The first mechanism is more likely unless specific conditions of the experiment dictate otherwise. In the presence of $\ce{CCl4}$, the addition of $\ce{Br2}$ takes place via the formation of a cyclic halonium ion as intermediate as follows:


Turn lead into gold via radioactive decay

Interesting idea, but it has already been done, and not cheaply - read on.

  How could we get a great quantity of $\ce{_{82}^{197}Pb}$ ?

There would be two problems with getting a large amount of $\ce{_{82}^{197}Pb}$.  First, the parent nuclide of $\ce{_{82}^{197}Pb}$ is $\ce{_{83}^{197}Bi}$ which is unstable and has a half-life of only 9.33 minutes - so you can't get a large quantity of $\ce{_{82}^{197}Pb}$'s precursor to begin with.  Second, once $\ce{_{82}^{197}Pb}$ is formed, it has a half-life of 8.1 minutes, so it transmutes quickly to $\ce{_{81}^{197}Tl}$.

  How long would it take for it to turn into gold?

$\ce{_{82}^{197}Pb}$ half-life = 8.1 minutes

$\ce{_{81}^{197}Tl}$ half-life = 2.84 hours

$\ce{_{80}^{197}Hg}$ half-life = 64.14 hours

After 10 half-lives, ca. 0.1% of the starting material will be left
$$\ce{(1/2)^10 = $0.0009766$}$$
The last step is the slowest by far, so after about 641.4 hours (26.73 days), you should have something around 99.9% pure gold.

  Can that be accelerated?

Unlike chemical reactions that can heated, catalyzed, etc., this type of nuclear transformation keeps a set schedule.

  Once most of it has been turned into gold, how can we extract the gold
  ? (there are still remaining Pb,Tl,Hg  molecules)

As noted above, you can get whatever purity you desire, just wait.

  What would be the yield of the process?

It would be high for the 3 nuclear transformations you listed.  Each of the elements you listed decays directly and only to the daughter isotope you've shown.  However, as noted above, you can't start with $\ce{_{82}^{197}Pb}$, you generate it from $\ce{_{83}^{197}Bi}$, the decay of which adds some impurity along with lead. And then since the bismuth isotope is not long-lived you'd probably start with its precursor, and so on until you find something that has a long enough life that you could assemble a reasonable quantity.

Back around 1980 Glenn Seaborg actually transmuted bismuth to gold, but only a few thousand atoms (see this reference also).

  How much money would it approximately cost compared to how much we win
  in gold?

The Wikipedia article I referenced directly above notes, "the expense far exceeds any gain."  There are other ways (fission and fusion) to produce gold, but at least with the methods available today, the cost would be astronomical.


Safe chemical reactions to start a campfire

Small magnesium metal bars are available commercially, packed with a little iron rod for starting fires. You smack the magnesium with the iron rod and get some sparks. If you have some light fluffy kindling, you are supposed to be able to get a fire going. I suppose a little steel wool would help.
If you soak some cotton balls with KNO$_3$, dry them completely and carry them in a sealed plastic bag, they would make excellent kindling.


Ice Crystal Matrix measurements of H2O molecules?

You can produce STL (and VRML) files directly from Mercury - see https://www.ccdc.cam.ac.uk/Community/blog/post-56/

I'm not sure how best you'd go about including female sockets for bike spoke segments, but be aware that for some 3D printing types you're charged for the total volume of material used, so you might be better off trying to create a model with multiple molecules in a smaller area with less empty space.


Why is the graph of fluorescence vs. wavelength continuous?

There are two main reasons for what you're describing.

The first is the fundamental properties of the molecules involved, as mentioned in the comments. Even for isolated molecules, the transition is not always between the same vibrational states of the excited and ground state. Since the vibrational states are quantized, we would expect a series of very closely spaced emission lines, each representing a different combination of vibrational states.

If the molecules are in a solution or in liquid or solid phase, then the interactions of the emitting molecule with the molecules around it can also cause small changes in the energy of the emission as the electronic structure of the molecule is perturbed.

The general term for these phenomena is "line broadening," and most reference books about spectroscopy will have a discussion of the various mechanisms that lead to it.

The second reason for it is the error in measurements. Most spectrometers cannot filter wavelengths perfectly, so, for example, light of wavelength 590 nm will contribute to a signal at 589.5 and 591.5 nm, less as you get farther away. Sometimes the slit width is an adjustable parameter, but even at the narrowest setting, most instruments will not give extremely high resolution. As a result, a series of closely spaced lines will appear instead as a smooth curve.


Elimination reaction on 2-fluorobutane using alc.KOH

E1CB elimination takes place because of a poor leaving group. This is because of the high $\ce{C - F}$ bond strength. Stronger bonds are more difficult to break, making them less reactive.

If the reaction was E2, then zaitsev product would have been formed. But since the mechanism is E1CB, the reaction proceeds via carbanion.

  From wikipedia: The first step of an E1cB mechanism is the deprotonation of the of β-carbon, resulting in the formation of an anionic transition state, such as a carbanion. The greater the stability of this transition state, the more the mechanism will favor an E1cB mechanism.

I hope you know that for carbanions, the stability order is primary > secondary > Tertiary.

Now that you know the stability order of carbanions, the preferred position for carbanion is a primary carbon. So base abstracts a proton from a primary carbon.


Why IUPAC name this amine is written as 2-methylpropan-2-amine and not as 2,2-dimethylethanamine?

The basic principle in nomenclature is to look for the longest carbon chain: here it is the chain with 3 carbon atoms so the suffix prop; since it is an amine connecting to the carbon 2 it gives the suffix propan-2-amine.
As you have a $ \ce {CH_3} $ group grafted in position 2 of the main carbon chain, you must add the prefix 2-methyl
So you have 2-methylpropan-2-amine.


In this case, the use of the word patina is a little misleading. Patina refers to the oxidation film on bronze or aged polish. I assume that you mean that the knife is coated with a substance.

First, when steel oxidizes, the iron oxide that forms does not stick well and will flake off the surface. This would not be useful for a knife since the edge will become dull quickly.

Second, if the knife is made with stainless steel, it does not need to be coated to prevent rust, only keep reasonably dry. Some stainless steel will actually rust since it was not properly formulated. Very few knifes are made of other metals since stainless steel is reasonably cheap.


Number of orbitals in full configuration interaction

In simple terms, Full CI uses all of the possible orbitals (a "full" active space). Thus the number of active orbitals should equal the size of your basis set. If you have 10 basis functions, and then perform a Hartree-Fock calculation, you get 10 orbitals out (some occupied, some virtual).

Complete Active Space (CAS) methods, on the other hand, only use some of the orbitals and then perform a Full CI on that subset of orbitals, leaving the other orbitals (inactive space) alone.

Full CI is CAS in the total orbital / total active space limit.


Gold is generally extracted by pressure oxidizing a refractory ore such as pyrite or chalcopyrite. They are common sulfide ores where finely divided nanosized gold are present. They are roasted to break the sulfide matrix and recover the metallic gold(Here). These refractory ores are considered low grade ores (where gold% is less) and are therefore roasted to extract the gold (Here).

Critically, the size at which one can efficiently liberate gold from the host mineral of the particle will determine the process to use for optimal gold recovery(Here). Gold(I) sulfide is not considered a refractory ore as because gold is not present in microscopic level and hence are generally not roasted (Here and Here).

But it is possible to roast gold(I) sulfide at 200-500°C(Here) but the yield will be less as compared to the gold  yielded when it is leached.

$$\ce{Au2S + O2 ->[200-500°C] 2Au + SO2}$$


Determine if the following change is an oxidation, a reduction, or neither

As pointed out in the comments and the other answer, it is neither an oxidation, nor a reduction.

Q: So, what is it?

A: It is a condensation: two molecules of an ortho acid combine under loss of one molecule of water to yield a diacid (pyro acid).

This is frequently found in oxoacids, see

$\ce{H3PO4}$ (phosphoric acid) and $\ce{H4P2O7}$ (diphosphoric acid)
$\ce{H3AsO4}$ (arsenic acid) and $\ce{H4AsO7}$ (diarsenic acid)
$\ce{H2SO4}$ (sulfuric acid) and $\ce{H2S2O7}$ (disulfuric acid)


Why does bond breaking in ATP release energy?

In most chemical reactions where bonds are broken, other bonds are formed. Take your example of the hydrolysis of ATP. A bond between two phosphate groups breaks, but one of the phosphate groups forms a new bond with the oxygen of water. Whether energy is released or taken up (whether the reaction is exothermic or endothermic) depends on the sum of the energies associated with breaking and making bonds in that reaction.

  How does bond formation release energy?

In the simplest case in the form of kinetic energy, heating up the reaction mixture. In the case of ATP hydrolysis in biological processes, the energy is sometimes converted into mechanical energy (muscle contraction), used to run pumps (transmission of signals in the nervous system), or other processes that would not go forward on their own.


Is consumption of chlorine, like fluorine, harmful to bones?

The OP uses the terms chlorine and fluorine. In the context of the question, fluorine doesn't make sense and it is assumed here that the the OP means fluoride. In the case of chlorine, it is likely, though not certain, that the OP means chloride. This answer will address the question primarily in terms of fluoride and chloride, with a note about chlorine.

Bones are largely composed of calcium compounds (mainly hydroxyapetite, $\ce{Ca5(PO4)3(OH)}$). The calcium salts of bone are very slightly soluble, and thus they are in an equilibrium between the solid form and the dissolved aqueous form. Because calcium fluoride is extremely insoluble, fluoride can displace the naturally occurring anions to form water-insoluble calcium fluoride. This results in weaker, more easily fractured bones, which is the condition known as skeletal fluorosis.

Calcium chloride is about five orders of magnitude more water soluble than is calcium fluoride, so an analogous displacement of the naturally occuring anions in bone by chloride does not take place. Thus the ingestion of chloride has no negative effect on bone material. The same is true of chlorine, in the case of chlorinated water, as there is no mechanism by which chlorine can be ingested (or inhaled), metabolized and then interfere with bone composition.


You won't find a general formula for the shelf life of a solution.  A silver nitrate solution will have a very short shelf life if it is kept in the sunlight, whereas potassium nitrate could last for years.  The shelf life is also going to be very strongly dependent on the quality of the container.  Think about what happens to the aging of a wine bottle if the cork does not provide a sufficient seal .  Interactions between the solution and the container have an impact on the solution shelf life as well.  To stick with the wine theme, corked wine is due to contamination of the wine with 2,4,6-trichloroanisole which is generated by natural fungi in the cork that have been exposed to chlorine.

So it is important to understand the chemistry of the solution with which you are working in order to determine its shelf life.


What you have shown in red is the second propagation step in the benzylic free radical bromination of ethylbenzene. The bond dissociation energy  (BDE) of bromine is +46 kcal/mol while the overall heat of reaction of the step is -12 kcal/mol. Thus, the formation of the C-Br bond must be -58 kcal/mol. The BDE for the C-Br bond in (1-bromoethyl)benzene is +58 kcal/mol.
However, the BDE for bromine is not listed in your tables. So how can one estimate the BDE of the C-Br bond in (1-bromoethyl)benzene using the data in the tables? A simple ratio and proportion will suffice. In equation A compare the primary C-H bond of ethane to that of toluene. This ratio is compared to the ratio of the BDE of the secondary C-H bond of propane and its counterpart in ethylbenzene, which is the unknown "x". An estimated value of
82 kcal/mol is obtained. In equation B, this ratio compares secondary C-H bonds with secondary C-Br bonds. The estimate for the BDE of the C-Br bond in (1-bromoethyl)benzene is the value of "y" which is equal to 59 kcal/mol. Pretty close, wouldn't you say?


Are methyl n-propyl ether and methyl iso-propyl ether metamers?

The term metamer is equivalent to isomer.  Isomers differ in connectivites of atoms between molecules with the identical number and type of atoms.

Methyl n-propoyl Ether and methyl iso-propyl ether have the same numbers of carbon, oxygen and hydrogen atoms.

As an aside, I have never heard the term metamer before, I expect it is archaic.


What are the main axis of research in Chemistry?

Even though I voted to close this question as not constructive, these are among the most important unsolved problems in chemistry:

Liquid Water - We still know so little about the molecular level interactions of water molecules with themselves and other species. The great extent of hydrogen bonding networks in liquid water make it hard to model. Just how many water molecules are solvating a proton in acid? Why are organic reactions between hydrophobic molecules accelerated in the presence of water?
Molecular origins of life, especially homochirality. The formose reaction is believed to be the origin of carbohydrates. Amino acids, lipids, and nucleic acids have similar hypothetical origins. However, how did these molecules come together to create living organisms? Why are most (not all) biomolecules present as one enantiomer and not the other?
Protein folding. Can we predict the three-dimensional shape of a protein simply from the order of the amino acids? If so, then we could design proteins to do whatever we wanted.
True designer proteins. Once we solve the folding problem, then we can design proteins with active sites of specific geometries and specific amino acids in those active sites to do whatever reaction we need.
High Temperature superconductors. Although room temperature superconductors would be nice, inexpensive superconductors that operate at liquid nitrogen temperature (77 K) instead of liquid helium (5 K) temperatures would make NMR, MRI, and a host of other important techniques much cheaper.
The Energy Crisis. We need a chemical solution to replace fossil fuels as the world's primary energy source. Hydrogen, fuel cells, biomass, and solar power all have serious drawbacks currently.
Replacement of petrochemical feedstocks. The petroleum industry remains the primary source of organic compounds. What will we do when the petroleum runs out?
Bonding theory. Modern bonding theories have developed into bloated mathematical formalisms that have removed any resemblance of intuition from the study of the structure of molecules. Can a unified modern bonding theory be developed that retains only the best features of valence bond and molecular orbital theories?
Improved ideality and efficiency of synthesis. Can the synthesis of compounds be improved so as to remove wasteful steps (like adding and removing protecting groups) or complex purifications to better adhere to the principles of Green Chemistry?
Structure-Property Relationship. All established relations between structure and activity is empirical, especially when the desired activity is biochemical. Can we design the best drug with the fewest side effects on the first try?

I find some of the unsolved problems on the to be too narrow (stability of norbonyl cation - this is a subset of the structure-activity problem) or trivial (origin of rotation barrier in ethane - the problem is important but the answer is useless).


How do I predict the solubility of a compound in a mixture of water and an organic co-solvent?

I think this is very much an open problem. As you may see from the Drexel pages, even solubility prediction in one solvent is still fairly hard, partly because of the lack of experimental data.

The only software that I know capable of making predictions about solubility in mixed solvents is COSMOtherm, based on the COSMO solvation model.

I have never used it, but the capability is mentioned on the website.


I don't understand why calcium carbonate decomposes... It's not soluble in water.

It's more soluble in water than you might think.  If it weren't soluble, we wouldn't have awesome things such as stalagmites in caves.

Calcium sulfate is more soluble in water than the carbonate, but that isn't saying much.  It's solubility is only about $\pu{2 g/L}$. So writing the product as $\ce{CaSO4(s)}$ is only correct if you are talking about dilute solutions.  Anything too concentrated would decompose to form precipitated calcium sulfate.

  I could understand the hydrogen ions oxidizing the calcium...

Now that would be a very curious reaction.  The calcium in calcium carbonate is already in the +2 oxidation state. Calcium has only two valence electrons and further oxidation would require disrupting an intact octet.  It would be very unfavorable.  $\ce{Ca^{2+} -> Ca^{3+} + e-}$ would be far too unfavorable a reaction for protons to carry out. $\ce{Ca^{3+}}$ would be a much stronger oxidant than aqueous protons.

  Why does calcium carbonate decompose in sulphuric acid?

The decomposition can still happen even with insoluble, solid-phase calcium carbonate because reactions can happen at surfaces. An acidic proton could be attacked by a solid-phase carbonate ion at the crystal surface, forming bicarbonate.  The bicarbonate could attack another acidic proton, forming a surface-bound $\ce{H2CO3}$ molecule, which could spontaneously transform into $\ce{CO2}$ and water, which would be very loosely bound to the surface, allowing them to escape and exposing fresh not-yet-reacted calcium carbonate molecules to the aqueous surface, and the process begins anew...


Can we use prefixes like iso, neo, etc in IUPAC nomenclature of organic compounds?

According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) various prefixes are retained for use in general nomenclature. However, many prefixes are no longer recommended.

  Trivial, common, and traditional prefixes have always been an integral part of organic nomenclature. However, as systematic nomenclature develops and becomes widely used, many of these prefixes fall by the wayside. Accordingly, each set of IUPAC recommendations contains fewer of these traditional prefixes.

For example, the prefixes “isobutyl”, “isopentyl”, “neopentyl” were contained in the 1993 recommendations but are no longer recommended as approved prefix. The prefix “isopropyl”, is still retained for use in general nomenclature; however, for the preferred IUPAC name (PIN), the preferred prefix is “propan-2-yl”. (The prefix “1-methylethyl” may be used in general nomenclature).

Anyway, the name “5-(1-ethyl-1-isopropyl)nonane”, which is given in the picture in the question, is clearly wrong since no parent structure is given for the substituents 1-ethyl and 1-isopropyl:
“5-(1-ethyl-1-isopropyl what?)nonane”

If you really want to use such a name including the substituents 1-ethyl and 1-isopropyl, the corresponding systematic name would be “5-(1-ethyl-1-isopropylmethyl)nonane”. However, such a name would be highly unusual and not in accordance with the basic principles of IUPAC nomenclature.

The name “5-(1-ethyl-2-methylpropyl)nonane”, which is proposed in the question, describes the correct compound and may be used in general nomenclature. However, the PIN for this compound is “5-(2-methylpentan-3-yl)nonane” since prefixes of the “alkanyl type” are preferred if the free valence is not in position 1 of the substituent group (see this similar question).


Complicated little rascal!
What we know:
1) The N2 is inert.
2) The oxygen is in excess.
3) 90% of the 7.8 mole % CH4 reacts; this is 7.02 mole % CH4.
4) This gives 8 moles CO2 per mole of CO, or 88.89% CO2 plus 11.11% CO.

The resultant output contains CO2: 7.02 mole % of the original CH4 x 88.89% = 6.24 mole % CO2.

The resultant output contains CO: 7.02 mole % of the original CH4 x 11.11% = 0.78 mole % CO.

The resultant output also contains 10% unreacted CH4. The 90% conversion refers to the starting material, not to the products. (Yield % would refer to one product.)

So the stoichiometry is 80% of the methane goes to CO2, 10% goes to CO, and 10% is unreacted.


Diastereoisomer of quinic acid and its proton NMR?

I see one major issue in your working immediately: The NMR has been recorded in $\ce{D2O}$, hence there will be no more $\ce{OH}$ groups present in your molecule, only $\ce{OD}$ groups. Since deuterium is not hydrogen, you cannot observe any coupling to the $\ce{OD}$ groups as if they contained hydrogen.

Also, you said:

  I made the assumption that the quartet is actually a doublet of doublets.

This is an incorrect assumption. They have multiple dd’s that they reported as dd which gives us less reason to suspect an incorrect report. Note that typical dd signals have two different coupling constants, e.g. the proton at 2.12. Now you could imagine that the two doublets fall together. But in that case, they would fall together to become a triplet, not a quartet. You can try it with splitting trees: Either a dd stays dd, or it collapses into a t. And finally, the quartet cannot be a set of two doublet signals from distinct hydrogens since it only has an integral of 1.

That said, the attribution of different hydrogens to their position in the ring is remarkably simple. Let’s label the quaternary carbon $\mathrm{C1}$ and then go around the ring in any direction. Our first stop is $\mathrm{C2}$.

   Since this is a $\ce{CH2}$ group, its chemical shift must be lower than $3~\mathrm{ppm}$. We have two choices. Either we take the two hydrogens that exhibit a $^2J$ coupling or the other two. Since we can easily identify the $^2J$ coupling, let’s take those.

With the hydrogens of $\mathrm{C2}$ in hand, we can check, which hydrogen must be on $\mathrm{C3}$.

   Again it’s pretty simple. The two coupling constants that are not $^2J$ constants must all be present in the hydrogen’s signal. There is only one in question.

Once we have gotten that sorted, we can go back and determine the stereochemistry that must be present here.

   We have one very large $^3J$ coupling and one mediocre one. The very large coupling is, of course, due to an axial-axial relationship between the two hydrogens.

We now have three hydrogens firmly in place: Those on $\mathrm{C2}$ and the one on $\mathrm{C3}$. Let’s move on to $\mathrm{C4}$.

   There is one coupling constant on the hydrogen at $\mathrm{C3}$ which is not yet assigned and which thus must belong to the coupling to this hydrogen. It’s magnitude again tells us pretty clearly which dihedral angle we expect and thus what the relationship between the two must be.

With $\mathrm{C4}$’s stereochemistry sorted out, $\mathrm{C5}$ is next.

   The coupling constant is rather different this time, indicating the opposite relative stereochemistry. Note that the hydrogen at $\mathrm{C5}$ is reported as a quartet of the same coupling. We’ll get back to that in a second. For now, we know what must be substantially different here.

And now we can move on to $\mathrm{C6}$, the one carbon we are still missing.

   As I just mentioned, we have an apparant quartet for the hydrogen on $\mathrm{C5}$. But that’s okay. It just means that there are three rather similar coupling constants to neighbouring hydrogens. One is taken from the coupling to $\mathrm{C4}$’s hydrogen, meaning the two of $\mathrm{C6}$ must be the other two. Thankfully, there is one signal with an integral of two left, which also has the same coupling constant. And also thankfully, if you did everything correctly, there should be no questions as to why this stereochemistry along the $\ce{C{5}-C{6}}$ bond be correct.

Finally, we have a hint in the question as to the overall structure:

   Namely that the two hydrogens on $\mathrm{C1}$ and $\mathrm{C4}$ are syn. Thus, the carboxylic acid must be in equatorial position and the hydroxy group axial.

A note on finding the diastereomers: Enantiomers are only enantiomers if all stereocentres are reversed. Thus, if you take one centre and keep it the same all the time and only exchange the direction of the others, you are bound to find diastereomers only. I suggest immobilising the $\mathrm{C1}$ carbon since it is rather prominent up there all alone; it gives the entire thing less hassle. Also, any achiral isomers will always have a plane of symmetry through it which helps, too.


Nuclear spin isomerism in molecules other than H2

Nucleons of the same type tend to pair up and cancel out each others' spins, so nuclides with an even number of protons and an even number of neutrons will have zero spin. Some common nuclides with this property include $^4\ce{He}$, $^{12}\ce{C}$, and $^{16}\ce{O}$.


Explanation of bond angles in the aluminium chloride dimer

The bonding in aluminum chloride ought to be considered largely ionic. The electronegativity of $\ce{Al}$ is $1.47$ vs $2.83$ for $\ce{Cl}$ (for comparison: $\ce{C}$ $2.50$; $\ce{Mg}$ $1.23$; $\ce{Na}$ $1.01$). Molecular orbitals and covalent bonding can be added for a fuller picture of electron densities.

Aluminum chloride solid (Melting point $\pu{193^\circ C}$, under pressure) has a cubic close-packed array of chlorine atoms with only $2/3$ of the octahedral holes between every other pair of chlorine planes occupied by the aluminum atoms. At low temperatures in the vapor state, aluminum chloride is a tetrahedrally coordinated dimer, and at very high temperatures it is a trigonal monomer. So aluminum can easily have $3$, $4$ or $6$ coordination. Whatever directionality the bonds have from hybridization of the $\ce{3s}$ and $\ce{3p}$ orbitals is mixed with plain electrostatic (ionic) attraction.

Imagine putting just the 6 chlorine atoms together. There will be two tetrahedral holes for aluminum (see image below). The radius ratio
$$\frac{\ce{Al^3+}}{\ce{Cl-}} = \frac{\pu{50 pm}}{\pu{181 pm}} = 0.276$$
is well within the stability range ($0.225 - 0.414$) for tetrahedral coordination. So you might expect all the $\ce{Cl-Al-Cl}$ bond angles to be $109^\circ$.

But while the void for the aluminum is big enough, the distance between the internal chlorine atoms would be only $\pu{300 pm}$ in this configuration; the chlorine atoms are squeezed (ionic radius $\pu{181 pm}$). In addition, the $\ce{Al-Al}$ distance is only $\pu{250 pm}$ and the $\ce{Al^3+\bond{-}Al^3+}$ repulsion would be extreme. (In the solid, the octahedral holes are farther apart) So both aluminum atoms and the internal chlorine atoms are pushed apart, consistent with imagining more $\ce{p}$ character in the central bonds and therefore less $\ce{p}$ character in the external chlorine atoms, so $\ce{Al}$ goes to $\ce{sp^2}$ (or nearly so). The internal $\ce{Cl-Al-Cl}$ bond angle is even less than $90^\circ$ (less than you would expect for two $\ce{p}$ orbitals), so π bonding needs to be considered.

Aluminum chloride is a strong Lewis acid. Its solubility in organic solvents is due to formation of complexes. It is not soluble in the way that carbon tetrachloride or hexachloroethane are soluble. The salt-like (ionic) character of solid aluminium chloride is still present in $\ce{Al2Cl6}$; the aluminum atoms are surrounded by a cocoon of chlorides. Solid aluminum chloride can be ground into particles and nano particles; $\ce{Al2Cl6}$ is like a pico particle of salt.


What is the correct definition of hydration enthalpy and why is it always negative?

Hydration is defined as the following process (with any salt but using copper(II) sulphate as an example):

$$\ce{CuSO4 + 5 H2O -> CuSO4 . 5 H2O (s)}\tag{1}$$

This reaction will only happen if the associated enthalpy $\Delta_\mathrm{hyd} H^0$ — or more precisely, the associated Gibbs free energy $\Delta_\mathrm{hyd} G^0$ is negative. For some compounds, that is not the case; for example sodium chloride does not form hydrates: reaction (2) does not occur spontaneously.

$$\ce{NaCl + n H2O} \nrightarrow \ce{NaCl . n H2O}\tag{2}$$

Thus, we cannot measure a hydration enthalpy. If we can measure it, the process must be spontaneous and thus the enthalpy negative. Thus, all measurable hydration enthalpies are negative.

Sometimes, salts can form multiple hydrates. However, not every hydrate is always possible. Hydration enthalpies only exist for those hydrates which are possible.

Your definitions are basically identical only that they do not measure hydration enthalpies but solvation enthalpies. That is the process as shown in equation (3).

$$\ce{CuSO4 (+ H2O) -> Cu^2+ (aq) + SO4^2- (aq) (+ H2O)}\tag{3}$$

For that process, the enthalpy does not have to be negative. In fact, quite a few ionic compounds dissolve endothermicly, e.g. $\ce{NH4Cl}$.


Why doesn't diffusion take place between immiscible liquids?

In gaseous phases, we generally assume that there's very little or no bonding between individual molecules (or atoms in, say, a noble gas). This isn't strictly the case, but it's a pretty good approximation, and many good approximate techniques rely on it. A gas that behaves entirely in this way is known as an 'ideal gas'. (See also: ideal gas law.)

In contrast, liquids have bonding between individual molecules, and their behaviour is strongly affected by molecular properties.

So, in the examples you gave:

1. Bromine and air

In this example, imagine that you have a bunch of ping pong balls bouncing around at a reasonably high speed. (Ping pong balls are often used as examples for ideal gases.) Since the balls are already travelling at high speed (given a sensible temperature), and they only get in each other's way by bouncing off each other, it's relatively easy to get the bromine and the air 'balls' to mix, and there's really nothing apart from gravity to make them separate again.

2. Water and oil

Water molecules form pretty good bonds to other water molecules. (See: hydrogen bond.) Bonding in oils is generally weaker, but also occurs via a very different mechanism to the bonds in water. (See: Van der Waals force.) This means that oil does not form good bonds to water, and any water to diffuse into the bulk oil would have to break the bonds to all the neighbouring water molecules, and any oil to diffuse into the bulk water would have to separate a few water molecules to make a space to do so. Not just that, but since the water and oil molecules can reform their bonds if they manage to connect to other water and oil molecules, if you mix them up, the droplets will tend to recombine into the separate phases.

As a comparison to this example, let's check out two liquids that do mix.

3. Water and ethanol

For the water, we have basically the same situation as before -- water molecules forming good bonds to each other. The ethanol, though, has an -OH group that can form bonds to the water in the same way that the water does (though not as well). This means that ethanol that mixes with water (and vice versa) will tend to stay mixed, and given that the liquids are being mixed around just by random motions, means that you'll get one mixing with the other just as a matter of statistics.


Since, from what I know, $\ce{CuO}$ doesn't react nor dissolve in aq ammonia, does this experiment suggest a trace amounts of $\ce{Cu(OH)2}$ present in the (assumed) $\ce{CuO}$ precipitate, later forming a low-quality Schweizer's reagent, or am I missing something here?

Well in general this is true, but because the fresh precipitate can be partially hydrated and amorphous, which does allow the ammonium hydroxide to dissolve it.

  was the precipitate really $\ce{CuO}$ (if so, how to avoid creating it, i.e. what are the necessary conditions for $\ce{Cu(OH)2}$ creation here), and what is the easiest way to check for $\ce{CuO}$ presence in general?

Your precipitate was likely a hydrated copper oxide. Heat causes the copper hydroxide to lose its water.  To prevent $\ce{CuO}$ from forming, you want to keep the solution cool, perhaps with an ice bath. The easiest way to verify this is by visual observation. Copper oxide is black and copper hydroxide is blue, there is no false positive/negative to this for your preparation.

  how efficient should be a properly done homemade Schweizer's reagent (using the aforementioned $\ce{CuSO4 + NaOH + 25\%  NH4OH}$ route) in dissolving various celluloses in practice?

This is not easy to answer and depends greatly on the concentration of your reagents, temperature, pH and so on. I've seen videos on YouTube however showing about 2g of cotton balls dissolving in 500mL solution as a reference


Why does gas not liquify at a temperature above the critical temperature no matter how much pressure is applied on it? Why?

The critical point is a point of convergence of all state properties of the respective liquid and gas. It can be considered as the degeneration point, where there is no difference between gas and liquid and this distinguishing does not make sense any more.
It can be also said the supercritical fluid near the critical point is neither gas neither liquid. It is both at the same time. Farther from critical point, like for permanent gases, either gas-like either liquid-like properties are dominant, depending of the state.
By changing  pressure of supercritical fluid, it gradually transforms between  states, where it behaves more like gas or liquid, without evaporation or condensation. It is similar, as if you reach an elevated level via a big step (liquid/gas or evaporation/condensation) versus  along a slope (supercritical fluid). For the latter, there is no upstairs (gas) nor downstairs(liquid), there is just the slope.
Imagine you figurally walk as the substance on the phase diagram ($T$ grows to the right, $p$ grows upwards) around the critical point. If you do it counter-clockwise, you will evaporate as many times as many rounds you do, without ever condensing. If you do so clockwise, you will condensate as many times as many rounds you do, without ever evaporating.


Two-electron Integrals over Gaussian Plane Waves

I'm not sure if anyone has done it, but the closest I can think of is the GPW (mixed Gaussian Plane Wave) method. I think that involves a mapping between Gaussian and PW basis. It's DFT, but just look for the Coulomb integrals.

General idea:

Lippert, Gerald, et al. "A hybrid Gaussian and plane wave density functional scheme." Molecular Physics 92.3 (1997): 477-488.

The implementation can be found here:

VandeVondele, Joost, et al. "Quickstep: Fast and accurate density functional calculations using a mixed Gaussian and plane waves approach." Computer Physics Communications 167.2 (2005): 103-128.


Difference between 'Preferred IUPAC Name' and 'Systematic IUPAC Name'

The problem with systematic IUPAC names is that many compounds can have more than one systematic IUPAC name. A compound may be named correctly in two or more ways in accordance with the several methods recommended by IUPAC. Thus, asking for the systematic IUPAC name does not make sense for such compounds.

Therefore, current IUPAC recommendations (currently only for organic chemistry) include the definition of a preferred IUPAC name (PIN), which is the name preferred among two or more names generated from two or more IUPAC recommendations including also the many retained names.

Whether the preferred IUPAC name is also actually preferred by chemists cannot be generalized and depends strongly on the context.


It seems that you would need information about how much of each reactant you have when conducting this. edit: "Excess" is the key word here; so it hints that Oxygen is NOT the limiting reagent. Considering how combustion reactions are structured, the answer seems to be the other reactant here.


Temperature of chlorine gas at given pressure and volume

Assuming at these conditions it's safe to use ideal gas law ($pV = nRT$) for chlorine, and given the same gas sample ($n = \text{const}$, hence $nR = \text{const}$), let's express the unknown temperature $T_2$:

$$\frac{p_1V_1}{T_1} = \frac{p_2V_2}{T_2} \quad \to \quad T_2 = \frac{p_2V_2}{p_1V_1} \cdot T_1$$

Generally, I would recommend not to do intermediate calculations, especially if you are not asked to do so. Construct the mathematical expression first, convert the units if necessary, and plug in the values when you are done. Note that in this case you only need to convert temperature units.

   $$T_2 = \frac{\pu{876 torr} \cdot \pu{15.00 L}}{\pu{1124 torr} \cdot \pu{9.22 L}} \cdot \pu{297 K} = \pu{377 K}~\text{or}~\pu{104 ^\circ C}$$


The list looks accurate. For example, the melting point of polypropylene ranges from ~130 C to ~160 C, so it's likely safe to autoclave.  Commercially available polyethylene melts ~100 -110 C, so is unsuitable for autoclaving.

Another common plastic, not listed above, polyethylene terephthalate (PETE), used for packaging and soft-drink bottles, softens below 100 C and is also unsuited for an autoclave.

Electrical and optical items used in endoscopic procedures are often heat-sensitive and cannot be autoclaved, posing difficulties in removing contamination.

As suggested, if you're unsure but must autoclave an item, place it in an open or perforated container so as to reduce risk of damage to the sterilizer.


Does RMS velocity depend on number of gas molecules in a gas sample?

No, it does not depend directly on the number of molecules. The expression is
$$c_{rms}=\sqrt{\frac{3RT}{M}}$$
where $T$ is the temperature in K, $M$ is the molar mass in kg/mol and $R$ is of course the gas constant in J mol-1 K-1.
However, it is possible to have a situation where the number of molecule affects the temperature of the gas, which would change the r.m.s. velocity. For example, in Joule-Thompson effect, increasing the pressure causes the gas to heat up. It might be possible to do this by pumping in more gas into an insulated constant volume vessel. (although it sounds a bit contrived to be honest!)


The synthesis of 1-phenylprop-1-ene from propiophenone

This can be done it 2 simple steps. You are not adding to the carbon count so all you need to do is operate on propiophenone.

Step 1 - reduce the carbonyl group to a secondary alcohol. Sodium borohydride in $\ce{MeOH}$ or $\ce{EtOH}$ will do this in high yield and the workup is straightforward.

Step 2 - dehydrate the secondary alcohol to give the double bond. As it is a benzylic alcohol, protonation of the $\ce{-OH}$ group and formation of the benzylic cation is strongly favoured and will give you the alkene you want. Catalytic para-toluenesulfonic acid in refluxing toluene (with water removal by Dean-Stark apparatus) will do this easily.


difference between solving and dissolving something

According to the Merriam Webster dictionary, dissolve originates from the Middle English word solve, meaning to loosen or dissolve. There is no official modern English definition for solve other than 1) to find a solution, explanation, or answer or 2)  to pay (as a debt) in full.

The IUPAC Gold Book actually gives no definition for either solve or dissolve, though it defines solvation:

  Any stabilizing interaction of a solute (or solute moiety) and the solvent or a similar interaction of solvent with groups of an insoluble material (i.e. the ionic groups of an ion-exchange resin). Such interactions generally involve electrostatic forces and van der Waals forces, as well as chemically more specific effects such as hydrogen bond formation.

When two miscible liquids are combined, they are said to form a solution.


How can collission theory of chemical reaction and Law of mass action be connected?

For exponentiation, we can say that it is actually repeated multiplication. Comparing two cases:

$$\ce{A + B -> AB}$$
Here, the rate expression is $r=K[\ce{A}][\ce{B}]$. This is because one molecule of $\ce{A}$ and one molecule of $\ce{B}$ is needed for the product to form.
$$\ce{2A + B -> A2B}$$
First, write it in this way: $$\ce{A + A + B -> A2B}$$
Now, rate expression is $r=K\ce{[A][A][B]}$.
That's it - $r=K[\ce{A}]^2[\ce{B}]$

An intuitive starting point for me to understand why concentrations are multiplied is probably to connect it with multiplication theorem in probability (for independent events).


What is the highest the boiling point of water can be raised?

The "particular constant" $k$ you refer to in the question isn't an arbitrary constant.  This "law" is usually written as

$\Delta T = km = \frac{M_1 RT_b^2}{1000 \Delta H_1} m$

where k is the "constant you refer to; $M_1$ is the molecular weight of the solvent, $R$ is the gas constant, $T_b$ is the boiling point of the pure solvent, the constant 1000 makes the formula applicable only to water, and $\Delta H_1$ is the heat of vaporization of the solvent, here assumed to be water.  $m$ is the molality of the solute.

This equation is derived in courses in physical chemistry and is itself a simplification to allow slide rule simplification (this is true for a lot of formulas in chemistry).

So what you want to do to make $\Delta T$ as large as possible is to find a solvent with a large molecular weight and a high boiling point and a low heat of vaporization.  The constant 1000, which is the number of grams of solvent in a liter of solvent will also have to be adjusted.  And of course $m$ can be played with.

I don't think anyone has ever worked out what works best for a large $\Delta T$ since boiling point elevation, once used for molecular weight approximation, isn't used for that purpose any more -- and thermometers are a lot more sensitive and accurate than they used to be.


Why does a larger DNA gel box require a higher voltage to move DNA at the same speed?

It's actually a physics question. The speed of the particle in the gel is proportional to the force that acts on it. The force that acts on the charged particle is proportional to the strength of the electric field $E$. The strength of the electric field is proportional to the voltage $U$ and inverse proportional to the length of the gel $d$.

$$E = \frac{U}{d}$$

So if you want to have the same strength of the electric field you have to double the voltage if you double the length of the gel.


Put them in a container. It could be open at the top, but should be enclosed in the bottom.

The dry ice will insulate itself with a layer of cold air/CO2. As long as they are not under direct heating illumination or the convection of air in the room is high, this should be sufficient for a suitable mass of dry ice.

We routinely keep dry ice in "open" unpowered freezers at my lab (with no refrigeration) for the duration of the day. If your room is very humid you can get buildup of ice / condensed water, so some insulation around the container might be prudent.


Compressibility Factor and Generalized Compressibility Chart

You start out with an initial guess for the temperature.  I suggest starting out using the value that you would calculate using the ideal gas law.  You then,

Calculate the reduced temperature and pressure
Get z at this reduced temperature and pressure
Use these to calculate a value for the specific volume.
If the value for the calculated specific volume does not match 0.013, adjust your guess for the temperature and go back to step 1.

Continue doing this until the calculated specific volume matches the prescribed value of 0.013


Is Astatine a non metal or a metalloid?

Unknown

According to the Wikipedia article on astatine, the longest-lived isotope has a half-life of 8.1 hours. The element is so radioactive that large samples would melt and vaporize under the heat generated by its own radioactivity. Such behavior makes it difficult to assess whether astatine has the appropriate traits to be a metalloid.

Metalloids tend to be lustrous, brittle, semiconductors with other properties intermediate between metals and nonmetals. Nonmetals are dull/matte, brittle, and poor conductors of electricity and heat. All of these properties require a visible solid (or liquid) sample to assess, and thus astatine is impossible to assess and classify.

Astatine is usually placed with the metalloids or the nonmetals by different authors based on its position in periodic table, but in truth, astatine is so unstable in its elemental form, we have no idea which one it is.


Why does adding water to a saturated solution increase the number of ions present in the solution?

Consider a saturated solution of $\ce{Ag2SO4}$. The following equilibrium is attained.
$$ \ce{Ag2SO4 <=> 2Ag+  + SO4^2-}$$

Adding water accounts to increasing the volume of the solution. Hence more amount of solute can be dissolved since solubility of a salt depends on the amount of salt dissolved per unit volume. Therefore, there is an increase in the no. of ions produced in the solution. But as Avnish Kabaj mentioned, $[\ce{Ag+}]$ remains the same.

Also as MollyCooL said, $\pu{K_{sp}}$ doesn't change as it is a constant at a particular temperature.


Dependence of carbocation stability with respect to the length of alkyl group

Carbocation stability is widely known to increase going from primary -> secondary -> tertiary.

In the absence of other factors (such as lone pair donation etc.), the major contribution to the greater stability of tertiary carbocations vs primary is due to hyperconjugation between the vacant p-orbital (carbocation) and the C-H sigma bonding MO's.

Based on hyper conjugation alone, you can predict that length of the alkyl chain will have no effect on the stability by itself, only the number of neighbouring C-H bonds with which hyper conjugation can take place.
In specific regard to ethyl vs n-propyl cation, neither are considered stable.


Does it make sense to differentiate the Arrhenius equation with respect to temperature?

Note that $A$ is a kind of a collision frequency factor, which is temperature dependent via the mean molecule speed, even if far less then the Boltzmann exponential factor.
$$\mathrm{d}k/\mathrm{d}T = \frac {\mathrm{d}A}{\mathrm{d}T} \cdot \exp{(\frac {-E_\mathrm{a}}{RT})} + A \cdot \exp{(\frac{-E_\mathrm{a}}{RT})} \cdot \frac {E_\mathrm{a}}{RT^2} \tag{1}$$
$$A = A_0 \cdot {({\frac {T}{T_0})}}^{1/2} \tag{2}$$
$$\frac {\mathrm{d}A}{\mathrm{d}T} = A_0 \cdot \frac 12 \cdot {({\frac {T}{T_0})}}^{-1/2} \cdot \frac {1}{T_0}=\frac {A_0}2 \cdot {({\frac {T_0}{T})}}^{1/2} \cdot \frac {1}{T_0} =\frac {A_0}2 \cdot \frac 1{\sqrt{T \cdot T_0}}  \tag{3}$$
By subtitution in (1), using  (2) and (3):
$$\mathrm{d}k/\mathrm{d}T = \left( \frac 12 \cdot \frac 1{\sqrt{T \cdot T_0}}   +  {({\frac {T}{T_0})}}^{1/2}  \cdot \frac {E_\mathrm{a}}{RT^2} \right) \cdot A_0 \cdot \exp{(\frac{-E_\mathrm{a}}{RT})}\tag{4}$$
As AJKOER properly noted, temperature dependance  $A=A(T)$ can be usually neglected and we can use the simplified (1), as $\frac {\mathrm{d}A}{\mathrm{d}T} = 0$:
$$\mathrm{d}k/\mathrm{d}T = A \cdot \exp{(\frac{-E_\mathrm{a}}{RT})} \cdot \frac {E_\mathrm{a}}{RT^2} \tag{1a}$$
unless the activation energy is extremely low and reaction rate is driven by diffusion ( typical one is $\ce{H3O+(aq) + OH-(aq) -> 2 H2O(l)}$  ).
In solvents, we cannot usually ignore intermolecular interactions, so temperature dependence of $A$ may get quite complicated.


Why are all the orbitals that have the same principal number in Hydrogen degenerate?

The answer is right there in your question. It's only the interaction of multiple electrons in an atom like He, Li, Be, etc. that makes the different angular momenta wave functions differ in energy.

Consider this.. For the one electron system, why should a $p$ or $d$ orbital differ in energy from an $s$. What makes them differ?

In the multi-electron case, the $p$ orbitals have different spatial extent, different angular components, so the electron density caused by an electron in that orbital will interact differently with the other electrons. In other words, you need to have more than one electron for the "shape" of the $p$ and $d$ and $f$ orbitals to matter to the other electrons.

In the H atom, there's only one electron, so there's no electron-electron repulsion to differentiate the $s$, $p$, and $d$ orbitals.


Why is the formation of NaCl an exothermic process?

The negative standard enthalpy of formation of sodium chloride (i.e. ‘the formation is exothermic’) is primarily due to the formation of a crystal lattice. This crystal lattice is a macroscopic arrangement of ions in a cubic fashion with each sodium ion surrounded by six chlorides and each chloride ion surrounded by six sodium ions.

If it were not for the generation of this lattice and its inherent stabilising properties (a very good charge distribution, charges neatly alternate, contact between species of identical charge is strongly inhibited) the formation of sodium chloride would be much less exothermic.

Note most importantly, that it is not merely the generation of sodium cations (that process is endothermic) or the generation of chloride anions (that process is only weakly exothermic) that causes the overall exothermicity of the entire reaction.


Aluminium in homemade liquid castile soap?

The question is actually is skin absorption of Al3+ a health risk. The answer from a legal perspective varies depending on where you live. In the USA, for example, the answer is no, however, there appears to be a growth industry in selling Aluminum-free products, but not in Europe, where apparently stick deodorants containing the Aluminum ions are banned.
But what is the suspected chemistry possibly suggesting health problems with the Aluminum ion? My prior exploration of this topic points to some recent research. Based upon a review of two articles noting the pro-oxidant property of the aluminum ion, a potential health risk pose by the ability of aqueous Al3+ to recycle transition metals (like Fe, Cu, Mn,..) to lower valence states (via the superoxide radical anion) in Fenton and Fenton-like reactions, contributing to so-called oxidative stress related diseases (including Alzheimer's disease, various cancers,...). More explicitly, with iron ions, for example, recycling between ferric and ferrous ions feeding a fenton reaction via superoxide radical anion (produced in our bodies):
$\ce{ Fe^{3+} + O2^{•−} <=> Fe^{2+} + O2}$
Source: See Eq (1.xxxiii) which can feed a fenton reaction forming the hydroxyl radical:
$\ce{ Fe^{2+} + H2O2  -> Fe^{3+} + •OH + OH-}$
The first alluded to article is Oxidative Stress Gated by Fenton and Haber Weiss Reactions and Its Association with Alzheimer’s Disease by Tushar Kanti Das, et al, published in Archives of  Neuroscience, July 2014 2(3): e20078, DOI: 10.5812/archneurosci.20078.
The authors cite, in Figure 4, “Formation of Aluminum Superoxide Semi reduced Radical Ion and Aluminum Superoxide Complex (43)”, with described reactions proceeding as follows (also adopting notation and water complexing from the second article):
$\ce{ [Al(H2O)4]^{3+} + O2^{•−}   <=> [Al(O2^{•−})(H2O)4]^{3+} }$
$\ce{ [Al(O2^{•−})(H2O)4]^{3+} + Fe^{3+} -> O2 + [Al(H2O)4]^{3+}  + Fe^{2+} }$
And, in the presence of H+ :
$\ce{ [Al(H2O)4]^{3+} + O2^{•−} + H+  <-> [Al(O2^{•−})(H+)H2O)4]^{3+} }$
$\ce{ [Al(O2^{•−})(H+)H2O)4]^{3+} + [Al(O2^{•−})(H+)H2O)4]^{3+} -> 2 [Al(H2O)4]^{3+} + H2O2 + O2 }$
Note per below, the action of hydrogen peroxide on ferric also leads to ferrous:
$\ce{ H2O2 <=> H+ + HO2- }$
$\ce{ Fe^{3+} + HO2- -> Fe^{2+} + HO2• }$
or, pH> 4.88, we have:
$\ce{ H+ + O2^{•−} -> HO2•}$
Another work: Pro-oxidant Activity of Aluminum: Stabilization of the Aluminum Superoxide Radical Ion by J. I. Mujika, F. Ruiperez, I. Infante, J. M. Ugalde, C. Exley, and X. Lopez in J., published in Phys. Chem. A 2011, 115, 6717–6723, American Chemical Society, dx.doi.org/10.1021/jp203290b .
In the Mujika article to quote “In addition, the presence of LMM ligands such as citrate could also have an indirect effect in the oxidation capacity of aluminum by augmenting the bioavailability of Al3+ species, shifting the formation of Al(OH)4- to higher pH’s. However, one should also take into account the effect of citrate chelation itself in the thermodynamic equilibrium of [AlO2•]2+ formation.”
So, my short takeaway, our body produces the short-lived superoxide radical anion capable of recycling problematic transition metals which can engage in fenton (and fenton-like) reactions forming powerful radicals. The latter can attack your DNA leading to associated diseases. The role of Al3+ is that it can form a complex with the superoxide, likely increasing its effective half-life and augmenting destructive radical activity in one's body.
To answer the question: "Is the soap safe to use?", likely that it is a personal decision based on one's understanding of the proposed underlying chemistry, as jurisdictions apparently vary in opinions.


Monoepoxidation of 1,2-dimethylcyclohexa-1,4-diene using m-CPBA

The interaction of the alkene with m-CPBA is primarily nucleophilic/electrophilic in nature, i.e. the π MO of the alkene attacks the O–O σ* orbital. Therefore, the more electron-rich alkene (i.e. more highly substituted alkene) preferentially reacts.

With methyl groups there probably isn't too much steric hindrance. m-CPBA approaches above/below the alkene and doesn't directly attack over the methyl groups.

Indeed, the monoepoxidation of 1,2-dimethylcyclohexa-1,4-diene is reported to proceed selectively at the more electron-rich tetrasubstituted double bond (J. Org. Chem. 2006, 71 (4), 1653–1657):


Why do the units of rate constants change, and what does that physically mean?

Maybe it would make more sense if rate constants would have different symbols depending on whether it is a zero, first or second order reaction.
As an analogy, we might be asking "how big" an object is depending on the length of it. Depending whether the object is, say, a square, a cube or a hypercube, the answer would be in terms of its area, volume or 4D-volume, and the units (and dimensions) would be different in each case.
For the equilibrium constant, we avoid the issue by switching from concentrations to either concentrations divided by a standard concentration (this is a simplification) or to activities (the former multiplied by an activity coefficient that accounts for non-ideality).

Why do the dimensions of the rate constant change depending on different reaction orders, and what does that physically mean?

They have to change with the chosen system of units so that the equations are consistent. We could choose instead to have a dimensionless measure of concentration when we set up the kinetic equations. This would describe the reality equally well, but the kinetic constants would always have the same dimensions (1/time).
According to this argument, there is no physical meaning in rate constants having distinct dimensions depending on the order of reaction.


Criterion for gravimetric analysis of silver

The solubility of $\ce{AgCl}$ is equal to $\sqrt{K_\mathrm{s}} = \pu{1.3E-5 M}.$
The solubility $s$ of $\ce{Ag2CO3}$ is such that  $K_\mathrm{s} = 4s^3.$ So that its solubility $s$ is equal to $s = \pu{1.2E-4 M}.$ This is ten times more than the solubility of $\ce{AgCl}.$
For gravimetric purposes, $\ce{AgCl}$ is a better choice.


Open Source SDF (Chemical Table File) Parser in any language

The free, open-source RDKit chemoinformatics software can read .sdf files.  This software is most often used via it's Python API.

You can learn more about RDKit at the project's home page.

I recommend installing it via conda.  (If you use python regularly but don't use conda for controlling environments and package distribution, you probably should.)  The installation command is conda install -c https://conda.anaconda.org/rdkit rdkit
.

The python code to parse an sdf file and write to a csv file is very simple.

import pandas as pd
from rdkit import Chem
from rdkit.Chem import PandasTools

my_sdf_file = '/Users/curt/Desktop/sdf-isothiocyanates.sdf'

frame = PandasTools.LoadSDF(my_sdf_file,
                            smilesName='SMILES',
                            molColName='Molecule',
                            includeFingerprints=False)

frame.to_csv('/Users/curt/Desktop/excel_isothiocyanates.csv')


Aromaticity in Organic chemistry textbooks covers the Huckel version, which abide by the rules you have correctly listed above. However, these rules are mainly a tool to identify an organic molecule which has a high delocalization of electrons across a ring structure (a valuable tool in identifying a molecule's stability/reactivity/etc...).

So, for 3-dimensional aromaticity, the molecule must reach a certain threshold of delocalization in the ring structure (which, in 3D, becomes more of a sphere). Rules such as trigonal planar hybridization in 3D are no longer useful, and thus different approaches are taken to measure the resonance state of a system, and thus its level of aromaticity.

Nozawa et al. (2016, cited below) discusses use of "[t]he bond length alternation (BLA) in cyclic compounds" as "a good indicator of aromaticity, as it allows an evaluation of the degree of effective $\pi$-electron delocalization." These values, using a "harmonic oscillator model of aromaticity (HOMA)," are close to 1 for aromatic molecules by definition, and Nozawa obtains these from a variety of experimental and computational techniques to compare aromaticity in normally anti-aromatic structures.

For your above structure, as some comments pointed out as well, they are mesomeric, or resonance structures of the molecule illustrating the ability for the electrons to delocalize to an aromatic state. Firme et al's article mentions "delocalization indexes among bridged atoms," what I assume to be yet another similar measure of aromaticity related to the BLA using HOMA quantification.

Hope this helps! Please let me know if there is any confusion, as this response is a bit packed with summary detail.

References:
1. Nozawa, R. et al. Stacked antiaromatic porphyrins. Nature Communications 7, 13620 (2016).


When water boils at its boiling point how do we call it as an equilibrium?

An isolated system consisting of liquid water and vapour is always in equilibrium.

If you are applying heat to a system consisting of liquid water and vapour the system is not in equilibrium.


Indeed, the very small signal at 12.9 ppm is due to the carboxyl group. The reason why it's so low is because, as explained here, at room temperature the proton exchange makes it nearly undetectable. At higher temperatures, you would probably not even see that spike.

The methoxy group, on the other hand, is not affected by such effects.

There will always be a tiny amount of "noise" on experimental NMR spectra, because a solution always contains unwanted elements; but usually anything that looks like a well defined spike is a significant signal.


What changes does zinc chloride undergo when ‘fusing’ on an atomic level? How does this influence reactivity?

One purpose is to eliminate water as @Waylander has already pointed out.

Another is to have a clean crystal surface. Aging is a common process of crystals. The surface is transformed slowly. The crystalline structure changes since it is not the same to be surrounded with other atoms (molecules, ions) of the same composition than with nothing or a gas.

Finally, another purpose is to have a crystal size as small as possible. The smaller the crystals the bigger the surface area, so more is available and the reaction proceeds faster.
 The small crystals are obtained in the rapid cooling. The faster, the smaller.

The zinc chloride solidifies in one lump, but there are still crystals.
They only have grown together.
If it were a real glass, the reactivity would be the highest since there would be no packing.
This can only be achieved with extreme rapid cooling.
When cooling is fast but still takes some time, crystals grow.
When the cooling is slower, the crystals grow bigger.
The smaller they are, the more surface the lump has.

Melting must be done for the other reasons, but when you cool it must be done fast because, if not, you will get a less reactive mass.
It is not a comparison between molten and non molten product, but between fast or slow cooling of the molten salt.


Are all lone pairs considered as π-electrons?

I am quoting from http://www.hutters-online.de/publikationen/dmchdo.html.

  One requirement that is basic to this concept and is agreed upon by most authors is that there have to be two double bonds that are in conjugation with a third one, but not in a linear arrangement. The p electronic system thus forms a bifurcation. So the simplest cross-conjugated hydrocarbon would be 3-methylene-1,4-pentadiene (I) that can be viewed as an ethene disubstituted geminally with two vinyl groups:

Based on the above the following resonance structures are possible. These have 8 $\pi$ electrons in conjugation. You are correct.


Tendency of formation of complexes in Group-1 elements

The fallacy is the assumption that "inner $d$ orbitals" become "available".  Generally they are not, in the Group 1 and Group 2 metals.  There are rare occasions where some evidence of $d$-orbital bonding is found for heavier G2 elements, but the impact is small and not widespread.
In Groups 1 and 2, where there are inner $d$ orbitals they are so well shielded that an electron placed there is not tightly bound and thus in general tends not to stay put.  Somewhat ironically, $d$ orbitals become available to accept electrons from ligands just at the point where in neutral atoms they are being filled (i.e., the transition elements).  This is because we add protons to the nucleus as we add electrons to the subshell, and the electrons we are adding to the $d$ orbitals do not fully shield the remaining $d$-orbital states from the increasing protic charge.  It's as if the ligand electrons need to see their doppelgangers in the swimming pool before they figure out the water must be fine.
In truth, without readily available $d$ orbitals and with low charge density (except maybe lithium), the alkali metals are poor at forming complex ions.  That includes complexing with hydroxide ions, thus explaining the unique basic strength of the hydroxides of these metals.


From Synthesis of Succinimide on Thomas' Chemistry:

  Add 186 g of succinic acid to RBF, place the beaker in an ice bath and start adding 220 ml of 26% ammonia slowly. Assemble short path distillation without a condenser, add some boiling chips. Heat the mixture under a small flame to distil of all the water. When the temperature reaches over 100°C the mixture darkens. Switch the receiver and collect everything that comes off before yellow fumes temp 280-300°C. Cool down to room temperature and place the receiver into a freezer for 24h, recrystallize the crude succinimide in hot ethanol. Then filter the crystals and wash with 40 ml of cold ethanol.

(Note that RBF = round-bottom flask)

The catch is that the amide can attack itself to close the ring, forming succinimide.

There are two rounds of amidification via a nucleophilic addition/elimination mechanism.


What is it called when a benzene ring loses a hydrogen atom?

The name of the benzene minus a hydrogen is called a phenyl group (C6H5). https://en.wikipedia.org/wiki/Phenyl_group Look this page up for more information. The question you've asked is very similar to the one that @Gaurang has commented, so just be careful when you ask questions on here. Hope this helps


Is it possible to normalize the concentration equilibrium constant between different reactions for comparison?

Since the equilibrium constant depends on the concentration of the species involved, there is no way to normalize it in the way I think you imagine it. A fair comparison of equilibrium constants is already given. The equilibrium constant is, as its name gives away already, constant. As such there is no normalization to be made.

Also, the quote is slightly off the point where it says that

  the equilibrium constants can vary by a huge amount due to the presence of a second A atom in the reactants

because the equilibrium constant might actually be the same (except for the unit, of course). The way you calculate the constant changes, but inferring from that that the value changes drastically is simply overreaching.

If $K > 1$ then the equilibrium lies on the side of the products, if $K<1$ then the equilibrium lies on the side of the reactants.


How do I solve such questions on paramagnetism and ferromagnetism?

I believe GATE is a university entrance exam in India, so they will not expect you to solve an extremely complicated equation to predict ferromagnetism or to memorize an infinite list of substances. Sadly, there is some element of rote memorization still lingering like a pest in the educational testing system.

Good news for you is that relatively few compounds and metals show ferromagnetism or ferrimagnetism. Find out the list of ferromagnetic metals from the web, you might end up with very few elements and oxides, such as iron compounds, iron oxide containing materials. Later you can attempt the multiple choice questions by the process of elimination. For example, in the given choices, choose the answers which show that magnetite is ferrimagnetic.

Lo and behold, there is only ONE choice which matches with Q-3!


Why is solid sodium hydride a base and not an acid when reacted with water?

The answer lies in electronegativity. Sodium has a significantly lower electronegativity ($\approx 1.0$) than hydrogen ($\approx 2.1$) meaning that hydrogen pulls electron density towards itself, away from sodium to generate a sodium cation and a hydride anion.

For a compound to be a Brønsted acid, you need to draw electron density away from the hydrogen, i.e. connect it to an electronegative atom such as oxygen, fluorine, nitrogen etc. Only then can it formally be described as $\ce{H+}$ and can dissociate as such.

A hydride is much better described as $\ce{H-}$ and has a free lone pair. As such, it is a Brønsted base, not an acid. In fact, if you extend the Brønsted acid/base definition in the way Lewis did, you will arrive at the sodium ($\ce{Na+}$) being the acidic species here.

The Brønsted acid/base reaction product of the base $\ce{H-}$ and the acid $\ce{H+}$ happens to be $\ce{H2}$, and because the acidic hydrogen is abstracted directly from water (hydrogen gas can bubble away, shifting the equilibrium even if the reaction were not thermodynamically favoured), $\ce{OH-}$ ions are left which can be written with the remaining $\ce{Na+}$ cation to give sodium hydroxide.


Chemical Equilibrium - finding equilibrium constant given degree of dissociation

It looks like alpha $\alpha$ is the degree of dissociation. The way it is explained in your image is a little odd, so let me see if I can do a better job. What seems like it might be problematic for you is that the explanation you have been given skips some algebra along the way. Below, I am showing all of the algebra up until your points of confusion.

For your reaction: $\ce{N2O4 <=> 2NO2}$

In the first part of your question, we need to determine the equilibrium constant

$$K_p=\frac{P^2_{\ce{NO2}}}{P_{\ce{N2O4}}}$$

We know the total pressure, but we do not know the partial pressures of each species. Fortunately, thanks to Raoult's Law, we can substitute in mole fractions since $P_i =P \chi_i $.

$$K_p=\frac{P^2 \chi^2_{\ce{NO2}}}{P\chi_{\ce{N2O4}}}=P\frac{\chi^2_{\ce{NO2}}}{\chi_{\ce{N2O4}}}$$

We aren't given any number of moles, but let's say that initially there was some number of moles of $\ce{N2O4}$ before any dissociation occurred. We'll call that number $n_0$. We can write expressions for the number of moles of $\ce{N2O4}$ and $\ce{NO2}$ after dissociation using $n_0$ and $\alpha$. Since $\alpha\%$ of $\ce{N2O4}$ is dissociated, $(1-\alpha\%)$ remains undissociated:

$$n_{\ce{N2O4}}=n_0 (1-\alpha)$$

The moles of $\ce{NO2}$ are $n_0$ times $2\alpha$, since $\alpha\%$ of $\ce{N2O4}$ was converted into $\ce{NO2}$ and there are two $\ce{NO2}$ molecules for every $\ce{N2O4}$ molecule:

$$n_{\ce{NO2}} = 2n_0 \alpha$$

The total moles (need this for mole fraction) is

$$\sum{n}=n_{\ce{N2O4}} + n_{\ce{NO2}} = n_0 (1-\alpha) + 2n_0\alpha = n_0 (1-\alpha +2 \alpha) = n_0 (1+\alpha)$$

When we calculate the mole fractions of each species, the $n_0$ term cancels out, which is good since we do not know what its value is:

$$\chi_{\ce{N2O4}}=\frac{n_{\ce{N2O4}}}{\sum{n}}=\frac{n_0 (1-\alpha)}{n_0 (1+\alpha)}=\frac{1-\alpha}{1+\alpha}$$
$$\chi_{\ce{NO2}}=\frac{n_{\ce{NO2}}}{\sum{n}}=\frac{2n_0\alpha}{n_0 (1+\alpha)}=\frac{2\alpha}{1+\alpha}$$

  Now you know where $1+\alpha$ came from. These values are plugged into the law of mass action for $K_p$ to determine the equilibrium constant.

In the second part of the question, you use the value of the equilibrium constant and the new pressure to determine a new value of $\alpha$. It all starts with the law of mass action and we manipulate some algebra from there. First the law of mass action using mole fractions:

$$K_p = P\frac{\chi^2_{\ce{NO2}}}{\chi_{\ce{N2O4}}}$$

Next, substitute in the expressions for mole fraction in terms of $\alpha$.

$$K_p = P\frac{\chi^2_{\ce{NO2}}}{\chi_{\ce{N2O4}}}=P\frac{\frac{(2\alpha)^2}{(1+\alpha)^2}}{\frac{1-\alpha}{1+\alpha}}$$

The term $1+\alpha$ appears in both the numerator and the denominator and can be canceled.

$$\require{cancel}
K_p = P\frac{\frac{(2\alpha)^2}{\cancel{(1+\alpha)}(1+\alpha)}}{\frac{1-\alpha}{\cancel{(1+\alpha)}}}=P\frac{(2\alpha)^2}{(1-\alpha)1+\alpha}=P\frac{(2\alpha)^2}{1-\alpha^2}$$

Now we move $P$ over to the left side of the equation to separate the values we know ($K_p$ and $P$) from those we do not know ($\alpha$).

$$\frac{K_p}{P}=\frac{(2\alpha^2)}{1-\alpha^2}$$

  Now you have something like your second equation that you are confused about. Perhaps there is a mistake in the original source?


Short answer: nuclear neutrons greatly affect chemistry involving hydrogen.

Longer answer

Hydrogen is the lightest element, and deuterium is just less than twice as heavy. This makes a pretty massive (pun intended) difference to any vibrational modes involving moving hydrogen/deuterium. This means that, for example, the deuterated equivalents of weak acids are less acidic, and any reaction involving the loss of a proton will be significantly slower with the deuterated version.

Since helium is pretty unreactive, and by the time you're at lithium an extra neutron is only increasing the mass by about 14%, the effects on reaction energetics are much, much smaller for anything else.

I'd suggest looking up the "Kinetic Isotope Effect" for more information and more complex effects the mass change on substituting isotopes can cause to chemical activity.

As Eric Brown said in his answer, different isotopes may also have very different spectroscopic properties if the spectroscopic technique depends either on vibrational or nuclear properties.

As a minor curiosity, I'm given to understand this also means that heavy water is mildly hazardous to your health, since deuterated water and deuterated analogues of various vital chemicals have a nasty tendency to react mostly irreversibly or very slowly with vital enzymes that are "tuned" for hydrogen.


What is the electron dot structure of an alkyne?

Carbon needs four electrons to fill its octet.  In a $\ce{C-H}$ covalent bond carbon shares one electron with hydrogen.  A triple consists of one σ covalent bond and two π covalent bonds.  The carbons share two electrons in each of these bonds; in total six electrons are shared between them.  You can draw this as a dot structure:

$$\ce{H:C:::C:H}$$


Roothaan matrix equation equation orbital coefficients

The $\mathbf{C}$ matrix will always be $K\times K$, where $K$ is the size of the finite basis set you are using. That is to say, with a basis set of size $K$, you will produce exactly $K$ MOs. This means you will never wind up with $\mathbf{C}$ being a vector. In the example you linked, they use a basis set of size 2, so they get 2 MOs for helium and thus the $\mathbf{C}$ matrix is $2\times 2$. While helium has only one occupied spatial orbital, it will have $K-1$ virtual orbitals depending the size of the basis set you use to solve the restricted Roothaan-Hall equations.


Why are metals conductive in terms of orbitals?

It's not that the electrons are completely unbound from the atoms - they still attract each other and form the bonds with other atoms. But the attraction between the atom and its electron is pretty weak so it's easy to detach the electron from the atom.

The properties of the metals are better to be described via ionization energy - the energy needed to remove an electron from an atom. Elements to the right of the periodic table have high ionization energy - they are not willing to lose electrons. Therefore these elements are bad conductors (and are not metals).

Metals have lower ionization energy - their valence shell just started to be filling and the strength of the proton is not enough to keep them tight. Therefore these electrons are free to flow with minimum efforts applied - which would form electrical current.

The intention of the "sea" model is to describe (in simplified manner) the fact that electrons are floating around from one atom to another. And while they participate in bonding, this is a dynamic structure that breaks and forms all the time.

Here is a nice video from Khan Academy on the metallic nature.


Why is this chair conformation the most stable and which leaving group is best?

In cyclohexane, the equatorial position is energetically favored over the axial position. So the lowest energy conformer is the one where the most substituents are in equatorial position. (Or rather: Where you minimize the energy according to the A Value).

The axial Cl is favored as leaving group because of the elimination reaction mechanism. Normally, we see $\beta_{1,2}$ elimination in an anti-fashion. Which is what is proposed by your image.


What is the use of both mol and mass representing the same thing i.e. amount of something?

Mole itself doesn't have a magnitude, but the value of mole does, in fact it is the magnitude. Any measurement has a magnitude (ie for the mass of 5 kg the magnitude will also be 5).

I think you are confusing it with scalar measurements (ie forces) which also have a direction. For example 5 Newtons 'in that direction'.

Mole is similar to a dozen (just a bit larger). It is used to count how many atoms, molecules, atoms or anything else there are.

If you think about mass as "the amount of stuff there is in an object" you can confuse it with mols

but the actual definition of mass is:

  A property of a physical body and a measure of its resistance to acceleration (a change in its state of motion) when a net force is applied.

which is different to mols (which is used to count objects).


Why does changing the temperature shift the equilibrium?

Suppose you have some reaction
$$\ce{A <=> B}.$$
The equilibrium constant for the reaction is $K$ and the Gibbs free energy change is $\Delta G$. The equilibrium constant is
$$K = \frac{[\ce{B}]}{[\ce{A}]}.$$
Increasing value of $K$ shifts the equilibrium towards the right, i.e. more $\ce{B}$, and reducing the value shifts it to the left, i.e. more $\ce{A}$. The equilibrium constant and the Gibbs free energy are related by
$$ K = \exp\left(-\frac{\Delta G}{RT}\right)$$
Putting in the expression you give for $\Delta G$ gives us
\begin{align}
 K &= \exp\left(-\frac{\Delta H - T\Delta S}{RT}\right) \\
   &= \exp\left(-\frac{\Delta H}{RT}\right)\,
      \exp\left(\frac{\Delta S}{R}\right).
\end{align}
If we assume the variation of $\Delta H$ and $\Delta S$ is small we can ignore the entropy term because it's a constant, and we get
$$K \propto \exp\left(-\frac{\Delta H}{RT}\right).$$
If the reaction is endothermic $\Delta H$ is positive, so we have the exponential of a negative number and this is less than one. If we increase the temperature we decrease $\Delta H/(RT)$ and the exponential increases. So increasing the temperature makes the equilibrium coefficient bigger, i.e. it drives the endothermic reaction.
To understand why this happens consider what happens when we make the temperature very high.  If you do this the value of $\Delta H/(RT)$ approaches zero and the exponential approaches unity, i.e. the concentrations of $\ce{A}$ and $\ce{B}$ are the same. What we've done is make the thermal energy much greater than the activation energy in either direction, so the rates of the forward and backward reactions are the same.


After some more research I'm pretty much convinced in its existence. As Mithoron noticed in the comments, the name dinitratomethane is somewhat confusing, and is indeed not preferred.

I suppose the most accurate IUPAC-compatible one would be (nitrooxy)methyl nitrate. It also turned out that in the literature (mostly on regulated and hazardous chemicals) this compound is generally known as methylene glycol dinitrate.

The second volume of a classic handbook "Chemistry and Technology of Explosives" [1, p. 155] gives the most extensive description I discovered (reference numbers are updated):

  Methylene glycol dinitrate, or "nitromethylene glycol", is an oily liquid having a boiling point of $\pu{75-77 ^\circ C}$ at $\pu{20 mm Hg}$.

  It is a strong explosive, capable of dissolving collodion cotton. However it is not stable, being readily hydrolysed and therefore it is of no practical significance.

  A method of obtaining this explosive, by the nitration of moist
  paraformaldehyde was patented by Travagli and Torboli [2]. They utilised the well-known property of formaldehyde of reacting like methylene glycol in the presence of water.

  According to Travagli [3,4] methylene glycol dinitrate is obtained as follows:
  paraformaldehyde is dissolved in sulphuric acid and added drop-wise at $\pu{3-5 ^\circ C}$ into a mixture of nitric and sulphuric acid. The resulting oil is separated and washed with water to give a final yield of $35\%$.

The recent handbook of liquid explosives from 2015 [5, p. 227] contains virtually the same information without any updates, which indicates there simply hasn't been much of an interest in this compound due to its instability and lack of possible applications.

References

Tadeusz Urbanski. Chemistry and Technology of Explosives, 1st ed., Pergamon Press GmbH, 1965; Vol. 2.
G. O. Travagli and Torboli, Ital. Pat. 338080 1935.
G. O. Travagli, Gazz. Chim. Ital. 1938, 68, 718.
G. O. Travagli, Chimica ed Ind. 1938, 20, 389.
Jiping, L. Liquid explosives; Springer Berlin Heidelberg: New York, NY, 2015. ISBN 978-3-662-45847-1.


It is not possible to separate iron from blood without damaging the haemoglobin. Iron is a part of haemoglobin, and if iron is removed, the haemoglobin becomes something else. The blood then becomes incapable of transferring oxygen.


Valency is the combining power of an atom; so, when we talk of iron having a valency of 2, it can combine either with two atoms of valency of one; or with one atom of valency of two.  Oxygen has a valency of two and so the formula for iron (II) oxide is FeO.

In the history of chemistry valency pre-dates the idea of electron configuration which looks like the source of your confusion.

Hope that is clearer.

Edit: you mentioned that oxygen has a valency of minus 2; that's not quite right.  It has a valency of 2 but it can form an oxidation state of minus 2.  You need to separate out the ideas of valency and oxidation state.  Although the ideas are related, they are distinct.


Your thinking is correct. Within a set of degenerate orbitals (i.e. equal in energy), the orbitals will fill one electron at a time, and only after all have one electron will they start to pair up.

The situation with $\ce{[Ni(CN)4]^{2-}}$ is that it has square planar geometry, so the two orbitals that are $e_g$ in an octahedral complex are separated in energy. This diagram from Chemistry LibreTexts shows it nicely:

Because of the separation in energy, the $\mathrm{d}_{xy}$ fills completely (two electrons) before any electrons fill the $\mathrm{}_{x^2-y^2}$.


Fluorochrome (fluorescein) emission decay when linked to macromolecule

There should be no FRET for the free fluorescein unless you have specifically added an acceptor and in any case this would reduce the fluorescein emission.  When bound to the protein fluorphores are in a different environment (e.g. pH  change, hydrophobic pocket, tight binding, polar/non-polar nearby residues etc.)  and this can affect their fluorescence yield and lifetime up or down, ANS (anilino naphthalene sulphonate) is a good example.  The same type of thing is probably happening in your case, you know the motion is restricted as the polarisation is greater so there has to be some interaction. Fluorescein emits at a relatively long wavelength so aromatic protein residues cannot act via FRET as they absorb in the UV. You would need a chromophore (such as heam/chlorin/porphyrin) that absorbs in the green to red region to be a FRET acceptor from the fluorescein and so reduce its intensity.


"Drinking water" plastic - can I store beer in this?

I doubt that this is a chemistry problem. Food safe plastics which can hold wine and coffee are so abundant and cheep, there is no reason that they went to the trouble of finding one which is not suitable for these, but suitable for water, and made the bladder out of it.

All beverages beside water are a good growing medium for one microorganism or another. And the bladder is hard to get clean, the sucking straw even more so. So, it looks to me that they are warning you to not put in anything beside water because you'd risk starting your own bacterial or fungal colony. Beer is as bad as other drinks - don't store it there.


What is the true depiction of the periodic table?

La and Ac have $d^1$ electrons in their valence shells, rather than $f^1$ electrons.

The long table you found looks like that for a several reasons.

The trends going down Sc-Y-La are like those seen in groups 1 and 2. The trend going down Sc-Y-Lu is like that of groups 4 to about 10. Since lanthanide chemistry is basically that of trivalent alkali or alkaline earth metals, this tips the balance in favour of group 3 as Sc-Y-La.

Also helping is that group 4 is the first in which the really characteristic properties of transition metals (variable oxidation states; colour; paramagnetism) are seen. Ditto, same thing happens with group 12, which is why they're shown as post-transition metals.

And there's the lanthanide contraction, which starts at Ce and finishes at Lu. If you show Lu as being a transition metal it mushes the end of the Ln contraction into the d-block.

And there's the periodic law, which says that the chemical elements, if arranged according to their atomic numbers, show an approximate repetition of properties after certain regular but varying intervals. Here, La is the first element after Y that shows the approximate repetition in properties.

This research supports the periodic law outcome:

Glawe H, Sanna A, Gross EKU & Marques MAL 2016, “The optimal one dimensional periodic table: a modified Pettifor chemical scale from data mining”, New Journal of Physics, vol. 18, 093011, https://iopscience.iop.org/article/10.1088/1367-2630/18/9/093011/pdf
Restrepo G 2017, "Building classes of similar chemical elements from binary compounds and their stoichiometries", in MA Benvenuto (ed.), Elements old and new: Discoveries, developments, challenges, and environmental implications, American Chemical Society, Washington DC, pp. 95-110

Both sources point to La being more distinct from the Ln than is the case for Lu.

Finally, if you count the differentiating electron discrepancies in each block of the periodic table, noting the above table has a split-d block, you'll find that an Sc-Y-La table has 12 such discrepancies whereas an Sc-Y-Lu table has 13. The current agreed-within-IUPAC table has 14 differentiating electron discrepancies!

As many people have a hard time accepting the idea of a split-d block, I personally feel that a better solution would be a Sc-Y-Lu table with group 3 being shown as Sc-Y-Lu-Lr and La-Ac. That would have the extra benefit of resolving long-standing discussions about the composition of group 3, which have been going on since about 1982, when Jensen published his paper in JChemEd, arguing for Sc-Y-Lu:

http://www.che.uc.edu/jensen/w.%20b.%20jensen/reprints/018.%20la%20vs%20lu.pdf

Nothing much has to change. Just add a 3 above La-Ac in the f-block, in addition to the one over Sc-Y-Lu-Lr in the d-block.

A chemistry book chapter on this group 3 would make fascinating reading (a good thing given, to date, that group 3 is supposed to be the least studied group).

Hope that helps.


Paper Towel Soaked with Isopropanol/Water Mixture Burns without Damaging the Paper. Why?

I think your professor is incorrect, and it has to do with the thermodynamics and kinetics of the combined vaporization/combustion process.

The heat of vaporization of water $\left(40.7\ \mathrm{kJ}\ \mathrm{mol}^{-1}\right)$cite is considerably less than the heat of combustion of isopropanol $\left(1830\ \mathrm{kJ}\ \mathrm{mol}^{-1}\right)$cite.  These numbers indicate that approximately $45$ water molecules must evaporate for each isopropanol molecule combusted in order to hold the temperature constant. This  corresponds to a $\sim2\%$ mol/mol solution, which (a) is not consistent with the $\sim50\%$ v/v solution he actually used, and (b) would probably be too dilute to burn at room temperature the way it does in the video. (Someone more motivated than I could probably confirm this using flash point data.)

I think, instead, the behavior depends on the following:

Vapor-liquid equilibrium stipulates that a liquid phase can never be at a temperature greater than its boiling point$^\dagger$.
It is the vapor of combustible substances that actually burns, not the liquid (see, e.g., this OSHA training document)
Up to a concentration of about $60\%$ alcohol, the vapor phase in equilibrium with a water-isopropanol liquid mixture has a considerably higher concentration of isopropanol as compared to the liquid (source):

For the present purpose, to read the diagram choose a temperature on the $y$-axis and track across the $x$-axis, noting where you cross the red and blue lines.  These correspond to the equilibrium liquid and gas phase compositions, respectively, at that temperature, given a mixture that's not too lean or too rich in isopropanol.  For example, take $82\ ^\circ\mathrm C$: you hit the red line (liquid) at about $20\%$ isopropanol, and the blue line (vapor) at about $53\%$ isopropanol.

Since the boiling point of isopropanol is $82.6\ ^\circ\mathrm C$, #1 means that as long as there's liquid left soaked into the paper (even $100\%$ isopropanol!), the paper can't burn.  You can actually see this in the video: he ignites the paper at around 0:57, but it doesn't actually start to char until about 1:08, ten seconds later.

The reason why the paper does then start to char in spots is that those spots have locally run out of liquid isopropanol to keep them cool.  There remains a cloud of burning gaseous isopropanol all around the paper (clearly!), which is more than enough to burn the now-dry paper.

When there's water mixed with the isopropanol, though, the vapor-liquid equilibrium comes into play.  Per point #3, based on the phase diagram, the isopropanol likes to be in the gas phase a lot more than the water, so once the vapor is ignited and burning (point #2), the heat produced tends to evaporate more of the alcohol than the water.  This feeds the flame for a while, but eventually it sputters out, with plenty of water left on the paper to keep it cooled well below its char point.

$^\dagger$ Mixtures of liquids technically don't have boiling points, as their behavior is more complex, but the concept is useful for making the point.


Is equilibrium constant at a given temperature derived from Gibbs free energy of reaction valid for doing simple kinetic modelling?

In general I would say 'no', thermodynamics only tells us about starting and ending points, thus you know the $\Delta G^\text{o}$ for the reaction but nothing about its actual mechanism, (since time does not come into thermodynamics). As $K_\mathrm{eq}$ is a ratio of rate constants these can take on many different values and still have the same ratio. To study the mechanism you will need to monitor species and intermediates vs time as the reaction proceeds.

Before doing any experiments you can only make informed guesses as to what the mechanism might be, even apparently simple reactions, such as between hydrogen and oxygen, is not simple but by experiment found to be a chain reaction.


Is there any software package for quantum chemistry that includes CAMB3LYP?

CAM-B3LYP is present in Gaussian, Q-Chem, GAMESS, NWChem, ORCA, DALTON, DIRAC, and perhaps other major software packages, either as cam-b3lyp or camb3lyp, however keywords are entered.

Notably it isn't available in TURBOMOLE as of version 7.1.


How to calculate the height of an hcp lattice?

To calculate the height of a unit cell, consider a tetrahedral void in an hexagonal closed packing arrangement. It can be imagined as a 3 solid spheres touching each other and at the center-point, you have another sphere stacked over them. An interactive version can be viewed on this site. The situation looks like this:

If you join the centers of these four spheres, you'll get a tetrahedron. That's basically a pyramid with a triangular base. I'm assuming each edge of our tetrahedron to be equal to $a$.

Now, you have a pyramid ($ABCD$), with an equilateral base ($\Delta BCD$), I would like you to drop a perpendicular from the highest point ($A$) to the center ($G$) triangular base. If you're following me correctly, you'll have a figure like this:

All we have to do now is to calculate the length $AG$. For this, simply use the Pythagorean theorem in $\Delta AGD$.

$$ \begin{align*} AD^2 &= AG^2 + GD^2 \tag{1} \end{align*} $$

Although we know that $AD=a$, the side $GD$ remains unknown. But that's easy to calculate. The point $G$ is the centroid of $\Delta BCD$. Thus, the length $GD$ equals $a/\sqrt{3}$. Pluggging in the values in our first equation, we get $AG=a \sqrt{\frac{2}{3}}$. But note, this is half the height of our unit cell. Thus, the required height is $2a \sqrt{\frac{2}{3}}$.


Iron and sulfur ions bond by electrostatic attraction.

This is the right answer. I see a hard time maneuvering around it.

  Breaking an ionic bond between $\ce{Fe^2+}$ and $\ce{S^2-}$ releases energy.

Around the basic definition of forces and bonds, there are usually two big beliefs. One is a big misconception, and the other one is right.

The misconception: Some students fall in the trap of believing that bonds are like boxes of energy. So, breaking these boxes of energy should release energy, right? No! In reality, ionic bonds are nothing but forces. This is like imagining that when you pull a box, you're applying or somehow putting some kinda a box.

The reality: Ionic bonds are just forces that are a result of attraction between two species (such as two atoms). Why do these happen in the first place? Because for instance in \ce{NaCl} the sodium atom really likes to lose an electron and chlorine wants to earn one. This love of giving and/or taking an electron is because of reaching the stability. Stability in this case means lesser energy. So, ionic bonds happen because atoms want to have lesser energy.

Finally, what would breaking this bond result in? If the atoms bonded together to lower their energies, then what would removing the bond result in?

  I don't think the ionic radius of an iron ion is equal to its atomic radius.

You're right! Think of the sodium chloride example again. The new sodium cation used to have 11 protons and 11 electrons. Now, after the lattice builds up, sodium crudely has only 10 electrons, with the same 11 protons. The electrons are outnumbered! They can't overpower protons anymore. So, what will happen to the distance electrons were able to travel to? (The distance they could "travel" further from the nucleus was called the atomic radius, and it's the ionic radius now)
So, is the atomic radius of iron equal to its ionic radius?

Rather pedantic and irrelevant, but there's no clear distinction between an ionic bond and a covalent bond. There are no "pure" ionic bonds.


Angular momentum of an electron in an atom

The question does not state what kind of atom, but I assume this is about the Hydrogen atom. In this case you can combine orbital and spin angular momentum according to Russell–Saunders coupling.

Since we are talking about the Hydrogen atom, the ground state is essentially just the $1s$ orbital. In general however, for many-electron systems, electronic states and orbitals are different things!


Bromination with NBS on cyclic structures

As a follow-up to the excellent comments provided thus far, the following thoughts are added. Electrostatic Potential Map of azulene (1a) shows that there is high electron density in the 5-membered ring (red color signifies high electron density). The 5-membered ring is reminiscent of the aromaticity of cyclopentadienide anion and the 7-membered ring bears a positive charge and has aromaticity similar to cycloheptatrienyl cation. This phenomenon is represented in structure 1b. The reaction does not involve bromine atom (a radical) because a bromine atom would be expected to attack the electropositive ring, not the electron-rich ring. The reaction is ionic and likely involves bromine rather than N-bromosuccinimide (NBS, 4) directly. Bromine is liberated from NBS by trace amounts of HBr formed by the slow decomposition of NBS. In the event, bromination of azulene (1c) produces a resonance stabilized cation 2a. Its canonical structure is 2b that is identical to 2b'. Loss of a proton from 2b (or 2b') provides bromoazulene 3.


In order for tautomerism to occur, the enol must be deprotonated, so as to allow the negative charge to come down, kick the pi bond of the C=C bond, which will bond to a proton. Hydrogen atoms on oxygen are much more acidic than hydrogens on carbons because of the electronegativity differences between O and H. Thus, the deprotonation does not happen in alkenes.


For the concepts underlying General Chemistry find that Huheey's book gives very lucid explanations.  As far as Organic Chemistry goes, vol. A of Carey-Sundberg's opus might be what you are asking for.  What level of text are we looking at?


I guess that 1.081 refers to the distance length (in Angstrom) separating the C and N atoms. You obtain this value by applying by using the values listed in the table; this is quite tedious, takes some time and require quite good knowledge about crystallography. Otherwise (and much more easy), you can use the crystallographic data listed in the table to build the structure using a proper crystallographic software (such as Vesta); then the software will calculate bond distances for you.


How is rotenone a polyphenol?

It is not. Wikipedia says: Polyphenols are $\ldots$ characterized by the presence of large multiples of phenol structural units.

And if you're to pull up an image which gives you the structural formula of rotenone you get (Image source: https://upload.wikimedia.org/wikipedi...)

Do you see any phenols in this structure?

  If it isn't which subclass would you put it in

According to this article I've found:

  Article source

    In the late $1920$’s, the chemical structure of rotenone was determined. Rotenone is a member of a class of compounds of related molecular structure referred to as isoflavones.

And finally,

  which functional groups does it contain?

With a keen eye, if you look at the structure carefully, you'll at least be able to spot a ketone.


One answer is that the ancient Greeks, specifically Plato, devised the term "element".  Quoting from wikipedia article on Classical elements:

  Plato seems to have been the first to use the term "element (στοιχεῖον, stoicheion)" in reference to air, fire, earth, and water. The ancient Greek word for element, stoicheion (from stoicheo, "to line up") meant "smallest division (of a sun-dial), a syllable", as the composing unit of an alphabet it could denote a letter and the smallest unit from which a word is formed. A similar alphabetic metaphor may be the origin of the equivalent Latin word elementum (from which the English word comes), possibly based on the names of the letters 'l', 'm', and 'n', though the validity of this idea is debated.

If you had a more modern, chemical definition of "element" in mind for your question, then another possible answer is Robert Boyle.

Quoting from wikipedia again, this time from its article on chemical elements:

  In 1661, Robert Boyle proposed his theory of corpuscularism which favoured the analysis of matter as constituted by irreducible units of matter (atoms) and, choosing to side with neither Aristotle's view of the four elements nor Paracelsus' view of three fundamental elements, left open the question of the number of elements. The first modern list of chemical elements was given in Antoine Lavoisier's 1789 Elements of Chemistry, which contained thirty-three elements, including light and caloric.

I can't resist adding a reference to my favorite English-language modern chemistry text that omits the word "element" entirely.

  For most of its being, mankind did not know what things are made
  of, but could only guess. With the growth of worldken, we began
  to learn, and today we have a beholding of stuff and work that
  watching bears out, both in the workstead and in daily life.

  The underlying kinds of stuff are the firststuffs, which link
  together in sundry ways to give rise to the rest. Formerly we
  knew of ninety-two firststuffs, from waterstuff, the lightest and
  barest, to ymirstuff, the heaviest. Now we have made more, such
  as aegirstuff and helstuff. [...]

If the Normans had never invaded England in 1066, we might not have "elements" at all, but "firststuffs" instead.  Either way, it's just a debate about the right translation for "stoicheion", a concept which Plato seems to have given us.