Cyclic Voltammetry - HOMO and LUMO levels

Yes, you can "convert" this way, but you're correct to be skeptical.

Let's start with interpreting the cyclic voltammetry curves themselves.

(figure from Wikipedia)

Note that the "peak" actually has two sides. When you sweep the potential with CV, the cathodic peak ($E_{pc}$) and the anodic peak ($E_{ac}$) won't exactly line up. If you have a well-behaved, reversible redox species, the two peaks will be separated by $E_{pa}-E_{pc}=\frac{56.5\text{ mV}}{n}$ for $n$ electrons.

  Do I measure to the onset of the oxidation/reduction curve or the peak

You do neither. Instead, you take $E^0 = (E_{pa}-E_{pc})/2$ - the average of the anodic and cathodic peaks. This is the standard potential for that redox reaction.

You mention that you're including ferrocene as an in-situ reference potential, so you would compare the $E^0$ potential of your redox couple to the $E^0$ of ferrocene.

  What justification do I use to convert the voltage to an energy? How can I say that a difference in meV corresponds to eV?

In principal, you're monitoring an oxidation or reduction. So you measure the oxidation potential relative to ferrocene.

The correspondence to the HOMO energy is based on an interpretation of Koopmans' theorem. In short, the oxidation potential involves removing an electron from the molecule, and the lowest ionization potential should come from the HOMO.

So there should be a correlation between the oxidation potentials from electrochemistry and the orbital energies. Some people take this as fact, but it's good that you're skeptical. I see many papers that do this "conversion" without any warning that these are different things.

You asked about units - as you guessed, the conversion between potential (V) and eV (for orbitals) occurs, because in most cases, one electron is transferred, and so the charge is known. In the general case, this is certainly a cause for concern.

  Maybe this might make sense if it was a one electron process, but how do I know that?

Heh. Generally, it's one. That's a good question in electrochemistry. There are multiple ways of determination. If you integrate the peaks, the area should correspond to the number of charges transferred. But usually, you don't oxidize or reduce the whole solution.

You can estimate the number of electrons based on the peak spacing, as indicated above $\frac{56.5 mV}{n}$.

You can change the scan rate and use the Randles-Sevcik equation

$$i_p = 268,600 \ n^{\frac{3}{2}} AD^{\frac{1}{2}} Cv^{\frac{1}{2}}$$

  Are these peaks even the HOMO/LUMO levels at all?

They are not. There should be a correlation between oxidation and reduction potentials and HOMO and LUMO energies.

But step back and think - in one case, you have electrochemistry in the electrolyte solution at an electrode surface.

Can the solvent change the observed redox potential? Yes.
Can the electrode change the observed redox potential? Yes.
Could the molecules aggregate and change the observed redox potential? Yes.
Could the oxidized or reduced species undergo reactions that change the observed redox potential? Yes.
Can the sweep rate change the observed redox potential? Yes.


We could explain it from an electrolytic point of view.

Iron has a certain tendency to dissolve in water according to the following equation:

$\ce{Fe (s) -> Fe^2+ (aq) + 2e^-}$

The release of electrons causes a small current to flow in the metal (Remember that metals conduct electricity). This turns the point where iron dissolves into an anode, and the region around this area becomes rich in electrons, a cathode. Now, we have to remember that water dissociates to small extents according to:

$\ce{H2O (l) <=> H^+ (aq) + OH^- (aq)}$

If it was an acid, the $\ce{H^+}$ would have quickly taken all the electrons liberated from the dissolution of iron earlier, but the concentration of hydrogen ions is not large enough in water, so that we get another reaction taking place at the cathode [The hydrogen ions actually 'gather up' near to the surface but cannot do anything, creating a thin protective layer of hydrogen ions around the cathode, but not strong enough to prevent further reaction]. This reaction involves oxygen and other water molecules:

$\ce{2H2O (l) + O2 (g) + 4e^- -> 4OH^- (aq)}$

Now this is the $\ce{OH^-}$ that reacts with the earlier liberated iron (or ferrous) ions:

$\ce{Fe^2+ (aq) + 2OH^- (aq)->Fe(OH)2 (aq)}$

Now more oxygen will react to oxidise the iron (II) hydroxide...

$\ce{Fe(OH)2 (aq) \xrightarrow[O] Fe(OH)3 (aq) \xrightarrow[O] Fe2O3 (s)}$

You probably can recognise the last product.

As the $\ce{Fe^2+}$ are consumed, more iron will get dissolved (Le Chatelier's Principle) and keep the whole process going.

Maybe a little picture to go along with this :)


Why is Na+ reduced instead of H+ in the electrolysis of dilute NaCl(aq) with Mercury cathode?

The mental conflict occurs because two processes could theoretically happen (electrodeposition of sodium, or electrodeposition of hydrogen), and the theoretically unfavorable one occurs - and not only occurs, but is so favorable that it was used commercially to produce NaOH as EdV noted.
So why is the theoretically favored process not favored in this instance? Hydrogen evolution is a multi-step process, and if you wave your hands a lot, you can call it overpotential and be off to other projects.
When (or if) sodium is deposited (on mercury), it can dissolve and be removed from contact with water. So this process is a slam-dunk if you push with a high enough voltage.
But the hydrogen evolution should take place at a lower voltage. Why not? If/when hydrogen is deposited, its first appearance is atomic: one H atom sticks somewhere on the mercury surface; the second step is deposition of another H atom somewhere else. The next step, for the two atoms to form a molecule, requires some mobility on the part of the hydrogen and the mercury surface. At this point, conventional wisdom suggests that the molecule could dissolve into the solution, but eventually the aqueous solution becomes saturated. Then the first molecule must meet another molecule, and another, to begin the process of forming a bubble which will eventually rise to the surface of the aqueous solution. Somewhere in there is an impediment to the formation of bubbles, and it may involve mobility of atoms or molecules, or attachment of hydrogen to mercury, but the process is difficult enough to allow the voltage to be raised high enough to allow deposition of sodium to proceed.
And then we say that mercury has a high overpotential for deposition of hydrogen.
Now suppose we do the same, but with gallium. Sodium should deposit and become dissolved as with mercury, if the voltage is high enough. (Sodium alloys with gallium; amalgamation is a chemical term that should be reserved for mercury alloys.) But the hydrogen overpotential on gallium is less than on mercury (Ref 1), and gallium and hydrogen are deposited together from a solution of $Ga^{+3}$ (Ref 2). So, in the case of liquid gallium, it seems that hydrogen would be more easily deposited than sodium (unless, perhaps you used a very high voltage, which would override the thermodynamic preference because diffusion couldn't keep up with the current demands).
Ref 1. https://pubs.rsc.org/-/content/articlelanding/1966/tf/tf9666203524/unauth#!divAbstract
Ref 2. https://link.springer.com/article/10.1007/BF01016217


It is possible due to the fact in NO oxygen has an electron configuration of a noble gas which makes it less energetic. And N shares 2 electrons and it gets close to a noble gas electron configuration.


Method for regioselective hydroalkylation of terminal alkyne

Here is a potential solution using old-school chemistry, though with several steps.

Hydroborate the alkyne and work up with $\ce{H2O2}$ to give the aldehyde.

Alkylate the aldehyde enolate with $\ce{RI}$, either directly (LDA or LiHMDS) or via enamine/silyl enol ether

Reduce aldehyde to alcohol ($\ce{NaBH4/EtOH}$)

Form triflate/mesylate and eliminate (DBU/Toluene, reflux)

If you wish to explore @orthocresol's suggestion of enol triflate coupling then coupling the triflate with the cuprate of your alkyl chain looks to be viable according to this Tet. Lett. article from McMurry and Scott1
References

John E. McMurry, William J. Scott, “A new method of olefin synthesis. Coupling of lithium dialkylcuprates with enol triflates,” Tetrahedron Letters 1980, 21(45), 4313-4316, DOI: 10.1016/S0040-4039(00)77845-2


Why do transition metals tend to form positive ions instead of negative ions?

See the outer configuration of some transition elements:

Sc:      $\mathrm{[Ar]~3d^1 4s^2}$
Cr:  $\mathrm{[Ar]~3d^5 4s^\color{red}{1}}$
Co:  $\mathrm{[Ar]~3d^7 4s^2}$
Cu:  $\mathrm{[Ar]~3d^{10} 4s^\color{red}{1}}$
Zn:  $\mathrm{[Ar]~3d^{10} 4s^2}$

If you are wondering about configuration of Cu and Cr read why this happens!

As you may notice, they can form ions by either losing or gaining electron in 4s orbital. In the chemistry of the transition elements, the 4s orbital behaves as the outermost, highest energy orbital. When these metals form ions, the 4s electrons are always lost first, leading to a positive charge on ion. Gaining any electron in 4s orbital would decrease the stability of anion thus formed. So, they form only cations (positive ions).

In general, the outer electronic configuration of transition elements is $n~\mathrm{s}^2~(n-1)\mathrm{d}^{1-10}$. Because of reasons above, the electron removes from  the $\ce{ns}$  orbital, where n = principal quantum number of atom.

See $\ce{Zn^{2+}}$ configuration:  $\mathrm{[Ar] (3d)^{10}}$ as an example. The electron exits from 4s orbital, leading to the formation of a positively charged ion.


R and T conformations in relation to Activators and Inhibitors; Km of reaction in relation to uncompetitive and non-competitive inhibitor

Allosteric inhibitors changes the conformation of enzyme enzyme for preventing the substrate binding, and allosteric activators modify the enzyme conformation so that  substrate binding can take place.

Competitive inhibitors don't change the enzyme conformation, as these inhibitors compete with the substrate to bind at enzyme's active site.

Non-competitive inhibitors change the enzyme conformation, as they inhibit enzyme activity by binding to site other than active site and prevent the substrate binding by changing the active site's structure so that substrate can not bind to enzyme.

Uncompetitive inhibitors inhibit enzyme activity by binding to enzyme substrate complex and these inhibitors change the conformation of the enzyme which reduces the affinity of the substrate for the enzyme's active site.

Km decreases in uncompetitive inhibition because inhibitor binds to the enzyme-substrate complex and it becomes incredibly difficult for the substrate to leave the enzyme. This results in increased affinity and Km represents an increase in affinity.

(Via: https://courses.lumenlearning.com/boundless-microbiology/chapter/enzymes/

https://en.wikibooks.org/wiki/Structural_Biochemistry/Enzyme/Reversible_Inhibitors)


How many valence electrons does a copper atom have?

Yes, copper only has 1 valence electron. Remember: valence electrons only include the electrons in the highest energy (n) shell.


Organic chemistry - alkanes step up reaction

I think the reagent you're talking about is ketene (R=H in the figure below).  It is a linear molecule (the central carbon is $\ce{sp}$ hybridized) and very reactive.

When mixed with an alkane and heated it dimerizes to form diketene rather than react with the alkane.

However, when irradiated with light, ketene will produce the reactive carbene "methlylene" ($\ce{CH2:}$).  When this photolysis is performed in the presence of an alkane, the methylene will insert itself into the various $\ce{C-H}$ bonds like you've drawn.  However the reaction is not synthetically useful for a number of reasons.  For example, the $\ce{C-H}$ insertion reaction is very indiscriminate and a variety of products will be formed if different $\ce{C-H}$ bonds exist in the molecule.


Adding lemon to lentils changes color from black to pink, why?

In the black lentils used in Indian cooking (Vigna mungo), the natural water-soluble dye that causes the color change has been identified as quercitin-3-O-pentoside:

which belongs to a class of compounds called flavonols. When you pop hydrogen ions on and off some of the -OH groups, you can get a pure benzenoid form (colorless), a quinonoid form (pink), and an extensive quinonoid form (yellow). The color changes
are pH dependent, and you get pale pink at pH < 7, yellow at pH > 7, and
colorless at pH 7. When you add your lemon juice to the lentil water, you're getting the quinonoid form.

Other species of black lentils probably do something similar.

The pH-dependent color change occurs because when you have $-\mathrm{OH}$ groups on the rings instead of $-\mathrm{O}^-$ groups, the way electrons are confined changes; that changes the spacing between the molecule's energy levels, and changes the colors of light the molecule can absorb. (See my article here for more about how this works).

Many plants and household materials contain molecules that can change color or otherwise indicate changes in pH (See my article here for a list of some of them). Almost any flower, fruit, or plant part that is red, blue, or purple contains a class of chemical compounds called anthocyanins that change color with pH. The color of a flower or fruit depends on which anthocyanins are present, the pH of the pigment-bearing tissues, and the presence of other pigments, like yellow flavones.


There is resonance between various Lewis structures as:

$$\ce{:\!\overset{\ominus}{C}#\overset{\oplus}{O}\!: ~<->~ :\!C=O\!:: ~<->~:\!\overset{\oplus}{C}-\overset{\ominus}{O}\!:::}$$

The term "valence" is not used much nowadays, and "oxidation state" is used but has a different meaning.

The IUPAC definition of oxidation state requires the oxygen have a -2 oxidation state except in peroxides.  All the oxidation states must add up to 0 for a neutral molecule.  Therefore the oxidation state of carbon in CO must be +2.

There is an IUPAC definition of valence:

  The maximum number of univalent atoms (originally hydrogen or chlorine atoms) that may combine with an atom of the element under consideration, or with a fragment, or for which an atom of this element can be substituted.

This imposes a unique valence upon each element, as explained in this article, in which case valence is always 4 for carbon.

There are other definitions of "valence" such as:

  The number of hydrogen atoms that can combine with an element in a binary hydride or twice the number of oxygen atoms combining with an element in its oxide or oxides

(Chemistry of the Elements by Greenwood et al.)

According to this definition, carbon would have a valence of 2 in CO.


Oxidative chlorination mechanism (sulfide to sulfonyl chloride)

As Ben Norris already pointed out, these processes are typical a bit complex. It is however conceivable that the direct oxidation of the thioether 1 is among the first steps involved.

While the oxidative cleavage of PMB (= 4-methoxybenzyl) ethers, often applied as protecting groups, is assumed to take place in the methoxy-substituted arene, here the electron transfer mostly likely proceeds from the sulfur atom, leading to the sulfur-centred radical cation 2.

In the presence of a nucleophile, this species is likely to break down en route to the radical 3.

Whether a disulfide, formed upon dimerization of 3, will take part in the further course of the reaction is beyond these musings.


Why are enynes named as -en-yne and not -yn-ene?

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

  P-31.1.1.1 The presence of one or more double or triple bonds in an otherwise saturated parent hydride (…) is denoted by changing the ending ‘ane’ of the name of a saturated parent hydride to ‘ene’ or ‘yne’. Locants as low as possible are given to multiple bonds as a set, even though this may at times give ‘yne’ endings lower locants than ‘ene’ endings. If a choice remains, preference for low locants is given to the double bonds. In names, the ending ‘ene’ always precedes ‘yne’, with elision of the final letter ‘e’ in ‘ene’. Only the lower locant for a multiple bond is cited, except when the numerical difference between the two locants is greater than one, in which case the higher locant is enclosed in parentheses.
  (…)

Since, in names, the ending ‘ene’ always precedes ‘yne’ (with elision of the final letter ‘e’ in ‘ene’), the given compound is a ‘hept-x-en-y-yne’.

Since the lowest sets of locants in ‘hept-1-en-6-yne’ and ‘hept-6-en-1-yne’ are equivalent (1, 6), preference for low locants is given to the double bond. Therefore, locant 1 is assigned to the double bond and locant 6 is assigned to the triple bond.

Thus, the correct name is hept-1-en-6-yne.


How to explain covalent bonds with Bohr's atom model?

electrons keep on revolving around the nucleus of the atom, much like the planets around the sun.

Electrons are quantum particles. Trajectories are not applicable for them. One have to use quantum wave equation to explain "movement" (note the quotes, you cant say anything about electron movement, you can only determine the probability of electron presence in some place) of electron. Planetary (Bohr) model of the atom can help to describe some effects for isolated atoms using Newtonian physics' equations with some amendments (Bohr quantum rules), but it does not describe bonding.

So for bonding you can use valence bond theory (which exactly the case with sharing electrons you described) or some more complex theories (molecular orbital (MO) theory, as widely used example) which consider quantum wave equations for describing electrons. But you surely can't mix in Bohr atom model in this case, because it is too simple to describe bonding.

Hope my explanation helps.


What is the reason behind the characteristic colours of elements?

Colour is a property of the light reflected or transmitted by any substance. Depending on the substance's characteristics (energy levels of constituent levels, bonding, intermolecular forces), there are certain available energy levels (molecular orbitals, bands etc) for the electrons. Whenever any wide spectrum energy is available to a compound (considering only the visible spectrum), then depending on the various energy gaps between the available to the electrons, different wavelengths will be selectively absorbed and re-emitted, but now in any random direction, therefore the reflected and the transmitted light will be partially deficient n these wavelengths. The absorption and re-emission is very characteristic of all substances, and hence, the wavelengths absorbed and the resultant spectrum (transmitted or reflected) will be characteristic. Depending on this spectrum, various colours can be seen for different compounds.

Sometimes, the energy gap between the electron energy level is very high and therefore the light in the visible spectrum cannot excite the electron. No wavelength will be absorbed and it will act as a transparent glass. If say, the gap corresponds to a energy gap of $E=h(4000A)$, then the blue light will be absorbed and the remaining light will be white-blue=yellow. This is the general way in which colors are produced. The energy gaps can be anything, the band gaps in metals, molecular orbital energy difference in covalent compounds, orbital energy difference in isolated elements, or simply the energy difference between the various vibrational modes in a gas/solid.


A function such that the equilibrium state of a system is determined by its maximum (or minimum) is called a characteristic function. There is a whole bunch of those. All they are ultimately derived from the Second law after some fiddling around with differentials.

The said law basically postulates that entropy is one such function. But it is only good for closed systems, that is, those with constant $U$ and $V$. Other kinds of systems use other kinds of characteristic functions. To derive them, we typically go along the lines of Legendre transformations. Say, you have $f(x,y,z,\dots)$, which is a characteristic function in the variables $x,y,z,\dots$, and you want to switch from $x$ to $f'_x$ to use it as your independent variable. Then $f-xf'_x$ is a characteristic function in your new variables.

So let's begin with switching from $U$ to another variable. The new characteristic function is
$$S-U\cdot\left({\partial S\over\partial U}\right)_V=S-{U\over\left({\partial U\over\partial S}\right)_V} = S-{U\over T}$$
(Cryptic as it might seem, the identity $\left({\partial U\over\partial S}\right)_V=T$ is true, and moreover, is the definition of temperature.)

To bring this characteristic function to a more familiar appearance, we switch the sign (so we'll be looking after a minimum, and not maximum, as we did with entropy), then multiply it by $T$, since it is only going to be used in systems with $T=const$ anyway, and get the Helmholtz energy $F=U-TS$. That's what determines the state of equilibrium for fixed $(T,V)$.

Now let's make the final step. Luckily for us, $-S$ is the thermal derivative of $F$, hence the characteristic function in variables $(S,V)$ is
$$F-T\cdot\left({\partial F\over\partial T}\right)_V=F-T\cdot(-S)=F+TS=U$$
just as we were seeking to prove.

So it goes.

One more variable change from $V$ to $P$ will give us Gibbs energy, but that's another story.


Does aldol condensation give different products in different conditions?

I believe your conception of the need for deprotonation under acidic conditions is misconceived.  Whenever you are under acidic conditions, your mechanisms must necessarily start with protonation.

The key concept here is the enol tautomerization.  The enol is more favored than for propanal as compared to ethanal because it results in more substituted alkene.  Thus there is likely a higher concentration of this nucleophile in the reaction mixture.  The mechanism of tautomerization is difficult to elucidate, but presumably starts with protonation of the carbonyl in acidic conditions.  Then, likely another carbonyl deprotonates the the recently protonated carbonyl at its alpha position, forming the enol.  This transformation has produced a nucleophile (enol -- 1-propene-1-ol ) and also a strong electrophile (a protonated carbonyl -- protonated ethanal).  The reaction occurs, leading to a protonated aldol product (proton on the carbonyl) which then is deprotonated by the next carbonyl and the process continues.

As far as the base catalyzed case, ethanal is more acidic because it does not have the extra methyl group donating electron density disfavoring formation of a negative charge.  Hence ethanal becomes the nucleophile and propanal becomes the electrophile.

In each case, these are only major products.  I don't expect the energy differences are large enough to give great selectivity.


How do scientist obtain the value of enthalpy of formation?

There is a consent to start with: the standard enthalpies of formation of elements (like $\ce{H2}$), at standard conditions ($\pu{1 bar}$, $\pu{298.15 K}$) are set to zero.  If there are several forms of one element, than the most stable polymorph is chosen as reference (except for phosphor, ref).

More generally, to determine the $\Delta{}_fH^\circ{}$ of other compounds, you may embark experiments by calorimetry. By means of a calorimeter, you are able to record the energetic changes that occur in the course of a reaction.  For example for the exhaustive combustion of carbon with oxygen:

$$\ce{C + O2 -> CO2}$$

all the heat recorded is due to the formation of carbon dioxide. Since this reaction equation was balanced that it equals only one mol of product ($\ce{CO2}$), the more general standard reaction enthalpy ($\Delta_{\text{R}}H^\circ{}$)  equals to the standard enthalpy of formation ($\Delta_fH^\circ{}$).  And by one experiment, after an other, the body of experimentally determined enthalpies grows.

Based on other experiments than combustion, one was able to determine how much energy is necessary to break a bond between atoms which is called bond dissociation energy. Conversely, this value expresses how much energy over all is released if two atoms establish a bond with each other.  In some instances, these energies are (comparatively) small, for example single bonded $\ce{Cl-Cl}$ ($\pu{242 kJ/mol}$), in other instances (comparatively) large, like triple bonded $\ce{N#N}$ ($\pu{945 kJ/mol}$).

If one knows the structure of a compound, which atoms are connected with each other and how (single, double, triple bond), than these bond dissociation energies may be used to estimate the standard enthalpy of formation of the new compound, for example by an increment system. Alternatively, by virtually setting up for example a combustion reaction of exactly one mole of the new compound into $x$ moles of $\ce{H2O}$ and $y$ moles of $\ce{CO2}$.  Then, the calculation to perform is

$$\Delta_{\text{R}}H^\circ{} = \sum_{\text{all products}}{\Delta_fH^\circ{}} - \sum_{\text{all starting materials}}{\Delta_fH^\circ{}}$$

where $\Delta_{\text{R}}H^\circ{}$ again is the overall reaction enthalpy.  Since these estimations are based on some empiric data, they do not a substitute for an experimentally determined value for your newly prepared compound. Often, increment systems leave some of inter- and intramolecular interactions unconsidered (e.g., ring strain, possible existence of polymorphism) which then yields a difference between the estimated, and the actual value of $\Delta_{\text{R}}H^\circ{}$ or $\Delta_fH^\circ{}$.


If there is no concern over the quality of the food, I'm guessing the question is more specific to whether the microwave oven is a safe piece of equipment to use.

The U.S. Food and Drug Administration says:

  Manufacturers must certify that their microwave ovens comply with strict FDA emission limits. The emission limits are well below the threshold for risk to public health.
  Most injuries related to microwave ovens are the result of serious thermal burns from hot containers, overheated foods, or exploding liquids.

  There have been extremely rare instances of radiation injury due to unusual circumstances or improper servicing.

If we're not talking about the oven itself, and rather the nature of chemicals involved with its use, then we could talk about microwaving with plastics with foods. There are sources online that raise concerns over plastics leeching dangerous chemicals into the food.

Harvard School of Medicine says:

  If you’re concerned about plastic wraps or containers in the
  microwave, transfer food to glass or ceramic containers labeled for
  use in microwave ovens.
  Don’t let plastic wrap touch food during
  microwaving because it may melt.
  Wax paper, kitchen parchment paper,
  white paper towels, or a domed container that fits over a plate or
  bowl are better alternatives.
  Most takeout containers, water bottles,
  and plastic tubs or jars made to hold margarine, yogurt, whipped
  topping, and foods such as cream cheese, mayonnaise, and mustard are
  not microwave-safe.
  Microwavable takeout dinner trays are formulated
  for one-time use only and will say so on the package.
  Old, scratched,
  or cracked containers, or those that have been microwaved many times,
  may leach out more plasticizers.
  Don’t microwave plastic storage bags
  or plastic bags from the grocery store.
  Before microwaving food, be
  sure to vent the container: leave the lid ajar, or lift the edge of
  the cover.


Skin adhesive that can be used daily without damaging skin

what is the branch of chemistry that deals with skin adhesives?

Likely a combination of biomedical engineering and organic chemistry, with some materials science in there as well.

  Is there an adhesive that is not water soluble that is safe to use every day?

Yes. Mastisol fits your criteria.


Interesting question. It is much less studied and reported on than the case of non-classical carbocations, but I did find a few papers. Brown and Occolowitz (Ref.1) reported that deuterated bicyclo[3.2.1]octa-2,6-diene 1b, below, undergoes base-catalysed de-deuteration (potassium tert-butoxide in DMSO) to give 1a much faster (ca. $3 \times 10^4$) than the bicyclo[3.2.1]octa-2-ene counterpart, i.e. 2b $\rightarrow$ 2a. They suggested this is evidence for the intermediacy of a 6π-electron, delocalised (non-classical) carbanion 3.

Much more recently, Brown reviewed and provided additional DFT computational evidence for stabilisation and charge delocalisation in these species (Ref.2).

References:

J. M. Brown, J. L. Occolowitz, “A non-classical carbanion,” Chem. Commun. (London) 1965, (16), 376-377 (https://doi.org/10.1039/C19650000376).
John M. Brown, “Origins of Stabilization and Evidence for Charge Delocalization in the Bicyclo[3.2.1]octadienyl Anion and Related Species,” Australian Journal of Chemistry 2014, 67(9), 1296-1300 (https://doi.org/10.1071/CH14244).


The electron-electron repulsion potential is an interaction potential, therefore it has one term per interaction. If you have two electrons, you will have only one term, representing the interaction between them. The number of potential terms will be, in your case:

Attraction between electron 1 and the nucleus (one interaction)
Attraction between electron 2 and the nucleus (one interaction)
Repulsion between electron 1 and electron 2 (one interaction)

In total, you will have three interactions and therefore only three potential terms (two for electron-nucleus and one for electron-electon interactions).

In general, an interaction potential that only depends on the distance of two particles can be written as
$$
U = \sum_{i<j} u(|\vec{r}_i - \vec{r}_j|),
$$
where $u$ represents the interaction between only two particles. You see that for two electrons you will have only one term, since $i<j$. For three electrons you will have three terms (interaction between 1 and 2, between 1 and 3 and between 2 and 3) and so on.

Maybe your confusion comes from the fact that, sometimes, this interaction potential is written as
$$
U = \frac{1}{2}\sum_{i\neq j} u(|\vec{r}_i - \vec{r}_j|)
$$
In this case, for two electrons, you will have two terms. However there is the factor $1/2$ which cancels out the double counting, since
$$u(|\vec{r}_i - \vec{r}_j|)=u(|\vec{r}_j - \vec{r}_i|).$$


Why is there a difference in available energy in saturated vs. unsaturated fatty acids?

The total energy stored in macromolecules is commonly quantified by the amount of $\ce{ATP}$ produced, as $\ce{ATP}$ is used throughout the body to couple reactions and make them thermodynamically favorable.

First, as the oxidation of a double bond in an fully saturated fatty acid requires a reducing agent, you here produce $\ce{FADH2}$ from $\ce{FAD+}$ which is equivalent to a $\ce{1.5 ATP}$ gain.[1]

Also, unsaturated fat has double bonds in cis positions that are unfavorable to beta oxidation.[2] An enzyme coupled reaction is needed for the double bond shift. Therefore, an enzyme coupled reaction that requires energy (in the reductant form $\ce{NADPH}$) is needed to make the molecule a trans fat with a shift in bond position in order to continue oxidizing the molecule completely. $\ce{NADPH}$, usually is defined as $\ce{2.5ATP}$ equivalents[1] and this extra energy expenditure in shifting the double bond is why less energy is ultimately produced.  Note, this $\ce{NADPH}$ is only used when unsaturated fats have an even numbered bond position.[2] This happens with polyunsaturated fats when the first bond is shifted.

For example:

18:2 linoleic acid will yield $1.5 + 1.5 + 2.5 = 5.5$ less $\ce{ATP}$ than 18:0 stearic acid.

References:

[1]: http://www.saylor.org/site/wp-content/uploads/2013/04/BIO101A-6.2.3-CellularRespiration.pdf

[2]: https://en.wikipedia.org/wiki/Beta_oxidation


We know that the number of electrons in both the bonds are same. One of the bond involves a covalent bond $\ce{Al-Cl}$ where as the other is a 3-centred bond $\ce{Al-Cl\bond{->}Al}$. Thus while providing a dative bond, the effective atomic charge of bridged chlorine decreases in comparison with that at the terminal position. Due to this difference in atomic charge, the terminal ones attract more and closer that the bridges.


Biodegradation time for biodegradable shopping bags?

It depends on the environment and on the definition of degrade.
In an anoxic landfill, a very long time before any change is noticed. Even in a wet environment, open to air and sunlight, only the cornstarch additive decomposes.
Initially, it was thought that adding cornstarch would help speed decomposition (and provide an outlet for surplus farm products), but studies such as this show that though some speed in decomposition is gained (shown by water absorption), tensile strength of the plastic is decreased by adding starch, so yet more plastic is needed to make the material sufficiently strong (and the bags still rip).
Also, though the corn-starch additive can decompose rapidly, this study shows that the polyethylene, though now fragmented, does not further decompose. This leads to run-off of microparticles into lakes and oceans.
So if your definition of biodegrade means how long does it take for the bag to shred, according to yet another study performed in a heated environment with added oxygen, about a year. At the end of that time, though, perhaps 80% of the initial mass remains, mostly polyethylene fragments. Much of this eventually is washed into oceans, is eaten, and winds up in us, as well.


Need 28% acetic acid from 98% glacial acetic acid

As @Todd Minehardt already stated, a lot may be found in the internet.

Beside common sense (incl. appropriate equipment and protective clothing) and requesting help from trained persons (teaching assistant in lab class, chemistry teacher, to mention a few) in case of doubts:  If the chemical is well known and is (commercially) of importance, where available, chemical suppliers provide both the chemical as well as safety relevant data (in material safety data sheets, MSDS).  Often these data are provided in more than one language, too.  As two examples, look-up catalogues by Sigma-Aldrich or Acros Organics.  The wiki page mentioned above even mentions a dedicated search engine.

Sigma, by the way, compiled many useful data here, where the molarity calculator is one possible answer to your third question. In your case, the final volume of the diluted solution is 3.5 times the initial volume of the concentrated volume to be diluted in water.

Fourthly, the diluted acetic acid may be stored (at room temperature) in a bottle of PP.  Charts like this provide an overview. Do not forget to check it prior to use and then regularly for cracks or leaks, as for every container containing chemicals, and to store it safely according to your local law.


Placing of lone pairs in axial/equatorial position

There is no axial or equatorial position in an octahedral structure. All places are equivalent and you can put the lone pair on position.


Why is hydroxide ion the strongest base only in water?

If you add a superbase to water in small enough quantity that it still makes sense to think of water as the solvent, all (or essentially all) of your superbase will be converted to its conjugate acid by the reaction

$$\ce{B-}+\ce{H2O} \to \ce{HB} + \ce{OH-}$$

This will have a very large equilibrium constant (by the assumption that $\ce{B-}$ is a superbase) and will also be driven heavily to the right by the high concentration of water. If you add enough of your superbase, eventually you can have appreciable amounts of $\ce{B-}$ sticking around in the solution, but at this point it no longer really makes sense to call the solution "aqueous".


Confused about writing out electron configurations

For a given configuration, the order of writing the orbitals is not
  completely fixed since only the orbital occupancies have physical
  significance. For example, the electron configuration of the titanium
  ground state can be written as either $\mathrm{[Ar] \, 4s^2 3d^2}$ or $\mathrm{[Ar] \, 3d^2 4s^2}$.
  The first notation follows the order based on the Madelung rule for
  the configurations of neutral atoms; $\mathrm{4s}$ is filled before $\mathrm{3d}$ in the
  sequence $\ce{Ar, K, Ca, Sc, Ti}$. The second notation groups all orbitals
  with the same value of $n$ together, corresponding to the
  "spectroscopic" order of orbital energies that is the reverse of the
  order in which electrons are removed from a given atom to form
  positive ions; $\mathrm{3d}$ is filled before $\mathrm{4s}$ in the sequence $\ce{Ti^4+, Ti^3+, Ti^2+, Ti+, Ti}$.

Source: Wikipedia article on electron configuration.


Thank you for all the help, I've figured out the answer:

$$\ce{NaHCO3(aq) + HCl(aq) -> NaCl(aq) + CO2(g) + H2O(l)}$$

Since we're given $\pu{1.60 g}$ of $\ce{CO2(g)}$ we can use the mass to calculate the number of moles with the equation:

$$n = \frac{m}{M} = \frac{\pu{1.60 g}}{\pu{44.01 g mol-1}} = \pu{0.036355 mol}$$

Since it's a 1 : 1 ratio between $\ce{NaHCO3(aq)}$ and $\ce{CO2(g)}$, we can directly solve for the mass of $\ce{NaHCO3(aq)}$ using the same formula:

$$n = \frac{m}{M} $$

$$m = n \times M = \pu{0.036355 mol} \times \pu{83.42 g mol-1} = \pu{3.0509116 g}$$

Round to two decimal places: $\pu{3.05 g}$


Are there notable organic compunds which look very similar but have very different properties?

One example of disastrously different effects of very similar molecules is thalidomide. There are two isomers, only differing in their stereochemistry. The (R)-isomer is responsible for the sedating effect it was initially used for. The (S)-isomer lead to horrifying birth defects in the children of woman that took that drug.

It is somewhat more complicated, as the isomers are racemized quickly in vivo, so even if you administer only the (S)-isomer, it will still be converted to the (R)-isomer in the body.


First, you have to find out the limiting reactant in the reaction, in your case it's seems that cyclohexanone is limiting reactant. Theoretically, one equivalent of cyclohexanone (27mmol) react completely with 2 equivalent of benzaldehye (54mmol),but benzaldehye is used in excess (73mmol) to gave one equivalent of product (27mmol) and two equivalent of water (54mmol).
When you multiply 27mmol of product with mol. weight (274.36g/mol) which is equal to =7408mg/1000= 7.4g

Similar you can do for water as well..
I hope this would be helpful, right now i don't have pencil with me, otherwise i will post a picture to this problem.
anyway if you didn't understand let me know..
sincerely,


Competitive 'Wittig' vs 'Peterson' C=C bond formation

Seems like the silicon atom is more oxophilic as compared to the phosphorus atom. This could in principal be explained by the higher strength of the $\ce{Si-O}$ bond, as indicated by the dissociation energies $D$ (see http://www.wiredchemist.com/chemistry/data/bond_energies_lengths.html):
$$\begin{array}{ccc}
\hline
\textbf{Bond} & \textbf{Dissociation energy / kJ mol}\mathbf{^{-1}} & \textbf{Bond length / pm} \\
\hline
\ce{Si-O} & 452 & 163 \\
\ce{P-O} & 335 & 163 \\ \hline
\end{array}$$
The $\ce{Si-O}$ bond is stronger thus given an explanation why the Peterson olefination is favoured.
It seems that the bond lengths $r_b$ are the same, though.

By the way, a nice discussion of oxophilicity was given recently in Inorg. Chem. 2016, 55 (18), 9461.


No, sample volume will not effect quantitation if the following conditions are met: 1) you are comparing samples in tubes with the same inside diameter, 2) you have enough sample volume so that the sample liquid extends at least a few diameters past the ends of the RF coil, 3) your shim is good enough so that the peaks you want to measure do not significantly overlap other peaks. Field distortion due to susceptibility discontinuity at the ends of the sample liquid can degrade shim, and people often use samples that extend far beyond the RF coil for that reason.  However, provided the above conditions are met, shim does not change peak integrals. This is because the integral of the line shape is exactly equal to the initial amplitude that the line contributes to the FID, and is independent of how it decays in time.


we assume for no particular reason that dipole moments must be behaving like vectors

Ah, but there is a reason. Consider the interaction of a molecule with the scalar potential

$$
E_{\text{int}} = \int \rho(\mathbf{r})\phi(\mathbf{r}) \, \mathrm{d}^3 \mathbf{r}
$$

where the integral over all space is turned into an expansion:

$$
\int \rho(\mathbf{r})\phi(\mathbf{r}) \, \mathrm{d}^3 \mathbf{r} = Q^{[0]}\phi^{[0]} - \sum_{n=1}^{\infty} \frac{1}{n!} \mathbf{Q}_{j_{1}j_{2}...j_{n-1}}^{[n]}\cdot\mathbf{E}_{j_{1}...j_{n-1}}^{[n-1]}.
$$

Now to identify some terms. $\phi^{[0]} = 1$, so $Q^{[0]}\phi^{[0]} = E_0$, the interaction-free or field-free energy. $\mathbf{E}^{[n]}$ is the internal electric field or electrostatic potential generated by the atoms to the $n$th total order, and the part that would be the derivative term in a Taylor expansion is the set of electric multipole moments:

$$
\mathbf{Q}_{j_{1}...j_{n-1}}^{[n]} = \int r_{j_1} r_{j_2} ... r_{j_n} \rho(\mathbf{r}) \, \mathrm{d}^3 \mathbf{r}.
$$

The first-order term is the (electric) dipole moment:

$$
\mathbf{Q}_{i}^{[1]} = \int r_{i} \rho(\mathbf{r}) \, \mathrm{d}^3 \mathbf{r} = \mu_i.
$$

The second-order term is the (electric) quadrupole moment:

$$
\mathbf{Q}_{ij}^{[2]} = \int r_{i} r_{j} \rho(\mathbf{r}) \, \mathrm{d}^3 \mathbf{r}.
$$

Convert the continuous electric dipole integral back into a finite sum by discretizing all space into atomic positions and representing the interaction of the electrostatic potential at zero strength with the atomic density as the atomic charge:

$$
\mu_i = \sum_{a}^{N_{\text{atoms}}} \sum_{i\in\{x,y,z\}} r_{ia}q_{a}.
$$

The set of atomic coordinates is nicely represented by each atom as a vector pointing from the Cartesian origin (0,0,0) to the atom $a$'s position $\mathbf{r} = \vec{r} = (r_x,r_y,r_z) = (a_x,a_y,a_z)$ or however you want to notate it, leading to

$$
\vec{\mu} = \sum_{a}^{N_{\text{atoms}}} \vec{r}_{a}q_{a},
$$

which can be seen as a vector of length 3 from the reduction $r_{ia} \rightarrow \vec{r}_{a}$. The quadrupole moment can be seen as a matrix describing the non-uniformity of the internal electric field, or its gradient.

  Dipole moment is a concept introduced by man ryt? So why are they vector quantities? Or better: How did man know that they are vector quantities? How was that deduced?

The dipole moment is a concept "introduced" by us in only two ways that come to mind. The first is that the electric field interaction is truncated to first order in the summation, which is mathematically inexact, even if higher-order moments happen to be zero. Nature includes this interaction to infinite order. Related to the first point, take a molecular sample and apply a single, uniform, finite electric field to it. The above answer deals with a system's permanent or internal moments, but one can also consider the dependence of the energy on an external electric field

$$
E(\mathbf{F}) = \sum_{n=0}^{\infty} \frac{1}{n!}E^{(n)}(\mathbf{a})\cdot(\mathbf{F}-\mathbf{a})^{n},
$$

where the electric field is $\mathbf{F} = \vec{F} = (F_x,F_y,F_z)$ and $\mathbf{a}$ is the expansion point. In practice, we always expand around $\mathbf{a} = \mathbf{0}$, so it is a Maclaurin series:

$$
E(\mathbf{F}) = \sum_{n=0}^{\infty} \frac{1}{n!}\mathbf{E}^{(n)}(\mathbf{0})\cdot\mathbf{F}^{n}.
$$

Assume that $E(\mathbf{F})$ is well-approximated by the first 3 terms of this expansion ($0 \leq n \leq 2$) and truncate it:

$$
E(\mathbf{F}) \approx E^{(0)}(\mathbf{0}) + \mathbf{E}^{(1)}(\mathbf{0})\cdot\mathbf{F} + \frac{1}{2}\mathbf{E}^{(2)}(\mathbf{0})\cdot\mathbf{F}^{2}
$$

I've been rather sloppy, but this is another way of writing the first derivation, except we are using external applied fields $\{\mathbf{F}\}$ rather than the internal fields $\{\mathbf{E}\}$. For a single applied electric field, only up to the first term $\mathbf{E}^{(1)}(\mathbf{0})$ survives, which is the induced electric dipole moment. I bring this up because assuming the applied field is strong enough that the induced dipole is stronger than the permanent dipole, the total electron distribution will most likely move uniformly or linearly with the field direction, so it makes physical sense to consider just the dipole (up to a point). (Unrelated, $-\mathbf{E}^{(2)}$ is the molecular polarizability $\alpha$.)

I would also say the above shows that we don't define the dipole moment as a vector, it appears naturally that way from mathematics. I also wouldn't call these "thought inventions".

So yes, the dipole moment is represented as a vector, the quadrupole moment is represented as a matrix, the octupole moment is represented as a 3rd-order tensor, etc.


The mechanism of demethylation of a quaternary ammonium

As you point out, the tert-butyl group wants to remain in an equatorial position, else both the tert-butyl group and the trimethylammonium group end up being axial which is disfavoured for two bulky groups.

The obvious role of tert-butoxide:

The tert-butoxide would obviously rather act as a base than a nucleophile (in-fact under ordinary circumstances we would say that tert-butoxide is a non-nucleophilic base in the same way we refer to LDA).

In order for the tert-butoxide to act as a base, and do the elimination, the conformation of the ring would first have to put both groups axial, which we have already said isn't favourable.

There is an additional reason why this is unable to happen however, since the protons that you need to abstract to get elimination are 1,3-diaxial to the tert-butyl group. This would be highly sterically disfavoured and so is unlikely to happen.

The less obvious role of tert-butoxide:

Since the tert-butoxide can't act as a base, there is the possibility of it attacking one of the methyl groups on the trimethylammonium group to give the product as drawn in an SN2 type reaction. Whilst this seems a little odd, it does have the advantage of neutralising the charges on both the tert-butoxide and the trimethylammonium group to give two neutral molecules.


Furan and thiophene are going to be even less basic than their saturated counterparts, which can act as Lewis bases but aren't particularly well known for acting as Brønsted bases (very strong acids can significantly protonate ethers and sulfides, but not weak ones). They do, however, have a free $\mathrm{sp^2}$ lone pair with which they can bind a proton, but only one, unlike the saturated analogues, and an $\mathrm{sp^2}$ lone pair is going to be even less basic than a an $\mathrm{sp^3}$ lone pair because the electrons are closer to the nucleus (i.e. $\mathrm{sp^2}$ is "smaller" and closer to the nucleus than $\mathrm{sp^3}$).

Pyrrole, however, might be expected to be basic, but it also is not, as you rightly expected. This is due to delocalization of the nitrogen lone pair to create aromaticity. The $\mathrm pK_\mathrm a$ of the conjugate acid of pyrrole is $-3.8$, which is lower than that of even sulfuric acid, suggesting pyrrole is a very weak base.

All of those heterocycles are less basic than $\ce{Cl^-}$ or $\ce{H2O}$, so either your lecturer is incorrect or has been misunderstood.


Net ionic equations for an aqueous solution

Your answer is probably correct. There are some extra details one might wish to consider. If a short yes/no answer was enough, you must not read any further.

First, let us remind ourselves that in aqueous solutions metal ions coordinate with water molecules. Octahedral complexes are most common.

$$\ce{Ni^2+(aq) + 6 H2O(l) <=> [Ni(OH2)6]^{2+} (aq)}$$

Then $\ce{NH4OH}$, better described as a hydrate $\ce{NH3.H2O}$, is added.
$$\ce{NH3 (aq) + H2O (l)<=>NH3.H2O (aq)<=>NH4+ (aq) +OH-(aq)}$$

Now either $\ce{NH3}$ or $\ce{OH-}$ steals a proton from $\ce{[Ni(OH2)6]^{2+}}$. Note that the resulting complexes are equivalent.

$$\ce{[Ni(OH2)6]^{2+}(aq)+NH3.H2O(aq) <=> [Ni(OH)(OH2)5]+(aq) + NH4+(aq) +H2O(l)}$$

$$\ce{[Ni(OH2)6]^{2+}(aq) + OH-(aq)<=> [Ni(OH)(OH2)5]+(aq) + H2O(l)}$$

If another proton is taken, we reach the required precipitate $\ce{Ni(OH)2}$. (Due to the equivalency mentioned, I will write only one option.)

$$\ce{[Ni(OH)(OH2)5]+ (aq) + OH-(aq)<=> [Ni(OH)2(OH2)4](s) + H2O (l) \tag 1}$$

Often, however, the $\ce{Ni(OH)2}$ precipitate is not visible. The whole point of this excercise was to give background to the explanation that follows. When $\ce{NH3.H2O}$ is in excess, the following processes occur. The precipitate could directly react, in which case we might end up with

$$\ce{[Ni(OH)2(OH2)4](s) + 6 NH3.H2O (aq)<=> [Ni(NH3)6](OH)2 (aq) + 10H2O(l)}$$

Generally, the preferred explanation is through Le Chatelier's principle. $\ce{[Ni(OH2)6]^{2+}}$ ions react via $(2)$

$$\begin{align}
\ce{[Ni(OH2)6]^{2+} + 4 NH3.H2O (aq)} & \ce{<=> [Ni(NH3)4(OH2)2]^{2+}(aq) + 8H2O(l)}\\
\ce{[Ni(OH2)6]^{2+} + 6 NH3.H2O(aq) } & \ce{<=> [Ni(NH3)6]^{2+}(aq) + 10 H2O(l)}\\
\ce{[Ni(NH3)6]^2+(aq) +2 OH-(aq) } & \ce{<=> [Ni(NH3)6](OH)2(aq)}
\end{align}$$

Since the $\ce{[Ni(OH2)6]^{2+}}$ ions are used up, the equilibrium of $(1)$ tilts to the left, thus the precipitate disappears.

Once more, this was simply to explain why showing $\ce{Ni(OH)2}$ as a precipitate might not correspond with reality when there is (excess) dissolved ammonia.


Helium has two electrons in total, and according to the aufbau principle, it adopts the electronic configuration $\mathrm{1s^2}$. This means it has two electrons in s orbitals with a principal quantum number of 1.

The last (and only) level of helium's electronic configuration is $\mathrm{1s^2}$, and therefore $\ce{He}$ has 2 valence electrons.


What is the role of hydration enthalpy in oxidation?

Fluorine is a stronger oxidizing agent than chlorine. So fluorine oxidizes chloride ion as per the following reaction:

$$\ce{F2 + 2Cl- -> 2F- + Cl2}$$

It can be observed that low enthalpy of dissociation of $\ce{F-F}$ bond lowers the energy of activation of the reaction mentioned above and hence favours the formation of products. Due to the small size and high electronegativity of fluoride ion, it has a higher hydration enthalpy or in other words it's easily covered by a lot of solvent molecules. This factor greatly reduces the rate of the reverse reaction. This process also shifts the equilibrium towards the products. This is how hydration enthalpy plays an important role in the oxidizing capability of fluorine.


Why hasn't chemistry brought about a greater level of human enhancement(as of yet)?

This is not a limitation related to chemistry, but related to the lack of understanding of the human body and mind.
There are a couple exceptions, such as the Cochlear implant (wich basically transforms a deaf person into a hearing cyborg), pacemakers, and some drugs that can enhance mental capabiliyties.


Products of the o-Toluidine oxidation with glucose oxidase and peroxidase

The product most likely is 2-methylnitrobenzene (3), which means that the amino group ($\ce{-NH2}$) in o-toluidine (1) is oxidized to a nitro group ($\ce{-NO2}$). The oxidation proceeds via 2-methylnitrosobenzene (2).

The formation of 2-methylnitrobenzene can be detected photometrically at $\lambda$ = 400 nm, where the aniline does not absorb.


No buffer region in the titration of strong acids or bases?

Strong acids and strong bases essentially completely dissociate. So if you're adding the strong base to the strong acid, the solution stays very acidic, then neutralizes, then goes basic with just a small amount of the strong base.

With a weak-acid at the pK value you have as much $\ce{H+}$ as $\ce{A-}$. Thus a small amount of strong base or strong acid won't change the pH value much. That is the very notion of a "buffer solution."

So yes there is a rational chemistry explanation. :-)


Mercury has a high overvoltage for hydrogen evolution; that means a barrier exists so that molecular hydrogen is not formed easily. Evolving hydrogen requires two protons to be neutralized, get together to form a molecule and clump up with a few more hydrogen molecules to form a gas bubble. If a sodium cation is neutralized right next to mercury liquid, it can quickly amalgamate/dissolve into the mercury, where its activity is much reduced. Sodium amalgam reacts with water, but much slower than metallic sodium reacts with water.
Similarly, ammonia amalgam can be made by electrolysis, and the mercury will foam up - it's not very stable.


Why are bridge bonds weaker than terminal bonds and, H-B-H terminal bond angles larger than H-B-H bridging bond angles in diborane?

It depends on what you define as a bond.  A three-center two electron bond is in general stronger than an ordinary two-center bond because delocalizing the bond over three atoms instead of two makes the bonding MO more stable.  But, the bridge bond is also shared between two linkages so each individual linkage has less bonding than a single two-center bond linkage.
The question of bond angles is a bit more complex because it hides a couple other issues.  Superficially, the bond angles in the bridge region are forcibly reduced by forming a four-membered ring.  But that begs the following questions:

Why is the dimer formed instead of a trimer or tetramer, which would enable a larger ring and more natural bond anges?

Why is the bond angle through boron actually greater than the bond angle through hydrogen?  Since hydrogen forms smaller atoms steric factors would favor the reverse.

An answer to both questions may lie in the tendency of 3c-2e bonds to form closed structures.  The bonding orbital described above gets an extra bonding linkage by bringing the end atoms of a chain together.  Such a tendency is seen in the formation of a triangular trihydrogen cation in interstellar space, and in the intermediate stage of a carbocation rearrangement.
In the case of diborane this tendency shows up as boron-boron bonding. The dimer has a favorable geometry for forming such a boron-boron bond, whereas an oligomer with larger bond angles in the center would tend to hold the boron atoms too far apart.
The nature of this boron-boron bonding is revealed in molecular-orbital calculations reported by Prof. Bruce Tattershall at Newcastle Universitywinnoplus=2&outputype=HTML5.0_JSmol&basis=MO&orbital=B(p_x)_H_bridge*&plane=xz&orbno=10).  The diagram taken from this source is shown below to reveal six occupied valence MOs and defines the $x$ and $z$ axes in the molecule for later reference.

Below are the quantitative data behind this diagram for the occupied molecular orbitals.  Entries marked with a blue dot indicate boron-boron bonding, while those with a gold dot are boron-boron antibonding.

The boron $s$ and $p_x$ interactions along the sigma-bonding axes show a mixture of bonding and antibonding overlaps; but the $p_z$ interaction, using orbitals that overlap the hydrogen atoms, shows a single bonding interaction with large boron components. This indicates that the closure of the 3c-2e bonds in the dimer occurs through the $p_z$ lobes rather than along the actual boron-boron axis, forming a "$\pi$ bond over a gap" analogous to what an organic chemist might find in a homoaromatic molecule.
Note also that the 3c-2e bond closure occurs around a hydrogen atom. A larger group in the bridging position could leave the boron atoms too far apart to overlap well and thus inhibit formation of the b9ron-boron bond in a dimer. This would explain the greater tendency of trialkylboranes to exist as monomers compared with the parent (di)borane.


Why can't you use m-CPBA to synthesise isophorone oxide from isophorone?

When using MCPBA, the undesired by-product is probably formed due to Baeyer–Villiger oxidation of the ketone to the corresponding ester (here: lactone).


Does an irreversible reaction have an equilbrium between reactants and products?

The reaction you are interested in is the precipitation of barium sulfate, a relatively insoluble salt.

The reverse of this reaction is the dissolution of barium sulfate crystals.  The dissolution kinetics of barium sulfate have been studied in a variety of systems over the decades.  Here are a few links:

Kornicker et al. 1991 reported that barium sulfate dissolution into water reached equilibrium in less than 30 minutes at 25 °C and took about 5 minutes at 60 °C.  They also presented a rate law for $\ce{BaSO4}$ dissolution of $r=k A (C_{eq} - C)^2$, where $A$ is the specific area of the barium sulfate crystals, $C_{eq}$ is the equilibrium concentration of barium sulfate, and $C$ the instantaneous concentration of dissolved barium sulfate.  This rate law had been used historically (see refs.) but these authors advocate instead for a first-order rate law.
Dove and Czank 1995 also disagreed with second-order rate law of Kornicker et al. and proposed a first-order reaction for $\ce{BaSO4}$, i.e. $r = k A (C_{eq} - C)$.

I couldn't find any references for barite solubility in sodium chloride solution, but I think it is safe to say that the solubility is finite.  Probably the kinetics of dissolution would not be more than an order of magnitude different than in pure water.

Since this dissolution is the reverse of the precipitation reaction you are interested in, and since the literature is clear that dissolution happens at non-zero rates, then both "reverse" reaction and forward reaction must be happening in your system.  That sounds like the definition of a dynamic equilibrium to me.

Note the appearance of $A$ in the equations.  The surface area of barium sulfate solid involved in the equilibrium determines the kinetics.  Usually we assume that the affect of $A$ is the same for dissolution and precipitation kinetics, so that the effects cancel out.  This is why we say that the activity of a solid phase is 1 and does not vary.  However, the effects of $A$ do not always cancel out.  The solubility of nanoparticles of $\ce{BaSO4}$ is probably higher than the solubility of bulk $\ce{SO4}$ for this reason.

Experimentally, one could probe the dynamics of the equilibrium by mixing $\ce{^131BaSO4(s)}$, i.e. radiolabeled barium sulfate, with a saturated solution of $\ce{^138BaSO4(aq)}$.  This could be done even in a sodium chloride solution.  At regular time intervals, samples of the solution would be taken, which could be filtered to remove any traces of solid particulates, and the levels of the radiotracer that had entered solution could be measured.

I don't know, quantitatively, what the outcome of the experiment would be, especially in sodium chloride solution.  But I am 100% confident that at some time scale, probably within tens of minutes, radioactivity would appear in the liquid solution.  That indicates that there is an equilibrium.

The relevant equilibrium constant in this case is the $K_{sp}$ for barium sulfate, as was alluded to in both your question and in the comments.


Predicting the pH of a weak acid and weak base solution

Note: The expected $\text{pH}$ values refer to 1:1 mixtures as suggested in the question

One major feature of weak acids and bases is that they do not dissociate completely in water, producing less free oxonium or hydroxide ions than strong acids and bases do (Brønsted). For example in Sodiumacetate, a weak base, there is an equilibrium where the reactions left to right and right to left occur at the same speed, hence cancelling each other.
\begin{aligned}\ce{
   2NaOAc_{(aq)} + H2O &<=> 2Na+_{(aq)} + {}^{-}OAc_{(aq)} + HOAc_{(aq)} + {}^{-}OH_{(aq)}} &(1)
\end{aligned}

The same also applies to weak bases like ammonium hydroxide $\ce{NH4OH}$. Note, that ammonium itself is not a Brønsted base as it lacks to donate hydroxide ions.
\begin{aligned}\ce{
   NH4OH_{(aq)} &<=> NH4+_{(aq)} + {}^{-}OH_{(aq)}} &(2)
\end{aligned}

Weak acids can be expanded in the same way, example benzoic acid:
\begin{aligned}\ce{
   Ph-COOH_{(aq)} + 2H2O&<=> Ph-COO^{-}_{(aq)} + H+3O_{(aq)}} &(3)
\end{aligned}

In the Lewis definition, weak acids are not able to complex the hydroxide ions of water's autoprotolysis.

\begin{aligned}\ce{
   2H2O &<=> H+3O_{(aq)} + {}^{-}OH_{(aq)} &\\
   BF3_{(aq)} + 3H2O &<=> [HOBF3]^{-}_{(aq)} + 2H+3O_{(aq)} + {}^{-}OH_{(aq)} &(4)
}\end{aligned}

These equations are also in equilibrium, meaning all species are present in aqueous solution.

In the Lewis definition ammonia $\ce{NH3}$ is a base and it reacts analogous to the above.

\begin{aligned}\ce{
   2NH3_{(aq)} + 2 H2O &<=> NH4OH_{(aq)} + NH4+_{(aq)} + {}^{-}OH_{(aq)}} &(5)
\end{aligned}

Now the rate of dissociation is strongly dependent on the weakest bond. In $(1)$ that is $\ce{Na-O}$ and successively $\ce{H-O}$. Overall the reaction will produce excess hydroxide ions, hence the solution will have a $\text{pH}>7~(@25~^\circ{}\!C)$.
The same applies analogously to $(2)$ and $(5)$

In $(3)$ the opposite is the case. The reaction will produce excess oxonium ions, hence having a $\text{pH}<7~(@25~^\circ{}\!C)$. The same applies to $(4)$.

So when adding a weak base to a weak acid, one has to evaluate the whole equilibrium:
\begin{aligned}\ce{
   NH4OH_{(aq)} + HOAc_{(aq)} + H2O
   <=>&\\ [(NH4)(OAC)]_{(aq)} + NH4+_{(aq)} &+ {}^{-}OH_{(aq)} +  {}^{-}OAc_{(aq)} + H+3O_{(aq)}
} &(6)
\end{aligned}

It is therefore very much dependent on the chemical properties of all compounds involved in the solution. These systems are also very well known as buffer solutions.

So a good start would be comparing $\text{p}K_a$ and $\text{p}K_b$ values of the involved components, giving you a hint whether it will be more acidic or more basic. But you also have to pay attention to the amount of substance in particular in these cases. Dilution also plays an important role in this kind of chemistry.


The addition of ammonium tartrate turns lead tartrate into a water-soluble tartrato complex:
\[\ce{Pb(tart) + tart^{2-} <=>> [Pb(tart)2]^2-}\]


What happens when 0.02 mole NaOH is added to a buffer solution?

The second solution using the Henderson-Hasselbach equation is correct. While buffer solutions change pH very little near the pka of their constituent weak acid, it would not be correct to say that they do not change at all. Also, an addition of 20 mmol of NaOH is not insignificant to a 100 mM buffer—a 20% change in concentration. The Henderson-Hasselback equation in the form you've shown is an acceptable estimation of the buffer pH in a moderately concentrated buffer with an acid/base ratio close to 1. Deviations from it in that form can occur for very dilute or concentrated solutions or if it's well away from the pka.


$\ce{Li+}$ 's low standard reduction potential value applies to aqueous solutions only.

$\ce{Li+}$ has very high polarising power due to its small size (and in turn a high hydration energy). This is why $\ce{Li+}$ has a very low reduction potential compared to other alkali metals in aqueous solutions.

When you deal with a solution of molten metals, the idea of hydration energy makes no sense anymore. Therefore, the electrochemical series order for aqueous solutions isn't applicable to the molten solution of alkali metals.

As you go down the alkali metal group, the ionisation energy decreases. The tendency of an atom to give its electron (or get oxidised) increases. Therefore, potassium oxidises more easily than lithium or in other words, potassium has a lesser tendency to get reduced compared to lithium.

As lithium has a higher tendency to get reduced than potassium, lithium gets deposited instead of potassium.


Why there is charge on the elements, in the dative bonding?

To begin with, the term "dative bond" is not used anymore. It has been replaced by "coordinate bond" :

Let's have an example of this kind of bond.

As you can see, this kind of bond occurs when an atom is giving two electrons from itself to form the bond. The typical coordinate bond is the $\ce {N}$ already bonded with $\ce {3H}$ forming a bond with a proton. Since the proton has no electron, the $\ce {N}$ atom has to fully give one of his electron from his remaining doublet to form the bond. But to respect the charge equality, you had a positive charge at the begining, no electrons were added or removed from the system, so the positive charge should remains on the final molecular edifice.

The question is, where is located this positive charge ? One could start thinking this charge is located on the $\ce {N}$ atom since it is giving an electron. But actually, the $\ce {N}$ atom has a greater electronegativity than the $\ce {H}$ atom, that's say its hability to attract electrons is greater. Therefore the positive charge remaining should be equaly distributed over all the $\ce {4H}$ since they are equivalent.

But, altogether, the net charge of the edifice will be $+1$.


Why is the F-F bond in fluorine a p-p sigma bond?

Hybridisation is a mathematical concept chemists use to better understand spatial distribution of bonded atoms. It should only be used if we cannot explain the structure will without resorting to hybridisation.

The energy of a 2s orbital is lower than that of a 2p orbital in atoms with a fully populated 1s orbital. Therefore, there is a desire to keep its energy low. A bond between two p orbitals (of different atoms) will also lower the energy of one of these p orbitals (and create an antibonding σ* unpopulated one, too).

Furthermore, p orbitals extend into space in a directed manner: Along their coordinate axis, while s-orbitals are spherical in shape. The p orbital is already in a nice direction for bonding; any mixing with s orbitals will make it extend into space slightly less — potentially unfavourable for orbital overlap.

Finally, the lone pairs on each fluorine atom will be a set of two p type orbitals, each carrying a lone pair perpendicular to the bond axis, and an s type orbital. Experimentally, one should be able to detect two different energy levels for the two different lone pairs. (However, I didn’t go looking for papers and I can’t remember any lecturer mentioning any off the top of my head.)

So the $\ce{F-F}$ bond is indeed a $\mathrm{p^\unicode[Times]{x3c3}\ce{-}p^\unicode[Times]{x3c3}}$ bond because it makes the overall system energetically favourable.


Heat transfer dependence on temperature gradient?

Yes, it is very much dependent on temperature gradient. The stefan-boltzman law is only for radiative dissipation of heat and not for general heat transfer by conduction(contact).
The equation for steady state heat flow is
$$\vec h=-k \vec{\nabla T}$$ where $\vec \nabla$ is the gradient operator which is based on the temperature difference between the bodies and $k$ is the thermal conductivity. $\vec h$ is the heat per unit area.

If the system is not in steady state, then the heat diffusion equation becomes relavent:-
$$\frac{dT}{dt}=\frac{k}{c}\nabla^2 T$$ where $c$ is the specific heat of the body.
Lastly, a to a crude approximation, the rate of cooling in atmosphere at low difference of temperature of atmosphere and the body is given by Newton's law of cooling (partly by radiation):-
$$\frac{dT}{dt}=-K(T_{body}-T_{surrounding})$$ where $K$ is a constant.

In all these equation, it is the temperature gradient between the bodies in contact, which drives the rate of heat transfer (or temperature change).


You argument is largely based on enthalpy, whereas freezing point depression and boiling point elevation is (in the situation when the solute only weakly interacts with the solvent) mainly an entropic phenomenon.

Because of the increased entropy by adding a solute (i.e. there are more configurations by having two types of molecules instead of one), the solution will liquify at a lower temperature as compared to the pure solvent and with a similar reasoning, the boiling point will increase with respect to the pure solvent.


How to explain the products of the nucleophilic aromatic substitution of 1-chloro-3-methylbenzene with NaOH?

What is going on here is a benzyne process. Under the conditions of intense heat the hydroxide removes one of the protons either side of the Cl. Cl- then leaves to create a triple bond, giving a benzyne. As you might imagine this is a highly reactive species and reacts with the OH- with no regioselectivity; either end can get hydroxylated and you have two possible benzyne intermediates. Thus you get the mixture of products.


Fluoride concentration in LiF and CdF2 solutions

The problem as stated is not asking you which solution has the higher amount of fluoride but which one has the higher concentration of fluoride, i.e. the higher molarity (as this is usually the concentration unit more often used by chemists).
You need to calculate the fluoride concentration for the $\ce{LiF}$ solution in the same way you did for the $\ce{CdF2}$ solution. My answer comes to 0.04290 M.
Thus, the $\ce{CdF2}$ solution has a fluoride concentration of 0.2344 M while the $\ce{LiF}$ solution has a fluoride concentration of 0.04290 M. Therefore, the answer to the problem is: the $\ce{CdF2}$ has a higher fluoride ion concentration.


What is the chemical formula for the synthesis of Hydroxyapatite (Ca10(PO4)6(OH)2)?

First, let me state the standard formula of hydroxyapaptite:

The term "apatite" applies to a group of compounds  with a general
formula in the form $\ce{M10(XO4)6Z2}$, where $\ce{M^2+}$ is a metal
and species $\ce{XO4^3-}$ and $\ce{Z-}$ are anions. The particular
name of each apatite depends on the elements or radicals M, X and Z.
In these terms, hydroxyapatite (HAp) has the molecular structure of
apatite, where M is calcium ($\ce{Ca^2+}$), X is phosphorus
($\ce{P^5+}$) and Z is the hydroxyl radical ($\ce{OH-}$). This is
known as stoichiometric hydroxyapatite and its atomic ratio Ca/P is
${1.67}$. Its chemical formula is $\ce{Ca10(PO4)6(OH)2}$, with $\ce{39 {%}}$ by
weight of $\ce{Ca}$, $\ce{18.5 {%} P}$ and $\ce{3.38 {%}}$ of $\ce{OH}$.

The anion radical can be replaced with fluorine/chlorine to become fluoroapatite/chloroapatite. This formula can also be written as $\ce{3[Ca3(PO4)2]· Ca(OH)2}$ which indicates three moles of calcium phosphate and one mole of calcium hydroxide which is basically the same as the Sigma-Aldrich's formula. The confusion arises since they have omitted the brackets, you didn't considered calcium phosphate as single entity for which you had a wrong formula. Brackets are generally not given but you have consider them as a single entity or you can even use brackets conveniently for your purpose. But do note that this formula doesn't prove that it is a basic salt but is actually a product of polymerisation:

According to Bassett, the solid existing in stable equilibrium at $\pu{25 °C}$., with solutions of a range from faintly acid to nearly pure
lime-water, is oxy- or hydroxyapatite, but, although its composition
may be indicated by the formula $\ce{3Ca3(PO4)2.Ca(OH)2}$, it is not to be
regarded as a basic salt, but rather as the salt of a polymeride of
phosphoric acid, $\ce{H11P3O13}$ (or $\ce{3H3PO4.H2O}$), with one hydrogen atom
un-neutralised (like orthophosphoric acid or pyrophosphoric acid). The slowness of
formation of hydroxyapatite favours the view that it is produced by
polymerisation.

Reference

Hydroxyapatite-Based Materials: Synthesis and Characterization
By Eric M. Rivera-Muñoz, 2011, DOI: 10.5772/19123
http://calcium.atomistry.com/basic_calcium_phosphates.html


Identical compounds have the same chemical and physical properties (and other properties too).

Of the two types of stereoisomers, diastereomers (non-mirror image stereoisomers) are clearly nonidentical. Consider the isomers of tartaric acid. Meso-tartaric acid and (+)-tartaric acid have different melting points (165 $^\circ$C and 171 $^\circ$C, respectively). They also have different acidities.

The (+) and (-) isomers (enantiomers) are more challenging, but they are distinguishable by one physical property - the way they rotate plane-polarized light. If the two molecules were identical, they would not be isomers - we would not be able to distinguish them. If they are distinguishable, they are not identical.


Concept of Adjusted Retention Time in Chromatography

To find the time that the sample spends in stationary phase (adjusted retention time), we just need to subtract the time, in which it is in the column but not in stationary phase.

This is a misconception (a subtle point here). The adjusted retention time is the time an analyte spends in the column not the stationary phase. Remember that a chromatography column is made up of a stationary phase plus the mobile phase. The sample continuously partitions between the mobile phase and the stationary phase.

Dead time simply refers to the time it takes for a sample to travel from the injector to the detector without having any interaction with the stationary phase. Imagine a garden hose, when you open the tap, it take a while for the water to reach the other end. Dead time is exactly like that.

Calculating the fraction of the time spent by an analyte in the stationary phase is slightly more involved. Check this link on "Migration rate of solutes in a chromatographic separation"

https://web.njit.edu/~kebbekus/analysis/chromatography%20overheads.htm


Why is tin-112 stable, but indium-112 radioactive?

For isobars (i.e. nuclides with a constant mass number $A$), the binding energy $E$ as a function of the atomic number $Z$ describes a parabola, the so-called valley of β-stability (see also Weizsäcker’s formula or semi-empirical mass formula).

$$E = a \cdot Z^2 + b \cdot Z + c \pm d/A^{3/4}$$

The most stable nuclides lie at the bottom of the valley (but not necessarily exactly at the minimum of the parabola). Isobars with lower atomic numbers $Z$ are unstable to $\beta^-$ decay. Isobars with higher atomic numbers $Z$ are unstable to $\beta^+$ decay or electron capture. Therefore, at least one of two adjacent isobars must be radioactive (see Mattauch isobar rule).

Thus, since Sn-112 is stable, we expect In-112 to be radioactive.

The last term $(\pm d/A^{3/4})$ gives rise to three different parabola depending on whether the nuclides are even-$A$ (with even $Z$ and even $N$), even-$A$ (with odd $Z$ and odd $N$), or odd-$A$ (with even $Z$ and odd $N$, or with odd $Z$ and even $N$).
In the last case, in which the mass number $A$ is odd, the result is a single parabola with one stable nuclide at the bottom.
However, for even mass numbers $A$, we find two parabolas; the odd-odd-curve always lies above the even-even curve. Hence, it is possible that two or even three stable isobars exist.

The case with two parabolas also applies to your example. The stable nuclide Sn-112 has 50 protons and 62 neutrons (even-even). The radioactive nuclide In-112 has 49 protons and 63 neutrons (odd-odd). The next neighbour, Cd-112 has 48 protons and 64 neutrons (even-even) and is stable.


Stabilisation of carbocations through hyperconjugation

Hyperconjugation

Hyperconjugation, as you point out, is able to stabilise carbocation intermediates, IUPAC define this as:

  the interaction is ... between σ-bonds and an unfilled or partially filled π- or p-orbital

  Ref: IUPAC Gold Book

In a simple sense, we can visualise this interaction by drawing the carbocation and considering the adjacent σ bonds able to interact.

This can also be visualised in a more 'quantitative' way by considering a simple mixing of orbitals, in which we can see that the σC-H -> p+ lowers the energy relative to the vacant p-orbital.

Ref: Molecular orbitals and organic chemical reactions, Fleming

Stability of carbocations

If we now consider the various types of cation, we should be able to rationalise the order of stability.

The methyl group is completely unstabilised through hyperconjugation, due to the stereoelectronic requirement that the σ bond be located on the adjacant carbon (of which the methyl cation has none).

Primary carbocations have a maximum of 3 σ-bonds capable of hyperconjugation, secondary a maximum of 6, tertiary a maximum of 9– hence why the tertiary is the most stable cation out of the simply alkyl groups (the allyl and benzyl systems in the diagram above are stabilised through delocalisation of the charge onto the pi system).

Going back to your question of ethyl vs propyl carbocations, you can hopefully now see that they have the same number of σ-bonds in the correct position to participate in hyperconjugation.

An additional complication

Primary carbocations are sufficiently unstable that if they form, they tend to rearrange into more stable carbocations, such that the primary propyl cation will quickly undergo a 1,2-hydride shift to give the secondary propyl cation. A more complex example is shown below:


Is there a pH range for "neutral" detergents?

Neutral detergents are not neccesarily at pH 7, and I know of no legal restriction (at least in the EU) on labelling detergents as neutral based on pH. For example, this patent claims protection for a set of neutral detergent formulae with pH of "from about 5.5 to about 8.5".

An example of a commercial product is Proctor and Gamble's "Tide", which has a pH of 8.1-8.5 according to their safety data. This is advertised as "100%" neutral !


Why does a Fluoride ion only have a -1 charge and not a -2 charge or more?

The octet rule is usually used to explain the charges that atoms in groups I, II, VI, and VII have in ionic compounds, by saying something like "noble gases are stable, therefore having 8 valence electrons must be stable for some reason." This is true, but understanding why noble gas configurations are stable is important, and without knowing that, it does seem like magic that some atoms don't adopt a noble gas configuration in ionic compounds.

The octet rule is a useful way to quickly predict ionic charges and write Lewis structures, but it's not a magical law that atoms follow. They don't "want" or "try" to obtain specific electronic structures, although we often talk like they do. What is really important is the total energy of the electron configuration. The electron configuration with the lowest energy is the most stable one.

  "Stability" really means "lowest energy"

The total energy of the electron configuration is governed by two things:

The energy of the electron orbitals
The effective nuclear charge that is "seen" by the electrons

The orbital energies determine the order in which atomic orbitals will be populated by electrons in the ground state - in other words, which orbitals the electrons will exist in for a given atom. We know from empirical evidence and from quantum mechanical calculations that the order goes like this for most elements (not for some transition metals and heavier elements):

Electrons will occupy these orbitals from the "ground up" (called the Aufbau principle). We know, therefore, that for 4th-period elements, the $3d$ sub-shell will fill before the $4p$ sub-shell will.

The atomic orbital diagram indirectly tells us something else. The principle quantum number, $n$ (which for the valence shell is equal to the row or period number), roughly corresponds to the distance from the nucleus of the orbital's highest electron density region. In other words, any sub-shell with $n=4$ will have most of its electron density further away from the nucleus than a sub-shell with $n=3$. This means that $3d$ electrons are closer to the nucleus than $4s$ electrons, even though $3d$ electrons initially have  higher energy.

  The distance of the electron from the nucleus is roughly proportional to $n$

Since the electrons have a negative charge, and the nucleus is positively charged, there is an electrostatic potential energy between the electrons and the nucleus. If the electrons were uniformly distributed throughout the atom, we would expect that the nuclear charge "felt" by each electron would be equal to the number of protons - the atomic number. However, because of the "rules" of quantum mechanics (the allowed solutions to Schrodinger's equation), electrons are not uniformly distributed - they occupy orbitals in a specific arrangement. Since some electrons are closer to the nucleus (lower $n$) than others, this means that the nuclear charge is partially screened by the inner electrons, and the outer electrons experience an effective nuclear charge that is somewhat less than the full nuclear charge. This diagram from wikipedia demonstrates the idea:

And this one shows the idea using the Bohr model (note that this is a schematic - not an accurate picture of what electrons "look" like in orbitals - only the relative distances from the nucleus is right)

When the valence electrons experience a large effective nuclear charge, they are harder to remove (the energy cost for removing them increases), and when the effective nuclear charge is lower, they are easier to remove. The actual effective nuclear charge is very difficult to accurately calculate, but it is very easy to estimate - all you need to do is take the number of protons, then subtract the number of "screening" electrons. Basically, we are assuming that each electron between the valence shell and the nucleus cancels out one proton:

$$
Z_{eff} = Z - S
$$

where $Z_{eff}$ is the effective nuclear charge, $Z$ is the actual nuclear charge, and $S$ is the screening constant, here assumed to equal the number of "core" electrons.

To find $S$, all you have to do is count the number of electrons with $n$ less than the valence shell (highest $n$).

Taking fluorine as an example, the electron configuration is:

$$
1s^22s^22p^5
$$

We see that $Z=9$ and $S=2$, giving an effective nuclear charge of +7. Compare that to lithium: $3-2=+1$. The valence electrons on fluorine are roughly 7 times harder to remove than the valence electron on lithium. In fact, any element to the left of fluorine has a lower effective nuclear charge for the valence shell, and will give up electrons more easily than fluorine will. In other words, it's not so much that fluorine won't give up electrons, it's that nothing can take them from it.

What if we add an electron? For $\ce{F-}$, the electron configuration becomes:

$$
1s^22s^22p^6
$$

$Z_{eff}$ is still +7, since the number of core screening electrons hasn't changed. This means that it is energetically favorable for fluorine to pick up an electron, since the additional electron would still "see" a large positive nuclear charge.

What if we add another, to make $\ce{F^{2-}}$? Now the configuration looks like this:

$$
1s^22s^22p^63s^1
$$

$Z_{eff}$ is now $9 - 10 = -1$ - note that the number of screening electrons jumped to 10 because the "new" valence shell has $n=3$. In other words, because of the allowed orbital configurations the new electron would "see" a negative charge on the nucleus - it would require energy to keep it.

If you work through this process for all of the "fixed charge" elements, you find similar things - the effective nuclear charge is large for elements that gain electrons, until they gain too many (an octet), at which point it becomes negative. For fixed-charge metals, something else happens. Let's look at magnesium.

$\ce{Mg}$: $12-10=+2$

$\ce{Mg^1+}$: $12-10=+2$

$\ce{Mg^2+}$: $12-2=+10$

Once you remove two electrons, the effective nuclear charge jumps to +10 for the remaining valence electrons. The same is true for all the group I and II metals - once you get to the noble gas "core" - the octect - the effective nuclear charge seen by the valence electrons jumps way up.

This explains why "fixed charge" elements always have the same charge in ionic compounds, but what about transition metals?

Let's look at chromium. The electron configuration looks like this:

$$
1s^22s^22p^63s^23p^64s^23d^4
$$

Which electrons are the valence electrons? The ones with the highest $n$ - the 4s electrons. The effective nuclear charge they "see" is $24 - 22 = +2$ - note that the 4 electrons in the $3d$ sub-shell are counted as screening, since they have $n=3$ which is less than 4, putting them closer to the nucleus than the valence shell, despite their initially higher energy.

If we remove an electron, we get:

$$
1s^22s^22p^63s^23p^64s^13d^4
$$
(This isn't exactly correct, there would really be no $4s$ electron and 5 $3d$ electrons, but the reason is complicated and it doesn't matter to this explanation)

Note that we removed a valence electron - a $4s$ electron. Now $Z_{eff} = 24 - 22 = +2$ - it hasn't changed. We can remove one more to make $\ce{Cr^{2+}}$ and get:

$$
1s^22s^22p^63s^23p^63d^4
$$

Now the definition of "valence" becomes a little less clear - we know that the highest $n$ value corresponds to the valence shell, but what about $l$, the orbital shape? It turns out that d orbitals are larger than s and p orbitals, and so the s and p electrons end up screening the nucleus from the d electrons. This means that d electrons would be removed next, and the effective nuclear charge "seen" by them is $24 - 18 = +6$. This is a larger charge, but not so large that other elements (notably oxygen and the halogens) can't still pull the d electrons away. This gives us a range of chromium charges (oxidation states) from +1 to +6. For $Cr^{6+}$, the electron configuration would be:

$$
1s^22s^22p^63s^23p^6
$$

Now the effective nuclear charge is $24 - 10 =  +14$ - much too strong for anything else to "win" against.

As you can see, the rules of quantum mechanics determine the relative size of atomic orbitals and the order in which they are filled, which in turn determines the effective nuclear charge that valence electrons "see" for any given neutral atom or ion. The combination of the two leads to constraints on the charges that atomic ions can have, with the practical result being that group I, II, VI, and VII elements adopt a "filled octet" or "noble gas" configuration, while the other groups have more flexibility.


Why is molybdenum that important in inorganic synthesis as well as biosynthesis of ammonia?

Why Molybdenum is Used in the Manufacture of Ammonia?
Molybdenum tubing has particularly good resistance to corrosion by
mineral acids, provided oxidizing agents are not present. It is also
resistant to many liquid metals and to most molten glasses. Molybdenum
is relatively inert in hydrogen, ammonia, and nitrogen up to about
1100 C, but a superficial nitride case may be formed in ammonia or
nitrogen.(Ref. 1)

Also, molybdenum has a special capability to form stable complexes with nitrogen atoms and hence molybdenum goes very well in biological reactions as well(See the link @user79161 has posted in the comment section). Recently, a group of scientists has developed a method for synthesis of ammonia using a dinitrogen-bridged dimolybdenum complex as catalyst and under very mild conditions of room temperature and at atmospheric pressure. However, the reaction mechanism of this new method is still unclear.

Reference

https://samaterials.podbean.com/e/why-molybdenum-is-used-in-the-manufacture-of-ammonia/
https://www.u-tokyo.ac.jp/focus/en/articles/a_00259.html (Picture)


How can a sample that is in two different states be at a uniform temperature?

For a system in equilibrium: Think of it as a closed system, with a certain amount of energy shared between the solid and liquid components (discounting the container).  At any instant, some of the faster-moving molecules of the liquid could hit the ice, melting a bit, and thereby losing some of their own energy and freezing.

It takes ~80 cal/g to change ice at 0 °C to water at 0 °C. This enthalpy of fusion, aka latent heat of fusion , "absorbs" any additional heat applied until all the ice has melted. So a gram of water at 0 °C has ~80 calories more heat than an equal mass of ice at 0 °C. If you have a 1 gram chunk of ice and apply 40 calories as heat, then about half will melt.

BTW, pure water does not necessarily freeze at 0 °C... it may supercool, which will change the calculations a bit. See this video or this one for a supercool experiment!

As @MaxW states, in real life most systems are not in equilibrium (they need to be shaken or stirred, despite Ian Fleming's direction). To view a clearly non-equilibrium demo, where there is water in three states at once, see Red Hot Nickel Ball on Ice.


What does this lambda–sigma notation mean?

The notation refers to the valency (number of valence electrons involved in bonds, i.e., either 3 or 5) and coordination number (number of substituents attached) in organophosphorus compounds. For example, it could refer to the tautomeric forms of phosphonate esters:1

So the λ3σ3 refers to a P with 3 bonding valence electrons (the lone pair doesn't count) and 3 substituents attached while λ5σ4 refers to a P with all five of its valence electrons involved in bonds and 4 substituents attached.

Reference

Kraszewski, A.; Stawinski, J. H-Phosphonates: Versatile synthetic precursors to biologically active phosphorus compounds. Pure Appl. Chem. 2009, 79 (12), 2217–2227. DOI: 10.1351/pac200779122217.


Why does charcoal absorb smells from the air, and where do those smells 'go' once absorbed?

It is a form of carbon that has very large surface. The odors and impurities are held to the surface with Van der Waals forces. The capability of removing notable amounts of these impurities or odors is thanks to the extraordinarily large surface.


What is the correct net ionic equation for the neutralization of phosphoric acid?

Phosphoric acid dissociates into hydrogen ions and $\ce{H2PO4^-}$ ions in water.  This then can dissociate further, but $\ce{H^+}$ and $\ce{H2PO4^-}$ are the primary ionic species you'd find in a phosphoric acid solution.

$\ce{NaOH}$ dissociates completely into $\ce{Na^+}$ and $\ce{OH^-}$ ions.

The reaction would then be between $\ce{H^+}$ and $\ce{OH^-}$, your option 2.

The difficulty arises because adding more $\ce{NaOH}$ to the reaction pulls off the second then the third hydrogen from the phosphoric acid.  In all cases it is an $\ce{H^+}$ reacting with an $\ce{OH^-}$ ion with the phosphate ions as spectators, but that really doesn't tell the story of the reaction well.  This is a case where the net ionic equation really isn't your best option.emphasized text


Maximum number of coplanar atoms in difluorotrimethylphosphorane

Seven is correct (the OP found this in the comments).  The methyl carbons are in the equatorial plane and each one can rotate a hydrogen atom into this plane.  In fact the hydrogen atoms fit well with this conformation so it's actually favorable.


Stoichiometric defects in silver bromide

Fundamentally, although ionic radii may offer some guidance as to defect formation, the bottom line is that the silver halides have complex bonding energetics that have to be taken into consideration.  I might suggest 'Defect structures in the silver halides', D.J. Wilson et al., Phys. Rev. B77, 064115 (2008) (presuming you have APS journal access). There the authors present DFT calculations on the formation energies of both Frenkel and Schottky defects in AgCl and AgBr.


The rate of change in pH v.s Temperature

You are dealing with temperature variation of carbonic fizzy drinks. So, the situation is somewhat tricky.

Lets approach the problem step by step.

If you raise temperature of neutral water, the rate of decomposition of water will increase. So, the water will contain more protons at higher temperature than at lower temperature, although the neutrality is maintained. Therefore, the pH of neutral water will decrease with temperature. pH of water at 100°C is ~6.126.
Carbonic fizzy drinks contain sodium & potassium bicarbonate salts, which ionize completely in aqueous solution to yield alkali cations & bicarbonate anions. The initial pH is between 3 and 4. (1)
Bicarbonate anion can capture a proton from the system to yield carbonic acid. Carbonic acid is unstable and thus decomposes even at room temperature to produce carbon-dioxide, & thus the bubble! So, even at room temperature gradually the solution turns less acidic, then neutral & finally alkaline. So, here you are not increasing temperature & just letting time to flow, & the pH rises gradually. Therefore, I can expect 'linear relationship between change in pH of carbonic fizzy drinks over time'.
When you raise temperature of carbonic fizzy drinks, the pH increase gets faster because the decomposition of carbonic acid is very much temperature sensitive. It is worthwhile to mention that the rapid pH rise is somewhat countered by the pH lowering due to water's decomposition, but that effect is much less significant as water's decomposition is much less sensitive to temperature. In this case, I can certainly expect deviation from linear behavior (nonlinear or exponential increase) aided by temperature influenced decomposition of carbonic acid.

You may be able to arrive at a mathematical relationship by combining the temperature dependency of carbonic acid & water decomposition.


What will happen when a metal reacts with an acid which is neither dilute nor concentrated?

This question is a bit vague, but briefly, the reason some acids behave differently at high and low concentration is because changing the activities of components in the solution changes the redox potential of the various possible reactions.

Let's take the reaction of iron with sulfuric acid as an example. This is not necessarily the only factor and I'll make a few assumptions that are likely quite inaccurate to simplify the problem, but it will illustrate the situation:

Let's assume that the half-reaction with iron is always the same—iron is being oxidized:

$\ce{Fe_{(s)}<=> Fe^2+_{(aq)} +2e-}\ E^\circ=+0.44\ \mathrm{V}$

Now, if you look at a table of standard reduction potentials, you'll see that there are a couple of potential half-reactions that could be reduced to drive the iron oxidation:

$\ce{2H+_{(aq)} +2e- <=> H2_{(g)}}\ E^\circ=0\ \mathrm{V}$

$\ce{SO4^{2-}_{(aq)} +4H+_{(aq)} + 2e- <=> SO2_{(g)} +2H2O_{(l)}}\ E^\circ=+0.17\ \mathrm{V}$

If all these reagents were at unity activity, then the second reaction would be more favoured and we would see mostly $\ce{SO2}$, but the activities of the substances in solution are also important. By the Nernst equation, $E=E^\circ -\frac{RT}{zF}\ln Q$, we see that the potential of a complete cell or a half-cell reaction is dependant on the reaction quotient $Q$. Because we're assuming that only the one reaction happens to the iron, we can just look at the reduction half-cell. If we assume that the sulfuric acid dissociates completely to $\ce{SO4^2-}$, then $[\ce{H+}]\approx 2\cdot[\ce{SO4^2-}]$. If we assume the gaseous products of the reaction don't affect the equilibrium, for our two reactions, $Q$ is $\frac{1}{[\ce{H+}]^2}=\frac{1}{(2[\ce{H2SO4}])^2}$ and $\frac{1}{[\ce{H+}]^4[\ce{SO4^{2-}}]}=\frac{1}{(2\cdot[\ce{H2SO4}])^4[\ce{H2SO4}]}=\frac{1}{16\cdot[\ce{H2SO4}]^5}$, respectively.

Now, if we substitute these into the Nernst equation for different concentrations of acid, we can see which reaction is more favourable: In dilute acid (1 mM), we get −160 mV for hydrogen formation and −238 mV for sulfur dioxide. The picture changes as the acid gets more concentrated. In 1 M acid, the hydrogen formation half-cell has a potential of 18 mV, but sulfur dioxide is now up to 206 mV.

So you can see how the concentration of acid changes which reaction is more favourable. To answer your question about what happens in the middle, one need only do the calculation for a moderate concentration and find that at some point, the two half-reactions become close in potential and we may see a significant mixture of both reactions occurring.

This is a very simplified picture and the Nernst equation is only technically valid under conditions of no current flow, meaning that other factors may influence what is observed experimentally, but I hope it's apparent that there's not really a solid line between dilute and concentrated. Many different reactions can occur simultaneously, even if a single reaction is dominant under certain conditions.


I'd rather not put it in such terms. Dissociation of water is determined by the constant $\ce{K_w=[H+][OH-]=10^{-14}}$, which you can't change simply by adding acid (or anything else, for that matter). Acids dissociate easily into ions, just like you said, therefore adding acid to pure water (which is a poor conductor by itself) makes the latter more conductive and thus more prone to electrolysis.

Electrolytic refining of copper is a whole different story. Indeed, once dissolved in water, $\ce{CuSO4}$ would dissociate into ions completely, no matter whether or not you acidify it. The role of acid is just to prevent hydrolysis.

In general, as you change pH, some of your ions may undergo certain transformations (like $\ce{CO3^2- -> HCO3- -> CO2\uparrow}$, or $\ce{Fe^2+ -> Fe(OH)2\downarrow}$), which in turn might or might not affect the reactions on one or both electrodes. As it happens, being too general prevents us from drawing more specific conclusions.


How to draw protonated benzene (horseshoe) carbocation structure with ChemDraw?

Draw cyclohexane. Open Drawing Elements. Select 270o circle. Orient with lasso and adjust size. Insert in cyclohexane. Add H's and +.
ADDENDUM: For a centered structure. Draw cyclohexane, arc and + inside each other. Select them. Go to Object>align. Select L/R centered then T/B centered. Add hydrogens. See also Comments. PS: Clearly, ChemDraw does not recognize the structure as protonated benzene. Can't have everything ;).


Reason behind Opaque translucent transparent

Opacity or transparence is a matter of light reflection.

The way light is reflected or absorbed depends on severals parameters, first the composition of your material. If it contains molecules or crystals absorbing the visible light, you'll see it black, or colored with the color corresponding to the wavelenght the less absorbed.

There is also crystallographic considerations to take care of. If your material structure is well ordered, dimension of your latice will be set so to reflect the light, then you'll see it transluscent or opaque. For something without ordered structure, you'll probably see it transparent if the visible light isn't absorbed. We'll speak about "amorphous" material then like glass for instance.

edit : if you feel easy with those concepts, I can go further in the explainations, but since you asked for a really vulgarized answer I haven't developed more.


Why is chloroethene said to be a conjugated system?

Conjugation can be between pi-bond and another pi-bond, lone pair of electrons and a pi-bond(as in your case) and pi-bond and a vacant orbital(p or d).