Ionization energy comparison between K+ and Cl-

The general reaction we are looking at when discussion ionisation enthalpy is:

$$\ce{X -> X+ + e-}$$

With the electron removed into space (i.e. totally disappearing). An electron is a negatively charged particle, so removing it from something negative will be easier than removing it from something positive. Thus, you have a very stable $\ce{K+}$ ion and are trying to rip away another electron from this (bad) — and you need to rip it out of a core orbital — all valence orbitals are already empty, the last one when you removed that previous electron to generate $\ce{K+}$ (very bad). A combination of bad and very bad gives us a very high ionisation energy for $\ce{K+}$.

Discussing chloride, first of all we have an anion, so the removal of an electron is generally good for removing excess charge. Second, we again have something that has a full octet — but this time the octet is a valence octet, not a core octet, so removing the electron will be a lot easier. We have a combination of good and relatively good, so the ionisation energy of $\ce{Cl-}$ will be rather low.


Why are the C-C bonds in benzene closer in length to double bonds than single bonds?

one might expect that the bond lengths would be exactly intermediate between a single and a double bond

They actually are. The relevant reference structure, that can provide us lengths of single and double bond for $sp^2$ carbons with not conjugation involved is cyclooctatetraene. The article on the X-ray structure of cycloocta-1,3,5,7-tetrene gives lengths 1.47 and 1.33 angstrom, with their average (1.40) being the exactly value of C-C bond in benzene you provided.


Why and how does hydrogen peroxide decompose in the presence of light?

Chemical reactions in which, a single substance splits up into two or more simpler substances are called decomposition reactions. These reactions are carried out by energy, supplied by different sources. The required energy can be supplied by heat (thermolysis), electricity (electrolysis), or light (photolysis).

Let’s talk about photolysis reactions (not photosynthesis):

  Photolysis (also called photodissociation and photodecomposition) is a chemical reaction, in which a chemical (an inorganic or an organic) is broken down by photons and is the interaction of one or more photons with one target molecule. The photolysis reaction is not limited to the effects of visible light, but any photon with sufficient energy (higher than dissociation energy of the targeted bond) can cause the chemical transformation of the said (inorganic or organic) bond(s) of a chemical. Since the energy of a photon is inversely proportional to the wavelength, electromagnetic waves with the energy of visible light or higher, such as ultraviolet light, X-rays, and $\gamma$-rays, can also initiate photolysis reactions.

Like all other peroxides, hydrogen peroxide ($\ce{H2O2}$) also consists of a relatively weaker $\ce{O-O}$ bond, which is susceptible for light or heat. In the presence of light (the UV light from the sun catalyzes the reaction), $\ce{H2O2}$ spontaneously decomposes into water and oxygen.

The net equation for the reaction is:
$$\ce{ 2H2O2 -> 2H2O + O2}$$

The step-wise reaction mechanism is suggested as follows (Ref.1):

$$\ce{ H2O2 + h\nu -> 2 HO^.}$$
$$\ce{ HO^. + H2O2 -> HO-O^. + H2O}$$
$$\ce{ HO-O^. + H2O2 -> 2 HO^. + H2O + O2}$$

Using isotope studies ($\ce{^{18}O}$ labelled $\ce{H2O2}$), early work has been confirmed that the $\ce{O2}$ formed is cleanly derived from $\ce{H2O2}$ (Ref.2).

Notes: The rate increases rapidly in the presence of catalysts such as $\ce{MnO2}$ and $\ce{KI}$ (Ref.2). The rate of decomposition is slow at room temperature, but it increases with temperature. It is believed to be due to thermal decomposition of $\ce{H2O2}$, which seemingly accelerates the photolysis (Ref.3).

References:

J. P. Hunt, H. Taube, “The Photochemical Decomposition of Hydrogen Peroxide. Quantum Yields, Tracer and Fractionation Effects,” J. Am. Chem. Soc. 1952, 74(23), 5999–6002 (https://doi.org/10.1021/ja01143a052).
A. E. Cahill, H. Taube, “The Use of Heavy Oxygen in the Study of Reactions of Hydrogen Peroxide,” J. Am. Chem. Soc. 1952, 74(9), 2312–2318 (https://doi.org/10.1021/ja01129a042).
F. O. Rice, M. L. Kilpatrick, “The Photochemical Decomposition of Hydrogen Peroxide Solutions,” J. Phys. Chem. 1927, 31(10), 1507–1510 (https://doi.org/10.1021/j150280a004).


This is a really weird question.

Let's start with the general case for a FCC lattice, you'd count up the oxygen atoms per unit cell. In the $\ce{NaCl}$ example, as noted in the comment, there are four $\ce{Cl-}$ per unit cell (6 on the faces, 8 on the corners). Then you'd count up the number of Al atoms per unit cell and divide up the charge as you note above.

In the case of "fractional" oxidation state, it implies that there's a mixture of oxidation states.

Let's say, hypothetically, you had 3 cations and need a +8 charge to balance out the anions. That tells me there are two +3 cations and a +3 cation somewhere. The average is +8/3 but I need to assign formal charges as integers.

But...

Normal $\ce{Al2O3}$ is not an FCC lattice. It's triclinic.

Here, you would do the same thing - count up the number of Al and O atoms per unit cell, considering the fractional volume of the atoms inside the unit cell. It's a bit trickier, since the volumes aren't as obvious as the FCC case.

Instead

This question is about $\gamma$-alumina

Notice that the oxygen atoms actually form an FCC lattice. I have an oxygen at each corner of the unit cell, and one in the middle of each face. So that's 4 oxygen atoms per unit cell = +8 charge from the anions.

How many Al atoms are there?

There are 5 Al atoms completely in the unit cell, plus one per edge (12, each counting 1/4 = 3) so a total of 8 Al atoms!

So that's an average oxidation state of +1 per Al...


Chemical structure of xenon heptafluoride anion

About the geometry and shape:

Source: Housecroft, Sharpe, Inorganic Chemistry, Pearson Education Limited, 2005


Is this online fluorophore database reliable?

That is the correct domain name for the respected and accredited University of Arizona. I would feel confident using their data.

In general, that's the key: being certain that the source is who they say they are, and that they are a source that you have some reason to trust.

Regarding security, it's unlikely someone would spend the time, money and effort required to falsify some fluorescence spectra database.  If it were a database of controversial climate data, or clinical trials for a new medication, etc., you might want to investigate what the site does to secure their data.

I would also be cautious of using open wiki data like Wikipedia. Although I've personally gotten into the habit of trusting those sites more than I should, anybody can edit them. And while they try to police the sites for poorly documented data and such, it's a lot to keep up with. A good wiki article will have sources for their data, and reaching out directly to the source, if possible, is the better way of getting reliable data from a wiki site.


Going by the mechanism of hydrolysis under basic medium, the lone pair of hydroxide ion will attack on C=O carbon.

Due to the negative mesomeric (-M/-R) effect and negative inductive effect of the nitro group, the C=O will act as a more electrophilic substrate as compared to the one which is has ether substituent,

(which due to its +M and +I effect will reduce the electrophilic character of the C=O ),
will be easier to hydrolyse.

The effect of solvation on substituents is less likely to tell which compound will hydrolyse easily as it is not the Rate Determining Step of the reaction. So although the observation is not wrong, its still not the deciding factor.


TL;DR It depends on context and is unfortunately something that isn't going to change in the near future. However, the more you understand angular momentum, the easier it becomes to figure out the meaning from the context.

Quantum mechanical angular momentum is a tricky topic and it is not helped by some books which use confusing nomenclature. In QM, angular momentum is a vector quantity (much like in classical mechanics).* This is denoted with various letters;$^\dagger$ we will stick with J for the time being:

$$\text{angular momentum: }\mathbf{J} = (J_x, J_y, J_z)$$

Alternatively, you may also see vectors denoted with an arrow sign: $\vec{J}$.

All three components of the (QM) angular momentum are observable quantities. They are more commonly referred to as projections of the angular momentum onto an axis, which means the same thing. In any case, each one is associated with a Hermitian operator :

$$\text{ang. mom. projection operators: }\hat{J}_x, \hat{J}_y, \hat{J}_z$$

Apart from the $x$-, $y$-, and $z$-components of the angular momentum, the magnitude of the angular momentum is also an observable. This is similar to the length of the vector $\mathbf{J}$, and classically we would write $|\mathbf{J}|^2 = J_x^2 + J_y^2 + J_z^2$. The corresponding quantum mechanical operator is usually written without the boldface as

$$\text{ang. mom. magnitude operator: }\hat{J}{}^2 = \hat{J}{}_x^2 + \hat{J}{}_y^2 + \hat{J}{}_z^2$$

So, now, we have four unique operators: $\hat{J}{}^2, \hat{J}_x, \hat{J}_y, \hat{J}_z$. Because they do not all commute with one another, we usually focus on two of them, which are (conventionally) $\hat{J}{}^2$ and $\hat{J}_z$, and ignore the other two.

The quantum mechanical states of interest are those which are simultaneous eigenstates of these two operators, namely $\hat{J}{}^2$ and $\hat{J}_z$. It turns out that these eigenstates have the properties

\begin{align}
\hat{J}{}^2|\psi\rangle &= s(s+1)\hbar^2 |\psi\rangle \\
\hat{J}_z|\psi\rangle &= t\hbar |\psi\rangle \\
\end{align}

where $s$ is constrained to be a non-negative integer $0, 1, 2, \ldots$ and $t$ is constrained to be $-s, \ldots, -1, 0, 1, \ldots, s$. Different eigenstates will have different values of $s$ and $t$, which immediately suggests that we should label the eigenstates $|\psi\rangle$ by the values of $s$ and $t$ that apply to them. Constants such as $s$ and $t$ which can only take on certain values are called quantum numbers.

We can therefore associate the eigenstate $|\psi_{s,t}\rangle$ with the angular momentum $\sqrt{s(s+1)}\hbar$ (because the angular momentum squared was $s(s+1)\hbar^2$). We could write this as:

$$J_s = \sqrt{s(s+1)}\hbar$$

which says that the (magnitude of the) angular momentum $J$ of a state with quantum number $s$ is equal to $\sqrt{s(s+1)}\hbar$. In the rigid rotor system, we ignore $\hat{J}_z$ and instead look at another observable, which is the energy. The energy of a state with quantum number $s$ is:

$$E_s = \frac{\hbar^2}{2I}s(s+1)$$

where $I$ is the moment of inertia. So far, so good. But confusingly, instead of using innocuous labels like $s$, the convention is to use $J$ instead. And $t$ is replaced with $M$ (sometimes $M_J$), which at least is slightly less confusing.

\begin{align}
J_J &= \sqrt{J(J+1)}\hbar \\
E_J &= \frac{\hbar^2}{2I}J(J+1)
\end{align}

And therefore we have a situation where $J$ can mean several things, and it completely depends on context. To be extra clear, I'll colour the $J$ which means "the value of angular momentum" in red, and the $J$'s that refer to the quantum number in blue:

\begin{align}
\color{red}{J}_\color{blue}{J} &= \sqrt{\color{blue}{J}(\color{blue}{J}+1)}\hbar \\
E_\color{blue}{J} &= \frac{\hbar^2}{2I}\color{blue}{J}(\color{blue}{J}+1)
\end{align}

and the corresponding eigenvalue equations (recall that the operator for energy is always denoted $\hat{H}$, for the Hamiltonian):

\begin{align}
\hat{\color{green}{J}}{}^2|\color{blue}{J}\rangle &= \color{blue}{J}(\color{blue}{J}+1)\hbar^2|\color{blue}{J}\rangle \\
\hat{H}|\color{blue}{J}\rangle &= \frac{\hbar^2}{2I}\color{blue}{J}(\color{blue}{J}+1)|\color{blue}{J}\rangle
\end{align}

Here the green colour indicates an operator; the red in the previous equation was not an operator, but rather the quantity of the observable. Some texts will help you out by putting a hat on operators, but often it is omitted (either because authors are lazy or because it gets really cluttered after a while), so be careful!

* Technically a pseudovector.

$^\dagger$ $J$ is usually used for the total angular momentum of a system. In classical mechanics, there is only really one type of angular momentum, but in quantum mechanics there is orbital angular momentum and spin angular momentum.

Orbital angular momentum is the one which is much more closely related to the classical idea of angular momentum (for example, the orbital angular momentum operators are directly derived from $\mathbf{L} = \mathbf{r} \times \mathbf{p}$), and is typically denoted with $L$, not $J$. The quantum number is usually (helpfully) denoted $l$, although some books will also use $L$.

Spin angular momentum, on the other hand, is denoted with $S$. It has no classical analogue and has even weirder behaviour in that the associated quantum number $s$ can take on half-integer values, i.e. $0, \frac 1 2, 1, \frac 3 2, \ldots$

In quantum mechanics, the total angular momentum $J$ is the sum of both these two contributions. In the rigid rotor model we are not concerned with the spin of the particle(s), so $J$ is simply equal to the orbital angular momentum $L$, and there is no ambiguity about which is which. But if you start thinking about particles like nuclei or electrons, then it becomes important to distinguish between the two.


Why does the d orbital size decreases on addition of electrons?

There is not much to explain because the concept is wrong. There is no sulphur compounds in which the 3d orbitals (remember: they have an energy level similar to 4s) are involved in bonding in any noteworthy way, shape or form.

From what I believe I may remember about the outdated concept, ‘one d orbital’ refers to cases like $\ce{SO3^2-}$ with sulphur in the $\mathrm{+IV}$ oxidation state while ‘two d orbitals’ would refer to e.g. $\ce{SO4^2-}$ or an oxidation state of $\mathrm{+VI}$. In the Lewis formalism, the answer is easily deduced if you draw octet-abiding structures: $\ce{SO3^2-}$ will include a formal single positive charge on sulphur while in $\ce{SO4^2-}$ sulphur has a formal double positive charge. The higher an oxidation state, the more contacted an atom becomes.

A more correct answer would elaborate on ($\mathrm{sp}^n$-type, formed only by sulphur’s s and p atomic orbitals) molecular orbitals and the increasing number of electronegative substituents that need to bond to the central sulphur, reducing its charge density and leading to the same effect.


What additive can I add to Jelly Wax/Gel Wax to increase its cooling time/setting time?

At your own risk, you can try using plasticizers for waxes. Usually, these plasticizers are wax esters of palmitic acid (well-known example is spermaceti, which can be replaced with jojoba oil, or you can buy/obtain those esters). But it's hard to predict exact amount of plasticizer needed. Add too much - and it will be creamy I-will-never-cool substance. I hope we have chemical engineers here who know more about wax plasticizers and can add details to my superficial statement.


What is the oxidation state of C in this lewis structure?

This answer uses electronegativities for the calculation of oxidation states as proposed in the Expanded Definition of the Oxidation State by Hans-Peter Loock in 2011.

Foreword: As mentioned in the OPs comments, electronegativity and oxidation states are atom properties, not group properties.

Comparing Electronegativities

This table shows an excerpt from Pauling electronegativities ($\chi_{\mathrm{Pauling}}$):

$$\begin{array}{cc}
\hline
\text{Element} & \chi_{\mathrm{Pauling}}\\
\hline
\ce{H} & 2.20\\
\ce{C} & 2.55\\
\ce{O} & 3.44\\
\hline
\end{array}$$

Hence, the following applies:

$$\chi_{\mathrm{Pauling}}(\ce{H}) < \chi_{\mathrm{Pauling}}(\ce{C}) < \chi_{\mathrm{Pauling}}(\ce{O})$$

Looking at heteroatomic Bonds ($\ce{A-B}$, $\ce{A=B}$, and $\ce{A#B}$)

Since the oxygen atom has a higher electronegativity than the carbon atom, it attracts the bonds atoms more, resulting in an polarized bond. Presuming one would apply pulling force to the two atoms in a $\ce{C-O}$ or $\ce{C=O}$ bond until the atoms separate, it is assumed that the bond cleavage is heterolytic, thus allocating the bonding electrons to the oxygen atom.

The same applies to $\ce{C-H}$, $\ce{O-H}$, or any other heteroatomic bond ($\ce{A-B}$, $\ce{A=B}$, and $\ce{A#B}$).

Putting the jigsaw puzzle together

Now, having a look on the complete structure of the molecule in question, and applying the aforementioned rule to each bond:

Last thing to do is calculating the atoms hypothetical charge after separation, which is to be equatable with the oxidation state:

$$ \text{Oxidation state} = N_\mathrm{i}(\ce{e-}) - N_\mathrm{f}(\ce{e-}) $$

With $ N_\mathrm{i}(\ce{e-}) $ representating the number of electrons in a free atom, and $ N_\mathrm{f}(\ce{e-}) $ the one after separation (Never forget the lone pairs).

This will lead to the following oxidation states for each atom:

$$\begin{array}{cccc}
\hline
\text{Atom} & N_\mathrm{i}(\ce{e-}) & N_\mathrm{f}(\ce{e-}) &\text{Oxidation state}\\
\hline
\ce{H-1} & 1 & 0 & +1\\
\ce{H-5} & 1 & 0 & +1\\
\hline
\ce{C-2} & 4 & 2 & +2\\
\hline
\ce{O-3} & 6 & 8 & -2\\
\ce{O-4} & 6 & 8 & -2\\
\hline
\end{array}$$

The structure with assigned oxidation states:

The oxidation state of the carbon is $\mathrm{+II}$.


Can water be produced by mixing hydrogen gas and oxygen?

Indeed, it's called combustion! When you burn hydrogen and oxygen gas, the reaction is:

$$\ce{2H2_{(g)} + O2_{(g)} -> 2H2O_{(g)} + energy}$$

Basically, you could see the former U.S. space shuttle as a large machine that turns hydrogen and oxygen into water. (Note: the space shuttle used liquid hydrogen and oxygen, but otherwise the principle and formula is exactly the same)

Since the Gibbs free energy of the forward reaction is negative ($-242\textrm{ kJ/mol}$), you don't need to add energy to the system, the reaction is spontaneous (mind you, you generally need a spark to ignite the combustion because there's a kinetic barrier to the reaction called the energy of activation).

The reverse reaction, water to hydrogen and oxygen, has a positive Gibbs free energy, so you need to add energy in the form of electricity to the system to make the reaction happen.


Why is the methyl group electron-donating via the inductive effect?

According to http://www.chem.ucalgary.ca/courses/351/Carey5th/Ch04/ch4-3-2-1.html:

"Hyperconjugation is the stabilising interaction that results from the interaction of the electrons in a σ-bond (usually C-H or C-C) with an adjacent empty or partially filled p-orbital or a π-orbital to give an extended molecular orbital that increases the stability of the system."

How I see "hyperconjugation" is the formation of partial π bonds. Let's say we have a methyl group bonded to a positively-charged carbon atom with a vacant p orbital.

Ideally, it would be best if we could have an atom with a fully-filled p orbital, which is parallel to the p orbital of the this positively-charged carbon atom, to come and interact with this carbon atom. Such an interaction would stabilise the electron-deficient carbon atom. This is essentially the basis of hyperconjugation.

However, in our case, we do not have such an atom with a fully-filled p orbital around. Instead, what we have is a few C-H σ bonds. These C-H bonds are not parallel to the vacant p orbital of the positively-charged carbon atom but they are very close to parallel, around 19.5 degrees away, assuming that the geometry of the methyl group bonded to that positively-charged carbon atom is tetrahedral.

These C-H bonds are thus able to interact with the electron deficient vacant p orbitals of the positively-charged carbon atom, maybe not maximally, but sufficiently. These interactions would be what I would call "partial π bonds" and they stabilise the electron deficiency of the carbon atom. This interaction is also what is known as "hyperconjugation".

*Please note that the illustration is slightly inaccurate because the hydrogen atom would not pop off the methyl group for the formation of a full π bond between the two carbon atoms. This is only illustrated to better show the interaction of the C-H σ bonds with the p orbital of the positively-charged carbon atom.


When to use -ate and -ite for naming oxyanions?

"Most common" means the forms most frequently encountered in the entire scope of chemistry, which I'm pretty sure corresponds loosely to the most thermodynamically stable oxidation state. It is not always the case that the '-ate' ion is the most stable, however—as an example, see the links for chlorine below).

A detailed answer necessitates noting that the nomenclature extends beyond just '-ate' and '-ite'.  Most p-block elements that form oxo-anions form a whole series of them, each with the central atom in an oxidation state two away from each neighbor.  For all elements except the lightest ($\ce{C}$ and $\ce{N}$) and heaviest ($\ce{As}$, $\ce{Se}$, $\ce{Te}$, etc.) , usually the series is considered to contain four members, though it is not guaranteed that all are stable or able to be characterized (e.g., "bromite" and "hyposulfite").

As noted in the comments to another answer, in addition to the '-ite' and '-ate' suffixes, there are also the 'hypo-' and 'per-' prefixes, where 'hypo-' only is used with '-ite', and 'per-' is only used with '-ate'.  These four combinations are used to span the series of oxo-anions for each element:

Phosphorus: perphosphate ($\ce{PO5^{3-}}$), phosphate ($\ce{PO4^{3-}}$), phosphite ($\ce{PO3^{3-}}$), hypophosphite ($\ce{H2PO2^{-}}$)
Sulfur: persulfate ($\ce{SO5^{2-}}$), sulfate ($\ce{SO4^{2-}}$), sulfite ($\ce{SO3^{2-}}$), hyposulfite ("$\ce{SO2^{2-}}$")
Chlorine: perchlorate ($\ce{ClO4^{-}}$), chlorate ($\ce{ClO3^{-}}$), chlorite ($\ce{ClO2^{-}}$), hypochlorite ($\ce{ClO^{-}}$)
Bromine: perbromate ($\ce{BrO4^{-}}$), bromate ($\ce{BrO3^{-}}$), bromite ("$\ce{BrO2^{-}}$"), hypobromite ($\ce{BrO^{-}}$)
Iodine: periodate ($\ce{IO4^{-}}$), iodate ($\ce{IO3^{-}}$), iodite ($\ce{IO2^{-}}$), hypoiodite ($\ce{IO^{-}}$)

For carbon, 'carbonate' was given preference, per the 'most common' rubric, as the only known oxo-anion. For nitrogen, the use of the prefixes was avoided, for what I assume was the sake of simplicity.  (I would argue that nitrate and nitrite are both common enough in the natural world that without such a deciding factor there would have been stiff competition for the '-ate' suffix.)

Possible irregularities in structure within a series include peroxo-anions at high numbers of bound oxygens (e.g., I believe persulfate and "perphosphate" are both peroxo species) and variable numbers of bound oxygens for a given oxidation state in anions of heavier elements (e.g., metaperiodate, $\ce{IO4-}$, versus orthoperiodate, $\ce{IO6^{5-}}$, both of which contain heptavalent iodine). As well, there is the possibility for other, non-oxo-anion compounds of the central atoms with oxygen such as chlorine dioxide, $\ce{ClO2}$; nitrogen dioxide, $\ce{NO2}$; sulfur dioxide, $\ce{SO2}$, and trioxide, $\ce{SO3}$; and (of course) carbon dioxide, $\ce{CO2}$.

You can find references to oxo-anions of arsenic, selenium, antimony, and tellurium also (links are to the '-ates'), but to my mind, these tend to edge more toward the behavior of oxo-anion-forming metals, which either only have a single appreciably stable oxo-anion (chromate, molybdate, tungstate, etc.), or break this (hypo-)-ite/(per-)-ate paradigm pretty badly (see, e.g., permanganate, $\ce{MnO4-}$, versus manganate, $\ce{MnO4^{2-}}$).

If you really want to blow your mind, take a look at the polymeric oxo-anions like polyphosphate; or at the at-present theoretical-only orthocarbonate; or at iron, for which apparently all three of the known oxo-anions are referred to as 'ferrate'.


In a condensation reaction, which of the reactants loses their OH group?

Acid loses its OH and alcohol loses its H, that's the way it is.

Long story short, oxygen from alcohol's $\ce{-OH}$ attacks the carbon from the acid's $\ce{-COOH}$. That carbon, being linked to two oxygens, carries significant positive charge, so this is much more likely than the other way around.

Also, it has been proven rigorously with the use of isotope-labeled compounds. Say, you take some acid with "normal" oxygen ($^{16}\ce O$) and alcohol with $^{18}\ce O$, and behold, the ester comes out with one $^{16}\ce O$ and one $^{18}\ce O$, so the latter must have come from alcohol, so it is the oxygen from acid that was lost in the form of water.


The question was answered on the Open Babel mailing list

Suffice to say, there's no multiplication symbol in SMILES or SMARTS - only standard ASCII characters. It's likely a lower-case 'x' that's been mangled by a journal.


Your generation of electrophile is correct the subsequent primary carbocation formation by the attack of electrophile is incorrect.

  In my second step, that particular carbocation is formed since $\ce{Ph}$ is an electron-withdrawing group.

The benzene ring  stabilizes the carbocation through mesomeric effect.

The electrophile will attack at $\alpha$ position rather than $\beta$


Why do Allylic Halides prefer SN2 reaction over SN1?

You and your instructor are both correct, it is your premise that's wrong.

Your instructor's explanation about the allyl double bond stabilizing the $\mathrm{S_{N}2}$ transition state is correct. The following figure shows the stabilizing interaction.

image source

If we remove the attacking nucleophile and the leaving group and place a positive charge in the system, then we have the $\mathrm{S_{N}1}$ allyl intermediate.  As you point out, this intermediate is resonance stabilized, and therefore, the transition state leading to it is also stabilized.

Your premise,

  Why do Allylic Halides prefer SN2 reaction over Sn1?

is not true.  There are many cases where allylic halides react preferentially by an $\mathrm{S_{N}1}$ process.

Since both the allylic $\mathrm{S_{N}1}$ and $\mathrm{S_{N}2}$ reactions are stabilized, there is a delicate balance between the two pathways.  We can shift from one mechanism to the other by changing reaction conditions.

For example, allyl chloride reacts by an $\mathrm{S_{N}1}$ mechanism to produce allyl alcohol when we place it in a 50:50 mix of $\ce{H2O:EtOH}$ at 45°C.  If instead we run the reaction in the presence of ethoxide ion in ethanol at 45°C, the corresponding ethyl ether is formed by an $\mathrm{S_{N}2}$ process.


Concentration of hypochlorous acid and hypochlorite

You really overcomplicate things (as I did initially, because I thought you were talking about the dissociation of $\ce{Cl2}$ gas in water and the subsequent reactions).

Here are the basic pieces of information needed:

Reaction equation
Equilibrium equation
Mass balance equation

Reaction Equation

The reaction is a simple acid-base equilibrium reaction:
$$ \ce{HOCl <=>[K_\text{a}] H+ + OCl-} $$

Equilibrium Equation

The equilibrium equation follows easily from the reaction equation:
$$ K_\text{a} = \frac{[\ce{H+}][\ce{OCl-}]}{[\ce{HOCl}]} $$

Mass Balance Equation

The total amount of chlorine is calculated as follows:
$$ [\ce{Cl}]_\text{tot} = [\ce{HOCl}] + [\ce{OCl-}]$$

From the reaction equation also follows
$$ [\ce{OCl-}] = [\ce{H+}] $$

The Solution

Since the concentration of protons is known, we can easily calculate the concentration of the hypochlorite anion:
$$ [\ce{OCl-}] = [\ce{H+}] = 10^{-\text{pH}} = 10^{-8} $$

Since we know the total concentration, simply reforming the mass balance gives us the concentration (strictly speaking, the activity) of the hypochlorous acid:
$$ [\ce{HOCl}] = [\ce{Cl}]_\text{tot} - [\ce{OCl-}] = 0.1 - 10^{-8} = 0.09999999 \approx 0.1 $$

Slap on some units and we're good to go!

$$ \begin{align}
c(\ce{HOCl}) &\approx 0.1~\mathrm{mol\, L^{-1}} \\
c(\ce{OCl-}) &= 10^{-8} ~\mathrm{mol\, L^{-1}}
\end{align}$$


The cutoff keeps defaulting to 0.05 in Jmol

The syntax is

isoSurface id saSurface radius

So you should use

isoSurface id saSurface 0.008

You can use the optional id for giving it a name.

You may also want to see their reference page

Edit 1: Have you tried using DX files ?
I am not an expert in using DX to render the required results but here is the reference  :

  DX, as Cube, defines a three-dimensional grid of points in space. At
  each point is a number (a "scalar value"). This set of point values is
  then used by Jmol to define an "isosurface" -- the surface separating
  points having values greater than a given cutoff from those that have
  a value less than a certain cutoff. A typical application is molecular
  orbitals. The phase parameter of the isosurface command allows bicolor
  rendering: one color for "points greater than x" and another color for
  "points less than -x". Jmol can read DX files and re-export them to
  much smaller JVXL files.

Quoted from here


Why there is no bichloride or binitrate ion like bicarbonate or bisulphate?

Okay, you want simple, here's a real simple way to put it.

How many limbs does a dog have? 4 right?
Now if it loses one of its limbs, a prosthetic one can be placed,(We'll get to this: with its whole body, being able to cope with the leg). Now consider the case of a human being, same scenario, the same thing can be done, since we can place the limb in the "slot"(somebody put a better word here), where the leg previously was(although, the limb cant do everything a leg couldve done). Now consider the case of a creature with just one limb, if it loses the sole limb it had, life's gonna be hard, simply because of the amount of discomfort its going to face while adapting to the limb. Similarly, sulphate ($\ce{SO4^2-}$) and carbonate ($\ce{CO3^2-}$) have a valency greater than 1. Hence can accept an H+ ion and still exist as a stable specie, and further accept a metal ion;

In the case of $\ce{HCl}$, it has a valency of just 1, hence upon accepting an $\ce{H+}$ to get the "bi" tag to its name, it has lost its charge, and now is incapable of accepting a metal ion to form a salt(while keeping the $\ce{H+}$ to itself), as it has simply become hydrochloric acid.


Which electronic effects are responsible for the reduction of nucleophilicity of a hydroxy group orthogonal to an ester?

If we focus on the 2-oxo-3s-hydroxy-tetrahydrofuran the electronic  effects present are the hyperconjugation and the dipole minimization. "Both effects contribute to the preferred (Z)-conformation of esters over the (E)-conformation. In the (Z) conformation, the lone pair of electrons in the alpha oxygen can donate into the neighboring σ* C-O orbital. In addition, the dipole is minimized in the (Z)-conformation and maximized in the (E)-conformation. But this electronic effects occur between the carbonyl carbon and the oxygen of the ring but the hydroxyl group." [a]

In contrast, I calculated the PES of hydroxy group in the following two systems:

2-oxo-3s-hydroxy-tetrahydrofuran [S1]
(1S,2S)-isopropyl 1-hydroxy-2-((vinyloxy)methyl)cyclohexanecarboxylate

The latter one is:

The oxygen of iprO-(C=O)- group was pointed away from [S2] and towards to [S3] cyclohexane. The PES is: (It was calculated with am1 due to computational resources, it was done with my laptop.)

It can be observed that the intramolecular hydrogen bond (IHB) in [S3] is more stable than in the other systems. The order of intramolecular hydrogen bond preference is [S3]>[S2]>[S1]. In other words, the IHB formed with the oxygen SP$^3$ of iprO-(C=O)- group is preferred over the carbonyl oxygen. [b] This can be seen on the rotational barrier between both minima. This conformational preference allows the lone pair of hydroxyl oxygen to be more nucleophilic than in the other systems. As it can be seen in the following figures:

[S1]

[S2]

[S3]

Related to the electronic effects present in this IHB, it can be complemented that "this is a consequence of interacting the lone pair on the accepting oxygen with an empty hydride antibond of the donating O-H atoms, $n_O$ → $\sigma^*OH$. Occupancy of the antibonding orbital raises the energy of the O-H species while stabilizing the oxygen." As Albrecht et. al. concluded in her paper.[c]

Bibliography

[a] Cuevas, Eusebio Juaristi, Gabriel (1995). The anomeric effect. Boca Raton: CRC Press. ISBN 0-8493-8941-0.
[b] Rincon et al. J Solution Chem (2011) 40: 656.
[c] Albrecht et al. J. Phys. Chem. A. (2012) 116: 3946


Ordering the reduction potentials for Cr3+, Mn3+ and Fe3+?

Firstly, I'd like to say the explanations we propose here are merely our efforts trying to explain what we observed experimentally in the first place. Let's get some data:

$${\begin{array}{|c|c|} \hline \ce{Cr^3+/ Cr^2+} & \ce{Mn^3+ / Mn^2+} & \ce{Fe^3+ / Fe^2+} \\ \hline \pu{-0.41 V} & \pu{+1.57 V} & \pu{+1.97 V} \\ \hline \end{array}}$$

And while we're at it, let's ponder on electronic configurations as well:

$${\begin{array}{|c|c|} \hline \ce{Mn^2+} & \pu{[Ar] 3d5 4s0} \\ \hline \ce{Fe^3+} & \pu{[Ar] 3d5 4s0} \\ \hline \end{array}}$$

You were correct in thinking that $\ce{Fe^3+}$ will have the highest reduction potential due to the third ionization enthalpy coming into play, all we've got to do now is to think for manganese and chromium. Why don't we put it differently:

  Why is $\ce{Cr^2+}$ reducing and $\ce{Mn^3+}$ oxidizing when they both have the same $\pu{d4}$ configuration?

It's because if manganese acts as an oxidizer, it'll get converted into $\ce{Mn^2+}$, which is stable owing to the $\pu{d5}$ configuration. Chromium, on the other hand is notorious for showing the $+3$ oxidation state in its compounds. So, that's a win-win kind of situation we're dealing with.

The subtle role played by the solvent molecules when they're acting as ligands comes into limelight on asking yet another relevant question:

  Which is a stronger reducing agent: $\ce{Cr^2+}$ or $\ce{Fe^2+}$?

Answer: $\ce{Cr^2+}$. This is because of more CFSE stabilization of $\ce{Cr^3+}$ as compared to $\ce{Fe^3+}$ when water molecules behave as a ligand for our complex.


Spin orbital coupling and total angular momentum

J is the total angular momentum. L is the orbital angular momentum and S is the intrinsic total electron spin angular momentum. J = L + S
L = $\sqrt{L\left(L+1\right)}\frac{h}{2\pi }$
--> L can be n-1 where n is the Principal quantum number
S = $\sqrt{S\left(S+1\right)}\frac{h}{2\pi }$
--> S is an integer or half an odd integer, depending on whether the number of electrons is even or odd
J = $\sqrt{J\left(J+1\right)}\frac{h}{2\pi }$
J can be L+S ... L-S
For the  Na-D lines the J = 3/2 is caused by L=1 and S=1/2 for the 3/2 value and L=1 and S=-1/2
The values for $J,L,S$ have a range of values e.g. $-X, -X+1,\cdots X-1, X $, e.g. if $L=5/2$ the range is $-5/2,-3/2,3/2,5/2$


How to obtain nitrogen using only homemade utilities?

Crude process, but does the trick: You can use a combustion/oxygen consuming process, like a candle, to burn the excess oxygen. Eventually the candle will be too oxygen starved to sustain a flame, so some residual oxygen would be left (although this would already mean there is no fear of combustion of said tyres). Now to remove the remaining oxygen you'd use something like iron files (since you wanted to use only "household" stuff). Together with some humidity in that air, they will rust rapidly and should drive down the oxygen contents towards negligible. So now we have:

a small trace remainder of oxygen,
the equivalent of $\ce{CO2}$ from the combustion process,
some impurities from the candle wax probably,
trace gases such as helium etc. and ... mostly
nitrogen.

This already is not sustaining any combustion. But if you're a purist or want to use $\ce{N2}$ because it is lighter than $\ce{CO2}$, to remove the $\ce{CO2}$ you could use something like "roasted limestone", i.e. the stuff from you water kettle, which you heat so that you get lime, which then binds with the $\ce{CO2}$ to revert to limestone $\ce{CaCO3}$. Now you have something, I'd assume, like 98% nitrogen and surely to few traces of $\ce{O2}$ to cause any harm to your equipment. Of course, I'd buy bottled $\ce{N2}$, but for the sake of using household implements only, this might do the trick.


Negative ions and health: pseudo science or something more?

Charge separation at waterfalls with airborne ions, resulting in a potential between the base of the waterfall and the surrounding air, is a phenomenon examined by Philipp Lenard (Über die Elektrizität der Wasserfälle, Ann Phys (Leipz), 1892, 46, 584–636).

The effect is real and more recent field studies, such as Characterization of ions at Alpine waterfalls, as well as studies on the formation of charged droplets, such as Charge Separation in the Aerodynamic Breakup of Micrometer-Sized Water Droplets should be taken seriously.

However, the sources that you cited are pseudoscientific marketing bullshit.

The first quote links to a web shop that offers marvels like magnetic bands and ion-balance bands. While the silicone band probably isn't hamful to the skin, it will not emit anything. Nada, nothing, nix!

The webshop behind the third link sells Himalayan salt lamps. It's a matter of taste to have a lamp housing made from unrefined sodium chloride with traces of iron in it (hence the colour), but I wouldn't expect any health effects through the emission of whatever from these lifestyle items. Himalayan salt is being sold for quite a while, even for nutrition but if at all, the stuff has seen the Himalaya mountains only from a distance.

There are no halite mines in the Himalaya mountains.

Instead, the material is likely to come from the Salt Range in the Punjab province in Pakistan, or from Poland.

In summary, it's business as usual: Scientific facts are mashed up with esoteric mumbo jumbo to sell overpriced and useless stuff to well-meaning but naive clients.

Oh what a tangled web we weave,
When first we practise to deceive!
-- Sir Walter Scott


The Salt Room, is there salt in the air?

In short, no. The vapor pressure of salt is trivial compared to the grams of salt you would put on a steak when cooking it. I think your analysis is good- the most likely impact is that the salt modulates the humidity.


Yours is a common source of confusion. Reactions are can be selective but they are not specific. What's specific is a mechanism for the reaction. If the reaction only proceeds via one mechanism, then it is extremely selective.

Stereospecific. The mechanism provides a specific stereochemical outcome or proceeds from a specific stereochemical configuration. Examples:

a. Pd-catalyzed hydrogenation and hydroboration proceed via syn addition. These mechanisms are stereospecific because they indicate always adding to the same side of an olefin.

b. $\mathrm{E}_{2}$ elimination processes require that the hydrogen and leaving group are anti-periplanar. This mechanism is stereospecific.

c. Acid-catalyzed hydration proceeds via a planar carbocation intermediate where previous stereochemistry might be lost. This mechanism is not stereospecific.
Stereoselective. The reaction provides more of a product with a specific stereochemistry than any of the other possibilities.

a. Allylic alcohol directed epoxidation reactions frequently yield different amounts of the two product epoxides that result from reacting on the different faces of the olefin. This reaction is selective.

b. Asymmetric reactions that rely on pathways through diastereomeric transitions states are stereoselective.

c. Acid-catalyzed hydration could be stereoselective because the molecule can have other stereocenters that provide diastereoselectivity, even though the base mechanism disregards stereochemistry.
Regiospecific. The mechanism indicates a certain connectivity that should be obtained in the product.

a. Exo- versus endo- intramolecular reactions have regiospecific mechanisms.

b. A bit of a stretch, but carbonyl addition happens at the carbon. The mechanism is regiospecific.
Regioselective. The reaction favors one connectivity pattern over the other.

a. Hydrobromination favors the Markovnikov product, while hydrobromination in the presence of peroxides favor the anti-Markovnikov product.

b. In intramolecular ring closing reactions, 5-exo-tet products are preferred over 6-endo-tet products.

c. Carbonyl addition happens at the carbon. The only mechanism in operation is the one mentioned above, so the reactions are regioselective.

Importantly, reactions should not be stereospecific or regiospecific, even if they give only one product. This is a concrete case of a reaction proceeding exclusively (or almost exclusively) via a single mechanism that is stereospecific or regiospecific, respectively. The reaction itself is, respectively, stereoselective or regioselective.


Does gas chromatography follow the "like interacts with like" rule of thumb?

For GC, the most significant factor affecting separation resolution is differences in vapor pressures, boiling points if you will, between analytes.  While column phase selection, oven temperature programming, detector type and temperature, and other factors will have an effect on the efficiency of separation, IMFs are not generally important for the majority of separations.  The "like interacts with like" principle does apply to GC such that in the case you describe, nonpolar analytes will spend more time interacting with the stationary phase while polar will have little to nearly zero interaction and elute first.  The separation of the various nonpolar compounds will, holding all the other column conditions constant, "boil down" to varying vapor pressures.

Edited for clarification:

To begin, I think it beneficial to keep in mind that the question you pose is sufficiently clear we are dealing with polar analytes using a non-polar GC column.  Regarding proprietary column selection, temperature programming, and the choice of various other method parameters, these things are necessarily important in an efficient and successful separation (generally called method development) but they do not do much to answer the principle questions asked about whether 1) polar compounds are less retained on a non-polar column, and 2) does "like with like" apply to GC.  To that end, I will again address these two primary questions and then try to clarify another issue brought into the discussion that does not bear directly on answering the questions as asked.

The most direct answer to your first question is simply, yes, polar analytes do have shorter retention times than non-polar compounds when using a non-polar column because polar compounds have little interaction (partitioning) with the column's stationary phase.  The polar molecules of interest have little affinity for the non-polar stationary phase, i.e. hydrophobicity.  Had it been a polar stationary phase column then IMFs would be significant.  You mention making things comparable by considering compounds of similar boiling points which is precisely why a ramping temperature program is most beneficial for efficiently resolving such a separation.  This too, however, is a method development consideration and thus an example of the applied science of gas chromatography taking advantage of pure science principles.

Considering your second question, this, too, is most directly answered with another yes - the "like with like" concept does apply here, but as you point out, this is a rule of thumb, and as such, a conceptual aid not strict principle.  The qualification about hydrophobic effects in the stationary phase you make in your wording of the question actually explains, as well as I could, what is happening.  Also, as @Blaise points out, the statement with which you end your original question referencing a previous user (I [remember] reading that the interaction between a polar molecule and non-polar molecule is stronger than the interaction between two non-polar molecules.), is in error: the interaction between polar and non-polar molecules is not stronger than between two non-polar molecules.  Perhaps what this person meant is the interaction between two polar molecules is stronger, and relatively much stronger, than interactions between two non-polars.

With regard to your request I clarify why IMFs are not generally important (in this case), it must be remembered that the analytes in GC are not in "pure liquid form" but, rather, in the gas phase and they are completely surrounded by inert carrier gas molecules - little chance for IMFs to be of significance.  The polar molecules to be resolved do not spend a significant time partitioning into the stationary phase so they cannot form immiscible droplets as you brought up.

With respect to @SteffX, your question is about GC specifically and is not a question of what is the best separation method to use.  I take your question to be about using GC (on small sample sizes, right or wrong on my part) to effect a separation for analytical purposes, not as a preparatory separation like distillation.  Also, your original question(s) are with regard to gas chromatography in general and make no reference to column packing, bore, or length (other than being of a non-polar stationary phase) and I tried to answer in general terms.  However, since capillary GC columns were brought into the discussion, I think it deserves clarifying that, in fact, most capillary column separations are not isothermal.  Without going too deeply into something you were not asking about, it should be noted that isothermal temperature separations are a simpler and therefore preferred method development as long as you have good analyte resolution; the more common practice is to program a temperature gradient to resolve similar boiling point compounds - part of your original Question 1.  Lastly, Van der Waals force(s), a subject for much definitional discussion in its own right, have little importance to your polar analytes at GC conditions, and are moot almost before the sample hits the injector.

I hope this has been helpful and, in my opinion, I think you already have a good understanding of this.


Does more branching lead to lower boiling point in alkanes?

$A$ n-hexane, boiling point 69 degrees C.

$B$ 2,3 dimethyl butane, boiling point 58 degrees C.

$C$ 3-methyl pentane, boiling point 63 degrees C.

$B<C<A$


Can I just clarify, without going into the detail of the specific problem, if you have Cu(s) in solution (like you do here) can it ONLY be oxidised? So essentially you're looking for whether there's anything in solution that can oxidise Cu?

Let us take a longer approach to address your query. For electrochemical problems like these, first you have to make a list of starting materials and possible products. Do a thought experiment even before even looking up the tables. It is long way but it will help you in doing future problems.

Your question is: What happens when a piece of copper is placed in 1M HCl?

Do a thought experiment +  use some basic intuitive chemistry.

Making a list of starting materials, a) Cu metal, b) H$^+$, c) Cl$^-$.

Now there are only two possibilities for each substance.

a) Copper Cu can be oxidized to Cu$^+$ or Cu$^{2+}$ or reduced to Cu$^-$. Basic chemistry tells us that metals like to form cations so Cu$^-$ can be eliminated. Similarly, there is possibility for Cu$^+$ or Cu$^{2+}$. However, say, you are just interested in Cu$^{2+}$.

Now you have the problem well defined: Will Cu -> Cu$^{2+}$?

b) Follow the same reasoning: H$^+$, it can be oxidized to  H$^{2+}$ or reduced to H$_2$. Basic chemistry will tell you that H$^{2+}$ is not possible. Now your question is more well defined: Will H$^+$ -> H$_2$

c) In the same way, ask the same question for chloride ion. It can be oxidized to Cl$_2$ or reduced to  Cl$^{2-}$. Basic chemistry would tell you that Cl$^{2-}$ is not feasible. So your only concern is: Will Cl$^-$ -> Cl$_2$

Now you can ask only two questions:

Can copper metal reduce H$^+$ to H$_2$?

or

Can copper metal oxidize Cl$^-$ to Cl$_2$? This can be easily eliminated because in order for copper to oxidize it must reduce itself further, which is not possible.

At this stage you can utilize the electrode potentials. Recall cathode refers to reduction and anode refers to oxidation
Ecell= E$_{cathode}$-E$_{anode}$

Ecell= E(for hydrogen half cell bc it is being reduced) - E (copper half cell)

Ecell= 0.00- (+0.34) = - 0.34 V

The negative sign shows this not possible under these conditions.

Do the same for chloride ion.

Ecell= E(Reduction of Cu to Cu$^-$) - E (chlorine half cell)
Ecell= Undefined- (+1.36) = Undefined because Cu$^-$ does not exist in solution.

In short nothing will happen to copper in HCl.

P.S. Practically, HCl slowly dissolves copper in the presence of oxygen. This not relevant here.


Here's the B3LYP/6-31G* geometry. On something like this, I'd suspect this is fairly reliable, although I'll try a few other methods in a second. It gives the Br-C-Br angle as a healthy 113.4 degrees, and H-C-H as 112.5, both in agreement with experiment.

I don't see how the Br-C-Br angle is ever going to be smaller than 109.5. It's the Br-C-H angles that shrink (107.7 degrees), since breaking the symmetry forces the molecule into a "squashed" tetrahedron (point group $D_{2d}$ for the squashed tetrahedron, $C_{2v}$ for $CH_2Br_2$ in particular)

So your questions:

  1) Does Bent's rule work here (i.e. am I correct?)

Certainly not in the way you're thinking. As suggested in the other answer, the Br-Br repulsion opens that bond angle, and there's likely a Br-H electrostatic interaction shrinking that bond angle. (That's what I'd think.)

  2) Does Bent's rule always work?

I typically consider VSEPR and Bent's rule to be very useful heuristics. Indeed, it's used in one of my favorite force field methods, UFF.

I'm holding back on "always" because chemistry is incredibly varied. A quick Google search of "Bent's rule fail" turns up this interesting nugget from Weinhold's book:


How are turbostratic graphite phases formed?

"High-temperature phase transformation and low friction behaviour in highly disordered turbostratic graphite," doi:10.1088/0022-3727/46/39/395305

"Phase Transformation Mechanism of Graphite-Turbostratic Graphite in the Course of Mechanical Grinding," CHEM. RES. CHINESE U. 19(2), 216 (2003)

http://www.onxlti.com/product-divisions/contract-manufacturing-products/on-x-pyrolytic-carbon/


Finding the color of crystal violet theoretically

For a reliable and experimentally comparable answer, you really need to use a numerical method rather than a pencil and paper estimation. For an organic system this size you'll probably be safe using time-dependent density functional theory (TDDFT). Most computational chemistry packages have this built in. Off the top of my head, I know that Gaussian (commercial–standard for organic molecules like this), and GAMESS (free) both have TDDFT built in.

If you're intent on using a toy method you could use Hückel theory as orthocresol suggested. You could in principle tune the parameters of such a model until the excitation energy matches the experimental result, but I don't see what additional physical insight this would give you.


What is the reason behind choosing the specific elements used for the synthesis of heavier elements?

There are different strategies for synthesizing superheavy elements.

Two of the main strategies are referred to as "hot fusion" and "cold fusion".

The first two reactions are good examples of the "hot fusion" strategy. I will focus on these two reactions, as the other two are much older and I don't think they are good examples of current thinking.

Experimentally, a very heavy element target (the Es or Bk in the first two reactions) is bombarded by an ion beam of the second element.  The target element is choosen on the basis of being very heavy, availible and not rapidly decaying.

$^{48}_{20}\ce{Ca}$ is choosen for a very special reason.  This is not ordinary calcium.  The most common isotope of calcium has only 20 neutrons, but this isotope has 28!.  It is extremely neutron rich, espeically compare to other relatively stable isotopes in this mass-range of the periodic table.  $^{50}_{22}\ce{Ti}$ is also very neutron rich.

Why are neutron rich nuclei chosen? There are two reasons for this:

First, confining many protons in a small space causes electrostatic repulsion.  As atomic number increases, the stable nuclei deviate further and further from a 1:1 neutron:proton ratio, because having more neutrons keeps the protons further apart and decreases electrostatic repulsion energy.

Secondly, you will notice the asterisk (*) in the 1st, 3rd and 4th equations.  This denotes the compound nucleus which initially forms in the fusion reaction.  It is an extremely ($<10^{-21}$s) short lived intermediate state, not the ground state of a nucleus, but more like a highly excited state.  It's formation is not considered creating a new element.  It is critical to the strategy of forming an element that the intermediate state loss energy through neutron "evaporation".   Notice that in the second equation, the step of the neutrons evaporating is shown, while the intermediate state is not shown.  Generally speaking, in the hot fusion strategy, 3-5 neutrons will evaporate to cool the intermediate state.

For further reading see:

Future of superheavy element research: Which nuclei could be synthesized within the next few years? and

The discovery of the heaviest elements Rev. Mod. Phys. 72, 733


Redox reaction with tin and sulfuric acid solution

The sulfuric acid dissociates in water: $$\ce{H2SO4 -> H+ + HSO4-}$$$$\ce{HSO4- -> H+ + SO4^{2-}}$$

When writing redox equations the spectator ions are left out because they do not participate in the redox reaction.

In this case it is the hydrogen ions which are oxidising the tin and the sulfate and hydrogensulfate ions play no part in the reaction so they are left out. Including them in the hydrogen half equation would give (with sulfate ions): $$\ce{2H+ + SO4^{2-} +2e- -> H2 + SO4^{2-}}$$

Clearly the sulfate ions are unnecessary and so they are omitted from the equation.


What is advantage of creating elements that disappear in a flash?

Researchers are looking for the Island of Stability, a predicted group of heavy nuclei with halflives in the minutes or even days.

While we have some understanding that certain nuclear configurations and compositions are more stable, we are currently unable to reliably to a priori predict the stability, the half-life, or even the decay mechanisms of a nuclide. Thus, when a new element or isotope is produced, no one knows how long it will be around. Every time a new nuclide is produced, observations of its decay help refine the models of nuclear stability and decay.

The Island of Stability is believed to be centered around 120 protons and 180 neutrons. Element 118 $\ce{^194Uuo}$ has been produced, and while its most stable isotope has a half-life of only 0.89 ms, it is very close to the island of stability and there may be a heavier, more stable isotope of $\ce{Uuo}$ out there.

As in many areas of pure science, the benefits to society (or even other fields) may not be evident except for those few who are actually doing the work. The outcome of pure science research is knowledge and information. Consider, for example, that research on cathode rays led to the development of the television decades later.
The scientists studying cathode rays likely could not even envision their work being applied to the projection of moving pictures (especially those transferred over the airwaves as radio signals), as such technology did not exist or was very new at the time.


Standard electrode potential of copper disproportionation

Alas, the confusions related to signs in electrochemistry will never vanish. You mention that
$\ce{2Cu+ (aq) -> Cu (s) + Cu^2+ (aq)}$ is $\pu{0.36 V}$.
Let me start with a single equation, x-y = 10; There can be indefinite solutions if you can simultaneously change the value of x and y. However, the moment you fix the value of x, the value of y is fixed.
You stated that the overall cell potential is $\pu{+0.36 V}$. Electrochemically, this means that this reaction is spontaneous.
Now you also know that,
$$E_\text{cell} = E_\text{reduction} - E_\text{anode} \tag{1}$$
You are not supposed to change any sign of the half-cell from the electrode potential tables. People should stop teaching this nonsense to relatively innocent students. Suppose, I write
\begin{align}
&\ce{H2O (liquid) -> H2O (gas)} &T &= \pu{100 ^\circ C} \\
&\ce{H2O (gas) -> H2O (liquid)},    &T &=\pu{ -100 ^\circ C} ??
\end{align}
The equation (1), itself takes care of all sign flipping and all.
Your half-cell corresponding to the reduction is
$\ce{Cu+ + e- -> Cu}$ (potential $=  \pu{0.52 V}$)
And your half-cell potential for the oxidation is
$\ce{Cu^2+ + e- -> Cu+}$ (potential $=  \pu{0.16 V}$)
Using equation (1), what you do get (remember no sign flipping) = $\pu{+0.36 V}$


How does chloroform degrade nitrile gloves?

Nitrile gloves are made of nitrile rubber, or poly(butadiene/acrylonitrile). This polymer is highly soluble in chloroform, with some papers I found indicating that one can dissolve up to 18% in mass of nitrile butadiene rubber in chloroform. Moreover, it permeates easily through NBR, meaning we can expect the dissolution to be fast in addition to thermodynamically
favourable.

Finally, I would not expect the mechanism here to be any different from that of any polymer dissolution by a good solvent. The solvent will permeate through the polymer, intercalate between polymer chains, and solvate them (inducing swelling). Once solvated, the network of polymer chains looses its mechanical properties and they can fully separate. (I wish I could find a good existing illustration for that part, but I can't right now… If anyone can, feel free to edit!)


How to determine the order of a reaction with not a commensurate rate equation

To be honest with you, the term "order of reaction" is not an exact term. If I want to pull out the rigorous kinetician out of the closet, I'd say that only elementary reactions have orders, and we have first or second (or zero or maybe third) order reactions only, and they are obvious to spot.

What you have is not an elemental reaction, but an overall - somewhat empirical - net reaction rate for a complex mechanism. I personally do not like to talk about orders of non-elementary reactions, but some authors do this. In this context, I'd say your reaction is first order in $\ce{N_2O}$, simply because you have concentration squared divided by approximately concentration. Note that if $\ce{[N_2O]}$ is sufficiently large (so the $1$ in the denominator is negligible), the equation becomes $-r =\ce{\frac{k_1 [N_2O]^2}{k_2 [N_2O]} = k' [N_2O]}$, so in high concentration limit the unimolecular approximation becomes exact.


What is the proper way to prepare a solution from hydrate?

Originally I thought that all I have to do to figure how much $\ce{Ca(NO3)2}$ to weigh is

  and then just add  $10~ \mathrm{mL}$  water.

To properly make an aqueous solution of a specified concentration you should dissolve the desired amount of material in about $80\%$ ($8~ \mathrm{mL}$) the desired volume of water. Then dilute to the final solution volume (add water until the solution volume is $10~ \mathrm{mL}$).

  How can I verify that I got $\mathrm{1~M}$ $\ce{Ca(NO3)2}$ as opposed to $\mathrm{1~M}$ $\ce{Ca(NO3)2*4H2O}$?

Hydrates are for the solid state.  You do not have "hydrates" in the aqueous phase.  Therefore the two solutions are identical and indistinguishable.


Confusion in the order of precedence: alkyl side-chain vs. dimethyl at different positions

Your starting point is correct. According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), the name 5-(sec-butyl)-2,7-dimethylnonane is no longer recommended. (The prefix ‘sec-butyl’ was still contained in the 1993 recommendations.)

The names that are proposed in the question, 2,7-dimethyl-5-(1-methylpropyl)nonane and 5-(1-methylpropyl)-2,7-dimethylnonane, are unambiguous and describe the same structure. There may be many different ways to draw the structure of such branched alkanes. If you enter these names into the Open Parser for Systematic IUPAC nomenclature (OPSIN), which is mentioned in the question, the results are as follows:

2,7-dimethyl-5-(1-methylpropyl)nonane

5-(1-methylpropyl)-2,7-dimethylnonane

The structures may look different at first glance, but they actually describe the same compound. In this respect, there is no difference between 2,7-dimethyl-5-(1-methylpropyl)nonane and 5-(1-methylpropyl)-2,7-dimethylnonane. However, the name 5-(1-methylpropyl)-2,7-dimethylnonane is not in accordance with IUPAC nomenclature since ‘methyl’ is earlier alphabetically than ‘methylpropyl’. Therefore, the name 2,7-dimethyl-5-(1-methylpropyl)nonane is a correct IUPAC name.

Anyway, the preferred IUPAC name (PIN) according to current recommendations is 5-(butan-2-yl)-2,7-dimethylnonane.


Will CO2 bubbles ever fully dissolve in water?

To improve your CO2 intake and not lower dissolved oxygen levels, an apparent solution is to make the water in your fish tank more alkaline and also consider increasing the surface turbulence.
The logic is that there is an equilibrium based dissociation of CO2 in water creating Carbonic acid. However, in the presence of OH- ions, the formation of an insoluble carbonate (like CaCO3) will continue to drive the reaction to the right.
On the effect of turbulence from an air pump agitation of the water or perhaps a waterfall creating mist, will capture the carbon dioxide.
So, as solutions consider adding alkaline stones. There is also a supporting article: Limestone Rocks in Aquariums and Water pH, to quote:

Have you set up an aquarium and then noticed that the pH of the water continues to rise? Could rocks or gravel that were added as decor in the aquarium affect the water pH? Yes. If your rocks are actually limestone, they are the cause for the pH elevation in your aquarium water. Limestone is calcareous (contains calcium) and is known for its ability to both harden the water and increase the pH. In fact, if someone asks how to raise water pH, one method we suggest is to place crushed limestone, coral, oyster shell or any highly calcareous material in the filter.

On the waterfall idea a reference, to quote:

Speaking of pond health and waterfalls, they're a great way to aerate your pond, but you might not want to overestimate how much oxygen is getting into your pond from the waterfall. Specifically, the larger your pond is, the higher the waterfall would need to be to effectively aerate it alone.

Hope this helps.


Does presence of alkyl groups on alpha carbon increase yield of alkane in Kolbe's Decarboxylation?

I have done some electrosynthesis work using the Kolbe reaction, what I found was with straight chain carboxylic acids that my platinum anode was quickly fouled.

When I had some branching at the beta site, I had much less electrode fouling and I got a good yield of my alkane.


Mini Research Project Time

Updates

Added CCSD(T) $n_i$ and dipole moments and tweaked discussion (the delay was caused by a system-wide storage upgrade on the machines which took nearly a week to complete).

Preamble

This response is in no way meant to be contrary to what Geoff has already posted.  I happen to enjoy these types of questions and I like to tackle them as little research projects for fun.  Therefore, my answer will be brutally detailed and in-depth.  That said, this should be a good introduction to how I would approach this problem if I were going to do this at the 'production-level' of research but this definitely goes far beyond the purpose of any reasonable response for an SE answer.

Its All About the Dipoles Baby

Okay so I wouldn't use THAT heading in a paper but maybe a talk depending on who my audience was...

We can easily determine the dipole moment of cis-2-butene via electronic structure theory.  I have modeled two conformers of cis-2-butene as shown below.  I will refer to the geometry on the left as StructA and the geometry on the right as StructB.

A couple of things to note since I don't include any captions to tables.  Energies are always reported in $\mathrm{kcal\ mol}^{-1}$ and dipole moments ($\mu$) are given in Debye.

Computational Methods

Full geometry optimizations and corresponding harmonic vibrational frequency computations were performed with second-order Moller-Plesset perturbation (MP2) theory and a variety of density functional theory (DFT) methods using the Gaussian 09 software package for the conformers of cis-2-butene.  Both conformers were characterized in $C_{2v}$ symmetry.  The DFT methods implemented include B3LYP, B3LYP-GD3(BJ), M06-2X, MN12SX, N12SX, and APFD.  The B3LYP-GD3(BJ) method employs Grimme's 3rd generation dispersion correction as well as the Becke-Johnson damping function. All DFT computations employed a pruned numerical integration grid having 90 radial shells and 590 angular points per shell.  The heavy-aug-cc-pVTZ basis set was employed for these computations where the heavy (non-hydrogen) atoms were augmented with diffuse functions (e.g. cc-pVTZ for H and aug-cc-pVTZ for carbon).  This basis set is abbreviated as haTZ.  The CCSD(T) (i.e. coupled-cluster method that includes all single and double substitutions as well as a perturbative treatment of the connected triple excitations) was similarly employed using the CFOUR software package.  The magnitudes of the components of the residual Cartesian gradients of the optimized geometries were less than $6.8\times 10^{-6} E_h\ a_0^{-1}$.

Single point energies were computed with the explicitly correlated MP2-F12 [specifically MP2-F12 3C(FIX)] and CCSD(T)-F12 [specifically CCSD(T)-F12b with unscaled triples contributions] methods in conjunction with the haTZ.  These computations were performed with the Molpro 2010.1 software package using the default density fitting (DF) and resolution of the identity (RI) basis sets.

Natural bond orbital (NBO) analyses were performed for the MP2 optimized structures using the haTZ basis set and the SCF density.

All computations employed the frozen core approximation (i.e. 1s$^2$ electrons frozen in carbon).

Computational Methods in English

Two different conformations of cis-2-butene were characterized with a variety of cheap (but usually okay) approximations (i.e. DFT methods) as well as reliable (but generally more expensive) wave function methods (i.e. MP2 and CCSD(T)).  The wave function methods are necessary to validate the DFT results.  CCSD(T) is the gold-standard and gives very good results for single-reference closed-shell well-behaved systems so we will use this as our 'best estimate'.  We use a variety of methods in order to look for agreement in the results.  If we see good agreement across the board, we can be confident in our results.  If we see massive discrepancies then we will have to be careful when we analyze the results.  Geometries have been converged to a tight threshold (i.e. we have good molecules) and our DFT computations use a relatively dense integration grid (which leads to more accurate results).

Notice that I employ a heavy-aug-cc-pVTZ (haTZ) basis set.  Why leave the diffuse functions off hydrogen?  The purpose of diffuse functions is to describe electron density far away from the nucleus of an atom.  Therefore we slap these functions onto carbon which are relatively large atoms compared to hydrogen.  Hydrogen, on the other hand, has only one electron and therefore has a small electron density when isolated.  In cis-2-butene, hydrogen is bonded to carbon via a rather small bond distance.  The electron density around the hydrogen is even more reduced than an isolated hydrogen atom in the gas phase.  Therefore, it would be impractical to include diffuse functions on hydrogen.  Doing so may even lead to erroneous results since we will be trying to describe electron density far away from the hydrogen nucleus when in reality there is virtually none to be found.

Finally, we perform single point energies using explicitly correlated methods.  Because the CCSD(T) opts and freqs will likely not be done in time for this posting, we can gauge how the resulting geometry from each optimization procedure will vary from another geometry given by a different method.  If all of the energies are similar (within a few tenths of a $\mathrm{kcal\ mol}^{-1}$), then we can be confident that our geometries are not only very similar but that small deviations in the geometry will have little effect on the corresponding energies at least in this region of the potential energy surface (PES). Large deviations will usually mean that the method which produced the 'outlier' is not a good approximation for the system (I do not expect cis-2-butene to be problematic at all).  Explicitly correlated methods accelerate convergence to the CBS limit.  These methods have been shown to give results that a large basis set and a canonical method would provide but with a much smaller basis set.  For example,  the result that I get with a regular CCSD(T)/aug-cc-pV5Z basis set could be obtained using CCSD(T)-F12/aug-cc-pVTZ.  This makes the computations less intensive and much more feasible.

Results

The number of imaginary frequencies ($n_i$), relative MP2-F12 and CCSD(T)-F12 energetic  ($\Delta E^{\mathrm{MP2-F12}}$ and $\Delta E^{\mathrm{CC-F12}}$, respectively, in $\mathrm{kcal\ mol^{-1}}$) and dipole moment ($\mu_z$ in Debye) are given in the following table for both conformers of cis-2-butene for a variety of a methods.
The relative energies were determined by taking the difference of the respective geometry and the reference CCSD(T) geometry [e.g. E(CCSD(T))-E(MP2)].

StructB is a second-order saddle point ($n_i = 2$) on every single PES considered and therefore is not a minimum energy structure.  StructA is, however, a minimum ($n_i = 0$) on every PES considered.  The characterization of the nature of the stationary point is consistent between CCSD(T), our best estimate, MP2, and DFT methods.

Single point energies reveal negligible differences in the optimized geometries.  The energies associated with the MP2 optimized structures is the reference point for all other relative energies.  Deviations grow no larger than 0.27 $\mathrm{kcal\ mol}^{-1}$ for the MP2-F12 and CCSD(T)-F12 relative energies.  In addition, there is good agreement between the MP2-F12 and CCSD(T)-F12 relative energies, suggesting that higher-order correlation effects are small.

The dipoles for StructA and StructB are very similar with very small magnitudes, on the order of a couple tenths of a Debye.  To put these quantities into perspective, the dipole moment of water is 1.85 D.  We all know that water has a pretty large dipole moment so by comparison, cis-2-butene has a very WEAK dipole moment.  You can compare to the dipole moments of other molecules by referring to this NIST reference.  Clearly, the rotation of the methyl groups have a very small effect on the dipole moments of each conformer.  The MP2 and DFT dipole moments deviate from the best estimate by no more than 0.03 D, but remains in qualitative agreement for both StructA and StructB.

The figure below shows the directionality of the dipole.  The head of the (unscaled) arrow points toward the negative pole while the tail of the (unscaled) arrow is oriented to the positive pole.  The numbers on the atoms represent 'natural charges' from an Natural Bond Orbital (NBO) analysis of the MP2 optimized Struct A.  Clearly the carbon atoms have a small negative charge to them as these atoms are sucking (great scientific term here) electron density away from the neighboring hydrogen atoms.  This is because the nuclear charge (i.e. the number of protons) on carbon is much greater than that of hydrogen (6 vs. 1).

It was suggested that I add some data highlighting the energy difference between the two different conformers.  The following table presents the energy difference (where $\Delta E_{\mathrm{A-B}}$ is equivalent to E(A)-E(B)) of each optimized geometry using the MP2-F12 and CCSD(T)-F12 single point energies (in $\mathrm{kcal\ mol}^{-1}$).  We can see that StructA is about 1.5 $\mathrm{kcal\ mol}^{-1}$ lower in energy than StructB.  This makes sense because StructB is a higher order saddle point and StructA is a minimum on the PES. (I'm actually quite surprised that the energy difference is this large for a couple of methyl rotations...)

Conclusions

The dipole moment for two conformers of cis-2-butene has been examined using seven different computational approaches and a triple-$\zeta$ quality basis set.  The performance of these methods have been tested by evaluating the energies of each geometry with the 'gold standard' CCSD(T) method (the explicitly correlated variant).  Good agreement is seen across the board with respect to the Hessian indices, energies, and dipole moments. Only StructA was a minimum on each PES.  Both conformers of cis-2-butene have a very weak dipole moment on the order of 0.2 -- 0.3 D.  The positive pole is in the vicinity of the methyl groups whereas the negative pole is centered around the sp2 hybridized carbons.

FAQ

So you may be asking yourself (or rather, should be asking yourself) questions such as these listed below.  I will tackle them one at a time.

1.) Why did we look at two conformers of cis-2-butene and why does it matter?

StructA is a minimum which means that if you were to characterize this guy in the gas phase, you'd find StructA rather than StructB.  This is important because if we were to report these results to other scientists, they will want to know what they can expect to find without wondering.  Therefore, StructA is going to be our conformer of interest rather than StructB.  The latter still will provide insightful results but that is about it for the purposes of this examination.

2.)  Why did we use a variety of methods to characterize these molecular systems?

Computational methods are simply approximations.  They are not guaranteed to give the 'correct' answer.  So we try to address this by using a variety of methods (i.e. approximations) and we analyze the results accordingly.  If the results are in agreement, then we can feel confident that the results are correct since we tested them against a list of methods.  You can never rely on just one method unless it is rigorous, well-tested and has been shown repeatedly to perform well in the literature.  When we say 'perform well', this usually means that the computational results are in reasonable agreement with experimental results.  This is important because experimental results is the LAW (for all intents and purposes).  If the computations disagree with experiment, 99.9% of the time this means that your computational approach sucked and that your approximation was either flawed or misapplied.  By using a host of methods, we can put a little more faith into the results we observe because the chances of massive disagreement between the results of the methods is very unlikely in a normal situation.

3.) What is the point of doing a bunch of energy points?

Again, because we used a slew of methods to characterize the geometries of cis-2-butene, we end up with non-identical geometries each time we use a new approximation.  For instance, the methyl C-H bond lengths from B3LYP are going to be a little bit different from those obtained with MP2.  So then that begs the question, "How do these small differences effect the property of the system that we are interested in?"  Generally, minute differences will have little effect on the resulting energies of each molecule under consideration.  Energy is a very important property that chemists love to look at.  So, if we take each geometry (and each one is unique from the other) and we evaluate the energy of the molecule at the same level of theory (in this case, MP2-F12 and CCSD(T)-F12), then we can quickly see how 'resolved' each geometry was.  There should be very good agreement between the relative energies of each geometry (probably to within a few tenths of a $\mathrm{kcal\ mol}^{-1}$).

4.) Okay, so why MP2-F12 AND CCSD(T)-F12?

We use MP2-F12 AND CCSD(T)-F12 to test for 'higher-order correlation effects'.  MP2 methods are much cheaper than CCSD(T) methods but MP2 is not as rigorous and can be error prone in a host of molecular systems.  Therefore, we test the performance of MP2 by busting out our 'gold-standard' which is the CCSD(T) method.  If MP2 agrees well with CCSD(T), then we can feel confident in our MP2 results and never have to revisit the difficult, time-consuming CCSD(T) computations ever again.  ALso, CCSD(T) will also tell us how DFT performed as well.  DFT methods must always be calibrated against something more rigorous since DFT is known for 'getting the right answer for the wrong reasons' and it isn't always right.

The 'F12' bit just means that these are 'explicitly-correlated' methods.  Rather than give an introduction to what this means, you should understand why we use it instead.  You may have noticed that whenever we do a computational job, we specify a method AND a basis set (e.g. heavy-aug-cc-pVTZ).  These basis sets can be measured by how many atomic orbitals (or functions) are given in the set.  The more that are given, the better the 'basis-set approximation'.  Think of this in terms of Riemann sums where you try to approximate the area under a curve using a set of rectangles.  Each rectangle is a basis function and the number of rectangles you use forms a basis set.  The more rectangles you use, the better your approximation of the area under that curve will be.  Basis sets in computational chemistry behave the same way.  When you approach an infinite set of rectangles, you approach the exact answer.  When you approach an infinite number of basis functions, you approach what is called the CBS (complete basis set) limit.  At the CBS limit, you have an exact answer.  We cannot implement an infinite basis set in chemistry (for obvious reasons), and very large basis sets are cost prohibitive.  Therefore, people have devised these F12 approximations that are constructed in such a way to give results that are comparable to those that you'd get with a large basis set, but you can get them by using a relatively small basis set instead!  This is a powerful approach to convergent quantum chemistry that saves you a lot of time while maintaining a set of very good results.

5.) Why didn't you provide more pretty pictures?

That's just the nature of the beast.  Computational chemistry is usually short on graphics but very dense on spreadsheets.  Plus... I'm no artist.  I actually spent a good couple hours trying to get some electrostatic potentials posted but the new version of G09 hates the molecular viewer programs I currently use so I ditched that idea.


Why do different gases have the same molar volume?

You have the right thought process; if two real gases have different sized molecules, they won't have the same molar volume. The issue is that your statement about gases at STP refers to ideal gases which by assumption are just point particles that only interact via elastic collisions. To get a better sense of the difference that considering the finite size of molecules and their electrostatic interactions can make, you will want to read up on the various real gas models used, such as the Van der Waals, Redlich-Kwong, and Peng-Robinson models.

As a small side note, MaxW is correct that the definition of STP you are using is out of date so the molar volume at STP with the new definition is slightly higher. Its still quite common to see people accidentally continuing to use the old definition, so be on the lookout for that.


Yes, Lanthanum is classified as a lanthanide although it does not have any electrons in its f orbital.

La does not obey the general rule of lanthanides ([Xe]4f^n 6s^2), but is included in this group due to its characteristics (particularly trivalency) being in accordance with the rest of the group. Such as its atomic radius & ionisation energies.

Ce is another exception to the rule (as are Ga & Lu). It ought to be [Xe] 4f^2 6s^2, but the effective nuclear charge, which is still relatively small, is not great enough for f-orbital contraction to be sufficient for 2e in the f orbital. Thus the energy required to fill the f orbitals with 2e is still greater than the energy required to fill the d orbitals.
That being said, it is equally as difficult to fill the 5d orbitals with both electrons, as the f orbital contraction has lowered the energy required enough such that 1 electron can occupy the f orbitals and the other can occupy the d orbitals.

I understand if this isn't the exact answer you are looking for as many areas of periodicity have discrepancies, but reasons can essentially be derived back to a balance between the stability of the species & the energetics arising from the occupation of the electron in available orbitals.


Is the existence of chirality center equivalent to chiral?

The answer to both questions is yes.

In regards to the first question, meso compounds like tartaric acid have multiple chiral centers, but are not chiral overall because they have a mirror plane.

For the second question, a compound can be geometrically chiral without having a chiral center. A cool example class of compounds is helicenes, which are helical arrangements of aromatic rings. These compounds have axial chirality, where the stereoisomers are the left and right handed versions of the helix.


Washing stainless steel separator from organic residue

First try commercial cleaning solutions such as Alconox or Pierce RBS-35, with ultrasonic agitation (if you can sneak a transducer into the reactor).

Second, an oxidizer such as Godax Labs' Nochromix might work, but test it on an unimportant part of the reactor to see if it safe and effective. Read and understand its safety data sheet before use.

A last choice is chromic acid, $\ce{H2CrO4}$, which in some concentrations does not damage some types of stainless steel.  N.B. Chromic acid is toxic, carcinogenic and harmful to the environment. With chlorides, it can produce toxic and volatile chromyl chloride. In high concentration it can cause ignition of materials. Check a material safety data sheet before trying it! Also prepare to dispose of this as hazardous waste.
Even if the reference states it is safe for the particular steel in use, try it on an unimportant area... perhaps with some of the organic "gunk", to see if the gunk is removed and the reactor is unharmed.


Synthesis of trans alkene with organoborane

I think this might be a alkyl shift that results in an $\mathrm{S}_{N}2$ type substitution at an $sp^{2}$ center. Cool. I've only ever seen $\mathrm{S}_{N}2$ type substitution at an $sp^{2}$ center for vinyl iodides. Unfortunately, I don't have my teaching materials on me at work, so I can't find the reference for this for you.

EDIT:

Note the "inversion" at the $sp^{2}$ center.

Greg Fu had this great example of a substitution on a vinyl iodide, like I mentioned above. I can't seem to find this reference, but I just emailed him, so I'll update when he writes me back.


This question has been recently raised in Chemistry & Engineering Letters. The CRC Handbook (2017) in section "MELTING, BOILING, TRIPLE, AND CRITICAL POINTS OF THE ELEMENTS" [1, p. 4-117] lists the following values for gallium, quoting original publication [2]:

\begin{align}
t_\mathrm{tp} (\ce{Ga}) &= \pu{29.7666 ^\circ C} \\
t_\mathrm{m} (\ce{Ga}) &= \pu{29.7646 ^\circ C} \\
t_\mathrm{b} (\ce{Ga}) &= \pu{2229 ^\circ C} \\
\end{align}

  [...] values are given for the solid-liquid-gas triple-point temperature $t_\mathrm{tp}$ and pressure $p_\mathrm{tp}$; normal melting point at $\pu{101.325 kPa}$ pressure $t_\mathrm{m}$; normal boiling point $t_\mathrm{b}$; and critical temperature $t_\mathrm{c}$ and pressure $p_\mathrm{c}$. All temperatures are on the ITS-90 scale.

References

Haynes, W. M.; Lide, D. R.; Bruno, T. J. CRC Handbook of Chemistry and Physics: A Ready-Reference Book of Chemical and Physical Data.; 2017; Vol. 97. ISBN 978-1-4987-5429-3.
Geiger, F.; Busse, C. A.; Loehrke, R. I. Int J Thermophys 1987, 8 (4), 425–436 DOI: 10.1007/BF00567103.


Why does activated carbon preferentially adsorb anions?

This answer applies to carbon filtration of water.

According to most of the sources I found, activated carbon binds to most substances through London dispersion forces (from the Wikipedia article). This should mean that it adsorbs larger molecules and non-polar molecules preferentially since they would have larger dispersion forces. As your quote indicates, it does not bind polar molecules like "alcohols, glycols, strong acids and bases" well. This source also lists neutral, non-polar compounds, along with organics and compounds with low water solubility as being effectively removed.

I think that you may have misread or misinterpreted the sentence about binding "chlorine and iodine" as saying that it traps chloride, iodide, and (by extension) other anions well. I can't find data to support this statement. It traps molecular iodine well, and as a large (in terms of electrons), non-polar molecule, that makes sense. But it removes molecular chlorine $\ce{Cl2}$ through a chemical reaction by reducing it to the chloride ion $\ce{Cl-}$ which is then soluble in water and flows through the filter.

I did find a couple of sources (1, 2) that said "activated charcoal" has a slight electropositive charge that helps it to attract negatively charged species, which would fit with your statement that it preferentially attracts anions. (However, they don't list what kind of negative species they're talking about and nitrate, sulfate, and fluoride ions were specifically excluded). The idea that it will remove ionic compounds seems to be contradicted by the statement in at least a couple of sources (one given here) that carbon filters don't remove minerals, salts, and dissolved inorganic compounds


$\Delta H$ is a function of two states, the initial state and the final state.

For a given final state, there can be infinite $\Delta H$ values depending upon what the inital state was.

For a given inital state, there can be infinite $\Delta H$ values depending upon what the final state is.

Therefore, $\Delta H$ is not a state function.


Alphabetical priority of the isopropyl group

The iso prefix is included in the alphabetic order.

For eg:- The compound Thymol has the following IUPAC name :- 2-Isopropyl-5-methylphenol and not 5-methyl-2-Isopropylphenol.

Therefore the name of your compound should be 4-isopropyl-2,3,3-trimethylheptane.


Why are alcohols with longer chains less polar?

If you're looking at the overall polarity of the solvent, you're not only consider the value of the individual dipole moment on a single molecule, but also their “density”, i.e. the number of these dipoles per unit of volume. You are right that the dipole moment of an individual linear-chain alcohol will, in first approximation, be independent of the chain length. However, in a given volume, you'll fit fewer of those long molecules, and the polarity of the overall solvent is thus lower.


You determined the number of double bonds incorrectly, probably due to structure representation with aromatic rings, which, as Loong pointed out, is not entirely correct.
You can redraw the structure of naphthalene as such:

and make sure that there are five double bonds and two rings giving degree of unsaturation

$$\text{DU} = n_\text{rings} + n_\text{π-bonds} = 2 + 5 = 7$$

You can also compare formula of naphthalene $\ce{C10H8}$ with fully saturated alternative $\ce{C10H22}$ and proceed from there:

$$\text{DU} = (H_\mathrm{sat} - H)/2 = (22 - 8)/2 = 7$$

You again arrive at $\text{DU} = 7$ using a simplified formula:

$$\text{DU} = C - H/2 + 1 = 10 - 8/2 + 1 = 7$$


Try heating the water.

Sugar is also very water soluble, but just dumping sugar into cold water is going to take forever to dissolve. You'll age before you get a clear solution. So, add heat and stir it, and it should dissolve.

PS: Some dextrins are only partially solvable. If sigma aldrich says that your particular dextrin is fully solvable, then it should be. Check the label.


Why does a carbonyl reform in ester reactions

Alkoxide ion is only a "bad leaving group" in a relative sense (say compared to chloride). In this case, the tetrahedral intermediate resulting from the initial reaction between the Grignard reagent plus the ester is itself an alkoxide ion, and when the OR group leaves, that then becomes an alkoxide ion - so that step should be close to energetically neutral.

Saponification of esters by sodium hydroxide is another situation where alkoxide anion is a leaving group - again because alkoxide and hydroxide ions are close in energy.


Significance of sign of standard reduction potential

There are several questions here so let me start by defining the different types of cells and standard reduction potentials.

Definition of Electrochemical (Galvanic) cell: A cell that converts chemical energy into electrical energy. In these cells a redox reaction creates electrons that can do work.

Definition of Electrolytic cell: A cell that converts electrical energy into chemical energy. In these cells the electrical energy source provides the electrons to perform a reaction.

Standard reduction potential: The tendency for a chemical species to be reduced and is measured in volts at STP. The more positive the potential is the more likely it will be reduced.

Now, here are your two reduction equations for $\ce{Zn}$ and $\ce{Cu}$:

$$\begin{alignat}{2}
\ce{Zn^2+(aq) + 2e- &-> Zn}\qquad&{-}0.76\ \mathrm V\\
\ce{Cu^2+(aq) + 2e- &-> Cu}\qquad&{+}0.34\ \mathrm V
\end{alignat}$$

The more positive the potential the more favorable the reaction as it is written will be. Remember that  $\Delta G = -nFE^\circ$ and that when $\Delta G$ is positive, the reaction is non-spontaneous, and when $\Delta G$ is negative, the reaction is spontaneous.  Positive values of $E^\circ$ will lead to negative values of $\Delta G$ and vice versa.

So, the reduction of $\ce{Cu^2+}$ to form $\ce{Cu}$ is more favorable than the reduction of $\ce{Zn^2+}$ to form $\ce{Zn}$. This means that  $\ce{Cu^2+}$ is a better oxidant than $\ce{Zn^2+}$. For an electrochemical cell, the cell potential can be calculated by the following equation:

$$E^\circ_\text{cell}=E^\circ_\text{cathode}-E^\circ_\text{anode}$$

For a working electrochemical cell we need $E^\circ_\text{cell}$ to be positive. I would use $\ce{Zn}$ as the anode (oxidation) and $\ce{Cu^2+}$ as the cathode to give an $E^\circ_\text{cell}$ of $+1.10\ \mathrm V$.


How can a ketone be enantioselectively reduced, in the presence of an ester?

I have had great results for enantioselective ketone reduction using the Noyori Ru-TsDPEN system.
This has advantages through being catalytic with low stoichiometry and doesn't require high pressure, or molecular hydrogen for that matter. The downside might be the initial cost of the catalyst / chiral ligand but you don't need very much. The typical reduction reagents are triethylamine and formic acid, making this a very mild system to work with in terms of conditions.
It might be cheaper to purchase the RuCl2(p-cymene) and chiral TsDPEN separately. That way you can also play with the ligand-metal ratio if you need to.
This saves a lot of hassle rather than adding on and removing chiral auxiliaries


Question regarding chemical equilibrium when adding a salt to acid solution

the second relation is wrong to : $pK_{a} = -\log\frac{[\ce{H+}]_{eq}[\ce{A-}]_{eq}}{[\ce{HA}]_{eq}}=-\log[\ce{H+}]_{eq} -\log\frac{[\ce{A-}]_{eq}}{[\ce{HA}]_{eq}}$ so $pH=pK_a+\log\frac{[\ce{A-}]_{eq}}{[\ce{HA}]_{eq}}$
the acidity constant is the relation which links the concentrations of the species present (and not the reverse): if we add the salt $ \ce {A^-} $, then according to the relation you see that the pH increases. You can also see it by reasoning thus: you add a base so the pH can only increase.


Retrosynthesis of spirocyclic compounds?

Step 1 - Form the anion of your starting material ($\ce{NaOEt/EtOH}$ will do this nicely) and react with phenethyl bromide ($\ce{PhCH2CH2Br}$).

Step 2 - Add the product to hot polyphosphoric acid$\ce{^{[1]}}$ to cyclize it to the required product.

Reference

Cyclization of Aryl-Aliphatic Esters with Phosphorus Pentoxide in Phosphoric Acid by Richard C. Gilmore, Journal of the American Chemical Society, 1951, 73 (12), 5879-5880, DOI: 10.1021/ja01156a521


Computed partial charges for protonated amine: discrepencies between Gasteiger and other methods

tldr - atomic partial charges can be non-intuitive

Formal charges in a Lewis structure don't always match up with partial charge assignment methods.

Typically, one wants 'accurate' partial charge models, for some definition of 'accurate' - but what that means can vary considerably. I explained some of this in a related question.

While I haven't worked extensively with AM1-BCC, most of the common partial charge methods either attempt to divide the electron density, assign charges based on best fits to the electrostatic potential at the van der Waals surface, or approximate one or both of these concepts. The electronegativity equalization method (EEM) server you linked mentions fits to electrostatic potential models.

I think you have two parts to your question:

1) Why does Gasteiger give a different answer than AM1-BCC and EEM?

This one is easy. The Gasteiger model uses the molecular connection table (i.e., the valence bond diagram) and thus considers the initial formal charge on the nitrogen. AM1-BCC begins with a quantum calculation, and EEM is attempting to approximate a quantum calculation. You can't specify a formal charge on an atom to quantum mechanics -- the best you can do is express that the molecule is a cation and let the electrons go wherever.

2) Why do AM1-BCC and EEM methods give a negative charge to the nitrogen?

This is less clear to me, but consider that most of these models are attempting to fit point charges at an atomic center to best fit the electrostatic potential on the molecular van der Waals surface. In this molecule, the nitrogen is buried beneath four bonds - two to external hydrogen atoms that are assuredly positive partial charges.

There's ample literature that assigning the point charge to such a buried atom is very difficult because the nitrogen doesn't really have exposed VdW surface.

I haven't taken much time to look at other charge models, but I can tell you that the N+ in your structure does have a more positive charge  than the corresponding neutral N when you strip off the extra proton (i.e, B3LYP RESP gives -0.194 for NH2+ and -0.755 for NH).

That's a very quick, rough calculation, but it should indicate that relative partial charges are often better considerations. In other words, this nitrogen is more positively charged than the neutral analogue.


Does gravity affect the rate of a chemical reaction?

It sort of depends what you mean. If by "affect the rate of a chemical reaction", you mean affect the individual steps of some overall reaction as to slow them down, then no, gravity definitely has no effect at all.

This is simply because the gravitational force is so much weaker than the electrostatic force, which will dominate all interactions relevant in chemical reactions, that it does not have sufficient time to make any appreciable difference in the behavior of molecules or atoms.

A kind of standard way of demonstrating this is by imagining a situation in which it is guaranteed that electromagnetism will be stronger than gravity, which is all normal situations in chemistry, and then plugging the numbers into Coulomb's law or Newton's law of gravity.

This, however, is contrived because there are situations where gravity is stronger than electromagnetic interactions. Just never in chemistry. This article does a pretty good job of explaining this, and pointing out that the ratio of forces can really be anything at all.

The key reason gravity will never be relevant to individual chemical processes, and hence never affect the rate, is that the external gravitational field will affect the molecules equally once they are close enough to interact electromagnetically. And once they are interacting electromagnetically, it is guaranteed this force will be much, much stronger than the gravitational interaction between them because both follows the same $1/r^2$ law, so it is just a matter of comparing the proportionality constants which go in front of the law, and inputting some masses and charges characteristic to the situation. The article I linked does provide a context in which the gravitational force dominates the electromagnetic force, so this is possible.

So, no individual rates will not be affected. It's fun to imagine how gravity could affect macroscopic equilibria though, so let's think about that.

Given sufficient time, gravity is very good at organizing things in a particular way. For instance, if you imagine two planets which have atmospheres that at some time are exactly identical, but one planet is much more massive than the other one, I don't think anyone would contend that these atmospheres will look exactly the same after a long time. For instance, the escape velocity of an object is $\sqrt{2rg}$, $r$ being the radius of the planet and $g$ the gravitational constant of acceleration. If $g$ is increased significantly, then you would not lose nearly as much hydrogen from the atmosphere, which could affect the location of certain chemical equilibria in the atmosphere. The atmosphere as a whole would be kept much closer to the planet, which could also be relevant.

All of this is kind of contrived, however. The point is that gravity will only really affect chemistry in how it organizes materials, but cannot affect the individual rates.