How accurate would be the following conclusion on 'Critical Temperature'?

Ok, I added a point Z to the diagram which I've added below, and I'm going to quote you text in italics below the image...

Just consider any point in the Pressure-Volume isotherms of $\ce{CO2}$ given above, which is in the region of gas-liquid equilibrium (within the shaded dome shaped curve) but has a volume greater than the critical one. For e.g a point that lies:

Within the shaded region. and
On a line passing through B and parallel to the pressure axis.

Let that point represent the state of liquid inside the vessel.

Question 1 - My point Z meets your two conditions doesn't it?

... From there if we increase the temperature, we would arrive at a point on the boundary of the dome shaped curve, (viz. B) but that point would be on a lower isotherm than the critical isotherm.

Assuming you agree with point Z, I agree with the idea above. The next two sentences is where you're wrong.

And it looks like B is also a point where the two densities becomes equal.(isn't it?) Based on that premise, we can't say that the temperature arrived at in the above (heating liquid in a closed vessel) situation would necessarily be the Critical temperature.

Let's start at point A, and follow the purple isotherm. At point A all the carbon dioxide is gas. At point B all of the carbon dioxide is still gas. But as you start to goto point C from point B the carbon dioxide vapor becomes supersaturated and liquid carbon dioxide forms. Now when you get to C there is so little volume that all the carbon dioxide is liquid. Thus only in the blue shaded area under the curve do you have two phases gas and liquid.

Liquid carbon dioxide is denser than gaseous carbon dioxide. So the gas will be on top and the liquid below.


Is this rearrangement in carbocation possible?

yes,  the rearrangement in carbocation is possible depending upon the condition of reaction and the stability of the new carbocation formed
 eg- the general order of stabilty for carbocation is 3degree >2degree > 1degree
NOTE- the stability of a carbocation depends on aromaticity resonance hyperconjucation inductive effect and sometime ring expansion also occurs


What does the phrase "reducing atmosphere" mean in quantitative terms?

I’m an environmental chemist and we frequently refer to ‘oxidizing’ or ‘reducing’ conditions in the environment. I’ll use soils as an example. The ‘oxic zone’ is the area of soil that oxygen can reach by diffusion and is seen as being under ‘oxidizing’ conditions because the reduction of $\ce{O2}$ is the dominant redox reaction that occurs in that zone. Elements that have multiple oxidation states tend be in the higher oxidation states. For example, sulfur will exist primarily as sulfate ($\ce{SO4^2-}$). As you get further away from the Earth’s surface, the oxygen is depleted (by microbes) and the redox conditions change: the dominant reduction reaction changes. Usually in this order: $\ce{O2}$, nitrate, manganese(III,IV), iron(III), sulfate and finally $\ce{H2}$. When the conditions become so reducing that the microbes are using $\ce{H2}$ as an electron acceptor, they making methane via a process called methanogenesis. This order arises because of reduction potentials. Microbes get the most energy for the least effort from reducing $\ce{O2}$, so they do that until it runs out, which is when they switch to the next best thing available.

Ultimately what it means to be a ‘reducing environment’ is that if you added a compound with a higher reduction potential then the dominant redox pair, it would become reduced. For example, if you could put $\ce{SO4^2-}$ in a methanogenic environment, it would be reduced to $\ce{H2S}$ very quickly.

I’ve read that the composition of Earth’s atmosphere in pre-life earth was very different from today. If I recall correctly, there was no $\ce{O2}$ as it was formed by microbes later on. Without $\ce{O2}$ in Earth’s atmosphere, it makes sense that conditions would be very reducing based on an analogy to the soil redox zones.


Chelation and its relation to transition metal ligands

I'm giving the answer thinking that you are familiar with ligands and Denticity.

Polydentate ligands form two or more coordinate bond with the central metal atom. For eg:EDTA can form up-to 6 coordinate bond with the central atom. In these situations a single ligand encloses the metal ion like enclosing it in a cage. This is called chelation. Co-ordination compounds with chelated metal ions are more stable then those in which the metal ions are surrounded by many monodentate ligands.If you want more information read the Wikipedia article on this http://en.wikipedia.org/wiki/Chelation.


Protonation of a compound having alcohol and ethereal linkage

The alcohol group of your educt (tetrahydrofurfuryl alcohol) is protonated, and after subsequent loss of water, the resulting primary carbocation can rearrange to a more stable 6-membered ring cation, which is stabilized by resonance. Initial protonation at the ether oxygen is therefore disfavored, as it will not result in the formation of a similarly stable carbocation (with positive charge adjacent to oxygen). After final deprotonation, 2,3-dihydropyran is obtained.


Can you get oxygen form silica or alumina?

If you don't care too much about practicality, ionic compounds (i.e. salts) (which most minerals are) can be separated by melting them and then electrolysing. The only problem is that most salts have very high melting points. The needed temperature can be lowered by dissolving the salt in another lower melting point liquid, as happens in the Hall-Heroult process for aluminium production, but that only works if the dissolving liquid is more resistant to electrolysis than the stuff you're trying to electrolyse.

An even more general method of separating basically any compound is by heating it into a hot enough plasma to atomise it. The atoms can be separated using a calutron (which works similarly to a mass spectrometer). In reality such separation is only used with very small amounts of materials for analysis and historically for uranium enrichment, but there is no reason in principle why this could not be scaled up to an industrial scale.


Why did they ‘split’ this H2+ ion in the ionic equation?

$\ce{H2+ (aq)}$ ions don’t exist. Furthermore, even if they did, they would not make a neutral compound $\ce{H2SO4}$ with the sulfate ion, since the sulfate ion has a $2-$ charge.

So you must split $\ce{H2SO4(aq)}$ into $\ce{2 H+}$ and $\ce{SO4^2-}$ ions.


Can gamma radiation cause transmutation?

Yes.

See Photo-Fission in Heavy Elements (1947) Phys. Rev. 71, pages 3-10 :

  fission should be possible for all heavy nuclei which lie well
  beyond the minimum of the packing fraction
  curve, provided sufficient excitation is provided
  to produce the necessary deformation of nuclear
  fluid which precedes division of the nucleus.
  Such excitation can be provided by particle capture or by absorption of a $ \gamma $-ray into the
  nucleus.

"Packing fraction curve" is an older term for the nuclear binding energy curve, so the reference is saying that gamma ray induced fission is possible for nuclei that are substantially larger than nickel.

On the low end of the binding curve, there is a different process called photodisintegration.

  For example, magnesium-25, upon absorbing a photon of sufficient energy, emits a proton and becomes sodium-24.

Even the most stable nuclei such as iron and nickel can be disintegrated by energetic gamma rays.

See Photoneutron Thresholds Phys. Rev. (1951) vol. 84, pages 387-394 for quantitative information concerning how much energy a photon must have to photodissociate a given nucleus.


I do not like to post an answer for my own question, however I feel the need to clarify my mistakes ton anyone else. So here is the answer:

Calculate the moles of $\ce{NaHCO3}$
Then I need to plug in the moles to $\ce{PV = nRT}$
After finding the volume of $\ce{NaHCO3}$, now I plug it in a dimensional analysis to figure out the volume of $\ce{CO2}$. That is all and the answer would be $\ce{.77\ Liters\ CO2}$


The Wikipedia article Salt (chemistry) starts off with "In chemistry, a salt is an ionic compound that results from the neutralization reaction of an acid and a base." This seems like a pretty good Chemistry 101 definition. So, in essence, this means having an anion and a cation in the salt, which further implies ionic bonding. This seems like a pretty reasonable place to start.

I'd think that the definition could be tweaked to say "a salt is a solid ionic compound..." That would indicate that one "molecule" isn't a salt. So one molecule of NaCl in the gas phase is a molecule, not a salt. I also have trouble thinking of molten NaCl as a salt. So I'd describe it as a "molten salt."

  Kind of like way I think German works. I'd expect that German would have some compound word for "molten salt" which better denotes that there is one new idea. So in English if a "salt" is a solid that has a 3D crystal lattice how can a salt be "molten"?

In general, there are a lot more organic salts than inorganic, so dissolving in water isn't a good restriction.

These last paragraphs point out the sorts of problem in trying to make such a definition absolutely unambiguous. You get a one sentence definition, then two pages of qualifications which render the definition useless.

I can't think of a solvent in which you could dissolve lithium hydride and I don't see how you could get lithium hydride as a neutralization reaction, so I wouldn't think of it as a salt. I don't know a better word than "salt-like" to describe such a crystal structure of anions and cations.


What's the difference between RHF and UHF?

The names RHF and UHF refer to the invariance of Hartree-Fock wave functions with respect to certain symmetry operations. Fukutome has worked out the details. Also, Stuber gives the details in his PhD work. I.e., a rRHF ( = real Restricted Hartree-Fock) wave function is invariant under the $\hat{S}^2, \hat{S}_z, \hat{K}, \hat{\theta}$, where $\hat{S}$ is the spin operator, $\hat{K}$ the complex conjugation operator and $\hat{\theta}$ the time reversal operator. rUHF is invariant only w.r.t. the operators $\hat{S}_z$ and $\hat{K}$. So the operators form a group and the labels rRHF, rUHF, $\ldots$ are used to group wave functions.

Configuration Interaction (CI) is the interaction of a predefined set of Slater-Determinants (SD) or Configuration State functions (CSF) with each other. I.e., one forms a CI wave function
$$
\Psi_\mathrm{CI} = \Phi_0 + \sum\limits_{r}^{a} \Phi_{r}^{a} + \sum\limits_{rs}^{ab} \Phi_{rs}^{ab} + \ldots
$$
$\Phi_0$ is the reference wave function and $\Phi_r^a$ denotes the wave function constructed from the reference by exciting one electron from orbital $r$ to orbital $a$. The choice of $\Phi_0$, however, is up to you. Oftentimes a rUHF solution is chosen. But similarly, rRHF is possible.

Ground state in your nomenclature just refers to the lowest energy HF solution of a given symmetry. But in some cases, the wave function can be unstable w.r.t. the loss of some symmetries. E.g., you can calculate an rRHF wave function for the disscociated $\ce{H2}$ molecule and you will find it to be a (local) minimum. Using this wave function as a starting point for an rUHF calculation, you will quickly discover the local rRHF minimum to be a local rUHF maximum. Hence an optimization will lower the energy. This is closely connected to the terms singlet or triplet instability. For a more complete discussion please refer to books as, e.g., Molecular Electronic-Structure Theory by Helgaker, Jørgensen and Olsen.

In the mentioned case of $\ce{H2}$ dissociation it is also quite logical. As indicated already by others, RHF implies one set of spatial orbitals for all electrons of $\alpha$ and $\beta$ spin or, to put it differently, double occupation of each molecular orbital (MO). Triplet states cannot be represented with an RHF wave function! But while dissociating $\ce{H2}$ it is unreasonable to assume that the electron spins on the atoms stay paired forever. One would rather expect both atoms to be completely identical at infinite distance and thus to end up with two hydrogen atoms instead of a $\ce{H+H-}$ system. And this is what happens when you use UHF.

The complex versions of HF become interesting especially when dealing with relativistic effects. Due to the Pauli matrices the Hamiltonian becomes complex and hence the MOs become complex as well. By this the wave function is not symmetric w.r.t. to $\hat{K}$ anymore and hence belongs to a different group in the scheme of Fukutome and Stuber.


How does molecular collision excites the electron?

The general rule is that you do not need to absorb the photon to excite an electron. What you need is to provide some portion of energy.
It can be done by means of a photon which is absorbed by an atom, of course, but it is not the only way.

Alternatively, you could collide an atom with some other object and upon the collision some of the kinetic energy is converted into internal energy which is then redistributed by its components. One of the components of the internal energy is the electronic energy, so if enough kinetic energy is converted into this particular part of internal energy, an electron gets excited. This process is usually referred to as the collisional excitation. This is the process by which an electron in $\ce{He}$ atom is excited: $\ce{He}$ atom in its ground electronic state collides with an energetic electron which causes electron excitation.

A collision between $\ce{He}$ atom in an excited electronic state and ground state $\ce{Ne}$ atom also involves an energy transfer. But this time it is the excitation energy (possibly together with some amount of kinetic energy as well) which is transferred from $\ce{He}$ atom to $\ce{Ne}$ atom. Due to a near coincidence between the $\mathrm{2\,^3S_1}$ and $\mathrm{2\,^1S_0}$ energy levels of the $\ce{He}$ electronic states, and the $\mathrm{3s^2}$ and $\mathrm{2s^2}$ energy levels of $\ce{Ne}$, the transfer of excitation energy from $\ce{He}$ atom to $\ce{Ne}$ atom is quite efficient.

Image courtesy of Wikipedia.

Disclaimer: I'm a professional chemist, not a physicist, so I might be wrong in what follows...

... but as far as I understand the matter at the microscopic level the energy is transferred by means of electromagnetic radiation the quantum of which is indeed photon. But do not think about this process as a photon being somehow "produced" in a form of a "billiard ball" by one atom, then transferred to another atom and finally magically "disappearing" here. This classical way of thinking is the root of confusion.

Before the energy transfer we have two systems each described by its own wave function. In order for the energy to be transferred the systems should start to interact with each other, but once they do so they become entangled and we do not even longer have separate systems. Rather, we have a big composite system described by its own wave function. After the interaction we again can speak about the individual systems each of which is again in own quantum state, though, the final state of each system is different from the starting one due to energy transfer. And in order to predict the final state of both systems (to be more precise, the probabilities of all possible final states) one could use the rules of quantum mechanics.


What factors are important for quantitative analysis of a proton 1D-NMR spectrum?

The signals you get in a simple 1D proton spectrum are in most cases roughly quantitative, though there are some exceptions and some aspects you need to consider if you need high accuracy.
Acquisition
There are several things you need to keep in mind while setting up the experiment to ensure that the resulting NMR spectrum will be quantitative.
The spectrum should be of high quality, any artifacts and noise will affect the quantification.
Relaxation delay
The relaxation delay is the time between experiments, if it is too short some protons won't relax completely back to the equilibrium state. This affects the intensity of the NMR signal and will cause the integrals of those signals to be off.
The relaxation delay (including the acquisition time) should be around $5 \cdot \mathrm{T_1}$. You can either estimate the $\mathrm{T_1}$ or measure it.
13C satellites
The 13C satellites are around 1% of the total signal, if you need high accuracy you should use decoupling to get rid of those.
Other aspects
You also need to ensure that the pulses used in the experiment have a reasonably flat excitation profile for the area you're looking at. This is usually no problem for standard 1D proton experiments, but might be a problem for other kinds of experiments or if you have an unusually long pulse and/or and unusually wide distribution of your chemical shifts.
Processing
Baseline
The baseline should be absolutely flat and at exactly zero, any error there will lead to large errors in the quantification. Using a digital filter during acquisition is a good idea for that (baseopt option for Bruker spectrometers), this will ensure a very flat baseline at zero, as long as no other problems distort the baseline.
Phasing
The phase-correction should be exact, it might be necessary to perform a manual phase-correction.


Why is there a need of polar coordinates to solve the Schrödinger wave equation for the hydrogen atom?

Tl;DR

When solving the Schrödinger equation for the hydrogen atom, its potential energy is dependent on $r$. Using Cartesian co-ordinates $r$ is expressed as $\sqrt{x^2+y^2+z^2}$ which greatly complicates the process of solving the equation. Therefore we use polar co-ordinates as the variables are$~r,\theta,\psi$ which means that there is no nasty square root sign as we don't have to split $r$ into terms of $x,y,z$.

Long Version

This is because to solve the Schrödinger wave equation for the hydrogen atom you require to use the technique known as, separation of variables. Separation of variables is a method for solving partial differential equations. For a partial differential equation in a function $\Phi~(x, y~...)$ and variables $x, y,~...$ we assume that it can be factored into functions which is function of only one variable:
$$\Phi~(x, y,\dots) \equiv X(x)~Y(y)~\dots$$

By breaking the resulting equation into a set of independent ordinary differential equations, we are able to easily solve for $X~(x), Y~(y), ~...$ and then plug them back into the equation to get our original function, $\Phi~(x, y~...)$. For example, lets consider this differential equation of
$$\frac{\partial^2 U(x,y)}{\partial x^2} - \frac{\partial U(x,y)}{\partial y} = 0$$

To solve this partial differential equation we use the separation of variables method. We assume that $U(x,y)$ can be written as $X(x)~Y(y)$. Therefore the equation becomes:
$$\frac{\partial^2 (XY)}{\partial x^2} = \frac{\partial (XY)}{\partial y} $$

Now, the beauty of this is that $X$ and $Y$ are independent of each other meaning that $X$ is not a function of $Y$ and $Y$ is not a function of $X$. Therefore we can rearrange that equation to give:
$$\frac{Y~\partial^2 X}{\partial x^2} = \frac{X~\partial Y}{\partial y} $$

The equation can be rearranged to give:
$$\frac{X''}{X} = \frac{Y'}{Y}$$

Since each side of the equation is a constant with respect to the other since one is a function of x and the other is a function of y, we can write:
\begin{align}\frac{X''}{X} &= C, &\frac{Y'}{Y} &= C\end{align}

Now each differential equation can be solved independently of other to obtain $X(x)$ and $Y(y)$. From that, we can easily find $U(x,y)$.

So as you can see, for the method of separation of variables, it is crucial that you are able to split the variables apart so that you can solve each differential equation independently of other variables. This same technique can be applied to solve the Schrödinger equation for the particle in the $3$D box. However when we try to do this for the hydrogen atom we get a problem. The Schrödinger equation for the hydrogen atom using Cartesian co-ordinates looks like this:
$$\left(\frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} + \frac{\partial^2}{\partial z^2}\right)\psi + \frac{2m}{\hbar^2}\left(\frac{e^2}{r}+E\right)\psi = 0 $$

This is a partial differential equation in three variables and we suspect that we need use separation of variables technique to solve it. So lets assume that $\psi(x,y,z) \equiv X(x)~Y(y)~Z(z)$. Now all we have to do is substitute $\psi$ with that and then separate the $x$, $y$ and $z$ variables so that they are together then solve them independently.

But there is one problem. In the equation, $r$ which is the radius is equal to $\sqrt{x^2+y^2+z^2}$ That is a problem because due to the square root sign there is no way we can separate the $x$, $y$ and $z$ variables.

Therefore to solve this equation more easily, we use polar co-ordinates. By using polar co-ordinates, we take advantage of its spherical symmetry which is convenient as the entire system also has a spherical symmetry. So we now transform the Cartesian Schrödinger equation to polar co-ordinates, so that it is dependent on the variables, $r, \theta, \phi$. The Schrödinger equation using polar co-ordinates is:
$$
  \frac{1}{r^2}\frac{\partial}{\partial r}
     \left(r^2 \frac{\partial}{\partial r}\psi\right)
   + \frac{1}{r^2 \sin\theta} \frac{\partial}{\partial \theta}
     \left(\sin\theta \frac{\partial}{\partial \theta}\psi\right)
   + \frac{1}{r^2\sin^2\theta} \frac{\partial^2}{\partial \phi^2}\psi
   + \frac{2m}{\hbar^2}\left(\frac{e^2}{r} + E\right)\psi = 0
$$

This may seem really complicated at first, however with some manipulation, we will be able to separate out all three variables, $r, \theta, \phi$.

We assume that $\psi ~(r, \theta, \phi) \equiv R(r)~\Theta(\theta)~\Phi(\phi)$. For simplicity we will write them as $R, \Theta, \Phi$ without showing the functionality. The Schrödinger equation becomes:

$$
  \frac{1}{r^2}\frac{\partial}{\partial r}
     \left(r^2 \frac{\partial R\ \Theta\ \Phi}{\partial r}\right)
   + \frac{1}{r^2 \sin\theta} \frac{\partial}{\partial \theta}
     \left(\sin\theta \frac{\partial R\ \Theta\ \Phi}{\partial \theta}\right)
   + \frac{1}{r^2\sin^2\theta} \frac{\partial^2 R\ \Theta\ \Phi}{\partial \phi^2}
   + \frac{2m}{\hbar^2}\left(\frac{e^2}{r} + E\right)R\ \Theta\ \Phi = 0
$$
This simplifies to:
$$
  \frac{\Theta\ \Phi}{r^2}\frac{\partial}{\partial r}
     \left(r^2 \frac{\partial R\ }{\partial r}\right)
   + \frac{R\ \Phi}{r^2 \sin\theta} \frac{\partial}{\partial \theta}
     \left(\sin\theta \frac{\partial \Theta\ }{\partial \theta}\right)
   + \frac{R\ \Theta}{r^2\sin^2\theta} \frac{\partial^2  \Phi}{\partial \phi^2}
   + \frac{2m}{\hbar^2}\left(\frac{e^2}{r} + E\right)R\ \Theta\ \Phi = 0
$$
Divide both sides by $R\ \Theta\ \Phi$ and multiply by $r^2\sin^2\theta$:
$$
  \frac{\sin^2\theta}{R}\frac{\partial}{\partial r}
     \left(r^2 \frac{\partial R\ }{\partial r}\right)
   + \frac{\sin\theta}{\Theta} \frac{\partial}{\partial \theta}
     \left(\sin\theta \frac{\partial \Theta\ }{\partial \theta}\right)
   + \frac{1}{\Phi} \frac{\partial^2  \Phi}{\partial \phi^2}
   + \frac{2mr^2\sin^2\theta}{\hbar^2}\left(\frac{e^2}{r} + E\right) = 0
$$
Inspection shows that no other terms except for the $3$ terms has a functional dependence on $\phi$. Therefore with respect to the other variables, the third term can be treated as a constant, which we will call $-n^2$ for convenience. So we obtain: $$\frac{1}{\Phi} \frac{\partial^2  \Phi}{\partial \phi^2} = -n^2$$ The general equation to this can be easily solved. So we have separated one of the three variables. Now the equation can be written in the form:
$$
  \frac{\sin^2\theta}{R}\frac{\partial}{\partial r}
     \left(r^2 \frac{\partial R\ }{\partial r}\right)
   + \frac{\sin\theta}{\Theta} \frac{\partial}{\partial \theta}
     \left(\sin\theta \frac{\partial \Theta\ }{\partial \theta}\right)
   + -n^2
   + \frac{2mr^2\sin^2\theta}{\hbar^2}\left(\frac{e^2}{r} + E\right) = 0
$$
Dividing by $\sin^2\theta$ and rearranging:
$$
  \frac{1}{R}\frac{\partial}{\partial r}
     \left(r^2 \frac{\partial R\ }{\partial r}\right)
   + \frac{2mr^2}{\hbar^2}\left(\frac{e^2}{r} + E\right)
   + \frac{1}{\Theta\sin\theta} \frac{\partial}{\partial \theta}
     \left(\sin\theta \frac{\partial \Theta\ }{\partial \theta}\right)
   + \frac{-n^2}{\sin^2\theta}
    = 0
$$
Now by inspection, the first two terms depend on $r$ while the last two terms depend on $\theta$. As before, we will set the two terms equal to a constant, $\beta$, so that:
$$
  \frac{1}{R}\frac{\partial}{\partial r}
     \left(r^2 \frac{\partial R\ }{\partial r}\right)
   + \frac{2mr^2}{\hbar^2}\left(\frac{e^2}{r} + E\right) = -\beta
$$
$$
     \frac{1}{\Theta\sin\theta} \frac{\partial}{\partial \theta}
     \left(\sin\theta \frac{\partial \Theta\ }{\partial \theta}\right)
   + \frac{-n^2}{\sin^2\theta} = -\beta
$$
Now we have successfully separated all 3 variables into 3 separate single variable differential equations that can be solved relatively easily.


Reaction of 1,2-diketones with periodic acid

Yes it does, under basic conditions to give two carboxylic acids, according to this paper by Bunton & Shiner J. Chem. Soc., 1960, 1593-1598  here

Abstract of the paper:

  The oxygen atom of the acetone from oxidation of pinacol and 2-methyl- propane-1,Z-diol by periodic acid comes from the l,Z-diol, in agreement with the suggestion that an intermediate is formed by electrophilic attack of the periodate. In contrast to this, the additional oxygen atom of acetic  acid, formed by oxidation of biacetyl by periodate in basic solution, comes from the periodate. The mechanism is a nucleophilic attack by periodate upon the carbonyl-carbon atom,  and does not seem to involve the hydrates of the ketone as intermediates. In the oxidation of methylacetoin by periodic acid the oxygen of the acetone comes from the hydroxyl group of the hydroxy- ketone, and the additional  oxygen atom of the acetic  acid comes from the periodate. The mechanism  probably  involves  nucleophilic attack by the periodate  ion  upon the carbonyl  group, and co-ordination of the hydroxyl group on to the iodine atom of the periodate.


Some bases taste bitter, some do not. Between bitterness and basicity is correlation, not causation.
The strongest bases like hydroxides of alkali metals and alkali earth metals, or alkali metal carbonates, are not bitter.
Near all if not all bitter  bases are organic compounds with basic groups containing nitrogen, like aminogroups or heterocyclic nitrogen. They are often potentially poisonous alkaloids, so evolution came with unpleasant sensation for such compounds. The typical example is quite bitter caffeine.
OTOH, not all bitterness is caused by basic compounds. Grapefruits are bitter because of the glycoside naringin, which does not contain nitrogen and is not a base. Glycosides are frequently bitter, regardless of basicity.
Sensory impact is rather a topic for biology, not chemistry.


Why do ionic compounds have higher melting and boiling points despite having smaller radii?

As pointed out in a comment, there is a problem with your second sentence:

  "I thought that the closer the two ions are, the larger the repulsive force, and therefore it should rather require less energy to separate them..."

This statement is technically correct, in that the repulsive forces between the two atoms do increase with decreasing bond length. And it is true that two atoms having a charge of the same polarity would exhibit increasing repulsion as the distance between them decreases.

However, in the case of the oppositely charged atoms found in ionic bonds, the attractive electrostatic forces also increase with decreasing bond length, which is the reason for the strong bonding and thus high melting points of ionic compounds.

A discussion of the bond strength of ionic compounds is given in this Wikipedia article, which summarizes that the ionic bond strength can be determined as the sum of the attractive electrostatic energy and short-range repulsive energy.


USNCO question - Finding pH of salt solution from pKa of its parent acid

Sodium cyanide is a salt formed by a strong base and a weak acid, so the reduced ionic equation of hydrolysis is

$$\ce{CN- + H2O <=>[$K_\mathrm{h}$] HCN + OH-}$$

$$
\begin{align}
K_\mathrm{h} &= \left.\frac{[\ce{HCN}][\ce{OH-}]}{[\ce{CN-}]} \qquad \right|\cdot \frac{[\ce{H+}]}{[\ce{H+}]}\\
&= \frac{[\ce{HCN}][\ce{OH-}][\ce{H+}]}{[\ce{CN-}][\ce{H+}]}\\
&=\frac{K_\mathrm{w}}{K_\mathrm{a}} \label{eq:1}\tag{1}
\end{align}
$$

Sodium cyanide is highly soluble salt, so one can neglect auto-protolysis of water, and also express hydrolysis constant (assuming $[\ce{OH-}] = [\ce{HCN}]$ and $C$ – concentration of sodium cyanide) as follows:

$$K_\mathrm{h} = \frac{[\ce{OH-}]^2}{C-[\ce{OH-}]} \label{eq:2}\tag{2}$$

Equating \eqref{eq:1} and \eqref{eq:2}:

$$\frac{10^{-14}}{10^{-9.21}} = \frac{[\ce{OH-}]^2}{0.25-[\ce{OH-}]}\, ,\label{eq:3}\tag{3}$$

and solving the quadratic equation \eqref{eq:3} for $[\ce{OH-}]$, one can find that $[\ce{OH-}] \approx \pu{2e-3 M}$, $\mathrm{pOH} = 2.70$, hence

$$\mathrm{pH} = 14.00 - \mathrm{pOH} = 14.00 - 2.70 = 11.30$$


How to read electron energy levels abbreviations?

Those are Russel Saunders atomic term symbols and represent the quantum numbers of the many-electron wave function.

The notation scheme is $^{2S+1}L_J$ where $2S+1$ is the spin multiplicity, $L$ is the angular momentum quantum number and $J$ is the total angular momentum due to the coupling between the former two.

There are also molecular term symbols using greek instead of latin letters for the angular momentum. Those are commonly used for diatomic molecules.

You should find the Russel Saunders Coupling scheme in standard physical and quantum chemistry books. Molecular term  symbols are for example explained in detail in Molecular Spectra and Molecular Structure: I. Spectra of Diatomic Molecules by Herzberg and Huber (1950).
This link may give you a preliminary idea : http://wwwchem.uwimona.edu.jm/courses/RScoupling.html


Do atoms form either a positive or a negative charge, but not both?

Actually, in theory almost all of the elements can be found with both positive and negative oxidation numbers: it's just a matter of finding a system with the proper reagents and conditions to force it. If you isolate chemical species which have a very strong tendency of displaying some specific behaviour (accepting electrons, donating electrons, coordinating ions, releasing a leaving group, bonding to metals, releasing a proton, adopting a specific molecular geometry, or any other myriad properties), then you can often obtain strange results by clashing them with substances which also have the same tendency, but not quite as strong. This often causes the substance with the weaker behaviour to "operate in reverse".

Let me give a vivid and related example. As we all know, the alkali metals (group 1 elements) are exclusively present as cations with an oxidation number of +1, except in the pure metals, where it's zero, right? Well, here's something which might shatter your world: most of the alkali metals (with the exception of lithium, for now) also form alkalides, that is, stable salts containing discrete, clearly observed $\ce{Na^{-}}$, $\ce{K^{-}}$, $\ce{Rb^{-}}$ or $\ce{Cs^{-}}$ anions, with alkali metals displaying oxidation number -1.

How is this done? All you need is to find a neutral substance with a much stronger tendency to donate an electron than a neutral alkali metal atom (easier said than done). Since neutral alkali metals atoms form quite stable cations upon loss of an electron, this implies you need to search for a neutral substance capable of donating an electron and forming a cation with exceptional thermodynamic and/or kinetic stability. This can be achieved, for example, by the use of cryptands, which are cyclic molecules capable of coordinating very strongly to cations, strongly enough that they even coordinate alkali metal cations very well. The cryptand-coordinated cation is both thermodynamically and kinetically stable enough that alkalide anions, which would be extremely reactive otherwise, are not reactive enough in this case to immediately cause charge transfer and neutralize the negative charge.

Amusingly, it is actually possible to prepare a single compound which contains both alkali metal cations and anions, as exemplified by $\mathrm{[Na(2,2,2- crypt)]^{+} Na^{-}}$, which contains a sodium cation coordinated by a cryptand as a counterion to a natride/sodide ($\ce{Na^{-}}$) anion. One can imagine this compound being made by putting together a neutral sodium atom ($\ce{Na^{0}}$) and the neutral cryptand species $\mathrm{[Na(2,2,2- crypt)]^0}$. As Brian mentions in the comments, this latter species is actually an electride, which can be written as $\mathrm{[Na(2,2,2- crypt)]^{+}e^{-}}$ and thought of as a salt where the anion is a lone electron (!). Both the neutral sodium atom and the electride have a strong tendency to lose an electron in chemical reactions, but this tendency is much stronger for the electride. Thus, the electride ends up having its way, forcing its very loosely bound electron onto the neutral sodium atom, causing the neutral sodium atom to "operate in reverse" and accept an electron rather than donate it, resulting in the $\ce{Na^{-}}$ anion.

This table on Wikipedia is far more complete than most "common oxidation number" tables out there, and it lists many negative oxidation numbers for elements, including iron! For many of the transition metals, negative metal oxidation numbers are achievable by using the carbonyl ($\ce{CO}$) ligand, which removes electron density from the metal atom via backbonding. This stabilizes negative charges on the metal atom, once again allowing the resultant species to survive with the appropriate counterion.


How to prove that chloroform is more acidic than fluoroform?

My Views (I'm not entirely sure about these reasons)

(1) $\ce{CF3-}$ is highly unstable due to huge repulsions between the negative charge on carbon and the lone pair of fluorine.

(2)While in $\ce{CCl3-}$ chlorine has an empty d orbital thus the negative will be more dispersed into the this empty orbital  hence the repulsions will be quite less as compared to $\ce{CF3-}$.


@ Dhamnekar. Your calculations are correct, provided your assumption (that the volumes are additive) is correct. Nevertheless note that the last value ($x = 0.0097$) has no unit : it is not in moles. And we must also admit that, due to your assumption, x is probably simply $x = 0.010$.
Anyway, next time try to trust your calculations yourself. We are not here to verify correct calculations. We are here to HELP THOSE WHO DO NOT SUCCEED in solving their problems.


How (in)flammable is "isopropyl alcohol" really?

The Flash Point of a liquid is the minimum temperature at which a liquid gives off vapor within a test vessel in sufficient concentration to form an ignitable mixture with air near the surface of the liquid. The flash point is therefore dependent on the boiling point and consequent vapor pressure of the liquid (Source of Vapor Pressure):

$$
\begin{array}{l|c c c}
\text{Flammable Liquid} & \text{Boiling Point at $\pu{1 atm}$} & \text{Flash Point} & \text{Vapor pressure at $\pu{25 ^\circ C}$} \\
\hline
\text{Diethyl ether} & \pu{35 ^\circ C} & \pu{-45 ^\circ C} & \pu{532.7 mmHg} \\
n\text{-Pentane} & \pu{36 ^\circ C} & \pu{-40 ^\circ C} & \pu{434.3 mmHg} \\
\text{Hexane} & \pu{69 ^\circ C} & \pu{-26 ^\circ C} & \pu{151.2 mmHg} \\
\text{Ethyl acetate} & \pu{77.1 ^\circ C} & \pu{-4 ^\circ C} & \pu{94.7 mmHg} \\
\text{Methanol} & \pu{65 ^\circ C} & \pu{11 ^\circ C} & \pu{126.89mmHg} \\
\text{Ethanol} & \pu{78 ^\circ C} & \pu{12 ^\circ C} & \pu{58.8 mmHg} \\\text{Isopropanol} & \pu{82.2 ^\circ C} & \pu{12 ^\circ C} & \pu{43.6 mmHg} \\
\text{Water} & \pu{100 ^\circ C} & \text{N/A} & \pu{23.7 mmHg} \\
\end{array}
$$

By definition, any liquid with a flash point less than $\pu{37.8 ^\circ C}$ ($\pu{100 ^\circ F}$) is considered to be a flammable liquid. Any liquid with a flash point between $\pu{100-200 ^\circ F}$ is considered combustible. In general, the relative hazard of a flammable liquid increases as the flash point decreases.

However, OP's isopropyl wipes are soaked with about 70% isopropanol and DI water (the best combination for better sanitation). For example, see these wipes on market. As a consequence, the flash point ($\approx \pu{18 ^\circ C}$) is increase on these wipes. Also, since vaporizer has good ventilation system (I assumed), these vapors never would get to saturation point in addition to combination of water and isopropanol vapors. That would reduce the flammability of vapors even with a spark (here are heating coils, I assume). That's probably why the user manual did not warn fire hazard.

OP's question about evaporation from an open bottle: As you can see in the given table, vapor pressure of ethanol and isopropanol is quite high in room temperature ($\approx \pu{25 ^\circ C}$) compared to that of water. If bottle is open, the vapor escapes without ever getting saturated like wipes in closed box. Thus, alcohol in the bottle keep evaporate and, as a result, volume in the bottle reduces with time. On the other hand, the closed space in the box of wipes eventually get saturated with alcohol/water vapors, and evaporation stops at that point.


Is it a redox reaction if the oxidation state of the element has changed only in one of the products?

One sulphuric acid is used to form sodium sulphate and the other one is used for oxidizing.
So you have a reduction:

$\ce{4H+ + SO4^2- +2e- -> SO2 + 2H2O}$

and an oxidation:

$\ce{2Br- -> Br2 +2e^-}$

The sodium sulphate is formed with:

$\ce{H2SO4 + 2NaBr -> Na2SO4 + 2HBr}$

This last one is not a redox reaction.


Is the resonance depiction of benzene typically portrayed incorrectly?

Starting with a short and blunt answer to the title’s question: Yes. But you likely already knew that, otherwise you wouldn’t have written the rest.

  One might argue that the reader understands what is being portrayed or that this is done in aims of saving time, but that doesn’t sit right with me. The entire point of nomenclature, drawing a chemical, or anything scientific is that it can be interpreted by anyone in exactly the same way. I feel that if someone is deciding to draw a chemical using the resonance structures of benzene, then they should draw both resonance structures, lest it be mistaken for 1,3,5-cyclohexatriene.

Well, there is a lot more time saved that you seem to be imagining according to your quote.

You seem to suggest that there are only two correct resonance structures for benzene, which is incorrect. Even if you only consider the $\pi$ system, there are a lot more possible resonance structures which contribute to the final picture in a non-zero manner.

The ones I displayed in the picture are just a very small subset. In fact, solely for the $\ce{H2}$ molecule, a professor of mine claimed to have drawn some 18 (or was it 80?) different resonance Lewis structures and also claimed to be able to accurately calculate the energy of the molecule by adequately weighting all those structures. (‘Accurately’ meaning getting an energy comparable to one calculated using molecular orbital theory.)

This actually moves us towards another question: What actually is a bond? There is a lot more to a bond than simply drawing a line between two atoms and saying that ‘they are now sharing an electron pair.’ If you calculate simple-looking molecules like formaldehyde using a program like Gaussian, you are able to see a multitude of molecular orbitals — at least twelve. Two are defined as ‘core orbitals’, namely they represent the 1s orbitals of carbon and oxygen, respectively; all the others somehow contribute to bonding or antibonding interactions somewhere across the whole molecule.

So what would an accurate representation be? Drawing the atoms geometrically correctly, a cloud around them and writing a number 12 inside or next to the cloud? That way, it would be made clear that there are twelve electrons delocalised across the whole of the molecule and somehow keeping it together. But what would be gained from this type of more accurate representation? Or rather: What would be lost? You would lose the possibility to easily discuss lone pairs on the oxygen, bond order between carbon and oxygen and a lot more.

Similarly for benzene and all molecules that contain a six-membered ring of carbons with a $\pi$ system with six electrons (a.k.a. the benzene derivates you’re asking about): If you insisted on drawing ‘the two’ resonance structures, you would still be far from the truth. Furthermore, for most molecules that are not benzene but contain the same ring, the bonds are no longer equal and their bond orders are no longer exactly $1.5$ (true for any molecule that has lost the $\mathrm{D}_{6\mathrm{h}}$ symmetry). So drawing two sets of double bonds and claiming the structures to be equal would be outright wrong.

Finally, though, every chemist has learnt to get a more or less thorough understanding on the concept of a bond and on the concept of aromaticity that it can reasonably be assumed that what I presented is accessable somewhere at the back of the head if really required. Thus, every chemist should have a rather good understading that a line on paper is a line, a bond is not just a pair of electrons and a benzene ring can be drawn in multiple ways. When it really matters, we can either expect the chemist to instantly recognise the mattering due to a special case, or we can bring the mattering matter to the chemist’s attention by explicitly drawing it.

So there is a lot of time gained by simplifying drawings of molecules in the way they are and also cases where it matters can easily be illustrated by drawing more elaborate drawings which aren’t skipped in one’s attention because they are rarely encountered. And we also keep the ability to easily recognise and discuss simple concepts.

P.S: It’s not necessary for anyone to interpret these concepts the same way. In fact, it’s a good thing if an artist takes the picture of benzene and interprets it completely differently. If anything, ‘anyone’ should be restricted to the chemical — or maybe the entire scientific — community, but there I can see the interpretation safely, regardless of whether aromaticity is drawn or not.


Yes, this is a traditional method of gold purification from platinum group metals.


How to choose alcohol solvent for organic extraction?

The optimal solvent will extract the greatest amount of desired product and the least amount of difficult-to-remove substances.

The extraction is only the first step; separation from solvent and other extractives follows. Try one, then another solvent to see if there is a difference. Maybe number of carbons or hydroxyls is important; you can also mix alcohols, and/or add water.

Capsaicin has a phenolic group; perhaps it would be extracted from its source with alkali (NaHCO3 or Na2CO3).


Is NaCl the only precipitate when you add HCl to concentrated NaCl

$\ce{HCl}$ is at ambient conditions a gas, very well soluble in water, dissociating as a strong acid.

When there is excess of it in solution, that cannot be kept solved, it does not precipitate, but evaporates or even boils out, depending on scenario.


Considering your doubt, from Born-Lande equation:

  $$E = -\frac{N_\text{A}Mz^+z^- e^2 }{4 \pi \varepsilon_0 r_0}\left(1-\frac{1}{n}\right)$$

  $N_\text{A}$ = Avogadro constant;
      $M$ = Madelung constant, relating to the geometry of the crystal;
      $z^+$ = charge number of cation
      $z^-$ = charge number of anion
      $e$ = elementary charge, $1.6022\times10^{-19}\ \mathrm{C}$
      $ε_0$ = permittivity of free space
      $4πε_0 = 1.112\times10^{-10}\ \mathrm{C^2/(J\cdot m)}$
      $r_0$ = distance to closest ion
      $n$ = Born exponent, typically a number between 5 and 12, determined experimentally
      by measuring the compressibility of the solid, or derived theoretically.

And what you are assuming the radius is in fact the distance between the two atoms.
As the distance increases, the attraction decreases from Coulomb's law :

  $$F=\frac 1{4\pi\varepsilon_0r_0^2}Q_1Q_2$$

and thus there is less energy involved in breaking the lattice, so lattice energy decreases.

As you move down, the size increases and the valence electron in the outermost shell experiences lesser repulsion and thus it can be easily removed, therefore the elements at the bottom are highly electro-positive or conversely very less electro-negative. So when forming bonds, the elements at the bottom easily remove the electron(s) and gain more developed charged rather than the ones on the top who just release it just partially. This charge variation varies the dipole moment $\vec \mu=q\times\vec d$, by altering the partial charge $q$. Thus downwards as $q$ increases, $|\mu|$ also does increase and make the compounds formed by the element more ionic. Also the ionic/covalent character may be varied depending upon the bonding partner, in sense of, elements or other entities.

Also as charges $z^+$, $z^-$ increases, the magnitude of lattice energy does also increase as can be seen clearly from the Born-Lande equation.

$$\begin{array}{r|c|l}
&&\text{Ionic Solid}&\text{Lattice Energy (kJ/mol)}\\\hline
&\text{Variation of}&\ce{NaF}&926\\
&r_0 \text{ i.e. distance}&\ce{NaCl}&786\\
&\text{between ions}&\ce{NaBr}&752\\
&&\ce{NaI}&702\\\hline
&\text{Variation of}&\ce{NaF}&926\\
&z^+,z^- \text{ i.e. charge}&\ce{MgO}&3800\\
&\text{on ions}&\ce{Al2O3}&15900\\
\end{array}$$


Why does the unstabilised Wittig reaction selectively form cis alkenes?

To my knowledge the mechanism of the Wittig reaction isn't fully resolved yet. But maybe I can give you some ideas about why the Wittig reaction with unstabilized ylides is (Z)-selective (well, with the exception of the Schlosser modification) instead of (E)-selective.

In the excellent book by Clayden, Warren, and Greeves, there is a section beginning p. 690 that describes the oxaphosphetane formation as a concerted, antarafacial [2+2]-cycloaddition reaction (similar to your "puckered transition state").

Here, the phosphonium ylide and the carbonyl compound approach each other at right angles. The substituents are arranged in order to minimise steric repulsions in the transition state. In particular, the phenyl group (Ph) points away from the PPh3 group, and then the R group on the ylide points away from the Ph.

The formation of the oxaphosphetane is irreversible and kinetically controlled. Hence, even though the trans oxaphosphetane is more stable, it is not formed.

(By the way, the book also has some justification why the Wittig reaction of stabilized ylides is E-selective.)

Edit:

Since the OP asked for it I will try to give some justification on why the ylide and the carbonyl group approach one another perpendicularly. For a proper description one would need the frontier orbitals of both compounds and then reason about the orbital interactions that bring about the [2+2] cycloaddition. Since I don't have those I will try to give some oversimplified description that should convey the general principle at work here. In the approximate orbital pictures I'm going to use I'll leave out the substituents for clarity.

To begin with, if one imagines the orbital interactions between the ylide and the carbonyl group, it is clear that the most important (primary) interaction for the reaction is the one between the HOMO of the ylide (nucleophile) and the LUMO of the carbonyl group (electrophile). What happens when both groups approach one another head-on or at right angles is shown in the following picture (for the perpendicular approach a top-view is used).

In both cases there are as many bonding interactions (black) as there are antibonding interactions (red). These situations are overall non-bonding and cannot lead to the desired product. This is the reason why [2+2] cycloadditions are usually thermally forbidden (by symmetry).

However, some compounds have additional orbitals that: (1) are similar in energy to the frontier orbitals drawn in the picture, and (2) possess the right symmetry to interact with the other orbitals. These "secondary orbital interactions" are weaker than the HOMO-LUMO interaction, but might provide enough bonding to tip the balance from an overall non-bonding to a bonding situation.

So, how does this translate to the situation of the Wittig reaction? Since O is more electronegative than C, the HOMO of the carbonyl group will have a larger orbital coefficient at O, while the LUMO will have a larger orbital coefficient at C. The situation is similar with the ylide. Since C is more electronegative than P, the HOMO of the ylide group will have a larger orbital coefficient at C. The ylide group is the nucleophile so the primary interaction will be between its HOMO and the LUMO of the electrophilic carbonyl group. Since the orbitals interact best where the coefficients are large, this primary interaction will be biased towards C–C bonding.

The question is now, is there a suitable secondary interaction? One can argue that P has a lot of orbitals (valence shell extension) and that the LUMO of the ylide might be some empty phosphorus orbital (or at least an orbital with a big coefficient on $\ce{P}$) - this could be a p-orbital perpendicular to the ylide's C=P π-orbitals, or a d-orbital, or something like that. This phosphorus orbital can then interact with the carbonyl group's HOMO, which has its biggest coefficient on O. So this bonding interaction will be biased towards P–O bonding. The situation is shown in the following picture:

The biases in the primary and secondary orbital interactions will lead to some distortions from the pure perpendicular approach which will become more pronounced the nearer the two species get to each other. For example, the ylide might rotate a bit so that its C and P atoms are closer to the carbonyl's C and O atoms respectively, in order to maximize the bonding overlap. And of course the perpendicular approach of ylide and carbonyl group will proceed in such a way that their respective biggest substituents are as far away from each other as possible (see diagram above).

After this quite lengthy answer I want to make clear that this description of mine is only a guess. The real reaction path is still a matter of debate and there are some compounds that rather react via a radical or ionic pathway. But I think it gives a good explanation for most of the observed behaviour and hope it helped the understanding. A more exact frontier orbital description of another thermal [2+2] cycloaddition, namely that of ketene with ethene, can be found in the textbook by Brueckner on p 653.


What term should I use when I am quantifying the amount of matter, mass or weight?

Because of the reasoning provided by you, i.e. mass is universal, and weight is the product of mass and magnitude of the local gravitational acceleration, I suggest to stay at mass.  (A purist's view, taking @Ivan's comment into account.)

In addition, referring to the International System of Units, their brochure not only mentions the basic units, but the relevant section 2.1.1.2 starts with:

Unit of mass (kilogram)


The quick answer to which ion, $\ce{H-}$ or $\ce{F-}$, is bigger is "it depends". The general reasoning behind the claimed fact the hydride anion is larger than the fluoride anion typically comes down to charge density. In $\ce{H-}$, there are two electrons for the one proton, giving it a charge ratio of 2, while in $\ce{F-}$, there are ten electrons and nine protons for a charge ratio of roughly 1.1. The significantly higher charge ratio of hydride leads to a much more diffuse orbital, and a potentially larger ionic radius.

However, there are a few issues that arise, including:

How do we measure the ionic radius? Ions and atoms don't have a hard boundary; the probability of finding an electron a certain distance from the nucleus drops, but there's no firm boundary.
Ionic radii depend on the nature of the counterion

One potential way to measure the ionic radius is to look at lattice parameters of solids, determined by X-ray crystallography. This does not give much information about what the ions look like in solution, but this is not of huge importance in this case as formal hydride is unlikely to exist in solution. More importantly, X-ray crystallography gives the distance between the two ions, but it does not give the ionic radii. However, we can directly compare the lattice parameters of two solids sharing the same cation with the idea that a larger lattice parameter would be indicative of a larger anion.

For the alkali metal binary salts, the fluoride ion does typically lead to a smaller lattice constant than hydride (e.g. 4.027 and 4.083 Å for lithium salts, respectively). However, for calcium hydride and fluoride, the latter gives a larger orthorhombic cell, implying that fluoride is a larger anion.1 Indeed, for most of the analogous binary hydrides and fluorides in this reference, the fluorides produce larger lattice constants. Using different approaches to measuring ionic radii, including the soft sphere and hard sphere approaches, Lang and Smith succinctly declare that "the ions maintin the following order of size: $$\ce{Li+}<\ce{H-}<\ce{Na+}<\ce{F-}<\ce{K+}<\ce{Cl-}<\ce{Rb+}<\ce{Br-}<\ce{Cs+}<\ce{I-}$$
All the anions are larger than isoelectronic cations, but the Pauling order of ion size is not maintained."2

Therefore, while it may be true that the hydride anion appears larger than the fluoride anion in a few cases, this is not true in general.

1 J. Solid State Chem., 1970, 2 (2), 144

2 Dalton Trans., 2010, 39, 7786


Why do cookies prevent toothpaste from foaming?

My first take on this is that the cookies contain fat, which "consumes" the surfactants in the toothpaste (like sodium lauryl sulfate) so it does not foam up the way it normally would. You  might want to see what Wikipedia says about micelles and how they interact with fats. A similar effect can be seen when washing really greasy dishes with an insufficient amount of dishwashing liquid - in which case no suds will form. In your case I don't understand why the effect would not also be observed with other fatty foods such as chocolate or popcorn.


Measuring the colloidal nanoparticles stability by dynamic light scattering

Dynamic light scattering is a sizing technique; that is, it provides an indirect measurement of the size of nanoparticles (details may vary depending on experimental set-up).

However, it's important to note that it doesn't provide a direct measurement of the solid particle, but of its effective size when moving in solution. In other words, it measures the hydrodynamic radius of the nanoparticles - which, by the way, is not a single value but an ensemble average, which you usually measure in statistical fashion.

Now, at the nanoparticle-solution interface, you have several layers of material that is bounded to the surface and will move around with the nanoparticle, contributing to the hydrodynamic radius. In this particular paper, you have a CTAB/C monolayer (at least) which will be more or less tightly bounded to the gold particle, and also counter-ions to provide charge balance and several shells of solvation water (progressively less ordered and more weakly bound as you move away from the particle). It's the size of that whole entity that you measure (indirectly) with DLS.

As you can imagine, reducing the concentration of CTAB/C in the solution will change nanoparticle coverage - and will change the size, shape and strength of the layers surrounding the nanoparticle. As with colloidal systems, when synthesising nanoparticles there's a real concern that they can aggregate and precipitate or otherwise change their properties - which is the reason that you produce them in the presence of an stabiliser such as CTAB/C in the first place. If you look at figure 4, the histograms show hydrodynamic size for DLS measurements in different conditions and you can see they show that this aggregation isn't happening: if it did, you'd see far larger objects moving around in the solution. And as the body of the article explains, the small changes in hydrodynamic radius are probably due to the removal of CTAB/C from the nanoparticle surface, which affects solvation shells and changes the DLS-measured radius.


What do the dots mean in HN.CO.CH3 connected to the second carbon of N- Acetylglucosamine? And what do the N and acetyl indicate?

These are not dots. They actually are single bonds though I do not know why they are printed that way. As such, they should be represented by a half-long dash (called EN DASH in Unicode).

In the name N-acetylglucosamine, acetyl means that a hydrogen atom has been replaced an acetyl group (i.e. –CO–CH3) and that the replacement occurred on the nitrogen (aka "N") atom.

Overall, it means something like: "consider glucosamine but replace one hydrogen on the nitrogen atom by an acetyl group". This is the standard "substitutive naming" defined by the IUPAC.


Calculation of Atomic weight percentage in AAS

Relating mass fraction (ppm) with concentration (μg/ml) without knowing the density of the solution is impossible. You have to create your solutions with given concentrations (μg/ml). You have to weight a given mass of your substance and fill with water to a certain volume. You have to use volumetric flasks therefore. Consider the rules for applying volumetric flasks for getting correct results.


What is the phase of matter that has flexible volume, but constant shape?

There is a mathematical argumentative approach called reductio ad absurdum...i.e., bring your own logical reasoning to the point where the result is an absurd statement if you accepted the original statement as true. In this case you can accept that there is a phase which can change its volume but keeps it geometrical shape intacr. You can try disproving your own statement using this method for solids, liquids, and gases.

Do we have a phase of matter that fits in the red cell? In other words, do we have a phase of matter that keeps its shape, and only changes its volume? If so, what is the name of that phase?

a) When solids experience stress, there is strain = deformation, change in volume.
b) Liquids are almost incompressible and they have no shape, but under extreme extreme pressures their volume does change, typically, > 100,000 psi they can become solids. The solid will have the shape of container in which it became a solid.
c) Gases acquire the shape of the container (actually a bad statement, i.e., gases occupy the space provide by the container), and their volume can be easily changed by stress.
d) The fancier state of matter, plasma, it is also an ionized gas, so it does not have a geometrical shape. However, it can be confined by using magnetic fields so that it does not touch the walls of the container, and perhaps its "volume" and "shape" can be manipulated independently. A true physicist can better answer it.


Here are some more options:

Pd-catalyzed oxidation of dilute hydrogen gas in air: $\ce{2H2 + O2 -> 2H2O}$  If you use a dilute hydrogen gas mix (say 95% nitrogen and 5% hydrogen), no flaming combustion can occur in air.  But oxidation still happens, at least when catalyzed by palladium particles. This is how microbiological anaerobic chambers maintain a zero-oxygen environment.  You can probably buy some catalyst fairly cheaply from a microbiology anaerobic chamber company like Coy labs.
Phosphorus pentoxide hydration.  Wikipedia says this reaction, $\ce{P4O10 + 6 H2O -> 4 H3PO4}$, has $\Delta H_{reac}=-177\text{ kJ/mol}$.  There is no flame and no fumes, but the reaction does create phosphoric acid, which is corrosive and can be dangerous.
A modification of your sulfite / bleach idea: use sodium dithionite instead of sodium sulfite.  Sodium dithionite ($\ce{Na2S2O4}$) is air-unstable but if you can work out a way to handle without air exposure it will be more exothermic than sulfite.  Instead of dithionite itself you could use the much easier to handle Rongalite and hypochlorite.  Possibly, using calcium hypochlorite suspensions in water (mostly insoluble) will allow you to use higher concentrations of hypochlorite and Rongalite without the reaction rate and temperature skyrocketing (since the rate will be controlled by $\ce{Ca(OCl)2}$ dissolution).


Finding the repeating unit of polymerisation given two constituent molecules

You are starting out with amino groups on one monomer and carboxylate groups on the other monomer. As William R. Ebenezer states in the comment, these combine via condensation (to form an amide, which is written as -NHCO- or -CONH- in condensed formulas).
The linkages in the answer choices between repeating units are not visible until you write another repeating unit next to the ones shown. A. and B. show amide linkages between repeating units, while C. and D. show something else.
Here is the exercise rewritten to emphasize the amide bonds in the repeating units:

The compounds $\ce{H2NCH2CH2NH2}$ and $\ce{HOOCCH2COOH}$ react to form a polymer. What is the structure of the repeating unit of the polymer?
A. $\require{enclose}\ce{\enclose{horizontalstrike}{(}HNCH2CO-NHCH2CH2NH-CO\enclose{horizontalstrike}{)}}$
B. $\ce{\enclose{horizontalstrike}{(}HNCH2CH2NH-COCH2CO\enclose{horizontalstrike}{)}}$
C. $\ce{\enclose{horizontalstrike}{(}OCCH2CO-NHCH2NH-CO\enclose{horizontalstrike}{)}}$
D. $\ce{\enclose{horizontalstrike}{(}HNCH2CH2NH-COCH2NH\enclose{horizontalstrike}{)}}$

Now that we emphasized the amide linkages, you can see that the first monomer in A, the second monomer in C and the second monomer in D don't match your starting materials. We can also exclude C. and D. because repeating units are not linked via amides. The answer has to be B.


In NMR Spectroscopy why use absorption and dispersion components of the FID spectrum rather than magnitude and phase?

The FID signal itself is represented/considered as a set of complex points in time. (Keeler mentions this in chapter 4, section 4.1. - same link as you posted in your question.) Upon Fourier transform, the FID will therefore transform correspondingly to a set of complex points in frequency space.

As Keeler said in section 4.2 of the same chapter, the real part of the spectrum is the absorption mode lineshape, and the imaginary part of the spectrum is the dispersion mode lineshape.

From your last question I think you understood this already, but it is worth repeating.

Aside: Why is the FID a set of complex points?

While all the values that we can detect are real (we cannot detect imaginary current in the receiver coil), we can consider values detected in different directions. Taking the static magnetic field of the spectrometer as the $z$ direction, we can measure the $x$ and $y$ components of the NMR signal. Mathematically, the $x$ and $y$ components can be equivalently expressed as the real and imaginary component (and you gain some conveniences with the Fourier transform if you do so)

Detecting the $x$ and $y$ components of the NMR signal is known as quadrature detection (Keeler, section 5.6), and it is used in order to be able to distinguish between frequencies that are offset from the carrier frequency with different sign. As Keeler describes, instead of fitting two coils to detect the $x$ and $y$ components of the FID directly, it is achieved by phase shifting the receiver reference frequency in real spectrometers.

What is the physical interpretation of absorption mode and dispersion mode?

The pure absorption part represents resonant frequencies of the nuclear spins in your sample.

The pure absorption part of the spectrum is what NMR spectroscopists would use as the NMR spectrum. Signals in the absorption mode occur at distinct frequencies for a particular isotope within a particular sample at a particular magnetic field, which is compared to a frequency of a known reference substance (same type of nucleus, same field/spectrometer, etc). The difference between these two frequencies is the chemical shift, in units of parts per million (ppm) Hz. The chemical shift values for most/all of a particular isotope is the parameter that is reported most commonly in NMR experiments.

Just as the imaginary part of the FID is just the real part of the FID with a phase shift, after Fourier transform, the dispersion mode spectral lineshape has in principle the same information as the absorption mode lineshape. Its presence is mostly for signal processing purposes, but mostly done in the background and not displayed to the user. It is also not commonly used as the "result" of the NMR experiment because it can be hard to determine the accurate chemical shift if the signal does not return to baseline on both sides of the signal, which is exacerbated by the long (positive and negative) "tails" either side of the dispersive mode lineshape. The long "tails" also mean that signals are more likely to overlap each other and the resolution is therefore worse. Some more advanced NMR experiments generate a mix of absorptive and dispersive lineshapes "by default" (called a phase twist), and much effort goes into coming up with strategies to recover the absorption mode lineshape. (Example: 2D NMR spectroscopy, see Keeler chapter 7, section 7.6.2)

Why not just rewrite the spectrum in polar coordinates and consider the magnitude and phase separately?

It is possible to do this. But you will get back magnitude/phase lineshapes that is very similar to the real/imaginary (phase-sensitive absorption/dispersion) lineshapes.

The magnitude spectrum can be calculated by $\operatorname{Re}^2 + \operatorname{Im}^2$. We know that the absorptive lineshape should be completely positive and the imaginary part is more or less the same information but half positive, half negative with long tails. The effect of squaring both and adding them together means that you get something that looks mostly like the absorption lineshape with more signal to noise, at a cost of generating positive long tails from the squared imaginary part, which reduce the spectral resolution.

This is actually a (now somewhat old-fashioned?) technique to address the problem of phase twist in 2D spectra.

The phase spectrum can be calculated by $tan^{-1}\frac{\operatorname{Im}}{\operatorname{Re}}$. No one uses this in NMR (as far as I know!), but we can visualising it by considering what it looks like at different points.

We plot the $tan^{-1}(x)$/arctan function. (Axes cross at the origin.)

At absorption lineshape peak, dispersion lineshape is zero: $\frac{\operatorname{Im}}{\operatorname{Re}}$ is zero, arctan is zero.
At large values away from the peak, we assume everything goes to zero:
$\frac{\operatorname{Im}}{\operatorname{Re}}$ is zero, arctan is zero.
At a short distance from the peak on the right hand side, absorption lineshape is positive, dispersion lineshape is positive:
$\frac{\operatorname{Im}}{\operatorname{Re}}$ is positive, arctan is positive.
At a short distance from the peak on the left hand side, absorption lineshape is positive, dispersion lineshape is negative:
$\frac{\operatorname{Im}}{\operatorname{Re}}$ is negative, arctan is negative.

Assuming that the signal in the absorptive/dispersive lineshapes is also centered at zero, you get something like this:

Joining the points all together gives you a curve that looks like the dispersion lineshape (but not completely the same. My guess is that it is it is also a little broader than the phase-sensitive dispersive lineshape).


"Alkanes must be vaporised before they will burn" - what does this mean?

Burning an alkane means reacting it with oxygen (in air). The larger the contact surface between the alkane phase and the air, the better the reaction. That's why cars have carburators: there's no larger "contact surface" than between the molecules in a gas mixture.


Why does silver plating require undercoats of copper and nickel?

I have read that the nickel plating helps removing oxide layers when working with stainless steel which is beneficial for the silver plating process.

Sadly I do not know why the copper is used but I suspect the context could explain the copper.


Intermolecular forces in Carbon Nanotubes?

There is known, but poorly understood $\pi-\pi$ stacking. While weak compared to covalent, dipole or ionic interactions, it still beats usual van-der-vaals interactions.

Carbon nanotubes are, essentially, polycondensed aromatical systems, so they can and would interact with each other as aromatic systems via $\pi-\pi$ stacking


While there are examples of 1,2-methide shifts, this question has an uncanny resemblance to the dienone-phenol rearrangement whose mechanism was first elucidated by Woodward and Singh in 1950. Dienone 1 under acidic conditions undergoes rearrangement to phenol 6 and not, based on earlier speculation, to phenol 4. The direct 1,2-methide shift (2b --> 3) does not occur but rather the reaction proceeds through the spiro carbocation 5. continued

Applying this approach to the carbocation 7 generated from the alcohol in this question, spiro carbocation 8 forms the predicted tetrahydronaphthalene 10. One way to distinguish between the two mechanisms is via a labeling experiment. All of the carbon label in 7 will retain its location as the red star in 10 if the 1,2-methyl shift mechanism applies. The spiro mechanism will partition the label ~50:50 between the two ring benzylic carbons. For related studies, see reference 2.
1) R. B. Woodward and T. Singh, J. Am. Chem. Soc., 1950, 72, 494.
2) A. J. Waring, J. H. Zaidi and J. W. Pilkington, J. Chem. Soc, Perkin Transactions I, 1981, 1454.


How does shock trigger nucleation of gases dissolved in a liquid?

How is nucleation triggered?
There are, as I see it, three major pathways for nucleation of supercritical carbon dioxide in water.
Bottles: Cavitation
As @Michiel points out correctly in the comments, tapping a glass bottle may cause cavitation, which is then the initial nucleation site for the (runaway) bubble formation. However, I think it is noteworthy that not every type of tapping leads to this; It is when you tap on top of an open bottle with a hard object (such as another bottle) that the gas formation is strongest. This is because the shock waves propagating through the glass walls of the bottle meet up at the bottom of the bottle for a huge impulse into the liquid, forming a larger cavity than other methods (don't do this at parties, people will hate you and their beer will be stale). The relative difference of impulse can be immediately grasped from the fact that the tapping bottle will not start foaming, while the tapped bottle most definitely will.
Cans: Deformation
There are however alternative means of containing beverages, for example aluminium cans. Due to the form of the cans, cavitation formation is highly unlikely. More likely, deformation of the can upon dropping will provide the nucleation sites for the carbon dioxide to form bubbles; think jagged edges on the inside, reactive defect sites in the lattice, and a possibly broken oxide layer of the aluminium, exposing highly reactive elemental aluminium for a short time.
Shaking: Air Dissolution
As @Nikolau correctly pointed out in the comments, shaking a bottle will lead to small bubbles of air getting trapped in the liquid and providing nucleation sites for the gas. Nothing magical about that, but I gave it a heading anyway.

Onwards to the second question you had:
Waiting: Why no foam?
The reason that waiting leads to less foam and spewing around of your favourite carbonated beverage is twofold.

As explained in the linked question on Physics.SE, the bubbles will rise to the top of the liquid level. As such, releasing the pressure will cause further nucleation primarily at the surface and not within the liquid, leading to less liquid propagation by gas evolution. In plain English: This means you don't get soaked by the beverage as it is propelled out of the opening, because the gas evolution happens at the surface and "forces" the liquid to stay in place.

The pressure inside the container rises as more gas is formed. If you wait, a new equilibrium state will be reached by redissolving some of the carbon dioxide that nucleated out of the solution, subsequently lowering the pressure again. (I don't think that the pressure goes back to the initial level within a finite amount of time, but I may be wrong here. If someone could clarify this I would be grateful.)
As there are now less nucleation sites, combined with the fact that they rise to the top (see point 1) foam production is severely hindered.


In a galvanic cell where the two electrodes are in the same electrolyte solution, why do reduction and oxidation occur separately?

Magnesium blocks are connected to iron structures to serve as cathodic protectors. A key element is that the magnesium should react slowly enough (i.e., selectively) to last for a long time (years), while providing enough electron flow to keep the iron negative in the midst of all its electron acceptors (oxidizers). If the iron were in a severely corrosive situation and you needed more current, and you think of using sodium as a sacrificial anode to provide a faster reaction, it wouldn't be practical for many reasons.

The sodium situation is where the electrons wouldn't bother traveling to the iron cathode; things are happening too fast because of the energetics of sodium reacting with water (forget the acid!). This situation does not permit the selectivity to form separate anodes and cathodes.

Dry cells are constructed of zinc anodes and carbon cathodes; impurities in the zinc surface can short-circuit the cell: the zinc becomes both anode and cathode, and is consumed inefficiently. This condition was ameliorated by amalgamating the zinc surface with a tiny bit of mercury. Mercury has a high hydrogen overpotential (i.e., it takes a higher potential to force enough protons to hydrogen atoms and then to molecules and then to bubbles which rise up.) The mercury prevented the initiation of local cathodes, and did not prevent the migration of zinc to its surface and thru it. So the selectivity was improved by the mercury amalgamation, allowing the zinc to be a purer anode. (The next step was to improve it further so as to eliminate the mercury!)

A single iron nail can have electrochemical corrosion all over its surface with anodes and cathodes that switch around and/or are so small that you can't distinguish them. You see only a rusty nail. One way to separate the anodes and cathodes and to make them visible is to put the nail in an agar gel with a little NaCl, phenolphthalein and potassium ferrocyanide. The nail begins to corrode at a random spot. This first spot will lose Fe++ ions to the gel, where they will react with ferrocyanide ion to turn blue. Electrons from the reaction will flood the iron nail; somewhere else, O2 will be reduced to hydroxide ions, which turn the phenolphthalein red. Typically, the head and point of the nail will become anodes (because they are more highly stressed, therefore a little more anodic) while the shaft will be cathodic. Pieces of rebar were more random in their reactivities.

So, oxidation and reduction can occur at the same electrode; it's just more illuminating (and useful) to see them separated instead of all mixed up.


I think the comments basically have the answer - if you use only 2 equivalents of sodium amide, you will use 1.33 equivalents creating the alkyne and the remaining 0.67 equivalent will react with the alkyne after it is formed (an acid-base reaction) to form the acetylide salt -- probably a much faster reaction than the dehydrobromination reactions. So about 1/3 of the original dibromide will remain unreacted. You need the third equivalent of sodium amide to drive the reaction to completion.


Why is the electrode with the lower reduction potential the cathode?

No you misunderstand the meaning of reduction potential. Reduction potential refers to the voltage required to reduce a material under standard conditions. If a material has a higher reduction potential it takes more energy to reduce it than a lower reduction potential material. Therefore the higher reduction potential material is actually oxidized to reduce the lower reduction potential material.


How is the triferric tetroxide of iron possible?

Iron, like many of the Transition Metals does have a variable valency, it can valencies of 2+ (ferrous iron), 3+ (ferric iron) and in some cases, 4+ (tetravalent iron).

In regards to $\ce{Fe3O4}$, according to the Wikipedia page Iron(II,III) oxide,

  contains both $\ce{Fe^2+}$ and $\ce{Fe^3+}$ ions

A 2-dimensional diagram of $\ce{Fe3O4}$ is below, from the relevant ChemSpider page:

A 3-dimensional model is below (from Wikipedia):

red indicate oxygen


To prevent the condensation, you need to create the conditions under which the vapor does not become saturated. The way to do that is to heat up the whole transfer line (the roof of the first flask and all the tubes that follow) - the higher the temperature, the higher the saturated vapor pressure. Heated transfer lines are often employed in commercial equipment when there's need to get condensible gases from one setup (e.g. thermogravimetric) to the other (e.g. mass spectrometer).


The yellowish mustard color I would have to say was caused by your efforts to de-color this polyester, polyester is a decently strong polymer and thus is resistant to many forms of chemical tamper, and in all effects the bleach is damaging the fabric and leaving behind that yellow color, I would suggest using Sulfur Dioxide if you can get your hands on it as that can bleach polyester, you should be left with a white translucent fabric if the bleach hasn't already damaged the fabric too terribly.

Sulfur Dioxide part is iffy, you may choose to disregard that.


Spiralling of electrons into the nucleus

As a matter of fact, Bohr literally went past around this issue. In 1913, he proposed his model postulating electrons could only have certain classical motions:

Electrons in atoms orbit the nucleus.
The electrons can only orbit stably, without radiating, in certain orbits (called by Bohr the "stationary orbits") at a certain discrete set of distances from the nucleus. These orbits are associated with definite energies and are also called energy shells or energy levels. In these orbits, the electron's acceleration does not result in radiation and energy loss as required by classical electromagnetic theory (This was based upon Planck's quantum theory of radiation).
Electrons can only gain and lose energy by jumping from one allowed orbit to another, absorbing or emitting electromagnetic radiation with a frequency ν determined by the energy difference of the levels according to the Planck relation:
$$ \Delta {E}=E_{2}-E_{1}=h\nu \ $$

where h is Planck's constant.

As remarked above, these are postulates (statements assumed to be true from which further reasoning follows), and (even though that part was left unexplained), it's acceptance was due to the fact it explained correctly many experimental results. It was further displaced by the modern quantum theory.

Hope it helps, it also puzzled me a few years ago! :P

Reference: https://en.wikipedia.org/wiki/Bohr_model


Getting 3,4-dimethylhex-3-ene from but-2-ene

Your approach, which uses $\ce{Br2}$ to convert 2-butene into 2,3-dibromobutane and react that with ethylmagnesium bromide $\ce{CH3CH2MgBr}$ has two flaws:

This approach removes the alkene and there is no obvious way to get it back (there is a way)
More seriously, Grignard reagents are pretty terrible for $\text{S}_N 2$ reactions. They are very strong bases, so elimination can compete, but more importantly, they can undergo metathesis:

$$\ce{R-Br + R'-MgBr <=> R-MgBr + R'Br}$$

Assuming your approach worked to generate 3,4-dimethylhexane, you could carefully radically brominate it to produce 3,-bromo-3,4-dimethylhexane, and I am sure you can get back to the alkene from there.

A better approach uses two molecules of 2-butene. Do you see how you can break your target molecule up into two 4-carbon pieces? Convert one into an electrophile and the other into a nucleophile. It might take two reactions in each case. If you choose your electrophile correctly (think about what makes a great combo with a Grignard reagents), you will form the $\ce{C-C}$ bond between the two pieces and you will still have a functional group left that you can use as a handle to direct the elimination reaction to regenerate the alkene.


Why is the probability of finding an electron in an orbital not 100 %?

There is a nonzero probability of finding an electron anywhere except for at the nodes, where the probability is 0 by definition. An orbital might be better thought of as an infinitely large cloud that is more dense in some areas than others. Integrated over all space, there is also by definition a 100% chance of finding an electron. This should make sense intuitively (if if doesn't exist somewhere then where is it and what does the integral even mean?).

Orbitals really aren't so well-defined as the nice little shapes we represent them with (think cloud), but we can make them well-defined by arbitrarily deciding how much space we wish to enclose. It is often decided for the practicality of the representation to show a space such that the probability of finding an electron in it is 95%, though mathematically it's true that there is a chance of finding it outside the space.


To answer your question, what if we did utilize the method you propose?
Before going into the details, it should be noted that the Ion-Exchange method is used to remove hardness of water at an industrial scale. The principal reason that this method is preferred over any other hardness removal process is that it has a unique advantage:
All exchanges take place in an aqueous medium in the form of ions alone.
Keeping that in mind, let's come back to the adjustment you suggested. In the Anion Exchanger, the usual reactions are
$$\ce{R'OH- +Cl^- -> R'Cl- + OH-}$$
$$\ce{2R'OH- + SO4^2- -> R'2 SO4^2- + 2OH-} $$
$$\ce{2R'OH- + CO3^2- -> R'2 CO3^2- + 2OH- }$$
Given that you still haven't removed the cations, the ensuing process would result as follows:
$$\ce{Ca^2+ + OH- <=>> Ca(OH)2} \   \\K_{-1}=\frac{1}{K_{spCa(OH)_2}}$$ and $$\ce{Mg^2+ + OH- <=>>Mg(OH)2}\\K_{-1}=\frac{1}{K_{spMg(OH)_2}}$$
Note that $K_{spMg(OH)_2}=1.8\cdot10^{-11}$ and  ${K_{spCa(OH)_2}}=6\cdot 10^{-6} $
. Further, due to the high concentration of $\ce{OH-}$ ions in the medium, the equilibrium shifts forward, meaning that a high amount of salt is precipitated and deposited within the exchanger, also called scaling. This is detrimental to the industry from any perspective.
Now, let's go back to the part where we discussed advantage of the catex. Do you get it now?
In simpler terms, the fundamental phenomenon behind this process (ION EXCHANGE) is challenged and its efficacy is vastly downgraded.
For future references:

Detailed exchange process
$K_{sp}$ values


Suitable reagents for nitration of thiophene

According to this reference by Katritzky et. al (2005) (see below):

  Thiophenes  are  easy  to  nitrate  compared  to  other  five  membered  heterocycles.  They  react  with  mild  nitrating  agents  such  as  copper  nitrate.

The reference further reports the nitration of thiophene to 2-nitrothiophene by nitric acid/Trifluoroacetic anhydride in 78% yield.

The classic $\ce{c HNO3/c H2SO4}$ conditions are too strong and likely to result in substrate degradation.

Reference:

Alan R. Katritzky, Eric F. V. Scriven, Suman Majumder, Rena G. Akhmedova, Novruz G. Akhmedov, Anatoliy V. Vakulenko, "Direct nitration of five membered heterocycles," ARKIVOC 2005, (iii), 179-191 (https://doi.org/10.3998/ark.5550190.0006.320).


Can Na2CO3 remove temporary hardness of water?

Definition of hardness per a source:

The simple definition of water hardness is the amount of dissolved calcium and magnesium in the water. Hard water is high in dissolved minerals, largely calcium and magnesium. You may have felt the effects of hard water, literally, the last time you washed your hands.

Adding $\ce{Na2CO3}$ introduces the carbonate ions:
$$\ce{Ca++/Mg++ + CO3(2-) -> CaCO3/MgCO3 (s) }$$
which results in a precipitate of insoluble carbonates.
So, yes it is a temporary solution to address water hardness.
[EDIT] I recommend relatedly to hard water treatment with Washing Soda, a very interesting experiment of adding Na2CO3 (aq) to MgSO4(aq) as the entire solution, at times (it depends on respective concentrations),
can convert to a wet hydrate creating so-called Magnesium alba. Cited reaction per Wikipedia:

If magnesium chloride (or sulfate) is treated with aqueous sodium carbonate, a precipitate of basic magnesium carbonate—a hydrated complex of magnesium carbonate and magnesium hydroxide—rather than magnesium carbonate itself is formed:
$\ce{5 MgCl2(aq) + 5 Na2CO3(aq) + 5 H2O(l) → Mg(OH)2·3MgCO3·3H2O(s) + Mg(HCO3)2(aq) + 10 NaCl(aq) }$

where the magnesium mix oxide/carbonate is known, but not uniquely in name or precisely in chemical composition, as Magnesium alba. Now, MgSO4 aqueous per a source:

Magnesium sulfate salts can form hydrated compounds with up to seven ... Magnesium sulfate hexahydrate is formed by coordinating 6 water molecules to Mg2+ ...

So, Magnesium sulfate forms $\ce{MgSO4(H2O)6.H2O}$ with six coordinating water molecules (which could be created upon concentration of the aqueous MgSO4 solution, further extracting water from the mix). Further, if Magnesium alba similarly also extracts water, with excess MgSO4, it may be understandable as to why the entire solution can apparently turn into a voluminous wet precipitate, somewhat surprisingly.


why does FeCl₃ have such a specific smell?

The only clue I have been able to find is a one-off sentence in the Wikipedia article for $\ce{FeCl3}$, for which no reference is given:

  Iron(III) chloride undergoes hydrolysis to give an acidic solution.

If so, then the reactions are probably those along the conversion of $\ce{FeCl3}$ to $\ce{Fe2O3}$, producing hydrochloric acid $(\ce{HCl})$. The hydrogen chloride gas that escapes from the solution has a pungent acidic acrid smell.

Possible reactions include:

$$\ce{FeCl3 + H2O <=> FeOCl + 2HCl}$$
$$\ce{2FeCl3 + H2O <=> Cl2FeOFeCl2 + 2HCl}$$
$$\ce{2FeOCl + 2H2O <=> Fe2O3(s) + 2HCl}$$

Because these reactions are in equilibrium, solutions of $\ce{FeCl3}$ can be stabilized by addition of $\ce{HCl}$. Addition of $\ce{HCl}$ also promotes the formation of the tetrachloroferrate ion $\ce{FeCl4-}$, which is more resistant to hydrolysis.

$$\ce{FeCl3 + HCl -> FeCl4- + H+}$$

Ferric oxide (Iron(III) oxide $\ce{Fe2O3}$), is insoluble. Does your solution develop reddish precipitate as it ages? That would be  evidence in favor of hydrolysis as the source of $\ce{HCl}$.


Does gravity affect the trajectory of gas particles?

Yes gravity pulls on gas molecules. That is why the atmosphere doesn't just float off into space.

The gist is that the time between collisions is very short in the lower atmosphere, and the distances very short. The mean free path at atmospheric pressure is only about 70 nanometers. So the assumption is that gas particles travel in a straight line between collisions.


Determine polarity of covalent bond with formal charge

All bonds between two atoms of different elements are polar to some extent. To determine how polar, you need to check the elements’ electronegativities and only them.

Formal charges are exactly what their name implies: a formality. When drawing Lewis structures you should attempt to generate all-octet structures but often this isn’t possible without moving electrons entirely between atoms. In that case, formal charges are assigned to rationalise e.g. a nitrogen with only four electrons.

Check out these structures of nitric acid, sulphur trioxide and carbon monoxide. (Resonance structures applicable to $\ce{HNO3}$ and $\ce{SO3}$ not displayed.)  All bonds are polarised towards the oxygen, since oxygen’s electronegativity is $3.5$; the second highest amoung the non-noble-gas elements. Note especially that the bonds in $\ce{CO}$ are polarised towards oxygen despite oxygen bearing a formal positive charge. (There is a lot more to $\ce{CO}$ which is shown well in Martin’s answer including the molecular orbitals)


SN2: Neopentyl halides and methyl halides

Carey-Sundberg has some answers (vol. A, chapter 4).  The text is a bit advanced for you at the present stage, but you should give it a try.

It gives as reference JACS 1975, 97, 3694.  That paper has rate constants for nucleophilic substitution of compounds of type $\ce{RCH2Br}$, with $\ce{R=H, Me, Et, i-Pr, t-Bu }$ with $\ce{LiCl}$ in acetone at room temperature.

Rate constants in 105 s-1:
$\ce{R=H}$: 600, $\ce{R=Me }$: 9.9, $\ce{R=Et}$: 6.5, $\ce{R=i-Pr}$: 1.5, $\ce{R=t-Bu }$: 2.6*10-4

Nucleophilic substitution does take place in neopentyl halides, but at a rate of five orders of magnitudes less than in less hindered compounds.


Confused about weak/strong acids, conjugated acid-base pairs and buffers

Your statement a) isn't always true.

Water dissociation is represented by:

$$\ce{H2O + H2O <-> H3O+ + OH-}$$

$$ K_\mathrm w=[\ce{H3O+}]\cdot [\ce{OH-}] = 1\times 10^{-14}\ (\textrm{at}\ 25^\circ ~\mathrm C) $$

Note 1: We don't write the $\ce{H_2O}$ activity, since it can usually be rounded to 1 and the ions activities can be rounded to their concentrations.

Note 2: This value was obtained experimentally, considering the concentration of $\ce{H_3O^+}$ and $\ce{OH^-}$ in the medium was the same and measuring the ionization of water.

On a weak acid dissociation:

$$\ce{HA + H2O <-> A- + H3O+}$$

$$K_\mathrm a=\frac{[\ce{A^-}]\cdot [\ce{H3O^+}]}{[\ce{HA}]}$$

On its conjugate base:

$$\ce{A- + H2O <-> HA + OH-}$$

$$K_\mathrm b=\frac{[\ce{HA}]\cdot [\ce{OH^-}]}{[\ce{A^-}]}$$

If we 'add' both reactions we expect that both equilibriums will happen in the mixture, so we have:

$$\ce{HA + H2O <-> A- + H3O+}\ \ \ \ (K_\mathrm a)$$

$$\ce{+}$$

$$\ce{A- + H2O <-> HA + OH-}\ \ \ \ (K_\mathrm b)$$

$$\ce{============================}$$

$$\ce{H2O + H2O <-> H3O+ + OH-}\ \ \ \ (K_\mathrm w)$$

$$K_\mathrm b\cdot K_\mathrm a=[\ce{H3O+}] \cdot [\ce{OH-}] \cdot ([\ce{A-}]\cdot [\ce{HA}])/([\ce{A-}] \cdot [\ce{HA}]) = [\ce{H_3O^+}]\cdot [\ce{OH^⁻}] = K_\mathrm w$$

$$K_\mathrm b=K_\mathrm w/K_\mathrm a~~~~~~ K_\mathrm b=10^{-14}/K_\mathrm a\ \ \textrm{at}\ 25^\circ ~\mathrm C$$

Another way to write this:

\begin{align}-\log(K_\mathrm b)&=-\log(10^{-14}K_\mathrm a) \\ \implies \mathrm pK_\mathrm b &= -\log(10^{⁻14}) - (-\log(K_\mathrm a))\\ \implies \mathrm pK_\mathrm a+\mathrm pK_\mathrm b &=14\;.\end{align}

This is the relation between a conjugate base strength and its acid strength. A very strong acid has a weak conjugate base, but a weak acid doesn't necessarily have a very strong base. Say you have a $\mathrm pK_\mathrm a=5$, which is a weak acid, with $K_\mathrm a=1\times 10^{-5}$. The conjugate base would have a $\mathrm pK_\mathrm b=14-5=9$ or a $K_\mathrm b=1\times 10^{-9}$, which is not a strong base.

However, if we have a strong acid, like $\ce{HCl}$ with a $\mathrm pK_\mathrm a=-6.3$ and $K_\mathrm a=10^{6.3}$. Its conjugate base would have a $\mathrm pK_\mathrm b=14-(-6.3)=20.3$ and $K_\mathrm b=10^{-20.3}$ which is a really weak base.

Hopefully I didn't make it even more confusing for you! It's mostly an reaction equilibrium issue.


What will be the reaction between tin (II) chloride and copper (II) chloride?

This is indeed a redox reaction between copper and tin, but you should instead use reduction potentials for the $+2$ ions to calculate the potential.

This might be quite an exothermic reaction. http://www.periodicvideos.com/videos/mv_copper_chloride.htm


The preferred IUPAC name (PIN) of the compound given in the question is butan-2-yl hypofluorite. It is named as ester of hypofluorous acid. In the seniority order of classes, esters are ranked between anhydrides and acid halides.
The corresponding subsection of the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) reads as follows:

P-67.1.3.2 Esters of mononuclear noncarbon oxoacids
Esters of mononuclear noncarbon acids are named in the same way as esters of organic acids (…). Alkyl groups, aryl groups, etc. are cited as separate words, in alphanumerical order when more than one, and followed by the name of the appropriate anion. Partial (acid) esters of polybasic acids are named by citing alkyl groups, aryl groups, etc. as separate words, in alphanumeric order if more than one, followed by the word ‘hydrogen’ (with the appropriate multiplying prefix, as necessary) also cited as a separate word, and the name of the appropriate anion. Salts of partial (acid esters) are named by citing the name of the cation before the name of the organic group; remaining acids groups are denoted by the word ‘hydrogen’ as described above. Structural specificity for esters of chalcogen analogues of mononuclear noncarbon oxoacids is provided by the appropriate italic element symbols O, S, Se, and Te, prefixed to the name of the group, as needed.


Why water can't mix with oil or cooking oil?

The basic answer is that water molecules attract each other and so clump together forcing almost all of the oil out of the clump.

The attraction exists because water is a polar molecule.  That is, it has a positive end and an negative end.  Thus the molecules tend to clump together just as a bag of small magnets tend to clump together.  Oils, on the other hand, are not polar.

This principle can be expanded.  In general polar materials tend not to dissolve in oils or, if they are liquids, dissolve in them.  The rule in chemistry is "like dissolves like".


Are there biodegradable pigments that are soluble in oil, but not in water?

There are several such pigments which are bio-degradable by some microbial action (mostly bacterial action) and are oil soluble (lyophilic) and water insoluble (hydrophobic).

Some are:

chlorophyll : certain bacteria can decompose chlorophyll.
A paper citing zoo-plankton activity on chlorophyll is here.
Xanthophylls : This is a general category of several pigments whose degradation by marine bacteria has been widely  studied. (You can search for many relevant papers online).
Carotenes : Certain carotenes are also reported to be degraded by fungi and mushrooms. For example here

Anthocyanins (and related polyphenols) which have dark red to purple color are mostly water soluble.


An E2 reaction will not occur with this reagent.

An E1 reaction, on the other hand, could do so (though quite slowly: E1 is an all around pathetic reaction mechanism, and this molecule does nothing special to stabilize it). The chlorine will leave and there will be a carbocation on the end of the alkene, whereupon the proton from the other side of the double can be extracted with a base in a second step, leaving you with an alkyne.

That said, if you want the alkyne, it would be infinitely infinitely easier to start with the cis isomer of the molecule you have and just run it through very strong base. E2 is far far better than E1 in terms of reaction rate and yield.


Both names that are proposed in the question, ‘4-isopropyl-2-cyclohexyl-3,5-dimethylheptane’ and ‘2-cyclohexyl-3,5-dimethyl-4-isopropylheptane’, are unambiguous and describe the correct structure. However, they are not in accordance with IUPAC recommendations.

According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), simple prefixes (simple substituent groups consisting of just one part that describes an atom, or group of atoms as a unit, for example methyl, cyclohexyl, and isopropyl) are arranged alphabetically disregarding any multiplicative prefixes. For example, ‘3,5-dimethyl’ is considered to begin with ‘m’. Any multiplicative prefixes are inserted later and do not alter the alphabetical order.

  P-14.5 ALPHANUMERICAL ORDER

  Alphanumerical order has been commonly called ‘alphabetical order’. As these ordering principles do involve ordering both letters and numbers, in a strict sense, it is best called ‘alphanumerical order’ in order to convey the message that both letters and numbers are involved

  Alphanumerical order is used to establish the order of citation of detachable substituent prefixes (not the detachable saturation prefixes, hydro and dehydro), and the numbering of a chain, ring, or ring system when a choice is possible.

  (…)

  P-14.5.1 Simple prefixes (i.e., those describing atoms and unsubstituted substituents) are arranged alphabetically; multiplicative prefixes, if necessary, are then inserted and do not alter the alphabetical order already established.

Therefore, the correct alphanumerical order for the example given in the question corresponds to the name 2-cyclohexyl-4-isopropyl-3,5-dimethylheptane since ‘cyclohexyl’ begins with ‘c’, isopropyl begins with ‘i’, and methyl begins with ‘m’.

However, note that the prefix ‘isopropyl’ is retained for use in general nomenclature but the preferred prefix is ‘propan-2-yl’. When using this prefix, the correct alphanumerical order  corresponds to the name 2-cyclohexyl-3,5-dimethyl-4-(propan-2-yl)heptane.

Anyway, to construct a preferred IUPAC name (PIN), the ring is selected as the parent hydride. Therefore, the PIN for the example given in the question is [3,5-dimethyl-4-(propan-2-yl)heptan-2-yl]cyclohexane. (In general nomenclature, the context may favour the ring or the chain, so that, for example, substituents may be treated alike or an unsaturated acyclic structure may be recognized, or the one chosen has the greater number of skeletal atoms in the ring or in the principal chain of the acyclic structure.)