Which is more basic, hydrazine or ammonia?

Obligatory link to Evans' pKa table.

Ammonia is more basic than hydrazine, by about one order of magnitude.  This is expected, because the -NH2 group is more electronegative than -H or -CH3.

The second lone pair is not involved in the acid-base reaction, it does not point towards the -NH4+ group.  You can, however, force two lone pairs into close proximity.  This destabilizes the unprotonated form.  The keyword is "proton sponge".

1,8-Bis(dimethylamino)naphthalene has a pKa of 12.3, it's one of the strongest known amine bases.  Compare that to the pKa of aniline, which is something like 4.5.


Both are correct to some degree.

The first book predicting 12 lines is being very theoretical.  In principle with narrow enough line widths 12 lines would be observed.  However, due to coupling constants (J) being similar from the beta protons to the alpha and gamma protons, in practice with reasonable resolution, only 6 lines are observed (the first book is actually explaining this in fine print).

(source: ucalgary.ca)

Credit: Univ. Calgary


What are the steps of Hartree-Fock Theory?

Yeah, I think it asks you to write the energy as expectation value, vary one orbital and get the Roothan equation.
The Hartree Fock model is the extension of the Hartree model where the wavefunction was schematized by the following guess:
\begin{equation}
\psi=\phi_{a}(1)\phi_{b}(2)...\phi_{n}(n)
\end{equation}
If you work out the equations, you'll see the equations look a lot like the HF ones, but you'll miss the exchange term. This is because the wavefunction is not in the correct fashion to describe a fermionic system not made by infinitely distanced parts.
So you need to plug in the antisymmetric form of the wavefunctions (Slater determinants are a basis in the space of antisymmetric functions). Also, the HF approximation requires you to just approximate the wavefunction with just one determinant, just one piece of the expansion. This is clearly a simplification but it can be correct enough if there is a significative HOMO-LUMO gap and the system is closed shell.
So the HF model gets some correlation (known as fermi hole) meaning that electrons having same spin do avoid each other (you enforced it using antisymmetric wavefunctions) but the electrons don't avoid themselves respecting the Coulomb repulsion (two electrons cannot be in the same place even if they have not the same spin) this is a first piece of correlation you lose, the other being the fact you simply are using one determinant. Unfortunately correlation is described as everything going beyond HF because of Quantum information theory, so you have to be careful when you talk of correlation of HF. As it regards DFT, you'll plug in it functionals, so you may include functionals in it that allow you to recover some more correlation, but remember that DFT functionals are, most of the time, semiempirical.


Volume change inside a balloon upon decreasing the the outer pressure

The relationship between pressure and volume is given by Boyle's Law.
$$\mathrm{P}_1\mathrm{V}_1 =  \mathrm{P}_2\mathrm{V}_2$$

      P       V
     0.199    11.41
     0.20     11.35
     0.201    11.29

     1.0      2.27

So yes, you're right. Using Boyle's Law, the volume should be more than 11.35L so the pressure stays below 0.20 bar.

However this doesn't make real world sense. When you blow up a balloon, as the pressure increases the volume increases.

Edit - Thanks @Mithoron here is a version of the problem that works...

    A balloon is filled with hydrogen at room temperature (25 C) and pressure (1.00 bar) to a volume of 2.27 L in a chamber. The pressure in the chamber is then reduced isothermally. The balloon will burst when the pressure in the chamber is reduced to 0.200 bar. What will the volume of the balloon be just as it explodes?


Why do better π-acceptor ligands cause smaller Δ(T) d-orbital splitting?

Since not many people know the ligand formazanate (I hadn’t heard of it before this question), let me quickly supply an image of what it looks like:

Scheme 1: General formazane structure.

Coodination happens from the lower two nitrogens, one of which is deprotonated (hence -ate) and both of which chelate the metal with $\mathrm{sp^2}$-type orbitals. The five formazane atoms form a π system, which can be extended by appropriate $\ce{R}$-groups. And one of these $\ce{R}$-groups is now modified to include an electron-withdrawing group.

I already previously gave a qualitative orbital scheme for a tetrahedral complex including σ and π interactions as given below.

Figure 1: Qualitative MO-scheme of a tetrahedral $\mathrm{d^0}$ complex with σ and π interactions. Originally created for this answer.

The σ donor orbitals transform according to $\mathrm{a_1 + t_2}$, leaving $\mathrm{e + t_1 + t_2}$ for the π ligands. Unfortunately, the picture is a tad clouded, since both σ and π interactions have a $\mathrm{t_2}$ contribution.

Adding an electron-withdrawing group to your ligand will leave the σ interaction unchanged in first approximation, since mainly the ligand’s π system would be affected. The orbital energies of the π system would be reduced. Since we are talking about a metal with a low oxidation state, its orbitals are higher than drawn in the qualitative scheme (which was designed for permanganate). And thus also, the most important π interaction of the ligand will be, as you stated, accepting electrons into its π system.

The exact nature of any effect observed will depend on the original orbital energies. However, the $\mathrm{e}$ orbitals are only affected by the π interaction while the $\mathrm{t_2}$ orbitals are also affected by the σ interaction. The magnitude of π interaction is expected to be the same for both groups of orbitals, since the start from the same value on each side. Thus, the difference between $\mathrm{e}$ and $\mathrm{t_2}$ stems mainly from σ interactions which only affect $\mathrm{t_2}$ due to symmetry.

Because the σ interaction starts with the ligand orbitals being much lower in energy than the metal orbitals, reducing their energy even more will result in a weaker interaction, therefore a lowering of the $\mathrm{t_2}$ level and thus a smaller $\Delta_\mathrm{T}$ or $10~\mathrm{Dq}$ value.


How does functionalized activated charcoal work?

No. It is not a precipitation reaction. It is physical reaction. The pores are made of the rest of the membranes of the cells. When wood is overheated in air, it burns. When this is done without too much contact with the air, the cellular water gets evaporated. The content of the cell (cytoplasm, proteins, etc.) is pyrolyzed, destroyed and mostly transformed into gases. The membrane is carbonized or charred, but remains here, forming charcoal. The rest of the cell being eliminated, the charcoal contains a huge amount of the previous cells which behave like pores. Big molecules can enter these shells, and have difficulty to get out later on.


Restricted synthesis; formation of nucleophile and electrophile from one carbon source

Your proposed synthesis looks reasonable. Dehydration of the primary alcohol is indeed possible, it proceeds via the E2 mechanism (although primary alcohols are less reactive than secondary or tertiary ones). Chlorinated solvents, like methylene chloride and chloroform, are commonly used for the bromination of alkenes, for the reason you have provided (inertness).


If copper has 2 valance electrons and sulfur 6, why don't they bind in pairs?

the average neutral copper atom has 2 valance electrons

Nope. Only one:

In the transition metals, the previous shell is being filled. Generally, the valence shell can't have more than 8 electrons, but lower shells can. So, you have the two outer electrons filled in in Group 1 and 2, and the rest are filled in in the inner shell. With copper, we have 17 electrons in the inner shell, and 2 on the outer shell by this rule. But--it is sometimes more favorable for the inner shell to be "complete" with 18 electrons, leaving the outer shell with only one.

This is much better explained if you understand what atomic orbitals are. Think of these as as "subshells". In copper, we have two possible shell configurations: 10 electrons in $3d$ and one in $4s$, or 9 in $3d$, and 2 in $4s$. The numbers signify the shell, and the letters signify the orbital type. There are $5$ d orbitals per shell (not all shells have $d$ orbitals, though), and 1 $s$ orbital. Each orbital holds two electrons. Here, we have the trade-off between having a fully filled $3d$ orbital set and having a fully filled $4s$ orbital. Fully filled orbital sets lead to stability. Anyway, since both choices lead to a fully filled orbital set, a lot of other external factors decide which electronic configuration copper uses. So you have both $\ce{Cu^+}$ and $\ce{Cu^{2+}}$.

As a side note, $\ce{CuS}$ exists as well.


How to quantify amount of pigments in plant leaves?

Once you've found a suitable mobile phase in analytical TLC, you could upscale the separation and transfer the results to preparative TLC or column chromatography.

In the case of column chromatography, collect the eluted material in test tubes, check the content by analytical TLC and/or UV-VIS spectroscopy, unite identical fractions, evaporate the solvent and determine the weight of the respective fractions.

In the case of preparative TLC, scratch the separated bands off the plate, elute with a solvent of your choice and proceed as above.

If necessary, repeat the chromatographic separations for the different compound classes with other solvent mixtures.


There are often some unstated simplifying assumptions in questions of this kind. For example, in both cases, no account is taken of heat transfer to the surroundings (the air?). Unless the system is isolated, one expects that eventually the system would reach the temperature of the surroundings. However, it is hardly worth stating this explicitly: our common experience is that we can neglect the transfer of heat to the surroundings, on the timescale of the experiment, because it happens comparatively slowly.

In the examples you give, some similar assumptions may apply. If you have a metal can, perhaps not very thick (low overall thermal capacity) and of high thermal conductivity, it's a reasonable assumption that it comes to thermal equilibrium with the contained liquid fairly quickly. On the other hand, for a glass, with lower thermal conductivity, it may be a reasonable approximation that no heat is transferred to it on the timescale of the experiment.

It is, as you say, situation-dependent: it depends on the material properties of the container. It would be most helpful if problems of this kind made it clear what you should assume. Alternatively you may be able to deduce what to assume, from the data provided (or not provided). Or indeed, perhaps it is expected that you can make the right assumptions, knowing something about the materials concerned.


Let's start by deciding what compounds we're talking about here. You reference an example by your instructor of benzoate, which is the conjugate base of benzoic acid.

What your instructor is suggesting is correct for benzoate. The most important resonance structures for benzoate are the two equivalent structures that place negative charge on the oxygen atoms. It is also possible to draw a resonance structure where electrons are removed from the ring and both oxygens carry a negative charge.

The "carboxylate cage" essentially means that there is no way to delocalize the negative charge of the carboxylate into the aromatic ring. The only way to move electrons into the ring is from the C-O pi bond, which leads to (what I would consider) an unreasonable resonance structure, as it renders an oxygen both positively charged AND non-octet.

Carbonic acids are either a subset of carboxylic acids or a closely related functional group to carboxylic acids. Depending on what property you are interested in, they may be useful to compare or the difference might be too great. In this case, starting from the conjugate base of phenyl carbonic acid (as drawn in the question), there will be no way to delocalize the charge into the aromatic ring.


Just as a note to anyone considering it as an antiseptic agent, it is toxic and probably absorbs through the skin.
Any compound that causes a change in osmotic pressure within a pathogen may kill it through plasmolysis, depending on how much of the compound is soluble in water.  There may be other mechanisms by which a hydrocarbon poisons pathogens like the interruption of metabolic pathways.
Anyhow petroleum distillates are dangerous to apply to the skin and nobody should ever do it on purpose.


After German chemists claimed to have discovered element 43 in 1925, the name masurium was used in the first publications. However, the German experiments could not be reproduced.

The official discovery of element 43 was accomplished by C. Perrier and E. Segrè in 1936. They proposed the name technetium (see Perrier, C.; Segrè, E. Technetium : The Element of Atomic Number 43, Nature, 1947, 159, 24.).

In 1949, the 15th IUPAC Conference adopted this name.

The name technetium was also used in the early American literature, for example in Radioactivity Applied to Chemistry; Wahl, A. C., Bonner, N. A., Eds.; John Wiley & Sons: New York, 1951.


Was Wöhler’s urea synthesis carried out with or without oxygen?

Wöhler’s synthesis of urea is not from 1828 but from 1825. We’ll get to that in a second.

In cases like these, one should always rely on the original publications most. Unfortunately, publications of that era did not follow the somewhat stricter standards of experimental sections that nowadays’ publications do. Still, one can make very educated guesses based on the presence (or absence) of statements. The relevant quotation from the 1828 publication, in which Wöhler notes that the substance he created is indeed urea reads:[1]

  Ich habe schon früher angegeben, daſs man die oben erwähnte krystallisirte weiſse Substanz am besten erhält, wenn man cyansaures Silberoxyd durch Salmiak-Auflösung, oder cyansaures Bleioxyd durch flüssiges Ammoniak zersetzt.

Roughly translated:

  I have previously disclosed that the best method of obtaining the above-mentioned white crystalline substance is by decomposing cyanacidic silver oxide[2] with salmiak or cyanacidic lead oxide with liquid ammonia.

It took a while to locate the previous reference which — in accordance with common custom at the time — was given as ‘In einer früheren kleinen Notiz, die in dem III. Bande dieser Annalen abgedruckt ist, …’ (‘In a previous little note that was printed in the IIIrd volume of these annals …’). In the end, I located the most probable candidate to be the artice Ueber Cyan-Verbindungen (On cyan compounds) also published in the Annalen der Physik.[3] Therein, Wöhler reports two syntheses of a novel crystalline compound; the second one being the interesting one for us:

  Rein erhält man [die eigenthümliche kryſtalliſierte Materie], wenn cyanſaures Blei durch kauſtiſches Ammoniak, oder cyanſaures Silber durch Salmiak zerſetzt wird. Sie kryſtalliſirt in weiſsen durchſichtigen, ſtrahligen Kryſtallen, iſt leicht in Waſſer und Alkohol auflöslich.

Translated roughly:

  [The weird krystalline material] can be acquired in pure form if cyanacid lead is decomposed with caustic ammonia or cyanacidic silver with salmiak. It crystallises in white transparent spear-like crystals and is readily dissolved by water and ammonia.

Wöhler then goes on to describe how different substances interact (or rather: don’t) with what he newly synthesised as was common in the days.

Here, we cannot really make any educated guesses on the reaction procedure by what is written because the fragments are extremely short. However, we can make educated guesses by what is not written. Wöhler made no reference to using sealed glassware or any specific atmosphere. He does not not any specific purification of the reactants. And he does not indicate any special care that may be required. Therefore, it is reasonable to assume that all reactions were performed at ambient pressure in the presence of oxygen and water (although water content will have been neglectable in the case of liquid ammonia).

References and Note:
[1]: F. Wöhler, Ann. Phys. 1828, 88, 253–256. DOI: 10.1002/andp.18280880206.

[2]: My note: Wöhler (incorrectly) assumed the acid anion to be $\ce{(CN)2O}$ which meant that the other constituent must have been silver(I) oxide giving $\ce{[Ag2O][(CN)2O]}$ — which gives the correct sum of $\ce{Ag(OCN)}$.

[3]: F. Wöhler, Ann. Phys. 1825, 79, 177–182. DOI: 10.1002/andp.18250790208.


but how do ions have unpaired electrons when an ion has all its electrons paired either by loosing or gaining electrons to attain the nearest noble gas configuration?

(Quote from your question, emphasis mine)

Your fallacy is the word I emphasised. Surely it is often taught like that—especially in introductory chemistry—that when ions are generated the noble gas configuration (vulgo: octet or dublet in the case of He configuration) is reached. That is not the whole truth, though.

Especially when you form a polycharged anion such as oxide $\ce{O^2-}$, you can (and should!) consider the electrons being added (or removed) stepwise. Thus, going from an oxygen atom you would initially generate an oxide radical anion $\ce{O^.-}$ before arriving at the classical oxide anion $\ce{O^2-}$. If you look at it energetically, the first electron addition is sometimes (but not always!) exothermic but the second one is always endothermic, so it makes energetic sense to separate like this.

These monoatomic radical ions are not commonly encountered as they typically have very short half-lifes. They often rapidly react to either give the full noble gas ion or the original, neutral atom. However, the vast majority of known ions are molecular or polyatomic ions such as phosphate. Within these, one very common species of radical anion forms rather stable compounds: the superoxide anion $\ce{O2^.-}$.

Structure-wise, it is similar to the peroxide anion $\ce{O2^2-}$ which can be written as $\ce{^-O-O-}$ except with one electron removed. However, it is usually generated  starting from dioxygen ($\ce{O2}$) with one electron added, which turns dioxygen’s double bond ($\ce{O=O}$ in Lewis formality) into a single bond. when the heavy alkaline metals (potassium, rubidium or caesium) are burnt in air, the corresponding superoxide is the product:

$$\ce{K + O2 -> KO2}\tag{1}$$

If lithium or sodium are reacted identically, the product is the oxide or the peroxide, respectivesly:

$$\begin{align}
\ce{4Li + O2 &-> 2Li2O}\tag{2}\\
\ce{2Na + O2 &-> Na2O2}\tag{3}\end{align}$$


Periodic trend in difference of energy between the s and p orbitals

In naive terms I think this could be explained as follows:

$2s$ orbital is lower in energy than $2p$ orbitals because it is closer to the nucleus, and thus, experiences a bigger attractive force to the positive nucleus. The so-called effective nuclear charge is bigger for an electron on $2s$ orbital than on one of $2p$ orbitals since in the later case there will be an additional shielding from electrons on $2s$ orbital.
As the nuclear charge increases, both $2s$ orbital and $2p$ orbitals become lower in energy due to the increased attractive force to the nucleus, but the $2s$ orbital is affected to a larger degree for the same reason: it is closer to the nucleus and experiences a bigger gain in attractive force to it.

Thus, the energy gap between $2s$ orbital and $2p$ orbitals becomes larger as the atomic number increases.

That is all what they can tell you in school. But, of course, the situation is more complicated, and i don't think it can be addressed without doing electronic structure calculations. Yes, effective nuclear charge experienced by an electron is often expressed as $Z_{\mathrm{eff}}= Z - s$, where $Z$ is the real nuclear charge and $s$ is the so-called screening constant; $s$ is different for electrons at orbitals corresponding to different values of the principal quantum number $n$ and there are some empirical Slater's rules which provide numerical values of $s$ for different values of $n$. But these picture is "not quite right".

First, as I already pointed out and as one indeed can see from the Wikipedia page linked above, $s_{2s} = s_{2p}$ (same value of $n=2$). But surely if one calculate these screening constant by solving the corresponding Schrödinger equation he/she will indeed found that $s_{2s} \neq s_{2p}$. But even without solving the Schrödinger equation the naive argument given above in (1) favour this inequality. So, $s$ is not just a function of $n$, but also of $l$.

Secondly, screening constants in Slater's rules does not change their values as atomic number increases which is also wrong. Loosely speaking, the more nuclear charge electron can screen, the more will it screen. Below is the related graph from the article with screening constant calculated at the Hartree-Fock level linked above.

So, $s$ is also a function of $Z$, and besides for different $n$ and $l$ it depends on $Z$ in different ways. The "real" picture (in quotes since orbitals aren't real anyway) as we can see is much more complicated, and, honestly speaking, I do not even believe that there is a trend you are looking an explanation for.


How to calculate the concentration based on dilution of a stock solution?

Well, I'll start by saying I redid your math, and you have rounding errors.  Using correct sig figs, your stock solution has 100. mg of Li in it, giving a concentration of 1.00E3 mg/L.  Those dilutions therefore should be 30.0 mg/L, 50.0 mg/L, 80.0 mg/L, and 100. mg/L.

You're correct aside from the rounding errors, and the answer key is wrong.  The question as you've written can't have those answers, as 300 μg/L is 0.3 mg/L, which is not the answer.


Are alkynes more electrophilic than alkenes?

Here are what my notes from Bruice's Organic Chemistry textbook says:

"The clouds of electrons surrounding the sigma bond makes an alkyne an electron-rich molecule. They are therefore nucleophiles that react with electrophiles.

Thus alkynes, like alkenes, undergo electrophilic addition reactions because of their weak pi bonds. The same electrophilic reagents that add to alkenes also add to alkynes.

Despite being less stable than an alkene, an alkyne is less reactive than an alkene, since reactivity depends on ΔG ‡ , which depends not only on the stability of the reactant but also on the stability of the transition state.

When a proton adds to an alkene, an alkyl cation is formed. When a proton adds to an alkyne, a vinylic cation is formed. A vinylic cation has a positive charge on a vinylic carbon. It is less stable than a similarly substituted alkyl cation because the positive charge is on an sp carbon, which is more electronegative than an sp 2 carbon of an alkyl cation. This makes it less able to bear a positive charge."


Reducing property in aqueous phase depends upon (1)  enthalpy of sublimation (2)enthalpy of ionisation (3)enthalpy of hydration.
First two processes are endothermic while last one is exothermic.
For lithium enthalpy of hydration is very high because of its small size hence Li acts as stronge reducing agent in aqueous phase. For Cs enthalpy of hydration is very small hence overall process is less exothermic .


Any false chemical element in history of chemistry?

The example that comes to mind is didymium, which turned out to be a mixture of praesodymium and neodymium. The term is still used, as far as I know, to refer to glass doped with a mixture of these lanthanides. This is discussed in one of the 'Chemistry in its element' podcast episodes.

As it turns out, wikipedia has a category called 'Misidentified Chemical Elements'. Especially notable are illmenium, dianium and pelopium, which were all likely mixtures of niobium and tantalum, two rather similar elements.


Do SN1 reaction always result in Racemisation

This doesn't qualify as racemisation, it is epimerization. Racemization would require both stereocentres to invert to form the enantiomer and - as you rightly suppose - the stereocentre bearing the methyl group is untouched by this reaction.

It is also unlikely to lead to a 1:1 mixture of products since the two diastereomers (and the transition states leading to them) are different in energy, so under either thermodynamic or kinetic control one would expect a non-1:1 mixture. This contrasts with the case of a molecule undergoing true racemization in which the two enantiomeric products will be equal in energy so will be formed in equal quantity.

Your title also poses an interesting and more general question - do SN1 reactions always lead to racemisation even in molecules containing only one stereocentre? I was winding up to write an answer to this question, before finding the outstanding answer previously given by NotEvans here - well worth a read. (hint: the answer is - not necessarily).


Can a zeroth order reaction be reversible?

A zeroth order reaction does not achieve equilibrium as a zeroth order reaction.  Instead the kinetics will go away from zeroth order when we get close to the equilibrium condition.

Zeroth order kinetics is an approximation we get to when some aspect of the reaction approaches a saturation condition, for example the reaction mechanism may involve adsorption on a surface and there is enough reactant to saturate all the adsorption sites.  Such conditions are perforce relatively far from equilibrium; "saturation" and "reversibility" do not come about simultaneously.


However what is so acidic about CaO and basic(no pun intended) about SiO2 while calculating basicity of slag?

What is so acidic...and basic...?

Nothing.

This derives from the historical perception that silica in geological systems and in melts was in the form of silicic acid ($\ce{H2SiO4}$) and the alkali and alkali earth elements were considered as bases.

We now know that in high temperature silicate liquids (that eventually solidifies into slag, or rocks, or glass) there is no acid–base chemistry, at least in the form that we know it from low-temperature aqueous acid–base chemistry. However, the usage of the terms acid and base in relation to rocks and slags still persists, primarily with older professors (who still teach) or in certain parts of the world, such as Russia. Nowadays geologists usually prefer the term "felsic" over the term "acidic" for silica-rich rocks.

Nonetheless, the measure of $\ce{CaO}$ and $\ce{SiO2}$ contents in slags, glasses and rocks is still useful for various reasons. Therefore, the ratio is still calculated and used, and the old name persists. A slightly better parameter, called optical basicity, takes into account the compositional variability other than $\ce{CaO}$ and $\ce{SiO2}$, by adding other oxides such as $\ce{Na2O}$ and $\ce{K2O}$ into the calculation. It is widely used and has many applications.


When diagonalising the overlap matrix, what happens if the square root of the diagonal matrix does not exist?

The matrix $\mathbf{s}$ in your question are the eigenvalues of the overlap matrix $\mathbf{S}$, further the matrix $\mathbf{U}$ are a matrix of the eigenvector of the overlap matrix.
Now looking at the equation that gives you concern:

$$\mathbf{X} = \mathbf{U}\mathbf{s}^{-1/2}\mathbf{U}^\dagger$$

Since $\mathbf{s}$ is a diagonal matrix then taking the power of $\mathbf{s}$ is just taking the power of every element:

$$\mathbf{s}^{-1/2}=\begin{bmatrix}s_1&0&...&0\\0&s_2&...&...\\...&...&...&0\\0&...&0&s_n\end{bmatrix}^{-1/2}=\begin{bmatrix}s_1^{-1/2}&0&...&0\\0&s_2^{-1/2}&...&...\\...&...&...&0\\0&...&0&s_n^{-1/2}\end{bmatrix}$$

The only way as you noted this can cause trouble is if any of the elements of $\mathbf{s}$ are negative.
Now here we have to remember that $\mathbf{s}$ are the eigenvalues of the overlap matrix, it is therefore guareenteed that all of the elements will be equal to or greater than zero!
For the proof of this see my answer to one of your other questions, Basis sets - negative contraction coefficients?


Are there any known chemical properties of tritium water that make it unusually different from protium water?

Yes, $\ce{T2O}$ has been prepared and is available in significant quantity.  When relatively pure, the energy released by the radioactive decay process is so intense that $\ce{T2O}$ will boil.  It must be transported in a shielded, cryogenic dewar.

A significant difference between compounds containing an element bonded to protium, deuterium or tritium is the strength of those 3 bonds.  The $\ce{X-H}$ bond will be the weakest and $\ce{X-T}$ bond the stongest.  We can compare these bond strengths by measuring the relative rates at which the $\ce{X-H}$, $\ce{X-D}$ and $\ce{X-T}$ bonds are broken in a given isotopically substituted compound. These rate differences reflect what is known as a primary kinetic isotope effect. In carbon systems (X=C) the maximum primary isotope effects are roughly as follows:
$$\frac{K_H}{K_D} \sim 6-7$$
$$\frac{K_H}{K_T} \sim 13-14$$
In other words, a $\ce{C-H}$ bond may break as much as 13 times faster than a $\ce{C-T}$ bond.

These same effects will also be seen with $\ce{O-H}$, $\ce{O-D}$ and $\ce{O-T}$ bonds.  The primary kinetic isotope effects will be slightly smaller here because the magnitude of the effect is mass dependent and oxygen has a larger mass than carbon.  Still, as you pointed out, the effects can be disastrous in biological systems.  Biological systems cannot survive if the rates for key reactions are slowed down by such large factors.

The blue color of protium water ($\ce{H2O}$) is due to red light absorption around 700 nm. The frequency of an absorption is given by the following equation

$$\nu_e=\frac{1}{2\pi}\sqrt{\frac{k}{\mu}}$$

Here $\ce{\mu}$ is the reduced mass of the system (e.g. the bond involved in the vibration that is producing the light absorption) and is given as

$$μ=\frac{m_1 \cdot m_2}{m_1 + m_2}$$

where $m_1$ and $m_2$ are the atomic masses located at both ends of the bond.  We see that the reduced mass for an $\ce{O-H}$ bond is 16/17, while it is 32/18 for an $\ce{O-D}$ bond. Since, as shown above, the vibrational frequency is inversely related to the reduced mass, we would expect the $\ce{O-H}$ vibration to occur at higher frequency (shorter wavelength) than the $\ce{O-D}$ vibration.  Indeed, while $\ce{H2O}$ absorbs around 700 nm, $\ce{D2O}$ absorbs at higher wavelength (~1000 nm) and is colorless (reference, see p. 82). Given the even larger reduced mass for $\ce{T2O}$ (48/19), it's absorption should be shifted even further out of the visible range, so it too should be colorless.


Wikipedia reads:

  The third edition of the Green Book (ISBN 978-0-85404-433-7) was first published by IUPAC in 2007. A second printing of the third edition was released in 2008; this printing made several minor revisions to the 2007 text. A third printing of the third edition was released in 2011. The text of the third printing is identical to that of the second printing.

So yes, maybe they uploaded the wrong file. But the content of 2011 and 2008 should be 100% identical.


Do lone pairs contribute towards the pi electron count of heterocycles?

Only count the lone pairs/ pi-bonds/ groups which are participating in conjugation and ignore them in all other cases.

For example, in compound 2 (thiophene), there are two lone pairs on sulfur.

One lone pair (brown) is in a p-orbital, and hence participates in conjugation with the two π-bonds. The other lone pair (blue) is pointing outwards from the ring in an $\mathrm{sp^2}$ orbital. This lone pair is orthogonal, or perpendicular, to the π-system and hence cannot take part in conjugation. In total there are six π electrons, and the compound is therefore aromatic.

See also: Conjugated system on Wikipedia.


Which chemicals can counter the corrosion of tamperproof hardware encryption chips?

Hydrogen fluoride protects the chip by converting into hydrofluoric acid upon contact with moisture or human tissue.

I was able to conclude this through this wiki article:
Hydrogen Fluoride

  Upon contact with moisture, including tissue, hydrogen fluoride immediately converts to hydrofluoric acid, which is highly corrosive and toxic, and requires immediate medical attention upon exposure.[31] Breathing in hydrogen fluoride at high levels or in combination with skin contact can cause death from an irregular heartbeat or from fluid buildup in the lungs

The substance that can neutralize the Ph levels of the hydrofluoric acid is calcium gluconate.

Calgonate 2.5% calcium gluconate gel neutralizes HF

This can be used to help clean up accidental contact with hydrofluoric acid and could possibly be used to prevent any chemical reactions from occurring upon tampering with the chip. The best bet in this situation is to conduct experiments in a dry lab environment and wear gloves. You want to have a fume fan as well as the caluium gluconate in the event of failure.


Yes, chemical reactions do take place in high-temperature plasmas. In fact, these reactions present some problems in the analytical technique known as Inductively Coupled Plasma - Mass Spectroscopy (ICP-MS), which typically uses an argon plasma.

As noted in the comments, the conditions within a plasma are very different from those in a laboratory at STP, for example. But that just means that the chemical reactions that are observed are different, not that they don't exist. In ICP-MS for example, one is trying to detect the masses of very low concentrations of various elements in the presence of relatively huge quantities of argon, oxygen and hydrogen (the latter assuming that an aqueous sample is being analyzed). Species that are formed within the plasma include $\ce{Ar2+}$, $\ce{ArO+}$ and more. When one of these species has the same mass as the trace isotope of interest, it only takes a very small percentage of these products to interfere with the analysis.

This example of argon plasma can largely be extended to your specific question "If hydrogen and oxygen plasma were combined, would the separation of electrons from nuclei prevent normal chemical reactions?". Depending on the conditions of the plasma (temperature, concentrations, degree of ionization, etc.) the answer is that the reaction would only form some small degree of transient water molecules, as compared to say molecular hydrogen and oxygen gas at high temperatures or in the presence of a catalyst, which would overwhelmingly form water. In other words, the former (plasma state oxygen and hydrogen) would form trace amounts of water, and probably hydrogen peroxide and other oxygen-hydrogen species, where under "normal" conditions the formation of water would be strongly favored (thermodynamically anyway). It's just the nature and definition of a plasma that whatever species you have in your plasma will want to be broken into it's elemental constituents and the formation of molecules will be strongly unfavored.

Summary:
The conditions within a plasma are very different from conditions typically encountered in a non-plasma environment, and thus the chemical reactions are different. The formation of otherwise exotic species like $\ce{Ar2+}$ are relatively commonplace within an argon plasma for example ("common" being a relative term of course, as concentrations of $\ce{Ar2+}$ may still be relatively low, just much higher than would be seen outside of the conditions present in the plasma). Plasma conditions themselves can also vary widely, from conditions where only a small portion of the gas molecules are charged to conditions where the majority are charged. Regardless, the high energy state of a plasma tends to result in the destruction of molecules to their atomic precursors, but occasionally there are reactions to form products that would not be seen outside of a plasma.


Rubbing alcohol gets warm after diluting with water

Ethanol does the same. It is the consequence of deviation from the Raoult law.

The other consequences are existence of azeotropes ( affects distillation ) and mixing volume deviations. If you mix 1 L of ethanol and 1 L of water, it warms up. When it cools down back, its total volume is not 2 L, but about 1.9 L.

It is closely related to total energy of inter-molecular bonds in pure solvents and in their mixture.


This phrase comes from Le Châtelier's Principle. Given a reversible reversible reaction, you may (arbitrarily) choose which chemicals to consider the reactants/products, since the reaction goes from both the reactants to the products and from the products to the reactants. Simply put, the forward reaction starts with what you consider to be the reactants. The reverse reaction starts with the products and ends with the reactants. If you flip the equation these names change.


How does biodiesel have so much energy density?

Where you go wrong is in thinking that cooking oil has less energy density than biodiesel. The Wikipedia page for biofuel energy content list a number of different types of fuel. Along with a long list of other fuels, it lists the following:

Fuel Type   Specific energy (MJ/kg) Energy Density (MJ/L)
Methanol    19.9 – 22.7             15.9
Fat         37.656                  31.68
Sunflwr oil 39.49                   33.18
Olive oil   39.25 - 39.82           33 - 33.48
Biodiesel   37.8                    33.3 – 35.7
Diesel      48.1                    40.3

As you can see, both the specific energy (MJ/kg) and energy density (MJ/l) are about the same as biodiesel and close to those of ordinary diesel fuel. To convert the oil into biodiesel, some methanol is usually added, but that does not increase the energy density.


Are you familiar with activation energy?  Consider a hole at the top of a large hill.  If a ball or some water was in that hole it would seek the lowest point it could find.  Clearly, the bottom of the hill would be preferable to the bottom of the hole at the top of the hill, however, in order to get there it must first be brought up over the top of the hole.

In this picture, the $H_2$ and $O_2$ are on the right side of the curve.  Just like water in the above analogy, they'd like to move to the lower area, but first must be brought up over the bump.  So some energy must be added to accomplish this, and then they end up in the much lower energy position on the left.

Once over the bump, they stay as $H_2O$ because that is the state they'd like to be in at any time, but were prevented from being in before by the activation energy bump.


How to refine manganese metal from manganese(IV) oxide found in batteries?

You may heat mixture of $\ce{MnO2}$ with charcoal to higher than $\pu{850 °C},$ in the absence of air. Manganese will be produced in the solid state by reduction of the oxide. But it is sensitive to air oxidation and easily reoxidized. So better use an excess of charcoal. In the industry, rough $\ce{MnO2}$ out of the mine is first reduced to $\ce{MnO}$ by $\ce{CO}$, then leached and reduced into $\ce{Mn}$ metal by a solution of $\ce{Fe^{2+}}$ in the absence of air. Metallic manganese is then dissolved in sulfuric acid and then reduced by electrolysis. See "Manganese metal" over Wikipedia.


What copper oxide is being created upon electrolysis?

It appears to be copper hydroxide or copper carbonate.  Copper oxide tends to be black/grey/brown.  Copper hydroxide is blue and copper carbonate is green.  Copper carbonate forms when the hydroxide is exposed to CO2.

If you would like to dissolve it, try an acidic solution.


How do I calculate lattice points per unit volume?

Atoms on the corners, edges, and faces of a unit cell are shared by more than one unit cell.

An atom on a face is shared by two unit cells $\implies$ half an atom belongs to each unit cell.

An atom on the edge is shared by four unit cells

and an atom on the corner is shared by eight unit cells

What does this have to do with the equations?

Well lets say we have an sodium atom that is crystallized in a simple cubic unit cell, then there would be an sodium atom on each of the eight corners of the cell. However, only $\frac{1}{8}$ of these atoms can be assigned to a given cell. Hence the simple cubic structure is

$$8 \, \text{(corners)} \times \frac{1}{8}=1 \, \text{atom}$$

if sodium formed a body centered cubic structure, there would be two atoms per unit cell (as seen in the top diagram). It isn't shared with an other cells. Therefore the body-centered cubic structure is given by

$$8 \, \text{(corners)}\, \times \frac{1}{8} + 1 \, \text{(body)}= 2 \, \text{atoms}$$

now finally if sodium was a face-centered cubic structure the six atoms on the faces of the unit cell would contribute three net sodium atoms, for a total of four atoms per unit cell. Hence the face-centered cubic structure is

$$8 \, \text{(corners)} \times \frac{1}{8} + \Big{(}6 \, \text{(faces)} \times \frac{1}{2}\Big{)}= 4 \, \text{atoms} $$

Hope this helps


There is a lot in chemistry that cannot simply be explained by Lewis structures, usually requiring a detailed molecular orbital description to rationalise observations.

For example considering a carboxyl group versus a simple hydroxy group. The carboxyl group has a $\mathrm{p}K_\mathrm{a}$ of approximately $5$ while the hydroxy group has a $\mathrm{p}K_\mathrm{a}$ of approximately $15$. In Lewis structures, these two look identical save the neighbouring carbonyl group.

Why can this hydrogen be abstracted easily at all? The hydrogen on a neighbouring carbon (e.g. $\ce{C\mathbf{H}3COOH}$) is not easily abstracted ($\mathrm{p}K_\mathrm{a} \approx 25$ if I recall correctly). The Lewis structure in itself provides no information. You need to additionally consider the concept of polar bonds to explain it: The $\ce{O-H}$ bond is polarised towards oxygen, that means the hydrogen already feels as if it were somewhat like a $\ce{H+}$ and therefore the transition to the true $\ce{H+}$ is easier.

While that explains the acidity per se, it doesn’t explain why one is more acidic than the other. The theory says, that an acid is stronger if its conjugate base is more stable. Since we cannot rationalise the different acidities from looking at the acids, let’s look at the conjugate bases. In one, we just generate an alcoholate much like a hydroxide ($\ce{R-O-}$). Nothing special-looking. The same for a carboxylic acid. No distinction. (However, this clearly shows why $\ce{C-H}$ bonds are usually not acidic: the carbanion is very unstable when compared to an oxyanion.)

Since we cannot find the origin of acidity, we need to expand our theory. Specifically, this is where resonance comes into play as a partial representation of some features of molecular orbital theory in Lewis structures. The carboxylate can resonate between two different structures as shown in the scheme below. Therefore, every oxygen atom only has to take the burden or half a negative charge ($\ce{O^{\frac{1}{2}-}}$). This is much more stable than having a full negative charge — even for an electronegative element such as oxygen. The third structure of the scheme simply displays a way of drawing ‘something in-between’: Both bonds are somewhere between a single and a double one and the charge is located on both atoms simultaneously (but that is not easily drawn with chemical drawing programs).


Why is it that only ozone absorbs UV radiation?

The ISO 21348 Definitions of Solar Irradiance Spectral Categories defines UV radiation as that portion of the electromagnetic spectrum having wavelengths from 10 to 400 nm.  Different atmospheric compounds absorb more or less strongly than others throughout the UV portion of the spectrum.

The following table was reproduced from this Wikipedia article:

Although not fully quantitative, this table indicates that nitrogen, oxygen and ozone all absorb different portions of the UV spectrum.  Water vapor is also an important UV absorber that is not discussed in the above table.

Also, regarding the UV absorption of (liquid) water and your comment that "It is widely accepted that life on Earth wouldn't be possible without ozone layer", it was in large part the protection from UV radiation afforded by liquid water that allowed aquatic organisms to survive prior to the existence of the ozone layer. If the ozone layer were to vanish today, although the nature of Earth's biosphere would change drastically, I suspect many aquatic species, particularly microbes, would adapt and continue to survive in the absence of an ozone layer. It would be bad news for you and me though ;)


How many kilograms of graphite are consumed while producing 4 kg of Al through Hall-Heroult process?

According to Aluminium Production1:

The production of 1 ton of aluminum typically requires 420 kg of carbon, 1920 kg $\ce{Al2O3}$, and 16 kg $\ce{AlF3}$ ....
The theoretical carbon consumption is only $\pu{333 kg C/t} \space \ce{Al}$. The excess carbon consumption is mainly due to air oxidation of the hot anodes in the hooded cells. The reaction occurs preferentially in the pitch binder matrix and leads to the physical loss of coke particles to generate dust that floats on top of the electrolyte.

Theoretically, your answer holds good. However, it fails practically. It's a stupid tricky question by the problem setters, who are apparently testing your knowledge of metallurgy rather than stoichiometry.
References:

Tabereaux, Alton T., and Ray D. Peterson. “Aluminum Production.” Treatise on Process Metallurgy, Elsevier, 2014, pp. 839–917. doi:10.1016/B978-0-08-096988-6.00023-7


Why are silver halides insoluble in nitric acid

Nitric acid (even concentrated) is essentially an aqeuous solution.

Effectively, silver halides will not dissolve in water (they  have incredibly low $K_\text{sp}$), so the reaction indicated is moot.... there is no appreciable concentration of bromide in solution to be oxidized.

$K_\text{sp}(\ce{AgBr}) = 5.0\times10^{-13} \Rightarrow [\ce{Br-}] \approx 700~\text{nM}$


tl;dr It's so simple it was likely never published as a "research result", since it stems quite directly from definitions.

It's a small step from the conservation of matter and the definition of concentration.
By definition: $C = \frac{n}{V}$, thus $n = CV$. If you dilute with pure solvent, you don't change the amount of solute that you have, in other words: $n_1 = n_2$. Using the previous equation, you get $C_1V_1 = C_2V_2$.


What are the dimensions of a and b in van der Waal's real gas formula?

Part of your confusion is in the formula you have written.  According to the Tanner's General Chemistry webpage The van der Waals Equation, if you are considering 1 mole of a substance, then the formula is

$$(P+a/V^2)(V-b) = RT$$

For $n$ moles

$$(P+an^2/V^2)(V-nb) = nRT$$

The Science HQ website Van der waal’s equation, the constants $a$ and $b$ represent pressure and volume corrections respectively.  The units are based on the formula for $n$ moles of substance (the 2nd equation above).

Taking each term on the left hand side individually:

$(P+an^2/V^2)$ is the van der Waals equation equivalent for $P$ in the ideal gas equation, thus this therm must have the same unit, so using $Pa$ as the unit for pressure would require that the units of $a$ to be $Pa L^2 mol^{-2}$  (the website uses the unit $atm$, but the same idea still applies.

In a similar manner, the term $(V-nb)$ is the van der Waals equivalent of $V$ in the ideal gas equation.  So to keep the unit as for volume in the second term, the unit for $b$ must be $Lmol^{-1}$.

One thing to always be careful with, as there are a number of units used for both pressure and volume, it is imperative to keep these constant throughout your calculations.

For example, volume can be expressed as $m^3$ or $L$ and several others (hence the $L^2$ term above would be the same as $(m^3)^2$ or simply $m^6$).

One final document is a master list of $a$ and $b$ constants for several chemicals (once again be careful with the units - unit conversion are provided in the document), van der Waals constants for gases.. Also further explanations of the above can be found on this Purdue University page entitled Deviations from Ideal Gas Law Behavior


Effects Guiding Electrophilic Aromatic Substitution

The first thing I notice is the hydroxyl group, a very powerful activating group and an ortho-para director.  The hydroxyl group is such a strong activator that if you take a solution of phenol and drip bromine into it, the phenol will react with the bromine so rapidly that the bromine color will be immediately removed from the solution as 2,4,6-tribromophenol is formed.

The next thing I notice is the very bulky t-butyl group, also an activator, but much milder than hydroxyl and also a mild ortho-para director.

Next I notice that the two substituents are arranged meta to one another.  There is one position in between the two groups that will be sterically blocked (double-ortho effect); the position on the other side of the t-butyl group will also be sterically blocked, but not as bad.  Also the directing effect of the hydroxyl group will overwhelm the directing effect of the t-butyl group.

Finally, in the halogen series size increases from chlorine to iodine.

In the reaction with iodine, the position ortho to the hydroxyl group and para to the t-butyl group is the only position that is electronically activated and sterically accessible to the bulky iodine.

In the reaction with smaller bromine, the position ortho to the hydroxyl and para to the t-butyl and the position para to the hydroxyl and ortho to the t-butyl are electronically activated by both groups and sterically accessible.  The "double-ortho" position is still sterically blocked.

With the smaller chlorine atom, all 3 positions - even the "double ortho" - position are sterically accessible to the smaller chlorine.

Based on this reasoning, I would select answers A, B, and C.  I would probably also include D because the results are (weakly) reinforced by the electronic effect of the t-butyl group.


The meaning of the answer you've gotten, "it is because there is no hydrogen to remove", is with respect to oxidizing the carbon of the tertiary alcohol itself. You could oxidize the rest of the molecule into any number of things, but there is no way for the tertiary carbon to be further oxidized than the alcohol.


I think that you should focus on what you wrote first, longer the hydrocarbon chain higher their entropy in solution. When dissolved, they can access a multitude of states both energetically and in spaces (think of all possible conformers).
This said, it does not mean that they must dissolve in, eg, water. Right because the H term.


What can I use to make evaporating fake snow?

The phase transition you are talking about (solid to gas) is called sublimation. There are a few compounds that sublime at room temperature and atmospheric pressure, but the one I think would best suit your purpose is dry ice. If spread into the air as a fine powder, it could sublime before it even touches the ground, especially if the air is warm.

It's important to consider the safety aspects of this, though. $\ce{CO2}$ sublimes at -$78º\mathrm{C}$, so if too much of this powder lands on people, you could end up with some cases of frostbite. Additionally, you would need to keep this space well-ventilated since you will be significantly increasing the concentration of $\ce{CO2}$. I wouldn't imagine it would be too hard to meet these standards, though.


How to find the amount of moles given mass percentage data


Why does the Heisenberg uncertainty principle work in an atom?

Electrons don't have a well-defined orbital radius even in a hydrogen atom

An electron "orbiting" in a hydrogen atom does not have a well-defined position or even just a well-defined radius just because it has a well-defined energy.

The orbital is more like a 3D probability density function covering a cloudy space of a defined shape (spherical in the case of the lowest energy orbital but not of a single radius and extending with a finite probability from the nucleus to a vaguely defined boundary we often regard as the radius of the atom).

Different orbitals have different, increasingly complex, shapes defined by solutions of the Schrodinger equation. Heisenberg's uncertainly relationship limits our ability to define the location and momentum of the electron at the same time. The fuzzy probability clouds defined by Schrodinger's solutions are the best we can do.


Calculate the second ionization energy for the Mg from the following data?

Thanks for showing your work.  We're given that
$$\ce{\Delta H_{f} (MgO (s))~ =~ -601$.$8 ~kJ/mol}$$
So any balanced equation we write that has MgO as the product will have this same heat of formation.  An equation that produces MgO and allows us to solve the problem is
$$\ce{Mg^{+2}(g) + O^{-2}(g) -> MgO (g)}$$
The heat of formation of $\ce{O^{-2}}$ is given by
$$\ce{\Delta H_{f} (O^{-2} (g))~=~1/2 the ~O2~ bond~ energy~ (only~ half~ because~ we~ only~ need~ 1~ oxygen~ atom) + the ~first ~ionization~ potential~ for~oxygen + the~second~ionization~ potential~ for~ oxygen}$$
or
$$\ce{\Delta H_{f} (O^{-2} (g))~=~1/2(498.7)~-~141+780~ = ~$888.4$~ kJ/mol}$$
Similarly the heat of formation of $\ce{Mg^{+2}}$ is given by
$$\ce{\Delta H_{f} (Mg^{+2} (g))~=~ \Delta H_{sublimation}Mg(s) + the ~first ~ionization~ potential~ for~Mg + the~second~ionization~ potential~ for~ Mg}$$
or
$$\ce{\Delta H_{f} (Mg^{+2} (g))~=~130 + $738.1$ + IP_2(Mg)}$$
Before solving for $\ce{IP_2(Mg)}$ there is one more adjustment to make.  Everything we've done so far has been for the gas phase except for our starting heat of formation for $\ce{MgO}$ which was for the solid, so let's adjust this heat of formation to its gas phase value.
$$\ce{\Delta H_{f} (MgO (g))~ =~ \Delta H_{f} (MgO (s)) + lattice~ energy~ MgO }$$
or
$$\ce{\Delta H_{f} (MgO (g))~ =~ $-601.8$+ 3800 ~=~$3198.2$~ kJ/mol}$$
Now, inserting all of these values into our equation we get,
$$\ce{130 + $738.1$ + IP_2(Mg) + $888.4$~=~$3198.2$}$$
or
$$\ce{IP_2(Mg)~=~$1441.7$~kJ/mol}$$


Why are many chemical relationships logarithmic?

Some logarithmic behaviour is built into the way the world works and some is a consequence of how we perceive things which span very large ranges.

The simplest chemical and nuclear reactions (first order ones) are, for example, inherently logarithmic. Take any first order process (like radioactive decay). The process has a fixed probability of happening (a radioactive atom or an unstable molecule might have a 10% probability of decaying in a minute). What you observe in the bulk with processes like this is that the rate of the reaction decays exponentially. You start with a rate of say 12 units per minute. After a certain time half the initial material has gone and the rate is now 6 after the same time again the rate is now 3 and so on. This is exponential decay emerging from a very simple process with a fixed probability. A logarithm emerging from a probability, if you prefer. This sort of thing is very common in chemistry and physics.

The other cause of logarithms is human perception. Our perception of light and sound are both inherently logarithmic. This is partially because there is no sensible way to map the range of magnitudes of many natural phenomena to a linear scale. The intensity of sound and light vary over huge orders of magnitude. The Sun is about 15 trillion time brighter than the faintest visible star and it is fairly inconvenient to divide this scale into equal bands. So perception tends to divide the scale logarithmically with each perceived difference representing a fixed ratio not a linear difference (each star magnitude, for example is about 2.5 times brighter than the next lower one, but represents a just-about discernible difference).


How does the reaction of HCl and KMnO4 proceed?

As others have pointed out, you don't have a "normal" reaction, you have a oxidation-reduction reaction (if you want to sound especially geeky, calls them RedOx (pronounced "ree-dox") reactions.

The electrons (which we don't normally balance in a equation) feature prominently because there are elements changing oxidation state.

First, split your reaction into its half reactions. Bring the "spectator ions" along for the ride and make sure they balance:

Oxidation: 2 $\ce{Cl-}$ gets oxidized to $\ce{Cl2}$ and gives up 2 electrons. $\ce{H+}$ is your spectator ion. $$\ce{2 H+ + 2 Cl- -> Cl2 + 2e- + 2H+}$$
Reduction: $\ce{Mn^7+}$ needs 5 electrons to reduce it to $\ce{Mn^2+}$. $\ce{K+}$ and $\ce{O^2-}$ are your spectator ions. $$\ce{K+ + MnO4- + 5e- -> Mn^2+ + K+ + 4O^2-}$$
Now, just add water. Oxidation reactions get $\ce{OH-}$; reduction reactions get $\ce{H+}$:
$$\begin{align}
\ce{2OH- + 2 H+ + 2 Cl- &-> Cl2 + 2e- + 2H2O}\\
\ce{8H+ + K+ + MnO4- + 5e- &-> Mn^2+ + 4 H2O + K+}
\end{align}$$
Now we balance the electrons (remember least-common-multiple from math class? You thought you'd never use that, didn't you?)

LCM for 2 and 5 is 10, so multiply the Cl equation by 5 and the Mn reaction by 2.

$$\begin{align}
\ce{10OH- + 10 H+ + 10 Cl- &-> 5Cl2 + 10e- + 10H2O}\\
\ce{16H+ + 2K+ + 2MnO4- + 10e- &-> 2Mn^2+ + 8 H2O + 2K+}
\end{align}$$

Add everybody back together, make more water, and cancel out the electrons:
$$\ce{10H2O + 10 HCl + 6H+ + 2KMnO4 -> 5Cl2 + 10H2O + 2Mn^2+ + 8 H2O + 2K+}$$
Take out the excess water:
$$\ce{10 HCl + 6H+ + 2KMnO4 -> 5Cl2 + 2Mn^2+ + 8 H2O + 2K+}$$
Looks like we're short something to balance out the charges. We can add more chloride spectator ions to both sides of the equation. They're not part of the redox since they are $\ce{Cl-}$ on both sides of the equation:
$$\ce{10 HCl + 6HCl + 2KMnO4 -> 5Cl2 + 2MnCl2 + 8 H2O + 2KCl}$$
Collect terms one last time and we're done:
$$\ce{16HCl + 2KMnO4 -> 5Cl2 + 2MnCl2 + 8 H2O + 2KCl}$$


Can all compounds that plants and animals produce be synthesized in a lab?

I suspect, in its current form, this question is largely opinion based. Currently, as it stands in September 2015, there are definitely limitations. But never say never.

On the small molecule scale, Maitotoxin is a famous example of a molecule that has yet to been synthesised in the lab. But perhaps only for a matter of time (or money)

3D printing and automated synthesis is expanding the scope for small molecule synthesis, giving more flexibility to synthesising recalcitrant targets.

On the larger scale, human DNA contains up to about 300 million base pairs. Currently, the largest DNA synthesised in the laboratory is about 600,000 base pairs, so still a way off. However, with increasing capacity in areas like artificial gene synthesis, complete replication of entire DNA is not unforeseeable. One day.


Short version:

We don't call bonds "physical", there are chemical bonds and other types of interactions between particles. The chemical bonds are classified this way because they make up molecules, salts, polymers and such, which are the materials chemists are interested in studying, and not particularly because of their sub-atomic/electrostatic characteristcs.

Your question relates to two frequent sources of confusion:

How do we define a chemical bond? What makes a bond between two atoms ionic or covalent?
Where does chemistry (hence chemical interactions) ends, and consequentially where does physics (and hence physical interactions) begins?

Both questions have been pretty much answered here on Chemistry SE and also on Physics SE. I'll give some links to related questions and also a shrot summary.

1. How do we define a chemical bond? What makes a bond between two atoms ionic or covalent?

Your question didn't go much into covalent bonds, but anyone trying to understand anything past the surfaces of chemical bonds should know that the definition of a bond as covalent or ionic is simply an easier way for us to classify things, because when the bonds are categorized in groups we can say that one group behaves in a way, and the other behaves in a different way. The truth is, all bonds have some ionic and some covalent character to them.

Good answers here on Chemistry SE about that can be found here:

Can 100% covalent bonds exist?
What happens if the electronegativity difference is exactly 2.1?

2. Where does chemistry (hence chemical interactions) ends, and consequentially where does physics (and hence physical interactions) begins?

For a Physics pont of view, see: What distinguishes between physics and chemistry?

For a Chemistry point of view, see: Differences between chemical physics and physical chemistry?

Basically, it's clear that both sciences have a lot of overlapping, and perhaps the main difference is the approach taken and how the results will be used. While chemists usually focuses on understanding the structure and interactions between substances and materials, in order to be able to improve them or develop new ones, physicists are mostly interested in understanding the exact laws of nature and representing them with mathematical equations that can be used to explain and predict phenomena. This definitions do not seem so different, and that's not supposed to be wrong: We do have the Physical Chemistry and Chemical Physics fields.

Ok, but Why is an ionic bond a chemical and not a physical bond?

Two important things were mentioned by @Mithoron in the comments. The term "Physical bond" is not frequently used by chemists, (or anyone I suppose), pretty much because all interactions in whole universe are physical. This single quote by @Mithoron is related to the two points I highlighted above. Both "types" bonds "follow" the same laws, and there is no hard line between them.

It's also worth noting that the definition you gave for Physical Bond was taken from an answer without a single upvote, which is an indicator that it isn't what the Chemistry SE community thinks (despite the fact that Google liked it very much and uses it in it's short-answer for the query). Another indicator that "Physical Bond" isn't really a type of bond is that there aren't definitions of it out there (Yahoo answers doesn't count), not even a Wikipedia article.

So why do we call them chemical bonds?

That's pretty much because the bonds we call chemical are those which make up molecules, salts and other materials which are mainly studied by chemistry. The definition of chemical bond varies a lot (as explained here), but they are the bonds which create what chemists study.

Other types of bonds which aren't called chemical are the intermolecular interactions, which are obviously of interest to chemistry, but also the Strong Force, which is the force that binds protons and neutrons (nucleons) together to form the nucleus of an atom. (I hope it's okay to call Strong Force a bond, I'm not sure to be honest).


Catalytic iodine in synthesis of phosphonium salts

Iodine is a stronger nucleophile here than bromine

I agree with your HOMO-LUMO reasoning, but this doesn't really matter too much.  As long as iodide is a strong enough nucleophile to displace bromide at least some of the time, then the reaction we want to happen, will happen.  Of course the more nucleophilic iodide is (here's where your HOMO-LUMO argument comes into play), the faster the desired reaction will happen.

I pretty much agree with the rest of what you say Jori, bit I might say it just a bit differently.

I'm not sure how triphenyl phosphine (TPP) is hybridized, I'd guess that it is unhybridized just like phosphine ($\ce{PH3}$).  If it is unhybridized than our lone pair on phosphorous is in an s-orbital.  Since the s-orbital has little "directionality", this lone pair on phosphorous will be a weak nucleophile.  On the other hand, if TPP is hybridized (like triphenyl amine), then although the lone pair is now in an orbital with high p-character (and therefore very directional) it will be extensively delocalized into the 3 aromatic rings (just like triphenyl amine which is a very weak base for just that reason), which again means that the lone pair isn't very available and consequently TPP will be a weak nucleophile.  So now matter how TPP is hybridized, we expect it to be a weak nucleophile.

Iodide is, as you note, a much better leaving group then bromide.  In fact, iodide is such a good leaving group that it can be readily displaced by even a poor nucleophile like TPP. Iodide is also a good nucleophile and can displace bromide.

The reaction is run with just a small (catalytic) amount of iodide.  As soon as iodide replaces a bromide (nucleophilic displacement), the iodide, being such a good leaving group, will now be quickly displaced by the weakly nucleophilic TPP.  During this last step, iodide is regenerated and the cycle starts again.


First of all, molar absorptivity is dependent on wavelength, so you either need to specify a wavelength or you need the entire spectrum.

When searching for absorptivity, it's often effective to search for the older name 'extinction coefficient'. Using that search term in Google, I found an absorption spectrum for thymol blue here (pdf).


The inventor, Richard M. Palmer, has filed at number of patents on "energy absorbing materials", one of which is US patent (7,381,460).

On the other hand, there does not seem to be any peer-reviewed publication by Mr Palmer.

There's always the option to ask the manufacturer D30 for further information and/or material samples.

The possible downside of such a cooperation is that you might not be allowed to report on the progress of your experimental work on conferences or in publications due to an NDA previously signed!


Ripping apart plastic: is this a chemical or physical change?

When you rip, tear or mechanically deform a polymer (for example a piece of plastic) you are putting energy into the material.  The energy from this deformation causes the polymer chains in the vicinity of the deformation to attempt to align.  To some degree this partial alignment makes continued deformation easier.  To continue tearing the polymer apart you just need to overcome the van der Waals forces holding the chains together.

In addition to the polymer chains being pulled apart, some chemical bonds will be broken.  I don't know how many bonds are broken per chain pulled apart, but I suspect it is low.  Low or not, some free radicals (or ions) will be generated when the chemical bonds break.  If the experiment is performed under ambient conditions, the radicals or ions will quickly react with oxygen or ions in the air.  If the experiment is performed in an inert atmosphere, then the radicals will persist (dangling bonds) and can be observed in an esr spectrometer.  It has been demonstrated that cross-linked polymers, which are less prone to flow deformation and therefore more likely to undergo bond breaking, do produce larger esr signals.

So most of what is going on is molecular deformation and pulling apart of the polymer chains - but some bond breaking does occur.

It would be an interesting experiment to cut two pieces of plastic and then cut one of them into finer shreds.  Place the two samples into separate tubes and then record their esr signals.  How much more intense would the signal be in the finely cut sample compared to the bulk sample?  Polycarbonate-A, straight out of the reagent bottle, produces an esr signal.


Are bridge bonds in diborane longer than the terminal H bonds?

As ocresol said, its not wise to conclude the bond strength from one reaction. Rather, considering the structure of the molecule is a more viable option.
Actually the bridged bond is weaker than the terminal $\ce{B-H}$ bonds. This is because of the presence of multielectron density region. Since it is a pair of shared electrons spread between two regions, so more volume and less the electron density.

Image Source: Quora.
Also, the bond order of terminal $\ce{B-H}$ bonds is 1, whereas the bond order of the bridged bond is half of it ie. 0.5. And it is intuitive that bond strength is directly related to bond order.


What happens at the inert anode in an electrolytic cell?

There is oxidized oxygen in water
$$\ce{2 H2O -> 4 H+(aq)  + O2(g) + 4 e-}.$$
$\ce{HSO4-}$ in sulphuric acid solution can get oxidized as well at high current density and therefore high enough anodic potential:
$$\ce{HSO4-(aq) -> S2O8^2-(aq) + 2 H+(aq) + 2 e- }$$
This is the industrial way to produce potassium peroxodisulphate.


During the process of distillation, we condense the vapours (B) of the mixture (A) to give a mixture (C) that has the same mole fractions as that of the vapour (B) of mixture (A).
In order to solve this question there would be three steps.

Firstly, finding the vapour pressure of the initial mixture.
Second, we find the mole fraction of components in the vapour phase of this mixture.
The third and final part would be to find the vapour pressure of the distillate leading to our answer.

However, finding the actual composition of the vapours here, would be the biggest challenge in this question.
Step $1$
In order to find the vapour pressure, we use Raoult's Law which mathematically states:
$$P_T = \sum{P^\circ_i\chi_i}$$
Since this is a two-component system, the total pressure would be expressed as a sum of two terms for each component A and B.
$$P_T = P^\circ_\mathrm a \chi_\mathrm a +P^\circ_\mathrm b \chi_b$$
Substituting the values given in the question, we get $P_T= \pu{85 mmHg}$
Step $2$
The next step is to find the composition in the vapour phase. For this, we use the  property that the partial pressure of a component in the vapour phase is equal to the contribution of the same component in the liquid phase. Mathematically, it is equivalent to stating:
$$P_Ty_\mathrm a= P^\circ_\mathrm a \chi_a$$
Here, $y_a$ is the mole fraction of component A in vapour phase. Solving for $y_\mathrm a$, we get that $y_\mathrm a = \frac{5}{17}$, which implies $y_\mathrm b = \frac{12}{17}$.
Step $3$
The final is exactly the same as the first step, wherein you use Raoult's law to find the final vapour pressure of the distillate (mole fraction of components equal to that mole fraction of vapour in initial mixture).
Solving for the vapour pressure, you get $P_T^{'} = \pu{85.88 mmHg}$, which is the answer provided.


Formation of Bonding and Anti-bonding Molecular Orbitals

First, remember that orbitals are a more or less theoretical construct, based on mathematical models. With that in mind, remember that an orbital in the strictest quantum-mechanical sense is a wave function of one and exactly one electron. (I am separating α and β spins here.) Since an orbital somehow nicely corresponds to one possible particle state, it may seem logical to postulate a ‘rule of conservation of orbitals’ — and exactly that happens in quantum chemistry.

So if we take one 1s orbital of one hydrogen atom and another 1s orbital of another hydrogen atom, we need to create two new orbitals by linear combination if we choose to do so. The two orbitals we create correspond to σ and σ*, respectively. (In a spintastic sense, we would have taken two orbitals of each hydrogen, one corresponding to α the other to β. We would receive four orbitals: $\unicode[Times]{x3c3}_\unicode[Times]{x3b1}, \unicode[Times]{x3c3}_\unicode[Times]{x3b2}, \unicode[Times]{x3c3}^*_\unicode[Times]{x3b1}$ and $\unicode[Times]{x3c3}^*_\unicode[Times]{x3b2}$. The electrons now occupy the lowest energy orbitals which are $\unicode[Times]{x3c3}_\unicode[Times]{x3b1}$ and $\unicode[Times]{x3c3}_\unicode[Times]{x3b2}$.)


Can the electrophile attack at the position of the NH2 group in nitration of aniline?

Maybe just after some time, the $\ce{-NH2}$ group will become $\ce{−NH3+}$ (acidic medium) and $\ce{NH3}$ would depart as a leaving group?

No! You are confusing the leaving group oxonium ion with the ammonium ion. Simply put it, $\ce{R-NH2}$ does not leave as $\ce{R+}$ and $\ce{NH3}$, as you might expect from the hydroxyl group leaving mechanism.

But that does not mean there is no other way to make an $\ce{-NH2}$ group (which is a very bad leaving group) leave. You can instead make it leave as dinitrogen gas!

The way this works is by the reaction of $\ce{R-NH2}$ with $\ce{HNO2}$. If $\ce{R}$ is an:

aromatic group: a diazonium ion, $\ce{R-N2+}$, will be formed (stable at ice cold temperature). Further reactions such as Balz-Schiemann, Gattermann, Sandmeyer, etc. may be employed to replace the $\ce{-N2+}$ with another group, and make it leave as dinitrogen gas.
alkyl group: $\ce{R-N2+}$ will be formed but it will, being highly unstable (given that it does not enjoy the resonance stabilization of an aromatic ring), instantly decompose to release dinitrogen gas and form a classical carbocation $\ce{R+}$. This carbocation may rearrange, if necessary, and then receive an attack by a nucleophile.

This is how you make the bad leaving group, $\ce{-NH2}$, leave any aromatic ring or an alkyl group.


Is there a case when an atom of some element (e.g. carbon) does not have its expected number of protons (e.g. 6)?

A carbon atom has 6 neutrons. How many protons are present in the nucleus of a carbon atom?

Obviously this wording would make any astute student say the answer is 6 since carbon by definition has 6 protons. You are correct, elements are defined by their number of protons. Key word being defined, changing the number of protons changes the element.

I believe the key here that your book is getting at is you only know that you have an atom from a carbon sample that has 6 neutrons but nothing else. I think the question is trying to evaluate if you are aware that proton numbers are independent of neutron numbers thus knowing there are 6 neutrons does not help you determine protons, especially given one of the options is:

  D. Can't tell; more information is required.

If this is what they wanted then it was a very poor choice of words to say a carbon atom in the question then ask for number of protons. They should have said:

  Given an atom (or nucleus) with 6 neutrons, how many protons are present in the nucleus of the atom?


How do α,β-unsaturated acids undergo decarboxylation?

Albeit carboxylic acids and their derivatives lose carbon dioxide under a variety of experimental conditions, the literature on the decarboxylation of $\alpha,\beta$-unsaturated carboxylic acids is not common. The reason for this is all $\alpha,\beta$-unsaturated carboxylic acids seemingly do not undergo decaboxylation under mild conditions, except for those containing substitutions at $\beta$-$\ce{C}$ position, which stabilize the intermediate carbocation.

A wide variety of experimental conditions undoubtedly make the reaction to undergo by many different mechanisms. For example, the role of acid catalysis in the decomposition of cinnamic acid (and related compounds) has been clearly demonstrated in Ref.1.
The mechanism for this acid catalyzed decarboxylation is illustrated in the following figure (see I in the diagram). The structures of some example of these susceptible substrates are also included (Ref.1) in the diagram:

Some carboxylic acids undergo decarboxylation by pyrolysis. It is an experimental fact that the olefins resulting from pyrolysis of simple $\alpha,\beta$-unsaturated acids (note that simple means at least no substitution on $\alpha$-$\ce{C}$) are homogeneous and contain terminal unsaturation (Ref.2). Two different mechanisms leading to the same olefin are formulated in Ref.2. The first one is essentially the same mechanism suggested for the decarboxylation of cinnamic acid and related compounds (see I in the diagram).
Before discuss about the second mechanism, let’s look at the proposed mechanism for $\beta,\gamma$-unsaturated carboxylic acid compounds (see II in the diagram). The decarboxylation mechanism of $\beta,\gamma$-unsaturated carboxylic acid compounds undergo the reaction through a cyclic transition complex to give alkene products. This mechanism mimics the one proposed for the thermal decomposition of $\beta$-keto acids.

As a consequence, an alternative mechanism for thermal decomposition of $\alpha,\beta$-unsaturated carboxylic acid compounds was also proposed if the substrate contains allylic $\mathrm{sp^3}$ carbon to extend the double bond (see III in the diagram). It was assumed that under thermal energy, substrate first isomerizes to the $\beta,\gamma$-isomer, which then decarboxylates through a cyclic transition complex to give the terminal alkene product as given by mechanism I.

References:

W. S. Johnson, W. E. Heinz, “The Acid-Catalyzed Decarboxylation of Cinnamic Acids,” J. Am. Chem. Soc. 1949, 71(8), 2913–2918 (DOI: 10.1021/ja01176a098).
R. T. Arnold, O. C. Elmer, R. M. Dodson, “Thermal Decarboxylation of Unsaturated Acids,” J. Am. Chem. Soc. 1950, 72(10), 4359–4361 (DOI: 10.1021/ja01166a007).


How can we balance one-half of an oxygen molecule, or any other diatomic, in a chemical reaction?

Whenever a chemical reaction is written with fractional coefficients, implicitly it's meant to be interpreted as the amount of substance in moles of each species that is involved.  So, it's not really one half of an oxygen molecule reacting, but 0.5 moles of $\ce{O2}$ reacting.

bon's comment is spot-on -- if you are trying to think about a reaction written this way on a molecular level, you need to multiply the reaction through by whatever constant eliminates all the fractions.


Fragmentation involving formation of enamine

@Laura: Both Waylander and Zhe are correct in their respective comments. Your example is from the work of Hendrickson and Boeckman.1 Enamine 1 and conjugated aldehyde 2 were mixed neat forming bridged aminoketone 3. At some point, as Zhe has intimated, exchange of pyrrolidine from the ketone to the aldehyde must occur before cyclization to 3 can be accomplished. The usual explanation is that traces of water are present to catalyze the reaction. Quaternization of the amino group in 3 affords the salt 4 which cleaves with hydroxide via a Grob-like fragmentation to 5. Note the important anti arrangement of the dotted bond and the proximate vicinal bold bond. The product contains a double bond as Waylander has stated. The first example of this reaction, 6 --> 9, was reported by Stork and Landesman2 in 1956.

1) J. B. Hendrickson and R. K. Boeckman, Jr., J. Am. Chem. Soc., 1971, 93, 1307.
2) G. Stork and H. K. Landesman, J. Am. Chem. Soc., 1956, 78, 5129.


Interesting question but the sad answer is we will never know the true story. Here is the original text (thanks to Google Books) dated 1863, right from the horse's mouth Adolf von Baeyer.

Dieser Gedanke hat mich bei der folgenden Arbeit geeitet (spelling?), und ich habe
mich zunächst bemüht, die Kenntnis des Materials zu vervollständigen.
Man wird sehen, wie sich dasselbe in einfachster Weise um die Substanz
$\ce{N2CO3H4}$, die ich Barbitursäure nennen will, gruppieren lässt und wie
also die Frage nach der Konstitution der Harnsäure und ihrer Derivate
auf die Untersuchung dieser Substanz zurückgeführt ist.

[DeepL Translation] This thought led (?) me to the following work, and I first tried to complete the knowledge of the material. One will see how the same can be grouped in the simplest way around the substance $\ce{N2CO3H4}$, which I want to call barbituric acid, and how the question of the constitution of uric acid and its derivatives can be traced back to the investigation of this substance.
He never explained his choice- no footnotes anywhere. However the unabridged Oxford English Dictionary (not open access) attributes it to a woman's name Barbara.
< French barbiturique (Ann. de Chimie et de Physique, 1865), < German barbitur in barbitursäure (Baeyer 1863, in Ann. d. Chemie u. Pharm.), < Barbara, a woman's name.
Cited from Adolf von Baeyer.


Friedel-Crafts Alkylation Using Alkenes: Mechanism

The Friedel-Crafts alkylation is another example of an electrophilic aromatic substitution. I've drawn the mechanism below showing the formation of cyclohexylbenzene.  The reaction is typically run in the presence of an acid and the aluminum chloride serves as a catalyst to enhance the electrophilic properties of the proton, at the end of the reaction the aluminum chloride has been regenerated.

In the insert box I've shown that when the reaction is carried out with n-propylchloride isopropyl benzene is produced, not n-propylbenzene.  A carbocation intermediate is generated in thr Friedel-Crafts alkylation and will rearrange to a more stable carbocation if possible.  In the cyclohexene case, rearrangement of the intermediate carbocation just regenerates the same cyclohexyl carbocation, so the rearrangement is "invisible" in this case (unless we label the starting cyclohexene).


How to find the Cp, difference of Cp and heat of fusion from the dependence of vapor pressure on temperature?

Use the equation for the vapor pressure of the solid to get the heat of sublimation.  The heat capacity of the vapor minus the heat capacity of the liquid is equal to the derivative of the heat of vaporization with respect to temperature. The heat capacity of the vapor minus the heat capacity of the solid is equal to the derivative of the heat of sublimation with respect to temperature.

ADDENDUM

$$\Delta H_{sublimation}=RT^2\frac{d\ln{P}}{dT}=(\ln{10})RT^2\frac{d\log{P}}{dT}$$where log P is given by Eqn. 1.


Is Gibbs energy change during osmosis positive? If so then why is it spontaneous?

Let the subscript A refer to the solvent and B refer to the solute, and let the subscript 1 refer to the pure solvent container and  2 refer to the mixture container.  Then the free energy of the combined system is given by:
$$G=n_{A1}\mu_A^0+n_{A2}\left(\mu_A^0+RT\ln{\left[\frac{n_{A2}}{n_{A2}+n_{B_2}}\right]}\right)+n_{B2}\left(\mu_B^0+RT\ln{\left[\frac{n_{B2}}{n_{A2}+n_{B_2}}\right]}\right)$$If the number of moles of solvent in container 2 increases by $dn_{A2}$ and the number of moles of solvent in container 1 decreases by $dn_{A2}$, the change in the free energy of the combined system changes by dG.  If we do the math correctly, we obtain:
$$dG=RT\ln{x_{A2}dn_{A2}}\tag{1}$$where $x_{A2}$ is the mole fraction of solvent in the mixture container at any point:  $$x_{A2}=\frac{n_{A2}}{n_{A2}+n_{B_2}}$$Eqn. 1 indicates that the change in free energy is negative (as expected).


Potential term in the hydrogen atom Hamiltonian

What is the $r$ there?

$r$ is the same in the potential energy term of the Hamiltonian as in the kinetic energy term of the Hamiltonian.

$r$ is the distance from electron to the proton.

  But isn't the position of the electron non-deterministic?

In the generally accepted Copenhagen interpretation of quantum mechanics, the position of the electron is not deterministic. However, expressing the kinetic and potential energy of the the electron at each position in space does not contradict this interpretation. It is only saying, for each point, if the electron is at this point, the energy is _.

  Shouldn't that thing be a probability distribution of the distance instead?

The wave function solutions represent probability amplitude in the Copenhagen interpretation.  The square of the absolute value of the wavefunction represents a probability distribution.

  Why isn't it squared?

Force would be inversely proportion to $r^2$, but potential is inversely proportional to $r$.


A metal, M, forms two fluorides containing 40.43% F and 50.44% F by mass. Identify the metal

You need to use the general concepts of the law of definite proportions and the law of multiple proportions to solve this:
Sample 1 is 59.57% $\ce{M}$ and 40.43% $\ce{F}$.
Sample 2 is 49.56% $\ce{M}$ and 50.44% $\ce{F}$.
Assume $\ce{M_{x}F_{y}}$ for sample 1 and $\ce{M_{a}F_{b}}$ for compound 2.
Let $m$ by the molar mass of $\ce{M}$. We have
$$\frac{19.00y}{mx + 19.00y}=\frac{40.43}{100}$$
and
$$\frac{19.00b}{ma + 19.00b}=\frac{50.44}{100}$$
We can rearrange the equations to:
$$m\frac{x}{y}=28.0$$
and
$$m\frac{a}{b}=18.7$$
Note that $$\frac{28.0}{18.7}\approx\frac{3}{2}$$
so we can start by trying:
$$m\frac{3}{1} = 28.0$$
and
$$m\frac{2}{1} = 18.7$$
This is weird because the compounds would be $\ce{M3F}$ and $\ce{M2F}$.
Fortunately, as long as the proportions work, it's fine, so divide both fractions by 6:
$$m\frac{1}{2} = 28.0$$
and
$$m\frac{1}{3} = 18.7$$
This gives $m=56.0$, fairly close to the molar mass of iron which is $55.9$.
Cadmium is a bad candidate, since that would be $\ce{CdF4}$ and $\ce{CdF6}$.


Acidic character and anion stability across periods and groups of the periodic table

There are two factors determining acidity of a compound, the electronegativity factor and the size factor. Down the period, the size factor dominates, and the reason $\ce{HI}$ is more acidic than $\ce{HF}$ is because of this factor, as in the former molecule, the orbitals of iodide participating in bonding are larger and more diffuse, hence bond is weaker.
Across the period, however, size does not vary as significantly as down the group. Hence, we compare using the electronegativity factor. If electronegativity difference is higher, there is larger ionic character, and hence it is easier to deprotonate.
Also, basicity of an ion is proportional to its charge density. So you misunderstood the first concept. You should compare fluoride with hydroxide ion, and not oxide.


Close. But each (different type of) group added to the heptane backbone is numbered separately: 4-(1-chloroethyl)-4-methylheptane. Also, you have an asymmetric carbon, so the chirality needs to be specified (I will leave that to you).


Calculating the viscosity of water using particles

The basic idea here is that the spherical particles, each of radius a, are being bounced around via Brownian motion. So the two relevant equations are
$$r(t)^2 = 2Dt \tag{1}$$
$$D = \frac{kT}{6πηa} \tag{2}$$
Note that the first equation uses r(t) rather than x, to avoid needless confusion with the x and y cartesian coordinates of the spherical particles.
Now consider particle 1. At t = 0 s, it is at the defined location 0,0, where all distances are in $\mu m$. At four subsequent times, the x and y coordinates are as given in the OP's table. The following figure shows particle 1's location at t = 0s and at t = 30 s.

At any given time,
$$r_1(t)^2 = x_1(t)^2 + y_1(t)^2 \tag{3}$$
To estimate D, an ordinary least squares fit is performed on the tabulated data for particle 1. The result is shown in the next figure:

Note that the origin is used: at t = 0 s, then, by definition, $x_1(0) = 0 \space µm$ and $y_1(0) = 0 \space µm$. So $D = 1.1486 \times 10^{-12} m^2 s^{-1}$. Rounding off to two significant figures will be performed at the end.
Solving equation (2) for $\eta $ yields:
$$\eta = \frac{kT}{6πDa} \tag{4}$$
where T = 298.15 K, $a = 2.12 \times 10^{-7} m$ and $k = 1.38064852 \times 10^{-23} J K^{-1} $. The result for $\eta $ is
$$\eta = 8.97 \times 10^{-4} Nt \space m^{-2} \space s \tag{5}$$
where the fact that $1 J = 1 Nt \times 1 m$ has been used. N.B. 1 Joule equals 1 Newton times 1 meter.
Other units for $\eta $: Since 1 Pascal of pressure is 1 Newton of force per square meter, i.e., $1 Pa = 1 Nt / m^2$, and since 1 centipoise (symbol: cP) is 1 milliPascal times 1 s, i.e., $1 cP = 1 mPa \space s$, then
$$\eta = 8.97 \times 10^{-4} Pa \space s = 0.897 \space mPa \space s = 0.897 cP\tag{6}$$
Finally, rounding to two significant figures, $\eta = $0.90 cP. The viscosity of water at 25 °C is reported to be 0.91 cP.


What type of bond would be formed if dyz orbitals overlap along the z axis?

Before I answer the question, you should understand that - since everything is symmetric - the answer would be the same if we took two $\ce{d_{yz}}$ orbitals along the x-axis, two $\ce{d_{xy}}$ orbitals along the y-axis, etc.

Have a look at this snippet I picked from Wikipedia Commons (by Jeremy77, public domain):

Here you can see the overlap of two $\ce{d_{xy}}$ orbitals along the y-axis constructively (in the same phase). There are four lobes overlapping, which would lead to a total of two nodal planes. Such a bond is a speciality of $\ce{d}$ orbitals and it is called the delta ($\delta$) bond.

Now, can you understand what the answer to your question would be?

See also: Which d orbitals can form Sigma, Pi, Delta bonds?


Do auto-catalytic aromatic compounds capable of forming self-replicating molecules exist?

I recall seeing a system that consisted of two planar components (and I am fairly certain those were aromatic) that could couple to form the product. The self-catalysis consisted of a product molecule acting as a template for two component molecules to absorb to, which was claimed to have increased the reaction rate.

Edit: I have found the system I was thinking about. The Asker may decide for themselves if this fits the bill. The example is given in K. Ariga, T. Kunitake: "Supramolecular Chemistry - Fundamentals and Applications", Springer, 2006, p. 31. doi: 10.1007/b137036. As of this writing, the page in question is available through Google Books and the reference list for the section (2.10) is on the publisher's page. I was not able to definitively select the correct reference and I do not have access to them, anyway. Note that Ariga & Kunitake seem to have up their mind on what to call it:

  In this system, [the product] acts as a template for self-replication. It may be surprising that a simple host-guest system like this mimics the fundamental activity associated with life, self-replication.


Why does the Lewis definition of acids specify a PAIR of electrons and not a single electron?

It partially has something to do with the history of acid-base definitions. We started with the Arrhenius definition which was based on the generation of $\ce{H+}$ or $\ce{OH-}$ in aqueous solution, and then to the Bronsted-Lowry definition which moved towards acceptance or donation of $\ce{H+}$. Finally came the Lewis Definition which removed $\ce{H+}$ and left us only with electron pairs, to be consistent with the past two definitions. We couldn't just say electrons because that would have included a large class of reactions called redox reactions which are clearly not the same as acid-base.

Now, with respect to your answer to the multiple-choice question:

  I thought the answer was $\ce{NH4+}$ as it could accept an electron, but the Lewis definition specifies accepting a pair of electrons.

There are two points that I would like to draw your attention to:

You're right in that $\ce{NH4+}$ is an acid (more specifically we call it a conjugate acid), but by the two older definitions.
However, $\ce{NH4+}$ cannot accept even one electron in this case. If you draw the Lewis structure, you'll find that although there is a net positive charge, all hydrogens have 2 electrons and the central Nitrogen has eight, fulfilling the octet rule.


Is there such a thing as an acid without a hydrogen?

It depends on which definition of acids and bases you are using.

According to the Arrhenius theory, acids are defined as a compound or element that releases hydrogen (H+) ions into the solution. Therefore, there are no Arrhenius acids without a hydrogen atom.

According to the Brønsted–Lowry acid–base theory, an acid is any substance that can donate a proton and a base as any substance that can accept a proton. Hence, there are no acids without a hydrogen atom according to this theory.

But according to the Lewis theory of acids and bases, an acid is any substance that can accept a pair of nonbonding electrons. In other words, a Lewis acid accepts a lone pair of electrons. According to this theory, acids without a hydrogen atom can exist. (A coordinate bond is formed between the Lewis acid and base. The compound formed by the Lewis acid and base is called a Lewis adduct)

A great example of this would be $\ce{BF3}$. It is neither an Arrhenius acid nor a Brønsted–Lowry acid but it is a Lewis acid. The boron atom accepts a pair of nonbonding electrons from another atom or ion to complete its octet. Here $\ce{BF3}$ is a Lewis acid as it accepts a pair of nonbonding electrons.

The fluorine ion here is a Lewis base as it donates a pair of electrons.

If you want to get more rigorous about the definition of Lewis acids: "a Lewis acid is a type of chemical that reacts with a Lewis base to form a Lewis adduct".

More about Lewis acids and bases here.


As of May 20th 2019, the Avogadro constant will be set to $6.02214076 \times 10^{23}\ \rm{mol}^{–1}$, i.e. the Avogadro constant times one mole will be an integer, but not the one the OP asked about. In preparing for the switch of definition, the BIPM looked carefully at two ways of determining the constant, and then set it to the best estimate from experimental values, truncated to nine significant figures.

This way, there will be minimal practical changes of units even though the definitions have changed. Also changed is what is defined vs. what is measured. For example, the molar mass of 12-C is now based on measurement, and could change (certainly in terms of how many significant figures are known) as measurement techniques improve.

Setting the Avogadro constant to a dimensionless number as the OP asked about would make lots of relationships between quantities and constant dimensionally incorrect, such as $$ R = k_\mathrm{B} N_\mathrm{A}$$

with $k_\mathrm{B}$ the Boltzmann constant, R the universal gas constant and $N_\mathrm{A}$ the Avogadro constant.

  Would any unforeseen complications arise from here?

Setting Avogadro's constant to $6 \times 10^{23}\ \rm{mol}^{–1}$ (I added the proper units) would have lots of complications, but they would not be unforeseen, as @Zhe discusses in the comments.


How to get a protein's dihedral angles ordered by variance?

If I understand you, I can't imagine how to solve the your problem with this approach. Also I don't get the physical meaning of the proposed solution.

If you have to deal with a molecule with more than 100 rotatable bonds, you are just in problems.
You could find the rotatable bonds with some tools (see for example OpenBabel). But the problem will just start.

At first, you can not just try one by one the $N$ angles ($\{\theta _i \}$) (by changing the angle value $n$ times and trying each of them), because for exploring all the conformational space you will just have $n^N$ points. And you just can't know a priori which are the most important angles (in case (I think not) that effectively exist the most important angles).

Also note that not every point ($\theta _1, \theta _2, \cdots,\theta _N)$ is acceptable, just because parts of the molecule can superpose with other parts.

  "Take a set of different conformations and use principal component analysis on the phase space consisting of the dihedral angles."

I think that this is not possible. The values of the rotatable angles will be just chosen by you.
Even more, suppose that you have two rotatable bonds very close each other, in a linear part of the molecule. If not a large superposition ocurrs, you will have no diference in the efect of them. But the effect of them will be very correlated. So, you can get useful info. with PCA.

A possible approach

As I know very few of biology, I would make a research about if there are prefered configuration, that are much more common than others (because of biological facts).

If you are forced to work with the entire configurational space, the computational approach that comes to my mind is:

Use very fast method for energy computations that allow you set the connectivity of the atoms. This will prevent changes of the covalent bonds due to superpositions.
Use the Monte Carlo method, this should explore the configurational space giving you a reasonable idea of this space.

Final note

I find extrange doing docking with a large protein and ¿without a ligand? If there is a ligand for binding to the protein that we are talking about, I won't take too much care of the protein. The scoring functions or energies are few accurate, I find hard to believe that just one found configuration would represent accurately the problem (considering the large degeneracies involved)


Gosh! You would think that electrolyzing NaHCO3 with stainless electrodes would be a clean reaction to produce H2 and O2.

Perhaps OH. is powerful enough to attach to some atoms at the surface and make it more insulating. So, significant degradation could occur at the positive electrode because of oxidation.

Perhaps lower voltage and currents would be more gentle with the electrodes. And perhaps a more resistant grade of stainless would show less degradation (316 is more resistant than 304, and knife blades are typically 400 series, harder but less corrosion-resistant).

At the cathode, hydrogen would be produced in a nascent (atomic) state, which can reduce the oxide coating on stainless steel, making it active. This activity would not result in corrosion while the electrolysis is going on - it is being cathodically protected. This electrode would be a good conductor even as its protective oxide coat is reduced, but the anode will have its oxide coat thickened till it becomes somewhat insulated - or degraded.

A similar effect can be seen with aluminum.


Statistical model of ligand substitution

I think the easy way out is to invoke $S_\mathrm m = R \ln \Omega$. If we assume that for a generic complex $\ce{MA_{n}B_{$N-n$}}$,

$$\Omega = {N \choose n} = \frac{N!}{n!(N-n)!} \quad \left[ = {N \choose N-n} \right]$$

and that for the individual molecules $\ce{A}$ and $\ce{B}$, $\Omega = 1$, then the equilibrium constant $K$ for

$$\ce{MA_{n}B_{$N-n$} + A <=> MA_{n + 1}B_{$N-n-1$} + B}$$

is given by

$$\begin{align}
K &= \exp\left(\frac{-\Delta_\mathrm r G}{RT}\right) \\
&= \exp\left(\frac{\Delta_\mathrm r S}{R}\right) \\
&= \exp\left(\frac{S_\mathrm{m}(\ce{MA_{n + 1}B_{$N-n-1$}}) + S_\mathrm{m}(\ce{B}) - S_\mathrm{m}(\ce{MA_{n}B_{$N-n$}}) - S_\mathrm{m}(\ce{A})}{R}\right) \\
&= \exp[\ln\Omega(\ce{MA_{n + 1}B_{$N-n-1$}}) + \ln\Omega(\ce{B}) - \ln\Omega(\ce{MA_{n}B_{$N-n$}}) - \ln\Omega(\ce{A})] \\
&= \exp\left[\ln\left(\frac{\Omega(\ce{MA_{n + 1}B_{$N-n-1$}})\Omega(\ce{B})}{\Omega(\ce{MA_{n}B_{$N-n$}})\Omega(\ce{A})}\right)\right] \\
&= \frac{\Omega(\ce{MA_{n + 1}B_{$N-n-1$}})\Omega(\ce{B})}{\Omega(\ce{MA_{n}B_{$N-n$}})\Omega(\ce{A})} \\
&= \frac{N!}{(n+1)!(N-n-1)!} \cdot \frac{n!(N-n)!}{N!} \\
&= \frac{N-n}{n+1}
\end{align}$$

The reason for ignoring $\Delta_\mathrm r H$ is because we are only interested in statistical effects, i.e. entropy, and we don't care about the actual stability of the complex or the strength of the M–L bonds. However, the exact justification for assuming this form for $\Omega$ still eludes me. It makes intuitive sense (that there are $N!/(n!(N-n)!)$ ways to arrange $n$ different ligands in $N$ different coordination sites), but I can't convince myself (and don't want to attempt to convince you) that it's entirely rigorous. In particular, I feel like symmetry should play a role here; maybe it is simply that the effects of any symmetry eventually cancel out.