One mole of solute (NaCl) is dissolved in 1 litre water. The molarity of solution is

The answer $<\pu{1M}$ is correct.

Suppose you have $\pu{1L}$ ($\pu{1dm^3}$) of water and then you add your salt. When you add your salt the volume of the solution increases so now you have $\pu{1 mol}$ of $\ce{NaCl}$ in more than $\pu{1L}$ of solution. That means that you can't have a $\pu{1M}$ solution anymore.

If you want a $\pu{1M}$ solution then:

Add $\pu{1 mol}$ of salt in a flask.
Add water until you reach $\pu{1L}$ of volume.

EDIT: when you add salt, the volume of the solution increases slightly and less than what you expect(volume of water + volume of dry salt).


Terminology of atomic spectroscopy: Difference Among Term, States and Level

The figure below shows the situation between configuration for a $p^2$ configuration, terms, levels and states. The word 'state' tends to be used colloquially to mean any of  Term, Level or State.

The Configuration such as  $(1s)^2$ or ...$(2p)^2$ etc. tell us which orbitals are occupied.

These are split with electrostatic (Coulomb) coupling to form Terms.

The Term Symbol informs about the angular momentum. The number of levels in each Term is $(2J+1)$, the number of States is $(2L+1)(2S+1)$ .

The term symbol is

$$\large ^{2S+1}L_J$$

where $S$ is the total spin and $L$ the orbital angular momentum.  $J$ represents the angular momentum as quantum numbers $(J=L+S)$ and is determined by adding individual spin numbers according to Glebsch-Gordon series, i.e. if $q_1,\,q_2$ are angular momentum quantum numbers  then the series is all terms (separated by unit steps) with values $q_1+q_2 , q_1+q_2-1,\cdots \to |q_1-q_2|$. (There may be only be term).

The Terms are split with a magnetic interaction such as spin-orbit coupling, i.e coupling of spin angular momentum with that of the orbital. It is not present for S orbitals because here $L=0$.

Levels are distinguished by total angular momentum $J$, and are $2J+1$ degenerate.

An external magnetic field will remove all remaining degeneracy to form States. These are distinguished by $m_J$ and there are $2J+1$ of them.


The intermediate before forming the alkoxide is a tetrahedral intermediate which is also unstable. You're going from a high energy species to another high energy species.


How many different possible structures (isomers) are there for the complex, [Mn(OH2)3(NH3)3] 3+?

Metal complexes with three each of two types of ligands $(\ce{MX3Y3})$ have two isomers: mer- and fac-isomers. For example, for $\ce{Co(NH3)3Cl3}$ there are two isomers:

  Fac for facial, in which all three ligands of each type are clustered on a single face of the octahedron and the three other three ligands are on the opposite face.

  Mer for meridional, in which the three ligands of each type are arranged meridianally in a row around the metal and the other three ligands are arranged likewise on the opposite side in a perpendicular meridian.


Are you asking experimentally or theoretically?
Theoretically speaking for general cases you should know what is the energy of the high-spin and low-spin configuration, and in given state, what is the approximate spin and vibrational entropy so you can estimate an equilibrium constant for given temperature. Now this either needs some calculation, or some sort of database from where you can make this estimations.

Experience (i.e. database or whatever series) says that water is a weak ligand, from that you generally guess that water complexes are generally in high spin state.

Experimentally you can determine spin state by measuring magnetic susceptibility, interpreting EPR, measuring 57Fe Mossbauer spectrum etc


Do atoms in a molecule actually keep rotating freely in case they have sigma bond?

You really are referring to molecular conformations:

  Conformation, any one of the infinite number of possible spatial arrangements of atoms in a molecule that result from rotation of its
  constituent groups of atoms about single bonds.

Below are given two possible (out of many!) conformations of the ethane $\ce{C2H6}$ molecule:

The single $\sigma$ bond allows free rotation around the inter-nuclear axis of the two $\ce{C}$ atoms. One consequence is that around the single bond no cis-trans isomers can exist.

But if we add a $\pi$ bond, so we have a double bond and ethene $\ce{C2H4}$, then the methylene groups can no longer freely rotate about the bond.

That's because the electron probability density added by the $\pi$ bond doesn't lie on the inter-nuclear axis, but above and below it. Double bonds, and by extension also triple bonds, are rigid, with respect to rotation of the atoms they bond. This can give rise to cis-trans isomers.

Below: highly schematised rendition of the electron density of a $\pi$ bond:


If you look carefully you'll see that your compound has stereocenters, yet is achiral.  Both stereocenters have the opposite configuration.  This is the meso-form, these are achiral, and the hydroxy groups are prochiral.  Look carefully for the rotoinversion center!


onset of oxidation for cyclic voltammetry

Please don't apply the reversible CV diagram on something which is clearly a non-classical cyclic voltammogram. When you experimentally don't get the shape as predicted by the theory, it means it that particular theoretical model does not apply in your case.
Look at three different types of voltammograms in this picture. Does your CV look like a reversible one? If not, then you were comparing an apple with an orange.

There is a very nice article which is relevant to you case:
Practical Aspects of Cyclic Voltammetry: How to Estimate Reduction Potentials When Irreversibility Prevails,  Journal of The Electrochemical Society, 2019,  166(5):H3175-H3187
This figure is relevant:

Examples of cyclic voltammograms showing chemically reversible and irreversible oxidation with designated potentials: E a = anodic potential, E c = cathodic potential, E (i) = inflection-point potential, E (p/2) = half-peak potential, E (e) = edge potential.


\begin{array} {|l|r|r|}
\hline
  &\mathrm{pK_a~in~water} &\mathrm{pK_a~in~DMSO}\\
\hline
\ce{H2O} &15.74 &31.4\\
\hline
 \ce{CH3CH2OH} &16 &29.8 \\
\hline
\end{array}

As we can see from the $pK_a$ data water is slightly more acidic than ethanol in water but in DMSO ethanol is notably more acidic. In DMSO hydrogen bonding is minimal so the acidity is primarily dependent on the acid and not the solvent it is in.

However, in water the hydroxide ion is hugely stabilised by hydrogen bonding (the ethoxide is as well but not nearly to the same extent) so this accounts for the greater acidity of water in this situation.

http://research.chem.psu.edu/brpgroup/pKa_compilation.pdf
http://www.chem.wisc.edu/areas/reich/pkatable/index.htm


As the name suggests the oxidizing flame is used to oxidize probes, while the reducing flame is used to reduce probes. Hereby oxidation is done by oxygen, while reduction is done by reducing agents in the flame, such as carbon monoxide or hydrogen.

In the charcoal block reduction test with a blowpipe one often will oxidize the probe in the first step to get rid of nonmetal elements in the probe, and then will use the reduction flame to reduce oxidic metal compounds in the melt to the elemental metal. Volatile metals will hereby be precipitated at the colder regions of the charcoal as oxides. Metal and oxide precipitate can then be further analyzed.


Why is sodium methanethiolate still a better nucleophile than sodium methoxide when the solvent is ethanol

Thiolates are generally more nucleophilic than alcoholates. So whenever the choice happens to be between a nucleophilic sulphur and a nucleophilic oxygen, the sulphur will win.

Both the methanolate and the methanethiolate will show reduced nucleophilicity due to partial protonation, though. And of course you have introduced ethanolate as an additional nucleophile, too (which should be the least reactive, though).


Does Magnesium metal react with carbonic acid?

First, thanks for reporting some obvious unexpected chemistry occurring. Others in your community may also have similar metal leaching issues, which especially for children, may be a health concern (see, for example, this article).

My speculation, on what is occurring, is that the Iron pipe in your shower is displacing (upon standing over time) some dissolved copper (possibly from created bicarbonate). The likely acidic rainwater may also contain chloride or nitrate salts, which may further complex with the fresh active copper precipitate. Here is a related illustrative source, per an article 'Electrochemical detection of nitrate and nitrite at a copper modified electrode' to quote:

  The development of a reagentless electrode system for the determination of nitrate is presented. The approach is based upon the deposition of a macroporous copper deposit which shows marked selectivity for nitrate ion under mildly acidic conditions (pH 3) with a linear range extending from 10 to 200 mM nitrate.

So, do not employ Magnesium, address the iron presence. The coloration means there is some, but in actual amounts, most likely limited copper ion presence along with nitrate.


I think your main question is related to the difference? Have you heard a famous poem, The Blind Men and the Elephant? It is a story, I'd rather copy from Wikipedia

It is a story of a group of blind men who have never come across an elephant before and who learn and conceptualize what the elephant is like by touching it. Each blind man feels a different part of the elephant's body, but only one part, such as the side or the tusk. They then describe the elephant based on their limited experience and their descriptions of the elephant are different from each other. The moral of the parable is that humans have a tendency to claim absolute truth based on their limited, subjective experience as they ignore other people's limited, subjective experiences which may be equally true.

So replace these men with these analytical techniques and the sample with the elephant. XPS /EDS can see what Raman/FTIR cannot and vice versa.
XPS/EDS are elemental techniques and they will tell you what elements are present. Sometimes XPS spectrum, for an expert, can also reveal what is the oxidation state of that element, if you do high resolution scans. Each sample may take hours and hours.
Raman and FTIR can used to probe surfaces (they are fancy surface vibrational spectroscopy techniques) and it will tell you the molecular information, and even the angle at which the molecule is standing on the surface.
From the Raman spectra of an unknown it will be very hard to guess what elements are present.
In short, these are complementary techniques. Someone who is trying to study surfaces would use all of them - if needed.


What is the chemical composition of the grignard reagent?

A Grignard reagent is an organomagnesium halide of the generic form $\ce{RMgX}$, where $\ce{R}$ is a hydrocarbon group and $\ce{X}$ is a halide, usually $\ce{Cl}$, $\ce{Br}$, or $\ce{I}$.

These reagents are produced by the reaction of magnesium metal with the corresponding organic halide in a non-protic (but coordinating) solvent, like diethyl ether.

For example, you would make phenylmagnesium bromide from bromobenzene and magnesium:

$$\ce{PhBr(C6H5Br) + Mg ->[][\ce{Et2O}]PhMgBr}$$

They are moisture and air sensitive, but not so much that you need to get crazy with inert atmosphere. These reagents are not stable outside of solution.

Grignard reagents can be made in most laboratories set up for organic chemistry. In fact, they are very common reactions to include in undergraduate organic chemistry lab courses because they are very well behaved.


What is the Natural abundance of Potassium

A). If you know the percentages of the abundance of $\frac{2}{3}$s of the isotopes then finding the percentage of the $3^{rd}$ is as simple as

$ 100\% - (\%_{\ce{^39K}}+\%_{\ce{^41K}}) = \%_{\ce{^40K}} $

B). Finding the mass of the isotope should be pretty simple. It would be

$(\mathrm{Mass\ of\ proton} \cdot \mathrm{{\#\ of protons}} )+(\mathrm{Mass\ of\ a\ neutron} \cdot \mathrm{{\#\ of neutrons}} ) \approx \mathrm{Mass\ of\ isotope}$

You neglect the weight of electrons because the mass is small relative to the others.


Which is more reactive, chlorine radical or chloride ion?

I think we can also approach this from an energetics perspective. Reactivity can also be seen to be related to thermodynamic stability in the following manner: The more thermodynamically stable a chemical species is, the less reactive it is. With this in mind, we can look at the enthalpies of formation of $\ce {Cl.}$ and $\ce {Cl^-}$.

Your question is slightly vague as you did not state the specified state of the substance. This can critically affect the answer. The reactivity of a heavily solvated $\ce {Cl^-}$ ion is certainly different from that of a free $\ce {Cl^-}$ ion in the gaseous phase.

By definition, the enthalpy change of formation, $\Delta H_f$, of an element in its standard state (e.g. $\ce {Cl2 (g)}$) is $\ce {0}$. $\ce {2 Cl.}$ are produced from the homolytic cleavage of the $\ce {Cl-Cl}$ bond in $\ce {Cl2}$. Knowing the bond dissociation energy of this bond to be $\ce {+242 kJ/mol}$, we can approximate $\Delta H_f$ of $\ce {Cl.}$ to be $\ce {+121 kJ/mol}$.

Similarly, the gaseous chloride ion can be produced from the gain of an electron by $\ce {Cl.}$. The first electron affinity, $\ce {1st E_a}$, of chlorine is the energy change when one mole of electrons is acquired by one mole of these gaseous chlorine atoms to give one mole of singly, negatively-charged gaseous $\ce {Cl-}$ ions. $\Delta H_f$ of $\ce {Cl- (g)}$ can thus be approximated as $\Delta H_f$ of $\ce {Cl.}$ + $\ce {1st E_a}$ of chlorine . Since this $\ce {1st E_a}$ is $\ce {-349 kJ/mol}$, this enthalpy change of formation for $\ce {Cl^- (g)}$ can be approximated as $\ce {-228 kJ/mol}$.

As can be seen from energetics data, $\ce {Cl^-}$ is more stable than $\ce {Cl.}$ in the gaseous phase. Thus, we may say that the $\ce {Cl.}$ is more reactive in the gaseous phase. Furthermore, being charged, $\ce {Cl^-}$ is solvated by water molecules in aqueous solution, further increasing its stability as it has ion-dipole interactions with the neighbouring water molecules. Electrically-neutral chlorine radicals would not experience this solvation effect, thus they would also be more reactive in the aqueous phase. In non-polar solvents, $\ce {Cl^-}$ and $\ce {Cl.}$ would have similar reactivity as in the gaseous phase since they do not experience starkly different solvation effects. Overall, based on energetics, we may say that $\ce {Cl.}$ is generally more reactive than $\ce {Cl.}$.

Additionally, it is important to point out the difference in the type of reactivity of the two different chemical species. Being negatively-charged, chloride ions participate in ionic reactions, involving charged species. These reactions include nucleophilic substitutions, as well as nucleophilic additions. On the other hand, chlorine radicals mostly participate in non-polar reactions, involving mostly radical species, such as free radical substitutions.


As always, writing a Lewis structure seems to be a trivial task, but sometimes it just isn't. The smaller the molecule becomes, more often than not the bonding situation becomes more complicated and complex. This can often be attributed to the high symmetry of such molecules. Prime examples are $\ce{O2}$, $\ce{CO}$, $\ce{CO2}$, $\ce{N_xO_y}$, etc..

I found no evidence anywhere that $\mathbf{\ce{NCO+}}$ is bent. I have tried calculations in the singlet and triplet states and all of them lead back to the molecule being linear. It would be nice to know which textbook you are using for general reference.

To be perfectly candid here, it is not even a trivial task to set up the right calculations. At the DF-BP86/def2-SVP level of theory, the triplet state is 151.8 kJ/mol lower in energy than the singlet state. A broken symmetry singlet (a biradicaloid electronic state) is still 111.3 kJ/mol lower in energy.
On the CASSCF(6,6)/def2-SVP level of theory the triplet wave function is 143.3 kJ/mol lower in energy than the singlet wave function.

Lewis structures were not meant to accommodate complicated bonding situations. Instead they are meant to be easily understood. Even the concept of resonance can be quite tricky in certain situations. Without knowing that the ground state of dioxygen is a triplet state, we would interpret its common Lewis structure wrong.

The same shall apply here. We need to treat the concept of Lewis structures within its limitations. Based on the triplet DF-U-BP86/def2-SVP calculation the best representation is probably a bit surprising: The positive charge would be located at the most electronegative atom, while the spin density is located at the nitrogen (1). The broken symmetry solution gives us instead two double bonds with the positive charge at the nitrogen (2).
In all cases we shall not forget, that the most electropositive element is carbon in this molecule, so the actual charge is most likely located there (3). Since the π-orbitals are stretched over the whole molecule, the spin density will not entirely be located at nitrogen alone (4). A meaningful interpretation in terms of resonance structures (NRT of NBO) is, however, almost impossible. I tried it based on the closed shell singlet DF-BP86/def2-SVP calculation and there are three contributing main structures (5).

For what it's worth, the structures you have proposed are valid resonance structures. They are not all of them, but they are a good start. As you can see, any of them capture only a tiny glimpse in the complicated nature of this molecule.


How can I figure out if my store bought chlorine is the percentage it says it is?

Bleach solutions can be analyzed by addition of iodide and back titration with thiosulfite - somewhat difficult for non-laboratory conditions. Kits are available, but expensive (~$250).

A less precise (but probably adequate) method involves measuring the heat evolved as bleach oxidizes acetone. https://www.flinnsci.com/api/library/Download/650c3e8ed0574ed18c7b37b8832cccae It seems that you might be able to modify this process to do a rough quantification (+/- 10%) quickly. It was also discussed briefly on Quora: https://www.quora.com/When-happens-when-acetone-and-bleach-are-mixed

Just realize that if you drop solid hypochlorite into pure acetone, it could explode, catch fire, and make you wish you had never tried it. Creep up on this reaction: add only small amounts of acetone (1-2 grams) to a dilute hypochlorite solution (~5% ?) until you get the feel of how fast and how hot this reaction will go.

Eye protection is mandatory. Also have some water available (like a hose). Just be safe by expecting danger.


Why is methyl group more electron-donating than tert-butyl group?

Methyl group is not more electron donating than tert-butyl group, as the video provided in the link suggests (11:20 of the video), which is true.

Inductive effect is a permanent effect which results in displacement of electrons along the chain of atoms in the molecule (here, carbon atoms).  The reason for this displacement can be worked out from the electronegativity difference between the atoms.

Consider the tert-butyl group.  In a C-H bond, the electron density is directed towards carbon.  Methyl group, which has three hydrogen atoms attached to a carbon, pushes the electron density towards $\ce{C2}$ of tert-butyl acetic acid.  So do the two other methyl groups.  The amount of partial negative charge accumulated on $\ce{C2}$ of tert-butyl group (has 3 methyl groups) is more than that on methyl group (only has 3 hydrogens) of acetic acid.

The more negative charge on $\ce{C2}$ in tert-butyl acetic acid destabilizes the conjugate base, compared to a simple acetate ion.  Hence acetic acid has less pKa than tert-butyl acetic acid.


Does CH3COOH + H2O (vinegar) lose its antibacterial and antiviral properties when exposed to air?

Acetic acid boils at higher temperature than water, at 118 °C. Acetic acid does not react with the air.
It is more likely that acetic acid antimicrobial properties will improve as more water evaporates, because it evaporates faster.


Why do people often capitalize element names?

This is a Google ngram of the usage of the words: iodine, carbon, nitrogen, zinc and Iodine, Carbon, Nitrogen, Zinc (not at the beginning of the phrase).
There is no historical trend, at least not an usage of an archaic form (we can see that the maximum  usage of the capitalized element name is from 1900 to 1960 but normalized to the overall usage is steady).

We can make some hypothesis about why someone could choose to write capitalized element names:

If we must write element abbreviation capitalized (Zn, Cu, C, N) I  can think I have to write the element capitalized even if is not abbreviated.
Someone (me too) may think that if I write Zinc it is more unambiguous and means I'm talking about the concept of element "all atoms with the same number of protons in the atomic nucleus" and not about the metal zinc.  If you write report with some elemental techniques (XRF, AES) sometimes you feel the urge to specify to the reader (that maybe is not a chemist): "when I write that I've found zinc I mean that in the sample could be present $\ce{Zn}$ but also $\ce{ZnO}$, $\ce{Zn+}$, $\ce{Zn^{2+}}$".
The fact the atoms are identical and the definition of a proper noun is: "a noun that in its primary application refers to a unique entity" could induce to think that elements may fit the definition, and so gain the privilege to be capitalized.
In alchemy elements such water and fire were capitalized.

However up to my knowledge the form not capitalized is preferred even in the others languages, in Italian I'm sure.


Could detergents dissolve cell membranes?

Yes they can dissolve membranes and are even used to isolate active forms of membrane proteins by solubilization. The product are micelles of a lipid-detergent mix and a protein-lipid-detergent complex. Maybe you find this short abstract interesting.

  A critical step in any in vitro analysis of membrane proteins is the
  solubilization of the membrane to extract the protein of interest in
  an active form to obtain an aqueous solution containing the membrane
  protein complexed with detergents and lipids in a form suitable for
  purification and further analysis. This process is particularly
  delicate as the aim is to maximally disrupt the lipid components of
  the membrane while putting the protein components in an un-natural
  detergent environment without perturbing them. Looked at this way, it
  is remarkable that it ever works. Although the process is difficult
  and hard to master, an increasing number of membrane proteins have
  been successfully solubilized in active forms, allowing some general
  principles to be established that we illustrate in the method
  developed in this chapter.

Duquesne K., Sturgis J.N. (2010) Membrane Protein Solubilization. In: Mus-Veteau I. (eds) Heterologous Expression of Membrane Proteins. Methods in Molecular Biology™ (Methods and Protocols), vol 601. Humana Press


What is the mole fraction of dihydrogen in a mixture, given the relative rates of effusion of its components?

If you are familiar with Graham's Law, you will know that the ratio of the rates of effusion of two gases is inversely proportional to the square root of the ratio of their molar masses.  That is:

$$\frac{R_1}{R_2}=\sqrt{\frac{M_2}{M_1}}$$

Here, $M_1 = M(\ce{H2}) = \pu{2 g//mol}$ and $M_2 = M(\ce{He}) = \pu{4 g//mol}$, so the lighter of the two, hydrogen, effuses at $\sqrt{2}$ the rate of helium, or conversely, the heavier, helium, effuses at $\frac{1}{\sqrt{2}}$ the rate of hydrogen.  That's where the fraction came from.

Since we know that helium is escaping three times faster than hydrogen, there must be three times more helium than would product an equal rate of flow for the gases.  An equal flow would occur when there is $\sqrt{2}$ times as much helium, and triple flow would occur with $3\sqrt{2}$ times as much helium.

The total amount of gas present is then:
$$\ce{H2} + \ce{He} = 1 + 3\sqrt{2}$$
and the mole fraction of hydrogen is:
$$x(\ce{H2}) = \frac{n(\ce{H2})}{\text{total amount of substance}}
= \frac{1}{3\sqrt{2} + 1}$$


Why only two atoms share an electron and not three?

The normal case for a covalent bond is indeed a 2-centre 2-electron bond.

There are however cases, where 3 centres (= atoms) either share 2 or 4 electrons.

In organic chemistry, the most famous case for an electron-deficient 3-centre 2-electron bond is the 2-norbornyl cation.
In inorganic chemistry, boranes with their banana bonds are the role models.

Hypervalent molecules with 3-centre 4-electron bonds, like $\ce{SF6}$ or the bifluoride anion $\ce{HF2-}$ have been discussed here too.


Is it really possible for a liquid to stay afloat a gas (vapor)?

Yes! Here's a video of Cody floating NaK (liquid) or ethanol/water (liquid) on compressed xenon (gas):
https://youtu.be/AsP4yMY-a6U


Rotation energy barrier for carboxypyrazolate

One of way to answer this question is to put it in some QM software. The system is small enough and has higher symmetry so computations shouldn't take to long, even on an older computer.

For co-planar geometry the energy calculated at B3LYP/6-31G(d,p) is –492.132206748 a.u.
The geometry with dihedral angle of 90° between $\ce{CO2}$ and $\ce{C3N2}$ is –492.116719625 a.u. That makes roughly 41 kJ/mol (10 kcal/mol).

Now if you really want to be sure if that's the right energy you should make frequency calculations on both structures to find out if they are minima or transition states. This can take a bit longer (I haven't done it) and I would expect that the better structure (with lower energy) should have no imaginary frequencies and the higher one should have one (TS). If by some accident both of them are minima than TS has probably a dihedral angel somewhere around 45°.

Of course this is only computation. It gives as clues but not the exact answer. You can go for higher level of theory, add solvent modelling and... or you can take your compound and try to make temperature depending NMR spectra. With a bit of luck you can freeze your TS and find out how high is your barrier.

p.s.

It would be good if you mention in your post what tools (in this case computational) are available to you. I really would prefer to explain how to get the number instead of giving just dry answer.


How to calculate pH of solution made of H3PO4 and Na3PO4?

You have the the following six independent equations in order to specify the unknown concentrations of six species as present in an aqueous solution consisting of $\pu{0.1 M}$ $\ce{H3PO4}$ and $\pu{0.05M}$ $\ce{Na3PO4}$:

\begin{align}
K_1 &= \frac{[\ce{H+}][\ce{H2PO4-}]}{[\ce{H3PO4}]}
     = 7.5 \times 10^{−3} \tag{1}\\
K_2 &= \frac{[\ce{H+}][\ce{HPO4^2-}]}{[\ce{H2PO4-}]}
     = 6.2 \times 10^{−8} \tag{2}\\
K_3 &= \frac{[\ce{H+}][\ce{PO4^3-}]}{[\ce{HPO4^2-}]}
     = 3.6 \times 10^{−13} \tag{3}\\
K_\mathrm{w}
    &= [\ce{H+}][\ce{OH-}]
     = 1 \times 10^{−14} \tag{4}\\
0.15
    &=[\ce{PO4^3-}] + [\ce{HPO4^2-}] + [\ce{HPO4-}] + [\ce{H3PO4}]\tag{5}\\
[\ce{H+}]
    &= 3[\ce{PO4^3-}] + 2[\ce{HPO4^2-}] + 2[\ce{HPO4-}] + [\ce{OH-}] -0.15\tag{6}
\end{align}


Can solid acids and bases have pH values? If not, how are they classified as acids or bases?

I think we should not mix the concept of pH, which is purely an analytical measurement, with the concept of acid and bases. It is a common misconception that pH does not exist in organic solvents.  As you already know, there are several views of an acid or bases. The current IUPAC version is "A molecular entity or chemical species capable of donating a hydron (proton) (see Brønsted acid) or capable of forming a covalent bond with an electron pair (see Lewis acid)." This definition does not require the presence of water or any pH value.

If we take pure dry HCl gas and ammonia gas and mix them. The reaction is a classic acid base reaction. Solid ammonium chloride is formed instantly. One may then ask a question of what is the pH of a gas? However gas phase acid base chemistry does exist and it can be studied by mass spectrometry but pH does not need to be invoked.

Now instead of asking the pH of solid acids/bases, a more realistic question is what is the surface pH of a given solid? I was interested in this type of problems sometime ago but didn't find much literature. Assume an ion-exchanger which consists of $\ce{SO3-H+}$ groups on a solid polymeric matrix, basically sulfonate styrene divinylbenzene. Assume that it is in equilibrium with 1 mM HCl solution. The solution pH is 3, but what is the surface pH? It is certainly way below zero, because sulfonic acid is a very strong acid and the surface concentration of sulfonic acid groups is pretty high yet the groups are not mobile in the solution!

Similarly, solid acid catalysts exist. Also see how solid acids are titrated [1].

References

Chai, S.-H.; Wang, H.-P.; Liang, Y.; Xu, B.-Q. Sustainable Production of Acrolein: Investigation of Solid Acid–Base Catalysts for Gas-Phase Dehydration of Glycerol. Green Chem. 2007, 9 (10), 1130–1136. https://doi.org/10.1039/B702200J.


Why does H2O electrolysis produce O2 on the anode instead of H2O2

It does, as an intermediate product. The peroxide formed is on the surface of the electrode which changes it's chemical behavior.

Electrolysis is a four-electron process. In the first step, an electron is removed from a water molecule, producing a neutral OH adsorbed to the surface of the electrode. The second step removes another electron from the adsorbed OH, producing a neutral oxygen atom adsorbed to the electrode. Third, another electron is removed from a water molecule leaving an adsorbed OOH. Then the fourth electron comes from the OOH to produce O2.

My understanding is that the interactions with the electrode partially stabilize the unstable intermediate states. This is the reason some electrodes work better than others as catalysts.

This paper has more than you'll ever need to know: https://doi.org/10.1002/cctc.201000126


What's that tangy smell on cheap plastic?

Without a sample and a gaschromatograph it is hard to say what exactly it can be but... there are at least three well-known sources of odour in plastics:

Some residual of the monomer that makes up the plastic (that is a polymer).
Some residual of some other substance used during the manufatcturing process (catalyzer, co-polymers, modifiers and so on). For example, PETcan sometime contain small traces of terephthalic acid.
Some odourizing substance that is added to the plastic just to cover others smells.

Usually, tangy smells are a symptom of some kind of  ester


How does aqueous ammonia react with hypochlorous acid in water?

Yes, aqueous ammonia will react with hypochlorous acid to form chloramine gas as suggested by @Waylander. But the reaction will also form an intermediate product i.e  ammonium hypochlorite($\ce{NH4ClO}$).

$$\ce{HClO + NH3•H2O → NH4ClO + H2O}$$

  Hypochlorous acid react with ammonium hydroxide to produce ammonium
  hypochlorite and water. Ammonium hydroxide - diluted solution.(chemiday)

The mechanism of this reaction is explained in this Reddit post. You will find the insight of the post useful. I am just pointing out one information.

  Ammonia solution and hypochlorite react (in solution) to produce toxic
  chloramine gas. Ammonium hypochlorite is also formed during the course of
  the reaction, but it decomposes somewhat rapidly to chloramine.

I am assuming this is an exercise question. Do not try to attempt this reaction if you are not an experienced chemist as it also produce deadly byproducts like chlorine gas, hydrazine, dichloroamine and probably nitrogen trichloride. Chloramine is itself poisonous.


How does sodium in ethanol reduce carbonyl compounds?

Sodium is a powerful reducing agent. It wants to donate an electron to things that are electron deficient.

The mechanism of the reduction of aldehydes and ketones by sodium would be similar to that of the Bouveault-Blanc reduction. Ethanol is the source of the protons (at least early in the reaction). The dots in the reactions below signify radicals (unpaired electrons), which are represented to the best mchem and MathJax's ability (which is not great).

  Step 1 - one electron transfer from sodium to carbonyl group:

$$\ce{Na. + RCH=O->Na+ + RCH.-O^-}$$

  Step 2 - abstraction of a proton from ethanol:

$$\ce{CH3CH2OH + RCH.-O^- <=> CH3CH2O- + RCH.-OH}$$
This step is an equilibrium step, and is driven toward products by the second electron transfer (which consumes the $\ce{RCH.-OH}$.

  Step 3 - one electron transfer from a second sodium atom to the organic radical:

$$\ce{Na. +RCH.-OH -> Na+ + RCH:^{-}{}-OH}$$

  Step 4 - second proton transfer from ethanol:

$$\ce{CH3CH2OH + RCH:^{-}{}-OH -> CH3CH2O- + RCH2OH}$$

This reaction, while technically an acid-base equilibrium, essentially goes to completion because of the vast disparity in acidity of $\ce{C-H}$ and $\ce{O-H}$.

The net reaction is:

$$\ce{2Na + RCH=O + 2CH3CH2OH -> RCH2OH + 2CH3CH2ONa}$$
$%edit$


Are all esterification reactions reversible? What is their general mechanism?

I'll start this answer off with saying that both of these answers were the first result when searching for Google.

Yes, The esterification reaction is both slow and reversible

and here is the mechanism from an image search


Is my reasoning for resonance structures correct?

Well, your answer seems to be correct but not your reasoning. In 2 and 4 there are 2 extra pi electrons in the system which seem to have come out of nowhere. The best way to solve is to draw all the resonance structures and see which ones match with the options.


In a monomolecular reaction, how can a reactant experience collision to react?

One reactant, e.g. O3, a pure chemical species, has many molecules that are in continual motion and can collide with each other (as well as with the walls of a container or with a catalyst), so you'll get some O2 generated. Or the converse, if you have UV photons colliding with pure O2 gas.


Orbital conservation in Molecular orbital theory

In the linear combination of atomic orbitals (LCAO) there would be a bonding orbital and an antibonding orbital when two atomic orbitals combine to form a bond. The bonding orbital would be lower in energy that either atom contributing an orbital, and the antibonding higher than either.

One twist comes from the number of electrons. If there are just two electrons, then they both go into a bonding orbital. If there are three electrons then two go into a bonding orbital and one into the antibonding orbital.

There are three p orbitals so two atoms could be bonding with 1-3 p orbitals. If two p orbital bonds were being used then both p molecule bonding orbitals should be lower in energy than either antibonding orbital.

The whole concept of having a bonding orbital and an antibonding orbital is necessary to explain for instance electronic transitions. Imagine an oxygen atom in free space. If an electron is added, the oxygen molecule will accept the electron to form an ion $\ce{O^{-}_2}$. The electron doesn't go to one oxygen or the other, it is shared between both in a molecular orbital. Look at the diagram below for $\ce{O2}$. We'd predict that the electron will fall into one of the antibonding pi orbitals ($\pi^*$). From this we'd predict that a free oxygen atom could actually take three extra electrons and be stable. However the fourth electron would fill all the orbitals and $\ce{O^{-4}_2}$ would have no more electron binding energy than two free $\ce{O^{-2}}$ atoms.


How can glycine act as a buffer at pH 6.00 and why?

Your equation is not correct for pH 6; it should be for pH 1. For pH 6, the equation should be:

If an acid is added, then the equation becomes:

and if a base is added, then the equation is:


Why does cesium have a positive electron affinity?

All elements have a positive (first) electron affinity, except noble gases, for which it has not been measured conclusively, but might be either slightly positive or slightly negative.

You can more clearly see the reason why electron affinity is positive if you think of the opposite process: detaching an electron from the negative ion X– to form neutral atom X. Because the anion X– is stable, it takes energy to take away the electron.


Pentazole exists - an $\ce{N5H}$ ring. It is stabilised by aromaticity, with 6 $\pi$-electrons in a cyclic, planar system.

Sulphur forms many allotropes which are rings but these do not contain hydrogen. $\ce{S8}$ and $\ce{S7}$ are the most common.

Phosphorus also forms cyclic allotropes such as white phosphorous, $\ce{P4}$.

DavePhD also mentioned in his answer that a silicon analogue of benzene has been synthesised recently but it is not $\ce{Si6H6}$ because it has other groups attached to some of the silicon atoms.


Why does benzene have 3 π–π* transitions in the UV range, but ethene, butadiene, and hexatriene each have only 1?

It seems that here we only need to consider the lowest energy transitions, i.e. HOMO–LUMO transitions. The difference between benzene and the linear polyenes is that both the HOMO and LUMO of benzene are doubly degenerate, whereas the linear polyenes do not have degeneracy.

If we think about this HOMO–LUMO excitation in the case of benzene, we can schematically represent it like this:

The ground state has overall symmetry $\mathrm{A_{1g}}$ because all electrons are paired, but the first excited electronic configuration above $(\mathrm{e_{1g}})^3(\mathrm{e_{2u}})^1$ corresponds to three different electronic states, each with a slightly different spatial distribution of electrons. The symmetries of these three states are obtained by taking a direct product of the relevant irreps:

$$\mathrm{E_{1g} \times E_{2u} = B_{1u} \oplus B_{2u} \oplus E_{1u}}$$

Because these three states arise from the same electronic configuration, their energies are going to be fairly close to each other, and all three electronic transitions from the ground state

$$\mathrm{A_{1g} \to B_{1u}, \quad A_{1g} \to B_{2u}, \quad A_{1g} \to E_{1u}}$$

should fall within the UV range.

For any linear polyene (including ethene, butadiene, and hexatriene), both HOMO and LUMO are nondegenerate, so the $(\pi)^3(\pi^*)^1$ configuration will only correspond to one electronic state.

This is a rather simplified analysis, because we are only considering transitions which conserve spin, i.e. singlet–singlet transitions. These are so-called "spin-allowed" transitions. The singlet–triplet transitions also occur within the UV range, but are technically "forbidden". On top of that, not all of the three singlet–singlet transitions in benzene are equal: two of them are also "forbidden", as the transition dipole moment $\langle i | \hat{\mu} | j \rangle$ vanishes. But that is a story for another day. The keen reader may wish to check out J. Mol. Struct. 1973, 19, 143–166 for an assignment of the electronic transitions of benzene.


Reference Request - bond lengths in PCl2F3

To complete your literature survey, include this publication cited at least 11 times so far: «Dichlorotrifluorophosphorane (PCl2F3): molecular structure by gas-phase electron diffraction and quadratic force field» by French et al. in Inorg. Chem. 1985, 24, 2774–2777, doi 10.1021/ic00212a014.  Despite the paywall, ACS's «In lieu of an abstract, this is the article's first page.» informs you about some key properties you search, values including ($2\sigma$ estimated) e.s.d.

How did I find this publication?  Because the English Wikipedia has an entry about it (don't overlook the Chinese including complementary information).
But how to know if Wikipedia indexed such a compound?  Way better than trial-and-error is to address this question to the Wikipedia Chemical Structure Explorer allowing to query there by structure or string, which may lead you to the corresponding article (in Wikipedia).  It is backed by a public database (GitHub repository) initiated by Ertl et al.
Source: Ertl, P.; Patiny, L.; Sander, T.; Rufener, C.; Zasso M J. Cheminf., 2015, 7, 10, doi 10.1186/s13321-015-0061-y, (open access).


An ion is created. Usually this is not noted down in the nuclear reaction because while talking of nuclear reactions we only concern ourselves about nuclei, not the entire atom. The reaction is to be read as "A 14-$\ce{C}$ nucleus decays into a 14-$\ce{N}$, an electron, and an antineutrino."

Also, note that the neutrino has no charge.


Understanding a passage related to salt buffer

There is no such thing as  $\ce{NH4OH}$ . They would react to form water and ammonia. Thus, to say that both exist in the solution at such high concentrations would be mistaken. They would react. Consider the reaction between ammonium and hydroxide.

$\ce{NH4+ + OH- -> NH3 + H2O}$

However, do note that at low concentrations, they can exist according to dissociation. However that is not what your passage is saying

*On your test, it may be good to assume the reaction goes to completion and then solve for the dissociation of ammonia. That is most likely what is expected of you.

*I thought of a more understandable analogy:

Think about $\ce{OH-}$ and $\ce{H+}$ in solution at 25 celsius. If one increases, the other decreases! According to ${Kw}$. Adding a lot of acid and base together would simply neutralize, they cannot exist at any higher than ${Kw}$


Origin of diastereoselectivity in Simmons–Smith cyclopropanation of allylic alcohol

Yes. Right after Clayden mentions this stereoselectivity, he proceeds to state that this is due to "coordination between the zinc atom and the hydroxyl group in the transition state" (pp. 1067-8, 1st ed):


Why is only monochromatic light used in polarimeter?

The reason is partly historical and partly scientific. Most of the polarimetric data exists with the sodium yellow emission at 589 nm because sodium lamps were among the most convenient ones and remember electricity is rather new. Early light sources were either flames or sunlight. It was easy to generate sodium emission by introducing salt in the flame. If you are familiar with Raman spectroscopy, it was discovered using sunlight (heliostats) not lasers in the 1920s-30s. Today nobody can think of using sunlight for spectroscopy.

At a more fundamental level, optical rotation is a function of refractive index and refractive index is a strongly dependent on wavelength. You have to see the optics of a polarimeter, the question of using polychromatic light does not arise. You will see a number of prisms are used. Impossible to use polychromatic light.


Why this redox equation occurs on basic medium?

You are given a net ionic equation, which is the reduced form of a complete ionic equation: that's why you see neither $\ce{H2O}$ nor $\ce{OH-}$.
You can deduce that the medium should be slightly alkaline a priori only if you:

either know that manganate $\ce{MnO4^2-}$ formed in alkaline medium hydrolyses when there is a lack of reducing agent (here, sulfite $\ce{SO3^2-}$) to manganese(IV) oxide $\ce{MnO2}$ and that $\ce{MnO2}$ is also formed near neutral $\mathrm{pH}$ values;
or you compose the redox reaction as you were asked to and notice that two $\ce{OH-}$
are produced:

$$
\begin{align}
\ce{\overset{+7}{Mn}O4- + 2 H2O + 3 e- &→ \overset{+4}{Mn}O2 + 4 OH-} & |\cdot 2 \tag{red}\\
\ce{\overset{+4}{S}O3- + 2 OH- &→ \overset{+6}{S}O4^2- + H2O + 2 e-} & |\cdot 3 \tag{ox}\\
\hline
\ce{2\overset{+7}{Mn}O4- + 3\overset{+4}{S}O3- + H2O &→ 2\overset{+4}{Mn}O2 + 3\overset{+6}{S}O4^2- + 2 OH-} \tag{redox}
\end{align}
$$

As for why your problem states the redox reaction takes place in basic medium, it's just an extra hint assisting you in balancing the oxidation and reduction half-reactions (number of oxygens in oxides and oxoanions, for instance), but you could've balance the equation all right even if there were no information about the alkalinity of the medium.


Why it is not yet possible to produce ethanol selectively in the Fischer Tropsch reaction?

in Fischer-Tropsch synthesis-like reaction there are (roughly) following processes:

$\ce{CO \rightarrow  Surf-CO}$
$\ce{H2 \rightarrow  2 Surf-H}$
$\ce{CO + Surf-(CH2)_{n}-H \rightarrow  Surf-CO-(CH2)_{n}-H}$

3'. $\ce{CO + Surf-H \rightarrow  Surf-CO-H}$
$\ce{Surf-CO-(CH2)_{n}-H + 4 Surf-H \rightarrow
    Surf-(CH2)_{n\,+1}-H + H2O }$
$\ce{Surf-CO-(CH2)_{n}-H + 3 Surf-H \rightarrow
    HO-(CH2)_{n\,+1}-H }$
$\ce{Surf-CH2-CH2-(CH2)_{n}-H \rightarrow Surf-H  +
    CH2=CH-(CH2)_{n}-H}$
$\ce{Surf-(CH2)_{n}-H + Surf-H \rightarrow H-(CH2)_{n}-H}$

All processes happen simultaneously, and tuning catalyst and reaction conditions it is possible to hinder some processes. However, as you can see, as long as you have chain growth at all (that is necessary to have $\ce{C2}$ products, like ethanol), you cannot stop $\ce{C2}$ surface particles to participate in further chain growth, producing $\ce{C_{3+}}$ products. So, if chain growth happens at all, a mixture of products with different chain length is always obtained, and there is no way around it. On the other hand, on proper catalyst in proper conditions it is possible to prevent chain growth, but still allow reduction of carbon monoxide, gaining either methane or methanol depending on catalyst. So, in practice it is possible to obtain one of the following

Methane
Mixture of terminal alkenes and alkanes
Methanol
Mixture of linear terminal alcohols (OK, this case is a bit more complicated, as other oxygenates may be produced as well, giving rather complex mixture, but that's beyond the point)

It is still possible to obtain pure ethanol indirectly (using hydration of isolated ethylene), or using other processes. It is also possible to obtain product mixture where some products, say C2-C4 alcohols, will be dominating.


Order of activating ability in electrophilic aromatic substitutions

The most likely reason for this is that an $\ce{OH}$ group can act as an acid and exists in an equilibrium with the $\ce{O-}$ anion, which is significantly more activating.
$\ce{NH2},$ on the other hand, is not really that acidic. It does not benefit enough in this manner to be significant.
The loss of the anionic form is making $\ce{OR}$ less activating despite the inductive effect. $\ce{NR2},$ however, does not have as much to lose and only gains in activating ability.


How to extract wavefunction from GAMESS output?

I'm not too familiar with the input structure of GAMESS. But, I found and input in which you can find how to call the AIMPAC keyword GAMESS input with AIMPAC keyword and this is how wfn file looks like. In principle, when you indicate the AIMPAC keyword, the GAMESS itself print the wave function in a file with wfn extension at the same folder of the input/output.

After you get the wfn file, you have to integrate it with (I recommend) 1. AIMPAC or 2. MULTIWFN (Both are free).


The pattern is neither a true quartet nor a doublet of doublets: it is two doublets. Many thanks to long for pointing out that it is also frequently described as an "AB quartet".

The peaks arise due to the benzylic protons, i.e. the $\ce{CH2}$ between the ether oxygen and the phenyl group. Due to the existing stereocentre on the epoxide, these two protons are rendered diastereotopic: due to their different position relative to the stereocentre, they experience a different net magnetic field, and hence have a different chemical shift. Therefore, in the absence of coupling, one expects to see two singlets.

If we add in spin-spin coupling, then we see that both protons are capable of coupling to each other. The magnitude of the geminal coupling $^2J$ is generally somewhere between 10 to 15 Hz. In this case the zoomed-in diagram shows that the coupling is roughly 12 Hz. So, we now expect to see two doublets.

Why, then, are the intensities of the two peaks in each doublet different? This is due to so-called "roofing": when the difference in the resonance frequencies of two protons is comparable to the coupling constant between them, the shape of the multiplets will be such that they have a "slant" towards each other. In this case, the difference in resonance frequencies is roughly 20 Hz (measure the distance between the centre of each doublet), which is on the same order of magnitude as the coupling constant.

This can, to the best of my knowledge, only be explained with quantum mechanics: see What is the origin of 'tenting' or 'roofing' in NMR spectra? (the question has more examples).


How to compare stability between 5‐methylcyclohexa‐1,3‐diene and 3‐methylenecyclohexene?

As far as I have read, in general, exocyclic double bonds (which are outside the ring) are stable for 3 or 4 numbered rings while endocyclic double bonds (which are inside the ring) are stable in 5 or 6 numbered rings.
As you note all the factors, more or less all factors are same in both cases, so it becomes a little handwavy thing.
Additional Reading:

Goumans, T.; Ehlers, A.; Lammertsma, K.; Würthwein, E. endo/exo Preferences for Double Bonds in Three-Membered Rings Including Phosphorus Compounds. Eur. J. Org. Chem. 2003, 2003 (15), 2941–2946. DOI: 10.1002/ejoc.200300120.

Fleck, B. Notes - Stability and Ease of Formation of Endo and Exo Double Bonds in 5- and 6- Ring Systems. J. Org. Chem. 1957, 22 (4), 439–439. DOI: 10.1021/jo01355a023.

Exocyclic vs. endocyclic double bonds in E1 elimination (although this one is not much relevant here).


Stereochemical outcome after elimination followed by addition

Upon heating, there is an intramolecular $\mathrm{S_N2}$ reaction where the nitrogen lone pair attacks the chlorine-bearing carbon, which gives an aziridinium ion. The stereoelectronic requirement for a "backside attack" results in the formation of a specific diastereomer, illustrated here for the first of the two compounds:

Since there is absolutely no difference between both carbons in terms of sterics or electronics, the free $\ce{Cl-}$ ion can now attack either carbon with equal probability to give a mixture of two products:

The second compound reacts in exactly the same way:

$\hspace{20 mm}$

However, no matter which carbon the $\ce{Cl-}$ ion attacks, the same product is obtained - you can verify this yourself. The product is the same as the starting material.


I have drawn a three step mechanism as opposed to the concerted step reaction proposed in the other answer.

This mechanism has the added benefit of not leaving any room for formation of any side products as there is only one source carbon of acidic hydrogens in each compound.


Chemical oscillators that behave like a double pendulum

Before considering a double pendulum, let us consider a simple pendulum, which is modeled by

$$\ddot \theta + \frac{g}{\ell} \sin (\theta) = 0$$

Choosing units that make $\frac{g}{\ell} =1$, we have

$$\ddot \theta + \sin (\theta) = 0$$

Let $x_1 := \theta$ and $x_2 := \dot \theta$. Hence, the 2nd order ODE above can be rewritten as follows

$$\begin{array}{rl} \dot x_1 &= x_2\\ \dot x_2 &= - \sin(x_1)\end{array}$$

The 2nd ODE is not merely nonlinear, it is non-polynomial (due to the sine).  However, the ODEs that model elementary chemical reactions should be polynomial due to the law of mass action. Hence, it should be impossible to simulate the simple pendulum using a chemical analog computer. Simulating the double pendulum should also be impossible.

What may be possible is to simulate an approximation. For example, if $x_1$ is sufficiently "close" to zero, then we could use the approximation $\sin(x_1) \approx x_1$, which produces a system of coupled linear (and, thus, polynomial) ODEs

$$\begin{array}{rl} \dot x_1 &= x_2\\ \dot x_2 &= - x_1\end{array}$$

which approximate the original pendulum provided that $x_1$ remains "small" (in absolute value).

Lastly, since concentrations cannot be negative, we have non-negativity constraints $x_1, x_2 \geq 0$.


What causes oxygen cylinders to explode when in contact with oil (oiled cloth)?

Summary
⤑ Oils are mostly made up of Fatty Acids ("FAs")
⤑ FAs oxidize and release a lot of energy
⤑ The reaction is proportional to the O2 content of the gas in contact with FAs
⤑ The reaction is also proportional to the contact surface area between FAs & surrounding gas
⤑ Rags offer a high contact surface area between the liquid soaked into them and the surrounding atmosphere
⤑ Maybe that increasing the surface area between air and oil (via rags) also increases volatile FAs which yields an extreme increase in contact area between FAs & O2

Tentative Explanation
I think that oils are made up of mon-/di-/tri-glycerides (at least natural oils): One glycerine bound to 1, 2 or (most often) 3 fatty acid molecules. Fatty acid chains can be oxidised, using up O2 and releasing water + a lot of energy (that's part of our metabolism, and yields much more energy per gram than the metabolism of carbohydrates). I think that the less saturated a fatty acid chain is, the more readily it can be oxidised, which is why linseed oil, which has a very high Alpha-linolenic acid content (a polyunsaturated fatty acid), can undergo spontaneous combustion. A rag exposes oil to a high contact surface with the surrounding atmosphere. Now if in addition to all of that the atmosphere is high in O2, you have perfect conditions for a high-energy-yielding reaction to massively occur.

The local temperature, oil-to-air surface area, air pressure, and air O2 content are the factors which most affect the likeliness of spontaneous combustion. The amount of oil will mostly affect how much of the reaction can occur / how long it can occur (it's the fuel in the reaction). It has a minor impact on the reaction speed only proportional to the fact that there has to be some correlation between oil volume and contact surface area.


How does the equilibrium shift when the products are solid and gas when increasing pressure?

Consider the equilibrium expression for this reaction.
$$\ce{CaCO3(s) <=> CaO(s) + CO2(g)}\quad K_c^{'} = \frac{[\ce{CaO}] [\ce{CO2}]}{ [\ce{CaCO3}]}$$
The concentrations of solids and liquids are constant.  They are the molar densities. Since $\ce{[CaO]}$ and $\ce{[CaCO3]}$ don't change, they are moved to the left hand side and "folded into" the equilibrium constant.
\begin{align}
K_c^{'} &\cdot \frac{\ce{[CaCO3]}}{\ce{[CaO]}} = \ce{[CO2]} \\
K_c &= \ce{[CO2]}
\end{align}
Therefore, as long as solid $\ce{CaO}$ and solid $\ce{CaCO3}$ are present along with $\ce{CO2}$ gas there will be an equilibrium.  Only changes to the concentration of $\ce{CO2}$ will cause a shift in the equilibrium.
You asked how will the amounts change if the pressure is increased.  The pressure of $\ce{CO2}$ is increased by either adding more $\ce{CO2}$ or by reducing the volume of the container.  Adding more $\ce{CO2}$ will increase the concentration of $\ce{CO2}$ momentarily, which will shift the equilibrium to the left, using up some $\ce{CaO}$ and making $\ce{CaCO3}$. The pressure of $\ce{CO2}$ can also be increased by reducing the volume of the container.  Again, the concentration of $\ce{CO2}$ is increased, which increases the reaction with $\ce{CaO}$ to make additional $\ce{CaCO3}$.  The $\ce{CO2}$ pressure gets closer to what it was originally, as predicted by Le Chatelier's principle. Since the molar densities of CaO and $\ce{CaCO3}$ are constant, they don't appear in the equilibrium expression. That is why only changes to the pressure (concentration) of $\ce{CO2}$ affect the position of the equilibrium.
If the pressure in the container is increased by adding an inert or non-reacting gas, nothing happens to the amounts of $\ce{CO2}$, $\ce{CaO}$ or $\ce{CaCO3}$.  The added gas won't affect the partial pressure of $\ce{CO2}$.


By far the largest source of lye nowadays is the electrolysis of concentrated solutions of $\ce{NaCl}$ or $\ce{KCl}$ in water, the famous and incredibly industrially important chloralkali process, producing millions of tons of $\ce{NaOH}$ and $\ce{KOH}$ every year. While electrolysis is an energy-intensive process, presumably the electricity could be supplied by a low-carbon energy source such as solar or wind power.

"Replacing soap", in the sense of halting the usage of surfactants, is not a good idea and likely not even possible. Surfactants are too useful a class of substances. In fact, soaps are likely the most environmentally friendly surfactants, given that they're based on a simple hydrolysis and acid-base reaction with common biomolecules, fatty acids and triglycerides. Other surfactants such as alkylbenzenesulphonates (dishwashing liquid), tetraalkylammonium salts (hair conditioner) and perfluoroalcoxylates, among others, can be toxic to aquatic lifeforms and take longer to degrade in the environment. The excessive usage and improper disposal of surfactants is a recognized environmental problem, but tackling soaps won't really get you far in solving the issue.


What, if any, are the differences between hemiwicking and wicking?

I'm not especially familiar with the detailed differences between the two much beyond the definitions you've posted here, but it seems to me that:

Wicking occurs in the bulk volume of a porous substrate, not just on the surface; whereas
Hemiwicking occurs strictly on the surface of an otherwise impermeable substrate

As you note, in both cases the fluid movement is driven by capillarity, however.

  My assumption would be that Wenzel wetting would actually be an example of wicking?

I disagree. All three panels of the figure you've posted look to me as though they are trying to illustrate strictly surface phenomena, and thus none of them are forms of wicking.  The main difference between (a) and (c) is that in (a) the structured surface is wetted only immediately adjacent to the liquid drop, whereas in (c) the surface is wetted in an area beyond the extent of the drop. Thus, the Wenzel wetting is neither wicking nor hemiwicking.


empty 30 Gal. barrel of Potassium Hydroxide and Sodium Hypochlorite - safe for collecting Maple syrup?

While I also consider the question somewhat off-topic, I would still like to answer here. In a time in which people eat tide pods for fun it seems kind of necessary...

So, for general information, regardless of what chemical substances were inside the container or what you plan to put into, you should never (re)use containers for chemicals to store food!

Here are some reasons why:

Standard food containers are readily available, do not cost much, and are much better suited to store and conserve food
you are sending a disastrous message to children if you are taking food from a container with chemical warning symbols on it
there is a chance that you mix up your food container with one that contains what is actually on the label
you might not be able to remove all the previous content from the container by rinsing it with water, e.g. if the previous substance is non-soluble in water or if it is greasy. Minuscule quantities of the remaining substance may be toxic or hazardous
the substance may have changed over time if it was stored for a prolonged time or under inadequate conditions, like exposure to heat, moisture or sunlight. Products may precipitate or form films on the inside of the container.
you do not know the history of the container. Maybe the previous owner used up the original content and then used the empty container to store chemical waste. This is also not a good practice, but still more common than to store food!


Why does the internal energy increase when ice melts to give water?

Potential energy refers in this case primarily to the energy associated with intermolecular hydrogen bonds between water molecules. The total strength of H-bonds for each water molecule is on average weaker in the liquid. The density is a function of the geometry of the solid or liquid required to accommodate the bonds. Water is odd in that it acquires a spacious bonding arrangement in ice, but this does not mean that bonding is any weaker for that matter. On the contrary, you might say that the geometry is special in order to accommodate those strong intermolecular bonds.
To complement a point made in another answer: the change in internal energy  during a thermodynamic tranformation is the sum of the heat transferred to and work done on the system. Note that work done by a system (here the water) decreases its internal energy. In the case of ice melting under atmospheric pressure, the volume contraction means that the internal energy increases due to work done on the system by the surroundings. This increase in internal energy is on top of the gain due to a transfer of heat from the surroundings.
Note however that the free energy is unchanged during the process (if carried out reversibly), or, equivalently, the joint entropy of system and surroundings is constant. So the transfer of heat to the system (which reduces the entropy of the surroundings) is compensated by an increase in the entropy of water with melting.


The text is simplifying, as what it says is valid for and only for 1s orbital.
The text author had probably in mind that there is no sharp end of electron density as misinterpretation of orbital shapes may suggest. There is an asymptotic exponential-like ceasing of the wave function values toward infinity.
All higher s orbitals with n>1 have n-1 radial nodes, having zero $|\Psi|^2$ value of wave function at n-1 different radii.
Similarly, 2p orbitals have the node plane passing the center between the two lobes and upper p orbitals have additional node surfaces. Similar for d and f orbitals.


According to the Brønsted-Lowry acid base theory, an acid is a compound that releases a $\ce{H+}$ ion to give a conjugate base. For most acids, this reaction exists in a dynamic equilibrium. The equilibrium constant of this reaction is what defines $K_\mathrm{a}$. Generally more stable the product, the more of the forward reaction takes place.
Using this last statement, when comparing two compounds, we can say that if the conjugate base of one compound is more stable, it would mean that the value of its $K_\mathrm{a}$ would be greater. This implies its $\mathrm{p}K_\mathrm{a}$ would be less, meaning that the compound would be more acidic than the other.
For example, taking the case of two compounds, $\ce{HCOOH}$ and $\ce{C6H5-OH}$: Here formate ion would have resonance via oxygen whereas phenoxide ion would have resonance via carbon. This makes phenol less acidic than formic acid.
$\mathrm{p}K_\mathrm{a} \ {\text{ of phenol} = 10.0}$
$\mathrm{p}K_\mathrm{a} \ {\text{ of formic acid} = 3.75}$
Now, if we consider polyprotic acids, we can only consider one hydrogen at a time. Successive $K_\mathrm{a}$s decrease in magnitudes of very large orders.
For example, $\ce{H3PO4}$ has $\mathrm{p}K_\mathrm{a1} = 2.12$ whereas $\mathrm{p}K_\mathrm{a2} = 7.21$ and $\mathrm{p}K_\mathrm{a3} = 12.68$.
Hope this answered your question.


The solubility of dichloromethane in water (as well as the solubility of water in dichloromethane) depends on temperature.
At a temperature of $T=20\ \mathrm{^\circ C}$, the solubility amounts to $17.6\ \mathrm g$ dichloromethane per $1\ \mathrm{kg}$ water; i.e. the mass fraction of dichloromethane is $w_{\ce{C2H2Cl2}}=0.0173$ or $w_{\ce{C2H2Cl2}}=1.73\,\%$.
[source: “Aqueous Solubility and Henry’s Law Constants of Organic Compounds”, in CRC Handbook of Chemistry and Physics, 90th Edition (CD-ROM Version 2010), David R. Lide, ed., CRC Press/Taylor and Francis, Boca Raton, FL.]
(See here for values at other temperatures.)

You cannot simply evaporate pure dichloromethane (boiling point: $40\ \mathrm{^\circ C}$) from such mixtures since dichloromethane and water form an azeotrope, which boils at $38.8\ \mathrm{^\circ C}$. The mass fraction of water in this azeotrope is about $w_{\ce{H2O}}=1\,\%$.


Acidity order of xenon fluorides,oxides and oxofluorides

You have to consider the oxidation state of the xenon.  Higher oxidation state makes the central atom a stronger Lewis acid in general, xenon is just one case.

So we should consider each oxidation state separately.  Look at the ordering when we do that:

+8:  $\ce{XeO2F4}>\ce{XeO3F2}>\ce{XeO4}$

+6:  $\ce{XeF6}>\ce{XeO2F2}>\ce{XeOF4}>\ce{XeO3}$

+4, +3: Only the fluorides are given, no oxygen vs fluorine comparison possible from the list.

The +8 and +6 orderings are consistent with the expected effect of oxygen versus fluorine within a given xenon oxidation state.


In the first phase of chlorophyll biosynthesis, the amino acid glutamic acid is converted to 5-aminolevulinic acid (ALA) . This reaction is unusual in that it involves a covalent intermediate in which the glutamic acid is attached to a transfer RNA molecule. This is one of a very small number of examples in biochemistry in which a tRNA is utilized in a process other than protein synthesis. Two molecules of ALA are then condensed to form porphobilinogen (PBG), which ultimately form the pyrrole rings in chlorophyll. The next phase is the assembly of a porphyrin structure from four molecules of PBG. This phase consists of six distinct enzymatic steps, ending with the product protoporphyrin IX.
All the biosynthesis steps up to this point are the same for the synthesis of both chlorophyll and heme. But here the pathway branches, and the fate of the molecule depends on which metal is inserted into the center of the porphyrin. If magnesium is inserted by an enzyme called magnesium chelatase, then the additional steps needed to convert the molecule into chlorophyll take place; if iron is inserted, the species ultimately becomes heme.
The next phase of the chlorophyll biosynthetic pathway is the formation of the fifth ring (ring E) by cyclization of one of the propionic acid side chains to form protochlorophyllide. The pathway involves the reduction of one of the double bonds in ring D, using NADPH. This process is driven by light in angiosperms and is carried out by an enzyme called protochlorophyllide oxidoreductase (POR). Non-oxygen-evolving photosynthetic bacteria carry out this reaction without light, using a completely different set of enzymes. Cyanobacteria, algae, lower plants, and gymnosperms contain both the light-dependent POR pathway and the light-independent pathway. Seedlings of angiosperms grown in complete darkness lack chlorophyll, because the POR enzyme requires light. These etiolated plants very rapidly turn green when exposed to light. The final step in the chlorophyll biosynthetic pathway is the attachment of the phytol tail, which is catalyzed by an enzyme called chlorophyll synthetase (Malkin and Nyogi 2000).

Source: Plant Physiology Online: Chlorophyll Biosynthesis


Solubility of p-nitrophenol in various substances?

p-Nitrophenol is actually modestly soluble in water at neutral pH (16 g/l at 25 °C) because of its weak acidity ($p\mathrm{K_a}=7.16$ at 22 °C) and the formation of intermolecular hydrogen bonds between the phenolic OH group and water molecules. The latter would also explain why the compound dissolves in diluted $\ce{HCl_{aq}}$ despite being protonated at lower pH values. The importance of intermolecular hydrogen bonds is also demonstrated by the significantly lower solubility of the ortho isomer, which forms intramolecular hydrogen bonds and therefore appears to be less polar to the solvent (source):

In basic solutions, p-nitrophenol is deprotonated to p-nitrophenolate. The anion is stabilized by delocalization of the negative charge into the benzene ring and the electron-withdrawing nitro group (only two of the possible resonance structures are shown):


Is it possible to find the volume of condensed water by temperature change without knowing the initial pressure of the flask?

This problem is sloppily worded and, technically, it is not possible to give a definitive answer based on the information provided.  But the missing information is not what you think -- you don't need to know the air pressure in the flask.  What's missing are two key pieces:  The degree of humidity ("humid" doesn't mean "saturated") and the volume of water in the flask.  If the flask contains no water, and air that, while humid, is sufficiently below 100% $RH$ at $\pu{30 ^\circ C}$, there might be no water condensation when cooling to $\pu{27 ^\circ C}$ (i.e. you'd just be increasing the $RH$ from some number <100% to a value that, while larger, is still <100%). Alternately, if the flask is initially half liquid water, then the volume of water that condenses would be about half that if the flask instead initially contained a very small volume of liquid water. Thus here is what I think the author meant to ask:

  A sealed container of $\pu{6.24 L}$ contains an equilibrated mixture of air and liquid water at $\pu{30 ^\circ C}$. The container temperature is then lowered to $\pu{27 ^\circ C}$. Find the volume of water that condenses. Assume the volume of liquid water is insigificant relative to the volume of the container.  The vapor pressure of water is $\pu{32 mmHg}$ at $\pu{30 ^\circ C}$ and $\pu{27 mmHg}$ at $\pu{27 ^\circ C}$.

As to how to approach the question, that's straightforward:  You know the partial pressure of the water vapor at each temperature.  You also know the total volume of the air.  Given this, you can determine the number of moles of water vapor at each temperature (using the ideal gas law).  The difference is the number of moles of water that condenses, from which you can then determine the volume of the water that condenses.


How to predict the color of a pH indicator using Le Chatelier's principle

You're absolutely on the right track!

The keys to the question are

to  order $\ce{HCl}$, $\ce{NH4Cl}$, $\ce{NaOAc}$ and $\ce{NaOH}$ in terms of strong acid, weak acid, weak base and strong base
to realize that in the presence of $\ce{HCl}$, the indicator exists as $\ce{H2Z}$, while it is fully deprotonated to $\ce{Z^{2-}}$ in the presence of $\ce{NaOH}$
in the presence of weak acids or weak bases, the indicator exists in form of each two species. Consequently, a mixed colour is observed.

In the presence of $\ce{NaOAc}$, a weak base, $\ce{H2Y}$ is completely deprotonated, thus no red colour is observed. In addition, a part of $\ce{HY-}$ is deprotonated too: the second equilibrium isn't completely on the left side. Some amount of $\ce{Y^{2-}}$ is present too. As a result, the green colour (yellow + blue) is observed.

It is all about the position of the equilibria.


Why does adding a charged group to an organic molecule decrease its stability?

The quick and general answer is electrostatic repulsion. Fructose-6-phosphate is already a dianion. Phosphorylating it a second time will mean adding another two negative charges to the dianion. Like charges repel each other, so the two dianionic phosphate groups are pushed apart electrostatically.

However, you cannot generalise this. Sometimes, the additional negative charge may serve to better balance charges overall. It could also be so far away from an existing negative charge that electrostatic interactions are minimal. Finally, there are many ions and positively charged organic molecules dissolved in the cell which altogether stabilise negative charges relatively with respect to a simple solution of fructose-1,6-bisphosphate. (The same can be said for the stabilisation of positive charges.)

Finally, please note that the additional negative charge does not really assist the fragmentation of fructose-1,6-bisphosphate. The reaction — a retro-aldol — is enzymatic and catalysed principally by a lysine and aspartate residue. The former is used to convert the keto functionality into an imine which has a greater tendency to form an enamine while the latter serves as a proton shuttle. The phosphate groups are merely there so the enzyme recognises the correct molecule; modifying the enzyme’s structure to allow for unphosphorylated fructose to be consumed would not alter the reaction rate significantly.


If pure hydrochloric acid is added to a sample of pure copper, there is essentially no reaction; the solution should not turn blue.

At a guess, you're seeing the copper turn pink because it has a significant oxide layer that is reacting with the acid, forming copper (II) chloride and leaving bare the pinkish metal.  The solution could be pale blue or pale green, and the time delay of the color would be a result of the amount of time required for the reaction to occur.  Note that if this is the case, you would not expect to see any bubbling on the copper, as copper (II) oxide will react with hydrochloric acid to form copper (II) chloride and water.

As Ivan Neretin pointed out in comments, large pieces of copper will react with hydrochloric acid over a period of several hours to days in the presence of air.  Permeakra added that a fine copper powder will react in minutes with a solution of hydrochloric acid in the presence of air.  In both cases it is the addition of oxygen that causes the copper to react, and in both cases the solution will be a greenish yellow color due to the excess of hydrochloric acid required.

As an example, a student of mine recently asked about this with regard to a penny.  We cut open a modern US penny (95% zinc, 5% copper by mass) to expose the zinc core, and placed it in 200mL of 6M HCl.  The oxidation layer on the penny was gone almost immediatley after it entered the HCl solution.  The zinc reacted quickly and by the next day the penny had been converted into a hollow copper shell.  The shell sat in the remaining HCl solution (still 5.7M) over the weekend with no visible reaction but the solution slowly turned green and the copper dissolved completely by the following Friday.

If your solution is turning pink, then I'm not entirely sure what's going on.


Electronic and geometric configuration of Fe(OH)₂

I apologize if there's something to this that I don't know about, but my understanding of it is...

We have $\ce{Fe(OH)2}$, also called iron(II)hydroxide (iron-two-hydroxide). The reason it's not called dihydroxide, is that the number (II) describes the oxidation state of the iron, so you can easily calculate how many hydroxyl-groups (which are always -1) it needs to have an overall charge of 0.

In iron(II)hydroxide you don't have covalent bonds (double bonds, single bonds)... You have ionic bonds. This particular compound consists of a $\ce{1Fe}^{2+}\ce{+2OH-}$. They don't SHARE electrons, they are simply held together by their charges!

Ionic compounds group together in lattices. They don't have "geometric" configurations like sulfur dioxide for example. Instead, they are clustered together like in the image below:

The negative dots would be the $\ce{OH-}$ and the positive dots would be $\ce{Fe}^{2+}$. Then, of course, you would have two negatives for each positive. Ion bonds are the strongest type of bonds there is.


What is this reference material in spectrophotometry?

As the reference is the solvent alone, does it mean the samples are dilute solutions in water ? if so, how they solved benzene in water?

I come to a different conclusion.

To me it reads as follows:

sample: solution of benzene, acetone, acetonitrile, etc. in water
reference: water

Benzene is soluble in water: about $1.8\,\mathrm{g\cdot L^{-1}}$ at 25 °C.

Edit

As a reference (blank), a solution of everything except the sample is used. In organic (photo)chemistry this usually means: just the solvent. The situation is different in biochemistry, medicinal chemistry, etc. when complex aqueous buffer systems are used. Here, everything except the probe really means that the whole buffer is used as a blank.


Is the vinegar and baking soda experiment safe for children?

A basic safety rule in chemistry is "don't breathe fumes". Maximum waft a sample of evolved gas to the nose with your hand. Given a controlled environment, adequate size vessels and small quantities and adequate ventilation, it could be a good introduction to chemistry. However - start in the way you should continue - safety glasses, safe operating procedures etc.. BTW, red cabbage makes a good indicator for this.


Tert-butyl esters are routinely cleaved by anhydrous acid conditions by loss of the tBu+ cation. TFA in DCM is a common system for this. Neat formic acid, HBr/AcOH and PTSA in refluxing benzene or toluene are also used (see Greene - Protective Groups in Organic Synthesis (2nd ed) p246).
Methyl esters usually need aq. acid for cleavage.


How closely do GCMS spectra typically agree to conclude a match?

I would never recommend making an assignment of identity based on Mass Spectrum alone - usually you'd want to get things like IR or NMR Spectra to further corroborate your identification but supposing that the underlying chemistry of the reaction is simple and that all you can do is mass spectroscopy you can use this to identify compounds.

  Unknown-1 is Compound B

Plausible, but the base peak of Compound B is 83, but the base peak of Unkonwn-1 is 55. Need more information, but reasonable guess.

  Unknown-2 is Compound A

Unlikely as the base peak of Compound A is 69, which is distinctly missing from the spectrum of Unknown-2. Unknown-2 also has significant peaks at 55-57 which are insignificant in Compound A's Spectrum.

  Is having the same set of M/Z fragments good enough or ought the abundance ratios also match mostly?

You would have a much stronger case if the abundance ratios also matched. Remember that the peaks are formed by the breaking up of the compound and the relative stability of each fragment. If the compound is the same and run in similar conditions, it should, in principle, break up in the same/similar way.


What should be the nomenclature of this organic compound?

Although this isn't disputed by anybody here, let's start by noting that in general, suffixes have greater "priority" over double and triple bonds in numbering. The rules are given in Nomenclature of Organic Chemistry: IUPAC Recommendations and Preferred Names 2013 (Blue Book):

P-14.4
When several structural features appear in cyclic and acyclic compounds, low locants are assigned to them in the following decreasing order of seniority:
[...]
(c) principal characteristic groups and free valences (suffixes);
[...]
(e) saturation/unsaturation

This means that the sulfonic acid suffixes should obtain the locants 1 and 2, which are the lowest possible for two adjacent groups. There are two possible ways of doing this.

With the sulfonic acid groups sorted, we can now turn to the thorny issue of the alkene. It is more complicated than I thought, actually. First, let's consider how to refer to the double bonds in the names. Again, the answer is provided by the Blue Book.

P-31.1.1.1
[...] Only the lower locant for a multiple bond is cited, except when the numerical difference between the two locants is greater than one, in which case the higher locant is enclosed in parentheses.

On the left, this is a "typical" double bond where the locants (2 and 3) differ by 1, and so only the lower locant is cited.
On the right, the locants (1 and 6) differ by  more than 1, so both locants are cited with the higher one in parentheses.

Lastly, we have to make a choice between these two. We turn to the last rule for today:

P-31.1.4.2 If there is a choice of names and numbering, the following criteria are considered in order until a decision is reached:
(1) a minimum number of compound locants. A compound locant is used for a double bond if the locants of the atoms at each end of the bond do not differ by a value of one. When a compound locant is required, the higher locant is cited in parentheses. [...]

Since the option on the left-hand side has no compound locants, and the one on the right has one compound locant, the left choice is preferred. Therefore, the preferred IUPAC name for the molecule is cyclohex-2-ene-1,2-disulfonic acid.


This reaction of ortho-alkylation of phenols was first indroduced by Nagata and coworkers in 1979 (Ref.1). Accordingly, for example, reaction of a phenol with a large excess of an aldehyde and a  slight excess of benzeneboronic acid in refluxing benzene (or toluene) in the presence of $\pu{0.1 mol}$ equivalent of propanoic acid (or trichloroacetic acid) gave about 97% of the expected product, a 2-phenyl-4 H-1,3,2-benzodioxaborin, as a crystalline product (Ref.1):

Note that the presence of propanoic acid is critical in the reaction (no propanoic acid, no expected product). The authors have suggested that mechanistically, the reaction would have taken place by coordination of an aldehyde molecule to the boron atom of a probable intermediate I (i.e. a substituted  phenyl benzeneboronate), followed by [3,3]-sigmatropic rearrangement (as indicated in II) to give the product, a substituted benzodioxaborin.

References:

Wataru Nagata, Kyo Okada, Tsutomu Aoki, “ortho-Specific $\alpha$-Hydroxyalkylation of Phenols  with Aldehydes. An Efficient Synthesis of Saligenol Derivatives,” Synthesis 1979, (5), 365-368 (DOI: 10.1055/s-1979-28683).


There are 2 deviation terms of the opposite sign.

The negative deviation term relates to the attractive forces between molecules, decreasing pressure,and is directly related to condensation ability of real gases.

The positive deviation term  relates to non-negligible own volume of molecules, compared to the total gas volume, leading to higher than ideal pressure, as with the higher own volume, the frequency of collisions with wall is higher.

These 2 effects sum to the overall deviation and in some circumstances may cancel each other.

For "permanent gasses" like hydrogen, helium, neon, nitrogen, the negative term is small and there is more or less steady increase of positive deviation with pressure.

For gases that are easy to turn into liquid ( see critical temperature and pressure ) like methane or carbon dioxide, the negative term is dominant for lower pressures and temperatures. If they are high enough, the dominance is overtaken by the positive term.

A lot of insight can be gained from van der Waals equation Wikipedia page


Lead's affinity for oxygen in the cupellation refining process

Indeed if you do the cupelation, the lead will form litharge and some parts of the litharge will volatilize.

But usually the cupel is made out of bone matter - this is very porous.

While the molten lead and the gold will stay in the cupel, the litharge will be absorbed by the cupel like a sponge.

If all lead is converted to litharge and all litharge is absorbed by the cupel only the gold shall be left - so far to the theory.


Is there any aqueous solution (or simple method) to remove trace level of sulfur dioxide/nitrogen monoxide and nitrogen dioxide in a gas stream?

Removal of $\ce{SO2, NO2}$ from solution

As mentioned by Maurice, you can remove $\ce{SO2, NO2}$ by bubbling the mixture bubble through NaOH or KOH solution, or washing it through a column with flowing NaOH or KOH solution at countercurrent.

Removal of $\ce{NO}$ from solution

You can bubble the mixture through $\ce{aq.FeSO4}$ solution.
I remember doing this in eudiometry experiments.


Is the phenyl group in benzophenone electron donating/withdrawing?

A keto or an aldo group with phenyl ring is less reactive towards nucleophillic addition then their aliphatic counterparts(acetaldehyde or acetone).With phenyl group the positive charge on carbonyl carbon is delocalised via resonance (structures  2 , 3 , 4 , 5 ) .

Resonance with phenyl ring is due to conjugation.

such conjugation is absent in aliphatic aldehydes and ketones.Therefore carbonyl compounds with phenyl rings bonded to carbonyl group are less reactive then their aliphatic counterparts.

With respect to + carbon on carbonyl group pi electrons on phenyl group are polarized towards it , in a way electron donating .


What happens if at equilibrium pressure is increased by decreasing the volume?

When a reaction is at equilibrium, the forward and reverse rates are equal. Decreasing the volume increases the concentration of all species (both reactants and products). This will result in a higher forward rate (because the concentration of reactants increased) and in a higher reverse rate (because the concentration of products increased). If forward and reverse rate increase by the same factor, the reaction remains at equilibrium.


Is it possible to make graphite an insulator?

Tempting, but no.  Graphite is of course less conductive perpendicular to the basal plane but still boasts a conductivity of 330 S/m in that direction, https://en.wikipedia.org/wiki/Electrical_resistivity_and_conductivity.  Electrons can "jump" fairly easily from one pi-electron "layer" to the next.


Finding electron configuration by following the Aufbau principle

The so-called rule is to fill the subshells in increasing order of $n+l$. For subshells with the same value of $n+l$, the subshell with lowest $n$ is filled first.

$$\begin{array}{cc} \hline
n+l & \text{Orbitals} \\ \hline
1 & \mathrm{1s} \\
2 & \mathrm{2s} \\
3 & \mathrm{2p,3s} \\
4 & \mathrm{3p,4s} \\
5 & \mathrm{3d,4p,5s} \\
\vdots & \vdots \\ \hline
\end{array}$$

For all intents and purposes, this is exactly equivalent to your heuristic of "go from the nearest noble gas", since to get the configuration of the noble gas itself you have to go through this same ordering of subshells. You can easily verify that the table above reproduces the order of filling of the Periodic Table.

Chromium and copper are exceptions to the rule (obviously there are many more).


Synthesis using Cationic Hetero-Cope Rearrangement


Is my analysis of quantum numbers describing the ground state of At correct?

Yes , there is another name for it (n+l) rule or Aufbau principle

http://en.wikipedia.org/wiki/Aufbau_principle

Even though it is not correct to check this rule at such higher values of Z where the trend sometimes changes due to very less differnce between higher values of shell and orbitals .But if you are a beginner you should not heed to this statement.


Why do full shell ions such as chloride form coordinate bonds, while noble gases do not?

Taking the general molecular orbital scheme of an octahedral complex — for the sake of discussion, let’s assume it to be $\ce{[FeCl6]^3-}$ — as shown in figure 1, there is an incentive for both the metal and the chlorides to undergo complex formation.

Figure 1: Molecular orbital scheme of an octahedral complex including σ and π interactions from the ligands to the metal. Based on the schemes given by Professor Klüfers in his coordination chemistry course at the LMU Munich.

Complexing a metal is especially favourable for the ligands since they are able to lower their electrons’ energy levels significantly by forming dative σ bonds — and in the cases of halides or other full shell ions, even the electrons in π-symmetric orbitals benefit a little by π forward-bonding.

Therefore, the question should better be rephrased as:

  Why are noble gases not known to form coordinate bonds?

Adding an electron to an atom means that all its orbitals are raised in energy: There is additional negative charge repulsing the electrons. This brings the energy levels of the valence electrons closer to those of the metal they are bonding with, and only for similar energy levels is the energy gain from interacting maximised. Note that electronegativity (that inversely correlates rather well with the energy level of the highest occupied orbital) rises from left to right across a period, so the noble gases have highest electronegativity and therefore the lowest valence energy levels. (The correct physical observable would be ionisation energy, but the same holds generally true for that.)

Therefore, the orbitals of noble gases should be substantially lower than those of halides, which are again notably lower than those of oxido/sulphido ligands. This gives a large energy difference between metal and potential ligand orbitals meaning that noble gases would easily be displaced by better ligands such as water.

However, I could imagine without having checked the literature that metals with high oxidation states may be able to be complexed by noble gases as their orbitals are again substantially lowered.


Calculating solubility in water with other species present

Since the solution is acidic you have to consider $\ce{[OH- ]}$ concentration to be $10^{-7}$

Then solubility product: $$k_\text{sp} = \ce{[Al^3+ ] [OH- ]^3}$$

And if $x$ amount dissociated: $$k_\text{sp} = x \cdot (3x)^3$$

Then we have,

$$10^{-33} = 27 \cdot x^4$$

So $$x = 2.466 \cdot 10^{-9}~\mathrm{mol/L}$$


The short answer is that it's not hybridized.

In benzene, the carbon atoms are sp2 hybridized. That means that each carbon's set of atomic orbitals is three sp2 orbitals and one p orbital. The sp2 orbitals are utilized to make sigma bonds to the adjacent carbons and hydrogens. The unhybridized p orbital contributes to the pi system.


Why is neodymium the most paramagnetic lanthanide?

This is from Shriver and Atkins' Inorganic Chemistry (p.586, 2009 ed) :

  The magnetic moment of many d-metal ions can be calculated by using the spin-only approximation because the strong ligand field quenches the orbital contribution. But, for the lanthanoids, where the spin orbital coupling is strong, the orbital angular momentum contributes to the magnetic moment, and the ions behave like almost free atoms. Therefore, the magnetic moment must be expressed in terms of the total angular momentum quantum number J:
  $$\mu = g_J{\{J(J+1)\}}^{1/2}\mu_B
$$
  where the Landé g-factor is
  $$g_J=1+\frac{S(S+1)-L(L+1)+J(J+1)}{2J(J+1)}
$$
  and $\mu_B$ is the Bohr Magneton.

Being a high school student, the only thing of matter for you is the key point at the top of the page:

  The magnetic moments of lanthanoid compounds arise from both spin and orbital
  contributions.


Mechanism for the intra-molecular exchange of an ester and a nitrile group

Following is a reasonable suggestion for the rearrangement:


Isotropic Fermi Contact Couplings vs. Dipolar Coupling Constant

The output sections you mention are the major components of the electron-nuclear hyperfine interaction. There is also an orbital contribution, but this tends to be small compared to the Fermi contact and spin dipole terms, unless you go down the periodic table.

These are the terms in the effective spin Hamiltonian that contribute to the hyperfine interaction (from Wikipedia):

$$
\begin{align}
\hat{H}_{\textrm{hyperfine}} = & 2 g_{I} \mu_{N} \mu_{\mathrm{B}} \frac{\mu_{0}}{4\pi} \frac{1}{L_{z}} \sum_{i}^{N_{\mathrm{elec}}} \frac{\hat{l}_{zi}}{r_{i}^{3}} \mathbf{I} \cdot \mathbf{L} \\
&+ g_{I} \mu_{N} g_{e} \mu_{\mathrm{B}} \frac{\mu_{0}}{4\pi} \frac{1}{S_{z}} \sum_{i}^{N_{\mathrm{elec}}} \frac{\hat{s}_{zi}}{r_{i}^{3}} \left\{ 3(\mathbf{I}\cdot\hat{\mathbf{r}})(\mathbf{S}\cdot\hat{\mathbf{r}}) - \mathbf{I} \cdot \mathbf{S} \right\} \\
&+ \frac{2}{3} g_{I} \mu_{N} g_{e} \mu_{\mathrm{B}} \mu_{0} \frac{1}{S_{z}} \sum_{i}^{N_{\mathrm{elec}}} \hat{s}_{zi} \delta^{3}(\hat{\mathbf{r}}_{i}) \mathbf{I} \cdot \mathbf{S}
\end{align}
$$

The first term is the orbital contribution due to spin-orbit coupling, the second term is the spin dipole term due to the electrons and nuclei behaving as dipoles in the presence of each other's magnetic fields, and the third term is the Fermi contact term arising from the non-zero probability of the unpaired electron(s) being inside the nucleus.

Isotropic Fermi Contact Couplings are the eigenvalues of the Fermi contact part of the effective spin Hamiltonian. Because of the presence of $\delta^{3}(\hat{\mathbf{r}}_{i})$, there is only one value, hence the "isotropic" descriptor.

Spin Dipole Couplings and Anisotropic Spin Dipole Couplings in Principal Axis System are two different presentations of the spin dipole term. Spin Dipole Couplings shows the eigenvalues of the spin dipole part of the effective spin Hamiltonian (in a.u.). Because this is a symmetric 3x3 tensor, only the 6 unique elements are shown. Diagonalizing the tensor gives the 3 principal values Baa, Bbb, Bcc as the eigenvalues and the vectors from the origin for each principal value as the eigenvectors. These are the values you're interested in, most commonly reported in MHz by the computational/theoretical chemistry community and in Gauss by the spectroscopy community.


Why don't trigonal S and P compounds undergo inversion at room temperature?

Azane all by itself has an inversion barrier of about $6\ \mathrm{kcal/mol}$, which is  low compared to phosphine's approximately $30\ \mathrm{kcal/mol}$ barrier. Pyramidal nitrogen inversion requires a planar transition state. One can use a classical theory to model the thermal rate and we could write $$k \propto \mathrm e^{-E_\mathrm a/(RT)} $$ and using models like these we would predict that ammonia at $300\ \mathrm K$ would have a rate on the order of $10^8\ \mathrm{s^{-1}}$, instead what is observed is something on the order of $10^{10}\ \mathrm{s^{-1}}$

Assuming the potential energy surface has two minima equal in energy and a higher energy TS, two routes are possible: a thermal process going over the maxima or quantum-mechanical tunneling. Determining a tunneling frequency requires writing down vibrational wave functions, finding their linear combination and overlap. If we were to do all this for a particular vibration, we'd see that the tunneling frequency decreases exponentially with a dependence on increasing $\mu$, thickness of the barrier/shape, and the barrier height. As an example, $\ce{ND3}$ has approximately an order of magnitude slower rate of inversion than $\ce{NH3}$ at a sufficiently low temperature – this has been attributed to a decrease in its tunneling frequency.

But why the difference in the energy barriers and corresponding rates depending on the atomic position in the periodic table? Perhaps the easiest/quickest explanation rests on considering what the TS requires for interconversion. Ideally the classical conversion requires passing through a trigonal planar structure with bond angles on the order of 120 degrees. If you consider the series of azane, phosphane, arsane then you'll see bond angles of 108, 94, 92 degrees respectively. In the absence of other effects, the barrier goes from smaller to larger. Indeed, asane-like molecules were the first of this series resolved, followed by phosphane-like. To the best of my knowledge the fastest have yet to be resolved, because rates on the order of $10^{-5}\ \mathrm{s^{-1}}$ are required at RT for a decent shot at isolating them. Naturally, this isn't the whole story, as I've previously alluded to above. Let me know if you would like to know more.


Why are PTFE and FEP named "ethylene" when they don't have double bonds?

Polymers follow a slightly different nomenclature system than you are used to in organic chemistry. Instead of naming the polymer after the monomers as they are, we name them after the monomers as they originally were. In other words, a polymer made from ethylene monomers is called polyethylene. Since PTFE and FEP are made from fluorinated ethylene and propylene, we just put "poly" in front of the monomer names. This is despite the fact that after the reaction, the bond order is reduced by one.


How to differentiate a metal from its ion?

Other than the excellent answer by F'x, there is also actual spectroscopy, specifically UV/Vis spectroscopy. This'll only work if the metal has transitions in the right range, though -- transition metals, lanthanides and actinides are usually fine, s or p-block metals and metalloids, probably not.

Then you're looking for the absorption/transmission spectrum of the ion and the metal separately, and how much of each there is in a combined spectrum. It's not strictly quantitative, but you can get a good idea of the ratios.


$\ce{CF3OH}$ is more acidic because of the inductive effect of the $\ce{-CF3}$ group: Fluorine is very electronegative and therefore it pulls the electron-density toward itself weakening the $\ce{O-H}$ bond in $\ce{CF3OH}$.
Chlorine is less electronegative than fluorine, thus $\ce{CCl3OH}$ is less acidic than $\ce{CF3OH}$.
Because $\ce{H}$ is even less electronegative, then $\ce{CH3OH}$ is less acidic than all its halogenated derivatives.

A related question can be found here:
Why is CF3COOH exceptionally acidic?

The negative hyperconjugation is a type of resonance effect where there is a donation of electron density from a filled π- or p-orbital to a neighboring σ*-orbital. It does not occur in your molecules.


What elements and/or substances without water are liquid at room temperature?

This question is a bit broad in terms of the sheer amount of chemical compounds and mixtures that are liquid at room temperature. Examples include:

Compounds
Acids, bases, many hydrocarbons (e.g. hexane) and many more

Mixtures
Crude oil, aqua regia and many more

In terms of elements, there are only two that are liquid at room temperature (say about 20 °C or 293 K):

Mercury (as you identified).
Bromine

Francium, cesium, gallium and rubidium are close, with melting points at 300 K, 301.59 K, 303.3 K and 312.46 K respectively.

LennTech provides a list The elements of the periodic table sorted by melting point


How do Invert Sugars deter crystallization in Sucrose solutions?

I would like to know of how Invert Sugars physically interact with the other sugar molecules and prevent them from coming together

Glucose and fructose molecules (being essentially parts of sucrose) can merge into sucrose crystals during their formation, slowing or even completely stopping their growth in certain directions (if their concentration is high enough).

  even though they're not really resistant to crystallization (as pointed out above)

Glucose and fructose crystallization process is much slower than that of sucrose. From confectionery process's point of view, once the fast growing sucrose crystal structure is disrupted and its growth hindered no crystallization occurs.

  Why does the rate of cooling of a sugar solution affect its crystallization?

Because crystals' growth takes some time. If cooled fast, neither existing crystals will have a chance to grow, nor new seeds will appear. If cooled really fast, saturated sucrose syrup can even turn into glass with no crystals at all.

[literature]: Sgualdino, G., Aquilano, D., Vaccari, G., Mantovani, G., & Salamone, A. (1998). Growth morphology of sucrose crystals. Journal of Crystal Growth, 192(1-2), 290–299.


Why doesn't the enthalpy change of a reaction affect the rate of reaction?

The rate of a chemical reaction is directly dependant to the activation energy of that particular reaction. These two values are inversely proportional. The higher the activation energy, the lower the rate. As it has been mentioned though, the rate can be increased, in such cases, by increasing the temperature. All of the above is described as kinetics, which shows how fast the reaction will reach its end phase.

Enthalpy is linked with thermodynamics, which shows what kind of entities will be present at the end of the reaction. At what degree the reactants will turn to products.

I don't know your chemical background, so I tried to make this as simple as possible.