Do I need to include lattice energy when calculating the energy required to vaporize NaCl? I understand that the lattice energy is the energy released when a crystal forms and that's the reason it takes the negative sign. But do we need to add this energy again when we are trying to vaporize an ionic compound ?

For example, let's assume we want to completely vaporize a mole of NaCl starting at room temperature, how much energy do we need ?

I don't know neither the specific heat nor the latent heat of vaporization of NaCl, but I understand we would need:

1- Energy to raise the temperature of mole of NaCl from 298 °K to its boiling point.

2- Energy equal to the latent heat of vaporization of a mole of NaCl.

I don't know whether NaCl would go through the liquid state or not, but if it does then we would need to add the latent heat of fusion too.

Now, according to wikipedia, NaCl has a lattice energy of −756 kJ/mol. So my question here is: do we need to add this lattice energy to the calculated steps above to account for all the energy required to vaporize a mole of NaCl ?

Lattice energy

  Now, according to wikipedia, NaCl has a lattice energy of −756 kJ/mol.

First, we have to understand the term lattice energy.
Here is the textbook explanation (Fleming: Physical Chemistry): The lattice energy is the energy required to separate the ions in an ionic lattice so that they are at infinite distance (but still ions). This would be difficult to do experimentally, but the value may be determined by using a Born Haber cycle. As the diagram below shows, we know the enthalpy of reaction for:

$$\ce{Na(s) + 1/2 Cl2(g) -> NaCl(s)}$$

We can get the same energy in a thought experiment (path 2 in the diagram), first turning all species into gas, then into atoms, then transferring the electron and - finally - forming the lattice from the individual anions and cations. All these processes are experimentally accessible except for the last, so you can determine the lattice energy this way.

Source: diagram adapted from the cited textbook.

Enthalpy of vaporization

  But do we need to add this energy again when we are trying to vaporize an ionic compound?

You problem is not well-defined. What are the initial and final temperatures, what is the pressure? What does it mean to vaporize solid NaCl or any ionic compound, i.e. what is the product? If we boil sodium chloride at 1 atm pressure, it produces mostly NaCl monomers in the gas phase, see https://chemistry.stackexchange.com/a/14560. If you are interested in this process, you could write it as:

$$\ce{NaCl(s) -> NaCl(g)}$$

So to get the required energy, you could first separate all ions (lattice energy) and then calculate the energy of forming individual pairs of sodium and chloride ions (assuming the interaction is purely ionic).

$$\ce{NaCl(s) -> Na+(g) + Cl-(g)}\tag{1}$$

$$\ce{Na+(g) + Cl-(g) -> NaCl(g)}\tag{2}$$

It should be obvious that process (1) cost much more energy than what you get out of process (2) because you are going from 6 nearest neighbors to just one. If you look up the atomic distance of $NaCl(g)$, you could estimate the enthalpy of (2) pretty accurately and get a numeric answer.

  [Comment from https://chemistry.stackexchange.com/a/14177]: Roughly, somewhere around half of the lattice enthalpy of a salt comes solely from the binding of the smallest electrically neutral agglomerate; that is, it takes about as much energy to break a macroscopic solid NaCl crystal into a gas of ion pairs as it does to break all the ion pairs and create a true plasma. Thus, it is rather unlikely that, at reasonable temperatures, an ionic gas will break down any further than the smallest possible electrically neutral aggregates. – Nicolau Saker Neto Jul 8 '14 at 16:39

According to the comment above, the heat of vaporization should be about half of the lattice energy (with opposite sign).

  I don't know whether NaCl would go through the liquid state or not, but if it does then we would need to add the latent heat of fusion too.

Because enthalpy is a state function, it does not matter if you choose the actual (realistic) path or take a different path. You are right that if you step through what actually happens, you need to know heats of fusion and vaporization as well as heat capacities of all phases for the relevant temperature ranges.


How does black CuO impart a green color to glazes and glass? I read from a source that cupric oxide (CuO) imparts green to blue colour to glazes and glass. But CuO is black in colour. How is this possible?

This is due to the oxidation of cupric oxide.

The main componds believed to cause this colour comprise a mixture of 3 compounds: $\ce{Cu4SO4(OH)6}$ (green); $\ce{Cu2CO3(OH)2}$ (green); and
$\ce{Cu3(CO3)2(OH)2}$(blue).

The following reactions are believed to take place:

Copper (I) oxidised to the black copper (II) sulfide ($\ce{CuS}$)  in the presence of sulfur impurities. Under accelerated conditions, the described process occur at a faster rate.

$\ce{CuO}$ and $\ce{CuS}$  reacts with carbon dioxide ($\ce{CO2}$) and hydroxide ions ($\ce{OH-}$) in water (in presence of air) to form $\ce{Cu2CO3(OH)2}$.

The extent of humidity and the level of sulfur have a significant impact on how fast the compounds develop, (under controlled conditions these reactants are varied to produce a favourable hue) as well as the relative
ratio of the three components.

$$\ce{2CuO + CO2 + H2O → Cu2CO3(OH)2~~~~~~~(1)}$$

$$\ce{3CuO + 2CO2 + H2O → Cu3(CO3)2(OH)2~~~~~~~~(2)}$$

$$\ce{4CuO + SO3 +3H2O → Cu4SO4(OH)6~~~~~~~~~(3)}$$

References

http://www.wskc.org/documents/281621/282063/ENGAGE_E3S_Chemistry_Statue+of+Liberty.pdf/e4f24c7e-3666-425e-9c41-7dbdd0065eb4


How does pressure influence the dehydration of salts? It's well known that the boiling point of a substance increases with pressure. I was wondering if something similar happens with hydrated salts and the temperature at which they dehydrate.

I mean, if the pressure is increased, does the temperature at which a salt dehydrates rise or is it not affected by the pressure?

Yes - the pressure will impact the dehydration temperature.
You have correctly identified that temperature and pressure are related by the Gas laws.
High pressure can also force water molecules out, as water is loosely bonded in salts.


Is the reasoning of spin contamination correct? The common explanation of spin contamination is like the following:

The non-relativistic electronic Hamiltonian commutes with $S^2$ and $S_z$. Hence, the exact eigenfunction should be an eigenfunction for $S^2$ and $S_z$ operators.

However, the non-relativistic Hamiltonian has no spin variables. Any wavefunction contains spin, the spin part will commute with the Hamiltonian anyway.

For instance, in addition of angular momentum,
http://www2.ph.ed.ac.uk/~ldeldebb/docs/QM/lect15.pdf
One can form $\{J_1^2, J_2^2, J_{1z}, J_{2z}\}$ set of variables. They commute with Hamiltonian, but no longer eigenfunction of $S^2$, i.e. $\{J^2, J_z, J_{1}^2, J_{2}^2\}$.

If my argument is correct, what would be the actual justification for spin adaptation?

Commuting operators indeed admit a set of simultaneous eigenfunctions, and since the non-relativistic electronic Hamiltonian commutes with total spin operators $\hat{S}_{z}$ and $\hat{S}^{2}$, the exact non-relativistic electronic wave function (which is an eigenfunction of the electronic Hamiltonian) is also an eigenfunction of the total spin operators. And this requirement can be also imposed on an approximate non-relativistic electronic wave function.

This is certainly true that the non-relativistic electronic Hamiltonian commutes not only with total spin operators $\hat{S}_{z}$ and $\hat{S}^{2}$, but with operators of the individual sources of spin $\hat{S}_{iz}$ and $\hat{S}_{i}^{2}$ as well, or to be more precise, with their trivial extensions on the state space of a many particle system. Thus, this is indeed true that one could adopt the uncoupled picture as well, rather then the usual coupled one, but what is the usefulness of such picture? In the absolute majority of cases the most important characteristic of a system is a property which describes it as a whole, rather that its components individually. For this particular case, recall, for instance, that the total spin of a molecular system is a very important characteristic of it. For instance, it is heavily used in electronic spectroscopy: important terms like term symbols, spin-allowed and spin-forbidden transitions, fluorescence, phosphorescence, etc. are all directly formulated in terms of total spin angular momentum of electronic states.

Now if you decide to use the uncoupled picture in which the total spin is unspecified you basically loose all the body of knowledge of electronic spectroscopy. But what do you gain instead? What is the point of using the uncoupled picture? You can specify the spin of each electron, yes, but but so what?


Can 4-aminophenol be used to produce 2,5-cyclohexadien-1-imine? If the hydroxl group on 4-aminophenol was protonated by a strong acid in aqueous solution, and the compound further to that dehydrated, could a rearrangement of the subsequent carbocation lead to formation of 2,5-cyclohexadien-1-imine and thus regenertion of the strong acid before SN1 was possible?

Is there a more common name for this sort of reaction (assuming it is possible)?

TL;DR

No

Explanation

As far as I understand, the reaction being suggested is as follows:

The first thing one should notice is that 2,5-Cyclohexadien-1-imine is not going to stay as it is. It will tautomerise to the more stable and aromatic aniline.

Let us go back to the starting material - 4-aminophenol. Both the oxygen and the nitrogen are directly attached to the aromatic $\pi$-system. Both can donate a lone pair and cause a +R effect. Their electronegativities would result in a -I effect, which is very small, compared to the +R effects. The point is, the starting material is electron-rich, i.e. nucleophilic. It would not leave a carbocation (if it does form) alone.

Also, the thing about the lack of a hydride donor...


How it was observed, that elements in columns 6 and 11 assume abnormal electron configuration? Reading this question about why do elements in columns 6 and 11 assume abnormal electron configurations I wondering, how this was observed.

How it was observed, that elements in columns 6 and 11 assume abnormal electron configuration?

I assume you to mean

  How it was first observed, that elements in columns 6 and 11 assume abnormal electron configurations?

Chemists (starting with alchemists) have were doing stoichiometric reactions for a very long time.

The point is that if there was only a single "normal electron configuration" for chromium, then there would only be one chromium chloride. However chemists have synthesized three.

Chromium(II) chloride, also known as chromous chloride.
Chromium(III) chloride, also known as chromic chloride or chromium
trichloride
Chromium(IV) chloride

Thus the fact that there must be "abnormal" electron configurations was first discovered by stoichiometry, not spectroscopy.

Other columns in the periodic chart do have elements with multiple electron configurations. So such an "abnormality" isn't unique.

If your question means

  How are the bonding energies measured for the elements in columns 6 and 11 in various electron configurations?

Then ESCA would be one technique.


Hydrogen and Helium ideal gas behavior in room temperature Which one of hydrogen and helium behaves more like ideal gas in room temperature...

I thought hydrogen gas would show more ideal behavior in room temperature because it has lighter mass (less no of electron) although it's diatomic, so there would be less attraction.
However, I saw one question on the book and they said that the helium gas is more likely to show the ideal gas behavior in room temperature, not the hydrogen gas..

So, can someone please give me an idea how my thinking is wrong?.

Thanks for advance

For inert gas molecules, intermolecular interactions would be by Van der Waals forces, which are based on the total volume and polarizability of the gas molecule.

The diatomic hydrogen is larger and more polarizable of the two.


What benefits do higher-rated NMR tubes provide? You can buy NMR tubes in a huge variety of qualities, with an equally huge difference in price between the cheapest and the most expensive NMR tubes. They are usually rated for a specific spectrometer frequency, e.g. 300 MHz+ or 600 MHz+.

What is the difference between these different grades of NMR tubes? And how large is the effect on the quality of the spectra if a lower-grade tube is used?

A general rule is: crap goes in, crap comes out. A large-sample low-field 1D NMR at room temperature is usually only minimally affected by using a cheap NMR tube. There are important differences though and I’ll highlight a few.

The first distinction between prices is what the tube is constructed from: quartz obviously costs more than borosilicate. Why would a chemist ever use the more expensive quartz? You can heat/cool quartz faster (nice for thermal studies), the UV cutoff is lower (think 190 nm opposed to 320 nm) which is important for photolysis, you can work with quartz at higher temperatures (around 1300 °C instead of 250 °C), and the purity of quartz is better controlled than your typical Pyrex. There are different grades of quartz, fused and synthetic, and there are different grades of borosilicate, such as the high-quality Pyrex or the lower-quality Class B, each comes with its own limitations as far as purities are concerned and so forth.

Three more important parameters have to do with the manufacturing of your tube are: concentricity, camber and wall thickness. Lower quality tubes will tend to have less precision and accuracy over each of these parameters and as a result your sample may wobble while spinning (introducing problems such as modulation sidebands). A particularly bad tube can hit your RF coils and cause damage to your probe over time slowly or quickly if it is ignoring any reasonable standard – even more apparent for a tube at this level of “quality” is that it may be easier to break while acquiring your sample and we all should be aware of how much fun that is for everybody involved.

Shimming can deal with impurities present in the glass (such as ferric oxide) and increased impurities in the glass/inhomogeneities will result in taking longer to get a good shim. Time is money.

A lot of these things have lower tolerances in more complex experiments and at higher fields. It really does depend on your particular experiment and what you’re hoping to get out of it.


Do all metal salts in aqueous solution contain metal aquo complexes? I know that nickel nitrate in aqueous solution contains the metal aquo complex ion $\ce{[Ni(H2O)6]^2+}$. But do all metal salts in aqueous solution contain such complex ions? If not, on what factors does the formation of metal aquo complex ions depend? Is the metal required to be a transition metal?

[OP] But do all metal salts in aqueous solution contain such complex ions?

This is a good question and in many cases, the exact structure and dynamics of aqueous metal ions is still being studied. In the simplest terms, if something is in aqueous solution, it will be surrounded by water. To start getting some more details, take a look at the Wikipedia article on metal aquo complexes. It says that metals surrounded by a first shell of six waters (with an octahedral geometry) is very common. They also point out that some of these complexes are long-lived while others are not:

  Rates vary over many orders of magnitude. The main factor affecting rates is charge: highly charged metal aquo cations exchange their water more slowly than singly charged species. Thus, the exchange rates for $\ce{[Na(H2O)6]+ and [Al(H2O)6]^3+}$ differ by a factor of $\pu{e9}$. Electron configuration is also a major factor, illustrated by the fact that the rates of water exchange for $\ce{[Al(H2O)6]3+ and [Ir(H2O)6]^3+}$ differ by a factor of $\pu{e9}$ also.[3] Water exchange usually follows a dissociative substitution pathway, so the rate constants indicate first order reactions.

Type of metal

  [OP] Is the metal required to be a transition metal?

For sodium ions in aqueous solutions, just calling them $\ce{Na+(aq)}$ is probably sufficient because the waters change places so rapidly. For a nickel complex with waters in the first shell not exchanging that quickly, it makes more sense to talk of a aquo complex and to study its geometry.

Other coordination numbers and geometries

If you dig deeper, you will encounter other coordination numbers and geometries. For example, according to this report, copper(II) likes to surround itself with five ligands in a square pyramidal fashion. Here is an excerpt of the abstract:

  Cu K-edge extended X-ray absorption fine structure (EXAFS) and Minuit X-ray absorption near-edge structure (MXAN) analyses were combined to evaluate the structure of the copper(II) imidazole complex ion in liquid aqueous solution. [...] This core square-pyramidal motif has persisted through [Cu(H2O)5]2+, [Cu(NH3)4(NH3,H2O)]2+,(1, 2) and now [Cu(Im)4Lax)]2+ and appears to be the geometry preferred by unconstrained aqueous-phase copper(II) complex ions.

I think for geometries that are not the optimal space-filling ones, you need transition metals, but I have not done a careful survey of the available literature.


Is it possible to detect the component of an unknown substances using FTIR? Currently I'm doing research involving photo-polymer which contains unknown substances. Is it possible to detect each component of the substance by using FTIR? Will it be able to separate the photoinitator component, the monomer, and the oligomer?

Not really... FTIR is more like a fingerprinting device. FT basically means identification is done with your computer. But you just have one line/graph that represents all three together. It doesn't separate the spectra; but with practice you can "see" things that are mixed, although some chemical tend to dominate the spectrum. Also, the FTIR device that you get will probably have a program that can help interpret peaks (which I have found to be utterly useless). The real way it (the spectrometer) works is by searching through the library. Again, the device you get will come with a library, but it's usually not worth anything. You will need to create your own library including all of the possible combinations (identify/name each sample that you make).

I would say that with enough practice and repeated sampling (building a library of known mixtures), that you could might be able to identify an unknown sample as a mixture of the components; and possibly the amounts of mixed components (depending on what they are, exactly).

In other words, let's suppose you have decane, benzene, and acetone. You will need to create a library like:

pure decane
pure benzene
pure acetone
5% Benzene in decane
25% Benzene in decane
50% Benzene in decane
75% Benzene in decane
95% Benzene in decane
5% acetone in decane
25% acetone in decane
50% acetone in decane
...
35% acetone 55% decane 10% benzene
...

And you should run each sample 3 times. Then, after you have exhausted the list of known culprits, you should be able to nearly identify an unknown sample which is known to be mixture of these three chemicals (supposing that the amounts of each are what is unknown)... but be sure to run/analyze the unknown sample at least three times (adding it to the library each time) for best results.

Now with regards to photoinitiators... if using metals (like silver), it might be more tricky. IR Spec is most useful for organic chemistry.


Why does de Broglie wavelength lambda = c / frequency? For a chemistry problem, I want the velocity of the de Broglie wave to agree with one of these values:

A) $c^2/v$
B) $h\nu/mc$
C) $mc^2/h$
D) $v\lambda$

We know that
$$\lambda = \frac{h} {mv}$$
But I am confused by the next step in my book:
$$\lambda = \frac {c} {\nu}$$
After which:
\begin{align}
\frac{h}{mv} &= \frac{c}{\nu} \\
\implies v &= \frac{h\nu}{mc}
\end{align}
Why is the de Broglie wavelength $\lambda = c/\nu$? In every context I've seen it used, $c$ is the speed of light. But I don't think the velocity of the matter wave could equal the speed of light, because the matter wave is only as fast as the particle – which, having mass, cannot travel at the speed of light.
What am I doing wrong here? Does $c$ moonlight in quantum chemistry as the velocity of the matter wave, separate from its connotations in physics?

$v$ is not the wave speed, it is the speed of the object with the given mass and de Broglie frequency. Also, there should be rather $p$ instead of $mv$, unless $v \ll c$.
As $$ E = \sqrt{(mc^2)^2 + (pc)^2}$$
then
$$p = \sqrt{\left({\frac {E}{c}}\right)^2 - {(mc)}^2}$$
For a massless object like a photon:
$$p = \frac Ec$$.
For an object with mass, there is also $$p  = \frac {mv}{\sqrt{1-(v/c)^2}}= \gamma mv$$
See Matter wave, paragraphs De Broglie relations and Special relativity, where is clear c is the phase speed of light in vacuum:

$$\begin{align}&\lambda =\,\, \frac {h}{\gamma m_0v}\, =\, \frac {h}{m_0v}\,\,\,\, \sqrt{1 - \frac{v^2}{c^2}}\\
& f = \frac{c}{\lambda} = \frac{\gamma\,m_0vc}{h} = \frac {m_0c^2}{h} \bigg/ \sqrt{\frac{c^2}{v^2}-1}
\end{align}$$


Thermodynamics at absolute temperature Assume I have kept a single molecule in vacuum and in absolute zero (0 K) temperature. There is no energy left in the particle, and hence any type of motion is absent.
Now, what happens when I increase the temperature of that system? Does the molecule start to move? Does the energy increase?
What does temperature even mean in vacuum?
I don't know I'm asking a good question, but this troubles me a lot.

The situation you described is not a vacuum because it has a single molecule in it, so what you're asking about is how the concept of temperature applies to an isolated molecule.

First, the idea that you can start at absolute zero even with a single molecule is not consistent because there will always be fluctuations/movement possible in the ground state of a quantum system (the zero point energy or ZPE). The cleanest example of this is the quantum harmonic oscillator: a particle in the ground state has an average position of zero ($\langle x \rangle = 0$), but the fluctuations are greater than zero ($\langle x^2 \rangle > 0$). Let's think about this in terms of the uncertainty principle. The uncertainty principle is given by $\Delta p\Delta x \geq \hbar /2$ where $\Delta x  = \sqrt{\langle x^2 \rangle - \langle x \rangle^2}$ and $\Delta p  = \sqrt{\langle p^2 \rangle - \langle p \rangle^2}$. Because $\langle x^2 \rangle >0$ and $\langle x \rangle = 0$, we can see that the uncertainty in the momentum is also non-zero. As we know, the kinetic energy is $p^2 / 2m$, so our energy uncertainty is non-zero as well.

Second, and onto the main point of your question, you can certainly define a concept of temperature with only one molecule. Consider that even a single molecule has many different energy levels (i.e. electronic, vibrational, and rotational degrees of freedom are all available). The partition function for our system is still defined as $Z = \sum_i \exp{\left(-\frac{E_i}{k_bT}\right)}$ where $E_i$ is the energy of the $i$th state.(See Note) So, if we start very near absolute zero we will see only contributions from the ground state, but as we increase the temperature of the system we will observe non-zero probabilities of populating excited states. Nothing is broken about the concept of temperature in this situation, the energy scales are just considerably smaller than those in an ideal gas.

Note:

Your initial reaction to applying thermodynamic principles to a single molecule might be to think that it doesn't apply because we only have one molecule and not an ensemble. However, within the density matrix formulation of statistical mechanics, the quantum uncertainty of a system is entirely indistinguishable from the statistical uncertainty of a system.


Butler-Volmer equation with concentration dependence? For a simple elementary reaction on electrode

\begin{equation}
O+e^-\rightleftharpoons R
\end{equation}

We can derive the Butler-Volmer equation. But it seems that the formula found on John Newman's Electrochemical System seems to be different from the one found on Allen Bard's Electrochemical Methods, both of which are considered classics. Below is my understanding, if not incorrect.

In the most general case, Newman writes

\begin{equation}
i = i_0\biggl[\exp\biggl(\frac{(1-\beta)nF}{RT}\eta_s\biggr)-\exp\biggl(-\frac{\beta nF}{RT}\eta_s\biggr) \biggr]
\end{equation}

where $\eta_s=V-U$. $U$ is the equilibrium which depends on the surface concentration, and so do $i_0$, the exchange current.

Bard, on the other hand, writes, in the most general case, the current-overpotential equation
\begin{equation}
i_n = i_0\biggl[\frac{C_O}{C_O^*}\exp\biggl(-\frac{\alpha nF}{RT}\eta_s\biggr)-\frac{C_R}{C_R^*}\exp\biggl(\frac{(1-\alpha) nF}{RT}\eta_s\biggr) \biggr]
\end{equation}
which only leads to Butler-Volmer equation if mass-transfer is not concerned. Here $\eta_s=E-E_{eq}$ (with its notation). $E_{eq}$ is the equilibrium potential established on bulk concentration, which is a constant taken to be initial condition. $i_0$ is dependent on the bulk concentration.

Aside from the anodic, cathodic sign convention different, what Bard has written explicitly depends on the surface concentration, which in a sense suggest that even if the over-potential is negative (which, in Bard's convention, leads to positive cathodic current), anodic reaction can still be established if $C_R$ is sufficiently high.

The more widely known equation (also found in Bockris's Modern Electrochemistry vol.2) is the one Newman wrote. But the equation will only lead to (in Newman's convention) positive anodic current if the over-potential $\eta_s$ is positive, even if $C_O$ dominates (which leads to cathodic current in Bard's formula).

It seems that the derivation differ on the reference potential: Newman sets equilibrium potential $U$ to be dependent on the surface concentration while Bard let it be referenced to the open-circuit potential with bulk concentration. What's right and wrong?

I just figure it out that both are right (of course!). It's just that the definition of over-potential $\eta_s$ is different. In the case of Newman's, positive $\eta_s$ will guarantee positive (anodic) current. Here $\eta_s=V-U=\phi_{metal}-\phi_{solution}-U$ is positive in the sense that $V>U$, where $U$ is the equilibrium potential corresponding to the surface concentration. Whereas in the case of Bard's, sometimes positive $\eta_s$ will lead to positive (cathodic) current if $C_O$ is sufficiently large (possibly with convention or precipitation). But here $\eta_s=V-E_{eq}$, and $E_{eq}$ is defined as equilibrium potential corresponding to bulk (initial) condition. So the value of $V$ here will result in negative $\eta_s$ in the case of Newman's

In short, it’s just that the sign of over-potential is different with their corresponding definition of equilibrium potential.


Tried charging a penny/dime battery. Yellow goo bubbled. What did I create? When I was a young child, I applied 9 volts across an electric cell I made using a penny, a dime, and a salted piece of paper napkin wet with saliva.
Immediately, an unknown mustard-yellow liquid began forming and bubbling from in between them. It discolored the coins.

What reaction took place? What was the yellow stuff?

Hahaha, I had to translate for myself what a penny and what a dime is. Always the same for me; Fahrenheit, miles, ...

A penny is a copper-plated zinc coin. A dime is a nickel-plated copper coin. Knowing this, I would assume that the yellow color arose from the nickel content. If it would be blue, it would have more likely been copper.
The goo. Such a structure often occurs with proteins, polymers, and similar compounds. It's most likely that the bacterial farm on the coins you used, lead to the formation of this goo. It's impressive how many bacteria and what thick biofilms you can find on a coin. Additionally, your saliva also contains a multitude of living organisms and their remains.

Therefore, you should definitely reproduce this experiment! Clean the coins beforehand using ethanol, acetone, and water while scrubbing. Polish them with a linen cloth and not with sand paper. Otherwise, you would damage the plated metal film and surface the underlying metal (makes things now complicated). And don't spit in it yet! :-)


Why can a face-centered cubic lattice not be redrawn as a body-centered tetragonal lattice? A base-centered cubic lattice can be redrawn as a primitive tetragonal lattice, therefore we do not include it in the list of Bravais lattices. Why is it then that a face-centered cubic lattice cannot be redrawn as a body-centered tetragonal lattice? If they can be redrawn as each other, why is it then that they are still listed in the Bravais lattice list? What is the ultimate condition to be in the list?

OK, let's get it straight. A lattice is just a periodic set of points. For the sake of brevity, I'll talk about 2D lattices; trust me, with 3D it is pretty much the same, only more complicated.

Initially the lattices were just like that: periodic patterns of one infinitely repeating parallelogram, called the unit cell. Other than that, they were without form, and void; and darkness was upon the face of the deep. The very word "centered" was not invented yet.

Then people started to categorize lattices by symmetry.

First of all, there are those with $90^\circ$ rotational, or 4-fold symmetry (we may call them "tetragonal", or maybe "square"), and those with 6-fold symmetry (we may call them "hexagonal"). We put them aside and look deeper at the rest.

You see that the first one of these has no symmetry to speak about (other than the $180^\circ$ rotational symmetry, which all 2D lattices have by birthright). The link above calls such lattices "oblique"; I don't like the word, but whatever. The second one has mirror symmetry with respect to the lines running along the sides of the cell; this is an example of so-called "rectangular" lattice. And the third one has mirror lines along the diagonals of the cell. This just does not sit well with us; we'd rather have them along the sides. (We will meet these diagonal mirror lines later anyway, when we'll deal with the tetragonal and hexagonal lattices; but that's another story.) So we select another unit cell, twice bigger in area than the original, and call it centered.

See, the lattice (i.e., the arrangement of blue dots) is still the same; we couldn't and didn't make it more symmetric, or less symmetric, or in any slightest way different, for that matter. We just started to think of it in a different way.

And so the centered rectangular cell entered the list of 2D Bravais lattices. See, it did not get there because it can't be decomposed into some primitive cell (it surely can, as any centered cell can), or because other cells can't be centered (they surely can, only it would make no sense). It is there because we wanted our symmetry elements to be aligned in a certain way.

Like I said, in 3D we have the same story, only somewhat more complicated. Indeed, the face-centered cubic cell can be thought of as body-centered tetragonal (and also as some primitive non-centered cell, too); it is just that the cubic cell has 4-fold rotation axes running along all sides, and in a tetragonal cell some of them would run diagonally, which is not considered comme il faut.

My apologies to those who think they'd be better off with the diagonal mirrors and without centering. Anyway, the convention has been in use for well over a century now, so maybe it's not that bad after all.


Heat involved in isenthalpic process Question 28 from [1, p. 4.4]:

An ideal gas is expanded irreversibly from $\pu{5 L}$ to $\pu{10 L}$ against a constant external pressure of $\pu{1 bar}.$ The value of heat involved $(q)$ in this isenthalpic process is
(a) $0$
(b) $\pu{+500 J}$
(c) $\pu{+5 J}$
(d) $\pu{-500 J}$

It is mentioned in the problem that the process is isenthalpic, which means the change in enthalpy must be zero. So, my conclusion was there is no heat involved here. But the answer given is (b).
References

Neeraj, K. Advanced Problem in Physical Chemistry for Competitive Examinations, 2nd ed.; Pearson India Education Services Pvt. Ltd: Dehli, Chennai, 2015. ISBN 978-93-325-4373-7.

After further reflection I realized an initial comment I made was flawed*. To quote the Wikipedia:

  Isenthalpic processes on an ideal gas follow isotherms, since $\mathrm dh = 0 = c_p\,\mathrm dT$.

That means that for an ideal gas isenthalpic is synonymous with isothermal, and this implies $\Delta U = 0$ and so $q = -w = P_\mathrm{ext}\Delta V$.

For the isobaric process described in the problem, the work done by the gas is $\pu{500 J}$, and therefore the heat expected for an isothermal process is also $\pu{500 J}$.

* The source of my error is simple: for an irreversible process the heat is not equal to $\Delta H$. In fact, since the external pressure is not balanced by that of the system, $P_\mathrm{ini}\neq P_\mathrm{fin}$ for the system (the process is not strictly isobaric for the system). For a reversible isobaric process $q_p = \Delta H$.


Enthalpy of neutralisation for weak acids and weak bases at standard state I recently read about why for weak acids and bases the enthalpy of neutralisation is less as compared to the strong ones. This was because the acids and bases being weak, they dissociate incompletely and even in this they absorb energy.

I have two questions regarding this:

At standard state, for a solute, we take the hypothetical case of unit molality behaving as one which is infinitely diluted, then shouldn’t even weak acids or bases act as strong acids/bases (because they are, at least in theory assumed to behave as if they were infinitely diluted)?
It's mentioned that during dissociation they (weak acids/bases) absorb heat. Wouldn't even strong ones absorb at least some heat to effect their dissociation.

I find the first question to be more important in terms of conceptual value.

Q1, well you are correct to a certain level. standard state of a solute in a solution of any specified solvent is the state having solution under pressure exactly equal to standard pressure 1 bar,composition of solute equal to 'a standard composition', solution must be an 'ideal solution (perfect solution)', and finally solution must show 'properties of an infinitely diluted solution'. here standard composition refers to either, standard amount concentration (1 mol/dm3), standard molality (1 mol/kg) or standard mole fraction (1). note that temperature is not a part of definition of standard state. this is a hypothetical state for any temperature we use or what standard composition you may chose. here you have chosen standard molality,its ok.(note that we can prove that for an ideal solution showing properties of infinite dilution at a given pressure and temperature molar enthalpy of solute is a constant what ever its composition may be.plug into somewhere that defines you exactly the terms 'perfect solution',infinite dilution and properties of infinite dilution to understand this, its easy). infinite dilution means amount concentration or equivalently any other composition term of solute approaching zero. however acid  composition is molality =1mol/kg as you prefer or any other standard composition, but yet it is not infinitely diluted although it shows properties of infinite dilution. key thing here to understand is that infinite dilution is one thing and showing properties of infinite dilution is another. this is not the only hypothetical state you are dealing with, standard state of a gas existing in pure form , as a solute or in a mixture is the pure form of gas at pressure 1 bar and gas is ideal or equivalently compressibility factor =1. going back to your question, we can talk about strong and weak acids or bases. in fact Ka and Kb values are based on this.  Q2; strong and weak are relative terms. the better word is strength. even strong acids or bases absorbs heat .


Why Freundlich adsorption isotherms fails at high pressure conditions? Experimentally it was determined that extent of gas adsorption varies directly with pressure, and then it directly varies with pressure raised to the power 1/n until saturation pressure Ps is reached. Beyond that point, the rate of adsorption saturates even after applying higher pressure. Thus, the Freundlich adsorption isotherm fails at higher pressure.

This is an extract taken from Wikipedia explaining limitation of Freundlich isotherm.But analyzing the freundlich adsorption equation we can reach the same conclusions

The Freundlich adsorption isotherm is mathematically expressed as:
         $$X/M=KP^{1/n}$$

Giving whole number values to 'n' we can explain adsorption at various pressure conditions.If so why then it is said that Freundlich adsorption isotherm fails at high pressure?

Edit

  It cannot be used for a straight linear isotherm part occurring either at low pressures as the
  value n =1 should be then assumed, or at high pressures as the curve increases
  unreservedly, whereas a surface has a limited value and it must be in the condition of
  saturation.

Why is it given here that it's not possible to assume a value of 1 for n?

There is a saturation level for the adsorption of any adsorbate on the adsorbent due to limited surface area of the adsorbent. So once that saturation level is reached , more pressure won't increase the number of particles adsorbed .


Which carbon(s) in cyclohexene have the highest electron density? I think that carbons with highest electron density will be both $\ce C1$ and $\ce C2$ because of the loosely held $\pi$-bond between them, but according to the answer given the carbons with most electron density should be both $\ce C3$ and $\ce C6$.

How is this possible?

:-( The book gives an answer, not necessarily the right one.

Looking at the $\ce{^13C}$ NMR of cyclohexene the carbons at positions 4 and 5, in your diagram, are shifted the least and therefore have the most shielding. Thus those carbon atoms also have the greatest electron density around them.


Dimensional Analysis in Chemistry I was once told on this site that it was incorrect form to use units that specify the chemical being referred to in dimensional analysis. For example:

$$150.~\mathrm{g}~~\ce{KNO3} \cdot \frac{1~\mathrm{mol}~~\ce{KNO3}}{101.103~\mathrm{g}~~\ce{KNO3}} \cdot{} \frac{1~\mathrm{mol}~~\ce{C7H4O}}{6~\mathrm{mol}~~\ce{KNO3}} \cdot \frac{104.106~\mathrm{g}~~\ce{C7H4O}}{1~\mathrm{mol}~~\ce{C7H4O}}=25.7~\mathrm{g}~~\ce{C7H4O}$$

This is the way I learned it, and I feel that this clears up confusion that might otherwise arise when from performing such a calculation.

My overall question is: is it acceptable to do dimensional analysis as I presented it? If not, please explain why it is not correct to do it this way and show me what I should be doing instead.

The way I would perform the dimensional analysis you give in your question would be slightly different. That is mainly due to the fact that I don't consider chemical names or formulae as units, but as designators.

As an aside, I think you were starting from a question like the following:

  How much $\ce{C7H4O}$ would you have to weigh in for it to fully react
  with $150~\mathrm{g}$ of $\ce{KNO3}$ if the reaction is described by
  the following equation? $$ \ce{C7H4O + 6KNO3 -> Fumes^  + Tar} $$

The result is $m_{\ce{C7H4O}}$, given is $m_{\ce{KNO3}} = 150~\mathrm{g}$.

The amount of $\ce{C7H4O}$ used per $\ce{KNO3}$ is
$$|\nu_{\ce{C7H4O}} / \nu_{\ce{KNO3}}| = \frac{1}{6}$$

As such, the result is composed as follows:
$$\begin{align}
n_{\ce{C7H4O}} &= \frac{1}{6} n_{\ce{KNO3}} \\
\tag{$n=m/M$}\frac{m_{\ce{C7H4O}}}{M_{\ce{C7H4O}}} &= \frac{1}{6} \frac{m_{\ce{KNO3}}}{M_{\ce{KNO3}}} \\ \hline
m_{\ce{C7H4O}} &= \frac{1}{6} \frac{m_{\ce{KNO3}}}{M_{\ce{KNO3}}} M_{\ce{C7H4O}}
\end{align}$$

If we now apply dimension analysis to what we think is the result, we can very quickly see how the two molar masses cancel each others units out and only the mass of $\ce{KNO3}$ remains to give the right hand side of the equation a dimension of $\mathrm{g}$. As the only symbol left on the left hand side is a $m$ we know that it needs a dimension $\mathrm{g}$ and as such we can happily plug in the numbers into our calculator for the result.

Frankly, I am so used to "my" method that I had to look at your example quite closely to see what was actually going on. To me that method just seems unwieldy, as you clutter up equation space with chemical formulae.

That being said; find the method that works the best for you (meaning it is robust enough to also solve your chemical engineering, biology and other problems using dimensional analysis) and stick to it.


Is the Hollywood portrayal of the effects of ingesting HCN correct? In the 2012 movie Skyfall, the antagonist Raoul Silva rues that the "hydrogen cyanide" he took ("Do you know what it does to you, hydrogen cyanide?") didn't kill him, leaving him horribly disfigured instead.

Firstly; Could it have actually been HCN that he took?

From what I know about this, HCN is a gas at room temperature, and gases aren't typically the poison of choice for suicide pills. To begin with, suicide pills are preferrably kept as small as possible to make it easy to conceal. Keeping the size constraint in mind, simply filling it with HCN gas wouldn't allow the pill to store enough to (seriously) harm you, much less kill you.

From what I gather, suicide pills (WW2 era) often contain a concentrated solution of potassium cyanide, and not HCN.

Well, I'm bringing this up since this is a James Bond movie, so... what're the chances the MI6 could've compressed a significant amount of HCN in the suicide pill (possibly pressurizing it until it's a liquid), without having the easy-to-crush-between-the-teeth pill exploding on account of pressure? I'm not sure how easy it is to liquify HCN, but if it is possible to do so at room temperatures without an exorbitant amount of pressure, could that have been the 'suicide pill' he had taken?

Are there any known instances ( some half-arsed experiments and such) where a gas at room temperature was liquified and stored within such a small volume.

Secondly: Could hydrogen cyanide, or any other simple cyanide compound for that matter, cause such a disfiguration?

More than having tried to kill himself with some sort of HCN pill, I was taken aback to see what it did to Silva.

Just look at what happened to the guy:

Seriously, just look at him:

This is what I found on Wikipedia, on cyanide poisoning:

  If cyanide is inhaled it can cause a coma with seizures, apnea, and cardiac arrest, with death following in a matter of seconds. At lower doses, loss of consciousness may be preceded by general weakness, giddiness, headaches, vertigo, confusion, and perceived difficulty in breathing. At the first stages of unconsciousness, breathing is often sufficient or even rapid, although the state of the person progresses towards a deep coma, sometimes accompanied by pulmonary edema, and finally cardiac arrest. A cherry red skin color that changes to dark may be present as the result of increased venous hemoglobin oxygen saturation. Cyanide does not directly cause cyanosis. A fatal dose for humans can be as low as 1.5 mg/kg body weight.

I don't seem to find anything on 'severe-tissue-damage-and-jawbone-dissolution', so I'm still skeptical.

I mean, sure, HCN is a weak acid, but that is not what a weak acid does to you.

So did Raoul Silva bite down on a cyanide pill (of sorts...) or did he gargle a bottle of drain-cleaner and just forget about it?

Is there anything realistic about that scene, or is it just another movie gaffe?

Is there anything realistic about that scene, or is it just another movie gaffe?

I have never heard about anything similar caused by $\ce{HCN}$ or $\ce{KCN}$.

However,

bleeding of the gums,
loss of teeth, and
osteonecrosis of the jaws (= the bone tissue literally rots away)

are known symptoms of certain intoxications:

Bisphosphonate-associated osteonecrosis of the jaw was observed in the treatment of patients with cancer
Radium jaw was observed in painters (Radium girls) that licked their paintbrushes while applying radioluminescent paint to dials
Phossy jaw was observed in (mostly female) workers in the matchstick industry of the late nineteenth century that worked with white phosphorus.

Back in the days, the lives of the patients often could only be saved by removing the jawbones surgically. (If you feel for it, look for some photographs online.) Since one of the tags is chemistry-in-fiction, you might be interested that this was partly covered in the episode Become Man of the BBC series Ripper Street.


Phosphate extraction from sewage I read that phosphate reserves will exhaust  in coming years wiki, why don't we extract it from sewage , is it really that difficult to extract phosphate ?

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

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

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

Krepo Process
Seaborne Process
Aqua Reci Process

In it Heinzman also mentions that

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

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


Converting 26.62 NOx µg/L into mM (millimolars)? I have to convert  a whole lot of nutrient data : $\ce{NH_4}$, $\ce{PO_4}$ and $\ce{NO_x}$ data that is in µg/L into mM (millimolars).

Could someone give me an example for 26.62 $\ce{NO_x}$ µg/L? or 7.73 $\ce{PO_4}$ µg/L? or 14.39 $\ce{NH_4}$ µg/L?

I have a large data set in Excel to convert and send to someone and want to get it right. Any help would be really appreciated.

I would make it easy for yourself and convert your µg/L to g/L. You can then divide it by the formula weight to obtain the molarity (in M). After this it is just a simple conversion to mM.

I suppose in Excel you can create a new column and use a formula something like this:

=((CELL/106)/Formula weight)*1000

So for your example of phosphate:

=((0.00000773)/94.97)*1000

=0.00008139 mM // 8.139x10-5 mM


How is it that earth's inner core is still solid? Earth's inner core is made up of an alloy of Iron and Nickel.

The important note here is that it is solid.

The temperature at the inner core is about $5,763K = 5,490\text{ degrees C}$.

Now the melting point of both solid iron and nickel is about $1,500\text{ degrees C}$, and even though they are combined, I still think that $5,490$ is much greater then the melting point of the alloy.

Regardless, how is it that the inner core still stays solid? I would expect that at such higher temperatures, a phase change to liquid, or maybe even gas would occur.

Why does it not?

The melting point of the vast majority of materials increases as more pressure is applied. At https://www.researchgate.net/figure/312054885_fig1_Figure-231-Experimental-pressure-temperature-phase-diagram-of-pure-iron-a is a phase diagram showing how this variation occurs for pure iron:

From https://physics.stackexchange.com/questions/184032/what-is-the-pressure-at-the-center-of-the-earth-or-a-neutron-star the pressure at the centre of the Earth is 365GPa. At this pressure we can see from the graph that the melting point of pure iron is well over 5000k. Thus, given the experimental uncertainties and the that the core is not pure iron, it is easily conceivable that the core of the Earth is solid, or at least a large proportion of it is.


How to determine the worst resonance structures out of a given set? So I was "happily" doing organic chemistry homework when I came across this question:

For 1-methoxy-1,3-butadiene, which of the following resonating structure is the least stable?
$$\begin{align}(\mathbf{a})\ &\ \ce{H2\overset{+}{C}-\overset{+}{C}H-CH=CH-O-CH3} & (\mathbf{b})\ &\ \ce{H2\overset{+}{C}-CH=CH-CH=\overset{+}{O}-CH3} \\
(\mathbf{c})\ &\ \ce{H2\overset{+}{C}-\overset{-}{C}H-\overset{+}{C}H-\overset{+}{C}H-O-CH3} & (\mathbf{d})\ &\ \ce{H2C=CH-\overset{+}{C}H-CH=\overset{+}{O}-CH3}\end{align}$$

I decided to take a look through my textbook to see what I was missing. The rules to compare resonance structure energies (and therefore determine relative contributions) were in my textbook.  It said:

The following structures are considered to contribute most:

Structures having filled octet a for second row elements (C, N, O, F) are stable.
Structures having minimum number of formal charges and maximum number of bonds.
Structure in which negative charge appears on the most electronegative atom (C < N < O).
Structure in which there is minimal charge separation while keeping the formal charges closer together.

Instead of answering my initial question, two more questions came to mind. They are (in order):

Why do the octets of only the second row elements have to be filled in for greater relative contribution?
I suppose it has something to do with the greater electronegativity values of the elements, but would still like some clarification.
From option (a) and (c) in my question, how did the two positive charges get next to each other?

Resonance structures, as stated in my textbook, can be represented by two or more Lewis structures that differ only in the positions of the electrons.
The options clearly do not just redistribute the electrons. Or at least I don't think so.
So if I had to guess, the answer would be (b) because the terminal carbocation is worse than an internal carbocation.
But it turns out, the correct answer is (c). And I think that it shouldn't even be considered a resonating structure.

(C) is technically a valid resonance structure. You can derive it from (B) by moving the electron pair from the double bond between the carbons 2 and 3 (numbering from the left: this may not be correct from a nomenclature standpoint, but I'm using it for this answer) onto carbon 2 and moving the electron pair from the double bond between the oxygen and the carbon onto the oxygen.

(C) is least stable because it is, simply put, absolutely disgusting. Charge separation (carbons two and three) is gross, you have two cations in close proximity (carbons three and four), a terminal carbocation (carbon 1), and three atoms with empty valence shells (carbons 1, 3, and 4). So while the resonance structure is valid enough for (C) to be considered a resonance structure, that resonance structure happens to be the spawn of the devil.

That said, the compound in the answers isn't the same as the compound named in the question, like ron said.

As to the first part of your question, the filled valence shells are more strictly necessary for the second row elements because they don't have d orbitals to confound the picture.


What is the role of hydration enthalpy in oxidation? My textbook says that the greater oxidising behavior of fluorine than chlorine can be attributed to the high hydration enthalpy of $\ce{F-}$ ions and the low dissociation enthalpy of $\ce{F-F}$ bond.
While the second reason makes perfect sense to me, I can't reason why hydration enthalpy comes into play. I know that these elements act as oxidising agents in aqueous media, but I still can't understand why we need to consider hydration enthalpies.

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

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

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


What is the hybridization of chlorine in vinyl chloride? Is chlorine in vinyl chloride sp2 or sp3 hybridized?

Calculating using steric number it is found to be sp3 hybridized. But according to I. L. Finar* it is sp2.

*I.L. Finar: Organic chemistry Vol.1 Fundamental principles. Sixth Edition. Orient Longmans: 1973. Page 328.

Disclaimer: An important point to remember is, that hybridisation always follows the molecular structure, it is never the cause for a certain structure. As such, hybridisation is an interpretation tool, nothing more.

As a rule of thumb, terminal atoms (excluding hydrogen) are almost always (maximally) approximately sp hybridised. While other hybridisation schemes can be applied, they are usually not a good representation.*

One popular example is water. In many texts the central oxygen is described as having (approximately) four sp3 orbitals, which essentially makes the lone pairs equivalent. That view, however, is not in agreement with the photoelectron spectrum, which clearly shows that the lone pairs are not equivalent. At this point I'd like to refer you to Michael Laing's article: "No rabbit ears on water. The structure of the water molecule: What should we tell the students?" (J. Chem. Educ. 1987, 64 (2), 124.).

The same principle described there applies to terminal atoms. Because of the local $C_\mathrm{\infty}$ symmetry the notion of three equivalent lone pairs is likely to be false. The most likely (main resonance contributor) structure for these atoms are (if hybridisation is at all feasible, i.e. mostly in the second period) one sp orbital forming the bond, one sp lone pair, two (perpendicular) p lone pairs.

In this particular example, I have calculated the molecule on the DF-BP86/def2-SVP level of theory and analysed it with the natural bond orbital theory (NBO6). Below you find the localised orbitals which best fit with the common Lewis structure of the molecule.
I have ordered the orbitals starting with the carbon-carbon σ- and carbon-chlorine σ-bonds on the bottom. Following that is the carbon-carbon π-bond. Continuing with the three carbon-hydrogen σ-bonds, and finally ending with the three chlorine lone pairs. The occupied orbitals are in blue and orange, while the corresponding virtual orbitals are next to it in red and yellow. The detailed analysis in numbers is at the end of the post.

In ast's answer there is the claim that chlorine must be sp2 hybridised because of a resonance contributor. While this explanation is extremely tempting because it easy to understand, you cannot judge from the possibility of resonance to the electronic structure. As another general rule of thumb you can remember: the lesser the orbitals are hybridised, the more likely the structure is.
The calculation also provides an analysis in terms of natural resonance theory. On the DF-BP86/def2-SVP level the contribution of the second configuration to the total electronic structure is only 6%. The main configuration contributes with about 90%, other configurations (ionic) make up the remaining 4%.

In summary, I'm afraid, but your book is incorrect, or at least incomplete. The description with two equivalent sp2 lone pairs is unnecessary complicated and does not accurately reproduce the electronic structure.

Truncated results of the NBO analysis (skipping core orbitals)

     (Occupancy)   Bond orbital / Hybrids
 ------------------ Lewis ------------------------------------------------------
   8. (1.99378) LP ( 1)Cl  6            s( 81.55%)p 0.23( 18.44%)d 0.00(  0.01%)
   9. (1.97199) LP ( 2)Cl  6            s(  0.18%)p99.99( 99.79%)d 0.18(  0.03%)
  10. (1.90237) LP ( 3)Cl  6            s(  0.00%)p 1.00( 99.95%)d 0.00(  0.05%)
  11. (1.99642) BD ( 1) C  1- C  2
               ( 48.61%)   0.6972* C  1 s(  0.00%)p 1.00( 99.94%)d 0.00(  0.06%)
               ( 51.39%)   0.7169* C  2 s(  0.00%)p 1.00( 99.95%)d 0.00(  0.05%)
  12. (1.99535) BD ( 2) C  1- C  2
               ( 49.24%)   0.7017* C  1 s( 40.16%)p 1.49( 59.76%)d 0.00(  0.08%)
               ( 50.76%)   0.7124* C  2 s( 44.47%)p 1.25( 55.47%)d 0.00(  0.06%)
  13. (1.96941) BD ( 1) C  1- H  3
               ( 61.15%)   0.7820* C  1 s( 29.81%)p 2.35( 70.15%)d 0.00(  0.04%)
               ( 38.85%)   0.6233* H  3 s( 99.91%)p 0.00(  0.09%)
  14. (1.98105) BD ( 1) C  1- H  4
               ( 61.33%)   0.7832* C  1 s( 30.09%)p 2.32( 69.87%)d 0.00(  0.04%)
               ( 38.67%)   0.6218* H  4 s( 99.91%)p 0.00(  0.09%)
  15. (1.98492) BD ( 1) C  2- H  5
               ( 62.36%)   0.7897* C  2 s( 32.79%)p 2.05( 67.18%)d 0.00(  0.03%)
               ( 37.64%)   0.6135* H  5 s( 99.89%)p 0.00(  0.11%)
  16. (1.99056) BD ( 1) C  2-Cl  6
               ( 43.14%)   0.6568* C  2 s( 22.75%)p 3.39( 77.00%)d 0.01(  0.25%)
               ( 56.86%)   0.7541*Cl  6 s( 18.31%)p 4.43( 81.12%)d 0.03(  0.57%)
 ---------------- non-Lewis ----------------------------------------------------
  17. (0.09516) BD*( 1) C  1- C  2
               ( 51.39%)   0.7169* C  1 s(  0.00%)p 1.00( 99.94%)d 0.00(  0.06%)
               ( 48.61%)  -0.6972* C  2 s(  0.00%)p 1.00( 99.95%)d 0.00(  0.05%)
  18. (0.01232) BD*( 2) C  1- C  2
               ( 50.76%)   0.7124* C  1 s( 40.16%)p 1.49( 59.76%)d 0.00(  0.08%)
               ( 49.24%)  -0.7017* C  2 s( 44.47%)p 1.25( 55.47%)d 0.00(  0.06%)
  19. (0.00909) BD*( 1) C  1- H  3
               ( 38.85%)   0.6233* C  1 s( 29.81%)p 2.35( 70.15%)d 0.00(  0.04%)
               ( 61.15%)  -0.7820* H  3 s( 99.91%)p 0.00(  0.09%)
  20. (0.01032) BD*( 1) C  1- H  4
               ( 38.67%)   0.6218* C  1 s( 30.09%)p 2.32( 69.87%)d 0.00(  0.04%)
               ( 61.33%)  -0.7832* H  4 s( 99.91%)p 0.00(  0.09%)
  21. (0.02582) BD*( 1) C  2- H  5
               ( 37.64%)   0.6135* C  2 s( 32.79%)p 2.05( 67.18%)d 0.00(  0.03%)
               ( 62.36%)  -0.7897* H  5 s( 99.89%)p 0.00(  0.11%)
  22. (0.03245) BD*( 1) C  2-Cl  6
               ( 56.86%)   0.7541* C  2 s( 22.75%)p 3.39( 77.00%)d 0.01(  0.25%)
               ( 43.14%)  -0.6568*Cl  6 s( 18.31%)p 4.43( 81.12%)d 0.03(  0.57%)

Summary of Natural Population Analysis

                                     Natural Population
             Natural    ---------------------------------------------
  Atom No    Charge        Core      Valence    Rydberg      Total
 --------------------------------------------------------------------
    C  1   -0.43992      1.99995     4.42868    0.01129     6.43992
    C  2   -0.22801      1.99996     4.20978    0.01827     6.22801
    H  3    0.22758      0.00000     0.76996    0.00246     0.77242
    H  4    0.22498      0.00000     0.77165    0.00337     0.77502
    H  5    0.23463      0.00000     0.76237    0.00300     0.76537
   Cl  6   -0.01926      9.99998     7.00353    0.01575    17.01926
 ====================================================================
 * Total *  0.00000     13.99989    17.94598    0.05414    32.00000

Optimised Geometry DF-BP86/def2-SVP

6
symmetry cs ( E = -538.027155015 A.U. )
C        1.309538928      1.048660302      0.000000000
C        0.000000000      0.759346920      0.000000000
H        1.630332304      2.102388803      0.000000000
H        2.083175437      0.265507033      0.000000000
H       -0.793639761      1.523771533      0.000000000
Cl      -0.633947150     -0.867041806      0.000000000

* It is also important to understand, that atoms are never hybridised, only orbitals are. The presence of an sp hybrid orbital does not exclude the presence of a sp3 hybrid orbital.


Convert from Pa to Pa to J/dm^3 Given the value 380 PaK^-1, how do I convert Pa to Jdm-3?

I know that 1 joule = 1 Nm, and that 1 pascal = 1 N/m^2.

But where does the dm^-3 (or L) come into this? I don't know where to start.

Derived units are defined as products of powers of the base units. Some of the coherent derived units in the SI are given special names and symbols.

The joule (unit symbol: $\mathrm J$) and the pascal (unit symbol: $\mathrm{Pa}$) are such coherent derived units.

Expressed in terms of SI base units, one joule is

$$\mathrm{1\ J=1\ \frac{kg\ m^2}{s^2}}$$

and one pascal is

$$\mathrm{1\ Pa=1\ \frac{kg}{m\ s^2}}$$

(see: BIPM: The International System of Units (SI))

Hence,

$$\mathrm{1\ \frac J{m^3}=1\ \frac{\frac{kg\ m^2}{s^2}}{m^3}=1\ \frac{kg\ m^2}{m^3\ s^2}=1\ \frac{kg}{m\ s^2}=1\ Pa}$$

SI prefixes may be used in forming the decimal multiples and submultiples of units. Note that, when SI prefixes are used with SI units, the resulting units are no longer coherent, because a prefix effectively introduces a numerical factor.

$$\begin{align}
\mathrm{1\ m}&=\mathrm{10\ dm}\\[6pt]
\mathrm{1\ m^3}&=\mathrm{1000\ dm^3}\\[6pt]
\mathrm{\frac1{m^3}}&=\mathrm{\frac1{1000\ dm^3}}=\mathrm{0.001\ \frac1{dm^3}}\\[6pt]
\mathrm{1\ \frac J{m^3}}&=\mathrm{0.001\ \frac J{dm^3}}\\[6pt]
\mathrm{1\ Pa}&=\mathrm{0.001\ \frac J{dm^3}}\\[6pt]
\mathrm{1000\ Pa}&=\mathrm{1\ \frac J{dm^3}}
\end{align}$$


Cement strength by moistening After cementing a wall, water is applied on the surface to keep it moist to maintain its strength for the next few days. Why is this necessary for maintaining the strength of the concrete?

I'm not sure what you mean by "cementing" a wall. However, cement (typically Portland cement) is the paste that binds together the aggregate (gravel and sand) in concrete. The better the bond between the cement and the aggregate, the higher the resultant compressive strength of the concrete.

When cement is mixed with water in concrete it binds to the aggregate through a chemical reaction known as "hydration."

http://en.wikipedia.org/wiki/Hydration_reaction

A hydration reaction consumes water as a reactant, which is why you must keep concrete wet for several days after it is placed. If the concrete dries out then it has no water to continue the hydration reaction. The end result then is that you're left with cement that is not fully adhered to the aggregate (less compressive strength).

Careful though - you only need to keep the concrete wet after it has set up and is curing. Adding too much water to the mix (giving it a high water to cement ratio, w/c) actually drastically reduces the compressive strength of the concrete.

Curing compounds are especially helpful for properly curing concrete. (speaking from a heavy civil general contractor's point of view) We use this product all all the time:

http://www.wrmeadows.com/1100-clear-resin-water-based-concrete-curing-compound/


Why there are more examples of carbonylate anions than metal carbonyl cations? I know that metal carbonyl complexes that do not follow EAN rule or 18 electron rule try to achieve that by gaining electron (forming carbonylate) or loosing electron (forming carbonyl cations). But this question is way too dicey. It came in a national level exam in our country. I am yet to find a solution.

Yes there are more carbonylate ions than metal carbonyl cations, but the latter do exist.  A salt $\ce{[Mn(CO)6^+][BF4^-]•SO2}$ has been synthesized in a superacid medium.

Metal carbonyl cations are often disfavored because they have to contend with counterions that are anions, which may compete with the carbon monoxide ligands.  Such a competition is tilted towards the anion winning out if the metal is positively charged.  Solvents with significant Lewis-base character are also apt to displace carbon monoxide from the coordination sphere, for a similar reason.  Carbonylate anions do not face this situation (or at least not as severely) because cations are more likely to bind with the solvent than with an an ionic metal center that is dispersing its charge through pi donation to the carbon monoxide ligands.

Going back to the referenced salt, we can see features that protect $\ce{[Mn(CO)6^+]}$ from such an ignominious fate.  First note the counterion, $\ce{[BF4^-]}$which is a very weak nucleophile and has its negative charge buried inside the relatively bulky fluorine "shield".  Then note the superacid medium; perforce the solvent is not competing as a Lewis base either.


Why does graphite conduct electricity? The answer i know is delocalized electrons. But how?this site tells me that:

My textbook  tells me that one carbon atom is bonded to three others and two of these bonds are single while the third is a double bond.So all the electrons are bonded and there are no free moving electrons right? Where do these delocalized elctrons come from?

Carbon has four spare electrons. In graphite structure, two of them make two single (sigma) bonds while the other pair form a double bond (consists of one sigma and one pi bond). Electron that forms pi bond is freer to travel between atoms compared to the others and is theoretically associated with all adjacent atoms (imagine the 3d structure of graphite) rather than a single atom. Because of that, it is called delocalized electron. In other words, electrons that form pi bonds continously replace each other causing temporary, local crystallizations within the structure. According to Wikipedia "It is increasingly appreciated that electrons in sigma bonding levels are also delocalized". But it is not the subject we are discussing here.

Electric current is basically the flow of charged particles (ions and electrons, but electric currents in everyday life can simply be defined as flow of electrons). If graphite is exposed to an electric current in a circuit, "the pi electrons" which are already traveling between carbon atoms that forms the graphite structure begin to flow through the circuit and leave the structure from one point of connection between graphite and the circuit. At the same time, other electrons come via the other connection point to replace them as new delocalized electrons. So, the new delocalized electrons simply come from the current.


Why is the standard atomic weight of chlorine not a whole number? Why is the standard atomic weight of chlorine, 35.5, not a whole number? Like for example, it could be exactly 35 or exactly 36. Please show the solution of formulae on how u reach to 35.5) with some English words so I can write the answer in my homework assignment and please make it small. Thanks

The process is explained (for silicon) here:

  The calculation is exemplified for silicon, whose relative atomic mass is especially important in metrology. Silicon exists in nature as a mixture of three isotopes: $\ce{^28Si}$, $\ce{^29Si}$ and $\ce{^30Si}$. The atomic masses of these nuclides are known to a precision of one part in 14 billion for $\ce{^28Si}$ and about one part in one billion for the others. However the range of natural abundance for the isotopes is such that the standard abundance can only be given to about ±0.001% (see table). The calculation is

  $$A_r(\ce{Si}) = (27.97693 \times 0.922297) + (28.97649 \times 0.046832) + (29.97377 \times 0.030872) = 28.0854$$

For chlorine, the idea is the same, but you only have two isotopes, $\ce{^35Cl}$ (76%) and $\ce{^37Cl}$ (24%).


How is Standard Gibbs free energy measured? I have seen hole theory of thermodynamics but almost nothing about the experiments that measure $\Delta G^\circ$. I know how to measure it by electrodes, when the reaction is a redox one. But what with other reactions?

What are the experimental ways to determine $\Delta G^\circ$? Some examples would be helpful

Generally you measure the equilibrium constant $K$ and then convert it to $\Delta G^\circ$ (note that is change in standard free energy)

$ \Delta G^\circ = -RT \ln K $

In the case of redox reactions, this can be done with electrodes as you mention. For other reactions you usually use some spectroscopic technique (NMR, IRC, UV, etc) that can distinguish reactants from products, but calorimetry can also be used in certain cases.

Here is a video describing one example using UV.


What causes spray cans to get cooler when shaken? I've been searching the web for this phenomenon and surprisingly I found only one post about it. And the reply to this question about canned air begins with words:

  I don't know about the shaking part (it is hard for me to see how that would make a difference)

This beginning has strongly shaken my faith, so for sure I picked up a can with deodorant and shaken it. And it does become cooler. Significantly! After standing still, the can doesn't seem to become warmer, so this reaction doesn't revert itself and maybe is irreversible.

As I imagine it, when being shaken the average pressure in the can as a whole can't possibly change. Also it's interesting how you add energy by shaking only to have more energy drained by cooling.

So please, can you explain this strange behavior? Could you also cover the energy flow? When does the energy return back out of the can?

TL;DR: Spray cans don't actually get colder when shaken.  However, shaking a can does increase heat conduction from your hand to the can, making it feel colder.

Humans don't actually sense external temperature directly; our thermoreceptors are located under the skin, and thus effectively measure the rate at which body heat is lost through the skin.  This is why, for example, touching a metal surface will feel noticeably colder than a wooden or plastic surface at the same ambient temperature.

When you pick up a spray can, it is typically at room temperature, i.e. significantly below body temperature.  The can itself is a thin metal cylinder, and thus conducts heat well, but has very little heat capacity in itself.  Thus, the rate of heat loss from your hands (and thus the sensation of coolness) mainly depends on how quickly whatever's on the inside of the can will absorb heat from the surface.

Now, a typical aerosol spray can contains a propellant (mixed with the actual payload fluid) which would be slightly above its boiling point (and thus in gaseous form) at room temperature and pressure, but is kept liquid by the elevated pressure inside the can.  As the spray valve is opened and the pressure reduced, some of the propellant fluid will boil, bringing the pressure back up until the liquid and gas phases are again at equilibrium.

What this means is that, in a half-empty spray can, the top half of the can will contain gaseous propellant (which, being a gas, will not absorb heat very effectively), while the bottom half will contain liquid propellant (mixed with the payload fluid) at just under its boiling point.  This liquid will absorb heat much better than the gas at the top, both because it's denser (and thus has higher heat conductance and capacity), and also because heating it will cause some of it to boil, converting part of the absorbed heat into its latent heat of vaporization.  (The boiling effect actually causes a half-empty spray can to act, to some extent, as a diode heat pipe.)

Thus, even without shaking the can, it's possible to observe that the bottom end of a half-empty spray can will feel cooler to your hand than the top end.

Now, when you hold the can and shake it, you cause the propellant liquid to slosh around inside, coating all of the inner surface of the can instead of just the bottom.  The sloshing also significantly increases the heat transfer from the can to the liquid, by causing the warmer liquid near the surface of the can to mix with the colder liquid in the interior.  (This is essentially the same reason as why moving air feels cooler on your skin than still air.)  Thus, a shaken can will generally feel even cooler to touch than either end of an unshaken can.


If the rate law order is three, does that mean that three bodies collide at some elementary step? I am doing a chemistry review about rates, and I get asked "Which of these statements is false?"

The answer is

  In reactions that are second order in one reactant and first order in another, the slow step generally involves a three-body collision of the reactants.

I imagine it looks like

$$
rate = k[A]^2[B]^1
$$

So at the slowest elementary step there should be
$$
2A + B = something
$$

Which is a three way collision between two A molecules and a B.

What am I doing wrong?

EDIT:

Question number 2 (http://www.mesacc.edu/~paudy84101/CHM152F2005/Exam1%20Key.pdf)

This is not my review, in case you are wondering.  I just have the same question.

Let's disprove that statement by counterexample. Here, the slow step is a collision of some intermediate and another thing, but the maths work out that the overall reaction order is third.

Let's have a two step reaction with the stoichiometry:

$$\ce{2A + B -> C}$$

The mechanism looks like this. The first step is fast and reversible and rapidly establishes equilibrium between $\ce{A + B}$ and an intermediate $\ce{I}$.

$$\ce{A + B <=>[k_1][k_{-1}] I}$$

The second step is slow, not so reversible and involves the intermediate $\ce{I}$ reacting with the other equivalent of $\ce{A}$.

$$\ce{I + A ->[k_2] C}$$

Now the rate of the reaction shall be written as the rate of appearance of $\ce{C}$, because it makes our lives a little easier.

$$\mathrm{rate}=\dfrac{d[\ce{C}]}{dt}=k_2 [\ce{A}][\ce{I}]$$

However, it's messy to have the intermediate in the rate law. Since the first step was fast and in equilibrium, we can get an expression for $\ce{[I]}$ in terms of $[\ce{A}]$ and $[\ce{B}]$.

$$K_{eq}=\dfrac{k_1}{k_{-1}}=\dfrac{[\ce{I}]}{[\ce{A}][\ce{B}]}\\
[\ce{I}]=K_{eq}[\ce{A}][\ce{B}]$$

A little substituting and we get:

$$\mathrm{rate}=\dfrac{d[\ce{C}]}{dt}=\dfrac{k_2 k_1}{k_{-1}} [\ce{A}]^2 [\ce{B}]=k_{obs}[\ce{A}]^2 [\ce{B}]$$


Grey "skin" on gallium I have a sample of gallium (a few grams). I've noticed every time I melt the gallium and refreeze it, it forms this grey skin on the outside. When I remelt the gallium, this portion doesn't melt and remains a dull grey color. I have skimmed it off every time and collected it. I looked it up and gallium doesn't seem to create an oxide or other compound unless under extreme conditions. I have noticed this when melting it in a plastic bag, a glass beaker, and a plastic container. I have been using hot water to melt it, but the water doesn't contact the gallium, only the container. It seems I am slowly losing gallium from my sample and have a small amount of this grey material saved in case it can be recovered. Does anyone know what this is and how it can be reverted to elemental gallium?

I have no real knowledge of it, and I've found almost no literature on its handling and storage. From Scientific American 1878:"Unlike lead, however, it acquires only a very slight tarnish on exposure to moist air..." and "In aerated water it tarnishes slightly." My very very old copy of Cotton and Wilkinson (Adv. Inorg. Chem) states that it is a moderately reactive metal and will react with sulfur. And of course we know it readily alloys (I'd say "amalgamates" except that term is specific for mercury alloys) with a large variety of metals. The (III) oxide is white, so unless your scum is white, it ain't that. The (I, sub) oxide is brown-black so that's a possibility. As are mixed oxide /hydroxides. Since the original investigation says it does tarnish in air, I'd take that at face value. There is the obvious possibility that your sample is either contaminated or was sold adulterated. If it's been in contact with any other metal (especially Aluminum) then it is almost certainly contaminated. There's no practical way you'll be able to recover the metal - the expense will far exceed the price to simply replace it. The best way to check for any large contamination is to determine its melting point, it should be very very close to its stated MP, shouldn't even be off by 1°.


Trying to Understand Entropy I'm currently looking at Principles of Chemistry by Atkins, and I have a few questions involving entropy. These questions probably arise from my poor understanding of reversible/irreversible processes.

Why are reversible processes called reversible? Is it necessarily true that reversible processes can be reverted spontaneously, or is this a common misconception that people have?
So my chemistry textbook gives the definition of entropy as
$$\Delta S = \frac{q_{rev}}{T}$$
where $q_{rev}$ is heat transferred reversibly. However, I get lost at how entropy could increase during a reversible process.
Why is the reversibility condition necessary for the definition of entropy?

They're called "reversible" because the processes can run backwards without additional input of energy. There aren't really any truly reversible processes in the universe, though (some come very close, but the construct of "reversible" is an ideal, not a reality).
Entropy of a system can increase just fine during a reversible process: expanding a gas "slowly enough" is reversible ("slowly enough" means in an infinite number of infinitesimal steps), but the entropy of the gas increases as it expands as it has more microstates to go into. The entropy that remains constant in a reversible process is that of the system plus the surroundings (in other words, the universe).
The process does not need to be reversible to have entropy. However, we measure entropy based on the heat of the reversible process because entropy is a state function (it depends on the beginning and end points of your process, nothing more) while heat is a path function (it depends on the path you take to get there).


Barium sulfide vs Ba2S2 I was predicting a very simple chemical equation
$$\ce{Ba + S ->}$$
And I had solved it as
$$\ce{Ba + S -> BaS}$$
The answer should make sense as barium has a charge of +2, and sulfide has a charge of −2; however, the book had said that the answer was
$$\ce{Ba + S -> Ba2S2}$$
I do indeed know how and why they got the answer, but I just want to know why my answer isn't correct, or if it is and I'm just acting silly.

As orthocresol pointed out, $\ce{Ba_2S_2}$ does not exist as it's own substance. Perhaps you've misread the question and it instead asks for something else? Otherwise it is wrong, or at least not simplified. Your answer is in effect the correct one, although I'm not sure if sulfur and barium would react like that exactly.

It might in addition generate multiple polysulfides, such as $\ce{BaS_2}$, $\ce{BaS_3}$, following the rule of $\ce{BaS_{$x$}}$. However, I believe your answer is correct in that the initial reaction between barium and sulfur would produce $\ce{BaS}$, while further reactions might add more sulfurs onto it.


Is isobutanol miscible with ethanol? I can only find info on sec-butanol (aka butan-2-ol) and ethanol. I need to know whether isobutanol forms an azeotrope with ethanol. Can you direct me to a legitimate info source too, please?

According to the Merck Index${^{[1]}}$

  isobutanol is miscible with ethanol

1) O'Neil, M.J. (ed.). The Merck Index - An Encyclopedia of Chemicals, Drugs, and Biologicals. Cambridge, UK: Royal Society of Chemistry, 2013., p. 953


Why aren't the copper(II) ions in CuSO4 precipitated by hydroxide ions from water? From my own experience and literature, I know that $\ce{CuSO4}$ is well soluble in water and dissociates into $\ce{Cu^2+}$ and $\ce{SO4^2-}$. $\ce{Cu(OH)2}$, however, is not: $K_\mathrm{sp} = 2 \cdot 10^{-19}$.

Therefore, in pure water which has a pH of $7$, $c(\ce{OH-}) = 10^{-7}$. This means that a concentration of $2 \cdot 10^{-19} / (10^{-7})^2 = 2 \cdot 10^{-5}$ of $\ce{Cu^2+}$ would be enough to cause precipitation. How come, when dissolving $\ce{CuSO4}$, the $\ce{Cu^2+}$ stays in solution, and doesn't precipitate with the $\ce{OH-}$ ions of water to form $\ce{Cu(OH)2}$?

In aqueous solution, the copper(II) ion forms the aqua complex $\ce{[Cu(H2O)6]^2+}$:

$\hspace{55 mm}$

Because the central $\ce{Cu^2+}$ ion is positively charged, it polarises the $\ce{O-H}$ bonds towards oxygen, which makes the hydrogens more acidic. Therefore, the ion acts as a weak Brønsted acid, with a $\mathrm{p}K_\mathrm{a}$ of approximately $8$ (J. Phys. Chem. A, 2015, 119 (12), pp 2926–2939):

$$\ce{[Cu(H2O)6]^2+ <=> H+ + [Cu(H2O)5(OH)]^+}$$

Suppose we have a $0.1~\mathrm{mol~dm^{-3}}$ solution of $\ce{CuSO4}$. To a first approximation,

$$\begin{align}
[\ce{Cu(H2O)6^2+}] &= 0.1~\mathrm{M} \\
[\ce{H+}] &= \sqrt{(0.1~\mathrm{M})(10^{-8}~\mathrm{M})} \\
&= 3.162 \times 10^{-5}~\mathrm{M} \\
[\ce{OH-}] &= \frac{10^{-14}~\mathrm{M^2}}{[\ce{H+}]} \\
&= 3.162 \times 10^{-10}~\mathrm{M}
\end{align}$$

Therefore,

$$\begin{align}
[\ce{Cu^2+}][\ce{OH-}]^2 &= (0.1~\mathrm{M})(3.162\times 10^{-10}~\mathrm{M})^2 \\
&= 10^{-20}~\mathrm{M^3} < K_{\mathrm{sp}}
\end{align}$$

and no precipitation of $\ce{Cu(OH)2}$ occurs.

If the pH of a $\ce{Cu^2+}$ solution is neutralised by careful addition of base, then precipitation will be observed as Ivan noted in his comment. The exact pH at which precipitation will occur is of course dependent on the concentration of $\ce{Cu^2+}$. The same argument here actually applies to all of the transition metals: all of them form rather insoluble hydroxides, but they also form acidic aqua complexes, which prevent the precipitation of the hydroxide.

The calculations could have been done more accurately by introducing fewer assumptions; however, they should not affect the final conclusion. (If this was a textbook, it would probably say: this exercise is left to the reader!)


Is it possible to calculate atomic radius with electron configuration? I need to know whether is it possible to calculate the atomic radius according to the number of electrons and electron configuration. Or is there any way to calculate the atomic radius using common characteristics which an atom has?

There's no measuring the radius of a single atom, mainly because electrons are around the nuclei: We can't define a radius for any single atom. Take a loom at the uncertainty principle:

  Introduced first in 1927, by the German physicist Werner Heisenberg, it states that the more precisely the position of some particle is determined, the less precisely its momentum can be known, and vice versa. The formal inequality relating the standard deviation of position $\sigma _x$ and the standard deviation of momentum $\sigma _p$ was derived by Earle Hesse Kennard later that year and by Hermann Weyl in 1928: $$\sigma _x \sigma _p \geq \frac{\hbar}{2}$$ Wikipedia

Thus, if you want to measure an atom's radius, you have to measure the distance between two nuclei, and then divide it by two. This results in different types of radius defined for atoms. The most common are covalent radius, ionic radius, and Van der Waals radius. Radii trends are what that are mostly studied in undergrad chemistry, rather than the how of measuring them with advanced techniques.

Chemguide is pretty informative when it gets to teaching the trends:

chemguide: atomic radius


Is there any mathematical or logical proof that carbon with 4 different groups would be chiral? In chemistry; chirality is generally defined in 2 ways.

Lord Kelvin's definition: "I call any geometrical figure, or group of points, chiral, and say it has chirality, if its image in a plane mirror, ideally realized, cannot be brought to coincide with itself" (source: 1. chirality.org , 2. Biochemistry book by Lubert Stryer.)

"If a tetrahedral (sp3) carbon atom contains a different chemical group on each of its 4 hands, the central carbon atom is called chiral. Though there are some exceptions where we do not found any chiral carbon still some compounds can be chiral. Chiral molecules cannot be superposed with their 3-dimensional mirror image" (common classroom-definition for chirality).

But it seems the first definition is the basic definition; and the second definition seems to be a corollary or application of the first definition.
Now I want to know; is it possible to 'prove' or 'derive' the second definition (better to say second statement or second description) from the first (Lord Kelvin's) definition mathematically or geometrically or logically? and if possible then how to prove the second definition from the first definition?
This question is related but different from this question

Yes, with a couple of simplifying conditions, you can easily show that a tetrahedral carbon with 4 different chemical groups (let's call them 1, 2, 3, 4) is chiral. Let's call the atoms directly bound to the carbon L1, L2, L3 and L4.

So the first condition I will impose is that none of the groups have another tetrahedral carbon with 4 different chemical groups attached.

Chemistry

In a tetrahedral carbon, the groups can't swap places. If that happens, we will call it a chemical reaction, and we are not allowing any chemical reactions.

Geometry

Tetrahdral should be taken loosely: For the plane defined by L1, L2 and the carbon atom, L3 is on one side of the plane and L4 on the other. The exact angles don't matter.

Showing that the carbon is chiral

If we put a mirror in the plane defined by L1, L2 and the carbon atom, the mirror operation swaps L3 and L4 (L3' on the side of L4, and L4' on the side of L3 - not sure we need a formal proof for that). I cannot superimpose that mirror image with the original (L1, L2 and the central carbon are already in the same position as the mirror image. I would have to rotate the molecule to get L3 and L4 superposed with L3' and L4', but then L1 would move away from L1', and L2 away from L2' - not sure we need a formal proof for that). Thus, the carbon atom changes configuration when taking the mirror image.

3D illustration

see https://proteopedia.org/wiki/index.php/R/S_nomenclature (you have to turn the molecules a bit to have L1 and L2 point in the same direction as L1' and L2')

Is this trivial?

No, not at all. For trigonal bipyramids there is sometimes pseudorotation, and for ammonia derivatives the trigonal pyramid can "umbrella flip", so these are typically not chiral. If there are four ligands on a central atom and the geometry is square planar, there is no chirality. So a couple of things have to come together to get a chiral center.


Do catalysts increase the rate of exothermic reaction more in a reversible reaction? I'm just learning chemistry for school and I've learnt that adding a catalyst reduces the activation energy for the forward and backward reaction by equal amounts, therefore the rate of forward and backward reaction increase by the same amounts so equilibrium stays the same but is reached more quickly.

I'm having some trouble understanding this. I know that the activation energy of the exothermic reaction is always smaller than that of the endothermic reaction. So thinking about the Maxwell Boltzmann distribution, a change in $E_a$ near to $0$ will have a greater effect on the rate of exothermic reaction than the endothermic reaction. Therefore, a catalyst should cause equilibrium to shift to the exothermic side?

Look at the form of the Arrhenius rate equation (this is fine since they mostly all have this form):

$$k = Ae^{-\frac{E_{a}}{RT}}$$

If you make the substitution, $E_{a}->E_{a}+\Delta E$, then you have:

$$k = Ae^{-\frac{E_{a} + \Delta E}{RT}} = Ae^{-\frac{E_{a}}{RT}}e^{-\frac{\Delta E}{RT}}$$

This means that you're just scaling the rate constant $k$ by another constant. This scaling applies to both the forward and reverse rate constants. If the concentrations were such that they were in equilibrium before, they're still in equilibrium now.

You'd have to prove this in general for complex reactions with complicated rate laws, but for simple reactions, this should suffice to help you see that the equilibrium is unaffected.


Who figured out that different isotopes have different numbers of neutrons? I reasearched about the history of the discovery of neutrons and couldn't find who found out that isotopes are a change in the number of neutrons. Can you please tell me who it was that made this discovery?

From Pauling's "General Chemistry" (ref. 1):

In 1921 the American chemist W.D. Harkins described nuclei as being built of protons and neutrons; he used the word "neutron" for a hypothetical particle with mass equal to that of the proton and with no electric charge. Ernest Rutherford made a similar suggestion during the same year. In 1932 the discovery of tne neutron was reported by the English physicist James Chadwick (born 1891).

The original article in which Harkins (ref. 2) reported this hypothesis is linked by the Wikipedia:
References

Linus Pauling, General Chemistry, second edition, 1970, p. 102
Harkins, William (1921). "The constitution and stability of atomic nuclei. (A contribution to the subject of inorganic evolution.)". Philos. Mag. 42 (249): 305.


How much volume of oxygen will be required for complete combustion of 40 ml of acetylene and how much volume of CO2 and H2O will be formed? How much volume of oxygen will be required for complete combustion of 40 ml of acetylene and how much volume of $\ce{CO2}$ and $\ce{H2O}$ will be formed?

After balancing the left-hand side produces 5 volume of oxygen and 2 volume of acetylene. However, the right-hand side produces 4 volume of carbondioxide and 2 volume of water. Why doesn't the volume on LHS and RHS match? Or how is it matched?

Is it correct that 100 ml of oxygen and 40 ml of acetylene produce 80 ml of $\ce{CO2}$ and 40 ml of $\ce{H2O}$? If so, what happened to 20 ml on the right hand side?

The volume can change in the course of the reaction. For example, if you add $\ce{Zn}$ to $\ce{HCl}$ the volume of the system changes. In the reaction $\ce{2H2 + O2 -> 2H2O}$ three volumes of gases turn into 1 (or even zero, if you allow water to condensate).

While mass remains constant, volume can change in chemical reactions.

Behavior of gases (under low pressure) is described by ideal gas theory. In gas a molecule is a small particle that has a large free space around it. The amount of free space is independent from of the molecular mass. lightest molecule $\ce{H2}$ occupies as much space as a heavy gas $\ce{SF6}$. As a result, density of $c\ce{H2}$ is much smaller than density of air (this is how helium baloons fly) which in turn is smaller than density of $\ce{SF6}$.

If two molecules form a dimer the density of the gas doubles, and the volume decreases by a factor of two (if the pressure is kept constant.

From 12 carbons and 24 hydrogens you can make 6 molecules of ethane ($\ce{C2H4}$) or 4 molecules of cyclopropane ($\ce{C3H8}$) or three molecules of cyclobutane ($\ce{C4H8}$). The amount of atoms is the same, the weight is conserved, but the volume and density are not.


Does resonance overrule hyper conjugation in terms of stability in alkenes? Which of the following is more stable ?

  $\ce{CH2=C(CH3)-CH=CH2}$
  $\ce{CH3-CH=CH-CH3}$

I was going for answer 2 because it has 6 hyper conjugative effects. I found the answer given to be 1, which has 3 hyper conjugations but more resonance structures. Will the R-effect meet the deficit of 3 hyper conjugations and make 1 more stable than 2?

We can hydrogenate unsaturated compounds and measure the heat given off, this is called the heat of hydrogenation. As the following figure illustrates, if we hydrogenate the various butene isomers they all wind up producing the same molecule, butane.

image source

By comparing their heats of hydrogenation we can see which butene was lowest in energy to begin with, that is, which butene is the most stable. The heats of hydrogenation reported in the figure tell us that trans-2-butene is more stable than 1-butene by about 2.3 kcal/mol. This extra stability can be attributed to the additional hyperconjugative resonance structures involving hydrogen present in trans-2-butene (6) compared to 1-butene (2).

Next let’s consider 1,3-butadiene. Here when we hydrogenate it, butane is again produced and we might estimate that it will have a heat of hydrogenation equal to twice that reported for 1-butene (1-butene is the best model because its double bond has the same substitution pattern as found in 1,3-butadiene) or 60.6 kcal/mol of heat given off. When we actually run the experiment we find that only 57.1 kcal/mol of heat is given off, 3.5 kcal/mol less than we estimated. In other words 1,3-butadiene is about 3.5 kcal/mol more stable than isolated double bonds and this is due to the resonance interaction between the two double bonds in 1,3-butadiene.

The resonance stabilization in 1,3-butadiene (3.5 kcal/mol) is larger than the hyperconjugative stabilization in trans-2-butene (2.3 kcal/mol).

Your question actually asked about 2-methyl-1,3-butadiene. In this case we have the resonance stabilization found in 1,3-butadiene plus some hyperconjugative stabilization from the additional methyl group. It will be even more stable than 1,3-butadiene. Further, 2-methyl-1,3-butadiene is more stable than any butene, so answer #1 is correct. Resonance stabilization is larger than hyperconjugative stabilization in this series of compounds.


Significance of 4π in the uncertainty principle What is the significance of the quantity 4π in the Heisenberg uncertainty principle?
$Δp \times Δx = \frac{h}{4π}$
The focus of the question is to learn why the ratio of a circle's circumference to its diameter, universally known as $π$, is used in the denominator in the term $\frac{h}{4π}$. To put it another way, why is this geometric number relevant in Heisenberg's formula?

The quantity on the right side of the expression for the product of uncertainties basically depends on the mathematical definition of "uncertainty" used. Without rigid definition of this quantity one ofthen just say that the product of uncertainties in position and momentum is of the order of Planck constant (or the reduced Planck constant; since they are proportional to each other it does not matter)
$$ \Delta x \Delta p \sim \hbar \, .$$
To make the statement more precise, one have to define what is actually meant by "uncertainty" and usually uncertainties are defined as the standard deviations.
$$
\Delta x = \sigma_x = \sqrt{\langle (x - \langle x \rangle)^2 \rangle} = \sqrt{\langle x^2 \rangle - \langle x \rangle^2} \, , \\
\Delta p = \sigma_p = \sqrt{\langle (p - \langle p \rangle)^2 \rangle} = \sqrt{\langle p^2 \rangle - \langle p \rangle^2} \, , \\
$$
where angle brackets $\langle \phantom{x} \rangle$ stand for the average value. And with these definitions one can indeed prove that
$$
\Delta x \Delta p \geq \hbar / 2 \, .
$$
Actually, we do even favour the usual notation for the standard deviation to avoid ambiguity
$$
\sigma_x \sigma_p \geq \hbar / 2 \, .
$$

So, from this perspective the 2 is nothing but the factor of proportionality between the product of uncertainties in position and momentum being thought of as standard deviations and the reduced Planck constant. $4 \pi$ is the factor of proportionality between the product of uncertainties in position and momentum being thought of as standard deviations and the Planck constant itself.

In fact, you can prove a more general Robertson uncertainty relation that for any two observables $A$ and $B$ represented by self-adjoint operators $\hat{A}$ and $\hat{B}$
$$
    \sigma_{A}\sigma_{B} \geq \left| \frac{1}{2i}\langle[\hat{A},\hat{B}]\rangle \right| = \frac{1}{2}\left|\langle[\hat{A},\hat{B}]\rangle \right|.
$$
Now, for $x$ and $p$ it is postulated by so-called canonical commutation relation that $[\hat{x},\hat{p}]=i\hbar$, thus, leading to
$$
    \sigma_x \sigma_p \geq \hbar / 2 \, .
$$

From that perspective, the Heisenberg uncertainty relation is equivalent to canonical commutation relation, and $\pi$ actually comes to the first one from the last one (as part of $\hbar$). So now the question really is why do we have $\pi$ in canonical commutation relation? And as user3371583 noted, it indeed has something to do with Fourier transform, but the connection of position and momentum through Fourier transform is not trivial.

Intuitively, however, you can get a sense of it by looking at the simpliest case of a one-dimensional de Broglie plane wave, in which $\pi$ factor in a connection between $x$ and $p$ naturally comes from de Broglie relation
$$
    p = \hbar k \, ,
$$
where $p$ is the momentum and $k = 2 \pi / \lambda$ is the wavenumber. So, wave function for plane wave can be written as follows
$$
    \psi(x) = \mathrm e^{\mathrm i(kx-\omega t)} \, ,
$$
and its derivative with respect to $x$ is
$$
    \frac{\partial \psi(x)}{\partial x} = \mathrm i k \mathrm e^{\mathrm i(kx-\omega t)} = \mathrm i k \psi(x) \, .
$$
If we substitute then $p / \hbar$ for $k$ from de Broglie relation we get
$$
    \frac{\partial \psi(x)}{\partial x} = \mathrm i \frac{p}{\hbar} \psi(x) \, ,
$$
or
$$
    p \psi(x) = - \mathrm i \hbar \frac{\partial \psi(x)}{\partial x} \, ,
$$
which suggest that
$$
    \hat{p} \psi(x) = -\mathrm i \hbar \frac{\partial \psi(x)}{\partial x} \, .
$$
$\hat{x}$ is simple multiplicative operator in this picture
$$
    \hat{x} \psi(x) = x \psi(x) \, ,
$$
so, the commutator between $\hat{x}$ and $\hat{p}$ is indeed $[\hat{x},\hat{p}]=i\hbar$ and through this commutator $\pi$, in a sense, propogates to Heisenberg uncetainty relation.


Where do the electrons not in the S1 orbital get this "extra energy" from? In this video, Sal Khan says that an electron reaches a higher energy state if the electron possesses more energy. So I assume that, say in the case of carbon, the electrons not in the S1 orbital have more energy. Where does this energy come from?

In simple terms: The further apart the negatively charged electrons are from the positively charged nucleus, the higher is the potential energy of said electrons. When increasing the quantum number $n$, you automatically get further away from the nucleus. That is why the $2s$ electrons have a higher energy than $1s$ electrons. (That, and adding that the $1s$ electrons shield some of the nuclear charge.)