How can the Bonding in IF7 be explained using LCAO method?

No 6-centre-10-electron bond.

The bonding situation in $\ce{IF7}$ is explained by Christe et. al. (see ssavec's answer) as a 6-centre-10-electron bond. In the article itself I found no real evidence that supports this statement and I believe it to be completely wrong. The following section is taken from Karl O. Christe, E. C. Curtis, David A. Dixon, J. Am. Chem. Soc. 1993, 115 (4), 1520–1526. and might as well be the culprit of this monstrosity.

  The bonding in the square-planar $\ce{XeF4}$, can be described by two semi-ionic, 3-center 4-electron (3c-4e) bonds for the four $\ce{Xe-F}$ bonds and two lone valence electron pairs on $\ce{Xe}$ ($\mathrm{s^2p}_z^2$ hybrids). The 3c-4e bonds involve the $\mathrm{p}_x^2$ and $\mathrm{p}_y^2$ orbitals of xenon. Addition of an $\ce{F-}$ ion to the equatorial plane in $\ce{XeF4}$, results in pentagonal-planar $\ce{XeF5-}$ and the formation of a semi-ionic, 6-center 10-electron (6c-10e) bond involving the delocalized $\mathrm{p}_x^2\mathrm{p}_y^2$ hybrid orbitals of $\ce{Xe}$ and six electrons on the five $\ce{F}$ ligands. The two lone valence electron pairs on $\ce{Xe}$ in $\ce{XeF5-}$ are analogous to those in $\ce{XeF}$.
  The planar $\ce{IF5}$ fragment of $\ce{IF7}$, has essentially the same bonding as $\ce{XeF5-}$, as shown by the atomic population calculations given in Table VII. [...]

They continue to compare the populations of the valence orbitals in question, without stating the what methods they used to determine that.

  The electronic structure calculations were done at the ab initio molecular orbital (MO) and local density functional (LDF) levels. [...]

A more accurate description of the bonding in in $\ce{IF7}$

Iodine heptafluoride is one of the classic examples when it comes to describing hypervalency and octet-expansion. Early attempts included d-Orbitals into describing that phenomenon (VSEPR), resulting in a description that yields the correct geometry. We now know that this theory is obsolete and more often wrong than correct.
A very principle problem in describing the bonding in this molecule is assuming that it is completely covalent bound. Christe et. al. acknowledge that this is wrong by stating "semi-ionic" bonds. This is - as we will see - already a stretch.

A more accurate terminology for these kind of molecules is hyper-coordinated; i.e. the central atom has a higher coordination number than common.

The following analysis utilises a calculation on the local density approximation level of theory. The used basis set is of triple zeta quality and employs a small effective core potential for iodine. The density fitting approximation was used to somewhat speed up the calculations. This is often denoted as DF-BP86/def2-TZVPP. The program used is Gaussian 09 rev. D.

Geometry

The performance of the method is not overwhelming. The bond lengths are overestimated by quite a bit. As cited by Christe, observed is $r_\mathrm{obs}(\ce{I-F_{ax}})=178.1~\mathrm{pm}$ and $r_\mathrm{obs}(\ce{I-F_{eq}})=185.7~\mathrm{pm}$. I calculated $r_\mathrm{BP86}(\ce{I-F_{ax}})=190.4~\mathrm{pm}$ and $r_\mathrm{BP86}(\ce{I-F_{eq}})=195.6~\mathrm{pm}$. Other methods, that I could still reasonably perform did not do much better. The MP2 result is $r_\mathrm{MP2}(\ce{I-F_{ax}})=186.6~\mathrm{pm}$ and $r_\mathrm{MP2}(\ce{I-F_{eq}})=191.9~\mathrm{pm}$.
For the purpose of the analysis the DFT method should be enough. (I had a look at the other results, too; they don't differ significantly.)

Quantum theory of atoms in molecules (QTAIM)

I would usually start with looking at the orbitals, but in this case it is more appropriate to look at the electron density itself. Better yet, we look at the Laplacian of the electron density. The depicted QTAIM plots are the Laplacian distribution within the molecule. Solid blue lines indicate charge depletion $\nabla^2\rho>0$, dashed red lines indicate charge accumulation $\nabla^2\rho<0$, red spheres are bond critical points, solid black lines are bond paths, dark red lines are zero flux surfaces (these separate the atoms). See an earlier answer of me for some more information on QTAIM.
We have two bond critical points to consider, one for the axial and one for the equatorial bond. We find low values of electron density at both; $\rho(\ce{I-F_{eq}})=0.125~\mathrm{a.\!u.}$; $\rho(\ce{I-F_{ax}})=0.137~\mathrm{a.\!u.}$. The Laplacian at both points is positive; $\nabla^2\rho(\ce{I-F_{eq}})=0.33~\mathrm{a.\!u.}$; $\nabla^2\rho(\ce{I-F_{ax}})=0.42~\mathrm{a.\!u.}$.
These values indicate predominantly ionic bonding. The graphs nicely show that the atoms appear as separate entities.

With that background it seems wrong arguing the bonding situation by forming covalent bonds, like they are assumed in MO or VB theory.
However, as wrong as it is to neglect ionic bonding, it is equally wrong to neglect covalent bonds.

The Orbitals of $\ce{IF7}$

There are a total of four orbitals with contributions from iodine and fluorine. There are not more than eight bonding electrons. I cannot imagine that Christe et. al. found a different solution, as his level of theory was not better (maybe worse). They either did not know how to interpret these results or simply overlooked it.
Here are the relevant lines from the output.

 Alpha occ 17 OE=-1.196 is I1-s=0.23 F7-s=0.18 F8-s=0.18
 Alpha occ 25 OE=-0.587 is I1-p=0.30 F7-p=0.28 F8-p=0.28
 Alpha occ 26 OE=-0.568 is I1-p=0.31 F4-p=0.19 F3-p=0.19
 Alpha occ 27 OE=-0.568 is I1-p=0.31 F2-p=0.21 F6-p=0.14 F5-p=0.14

(OE is the orbital eigenvalue in Hartree, the contributions missing from 1.00 are at the other atoms, but to small to trigger printing.)

In the LCAO bonding picture we have a totally symmetric bonding orbital from all atoms, an orbital referring to the axial bonds, and two degenerate orbitals for the equatorial plane.

Charges and Wiberg Bond indices

The ionic character of the bonds can also be seen in the charge distribution. While the iodine has a high positive charge $q(\ce{I})=3.24$, the fluorines have a negative charge $q(\ce{F})=-0.46$. These are AIM charges calculated with the MultiWFN program package.
The Analysis with natural bond orbitals give a similar result, $q(\ce{I})=3.20$, $q(\ce{F})=-0.46$.
Looking at the Wiberg bond indices matrix formed from natural atomic orbitals, we find a bond index of $0.54$ for the equatorial bonds and a slightly higher value of $0.56$ for the axial bonds.

Conclusions

It is not possible to determine a number for how many sigma bonds are in the molecule, since this is strictly only possible for covalent bonds. The bonds are here predominantly ionic, with small covalent contributions. The number of bonding electrons does not exceed eight and therefore obeys the octet rule.
VSEPR fails, as it was expected. The term hyper-valency should be avoided and replaced by the term hyper-coordination.


Sorry, but the accepted answer is wrong. The problem is indeterminate. Is there 1 liter of solution, or 5 liters, or 10 liters, or some other amount?

If the problem was to be solved using mole fraction, then the problem should have been written that way. e.g.:

  Benzine and chloroform are considered very toxic organic solvents that dissolve in each other and in a solution of benzine and chloroform the mole fraction of benzine is 0.45. Find the the percent of mass( percent concentration solutions) in this solution.

There is no point in posing such sloppy problem statements.


Limestone is the natural sedimentary rock, e.g., the top of Mt. Everest that began as sea floor.  "Storing" $\ce{CO2}$ in limestone requires heating calcium carbonate white hot to dissociate it to lime, then scrubbing "other" $\ce{CO2}$ with the lime.  This is, to coin an adjective, smartless, since you also burned fuel to heat the limestone. Curing concrete imbibes and traps $\ce{CO2}$, but it was initially dissociated to release it.

Your second equation is terrible. You obtain a strong base by adding acid to a salt.  $\ce{CaCO3}$ plus carbonic acid, $\ce{CO2}$ in water, makes water-soluble calcium bicarbonate $\ce{Ca(HCO3)2}$.  That is how limestone caves are dissolved out, and why karst topography is stark (surface erosion) and unstable (sinkholes).

To vastly store carbon at a profit, plant forests, harvest, and build wood-frame buildings.  The photosynthesis bottleneck, carbon fixation, depends upon the slowest least efficient enzyme on Earth, RuBisCO.  A smart (at least, not corrupt) society then

1) Gene-gineers a better RuBisCO.  Ten times the productivity leaves it the worst enzyme on Earth.  Everybody would have a full belly.

2) Gene-gineers a 10x fast-growing forest with huge lignin content in its wood.  It won't rot, it won't feed termites, it will be immensely strong.  If wood-frame buildings become unfashionable, bury the wood in deserts to store carbon.

There is no problem so easily solved that social advocacy will not make it intractable, disastrous, and expensive.  The unabled maximize self-esteem to be credible - effectively selecting for the deluded, hence advocacy.  A society inundated with rules from religious and secular political classes ignoring productive ends will collapse.  USSR to Pol Pot, reality is what it appears to be - thermodynamics.  There is no cheating possible.


A sample of tap water is found to be 0.025 M in Ca2+. If 105 mg of Na2SO4 is added to 100.0 mL of the tap water, will any CaSO4 precipitate?

$K_{sp} =[Ca^{2+}][SO_4^{2-}] = 7.10 \times 10^{-5} M^2$

$0.025M[SO_4^{2-}]  = 7.10 \times 10^{-5} M^2$

$[SO_4^{2-}] = 0.00284M$ (this is the maximum soluble concentration of $SO_4^{2-}$ in the presence of $0.025M Ca^{2+}$)

However, 0.000739 moles of $SO_4^{2-}$ are added to 100 mL of water, corresponding to a concentration of $0.00739M SO_4^{2-}$.

Precipitation occurs because 0.00739M is greater than 0.00284M.


Relation between acidity and reactivity?

The issue is that reactivity is not an inherent property of a compound. It can be reactive towards specific types of reactions and completely inert to others. Acyl chloride reacting with phenol doesn't rely on phenol behaving as an acid, the type of reaction for which you would describe it as more reactive.


Tang Yong Boon certainly is in the correct direction (please refer to his comment).

I'll just wrap things up:
$$\gamma > \beta > \alpha $$
$$\mathrm{(Basic \ strength)}$$

Reasoning:

The $\gamma$ nitrogen is unquestionably the strongest basic site.
$\beta$-nitrogen follows, as its tendancy of lone pair donation is greatly enhanced due to the presence of an adjacent (ie. alpha wrt. itself) electron donating group ($\ce{-NH2}$), otherwise known as the alpha effect. Some explainations include destabilization and elevated energy of the ground state, resulting in lower activation barrier.
The $\alpha$-nitrogen is best called an acidic site.


Why doesn’t aniline undergo Friedel-Crafts alkylation?

The answer lies in the fact that aniline is a Lewis base and $\ce{AlCl3}$ is a Lewis acid. The reaction between aniline and $\ce{AlCl3}$ hampers the catalytic activity of $\ce{AlCl3}$ required to perform the Friedel-Crafts alkylation and acylation.

Despite the activation of the $\ce{NH2}$ group, Friedel-Crafts alkylations and acylations fail because the $\ce{NH2}$ group acts as Lewis base and interacts with the Lewis acid catalyst. ... A way to overcome this problem is to convert the $\ce{NH2}$ group into an amido group prior to Friedel-Crafts reaction. ...The amide group can be hydrolysed back to aniline group after alkylation.
(Pg. 1167 An Introduction to Organic, Inorganic and Physical Chemistry. 4th edition. Catherine E. Housecroft, Edwin C. Constable. Pearson Education Limited)

The aniline forms a Lewis acid base adduct with $\ce{AlCl3}$ :
This prevents the interaction between alkyl or acyl chloride with $\ce{AlCl3}$.


Angular dependence of x-ray and neutron scattering factors

Scattering factors describe scattering of an atom

To predict how an object will diffract, you essentially have to do a 3D Fourier transform of the scattering density (https://chemistry.stackexchange.com/a/115815/72973). The scattering factor is derived from the scattering of a single atom (placed at the origin). For xray diffraction, the relevant density is that of the electrons. For neutron diffraction, it is that of the nucleus. See slide 11, https://www.embl-hamburg.de/biosaxs/courses/embo2012/slides/neutron-scattering-basics-trewhella.pdf

The fuzzier the scattering density, the higher the angular dependency

The Fourier transform of a Gaussian is a Gaussian again; a sharp Gaussian results in a broad transform, and vice versa. For X-ray diffraction, the electron density is fuzzy (same magnitude as the wavelength), so there is a large angular dependence. For neutron diffraction, the scattering density is very compact, leading to almost no angular dependence of the scattering factors.


LC-MS: Can a "bad" peak still be interpreted

This is low-quality data.  It's very difficult to say whether this signal is meaningful for several reasons.

It appears as soon as the LC-MS starts recording data at very low retention times.  That means it could be an artifact of column regeneration.  There is no sign that the data contains a "void volume" peak, meaning it is unclear whether the low-RT signal is from sample injection, from the previous sample, or some other artifact.

You don't say whether you randomized sample injection order.  Therefore the differences between the red and black sets of samples could arise because of changing conditions in the LC-MS instrument over time (e.g. column fouling, sample aging in the autosampler) or from legitimate differences in the samples that are biologically meaningful.

You don't show the full mass spectrum for your proposed analyte, so it is possible that the single m/z you showed here has an isotopic pattern consistent with your analyte, or inconsistent with it.

Thus, to answer your questions:

It basically wrong on the basis of info presented to claim that you detected a "biologically interesting" metabolite on the basis of the information presented.

It is also bordering on misleading to say you detected "a molecule" with the given m/z at all.  Maybe you could say you detected an unknown signal at m/z of xyz and are unsure of its origin.  That's about it.


How do we obtain the π molecular orbitals for allene via Hückel theory?

For p-orbitals $i$ and $j$ which are orthogonal to each other (i.e. different colour in your diagram), the off-diagonal matrix elements are simply zero:
$$\langle i | H | j \rangle = 0$$
If we ignore the overlap matrix,* the eigenvalue equation to solve therefore has the form:
$$\mathbf{Hc} = E\mathbf{c},$$
or in explicit form, using the fact that the matrix elements of $\mathbf{H}$ are $\mathbf{H}_{ij} = \langle i | H | j \rangle$,
$$\begin{pmatrix}
\alpha & \beta & 0 & 0 \\
\beta & \alpha & 0 & 0 \\
0 & 0 & \alpha & \beta \\
0 & 0 & \beta & \alpha
\end{pmatrix}
\begin{pmatrix}
c_1 \\ c_2 \\ c_3 \\ c_4
\end{pmatrix}
=
E
\begin{pmatrix}
c_1 \\ c_2 \\ c_3 \\ c_4
\end{pmatrix}$$
This matrix has block-diagonal form, so solving for the eigenvalues of the entire matrix simply reduces to solving for the eigenvalues of each individual $2\times 2$ block (lots of discussion can be found on Google). For a physical interpretation, this means that both π systems have no interaction with each other, so you can indeed solve them individually.
Each block gives rise to one MO with energy $\alpha + \beta$ ("bonding") and one with energy $\alpha - \beta$ ("antibonding"), so all in all you will have two degenerate MOs with energy $\alpha + \beta$ and two degenerate MOs with energy $\alpha - \beta$.

* If you don't ignore the overlap matrix $\mathbf{S}$, whose matrix elements are defined by $\mathbf{S}_{ij} = \langle i | j \rangle$, this becomes a "generalised eigenvalue equation"
$$\mathbf{Hc} = E\mathbf{Sc}$$
However, because the overlap matrix also has the same block-diagonal form, ignoring the overlap matrix doesn't change the fact that the problem is separable into two $2\times 2$ blocks. It merely changes the exact eigenvalues and eigenstates by a small amount.


One will need the Henderson Hasselbach equation which is given as

$$\text{pH}=pK_a+\log\frac{\ce{A-}}{\ce{HA}}$$

where HA and A- are the concentrations of the conjugate acid and conjugate base.

First we find our moles of each.

$\mathrm{{mol_ {acid}}=75~cm^3\times(0.10~mol~dm^{-3})\times(10cm)^{-3}\ dm^{3} = 0.0075\ mol\ \times \dfrac{1000\ mmol}{1\ mol} = \ 750~mmol}$

same thing for base: $\mathrm {mol_{base} = 375} $ mmol

Set up an ICA (not ICE) to handle your limiting reagent: (Write down the chemical equation if this is hard to see. One should react acid with the best base available)

$\ce{HA}$: $\pu{750 mmol -375 mmol} = \pu{375 mmol}$
$\ce{OH-}$: $\pu{375 mmol - 375 mmol} = \pu{0}$
$\ce{H2O}$: --------------------
$\ce{A-}$: $\pu{0 + 375 mmol} = \pu{375 mmol}$

In the Henderson Hasselbach equation moles can be used instead of molarity because both conjugates are in the same volume so they cancel:

$$\mathrm{pH}=3.75+\log\dfrac{375\ \mathrm{mmol\ \ce{A-}}}{375\ \mathrm{mmol\ } \ce{HA}}$$

$\mathrm{pH}=3.75$


What stable arrangements of hydrogen-bonded water exist in the range of 4-20 water molecules?

The Cambridge Cluster Database can give you a start with theoretical binding energies, but only one isomer for each size is shown.

http://www-wales.ch.cam.ac.uk/~wales/CCD/TIP4P-water.html


How is the degree of unsaturation derived?

This is the correct formula:

$$ DU = 1 + \frac{1}{2}\sum _{v=1}^{j} n_{i}(v-2),$$
where

$j$ = maximum valence of any atom in molecule
$v$ = valence of atom(s) 1 to whatever
$n_{v}$ = number of atoms with that valence

Wikipedia gives a good explanation. In simple terms the formula will give the number of double bonds in an alkene. For all alkanes the number would be 0 of course. Rings and ethynes would complicate the use of the formula. Think of using the equation on chemical formulas like $\ce{C_{x}}\ce{H_{y}}$ or $\ce{C_{x}}\ce{H_{y}}\ce{O_{z}}$.


Why does FeCl3 react with carboxylic acid, but not with sulphonic acid in this question?

It all made sense because $\ce{CH_3COOH}$ was an acid

$\ce{CH_3COOH}$ being an acid facilitates the complex formation as the chlorido ligands can be easily removed as $\ce{HCl}$. But, that's not the only reason why the red colored complex $\ce{[Fe(OAc)_3]}$ was formed.
You have to understand that the stability of the complexes is also determined by how well the ligand interacts with the central metal ion.Here, the ligand forming the coordination complex is $\ce{OAc^-}$ and this is associated with enough negative charge density to stabilize the complex formed.

So, I felt that the compound being acidic due to the sulfonic group would work similarly as the first compound and thus react with $\ce{FeCl_3}$. However on verifying with the answer key, I found that it would not react with $\ce{FeCl_3}$

This is because although as you have said has a sulfonic acid group, it is not the acidic strength of this ( or acidic strength of $\ce{CH_3COOH}$ in the first question) that drives the complex formation.It mainly depends upon the strength of the ligand and effective nuclear charge of the central metal ion.The ligand in this case, is associated with lot of electron withdrawing groups like $\ce{-NO_2}$ and that makes this a poor ligand.Also, the steric hindrance due to the bulky groups also has a role to play for the complex formation not to take place.
NOTE:
All my arguments made regarding stability of complexes is as per the Crystal Field Theory which assumes the interaction between metal ion and ligand to be purely ionic.That's why, I was comparing the charge density of the ligands in the 2 cases.


Cyclodextrin hydroxyls are the external surface (hydrophilic and hydrogen-bonding).  They circle the larger open end.  The carbon and hindered ether skeleton is internal (hydrophobic).  A cyclodextrin looks like a bonded hollow detergent micelle in water.

Before you believe advertising hype, remember that underivatized cyclodextrins are nephrotoxic, crystallizing upon concentration, probably as the cholesterol endohedral complex.  Look it up.

Ethanols insert their hydrocarbon tails into the center, then hydrogen bond their polar ends with each other and the rim.


The term superphosphate is really old, even well before the concept of atoms was proposed by Dalton. Therefore it is difficult to rationalize the choice of this terminology.

In the unabridged version of the Oxford English Dictionary, you can see the earliest usage dates back to 1798

  Chemistry. A phosphate containing an excess of phosphoric acid; an
  acid phosphate. Now disused except in superphosphate of lime, calcium
  superphosphate: cf. sense 2.

  1798   Philos. Trans. (Royal Soc.) 88 17   It was..Scheele who discovered, that the urine of healthy persons contains superphosphate,
  or acidulous phosphate, of lime.

Further information from the OED on the usage of "super" in chemical names confirms its use since antiquity. See antique examples

  (b) Denoting the highest proportion of a component, esp. owing to a
  high oxidation state. Now chiefly archaic or hist., except in
  superoxide n.   and in the names of certain substances used in
  industry and commerce, as superphosphate n., supersulphate n. Cf. sub-
  prefix 4b(b).  (i) Categories » [1788   J. St. John tr. L. B. Guyton de Morveau
  et al. Method Chym. Nomencl. 107   New names... Acetite of lead.
  Ancient names... Sugar of lead, Super-acetated lead.] 1811   Jrnl.
  Nat. Philos. June 78   The aqua lithargyri acetati is a saturated
  solution of the proper acetate of lead,..it is an essentially
  different salt from the super-acetate of lead. 1913   Brit. Med. Jrnl.
  4 Oct. 875/1   Dr. Latham..used the superacetate of lead in
  consumption. 1979   A. J. Youngson Sci. Revol. Victorian Med. i. 18
  Acetate or superacetate of lead combined with opium was prescribed for
  haemorrhage of the lungs.


Where to find a comprehensive list of chemical/biochemical etymologies?

It is a good question and many comprehensive works are available specifically on chemical etymologies. Many scholars have spent on portion of their life on chemical etymologies.  Ignore these useless down votes or close votes. The answer essentially boils down to how money are you willing to spend? I would say, if you are really keen, go for OED.
After going through so many etymological dictionaries of science (mainly chemistry & mathematics), I still feel that the Unabridged Oxford Dictionary  (> 20 volumes) is the best (https://www.oed.com/). It is not available for public use but one can buy subscription for a year at an affordable price. Most universities have it. There you can find the etymology for sure, the best part is that you can see the earliest use with original reference.
I will give you a taste of it. Say you want to find out where does the word alcohol come from?  You can extend this that the suffix -ol indicates alcohol group. Perhaps it is too much for you but you can see the extent of depth of analysis. It is sometimes surprising that there might be no connection of older meanings with current meanings.

post-classical Latin alcohol, alcool, alcol, alcofol kohl (galena
(lead sulphide) or stibnite (antimony sulphide)) (from 13th cent. in
British sources), spirit, essence obtained by distillation (a1527 in
Paracelsus in alcool vini , also alcohol vini ) (perhaps via Spanish
alcohol: see below) < Arabic al-kuḥl , Spanish Arabic al-kuḥul < al
the + kuḥl , Spanish Arabic kuḥul eye cosmetic, also denoting various
specific substances used as eye cosmetics or eye unguents (compare
kohl n.1) < the same base as Arabic kaḥala, Hebrew kāḥal (in the Bible
only in an isolated attestation in Ezekiel 23:40), both in sense ‘to
stain, to paint’, Akkadian guḫlu antimony (used as eye paint).
Compare Middle French alcohol, (rare) alcofol, French alcool,
†alcohol, †alcol, †alkol, †alkool, etc., the chief senses of which
are: ‘kohl, very fine powder’ (c1370 in a translation of Chauliac; the
precise sense is often difficult to determine in early quots.),
‘essence obtained by distillation’ (1620, originally and chiefly with
reference to spirit of wine), ‘ethanol’ (1792), ‘any of the members of
a similar class of chemical compounds’ (1834: see note below). Compare
also Spanish alcohol (c1200 as †alcofor; also †alcofol, †alcool,
†alcol, etc.), the chief senses of which are: ‘powder used as eye
cosmetic’ (c1200 as †alcofor), ‘galena, sulphide of lead’ (1541),
‘antimony, also any of various minerals containing antimony,
especially stibnite’ (a1555 or earlier; in a number of early instances
denoting minerals (from the second half of the 12th cent.) it is
impossible to tell whether the word denotes galena (sulphide of lead)
or antimony sulphide), ‘spirit of wine, ethanol’ (1730), ‘any of the
members of a similar class of chemical compounds’ (1865).

On your specific query
Carbonyl

< carbon n. + -yl suffix. Compare French carbonyle (1856 or earlier),
German Carbonyl (1854 or earlier).

Further probing of -yl

Formerly occasionally -ule, a terminal element of chemical terms, <
German -yl, < Greek ὕλη wood, matter, substance (see hyle n.), used
for ‘chemical principle, radical’. It was introduced by Wöhler and
Liebig ( Ann. der Pharm. (1832) III. 262), and first used by them in
the term benzoyl; other early names were éthyle (éthule), élayle
(Berzelius), dadyle, peucyle, citronyle, citryle (Blanchet and Sell).
Some fifteen in anglicized form, including acetyl, amyl, cinnamyl,
glyceryl, salicyl, appear in the Elements of Chemistry by T. Graham,
1842,

If you find OED too much to digest, the next best etymological dictionary of chemistry is

Elsevier's Dictionary of Chemoetymology (1st Edition) The Whys and Whences of Chemical Nomenclature and Terminology by Alexander Senning.
It has 400 pages. For example, the entry aniline shows

aniline C6H7N, derived from al-nil (Arabic: indigo plant), ultimately
from nila (Sanskrit: dark blue) – referring to the fact that aniline
was first obtained by degradation of indigo

The next one, which focussed on organic chemistry is

Organic Chemistry: The Name Game Modern Coined Terms and Their Origins (1st Edition)
by Alex Nickon and Ernest F. Silversmith,
This has plenty of fancy organic molecules from modern organic chemistry but at the end, they do list traditional names for example for benzene

benzene from Styrax benzoin, a tree native to Sumatra and Java. The
bark yields a resin, gum benzoin, from which "benzoic acid" was
obtained. Peligot (1833) and E. Mitscherlich (1834) heated benzoic
acid with lime to form C6H6; Mitscherlich named it "benzine." Liebig
preferred "benzol," but Laurent (1835) proposed "benzene."

Coming to free rides. Dr. John Andraos made a series of webpages by the name Modern Coined Glossary of Coined Names & Terms Used in Science to which I had heavily contributed more than a decade ago.
Here is a link: http://www.careerchem.com/NAMED/Glossary-Coined-Names.pdf
Another free source is Dr. William B. Jensen's lovely book "Ask the Historian". Simply Google "Ask the historian Jensen" and the first result is his book. He explain a lot of names, and even history of concepts. It was kind of him to make this book free of all of us. He has retired and he is no longer active.


What is the meaning of electrode potential?

The potential difference between two electrodes represents the voltage that is generated between the two electrodes that will allow the transfer of current/electrons. Physically it's basically a measure of how strongly one electrode attract electrons over the other which leads to the flow of current.


Does enthalpy equal heat when PV work is done?

I shall start off by mentioning that your question is not so much about p-V work being done, but rather about the pressure not being constant. It is perfectly possible for p-V work to be done when the pressure is constant, and indeed under these circumstances, the equation $\Delta H = q$ is still obeyed. However, this would necessitate a change in temperature.

Well, you can always try to come up with a counterexample.

Your hypothesis would be that if pressure is not constant and p-V work is done, then $\Delta H = q$. So, let's try calculating those two quantities for a reversible isothermal expansion of $1\mathrm{~mol}$ of an ideal gas, from an initial state of $(p_1, V_1)$ to a final state of $(p_2, V_2)$. Let's just arbitrarily set $V_2 = 2V_1$. Of course, since this is an ideal gas we are talking about, $p_2$ will be equal to $p_1/2$. And since this is a reversible process, $p_\mathrm{int} = p_\mathrm{ext}$ throughout, so let's drop the int and ext subscripts.

Let's calculate the work done (on the system) first.

$$\begin{align}
w &= -\int \! p\,\mathrm{d}V \\
&= -\int \! \frac{nRT}{V}\,\mathrm{d}V \\
&= nRT \ln{\left(\frac{V_1}{V_2}\right)} \\
&= -RT \ln 2
\end{align}$$

Now the internal energy change of the system has to be zero, since this is a closed system with an ideal gas and therefore $U = U(T)$. $\Delta T = 0$ therefore implies $\Delta U = 0$. So:

$$\begin{align}
q &= \Delta U - w \\
&= RT \ln 2
\end{align}$$

However, the enthalpy $H$ is also only a function of temperature: $H = H(T)$. Therefore,

$$\Delta H = 0 \neq q$$

Hypothesis: disproved!


In QM, properties that can be measured (such as energy, spin, position, etc) are each associated with an "operator". Every observation will always return an eigenvalue for the relevant operator. Eigen values are values for $\lambda$ for which the equation $$\hat{O}\Psi = \lambda\Psi$$ is true, where $\hat{O}$ is the operator of interest and $\Psi$ is the wave function of the entity of interest.
For example, let's consider electron "spin", by which we typically mean specifically the component of the spin in the z-direction, given by the operator $\hat{S}_z$. For the electron, this operator has two possible eigenvalues, which we abbreviate as $+1/2$ and $-1/2$. An electron having one of these values of spin is described as being in the eigenstate (or more simply just "state") associated with the given eigenvalue.
So we can say that having $+1/2$ spin and having $-1/2$ spin are two possible states of the electron.
What makes QM different from classical mechanics is that measuring an electron to have $+1/2$ spin does not mean that it was necessarily in the $+1/2$ spin state prior to our taking the measurement. An electron can (and often does) exist in a super-position of states, meaning that it simultaneously has both $+1/2$ and $-1/2$ spin, but our measurement causes it to collapse into one state or the other.
As you described in your question, some pairs of operators have the property that you cannot measure a defined value for both at the same time. Position and momentum are the most famous pair, since they are the example for the Heisenberg uncertainty principle, but many other pairs behave similarly. Energy and position are one such pair. So if we measure energy precisely, we cannot know where the electron with that energy exists. Since energy levels are the basis of the Schrodinger equation and define much of the behavior of atoms, chemists tend to treat electrons as if they are in a state in which the energy has been measured and other properties are unknown.
To address your point of confusion - "wave" and "particle" are descriptions of the behavior of electrons, most useful when understanding the concept of positional uncertainty, since "position" is most associated with particle behavior. When we learn about the Heisenberg uncertainty, we learn about experiments in which electrons behave as waves when their position has not been measured and like particles when their position has been measured, so it's easy to see why you associate wave and particle behavior with velocity and position, respectively, but those are just convenient examples to show the concept of duality. Operators generally should not be thought of as being associated with one description or the other, nor should we think of an electron as switching between particle and wave. It is better just to think of it as having specific position in space or not, and whether or not it has specific position it is neither a wave nor a particle, but rather just a quantum mechanical entity.
The results of any experiments we perform with electrons are predicted very well by the mathematical system that has been developed for quantum mechanics, and some of that math has parallels with classical wave behavior and some with classical particle behavior and some with neither of them. Beyond that, we have very little understanding of what an electron actually is and how its behavior accords with our understanding of the physics of the macroscopic world.
ADDENDUM For completeness, I'll add that in addition to $\hat{S}_z$, two other important operators that measure properties of electrons that can be known at the same time as energy is known are $\hat{L}^2$, the operator for the magnitude of total (quantum) angular momentum and $\hat{L}_z$, the z-component of that angular momentum. Thus, we can theoretically have an electron for which all four properties are fixed at one time, and we define the state described by the four eigenvalues by representing those eigenvalues with the "quantum numbers" (simplified forms of the eigenvalues) with which you are probably familiar:
n is the quantum number which represents the eigenvalues of the energy operator $\hat{H}$ (ie the Hamiltonian, which we associate with a specific shell)
l is the quantum number which represents the eigenvalues of the operator $\hat{L}^2$ (which we equate with s,p, d etc orbitals)
ml is the quantum number which represents the eigenvalues of the operator $\hat{L}_z$ (which we equate with the axial orientation of orbitals)
ms is the quantum number which represents the eigenvalues of the operator $\hat{S}_z$ (which we equate with the above-described spin orientation)


How does the energy released during a bond formation typically manifest itself on atomic level?

As pointed out by @Paul in case of the reaction between 2 hydrogens atoms you need a third body to take away the excess energy. This third atom(M) can be another hydrogen molecule or other inert gases(eg. Ar, He, N2).

Or you can have the presence of a wall(eg. gas-phase hydrogens atoms in a container):

Another possibility is adding a reactive gas(eg.oxygen) and in this case, the mechanism is much more complicated with many steps.

All these reactions are discussed in ref.1.

Comment: when two hydrogens atoms combine 4.5eV is released. In the case of an inert third body, I haven't found any quantitative information about how much energy is transferred to the third inert gas. If all of it is transferred to the third body(eg.H2 molecule) as kinetic energy you end up with an H2 molecule moving at 20'800m/s that corresponds to a kinetic temperature of 34'800K!

Another example is the CsBr formation in the presence of a third inert body(R):

From ref.2:

  Analysis of the detailed dynamics of recombination at small impact parameters showed that the stabilization of an ion pair occurs through the collision of the third body with the two ions and that the dynamics of energy transfer in this region is almost
  completely determined by the repulsive interactions of the Xe atom with both the ions.

  In the range of impact, the main mechanism of stabilization is energy transfer energy through repulsion between the third body and both the ions or through consecutive collisions of the third body with the ions approaching each other.

Regarding how much energy is transferred to the third inert body:

  The amount of energy transferred to the third body is apparently determined by the collision configuration and by the phase of the collision of the ions at the moment of energy transfer.

References

Bennett, J. E., and D. R. Blackmore. 1971. “Rates of Gas-Phase Hydrogen-Atom Recombination at Room Temperature in the Presence of Added Gases.” Symposium on Combustion 13 (1): 51–59.
Kabanov, D. B., and L. Yu Rusin. 2012. “Mechanism of the Direct Three-Body Recombination of Atomic Ions in a Central Collision.” Russian Journal of Physical Chemistry B 6 (4): 475–85.
Stepukhovich, Aleksandr Davidovich, and Viktor Markovich Umanskii. 1969. “Kinetics and Mechanism of Three-Body Recombination of Atoms and Radicals.” Russian Chemical Reviews 38 (8): 590.


Yes, ions are actually charged particles and neutral atoms are not. The TA's version is correct: the positive and negative charges cancel each other out. In a positive ion, we have one extra proton that gives the whole thing a net positive charge. In a neutral atom, electrons and protons cancel each other out exactly. How can they not to? After all, their electric charges are precisely identical, just with opposite signs.

Now when we speak about cancelling out like that, we only mean the net charge. Take a mole of neutral atoms and a mole of positive ions (just theoretically, of course, otherwise the consequences might be catastrophic), and even a crude electroscope will tell you the difference between the two right away.

If you specifically intend to ask about one ion, let's look at it from a different perspective. A neutral atom or molecule is, well, neutral, but that doesn't mean it would not participate in electrostatic interactions: there are those small positive and negative charges inside it. They seem to be moving, or at least not being all in one place, and you can feel tiny electric fields resulting from that. The corresponding electric potential decays with distance as $\frac1{r^2}$ for a dipole, and even faster than that for a quadrupole or other configurations of charges. The same applies to an ion, but it also has a net charge, and its potential behaves like $\frac1r$. When looking from a far enough distance, you will only feel the pull of the net charge. Look at a neutral atom from the same distance, and you will feel nothing.

Sure enough, if you look very closely, or even inside an atom (or an ion), you will see a lot of things going on around you, and the net charge will not be all that important.

So it goes.


Is it possible to determine the absolute value of entropy of a system?

Well, in practical I would say it is not possible. But there is a way to theoretically approximatly calculate an absolute value for entropy.

According to the third law of thermodynamics the entropy change can be calculated as

$$\mathrm d S = \frac{\delta q}{T}\, .$$

Per definition, a perfect crystal at $\pu{0 K}$ would have an entropy $S=0$.

So when you now heat that crystal and measure the change in entropy, since your initial $S$ was zero, the change is equal to the absolute value.

One problem is though, that the above equation only holds for constant temperatures. So the steps at which you measure the entropy changes have to happen at very (infinitesimal) small temperature changes.


How to connect proteins via disulfide bonds computationally?

If you have a structural model for the interaction, i.e. atomic coordinates, then you could use our software called Disulfide by Design.  See the following links.

http://cptweb.cpt.wayne.edu/DbD2/

http://www.biomedcentral.com/1471-2105/14/346


Memorize Transition Metal Electronic Ground State

Wikipedia has a comprehensive list on chemistry mnemonics, with which I'm now memorizing the position of the transition metals.
For example, for the first two periods of the elements (excluding neon):

H He Li Be B C N O F
Ha. Healthy Little Beggar Boys Catching Newts Or Fish.

Or my favourite for the first period of transition metals:

Sc Ti V Cr Mn Fe Co Ni Cu Zn
Saint Thomas Very Carefully Made Five Completely New Copies of Zarathustha.


Hydrogenation of pyridine compared to other heteroaromatic rings

When we reduce pyridine using the $\ce{H2/Pt/HCl}$ reagent-catalyst combination, we require a significantly lower temperature and pressure compared to other similar reductions (but without $\ce{HCl}$), because $\ce{H+}$ (and other Lewis acids for that matter) will go and add to the nitrogen atom, forming pyridinium salts.

This makes the compound more reactive towards reduction (as a partial positive charge will develop in the ring).


While filling liquid oxygen into a tank, why do they do open one end?

The end is opened to allow oxygen gas to escape. The flow of liquid oxygen into the tanker will not be pumped, just undergoing the natural tendency to flow from an area of high pressure to low pressure. As such, there would come a point when the residual gas in the tanker would start to provide back pressure on the flow input, slowing it down & potentially stopping it once pressures equalize between the tanker and the source.

By allowing oxygen gas to escape the tanker, the pressure will always be lower in the tanker, and the flow will be faster more adequately fill the tanker. The same technique is used when topping up liquid nitrogen in an NMR spectrometer, for example.


Is is valid to compare reaction quotient to $K_p$ as well as $K_c$?

The reaction of interest is:
$$\ce{XY(s) <=> X(g) + Y(g)} \tag1$$
Thus, $K_p = P_\ce{X}\cdot P_\ce{Y} = 4.1$ since $P_\ce{XY} = 1$ in given condition because it is a solid. If we assume $\ce{X}$ and $\ce{X}$ are real gases, the given conditions are such that $P_\ce{X} = \frac{n_\ce{X} RT}{V} = n_\ce{X}$ and $P_\ce{Y} = \frac{n_\ce{Y} RT}{V} = n_\ce{Y}$, no matter what units are used for $P$.For example:
$$P_\ce{X} = \frac{n_\ce{X} \ \pu{mol} \times  \pu{0.082 L atm K-1 mol-1} \times  \pu{273.15 K}}{\pu{22.4 L}} = n_\ce{X} \ \pu{atm}$$
Suppose, initial amounts of $\ce{X}$ and $\ce{X}$ are $x$ and $y$ and amount of $\alpha$ reacted to produce $\alpha$ amount of  $\ce{XY}$ at the equilibrium (to be realistic, $\alpha \gt 0$). Thus, at the equilibrium: $P_\ce{X} = x- \alpha$ and  $P_\ce{Y} = y- \alpha$. Therefore:
$$K_p = P_\ce{X}\cdot P_\ce{Y} = (x- \alpha)(y- \alpha) = \alpha^2 - (x + y) \alpha + xy = 4.1 \tag2$$

If $x = 5$ and $y = 0.5$: From the equation $(2)$, $$ \alpha^2 - 5.5 \alpha + 2.5 = 4.1 \ \Rightarrow \ \alpha^2 - 5.5 \alpha - 1.6 = 0 \\ \therefore \ \alpha = \frac{5.5 \pm \sqrt{5.5^2 + 4 \times 1.6}}{2} = \frac{5.5 \pm 6.05}{2} $$
Thus, $\alpha = 5.78$ (not realistic) or $\alpha \ne -0.28$ (because $\alpha \gt 0$). Therefore no reaction.
If $x = 2.0$ and $y = 2.0$: Again, from the equation $(2)$, $$ \alpha^2 - 4 \alpha + 4 = 4.1 \ \Rightarrow \ \alpha^2 - 4 \alpha - 0.1 = 0 \\ \therefore \ \alpha = \frac{4 \pm \sqrt{4^2 + 4 \times 0.1}}{2} = \frac{4 \pm 4.05}{2} $$
Thus, $\alpha = 4.025$ (not realistic) or $\alpha \ne -0.025$ (because $\alpha \gt 0$). Therefore no reaction.
If $x = 1.0$ and $y = 1.0$: Similarly, you can prove that there is no reaction.

Hence, answer is "$(d)$. none of the above."


Aluminum corrosion in most environment produces $\ce{Al2O3}$, corundum, which is extremely hard (9 on the Mohs scale, where 10 is diamond). Corundum is also quite resistant to most chemicals.

There are various products called "Aluminum Jelly", such as that from Loctite and from Henkel, which can remove some of the surface corrosion. They probably attack the aluminum at its point of contact with the oxide, rather than reacting with $\ce{Al2O3}$ itself. Regrettably, this probably means that though it might brighten the surface, the jelly will not substantially free the bolts.

The drill bits break when they encounter a mass of corrosion and suddenly seize, and this breakage can be minimized by running at slow speed with lots of lubricant. Applying gentle heat and cooling to the parts immersed in lubricant (e.g. Alumaslick or Vibra-TITE) allows it to penetrate better.

In any event, any threads in the aluminum have likely been ruined, so consider drilling oversize holes and re-tapping them (aluminum is easily tapped, but shavings tend to stick in the tool, so use lubricant and clear the tap frequently).


Why are palladium and platinum carbonyls unstable at room temperature?

Tl, dr:  It's not just a matter of size.  A combination of coordination geometry, node distribution in the $d$-orbital wavefunctions, and atomic size all conspire to weaken $\pi$ overlap in palladium and platinum carbonyls, not only relative to the nickel compound but also relative to earlier transition metal carbonyls in their respective periods.
Ziegler et al. [1] studied the relative stabilitie of metal-carbonyl bonds for $\text{G6(CO)}_6$, $\text{G8(CO)}_5$ and $\text{G10(CO)}_4$ carbonyl complexes, where $\text{G6}$ in the first formula is a Group 6 metal and similarly for the second and third formulae.  The following general trends were observed:

Within a period, the G6 hexacarbonyl is most stable.

Within each group, the 4th period carbonyl is most stable, the 5th period carbonyl is least stable and the 6tg period carbonyl is slightly more stable than the 5th.

From these trends it is seen that palladium and platinum carbonyls are literally in a bad place.
The authors attribute the stability order primarily to $\pi$ backbonding.  We may expect such backbonding to be strongest for the G6 hexacarbonyls where the octahedral geometry about the metal atom causes the metal-atom donor orbitals to have pure $\pi$ symmetry.  The five- and four-coordinate geometries (trigonal bipyramidal and tetrahedral, respectively) are progressively less favorable with regard to this symmetry.  This accounts for effect (1).
Effect (2) is related to relative atomic sizes within each respective group, but also to a more subtle effect.  In the $3d$ subshell, which is used as the valence subshell in the 4th period, the lobes have no radial nodes, a condition that allows maximum overlap with the molecular orbitals coming from the carbon monoxide.  The $4d$ and $5d$ orbitals used by the heavier metals have radial nodes so poorer overlap.  With the 6th period metals, relativistic effects intervene, but this effect (which I cannot tell from the abstract, and the full paper is behind a paywall) is less potent than having more compact orbitals with no radial nodes at all in the $3d$ series.
Reference:
1.
Tom Ziegler, Vincenzo Tschinke, and Charles Ursenbach, "Thermal stability and kinetic lability of the metal carbonyl bond. A theoretical study on M(CO)6 (M = chromium, molybdenum, tungsten), M(CO)5 (M = iron, ruthenium, osmium), and M(CO)4 (M = nickel, palladium, platinum)", J. Am. Chem. Soc. 1987, 109, 16, 4825–4837. Link


Are there stronger oxidizing agents than fluorine gas?

There are stronger oxidization agents than fluorine gas. I can name two in particular. But I don't think either would actually oxidize fluorine gas in it's pure state, for obvious reasons that you'll see shortly.

Dioxygen Difluoride. Affectionately named 'FOOF'. This is an extremely, extremely potent oxidizer. It also lights ice on fire. In fact, it lights essentially everything on fire or blows up on contact with said things.

Chlorine Trifluoride. Arguably, worse. This gas burns through practically everything, except for a few specific structural metals, but only because these form a very thin passivization layer. If this scratches, the tank burns. A notable incident with this gas, a 900kg spill of it burnt through 30cm of concrete, then 90cm of gravel underneath that concrete. It burnt over a meter into the ground.

As a foreword: I have no idea why you'd want to get within three miles of these things. Please don't light yourself on fire. It's not fun. (Trust me).


Core cutoff radius and energy cutoff radius in pseudo potential DFT

The reason pseudopotentials are use is because (at low pressures), core electrons do not participate in chemical bonding or reactions, and more over have high kinetic energies. If you expand and electronic wave function in a plane wave basis, high energy wave functions require high frequency basis functions, therefore in order to capture the wave function of a core electron accurately, you would need a high plane wave cutoff, which is computationally expensive.

To avoid this, the core electrons are replaced by a pseudopotential, which is by construction smooth. The core radius is to some extent a measure of the 'hardness' of the pseudopotential. A soft pseudopotential will generally have a larger core radius, and will be smooth. A hard pseudopotential will have a lower core radius, and in order to capture the rapidly varying (more jagged) potential near the core, will be less smooth, and therefore require higher frequencies in its Fourier transform to recover the correct electronic wave functions. Harder pseudopotential are necessary, for example, at high pressures, when the nuclei are squeezed together --- if the core parts of the pseudopotential overlap, then the calculation will fail catastrophically.


How can I go about obtaining the activation enthalpy of a reaction using the Arrhenius Equation experimentally?

Experiments at various temperatures will allow you to calculate $E_a$. This is clear from a putting the Arrhenius equation in a different form:

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

A plot (or linear regression) of $\ln{k}$ vs $\frac{1}{T}$ yields $\frac{-E_a}{R}$ as the slope. The same experiment with a catalyst will show the activation energy is decreased.


How many atoms of carbon are needed to produce 0.45 mol aluminium?

The key to this is by examining the mole ratio. Your balanced equation shows that 4 mols aluminium: 3 mols carbon

If you are familiar with ratios from maths, you can therefore find the specific number of moles of carbon in this reaction.

$4\ce{Al}: 3\ce{C}\implies0.45\ce{Al} :~?$

I think you should be able to get the answer as 0.3375 moles of carbon (again, it's just ratios)! From here, you use moles = mass / mr to get your answer.


Iron Filing Catalyzed Toluene Reaction with Bromine Solution

The iron and bromine must first react to form the catalyst (FeBr3) as the iron does not actually catalyze the reaction (this is known as in situ synthesis of the catalyst).  The FeBr3 catalyzes the reaction between toluene and Br2 to form ortho and para bromotoluene.  Catalysts are NOT consumed in the reaction though, which means the culprit of the rusty color is the catalyst (google iron(iii) bromide to view the color).

Also, ensure that you use a larger excess of the bromine.  The FeBr3 interacts with bromine to make it more electrophillic and able to react with the benzene ring of toluene.  If there is no bromine left after the catalyst has been formed, no reaction will occur.


extraction/isolation of Curcumin from Turmeric using ethanol

In their open access article Optimization of the extraction of curcumin from Curcuma longa rhizomes, published in Revista Brasileira de Farmacognosia, 2013, 23, 94-100, the authors systematically examined the effect of various parameters on the extraction yield, namely

drug to solvent weight ratio (1/6 and 1/4 g/g)
agitation speed (30 and 70 rpm)
extraction time (12 and 24 h)
extraction temperature (50 and 80 °C)
ethanolic strength (70 and 90 % v/v)

Based on their analysis, the following optimum conditions were seemingly predicted:

drug to solvent weight ratio: 1/6 g/g
agitation speed: 30 rpm
extraction time: (12 and 24 h
extraction temperature: 80 °C
ethanolic strength: 70% v/v

Please have a look at the full article for further details.

Edit

  I need to know how to safely set up a basic distillation to recover the ethanol for reuse.

That is a rather fundamental question which suggests that you don't have a lot of experience in organic benchwork.

You might want to have a look at the English wikipedia page on distillation for the general setup.

Note that you will not be able to fully separate ethanol and water by destillation. Instead, you will obtain an azeotrope, a mixture that has a boiling point lower than that of ethanol and water. In order to check the density of your destillate and to dilute again to an ethanolic strength suitable for extraction, a hydrometer might be a suitable tool.

In terms of lab safety, the following points should be considered:

Do not work alone!
Ethanol vapour is flammable!

Keep open flames and electric sparks away.
Run your still in a well-vented place.
Have a fire extinguisher at hand.
Have a work space with an unblocked emergency exit.

If you use standard lab glassware, a stirrer and an oil bath to heat the still pot, make sure that the tubes for the cooling water that cools your condenser are properly secured by wire or clamps. Remember that the heating bath is hot! The last thing that you want is cooling water splashing into the hot silicone oil.

On a lab scale, I would would not set up a classical distillation apparatus, but remove (and collect the ethanol) at reduced pressure using a rotary evaporator with a water bath!

In any case, use suitable eye protection!


At low concentrations for which the ideal bp elevation/fp depression expressions usually apply, molarity is linearly proportional to molality, therefore the statements are equivalent.

This webpage explains nicely why low concentrations are important:

  Raoult's law only works for low concentration solutions. Why? Well, in order for our approximation to work, the interactions between the solute and solvent molecules must be nearly identical. If the interactions are stronger, then the heat of vaporization of the solvent will change and thus our whole approximation falls apart. Since we know that intermolecular forces vary greatly between molecules, the affect of those forces must be kept at a minimum by keeping the solute concentration real low.

In any case (as explained by additional statements in the abovementioned website, which discusses dissociation of ionic solutes), the statement made in the answer to your exam would not be true generally even if molality instead of molarity was mentioned.

The point of the question seems to be that a student should associate a universal response of the solvent that is independent of solute with colligative properties as described using Raoult's law, which is behavior most often approximated by dilute solutions (the statement that Raoult's law only applies to dilute solutions happens to be false, btw), and that in this realm of application molarity is approx linearly proportional to molality.


Organic compounds that react with carbonate ions

Cook it up with ethylenediamine or diethylenetriamine- but don't bet on it.  What do you plan to have as the counterion for $\ce{K^+}$?  Check your carbon oxidation states for the second half of your question.

No, wait!  Add citric acid.  Yeah, that's the ticket.  Plop, plop, fizz, fizz.


Deprotonation of 2- and 3-methylquinoline

Yes, in contrast to 3-methylquinoline, the methyl group of 2-methylquinoline (1) is easily deprotonated by potassium ethoxide.

The anion has been examined by NMR spectroscopy (DOI), but its use in synthesis was reported long before. There's an article (in German) in Chem. Ber., 1909, 42, 1140 (DOI) where the reaction of 2-methylquinoline with diethyloxalate is described.


Influence of protic and aprotic solvents on SN reactions

As you know, $\mathrm{S_N1}$ and $\mathrm{S_N2}$ reactions have different mechanisms. $\mathrm{S_N1}$ occurs in 2 steps: formation of a carbocation and addition of the nucleophile. However, $\mathrm{S_N2}$ has a one-step mechanism where the addition of the nucleophile and the elimination of the leaving group happen simultaneously.

A protic solvent would participate in hydrogen bonding with the nucleophile since it can easily share one proton. A solvent shell is thus formed around the nucleophile, making him less able to undergo a $\mathrm{S_N2}$ reaction.


Why does bromine add to a double bond instead of a triple bond if both are present?

I'm going to take "more reactive" as in the alkene undergoes bromination at a faster rate than the alkyne, which seems to imply that the activation energy of the transition state is greater.

A Kinetic Perspective

The formation of the bromonium ion involves a cyclic transition state where the the Br-Br bond is attacked by the alkene/alkyne pi-bond, and the pi-bond donates to the sigma* antibonding orbital to form a sigma C-Br bond. In doing so, both carbons in the alkene/alkyne will have their p-orbitals hybridized with the sp/sp2 orbital to form the sp2/sp3 orbital respectively. However, the sp2 bonding orbital will have more ring strain, as it naturally exists with bond angles of 120 degrees away from other sp2 orbitals, while the sp3 orbital is somewhat less strained with bond angles ~109.5 degrees apart. Thus the higher energy transition state will make the alkyne less reactive, and bromination occurs on the alkene first.

In addition, (as was pointed out here : https://www.masterorganicchemistry.com/2013/05/29/alkyne-addition-pathways-the-3-membered-ring-pathway/) the p orbitals on Cl conjugate to the pi-bond on the alkene transition state, resulting in a 4-electron anti-aromatic system! This further increases the instability of the transition state and results in a slower reaction.

TL;DR: There's more ring strain on the alkyne's transition state because the carbons are sp2 hybridized, and its transition state involves the formation of an anti-aromatic ring, so the reaction occurs slower than bromination on the alkene.


Freezing all the water on earth with a seed crystal

A precise physical argument against the existence of an Ice-9-like state of water is probably not trivial (ab initio methods required?). However, there is a simple but powerful argument against it; we haven't found anything like it yet. Water is a very common substance, and humans have been applying the scientific method to it for hundreds of years, humans have manipulated it for hundreds of thousands of years, but most importantly, the entire Earth has been churning massive amounts of water around for billions of years. If an Ice-9-like state of water did exist, we should have seen it by now.

To be clear, superficially this argument may seem to be a fallacious argument from ignorance. However, we can qualitatively back it up with a bit more knowledge. Assuming that Ice-9 is the true ground state of water, and that all liquid water on Earth is currently in a false ground state, separated only by a kinetic barrier, and realising that the conversion of liquid water to Ice-9 is autocatalytic, we must ask ourselves, if it only had to happen once, why hasn't it happened yet?. The only reasonable conclusion is that, if it can happen, then the kinetic barrier to conversion must be enormous. The interconversion of diamond to graphite is a chemical process with a famously large kinetic barrier (around $\mathrm{540\ kJ\ mol^{-1}}$), yet we have plentiful evidence of it happening many times in geological time scales (and we can do it in the lab too!). Thus the kinetic barrier for conversion between liquid water and Ice-9 would have to be even larger, which is unlikely. There's also a limit to how high we can expect chemical kinetic barriers to be, which we can put very generously at $\mathrm{100\ eV \approx 10\ MJ\ mol^{-1}}$. Water molecules on Earth have had many, many, many interactions at energies way higher than this, as they have been incessantly exposed to cosmic radiation, nuclear decays, lightning strikes, high temperatures, etc. In short, the collective history of all water molecules on Earth has explored a vast region of the space of allowed states, and not once has an Ice-9-like state been found. It is thus fair to assume it simply does not exist.

Regarding your second question, I just remembered a great example: pure tin. At ambient pressure, solid tin can exist in two different polymorphs, and the transition between the most stable form is conveniently close to room temperature. Above 13.2 ºC, tin is most stable as a ductile metallic material, called white tin, whereas below 13.2 ºC, the most stable form is a brittle non-metal called grey tin. It is possible to sustain white tin below 13.2 ºC for a period of time, creating a metastable system, much like a supersaturated solution. However, an adequate disturbance will cause the white tin to fall to its ground state. Like Vonnegut's Ice-9, this transition is autocatalytic; once conversion begins, there's no stopping it below 13.2 ºC, creating the so-called tin pest. There are actually several great videos showing the process, which you can easily find on YouTube, such as this one.


What is the meaning of the term 'reserve acidity'?

Reserve acidity, more commonly known as Buffer capacity, is the amount of acid/base (in gram equivalents) that must be added to a buffer solution to change it's pH by one unit.

In your question, the $\ce{CH3COO-}$ is able to react with any Hydrogen ions that are added to the solution, decreasing the $[\ce{CH3COO-}]$ while increasing the $[\ce{CH3COOH}]$. It takes care of any excess acidity, hence, it is called as the reserve acidity. Similarly, the $\ce{CH3COOH}$ should be called the reserve basicity, as it can reach with any base to prevent much pH change.

For a mathematical approach, we use the Henderson-Hasselbach equation for calculating the final pH.


Bonding between rings and iron atom in ferrocene

The "real" structure of ferrocene is an iron atom sandwiched between two flat, parallel, pentagonal C5H5 rings. The diagram above shows the atomic positions from a crystal structure but the bonds are merely a convenience and don't accurately summarize the way the bonding works though the picture does accurately summarize the atomic positions in the crystal structure.
The real bonding requires some understanding of molecular orbital theory and how the d-orbitals of iron interact with the delocalized pi-orbitals in the unsaturated ring. Some pictures of the orbitals contributing to the bonding are shown here.


How to carry out cyclization of 1‐(2‐methylbutyl)‐2‐nitrocyclohexane?

The Hofmann-Loeffler-Freytag reaction as suggested by @user55119 in the comments above is the way to go.
First you need to reduce your nitro group. The reduction of alkyl nitro compounds is less simple than aromatic nitro groups and lower yielding in general. I would suggest the use of Al/Hg amalgam.
Acetylate the resulting amine.
N-chlorinate the amide - trichloroisocyanuric acid is good for this conditions
Irradiate the chlorinated amide to form the [5] ring similar to this example Baldwin gelsimicine synthesis
Finally hydrolyse off the acetyl group with cHCl.


Why is fluorine more reactive than iodine despite the weaker I-I bond?

In comparing the reactivity of fluorine or iodine, there are a couple of factors to consider. Part of it is the strength of the $\ce{X–X}$ bond, which is a barrier that tends to discourage them from reacting. Obviously, the higher this barrier, the less reactive the halogen is (in general).

However, unless you're trying to consider the relative reactivity towards homolytic cleavage of the bond, it isn't sufficient to look at the bond strength. The other bit is the driving force for the formation of the products – in other words, something that encourages them to react.

In this case, the formation of fluorine-containing products is generally much more thermodynamically favourable than that of the corresponding iodine-containing products. Some factors influencing this are:

The electron affinity of fluorine is much larger, which makes reduction to fluoride much easier
$\ce{E–F}$ bonds are stronger than $\ce{E–I}$ bonds, in both an ionic and covalent sense.
Fluoride has a larger solvation enthalpy than iodide

For example, factors 1 and 3 lead to the large differences in the reduction potentials: $E^\circ(\ce{F2}/\ce{F-}) = \pu{+2.87 V}$ versus $E^\circ(\ce{I2}/\ce{I-}) = \pu{+0.54 V}$.

Silica is readily attacked by $\ce{HF}$ not because it is a strong acid, but rather because of the strength of $\ce{Si-F}$ bond thus formed.

That deals with the thermodynamic parts of "reactivity", in the sense that it explains why some reactions with iodine require elevated temperatures for any significant conversion, whereas reactions with fluorine proceed readily at room temperature.

The same factors, though, apply to the kinetic aspects of "reactivity". The activation energy is largely dictated by bonds that one has to break or form, for example, and the rate of electron transfer processes usually increases with $E^\circ$. Exceptions to this arise when – for example – the reactants have solid state structures which must be broken apart before reaction can occur. However, that is not the case here.


You are right! There is an approximation here in only considering the vapor pressure of the gas. But let's consider the value of vapor pressure of water at $298\ce{K}$:  it's  $13.85\,\ce{mm Hg}$. So, if you neglect the vapor pressure of water, the relative uncertainty is  $$\frac{13.85}{760}\times100=1.83\%$$
As you can see, it's negligible.

P.S. To calculate the vapor pressure of water, I used the first equation in http://en.wikipedia.org/wiki/Vapour_pressure_of_water


Elimination reactions of vicinal dibromide

After the first elimination, you have a vinylic bromide, so there are two options for the second elimination. The two acidic hydrogens that the base can abstract are one attached to sp2 carbon and one to sp3. Since sp2 hybridized carbon is more electronegative, the base abstracts the hydrogen attached to it, and then bromine leaves being a good leaving group, which gives alkyne as the major product.


$S_{\infty}$ is equivalent to $\sigma_\mathrm{h}$. $S_{\infty}$ means rotate 0° then reflected by $\sigma_\mathrm{h}$, which is just same as taking $\sigma_\mathrm{h}$ itself.


What is the difference between Alkanes and Polymers?

Some points to note: You said polymers are saturated. This generalization is not true in many cases, e.g. polystyrene, nylon, and much more. Alkanes are always saturated.

Alkanes can have branchings. The ones you have in mind are called normal alkanes, in which no carbon has higher order than two. An alkane can have any kind of branching, as long as it is saturated.

Now, maybe the most important one: polymers are made of so-called monomers. All polymers can be constructed from exactly one monomer (even co-polymers). An alkane which is sort-of tree-like randomly branched, will not have such an unit.

A normal alkane has some similarities with polymers, tho. As others mentioned it, the size is the deciding factor there. A normal alkane with a carbon count of 10000 is for sure a polymer.


Can atoms other than hydrogen be reduced to a bare nucleus?

The fact that ionization energies have been measured for many single-electron ions is clear proof that bare nuclei, other than the examples mentioned in the question, have been produced. However, some might say that the process of ionization is more of a physical than a chemical one. This distinction is left to the reader.


How is the hydrazone formed in the Wolff-Kishner reduction?

Wolff–Kishner reduction is the method for the reduction of aldehydes and ketones to corresponding alkanes. First, the carbonyl compound condenses with hydrazine to form the corresponding hydrazone. The resulting hydrazone is treated with base to induce the reduction of the $\mathrm{sp^2}$-carbon to $\mathrm{sp^3}$-$\ce{CH2}$, and the oxidation of the hydrazine to gaseous nitrogen in the process to yield the corresponding alkane:

Source of the image
The following is the accepted mechanism of the reduction:

Source of the image

How does the formation of the hydrazone occur in basic conditions?

The formation of the hydrazone can be done in either acidic or basic conditions (recall imine formation of aldehydes and ketones with primary amine and 2,4-dinitrophenylhydrazone formation of  them with 2,4-dinitrophenylhydrazine). And, yes, as Wikipedia mentioned, sometimes it is advantageous to use a pre-formed hydrazone as substrate for the reduction. Yet, remember, Wolff–Kishner reduction is better suited for substrates with acid sensitive groups. Therefore, use of acid catalyst to pre-form hydrazone for the reduction is meaningless in that situation.


Why does the Rosenmund reduction stop at the aldehyde stage?

In general, the Rosenmund reduction isn't actually used very often, mainly since the catalyst is also very good at reducing other functional groups which may be present in the molecule (the catalyst is similar to that used for a Lindlar reduction).

To quote from Comprehensive Organic Name Reactions and Reagents:

  [...] the Rosenmund reduction is an unreliable and irreproducible method for organic synthesis, and it is often in competition with other side reactions. For example, alcohols are often the by-products of the Rosenmund reduction.

How the reaction works/why we get an aldehyde

As with many reactions involving catalyst systems in which palladium has been absorbed onto something (the classic one being palladium on carbon/charcoal), the mechanism of the Rosenmund reduction isn't well defined, and indeed there are likely to be different mechanisms operating depending on the quality/specification of the catalyst used.

One possible pathway has been proposed, and is given below. The critical thing here is that the acyl chloride is first activated by oxidative addition into the acyl chloride – this activation allows the reduction to take place.

The product of the reaction is often therefore the aldehyde, which should be kinetically inert to further reduction (this has been discussed on chemistry.SE previously, and can be found here), though the alcohol is often obtained in small amounts.


Why there is a non-linear trend on the melting points for period 2 elements, especially Na to Si?

If your textbook says that the melting point increases linearly, that is incorrect, as you point out. Generally high school textbooks can only make broad, qualitative statements about characteristics like melting points since the physics involved can depend to a larg degree on very complicated (quantum mechanical) interactions between the atoms (and their electrons) in a solid.

See for instance http://www.ptable.com/, and try playing with the slider on the top right to see melting points for different elements. There is a general trend of heavier metals in the same period having higher melting points, but there are some exceptions, such as manganese (Mn) around 1570 K.

As for why specifically Si's melting point is so much higher than Al's, the only truly correct answer is something along the lines of "that's just because of the differences in the way Si and Al atoms interact with each other". Because that's not a very satisfying explanation, I can try to paint a more intuitive picture: (UPDATE: it has been pointed out in the comments that even this simplified picture is wrong, see the linked wikipedia article for more details.) Aluminium has three valence electrons, which means that it would need to form five covalent bonds with its neighbours in order to reach the favourable octet of electrons. Unfortunately there is no regular crystal structure where every atom has exactly five nearest neighbours, so what ends up happening is that one electron from each atom delocalises, and the resulting postively charged ions are weakly (i.e. noncovalently) bound to each other, resulting in a low melting point (in other words, the atoms can be shaken apart easily).

Silicon on the other hand, has four valence electrons, only one more than Al, but this means that it only needs to bind to four neighbours to get an octet, and it turns out that there is a regular arrangement of atoms such that each atom has exactly four nearest neighbours (called the diamond structure). So, in solid Silicon, each atom is covalently (and thus strongly) bound to four neighbours, (and no electrons delocalised) which means that the atoms are hard to shake apart, which leads to a much higher melting point.

Now, this might be a more pleasing explanation (at least I hope so), but note that it is only meant for illustrative purposes: the same logic, dealing with 3 vs. 4 valence atoms can also be applied to for example Thallium and Lead, but these two metals have virtually the same melting point. Why is that? Well, there are other effects at play for these heavier atoms. In other words, it really does come down to how groups of the same atom interact with each other, and much of that can only be understood through the framework of quantum mechanics, and a rigorous treatment of solid state physics is probably too much to expect from a high school textbook :-)


Stability has several meanings. A compound can be stable on its own, under the given conditions such as temperature and pressure. This is the case for a lot of compounds. The other definition of stable could be a compound's inability to react with other molecules (like the inability of carboxylic acids to be oxidized). This is probably what you meant. A good base, like $R-O^-$ is unstable because of the +I effect of the alkyl group, which shifts the electron density towards oxygen, which in turn makes the ion a very good and strong nucleophile. This is the reason why such "good bases" tend to cause eliminations  and substitutions in organic molecules.

P.S.: This is the simplest explanation I can provide. If you have any doubts, ask them in the comments.

P.S.S.: Stability of a compound can also be explained on the basis of what @NotNicolaou said.


What happens when I use cyclohexanol to prepare cyclohexene?

The acid-catalyzed dehydration of alcohols involves a carbocation intermediate, which can act as a Brønsted acid; the loss of a proton yields the desired cyclohexene. Furthermore, the carbocation may undergo other side reactions. In particular, the carbocation may be subject to a reaction with other molecules to give substitution products; or the carbocation may undergo carbocation rearrangement.

The coloured residue of the distillation is probably the result of cationic polymerization of cyclohexene. This reaction in combination with rearrangements may yield various oligomers.


What is the crystal structure of bismuth oxyhydroxyphosphate (BOHP)?


Relation between molar mass and van der Waals force

From chemwiki.Ucdavis:

  The more electrons a molecule contains, the higher its ability to become polar. Polarizability increases in the periodic table from the top of a group to the bottom and from right to left within periods. This is because the higher the molecular mass, the more electrons an atom has. With more electrons, the outer electrons are easily displaced because the inner electrons shield the nucleus' positive charge from the outer electrons which would normally keep them close to the nucleus.

This seems totally legitimate. However, (as chemwiki subtly points out) the main factor in work here is the volume occupied by electrons. Polarizability increase is mainly due to a volume increase 1. So, all you need to take into consideration is whether a molecule is bigger than the other in terms of comparing the stronger London dispersion force.

Simply put, the ability of a molecule to get polarized is called polarizability. See Wikipedia for more info.


The idea has been proposed by several journal articles and Wikipedia suggests that it is being done already. In "Phosphate fertilizer from sewage sludge ash (SSA)", Franz proposes a method

whereby more than 90% of phosphorus can be extracted to make an adequate phosphate fertilizer

Also, the journal "Phosphorus recycling in sewage treatment plants with
biological phosphorus removal" by Heinzman lists three processes:

Krepo Process
Seaborne Process
Aqua Reci Process

In it Heinzman also mentions that

up to now no phosphorus recycling with a defrayal of costs is possible. The future importance of phosphorus recycling will depend on the market price for raw phosphate, the recycling costs and, furthermore, on the general political framework

So, it appears that it is cheaper to mine it than to recycle it. This may change in the future of course.


What is the relationship between the stability constant of a copper (II) electrolyte complex ion and the output voltage Ecell?

The redox potential of Copper does not depend directly on the stability of the complex Cu-ligand. It depends on the concentration of the residual $\ce{Cu^{2+}}$ ion in solution. And in any solution of $\ce{CuSO4}$, adding a ligand produces a decrease in concentration of the free ion $\ce{Cu^{2+}}$. So, according to Nernst's law, the potential $\ce{Cu^{2+}/Cu}$ becomes less positive.
For example, the standard redox potential of $\ce{Cu^{2+}/Cu}$ is + $0.34$ V. Let's consider the ligand chloride $\ce{Cl-}$. If enough chloride ions $\ce{Cl-}$ is added to a $1$ M $\ce{Cu^{2+}}$ solution so that at the end the excess of $\ce{Cl}$- ions is $1$ M, the complex $\ce{CuCl4^{2-}}$ will be formed, with a concentration $1$ M. The measured residual [$\ce{Cu^{2+}}$] concentration is $2.5 ·10^{-6}$ M. In this case, Nernst's law gives the following $\ce{Cu^{2+}/Cu}$ redox potential : $$\ce{E_{Cu^{2+}/Cu} = + 0.34 V + \frac{0.0592}{2}log(2.5·10^{-6}) = + 0.34 V - 0.0296·5.6 = +0.34 V -0.166 V = +0.174 V}$$ This +$0.174$ V is lower than the standard redox potential (+$0.34$ V).
The stability complex of the $\ce{CuCl4^{2-}}$ ion is : $\ce{K = \frac{[CuCl4^{2-}]}{[Cu^{2+}][Cl^-]^4} = \frac{1}{2.5·10^{-6}·1^4} = 4·10^5 M^{-4}}$
If another ligand is used, which makes a stronger complex with copper $\ce{Cu^{2+}}$, the stability constant is higher, and the redox potential will be still lower.


How to calculate the concentration of a K₂Cr₂O₇ solution?

First, you need a balanced equation for the oxidation.  The gives you the mole ratio of dichromate to substrate.  You know how many moles of substrate you have, you know how many moles of dichromate you need.  If you have 21.6 g of dichromate ions (as such)/liter, then you know how many moles/liter you have.  You know the number of moles you need, you know the number of moles/liter you have, you know the number of liters you need.

You will be doing your Jones oxidation in acid, of course.  Collins reagent is a sweeter deal for workup, since MEK is 27.5 g/100 ml water-soluble.


Baking Soda and Vinegar: Are there any secondary reactions?

It seems to me, according to the comments on the question, that there are NO additional reactions or products to worry about.

However, because of the foam, some of the original reactants can be tossed up, and find their way into the later parts of the apparatus as a haze/fog/cloud that is not completely inert.

Is that more-or-less accurate?  I'm trying to get an answer here, and not just a string of comments.


Do metals react with ammonia gas (not liquid phase)?

Wikipedia reports that magnesium nitride can be prepared:

By passing dry nitrogen over heated magnesium:
${\displaystyle{\begin{matrix}{}\\{\ce {{3Mg} + N2->[{\ce{800^{\circ }C}}]Mg3N2}}\\{}\end{matrix}}}$
or ammonia:
${\displaystyle{\begin{matrix}{}\\{\ce{{3Mg} + 2NH3->[{\ce{700^{\circ }C}}]{Mg3N2} + 3H2}}\\{}\end{matrix}}}$

No references are cited.


Which effect does adding hydrogen chloride have when determining the concentration of sodium hypochlorite in bleach via titration have?

The reaction is in equilibrium and if you don't add any acid, then some of the $\ce {ClO-}$ does not react. Adding acid will change the equilibrium to the right so most of the $\ce {ClO-}$ reacts.


Yes, your derived expression looks correct to me.

As to the closed form for the integral, WolframAlpha is most likely unable to find one because it is simply not there among the elementary and special functions. This has nothing to do with the equation being simple. Of course it is simple; nature "solves" it quite easily. It is just that the nature's definition of "simple" does not necessarily coincide with ours. Helium atom is simple, so good luck trying to solve the Schrodinger equation for it. Would you argue an atom is not actually that simple? Well, your concentration cell is made up of quite a few atoms.

So it goes.


Dissociation rates: trans-effect rule in square planar complexes

First off, there are two nice articles on the trans effect on Wikipedia and LibreTexts that are worth reading.

You need to remember that there are two factors leading to the trans effect:

Weakening the $\ce{Pt-X}$ bond trans to another group (i.e., a structural effect)
Stabilizing the 5-coordinate transition state, since substitution reactions in square planar complexes are generally associative. (i.e., a kinetic effect)

(a) I don't think the $\ce{I-}$ will substitute for the $\ce{NH3}$ group. Instead, I think the $\ce{CO}$ "wins" the trans effect "competition" and so the trans $\ce{CH3-}$ will be replaced by the $\ce{I-}$.

Next, substitution with the $\ce{Cl-}$ should replace the $\ce{I-}$.

Now your second question seems to be if you'll actually get a substitution with $\ce{Cl-}$ since $\ce{I-}$ has a greater trans effect. Yes, you will. All of these are in equilibrium, so if you generate a lot of $\ce{I-}$ that can substitute back (and get starting material). But you'll definitely get the Cl-substituted product too if you separate the reaction mixture.

Indeed, the prototypical example for the trans effect is synthesis of trans-platin or cis-platin, starting from either $\ce{[PtCl4]^{-2}}$ or $\ce{[Pt(NH3)4]^{+2}}$. Even though $\ce{NH3}$ has a smaller trans effect, you can synthesize cis-platin from $\ce{[PtCl4]^{-2}}$ and substituting two $\ce{Cl-}$ with $\ce{NH3}$.


Determining exponents in a rate law with different data

If you've correctly determined exponent $m$, then to ascertain $n$ you can proceed by first dividing rate equation $(3)$ by rate equation $(2)$ (note that the rate constant $k$ cancels immediately and we have that $m = 1$):

$$
\dfrac{\text{rate}_3}{\text{rate}_2} = \dfrac{\ce{[C2H4Br2]}_3}{\ce{[C2H4Br2]}_2} \bigg(\dfrac{\ce{[I-]}_3}{\ce{[I-]}_2}\bigg)^n
$$

Rearranging this equation, one obtains:

$$
\dfrac{\text{rate}_3}{\text{rate}_2} \cdot \dfrac{\ce{[C2H4Br2]}_2}{\ce{[C2H4Br2]}_3} = \bigg(\dfrac{\ce{[I-]}_3}{\ce{[I-]}_2}\bigg)^n
$$

Now, to isolate $n$, take the logarithm (in any base of your choice, though the book arbitrarily uses the natural log, and I'll do likewise out of concern for consistency) of both sides of the equation, and remember that one of the basic properties of logarithms is that $\log_b x^n = n\log_b x$:

$$
\ln\dfrac{\text{rate}_3 \ce{[C2H4Br2]}_2}{\text{rate}_2 \ce{[C2H4Br2]}_3} = \ln\bigg(\dfrac{\ce{[I-]}_3}{\ce{[I-]}_2}\bigg)^n = n\ln\dfrac{\ce{[I^{-}]}_3}{\ce{[I^{-}]}_2}
$$

Solving for $n$, one finds:

$$
\frac{\ln\dfrac{\text{rate}_3 \ce{[C2H4Br2]}_2}{\text{rate}_2 \ce{[C2H4Br2]}_3}}{\ln\dfrac{\ce{[I-]}_3}{\ce{[I-]}_2}} = n
$$

Which is the final equation from your textbook's solutions manual.


How come electrons in the d-subshell are considered valence electrons in transition metals?

The valence electrons are the ones responsible for engaging in chemistry and corresponds to the orbitals which are highest in energy. These electrons are the most easily accessed because they are not buried in the energy well of an atom. For main group elements, the highest principal quantum number p and s orbitals are the highest in energy, as you said, but for the transition metals, the partially filled d orbitals are the highest in energy. Generally, whichever orbitals are being filled last as you go across the periodic table are the ones that should be considered the valence electrons.

Another thing to note, as elements are ionized, the orbitals with higher orbital angular momentum drop in energy faster than orbitals with lower angular momentum. So, in the transition metals, the n d orbitals drop much more than the (n+1)s and p orbitals. Since, transition metals have <10 electrons (the number it would take to fill up all the d orbitals), all of them reside in d orbitals (and none in the s and p) and so these are the highest occupied orbitals and therefore the ones that engage in chemistry.


Has this way of comparing internal energy got any sense?

Yes. This is the binding energy.

In general, the binding energy is defined as the amount of energy released upon bringing infinitely separated particles in to contact. This would then be a negative number. Another sign convention is the binding energy can be viewed as the amount of work required to separate two atoms or molecules (or whatever) which are in contact. This binding energy is the negative of the first one (is a positive number).

For molecules, you might be more familiar with the term bond energy instead of binding energy. For a diatomic, these two things are the same. Notice that when we talk about the bond energy in a molecule, we treat that bond as being a diatomic molecule.

For the diatomics you mentioned, the electronic energy for the heavier diatomics will be artificially larger than the lighter ones simply because there are more electrons. For this reason, when comparisons between systems like this are made, people usually divide by the total number of electrons. So, in this case you can visualize putting in positive work to bring the protons near to each other and then adding the electrons one at a time to the system until it has no net charge. The binding energy per electron is then a measure of the average amount of stabilization the system gained for each new electron. The same thing can be visualized for the nucleus of an atom, but in that context you will read of the binding energy per nucleon.

This same concept is also quite useful for molecular clusters and other non-covalently bound systems as a measure of the strength of intermolecular interactions. In that case it is infinitely separated molecules which define the zero of energy.

This concept of energy relative to infinitely separated components of a system is a very useful one as it gives an unambiguous definition of the strength of interactions. As a fun thing to ponder, consider how to determine the strength of an internal hydrogen bond? As it happens, to the best of my knowledge, there is no general method for determining the strength of internal hydrogen bonds as there is no natural zero of energy to choose. Some people have tried using the red shift in vibrational frequency, but this too is still a bit ambiguous and hard to determine.


To quote Erik Rakovský · 30.73  · 54.96 · Comenius University in Bratislava (from this discussion on researchgate)

Shannon published simultaneously two sets of ionic radii. So called "effective ionic radii" were made consistent with Pauling's r($\mathrm{O_2^-}$)=140 pm. "Crystal radii" are using Shannon's value r($\mathrm{O_2^-}$)=126 pm.

https://dx.doi.org/10.1107%2FS0567739476001551


Finding the value of pressure equilibrium constant

The density is given.From here you will get a relation between initial concentration and volume.From $K_c$ you will get relation between volume,extent of dissociation and initial concentration.From $K_p$ you will get relation between pressure,extent of dissociation and initial concentration.After that use the formulas that Klaus Warzecha gave.I think that will be enough to solve this.


Is there an all-natural, non-volatile and low viscosity oil?

If the goal is to have the oil cover the water surface, then you might need to add a natural surfactant, such as lecithin, to cause the oil to spread, rather than form globules. You can try egg-yolk or soy beans as a source of lecithin, or you can buy refined lecithin.

As for the smoke point, or maximum cooking temperature for the oil, many common vegetable oils could do: safflower, sunflower or refined canola (rape seed) oil all have a good margin of safety.

One issue might be that if the water is at a "roiling" boil, the turbulence will mix the oil into the body of water. For that reason, you'd need to keep the temperature below 100° C and prevent local hot-spots in the heat source from causing strong convection currents.


In these three-membered ring reactions, it's an SN2 substitution reaction that happens. Hence, there is an inversion of the configuration associated.
Edit: Perhaps the mechanism is more complicated than SN2, contrary to Klein's book which states that it's SN2, this online source says that the actual mechanism is between sn1 and SN2 (leaning to SN2 side) to explain the preference of nucleophile to attack tertiary carbon though more sterically hindered.

Reference: See page-431 of Organic Chemistry David-R-Klein


Reaction between potassium permanganate, alum, and sodium hydrogen sulfite

The fact that the solution turned clear is a pretty good guarantee that this was a redox reaction: the permanganate oxidised some other species, and in turn was reduced, probably to manganese dioxide, the brown precipitate you noticed. Unfortunately, neither of the other reactants you mention are reducing agents!
Do you think it's possible that the third reagent was actually sodium hydrogen sulphite, NaHSO$_3$? If so, this would make a lot more sense: the (hydrogen)sulfite ion would have been oxidised to (hydrogen)sulfate. This reaction requires an acidic medium, and the reason the alum (potassium aluminium sulfate) was necessary was to lower the pH.
These individual steps have the following equations:

Al$^{3+}$ + 4 H$_2$O $\rightarrow$ [Al(OH)$_4$]$^-$ + 4 H$^+$
MnO$_4^-$ + 4 H$^+$ + 3 $e^-$ $\rightarrow$ MnO$_2\downarrow$ + 2 H$_2$O
HSO$_3^-$ + H$_2$O $\rightarrow$ HSO$_4^-$ + 2 H$^+$ + 2 $e^-$

(note that the last two are half-equations -- they can’t occur on their own, but I’ve written them separately for clarity) giving an overall reaction of

Al$^{3+}$  + 2 MnO$_4^-$ + 3 HSO$_3^-$ + 3 H$_2$O $\rightarrow$ [Al(OH)$_4$]$^-$ + 2 MnO$_2\downarrow$ + 3 HSO$_4^-$

The white crystalline deposit in this case would just be alum again, more or less. You have some sodium in your solution now, from the sodium hydrogensulfite, so you'd get some sodium aluminium sulfate as well, but the only anion in your solution will still be (hydrogen)sulfate, and my guess is that with aluminium this will crystallise out as the sulfate.


Yes. All the bonds are equivalent as in the 4 resonating structures, each with 1.5 S and O bonds. So the resonant hybrid has four 1.5 double bonds (sigma and pi) with -0.5 formal charge on each O. This is why, the ion has a perfect tetrahedral geometry and shape as it is very symmetric and equivalent.
Hope it helps.

The picture below on the right is what is thought to be the most accurate representation nowadays. However, the resonance structure on the left (and all of its equivalent representations) are more than likely acceptable for introductory chemistry classes.

Note that no matter which resonance structure you think is a better representation of the sulfate ion, the bonds are all equivalent and therefore sulfate ion has the ideal tetrahedral bond angle of 109.5 degrees.


Can you make a magnetic fluid that's drinkable

Read Nails for Breakfast? and see if breakfast cereal manufacturers agree with your assumption "I obviously can't use the traditional method of using iron as it is harmful"

  Most people know that breakfast cereals are fortified with iron and other essential minerals and vitamins. What they don’t
  know is that in some cases the iron is actually added in the form of metallic, elemental iron—nails for breakfast!

Like permeakra commented, only a small amount iron, whether metallic or as a compound, is safe to consume.  See iron poisoning for details.


Is there a way to search for adducts by exact mass?

I guess that would be what you are looking for:
http://www.chemcalc.org/mf_finder/mfFinder_em_new
This one gives you theoretical combinations of atoms, that fit the mass.

You could also check the sdbs database for organic compounds.
https://sdbs.db.aist.go.jp
This one gives you existing compounds but it is rounding to 1 decimal so it gives you everything with 217.1 which is not perfect.

I just used the frist calc and then checking all close results (green) at SigmaAldrich to see, which formula matches a compound they sell. There are just two from the List:
$\ce{C10H16ClNO2}$ and $\ce{C11H11N3O2}$

So I would focus on these.


3-index integral schemes for density fitting

I think in the early days one really used just the normal integral schemes Obara-Saika, McMurchie-Davidson, Rys where for one exponent just a Gaussian with exponent zero was used. Later the people examined the schemes for this special purpose and modified them for calculating 3 index coulomb integrals. A quite recent paper on this topic can be found in Gyula Samu and Mihály Kállay, J. Chem. Phys. 2017, 146, DOI: 10.1063/1.4983393.

One minor thing I want to point out is that you presented RI/DF in a slightly unusual way (notation). Using RI/DF the integrals are evaluated as

$$
(ij|kl) \approx \sum_{PQ}^{N_{aux}} (ij|P)[V^{-1}]_{PQ} (Q|kl)
$$

where

$$
V_{PQ} = \left(P|Q\right) = \int \int {{\phi_{P}({r_1}) \frac{1}{r_{12}}  \phi_{Q}({r_2})}}d{r_1}d{r_2}
$$

Of course you can rewrite it a bit with forming $V_{PQ}^{-1/2}$ to arrive at a similar expression as given by you.


Soap and rain water, why does it not rinse off?

Yes, this is a good explanation. Rain water is almost pure water, it lacks of bivalent ions such as $\ce{Ca^{2+}}$ and $\ce{Mg^{2+}}$ which helps soap to rinse of. You will get the same effect when a water softener is installed on water distribution.


What is the necessity of re-hybridisation of carbon in the transition state of SN2 reaction?

Check out this animation (for reference, this is material which accompanies the beautiful "Clayden - Organic Chemistry" book).

As you can see, there is indeed an intermediate step (the "transition state"), when the nucleophile and the leaving group are in a intermediate (and unstable, being highly energetic) state (nor attached or detached). Look closely at the geometry of the carbon and the hydrogens: that's the sp2 hybridization of the transition state carbon.

The mechanism (also shown on the page linked) involves an electron pair transfer from the nucleophile (here, bromide ion) to the carbon and eventually to the leaving group (chlorine).

The curved arrow formalism, of course, is just a representation of what goes on in the real scenario: as you can guess from the animation, the charged bromide transfers part of its electron density to the carbon (which carries a partial positive charge!). The chlorine atom withdraws the electron density as it is transferred from the bromide to the carbon, and eventually, when a carbon-bromine bond is formed (full orbital overlap between Br and C), the chlorine atom has gained a full electron pair, which allows it to leave as a chloride ion.

This is the how.

As for the why, it's a matter of energy. The reaction conditions are energetic enough to push the system towards the unstable transition state, and the equilibrium is shifted towards the products because the energetic state of the products is lower than that of the reactants (that is: the products are more stable than the reactants)


Why are 7-membered rings less likely to form than 5- and 6- membered rings?

There are two factors that need to be considered to answer your question - entropy and enthalpy.

Entropy favors the formation of smaller rings. If we consider a hydrocarbon chain, the number of possible conformations increases dramatically as we go from 3 carbons to, for example, 10 carbons.  Only a few of these conformations will be correctly arranged so that the two ends of the chain are close to each other, which is necessary if we are to form a bond. So, entropically speaking, it will be much more probable to form a 3-membered ring than a 10-membered ring. This is shown in the following graph where 3-, 4- and 5-membered ring formation is entropically most favored.

Note that a less negative $\Delta S$ favors the reaction since $\Delta G=\Delta H - T \Delta S$.

Enthalpy comes into play when we consider the strain in the transition state leading to the possible products. Here is a chart showing the relative ring strain as a function of ring size.

As expected, strain is the highest in 3- and 4-membered rings.

If we overlay (add) these entropic and enthalpic effects we would get a graph similar to the one pictured below.

In general, we find that 5-membered ring formation is usually the favored pathway.  This web article (J. Am. Chem. Soc. 1984, 106, 1051-1056.) is a nice review of this topic that may be useful for a more in-depth reading on the subject. The above web article is also the source of the above three images.


What is the 3D shape of an ethane molecule?

You need to know the terms "staggered" and "eclipsed", because that's what we call these two conformations. Now to the point: the real ethane molecule rotates and changes shape all the time, so both geometries are possible, albeit the top one is more likely, because it is the energy minimum. To make the molecule stop in the staggered conformation, you have to freeze it to nearly absolute zero, or maybe put some bulky groups in place of hydrogens.
This, BTW, is the reason why I dislike to use ethane molecule as an example when I teach space groups to kids. Better use something rigid.
So it goes.


This is only a partial answer:

I am not sure about which step you're looking for or the substrate of your reaction, but the Fenton-based chemistry I work with is extremely fast. I use it to (supposedly) form a methyl radical from DMSO which then reacts with an α-iodoester to form another radical.

Literature suggests I add brine immediately after I've completed the hydrogen peroxide addition and work it up from there. Truth be told, I usually give it a few minutes, but the reaction is certainly quite fast. The whole process takes me less than 5 minutes, and if I wasn't using DMSO and skatole (so smelly), it would be my favorite reaction of the ones I do.

I use iron(II) sulfate heptahydrate as my catalyst, and I use 0.5 equiv. relative to the iodide. It ends up being about 0.1 g per milliliter of solvent.

As for the peroxide titration, I use a decent excess, so I've never needed to titrate it. That is why I said this was a partial answer. I will say this: don't quench large amounts of hydrogen peroxide with sodium thiosulfate at room temperature. You end up forming sulfuric acid in aqueous solution. Very exothermic (like the water in the flask can start boiling in your hand).


Why can't we use fusion to make ourselves rich?

While it is possible to synthesize gold from mercury-196 (exists as 0.15% of natural mercury) the resultant gold will be radioactive and will not be gold for long!

There is actually the spallation neutron source which creates gold from liquid mercury.

Wikipedia


Yes, it's anionic sigma complex, more precisely Meisenheimer complex. They are much more stable then arenium ions (Wheland intermediates) and can be isolated much more easily.

There was actually synthesized a sigma complex which is zwitterionic combination of both Meisenheimer and Wheland complexes:

(picture created by Wikipedia User:V8rik)

The term "sigma complex" has also other meanings. Organometallic compounds with sigma bonds connecting with aromatic rings are called like that, too. There are also sigma complexes of cyclobutadiene with aluminium chloride.

Generally, molecules are called sigma complexes to stress the difference between them and pi complexes which can also be created from cyclic annulenes. First ones are bonded using sigma bonds, the other ones use pre-existing pi orbitals.