your numbered questions:

1)  The speed is being measured to a certain point.  However, the chance is that all of these reactions are complete at this point.  The hour timepoint is to normalize everything to the same time frame.  But as indicated, the reactions are likely complete at this point.  What is really being measured is your second point.

2)  Absolutely, the quantity of acid needed to dissolve the material is being tested.  As you use more and more volume of acid, you are providing larger amounts of acid to the reaction.  This helps it dissolve more of the powder (as there is more acid now present), and also increase the rate of the reaction.

They standardize the reaction to normality because if they used molarity, 1 mole of citric acid has 3 acid equivalents, while $\text{HCl}$ has only 1 acid equivalent.


Book Recommendation for Molecular Orbital Theory

If you are interested in an applications, look at MO theory in the context or organic reactions, which is a part of chemistry where the theory works well.  Check out Fleming's Frontier Orbitals and Organic Chemical Reactions:
https://onlinelibrary.wiley.com/doi/book/10.1002/9780470689493 .


Why is the decarboxylation of 2-hydroxybenzoic acid (and related compounds) done 'under nitrogen'?

[…] it seems even more important to get the right mixture of gases around it at 122℃

No, it's only important to keep the oxygen away while you are heating the sample.

Have a look at the structures: the thermal decarboxylation of the THCA isn't different from the decarboxylation of salicylic acid.

But with the carboxyl group or without, (upon heating,) phenols are easily oxidized by air.

In principle, the sample could be heated under vacuum too. In this case, you might want to use a cooling trap to collect volatile material. But usually, one does that anyway to protect the vacuum pump.


Why are allyl halides reactive towards both SN1 and Sn2 reactions?

In a reaction, SN2 as well as SN1 reactions take place. It only matters that which reaction pathway gives the major product. The answer to your question depends on the type of base used, the substrate and the overall stability of the product. Allyl halides are also reactive to SN2' reactions, i.e allylic nucleophilic substitution reactions. There are a lot of possibilities.

Keep in mind that the electronegativity order of hybridized carbons is $\ce{sp > sp2 > sp3}$. So in an allylic halide, such as $\ce{CH2=CH-CH2Br}$, the alpha carbon is much more partially positive than the alpha carbon in $\ce{CH3-CH2-CH2Br}$. Therefore, the attacking nucleophile will be attracted to the alpha carbon more in the allylic halide compound than in the primary halide. Hence, allylic halides tend to follow SN2 pathways more than primary halides.


Why does two phenylhydrazine molecule react with glucose in osazone formation?

As according to modern mechanism it has first reacted with formyl then goes for tautomers and to form an intermediate..which further form iminoketone ,as phenylhydrazine is in excess then it again react with the c=o formed on C2 and then with carbon first and give osazone .  And as further reaction stops by H- bonding between H and Nitrogen of each phenyl hydrazine which forms chelation and makes it stable.


Can any solute be used to lower the freezing point of water?

The key is in the terminology- the freeze point suppressant must be a solute. Oil won't suppress the freeze point of water because it doesn't dissolve. A small amount of sodium chloride or ethanol will suppress the freeze point of water to -5°C.

The first exception is when the solute stops dissolving. This is called the eutectic point. For example, you can only dissolve 23.3% NaCl in water (by mass). That mixture will freeze at -21.1°C. After that, adding more salt won't reduce the freeze point further.

The second exception is with large amounts. If you mix water and ethanol in a 50/50 mol ratio, which is the solvent and which the solute anymore? You can calculate the freezing point suppression of each with respect to the other and they won't match. Liquid/liquid mixtures also have eutectic points at the minimum freezing point, which is less than the freezing point of either component.

To get the actual freezing point for non-dilute systems, try searching for "solid liquid equilibria", "eutectic point", or "binary phase diagram".


Rate of reaction increases exponentially with concentration

Rate equation is experimentally determined. So, you can't directly write $r=k[A][B]$ (where r is rate, k is rate constant). Probably you might have performed first order reaction. For which $r=k[A]=\frac{2.303}{t}log\frac{[R]}{[R_0]}=\frac{2.303}{t}log{e^{kt}}$.

The above equation suggests that rate is proportional to exponential function.


How and why does this Cu change into C (graphite)?

The electrode in here has two parts – the Cu and the graphite (C). The Cu won't dissolve. Only the graphite (C) will react. So there is no changing from Cu to C.


There is a section in this Wikipedia link entitled, "Solid hydronium salts."  It says that any acid with an ionization constant of 10^9 or higher may form stable hydronium ion salts.  Examples they give include hydronium perchlorate and carborane-based hydronium ion salts.


Relationship between rate of reaction and pressure

The exponential term in the Arrhenius equation does not imply reaction with other molecules. It applies to ready to react unstable adduct, about to overcome the energy barrier $E_a$ to form the product.
The concentration of this adduct is driven by concentration of reagents and therefore by the pressure.

Therefore the reaction rate equation implicitly involve 2 multiplicative terms:

Concentration of reagent adduct

$$\mathrm{const}\cdot [A]^2 = \mathrm{const}\cdot \left(\frac{p_{\rm A}}{RT}\right)^2$$
Probability the adduct reaches the reaction energy barrier, from the Boltzmann distribution:

$$\exp{\left(-\frac{E_A}{RT}\right)}$$

By other words, the constant $A$ in your equation

$$v = A \exp\left(-\frac{E_A}{RT}\right)$$

implicitly involves concentration, respectively pressure of the reagents.


Why is acidic hydrogen peroxide a potent oxidizer?

Just consider the Nernst equation for the half-reaction
$$\ce{H2O2 + 2e- -> 2OH-}$$

You see the potential is clearly pH-dependent. Increased acidity will decrease the concentration of the reduced form, thus shifting the equilibrium to the right and consequently making peroxide a more potent oxidizer.

(BTW, even pH-neutral hydrogen peroxide is far from harmless. It can leave white marks on your hands and gnaw a hole in your jeans, albeit not instantly.)


What does the hydrolysis of a salt tell us about its basic/acidic character?

Think of the salt in terms of its components: for example, sodium sulfide can be thought of as derived from sodium hydroxide (a strong base) and hydrogen sulfide (a weak acid) - resulting in a net basic character. When you dissolve sodium sulfide in water, the sulfide ion partially reacts with water to form hydrosulfide ion and hydrogen sulfide, plus hydroxide ion. Aluminum chloride, on the other hand, is derived from aluminum hydroxide (a weak base) and hydrogen chloride (a strong acid), so it is an acidic salt. Sodium chloride, from sodium hydroxide and hydrogen chloride, is a neutral salt (fully ionized). Knowing the dissociation constants of the corresponding acid and base of a salt, one can calculate how acidic or basic the salt is.


How to explain that an increase in temperature favours an endothermic reaction?

I believe that collision theory in itself wouldn't be enough to explain this behavior. You can explain it by having Arrhenius equation and the activation energy in mind. What you got wrong was the effect of temperature on reaction rates depending on the activation energies. Actually, the speed constant of a reaction with higher activation energy will be more significantly affected by variation in temperature. This can be verified easily by manipulation of the Arrhenius equation. Apply ln to both sides to obtain:

$\ln{k}=\ln{A}-\frac{E_a}{RT}$

The graph of ln(k) as a function of (1/T) is a straight line with slope's magnitude directly proportional to activation energy.

Now, assuming that the endothermic reaction has higher activation energy than its inverse, it will be more affected by, say, an increase in temperature. While both reactions will become faster, the speed constant of the endothermic reaction will increase to a greater degree and that will shift the equilibrium favoring the products of the endothermic reaction.
If the temperature is decreased, the endothermic reaction will be slowed down to a greater degree, favoring the products of the exothermic reaction.


What substance(s) will prevent superglue from hardening in its tube?

Based on what I gathered from this Wikipedia article on super glue :

Super glue is typically composed of a class of compounds called
 cyanoacrylates.

Cyanoacrylates are capable of taking part in this process we call "Polymerization", where chains of cyanoacrylate molecules, called "polymers" ( Poly- Many, mere- parts) are formed. Each constituent cyanoacrylate molecule in the polymer is called a "monomer" ( Mono- One, mere - part).

The chemical structure of methyl-2-cyanoacrylate:

Polymerization is known to occur for a number of reasons, and can follow a number of pathways. (So to keep this brief, I'll just skim over how methyl-2-cyanoacrylate polymerizes). Wikipedia provides the following mechanism:

Over there, the Nu bit is a common abbreviation we use for a nucleophile in chemistry. It is the presence of a nucleophile, that catalyzes the polymerization of methyl-2-cyanoacrylate.

So when you squeeze out a little superglue, how does it harden? Air around you, as well as the surface you're applying the superglue on will, inevitably, have trace quantities of water. To a small extent, water splits to provide positively charged hydrogen ions (essentially, it's just a proton) and negatively charged hydroxide ions (OH). It's the hydroxide ions that play the role of the nucleophile in the above-depicted mechanism.

So ironically, when super glue "dries out", it just transiently combined with (a part of) a water molecule. That is, as @MaxW points out in the comments, water catalyzes the "drying" of superglue.

Since it's water that causes superglue to harden/set, manufacturers go all out to make sure the packaging is water/air-tight. Ever noticed, that when you buy a tube of superglue, it has a seal at the mouth which you've gotta pierce before you can use it? That seal is what prevents the glue from drying out while it's still in the (unopened) tube. Once you pierce the seal, it's only a matter of days before a sufficient amount of water (courtesy: Atmospheric moisture) manages to seep in and render render the remaining superglue useless.

Note that in case of superglue, the manufacturers don't (typically) add anything to keep the superglue from drying...they just put on a seal and cross their fingers. Whereas your typical liquid/PVA glue is a water based emulsion. When PVA glue is squeezed out of the tube and onto a surface it, literally, "dries out" by losing water.

Once superglue is set, it is set... there's no turning back O:)

You can't revert it to usable superglue through any (simple) means.

However, since you now know that it's water that's responsible for superglue "hardening", if you want to make the superglue left in the (opened) tube last longer, store it an an air tight jar that's been packed with a desiccant, like silica gel ;)


Reboiling water will do precisely three things:

Some volatile contaminants in the water that survived the first boiling might be driven out by the second boiling.
Some additional evaporation will occur, causing anything dissolved in the water (salts, contaminants) to be slightly more concentrated.
Some additional material from the surface of the vessel you're boiling the water in might dissolve into the water.

It's virtually impossible to imagine any scenario in which this would matter, unless you were taking a huge amount of barely drinkable water and boiling away most of it or using a vessel whose lining could produce toxins in your water. Needless to say, you really shouldn't be drinking barely drinkable water at all, nor should you be boiling water even once in a vessel whose lining can produce toxins. If you're doing these kinds of things, the additional risk from reboiling would be miniscule.

It will definitely not summon aresenic or lead from nowhere.


Is ammonium a strong or weak conjugate acid?

Both ammonia is a weak base and ammonium ion is a weak acid.  Many, even most, acid/base conjugate pairs are like that.  We should be using comparative instead of absolute adjectives in the rule about conjugate acid-base strengths:

A weaker acid has a stronger conjugate base, not necessarily a totally strong one.

A weaker base has a stronger conjugate acid, not necessarily a totally strong one.


Reaction rate unit conversion: how to convert from molarity to ppm

We can do it in two ways:

1) For dilute solutions, ppm and mg/L is equivalent. So, for dilute solutions, the straightforward conversion scheme should look like::
$$ppm=\frac{mol}{L}\times MW\times 1000$$

2) with no assumption::

first, remember that $ppm=y_i \times 10^{-6}$

$$ppm=\frac{mol}{L} \times \frac{x gm}{mol} \times \frac{1 L}{10^{-3} m^3} \times \frac{m^3}{y~gm} \times 10^{-6}$$

here, x is the MW of solute and y is the density of the solvent (for dilute solutions) or solution in $\frac{gm}{m^3}$


Between alkenyl and alkyl groups, alkenyl groups (double bonds) have seniority over simple alkyl substituents groups such as methyl and ethyl groups (See the new version of Nomenclature of Organic Chemistry: IUPAC Recommendations and Preferred Names 2013 (Prepared by Henri A Favre & Warren H Powell); Royal Society of Chemistry, Cambridge, UK, December 2013, ISBN: 978-0-85404-182-4:

Briefly, it suggests "When several structural features appear in cyclic and acyclic compounds,"  low locants (numbering) are assigned to saturated/unsaturated groups in the following decreasing order of seniority: (a) low locants are given to ‘ene’ and ‘yne’ endings; (b) low locants are given first to multiple bonds as a set and then to double bonds; and (c) detachable alphabetized prefixes, all considered together in a series of increasing numerical order.

Thus, ene-group on cyclohexene get the highest priority such as cyclohex-1-ene (or, 1-cyclohexene). Since the double bond has the highest priority, the substituents on  it (if any) should get the next priority such that it should have been 1-methylcyclohex-1-ene (or, 1-methyl-1-cyclohexene), if there are no other substituents. But in your molecule, there is an another methyl group present. Therefore that has the lowest priority. Since 1,2-numbers belongs to the double bond as mentioned above, that methyl can have been only 6 with numbering. Thus, the compound is:  1,6-dimethylcyclohex-1-ene (or, 1,6-dimethyl-1-cyclohexene).

Note that 1-methylcyclohex-1-ene has higher priority over 2-methylcyclohex-1-ene (lowest locant rule within double bond). Therefore, naming the molecule as 2,3-dimethylcyclohex-1-ene (or, 2,3-dimethyl-1-cyclohexene) is also not possible within IUPAC rules.


Why does rearrangement not take place in Wurtz reaction?

The Wurtz Reaction mechanism involves radical anions. Neither radicals nor carbanions rearrange the same way that carbocations do.
The first step of the mechanism is a single electron transfer from $\ce{Na}$ to $\ce{R-X}$:
$$\ce{Na + R-X -> Na+ +[R-X].^{-}}$$
The alkyl halide radical anion fragments to a carbon radical and a halide anion:
$$\ce{[R-X].^{-}->R. + X-}$$
The primary alkyl radical is the least stable radical, but is not as self-destructive as the primary carbocation. Most importantly, primary radicals decompose through disproportionation ...
$$\ce{R-CH2-CH2. +.CH2-CH2-R->R-CH=CH2 + CH3-CH2-R}$$
... or coupling:
$$\ce{R-CH2-CH2. + .CH2-CH2-R -> R-CH2-CH2-CH2-CH2-R}$$
The coupling pathway is a perfectly good explanation of the Wurtz reaction, but another pathway is possible. $\ce{Na}$ is a good reducing agent, and that primary radical is readily reduced to an anion:
$$\ce{R. + Na -> R:^{-} + Na+}$$
The carbanion then reacts with a molecule of alkyl halide in a SN2 reaction:
$$\ce{R:^{-} + X-R -> R-R + X-}$$
The formation of the anion is actually problematic, especially if $\ce{R-X}$ is secondary or tertiary, since E2 elimination will compete.
$$\ce{(CH3)3C:^{-} + (CH3)3C-X -> (CH3)3C-H + (CH3)C=CH2 + X-}$$


Solving for H+ given dissociation of weak acid, c and Ka Can the quadratic equation yield two nontrivial solutions?

What you have got here is a strong acid ($\mathrm{p}K_\mathrm{a} \approx 0.38$) that is not diluted, whereof the latter tells you that you can safely ignore the autoprotolysis of water.

The pH approximation to choose is therefor:
$$[\ce{H+}] = \frac{[\ce{HA}]_0 K_\mathrm{a}}{[\ce{H+}] + K_\mathrm{a}}$$

or in a quadratic form:
$$[\ce{H+}]^2 + [\ce{H+}] K_\mathrm{a} - [\ce{HA}]_0 K_\mathrm{a} = 0$$

which has two solutions:
$$[\ce{H+}] = -\frac{K_\mathrm{a}}{2} \pm \sqrt{\left(\frac{K_\mathrm{a}}{2}\right)^2 + [\ce{HA}]_0 K_\mathrm{a}}$$

From those two solutions, only one is chemically meaningful, which is the positive combination which yields one of your solutions ... $[\ce{H+}] = 0.0834$.


The problem with odors or taste is that it is a form of Qualia. Whether cyanide smells like bitter almonds or is odorless is subjective. Similarly, whether Phenylthiocarbamide taste extremely bitter or completely tasteless is also subjective (depending on genetics).

Even in everyday life, coffee/wine/spices taste different to different people. Even for fruits, the Durian fruit can taste and smell fragrant to certain people but utterly disgusting to others. This is probably why the perfume industry is making billions - if everyone agree that a certain smell is "nice", everyone would be using the same perfume. (Do Genes determine our perfume preferences?)

So I believe it is difficult to have a standard list of odors. Even in the event that the list is created, it will not be universally agreed upon. There are bound to be disputes. Colours on the other hand, (which is also a form of Qualia), are less subjective (everyone can agree that chlorine is green and bromine is brown).


Can fluorescent & phosphorescent molecules still be fluorescent or phosphorescent as ions and not as salts?

Fluorescence derives from an electronically excited state after the molecule has been excited by absorbing a photon, usually in the visible or uv parts of the spectrum. Almost all organic molecules fluoresce to a greater of lesser extent, most common are aromatics for example, benzene and its derivatives as well as most dye molecules such rhodamines or coumarins. They can be ions or neutral.

When a molecule is electronically excited an electron is transferred from the highest occupied molecular orbital (HOMO) into an anti-bonding orbital usually the lowest one (LUMO). From here the molecule, which now has excess energy, can emit a photon and return to the ground state. The lifetime of the singlet excited state  varies between different types of molecules but is generally speaking around a few nanoseconds and not longer than microseconds.

By definition the excited state produced has the same spin as the ground state, which is usually means spin paired electrons $\ce{^ (v)}$. They remain paired even in the excited state (called a singlet excited state). Fluorescence is emission from a molecule with spin paired electrons, the transition is called 'allowed' and happens rapidly.

Occasionally the molecule can suffer a perturbation which causes one of the spins to be flipped to produce 'parallel' spins $\ce{^ ^}$ (heavy atoms or paramagnetic species can cause this). In this case a triplet state is produced and now the transition back to the ground state is far more difficult as the spin-flip has to be undone. The emission from such a state is called phosphorescence, it is long lived vs fluorescence, microseconds and longer, and as the triplet state is lower in energy than the singlet phosphorescence occurs at a longer wavelength in the same molecule, as fluorescence.

The figure  shows the energy levels. Different process compete with one another, thus the singlet can loose energy by fluorescence (with rate constant $k_f$) or by non-radiatively crossing to a triplet state (intersystem crossing) or non-radiatively to the ground state (internal conversion).

continued after question below.

The breaking of formally ‘ forbidden’ transitions is very common indeed, possibly the best example is in the observation of phosphorescence from almost any aromatic or dye molecule. Absorption in these molecules is generally a spin allowed transition, (single to singlet), however, triplet states, from which phosphorescence is observed,  can only be produced by a spin forbidden process, singlet to triplet, and just as in phosphorescence, is a spin forbidden process and becomes allowed due to spin-orbit coupling. In the case of the manganese ions and in other transition metal the forbidden transition is in absorption.

The general treatment starts with allowing the two states to interact via some  operator. In the case of photons (absorption / emission) it is an electric dipole operator, i.e. dipole moment operator. If the operator is $H_0$, the rate of the transition from one state to another is

$$W_{if}^0 \approx \int \psi_f H_0 \psi_i d\tau = \int \psi_f\sigma_f H_0 \psi_i \sigma_i d\tau $$

where $\sigma$ represents the spin wavefunction and $\psi$ the spatial one. As the operator does not depend on spin then
$$\int \psi_f\sigma_f H_0 \psi_i \sigma_i d\tau = \int \psi_f H_0 \psi _i d\tau\int\sigma _f \sigma_i dtau = 0$$
and the  result is zero because the spin wavefunctions belonging to i and f are orthogonal (as singlet & triplet) and so the last integral is zero.

If we suppose that it is possible to mix a little of the initial state wavefunction into the final one and generate a new state with wavefunction $\psi_m$ then $$\psi_m=\psi_f + b\psi_i$$ where b is some mixing constant we can determine later on. Now using this wavefunction as the final state and repeating the calculation above and separating out terms then as orthogonal spin terms are zero the only term left is

$$W_{if} \approx \int \psi_m H_0 \psi_i d\tau = b\int \psi_i \sigma_i H_0 \psi_i \sigma _i d\tau$$
in which case the integral is no longer zero and evaluates to a number.(Note that all the indices are the same unlike the equation for $W_{if}^0$)  This equation shows that there is a finite rate for the process with operator $H_0$. In the case in point, spin-orbit coupling, the operator $\Omega$ depends on the mass of the atoms involved and the product of orbital and spin angular momentum. This is why it is also called a ‘heavy atom effect’ when applied to emission processes.

The constant b is deribed using perturbation theory and turns out to have the form
$$b=\frac{\int\psi _f \Omega \psi _i d\tau}{E_f-E_i}$$
where $\Omega$ is the spin-orbit operator. Thus the rate for the process finally becomes

$$ W_{if} \approx \frac{\int\psi _f \Omega \psi _i d\tau}{E_f-E_i} $$

and we can use symmetry to decide if this integral is zero or is finite. The dependence on energy gap between states is clear, rapid rate constants for small energy gap. (The degenerate case when the energy gap is zero needs special and a more complex treatment)


What is the origin of the bactericidal properties of silver in water?

Silver is not as inert as gold. Tarnish is the name we give to the phenomenon when silver metal is oxidized and becomes a salt. Surfaces made of silver tend to disinfect themselves pretty quickly. As for disinfecting water poured into a silver cup, I imagine that would take a little longer since you have to wait for silver to diffuse away from the surface and into the solution. But even very trace levels of silver can have strong antimicrobial effects.