text:
What is the chemical composition of the human body?
What is the average percent composition of each element in the average human body?

oxygen 61.3533
carbon  22.8291
hydrogen 9.9877
nitrogen 2.5683
calcium 1.4268
phosphorus 1.1129
potassium 0.20
sulfur 0.20
sodium 0.14
chlorine 0.14
magnesium 0.03
iron 0.01
fluorine 0.004
zinc 0.003
silicon 0.001
rubidium 0.001
strontium 0.0005
bromine 0.0004
lead 0.0002
copper 0.0001
aluminum 0.0001
cadmium 0.0001
cerium 0.0001
barium 0.00003
iodine 0.00003
tin 0.00003
titanium 0.00003
boron 0.00003
nickel 0.00002
selenium 0.00002
chromium 0.00002
manganese 0.00002
arsenic 0.000010
lithium 0.000010
cesium 0.000009
mercury 0.000009
germanium 0.000007
molybdenum 0.000007
cobalt 0.000004
antimony 0.000003
silver 0.000003
niobium 0.000002
zirconium 0.000001
lanthanium 0.000001
gallium 0.000001
tellurium 0.000001
yttrium 0.0000009
bismuth 0.0000007
thallium 0.0000007
indium 0.0000006
gold 0.0000003
scandium 0.0000003
tantalum 0.0000003
vanadium 0.0000002
thorium 0.0000001
uranium 0.0000001
samarium 0.00000007
beryllium 0.00000005
tungsten 0.00000003

Reference: Emsley, John, The Elements , 3rd ed., Clarendon Press, Oxford, 1998


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Can we predict with any reliability whether a chemical is carcinogenic in people?
I know that some chemical compounds have been known to be carcinogenic in humans.
But can we make a reliable prediction of how likely a substance is to be carcinogenic based on the chemical and physical properties of the substance ? How good will the prediction be ?
It would be good if the following methods were compared

Computational chemistry
Crystallography
Bacteria models such as the Ames test
Animal models such as rats and mice
Chemical kinetics experiments
Predictions based on the fact that a substance is from the same class as a known carcinogen or non carcinogen

The biological interactions that create cancer are very complex. It seems unlikely that any simple prediction exists. As an indication of how big and complicated this question could be, this old question and answer of mine on skeptics.SE is worth a look .

I would suggest we take the question off hold, I think that while a universal predictor of carcinogenic potential for all classes of compound is not possible. It is possible for some classes of compound to make a good prediction of how strong a genotoxic carcinogen it is.

https://www.sciencedirect.com/science/article/pii/B9780128046678000055
For toxicity, this kind of studies are being done and have also successful results.
I guess prediction part is way more difficult than toxicity due to the longevity of the carcinogenicity experiments. That may very well change. And I also think that this question should be off hold. https://omictools.com/qsar-toxicity-predictions-category
This is another good source for similar topics.

If the question is taken off hold, I am willing to have a go at giving an answer which relates to alkylation agents. I am sure that some simple kinetics experiments can be used to make a reasonable judgement as to the potential of an alkylation agent to induce the mutations needed to induce cancer.

As a short answer: NO we cannot predict if a substance will be carcinogenic to humans. A molecule may not be carcinogenic by itself but become so after being processed by the liver, e.g. benzene. And because a molecule is carcinogenic in mice/rats/dogs/etc... does not mean it will be in humans. That being said, we may sometimes consider that a substance is likely to be blank (i.e. carcinogenic, repro-toxic...) after tests on animals. It just means that we should be really careful because it has deleterious effects on animals.

Hmm, seems we don't have enough interesting questions, which aren't too broad :/ Maybe you guys would rather consider answering one of related: https://chemistry.stackexchange.com/q/95168 https://chemistry.stackexchange.com/q/7761 https://chemistry.stackexchange.com/q/39109 https://chemistry.stackexchange.com/q/24186 https://chemistry.stackexchange.com/q/75590

SteffX is not correct, we can make predictions based on things like Ames tests, computational work and chemical experiments but the quaility of the different predictions will differ greatly. I think it would make a great question

I really do feel that this is too broad - I have several entire books dedicated to this on my bookshelf, and none of them come close to providing an answer that will neatly fit into an answer on StackExchange. That said, I think the question is an interesting one. I'd favour splitting it into more specific issues ("can computational chemistry be used to predict whether a compound is likely to alkylate and cause cancer") which individuals might actually have the expertise to answer

One of the great problems is that there is no single mechanism by which a chemical carcinogen causes the damage which leads to cancer. There are a range of different mechanisms, I will mention some of them.
One mechanism is alkylation of DNA, the attachment of alkyl groups to DNA will prevent it being read correctly. In some cases the damage can be repared while in other cased the damage will cause the body to misread one base pair as another. There is an interesting problem with electrophilic alkylation agents, if they are too reactive towards water then they will not survive to reach the DNA.
There is an interesting paper in which this idea is discussed E.W. Vogel and M.J.M. Nivard, Mutation Research, 1994, volume 305, pages 13-32 considers it at length. If we consider the ratio of the rate constant for the reaction of an alkylation agent with thiosulfate and acetate, then as the ratio of these reaction rates changes in favour of the thiosulfate reaction something known as the Swain Scott constant (s) also increases. The Swain Scott constant is the ratio between reaction at two different sites (7-alkylguanine/O6-alkylguanine) in DNA. In general the higher the value of the s constant the stronger a carcinogen is. In the paper by A. Barbin and H. Bartsch in Mutation Research, 1989, volume 215, page 95 to 106 a series of directly alkylating carcinogens a good but not perfect relatioship is seen between the s constant and how carconogenic they are in rodents.
So it appears for one class of carcinogen that it is possible by doing some simple chemical kinetics experiments that it is possible to make an assessment of how carcinogenic a substance.
One of the great problems is that many "carcinogens" are in fact "precarcinogens" which require metabolic activation to make them carcinogenic. For example hydrazine reacts with formaldehyde in the liver before being oxidized to form diazomethane which then methylates the liver. This is why acute large hydrazine exposures in humans can cause liver failure.
This brings us to the Ames test, this is a test using bacteria which measures how able a substance is to alter DNA. The idea is that cancer is caused by altering the DNA of normal cells. So something which is able to randomly alter the DNA of cells is able to induce cancer. The great problem with the Ames test is that while it is a good test. There are both false positives and false negatives if it is used as a test for carcinogenic potential.
The normal Ames test protocol includes some sterile extract of rat liver to provide some metabolic activation. This will increase the ability of things like polycyclic aromatic hydrocarbons to alter the DNA of the bacteria.
If we consider an inorganic example, the chromate anion is known to be carcinogenic while chromium(III) is not carcinogenic. To understand this we need to know something about chromium biochemistry.
It is true that chromium(III) can form some very stable complexes which are very slow to react with DNA. But the chromium(III) is unable to reach the DNA in the cell. However chromate is able to pass through the sulfate transport channels in both the cell and nuclear membranes to allow it to reach the nucloplasm. There it is reduced to chromium(III) which then damages the DNA. As a result we can make a judegment about chromate and sulfate.
Sulfate is unable to react in the nucloplasm to form something which will form long lasting complexes with DNA, so even while sulfate is able to reach the DNA by the same route as chromate it is clear that sulfate is unable to act as a carcinogen by the same mechanism as chromate.
Also permanganate or magnanate is a stronger oxidant than chromate, I suspect that as it would be less able to survive the journey through the body to the DNA it is less able to incude cancer. This is likely to be one of the the reasons why permanganate is far less carcinogenic than chromate. Plenty of evidence exists for chromate being carcinogenic but little if any evidence exists for permanaganate.
Also the complexes of manganese with DNA are likely to be shorter lived than those of chromium. This is becuase chromium(III) complexes are slow to exchange ligands while manganese(II) are rather quick to exchange ligands. This is due to the fact that chromium(III) complexes are d3 while manganese(II) are d4.

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Is every compound that has a H-bond donating group soluble in water?
I am wondering if there is any general rule of thumb

There are hardly any such simple generalisations that can be made in chemistry for the solubility of compounds. If the compound has a large nonpolar domain, clearly it won't be water soluble despite having an -OH group.

If we are making generalisations, I would say they are harmful. ;-)

Definitely not.
For instance, palmitic acid with a pK$_\mathrm{a}$ of 4.75 has a water solubility of only 0.04 mg/L.
A general rule of thumb is that every oxygen can solubilize four carbon atoms but there are many exceptions.
I addressed a similar question here.
Nitrogen also makes the compound more soluble, but there it depends very much on the pH.

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Electroplating with iron and copper acetate
Im trying to do cheep electroplating using iron and copper acetate as in introduction to electroplating and working with acids. Each solution was prepared using heated vinegar and hydrogen peroxide. I notice two things. First, my iron plating happens much slower than the copper plating. Is this because for my copper plating solution i used a vinegar with a 20% acid concentration and for the iron i used a 10%? Voltage is the same at 12V
My next question about about the gas coming off the cathode. Is it hydrogen gas? I was under the impression that my solution would never be depleted. As copper is plated onto an object it is removed from the copper anode and everything just keeps going. But if gas is bubbling up that means something is being consumed and will have be replaced. Is it that on the anode, oxygen is produced which then reacts with the copper and thats why I dont see any oxygen bubbling off?

Good question, it raises a lot of things which are in an undergrad electrochemistry lecture course. These include things like exchange current density and coordination chemistry from the inorganic lectures.


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IUPAC nomenclature of chelates
I could finally get a copy of the Red Book 2005 from a library. In the chapter dedicated to nomenclature of coordinate complexes, while explaining the κ-system of specifying donor atoms, the following two examples were given (among many others):

Can anyone explain how the naming is done for the two compounds (labelled 2 and 3 in the picture)? The problem that I am facing is with the closed rings. From where is the naming started?

The image gives a thousand words, it is difficult to say how many words the colorized image gives, but might help.


text:
Why is the carbon atom omitted from the molecular formula?
In my chemistry textbook, the following reaction was given:

I have two questions related to the marked in orange boxes/parts of the image:

Why is the molecule symbolized by $\ce{(CH3)3Cl}$ instead of $\ce{(CH3)3CCl}$ or $\ce{C(CH3)3Cl}$? Why is the carbon atom omitted?
Why is the "inversion" molecule with a $\ce{CH2}$ instead of a $\ce{CH3}$ attached to it?

I would greatly appreciate an answer. This is high-school chemistry that I have just learned and I am a bit confused.
Thanks :)

Typos!  Just bad editing. You are correct in your thinking. You should point these out to your instructor

Often the publisher has a list of errata for a book. You might try and look that up. Kudos for you studying diligently enough to catch the mistakes.

Step 2 in your text should have water as the nucleophile, not hydroxide. Hydroxide would lead to elimination to isobutylene, not substitution.


text:
Effect of Electronegativity on Bond Angles in NH3 and NF3
I am reading a book: Concise Inorganic Chemistry by J.D.Lee (Fifth Edition)
In the chapter-The Covalent Bond, the author says:

$\ce{NF3}$ and $\ce{NH3}$ both have structures based on a tetrahedron with one corner occupied by a lone pair. The high electronegativity of $\ce{F}$ pulls the bonding electrons further away from $\ce{N}$ than in $\ce{NH3}.$ Thus repulsion between bond pairs is less in $\ce{NF3}$ than in $\ce{NH3}$. Hence the lone pair in $\ce{NF3}$ causes a greater distortion from tetrahedral and gives an $\ce{F-N-F}$ Bond angle of 102°30', compared with 107°48' in $\ce{NH3}$. The same effect is found in $\ce{H2O}$ (bond angle 104°27') and $\ce{F2O}$ (bond angle 102°)

I am not able to understand why the pulling of the bonding electrons (in $\ce{NF3}$) by F results in reduced repulsion between the bond pairs and how does this ultimately lead to a smaller bond angle ($\ce{F-N-F}$) in $\ce{NF3}$?

If the electrons are more pulled towards $\ce{F}$, the distance between the three bond pairs increase from one another and hence electrostatic repulsion decreases, so they can go a little bit closer to achieve the new equilibrium distance , which results in shrinking of bond angles.

It would be helpful if you could give a pictorial representation of what you are trying to say, your comment isn't quite what I am looking for.

Related: Why does NF3 have a smaller bond angle than NH3?

Thanks @Gaurang, that was helpful. I actually wanted to know how the author has reached the conclusion, which is still not clear to me. Also the posts mention 'Bent's rule' which I am not aware of.

https://chemistry.stackexchange.com/questions/15620/what-is-bents-rule


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De-excitation of a molecule
In our chemistry lecture today on UV/vis-spectroscopy, we discussed the origin of colour in conjugated molecules due to electronic transitions from lower-energy molecular orbitals to higher-energy molecular orbitals. One of the examples we talked about was the origin of the orange colour of carrot. The lecturer mentioned that carrots appear orange as it contains conjugated molecules that absorbs blue light, resulting in the complementary orange colour being emitted. This blue light is used to excite electrons from π MOs to π* MOs in the molecule. One of my classmates asked what happens to the excited electron when it falls back to the ground state, wouldn't it cause blue-light photons to be emitted?
To this, the lecturer replied that perhaps, there is de-excitation of the electrons through other pathways, such as through molecular vibrations (i.e. kinetic energy of the molecules). However, he also was not so sure about the matter. Thus, I would like to seek clarification on how these excited electrons de-excite after initial excitation.

After radiation is absorbed and the electron is at an excited state in the molecule there are several pathways for de-excitation to occur (see fig.). The pathway of choice depends on its rate, ie how fast it can happen. It turns out that the fastest de-excitation pathways are radiationless (wavy arrows in fig) such as internal conversion (IC) that happens through vibrational relaxation of the molecule and external conversion (EC) where the molecule returns to its ground state through collision with solvent molecules. This is why matter usually doenst emit light after absorbing it. There are instances however where this is actually possible because it happens that the radiationless pathways are not the most efficient way for the molecule to return to its ground state and that is when fluorescence (FL) and phosphorescence (Ph) occur.

Due to the vibrational relaxation, will the carrot heat up ever so slightly when the electrons de-excite via the internal conversion parthways?

@TanYongBoon Yes, i think this is correct. It can actually heat up quite a bit like anything does when left in the sun. See here as well: https://www.scientificamerican.com/article/how-exactly-does-light-tr/

Simply put, the reason the emitted color is different from the absorbed one is because there is a loss of vibrational energy that takes place in addition to the photon emitted by "falling back to ground state". This site has a good detailed explanation: https://micro.magnet.fsu.edu/primer/lightandcolor/fluoroexcitation.html
Essentially, the "π  MOs to π∗ MOs" description is a simplification that omits the existence of vibrational states. For another visual, here is a diagram taken from Wikipedia. The loss of vibrational energy is shown as the red arrows.

What your lecturer (as many lecturers, unfortunately) did not tell you is that the de-excitation pathways depend on the substance and the wavelengths you consider.
You can't know beforehand if a substance will be fluorescent, phosphorescent, or de-excitates with or without radiations. That is why only some substances can be used to make lasers (they need to support stimulated emission of radiations)


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Why are there as many molecular orbitals as atomic orbitals?
So molecular orbitals are represented as a weighted sum over atomic orbitals. Of course I reckon that the total sum of the weights per atomic orbital has to equal 1, when summed over all molecular orbitals. But why then should n atomic orbitals results in exactly n molecular orbitals? Is there a fundamental reason for this I am not seeing?

When you combine two orbitals you add and subtract them.

The notion is that two atomic orbitals combine to give you a molecular bonding orbital and a molecular antibonding orbital.

Possible duplicate of The “rules” for LCAOs in Molecular Orbital Theory - fundamentally this arises because an $n \times n$ matrix has exactly $n$ eigenvectors (MOs) with associated eigenvalues (energies)

The orbitals, both atomic and molecular, must be orthogonal, that is, they have no net overlap or correlation:  equal constructive and destructive interference. You can think of them as forming the basis set of an n-dimensional vector space. If you think of the atomic orbitals as the standard basis...
Example for a 4-dimensional space:
$\langle 1,0,0,0 \rangle$,
$\langle 0,1,0,0 \rangle$,
$\langle 0,0,1,0 \rangle$,
$\langle 0,0,0,1 \rangle$
...Then the molecular orbitals are just a rotated basis. In other words, they take linear combinations of the atomic orbitals to span the same space.
Ex:
$\langle \frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2} \rangle$,
$\langle \frac{1}{2},\frac{1}{2},-\frac{1}{2},-\frac{1}{2} \rangle$,
$\langle \frac{1}{2},-\frac{1}{2},\frac{1}{2},-\frac{1}{2} \rangle$, $\langle \frac{1}{2},-\frac{1}{2},-\frac{1}{2},\frac{1}{2} \rangle$
The same n-dimensional space requires at least the same number (n) of vectors to span it. This is exactly the same phenomenon as rehybridising individual s and p orbitals to form sp$^3$ orbitals.
You can find explicit formulae for the orbital vectors with the actual numbers (one possible representation), along with fantastic illustrations here Darmstadt, LCAO Theory.
There are two normalisation conditions when combining atomic orbitals to make either hybridised orbitals or molecular orbitals. The first is that every atomic orbital must be used completely. That is, the sum of its contributions to all the molecular orbitals must be 1. The other is that each of the new orbitals cannot have more than 100% orbital. Although I think that it is this second part that you were questioning the reasoning for.


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Pulsed NMR change in magnetisation vector
When a second, radio frequency magnetic field $\ce{B1}$ is applied perpendicularly to $\ce{B0}$, why does the magnetisation vector M rotate in a direction perpendicular to both $\ce{B1}$ and $\ce{B0}$?

I understand how electric charges move in magnetic fields, and I understand the classical electromagnetism magnetic moment, but I can't figure out how $\ce{B1}$ move M in this direction.
Any help would be appreciated.
Image from http://chemistry.mit.edu/sites/default/files/DCIF-IntroNMRpart1-theory-o07.pdf

In the nmr experiment the sample is placed into an intense but static magnetic field. By the (nuclear) Zeeman effect the spin energy levels that are degenerate in the absence of a magnetic field become split in energy and thus more spins reside in the lower energy levels that in higher one. The number of any level is given by the Boltzmann distribution at the prevailing temperature. The total spin angular momentum of the sample is called the magnetization. To make a measurement a magnetic field from a radio frequency source is applied at right angles to the large, static field. This tips the magnetization into the x-y plane making measurements possible. I try to explain how and what happens below.
The spins have a magnetic moment $\bar m$, which is a vector quantity meaning it has direction and magnitude, and as a result the static field causes magnetic induction $\bar B$ in the sample. This is also a vector. The magnetic induction (confusingly but commonly also called the magnetic field) is induced in the sample by the the large static magnetic field often labelled $H$.
If all the individual spin moments are added up these form the magnetization $ \bar M$. This is a magnetic field and is also a vector and so has both magnitude and direction.    All nuclear spin magnetic moments experiences a torque ( a rotational motion generating angular momentum ) in a magnetic field and is given by classical electromagnetism as $\bar m \times \bar B$ where $\times $ means this is a vector cross product; $\bar M \times \bar B=|M\|B|\sin(\theta)$ where $\theta$ is the angle between the vectors.  The change in angular momentum produced is given by $\displaystyle \frac{d\bar L}{dt}=\bar m\times B$. As the angular momentum is given by $\bar m=\gamma \bar L $,   where $\gamma$ is the magnetogyric ratio (these are known and for H is $27.75\cdot 10^{7}$ rad/T/s), then
$$ \frac{d\bar M}{dt}=\gamma \bar M\times B \tag{1}$$
This equation can be broken down into components in x, y and z directions. The z direction is conventionally taken as the direction along which $\bar B$ lies, and this has no x or y components. The three equations produced are
$$\frac{d M_x}{dt}=\gamma m_xB;\quad \frac{d M_y}{dt}=-\gamma m_yB; \quad \frac{d M_z}{dt}=0  \tag{2}$$
If there is no external perturbation then these equations show that the magnetization remains along the z axis and as we place the detector along the x or y axis then no signal can be observed. This is  because the magnetic field of the magnetization has to 'cut the coils' of the detector as so induce a signal that can be measured. It clearly cannot do this if it has magnitude only along the z axis.
The equations (2) can be solved if some initial conditions at time zero are chosen and physically important solutions are found only when the x and y components are not initially zero. In that case the solutions show that then magnetization in x and and y are proportional to $\sin(\omega_0t)$ and $\cos(\omega_0t)$ respectively where $\omega_0= \gamma B_0$ where $B_0$ is the z component of $\bar B$. The Larmor frequency is $\nu_0=\omega_0/(2\pi)$ and is the frequency measured in an nmr experiment.
As these solutions require that the magnetization has to be in the x-y plane this  condition has to be satisfied and is done so by exciting the sample with rf radiation directed along the x (or y) axis. This radiation has a magnetic  field that is weak compared to $\bar B$ and of magnitude $B_1$; its frequency we will call $\omega_r$. Classically the magnetic field applies a torque to the magnetization and rotates it away from the z axis.  In a molecular (quantum) explanation, if the rf energy is resonant with the energy gap, formed by the large static field $B$, then spins are excited from the lower to upper level and this changes both the size and direction of the magnetization $\bar M$.
In a classical  explanation the  polarised rf field $B_1$ is considered to be composed to two oppositely rotating fields. One moves in the same sense as the magnetization precesses, and so interacts with it, the other moves in the opposite sense and has little interaction and can be ignored. The experiment is now such that the spins experience both a static magnetic field $\bar B \equiv B_0$ in the z direction and a rotating field $B_1$ in the x-y plane. Clearly this is now a very complex problem but can be simplified if one assumes that the axes x and y rotate about z at the frequency of the $B_1$ magnetic field. This is called the 'rotating frame'. The result is that the equations similar to (1) are produced but with $B_{x,y,z} $ replaced by $B_{eff(x,y,z)}$ where $\displaystyle B_{eff}^2=(B_0-\frac{\omega_r}{\gamma})^2+B_1^2$.

A more precise wording is that the magnetisation vector rotates about the applied field $\vec{B}_1$. This can be derived either classically or quantum mechanically.
I will start with a word of caution: fully understanding the images you have shown requires much more than this because there are certain concepts which are hidden. Without going into too much detail, the $B_1$ field is actually not static, but oscillates along one particular axis. This is decomposed into a sum of clockwise- and anticlockwise-rotating magnetic field. One of those two, say anticlockwise, is neglected because it is "off-resonance", and the frame of reference is taken to be a rotating frame, rotating clockwise at the same frequency as the clockwise half of the $B_1$ field, such that the clockwise $B_1$ field is static in this frame of reference. The last paragraph of porphyrin's answer gets towards this, but does not completely explain. (In no way is this a criticism, as the theory is rather involved, and I suggest reading a proper book on NMR theory, e.g. Keeler's book.)

porphyrin's answer already guides you through the classical approach, and I suppose any text on electromagnetism should also cover it. Without restating the answer fully, the crucial bit is the set of differential equations
$$\frac{\mathrm d\vec{M}}{\mathrm dt} = \gamma (\vec{M} \times \vec{B})$$
or more explicitly
$$\frac{\mathrm dM_x}{\mathrm dt} = \gamma (M_yB_z - B_yM_z) ; \quad \frac{\mathrm dM_y}{\mathrm dt} = \gamma (M_zB_x - B_zM_x) ; \quad \frac{\mathrm dM_z}{\mathrm dt} = \gamma (M_xB_y - B_xM_y)$$
Given (as in your diagram) $\vec{B} = (0, B, 0)$ as well as the initial condition $\vec{M}_{t = 0} = (0,0,M)$ you can then solve for $(M_x,M_y,M_z)$ as a function of time. You will get some sort of $\cos(\omega t)$ and $\sin(\omega t)$ dependency, where $\omega = -\gamma B$, which reflects the circular precession of the vector $\vec{M}$ about the vector $\vec{B}$ at the Larmor frequency $\omega$.

Quantum mechanically one can obtain the same relation for the precession of a single nuclear spin in a magnetic field. What you need to do is to use the time-dependent Schrödinger equation
$$\require{begingroup} \begingroup \newcommand{\wf}[0]{|\psi\rangle} \newcommand{\al}[0]{|\alpha\rangle} \newcommand{\be}[0]{|\beta\rangle} \newcommand{\ca}[0]{c_\alpha} \newcommand{\cb}[0]{c_\beta}
\mathrm{i}\hbar \frac{\mathrm d\wf}{\mathrm dt} = \hat{H}\wf$$
In this case the Hamiltonian corresponding to a magnetic field along the $y$-axis is $\hat{H} = \omega\hat{I}_y$. The wavefunction for a single nuclear spin can most generally be written as a linear combination of up and down spin, $\al$ and $\be$ respectively:
$$\wf = \ca\al + \cb\be$$
where the coefficients $\ca$ and $\cb$ are time-dependent. Note that there should technically be a phase factor like $\exp(-\mathrm iEt/\hbar)$ in front of both $\ca$ and $\cb$, but in the rotating frame it so happens that $E_\alpha = E_\beta = 0$, so these phase factors disappear.
Now if we substitute this into the Schrödinger equation, and using the relations $\hat{I}_y\al = (\mathrm i\hbar/2)\be$ and $\hat{I}_y\be = (-\mathrm i\hbar/2)\al$ (these are most easily derived using the Pauli spin matrices. Note that $I_y$ must include an additional factor of $\hbar/2$ relative to the $\sigma_y$ defined on Wikipedia), we get the following coupled differential equations:
$$\frac{\mathrm d\ca}{\mathrm dt} = -\frac{\omega}{2}\cb ; \quad \frac{\mathrm d\cb}{\mathrm dt} = \frac{\omega}{2}\ca $$
and you can once again solve these (using the initial boundary conditions $(\ca,\cb) = (1,0)$ at $t = 0$) to find the coefficients
$$\ca(t) = \cos\left(\frac{\omega t}{2}\right) ; \quad \cb(t) = \sin\left(\frac{\omega t}{2}\right)$$
The factor of half is probably a little worrying. However, note that the coefficients $\ca$ and $\cb$ do not reflect the actual magnetisation and hence the direction of the vector. To find this we must do one more step and find the expectation values of $\vec{I}$. This is demonstrated for the $x$-component:
$$\begin{align}
\langle I_x \rangle &= \langle \psi | I_x | \psi \rangle \\
&= \frac{\hbar}{2}\begin{pmatrix}\ca & \cb\end{pmatrix}\begin{pmatrix} 0 & 1 \\ 1 & 0\end{pmatrix}
\begin{pmatrix}\ca \\ \cb\end{pmatrix} \\
&= \frac{\hbar}{2}(\ca\cb + \cb\ca) \\
&= \frac{\hbar}{2}\left[2\cos\left(\frac{\omega t}{2}\right)\sin\left(\frac{\omega t}{2}\right)\right] \\
&= \frac{\hbar}{2}\sin(\omega t)
\end{align}$$
Following the same procedure for $I_y$ and $I_z$ you can now readily demonstrate that the magnetisation vector for a single nuclear spin precesses in exactly the same fashion as a classical vector.

The only remaining issue is that NMR experiments are not performed on only one spin, but on a large number of spins, or an "ensemble" of spins. Intuitively this is not a huge problem, because one can imagine that each individual spin in the ensemble will behave exactly as described above, leading to an overall effect that behaves the same way (an overall magnetisation which precesses about the magnetic field $\vec{B}$).
Rigorously this would be shown using density matrices, denoted $\rho$. The entire explanation could well be another standalone answer, but suffice it to say here that the initial density matrix above is $\rho(0) = \hat{I}_z$. The Hamiltonian for a pulse applied along the $y$-axis, as before, is $\omega \hat{I}_y$. The effect of the Hamiltonian on the density matrix is described by the Liouville–von Neumann equation
$$\mathrm i\hbar \frac{\mathrm d\rho}{\mathrm dt} = [H,\rho]$$
or its integrated form
$$\rho(t) = \exp\left(-\frac{iHt}{\hbar}\right) \rho(0) \exp\left(\frac{iHt}{\hbar}\right)$$
After substituting in the relevant matrices and operators, one finds that
$$\rho(t) = \exp\left(-\frac{i\omega t I_y}{\hbar}\right) I_z \exp\left(\frac{i\omega t I_y}{\hbar}\right)$$
After a bunch of mathematics it is possible to show that (for details, see e.g. Levitt Spin Dynamics 2nd ed., Appendix A.2 – beware that in NMR texts, $\hbar$ is usually taken to be unity. They will all cancel out anyway)
$$\rho(t) = I_z \cos(\omega t) + I_x \sin(\omega t)$$
as desired.


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Multiple leaving groups and SN2/E2

In the first stage of the reaction i am confused in which place will the SN2 happen (we have aprotic polar solvent and 4 leaving groups). In my opinion SN2 will happen either to the carbon with F- attached or the carbon with $\ce{Br-}$, $\ce{MeSO3}$ attached as these are the primary carbons. The leaving group strength is $\ce{F- < MeSO3 < Br-}$ so I suppose the  Br will be substituted. The thing is that the nucleophile must attack from 180 degrees direction and I think steric hindrance in this carbon is bigger than the one with $\ce{F-}$. Another factor is that the oxygen atom from $\ce{MeSO3}$ can stabilize the intermediate with resonance.
$\ce{NaH (H^-)}$ activates the  nucleophile by taking the most acidic proton from $\ce{-SH}$. So I think compound X will be:

Then E2 will occur ($\ce{NEt3}$ strong base) and the leaving group will be $\ce{MeSO3}$ (best LG). forming Z :

Is my solution correct?

Are you sure that the -OH is the most acidic proton? Perhaps you should look up the pKa of thiophenol

The question here is more like which group is the most nucleophilic on the aromatic compound after deprotonation by NaH AND which is the best leaving group for an SN2 (F, OMe, Br, Ms). FYI, the best leaving groups are the ones which can distribute a charge in a wider volume.

I was wrong -SH  group is more nucleophilic

Ms and Br can stabilize the negative charge better... also Ms can stabilize with resonance this makes it better leaving group from Br. But comparing their pka HBr has -9 and MsOH -1.9 making conjucate Br- less basic than MsO- and therefore the weaker base is better leaving goup. Which criteria wins?

Are you sure about the E2 step? Where is the most acidic proton in X?

I think X will be the Br compound, on the basis that either OMs or Br can be displaced in the initial reaction with ArS-, but in NMP NaBr will displace OMs but NaOMs will not displace Br


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Creating an IR Spectrum using Gaussian ADMP Calculation
I have a Gaussian 16 trajectory for a small cluster of water molecules carried out with ADMP technique. I was wondering if anyone knows how to compute the IR spectrum from the information in the trajectory.
Thank you

Dear Victor, if you have an XYZ File, you can use Travis (by Brehm and Kirchner) to calculate power spectra. For an IR spectrum, you need to compute the dipoles - normally done using Wannier centers, the analysis to make the spectrum is also possible in Travis: http://www.travis-analyzer.de/


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Why do amides react differently with PCl5?
In a reaction involving $\ce{PCl5}$ and any reactant containing oxygen atom, it generally takes out the oxygen and put chlorine their due to formation of strong bond between phosphorus and oxygen.
But, in case of amides like $\ce{RCONH2}$, when they react with $\ce{PCl5}$, a cyanide is formed. Why?

PCl5 does react on oxygen which forms R(Cl4PO)C=NH. Elimination of Cl4POH forms the nitrile, RCN. Cl4POH decompoes to  POCl3 and HCl.

https://chemistry.stackexchange.com/questions/65852/dehydration-of-amide-with-pcl5


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Why is ammonia less acidic than terminal alkynes?
Here's a picture from Organic Chemistry by Solomons, Fryhle & Snyder :

Why is it that ammonia is less acidic than terminal alkynes even though water, and even alcohols, are more acidic than them?
Water and ammonia seem to be very similar to me:

Electronegativity difference of oxygen and nitrogen is only 0.4 (Pauling scale)
Both are very similar and differ by only number of hydrogen atoms.
Both are $\ce{sp^3}$ hybridized
Oxygen and nitrogen atoms have similar size

However, the difference in acidic strength is huge. If my understanding of $\mathrm{p}K_\mathrm{a}$ values is correct, water is nearly $10^{10}$ times more acidic than terminal alkynes, whereas ammonia is less acidic than terminal alkynes by a factor of $10^{13}$!
Shouldn't the list be like: water > alcohols > ammonia > terminal alkynes > terminal alkenes > alkanes or water > ammonia > alcohols > amines > terminal alkynes > terminal alkenes > alkanes ?

It is the alkyne that is not in line.
In the alkynes, the carbon is sp hybridized and this gives it much more s character.
Because of this, the electrons are closer to the carbon nucleus and the $\ce{C-H}$ bond is slightly polarized.
The carbon has some δ - and the hydrogen δ +.
Therefore the terminal alkynes are more acidic and have a lower $\mathrm{p}K_\mathrm{a}$ than ammonia.

Ok ... so, I googled the $pK_a$ of HF and it's $3.17$. Seems like a change of around $10$ $pK_a$ is normal when moving in the second period. So, my next question would be — how can we, theoretically, assign acidity order for terminal alkenes and alkynes? Is there any data on the electronegativity (if that's the right term) of carbon in $sp^2$ and $sp$ hybridisation? I read a question related to this on Quora which uses %s character to find the E.N. of $sp$ carbon, but the reasoning is not very convincing.

Please ask this as a new question, explaining that it follows up from this one.

Just a reminder: Atoms are not hybridised, only orbitals are. Therefore: Carbon utilises sp hybrid orbitals for the C-H bond.


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When can colourimetry be used to determine the rate of reaction?
Can colourimetry be used if a coloured reactant react to give a coloured product? For example, in the reaction of reacting dichromate ions with oxalate ions in acidic environment, where dichormate ions (orange) react to give chromium three ions (green)?

It can be used if reaction is slow enough.

In a more general sense, spectroscopy is used nowadays to measure reactions on all time scales from tens of femtoseconds ($10^{-15}$ s)  to hours. If a laser can be used to start the reaction these methods are now called 'pump-probe' but used to be called 'flash photolysis'. Rapidly mixing reactants that then react without using a laser to initiate it,  which is the most general case,  and using spectroscopy to monitor the colour change is called 'stopped flow' .


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What causes the "rotting fish smell"?
Rotting fish seem to give off the same (very pungent) kind of smell, regardless of the kind (salmon, seabreen, tuna, etc).
What exactly is it that's responsible for this unique smell?

(Though I've answered my own question below, do feel free to post an answer of your own if you want.)

How to make it.  -  https://www.youtube.com/watch?v=cIDWdpMhKII

The "fishy" odor that you're familiar with is brought about by a whole bunch of compounds, and not any single one. Then again, if we were to narrow this down a bit, we could say that simple nitrogen compounds are the main culprits.
But suppose we want to blame only a single compound for the delightfully pungent odor of rotting fish, and we couldn't be bothered with the nuances of olfactory physiology or food chemistry to go through a list of compounds running into the hundreds (if not thousands). In that case, we can pin the blame on trimethylamine (TMA).

With regard to fish, the TMA is a result of the degradation of trimethylamine oxide (TMAO) present in the body fluids (blood, lymph, excreta) of the fish.
Just so you don't get the wrong idea, TMAO is present in the body fluids of all vertebrates ̶I̶ ̶c̶a̶n̶ ̶t̶h̶i̶n̶k̶ ̶o̶f̶, not just fish (albeit in much lower concentrations, which is why you don't mistake ground chicken for tuna).

Now why is TMAO more abundant in fish?
Ever tried to pickle a cucumber (or anything for that matter)? Notice that the cucumber shrivels up over time? It's losing water to the pickling fluid by a process known as exosmosis: the water is drawn out from a less concentrated environment (cucumber's tissue) to a more concentrated one (pickling fluid).
 Disclaimer: As user @lly pointed out in the comments, this particular bit has the potential to cause a bit of confusion. The inside of a cell (the cytoplasm I mean) is technically an aqueous colloidal solution (the pickling fluid is an aqueous solution as well). In the context of aqueous solutions, "concentration" is an indication of how much solute is present, not water. So in this context, if something is more concentrated, it has less water in it

Now the same thing is going to want to happen to a fish in salt water, which isn't so nice from the fish's perspective.
To counter the outflow of water, a fish has a significantly higher concentration of the familiar metabolite urea in its body fluids (as compared to terrestrial vertebrates). This establishes a situation where the fish's body fluids have roughly the same "concentration" as the salt water it swims in (the more correct term here would be "osmolarity", in place of "concentration")
That urea, if left to its own devices, will seek to destabilize  macromolecular structures and inhibit cellular functions; the TMAO present alongside the urea helps counteract those detrimental effects.
Of course, going by this logic, saltwater fish should have a much stronger "fish odor" than freshwater fish (which is certainly the case).
Lets not forget the classic kitchen tip: Rubbing fish with something acidic like vinegar, b̶a̶t̶t̶e̶r̶y̶ ̶a̶c̶i̶d̶ or lemon juice helps reduce the odor (the TMA degrades under acidic conditions).

Also, some of you might know that a fish-like smell is observed in rancid food. That, apparently, is the result of the oxidation of omega-3 fatty acids.

This bit however--the water is drawn out from a less concentrated environment... to a more concentrated out--first off, you probably mean "one" there but, second, you have this backwards, right? or you're talking about the water being drawn to something with a more concentrated amount of sth-other-than-water?

@lly Thanks! Though I did take enough care to pick the right sources/references, there is of course always a possibility that I may have erred somewhere. And yes, I had a feeling someone or the other would raise the "concentration" problem :P

@lly I've edited the answer to address that last issue. I hope that works O:)

Shouldn't we be asking: Why have we evolved a sensitivity to the smell of rotting fish? If seagulls can smell it they certainly don't mind it.

@DDuck As far as I understand it, us evolving an aversion to the smell of rotten anything (not just fish) is Nature's way of telling us that some of the by-products of microbe induced decomposition of most foodstuff isn't meant to find its way into our mouths (much less our stomachs). Seagulls, vultures, flies, etc have mechanisms within them that nullify/greatly diminish detrimental effects of consuming such food (that would otherwise be seen in humans) ;-)

@DDuck Of course, yours is not an unreasonable question, so feel free to ask it at the Bio.SE (though I have a feeling I've seen something vaguely similar to this over there0 before) O:)

Putrescine, or tetramethylenediamine, is a foul-smelling organic chemical compound related to cadaverine, both usually produced during tissue decomposition. It is a toxic diamine formed by putrefaction from the decarboxylation of arginine and ornithine, and is largely responsible for the foul odor of putrefying flesh.
See reference.

Try to avoid using Wikipedia as a reference, instead please use something like a scientific paper, the CRC handbook of chemistry / physics "rubber book" or some other source which is better than wikipedia

@NuclearChemist Wikipedia is known to have varying degrees of accuracy, so it is not to be used blindly. But that article has fourteen references, do you have any reason to recomend not using that particular article?

@Jose Yeah, but all 14 references don't back up the premise of this answer ("Putrescine is the cause of the rotting fish odor"). I gave that article a cursory look, and it appeared that most of the references (if not all of them) are references to the properties and biosynthesis of putrescine, and not to the fish-odor claim. :-)

I upvoted this because I know putrescine to have a major role in the odor of rotting (generic) flesh. However, the distinctive odor of rotting fish is mainly attributed to TMA. In my answer, I did mention that other componds were involved in the odor as well, but TMA was the main factor.


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Why does the liquid in an Oralmedic stick flow uphill after I turn it upside-down?
I was experimenting with an Oralmedic stick, which is basically a cotton swab with a hollow stick that has liquid medicine inside, and observed something very unexpected.  I followed the usage instructions, which say to snap the cotton bud at one end. As expected, the liquid inside the stick began flowing down into the cotton bud at the other end.
Out of curiosity, I turned the stick upside-down, to see what would happen. I was really surprised to see that the liquid still kept flowing in the same direction, which was then uphill after I inverted the stick!
How is this possible?

@Justanotherchemist I think capillary force is the key  together with some non newtonian fluid.I'll keep the researching.

The most likely explanation is a combination of two capillary effects, where a combination of surface tension and adhesion of the liquid to surrounding solid surfaces causes macroscopic flow to occur.
For the first, the internal diameter of the hollow stick is small enough that if one were to pull off the cotton buds at the top and bottom, capillary forces would likely cause the fluid to flow only very slowly through it. This provides some resistance to flow, sufficient to keep the liquid from gushing up to the 'snapped bud' end when you invert the stick. The surface tension associated with this capillary effect also results in the liquid behaving like a "plug" in the hollow tube: gas cannot "slide" around the liquid meniscus, which is also an important component to the physics.
The other role of capillary action is active in the fine threads of the lower cotton bud itself.  Once the top bud is snapped, the liquid starts to slowly flow down toward that unsnapped bud. Once the liquid reaches it, it wicks aggressively into the finely structured cotton.  Since no gas can get past the liquid in the tube, conservation of mass requires that the rest of the liquid continue to flow down toward the unsnapped bud.
The reason why flow continues toward the unsnapped bud when you invert the stick is a combination of all three of the above factors: the capillary action drawing the liquid into the unsnapped bud is essentially unaffected by the inversion of the stick, and since no gas can slip around the remaining liquid in the tube, that remaining liquid has to continue flowing up toward the unsnapped bud.
Another way to think about it is this: as each microscopic parcel of liquid wicks into the unsnapped bud, it leaves a region of slightly lower pressure behind it.  Since the end of the liquid column facing the snapped bud is open to atmosphere, that results in a pressure differential against the snapped-bud side of the liquid column that pushes the liquid up toward the unsnapped bud.


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Reactants in high temperature chemical vapour deposition
I'm setting up a high-temperature electrolysis rig for the purpose of chemical vapour distribution.
The molten carbonate (750 degrees Celsius) will be lithium carbonate. The cathode will be galvanised steel with the anode being nickel. Nitrogen, carbon dioxide and water vapour will be fed in with the intention of splitting the carbon dioxide as a low voltage, high current is passed through the electrodes.
I'm anticipating that the carbon dioxide splits, but I am not sure how the nitrogen and water vapour will react under those conditions. I'm guessing the nitrogen won't react, but how would the water vapour react? Ideally, I'd rather not have to go to the trouble of using pure/near-pure carbon dioxide.
Thanks in advance!


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Non-uniform sampling (NUS) in 2D-NMR
The NMR technicians where I work have recently started replacing the standard 2D NMR experiments (COSY, HSQC) with NUS variants (25% sampling). For the uninitiated, the rationale behind the technique is summarised by Claridge (who does a better job of explaining than I would):

The classical sampling of 2D (and more generally multi-dimensional) NMR experiments requires the uniform sampling of data in the indirect dimension(s) that allows for the processing of the data by the discrete Fourier transform. This means a sequential, stepwise increment of the t1 period of a 2D data set is made to the limit t1 which dictates the resolution in this dimension. The number of such t1 increments employed ultimately defines the total duration of the experiment. The method of non-uniform sampling (NUS) seeks to reduce the number of data points collected in the indirect dimension(s) and so reduces the total experiment time.
Claridge, T.D.W. High-Resolution NMR Techniques
  in Organic Chemistry, 2016

My experience of these experiments has been overwhelmingly positive so far: shorter experiment times and greater resolution, even when acquiring spectra of small quantities (0.1 - 0.5 mg) of large molecules (ca. 1000 Da).
The challenge is knowing what to do when the spectrum doesn't come out looking quite right. With 'traditional' experiments, running more scans often fixes a multitude of sins (up to a point), but with random point sampling I can't see this holding up so well.
The processing of data is also causing some confusion - TopSpin and MNova give (often very) different results, despite both having NUS processing capability. This processing issue leads me to suspect that many of the issues I'm having with spectra from NUS pulse sequences are actually due to the processing (which isn't terribly well covered in the TopSpin manual).
My question, therefore, is what considerations should be taken into account with NUS spectra: are there instances where its better to take the hit and run a long HSQC? What do I need to be careful of when processing data to ensure spectra come out looking okay?

I never used NUS myself, but from what I heard the "CS" mode for processing in Topspin gives much better results than the "MDD" mode, but also takes longer.

If you suspect the NUS itself is the problem (rather than the processing), one can usually control the percentage of points sampled during NUS (e.g. collect 50% of the points).  If the problems disappear as you increase the number of points collected (eventually reaching 100% or uniform sampling) then you have your answer.

I think a really full answer will require delving into the theory of NUS, which (just like the Fourier transform) goes way beyond NMR as a field: the idea of undersampling a signal and then reconstructing it is of great interest in a variety of engineering areas. I can't give that level of detail, so I don't pretend that this is really an answer, but maybe having some of this written down will help.
I'm really not an expert in this, but I can backup Mad Scientist's comment that for 2D data, TopSpin's reconstruction mode (the MDD_mod processing parameter) generally should be set to cs and not mdd. I suspect this may require a license from Bruker, so it might be best to check with the relevant NMR manager. Mnova, on the other hand, uses a different reconstruction algorithm entirely, so it's not overly surprising that there might be some differences in performance.
The amount of undersampling you want to do will also affect the quality of your results. I think a good rule of thumb is to make sure the ratio of points measured in $t_1$ to number of peaks in each column of the spectrum is somewhere around 2 to 5. The reasoning is simple: if you need to detect $k$ peaks in the frequency domain, then you need to measure some multiple of $k$ signals in the time domain. In other words, the more data you want to pick up, the more signals you need to measure. For a HSQC where each column has very few frequencies, you can get away with very sparse sampling, and 25% is likely to be fine (as long as your TD1 is not too small). For other spectra like COSY, TOCSY, HMBC where there is less empty space in the spectrum, this may not work quite as well.
As far as I'm aware, increasing the number of scans should have a similar effect for both uniform and non-uniform sampling. The main changes are the increase in S/N, as well as possibly better artefact suppression achieved via phase cycling. However, with modern gradient-selected experiments (especially the standard experiments like HSQC, COSY, etc.) you should not really need extremely long phase cycles to achieve good artefact suppression. In my experience the difference in spectral quality between NS=2 and greater is already pretty minimal. Bumping up the number of scans is mainly useful for increasing S/N. [Obviously, if the reason why the NUS reconstruction is failing is because of too poor S/N, then increasing the number of scans will help.]
Lastly, this is a trivial point, but I'd also suggest double-checking the other processing parameters (window functions, zero-filling etc.) after the reconstruction is done. It's quite easy to miss something out (for example, changing QSINE to SINE) and inadvertently get ugly peak shapes. I learnt (and am still learning) it the hard way.
Having said all this, what should you do if you find your NUS data is not good enough? I don't think there are really many answers to this, although someone more knowledgeable will be able to make better suggestions. One is to play around with the reconstruction algorithm, but unless you are very well-versed with the theory, this is essentially limited to trial-and-error with a dropdown box. The other is to just reacquire the spectrum with less undersampling, especially if you suspect you may have pushed it a bit too much.

I am curious as to why non-uniform sampling rate enhance would resolution? Isn't the basic purpose of any sampling process to represent a continuous signal as accurately as possible? I can understand the logic the NUS can save time but resolution enhancement is something interesting. I didn't study the theory of NUS.

@M.Farooq It's not explicit, but (at least in the field of NMR) when people talk about using NUS to gain resolution in nD NMR ($n \geq 2$), they usually mean given the same experimental time. Thus, while under uniform sampling one may measure 32 increments in one spectral dimension, NUS allows one to measure 32 points in the same amount of time and then reconstruct 128 points from them, for example. [Of course, one could always try to reconstruct 128 points from the uniformly sampled 32 points to get the higher resolution; so actually it's a more subtle question....

... of whether the reconstruction of undersampled NUS data, is better than the reconstruction of a smaller amount of uniformly sampled data. That's right about where I'll draw the line and say I don't know enough to make a conclusion, although the NUS people will always tell you yes, it is better :-) and I am inclined to believe them.]


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Why ammonium salts can not undergo substitution
Why is ammonium salt, such as RNH3+, can not react with nucleophiles like HI to give RI and NH4+? The reaction is pretty much like the substitution of alcohol in acid condition.
In my opinion, NH3 (rather than NH2-) is a very good leaving group, so it can be substituted by good nucleophiles like I-. In order to prevent the resulting ammonia from attacking back, we can add more acids like HI to drive the equilibrium to the right direction.
I haven't found any signs of this reaction in textbooks, so I thought this reaction cannot happen. If it is a common reaction, where can I find more information? Thank you all in advance.

RNH3+ protonates any nucleophile that approaches it destroying the nucleophilicity.

@Waylander On the one hand, the scope of this reaction is indeed limited, but at least I-(Iodide) can act as a good nucleophile which cannot be protonated. On the other hand, If the ammonium salt is tertiary, such as R4N+,  the substitution is still possible, just as the cleavage of ether.

Quaternary ammonium salts are well etablished as good leaving groups but I can find no examples of a primary ammonium hydroiodide salt reacting to form a primary iodide as you suggest. Perhaps someone with database access can search this better than me.

@liwt31 Quaternary maybe?

@OscarLanzi yes, thank you for your correction.


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Why does lithium aluminium hydride appear grey?
Commercial lithium aluminium hydride is a grey compound, which loses a lot of reactivity the darker it is. Luckily you can clean it up to get the reactive white compound and the separated grey contamination. Over some time, however, it is said to slowly convert back to the grey form (at least this is what I read, I haven't watched clean samples turn grey again yet).
My question is, what is this contamination? I remember to have heard somewhere that it may be elemental aluminum, which forms by decomposition. And in fact, the thermal decomposition really yields aluminum as a side product. But I cannot find any reliable source I could quote for my thesis. Does anyone have a paper or book that actually mentions the grey contamination?

Reaction with trace water impurity and/or oxygen will possibly give Al2O3, Al(OH)3, and LiOH. All the above are white powders apart from Al2O3 which is gray. Other impurities could be LiH (which is normally white but also appears gray so the question becomes why is LiH gray when impure) or Al which can also look gray. So it is hard to say with certainty especially since it doenst seem that anyone analyzed the grey stuff.  Once, i had LiCl precipitating from a LiTol + ZnCl2 reaction as a grey powder (reaction done in schlenk) possibly due to formation of Zn so i would go with Al or Al2O3.


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What are the different types of catalysts used in catalytic hydrogenation of alkynes?
This is a question from Org. Chem. by Solomons, Fryhle & Snyder, adapted for JEE by M.S. Chouhan :

In the first step, I expected that the reaction would stop at alkene formation via syn addition of $\ce{H_2}$ since Lindlar catalyst is poisobed and heterogeneous. What are some other other poisoned catalysts that stop at alkene?
In the second step, I expected syn addition of $\ce{D_2}$ since $\ce{Pd-C}$ is a heterogeneous catalyst, forming product (a). But the answer given is (c) — does this mean that some heterogeneous catalyts can give anti addition, or is it just an error on the adapting author's part?

The "Lindlar catalyst" (Pd/CaCO3) is probably the most common way of effecting stereoselective reduction of alkynes to (Z)-alkenes. It is fairly accessible and the reaction is easy to run. But it is not the only way: there are other catalyst systems. One of the more notable ones is probably the Brown catalyst "P2-Ni" (Wikipedia).
There is a nice review on alkyne semihydrogenation here: Chem. Rev. 2013, 113 (3), 1313–1350, which covers both heterogeneous and homogeneous methods, as well as methods involving other hydrogen sources apart from H2 gas itself. I recommend it to interested readers and especially anybody considering carrying out such a reduction.

The Lindlar hydrogenation of an alkyne forms (Z)-alkenes stereoselectively. Syn hydrogenation of the (Z)-alkene gives product (c), not product (a).

Note that in the fully reduced alkane, there is free rotation about the central C–C bond. So, having both D atoms pointing "out of the plane" is irrelevant, as you can easily rotate to a conformation in which one D atom is pointing in the plane. However, the configuration of the product cannot be changed, and that is the difference between options (a) and (c).


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Naturally Occurring Sulfites or Sulphites form
In the following link, it says that the product

May contain trace amounts of naturally occurring sulfite residue

I asked the manufacturer about it and the explanation I got was:

The trace amounts of naturally occurring sulfites are present in hydrolyzed collagen (as are most other proteins as well). The sulfur that holds the bonds between the molecules is broken up from these processes and changes them to sulfite

Does anybody knows the exact chemical form of the sulfite that this product contain?
Is it in the form of an anion or in the form of a compound?

BTW, wines may have sulfites added as preservative,  but even without additives, there are enough naturally occurring sulfites from fermentation to require a warning on the label. https://naturalmerchants.com/organicwines/sulfites-in-wine/

When collagen is hydrolyzed, sulfite ions ($\ce{SO3^2-}$) are released. Usually manufacturers filter out the residues, but may some still remain behind, which is why they put the warning on the label.

Sorry I'm new, what is "/cf"?

@Tomsofty33 it is actually a chemical markup gone wrong. It should be \ce. They are used to write chemical formulae and equations.


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Explain the difference between reaction intermediate and reactive intermediate
Wikipedia definitions of the terms:
Reaction intermediate:
https://en.m.wikipedia.org/wiki/Reaction_intermediate
Reactive intermediate:
https://en.m.wikipedia.org/wiki/Reactive_intermediate
What I don't really understand :
Are they defined just based on their stability?
What I mean is that, if an intermediate is stable is it reaction intermediate and if unstable is it reactive intermediate?
(Or) is there any deeper difference between them?...After reading the entire page I have an intuitive feeling that there is more to them than just stability.

I think that you are correct in that the only difference is one of time-scale and this can mean different things to different people. A nanosecond is a long, long time for someone who works with femtosecond laser pulses to measure intermediates, but a very, very short time to someone who uses, say, stopped flow with microsecond resolution.

A reaction intermediate is nothing more than a plain old intermediate, which you probably understand. There is no stipulation as to how stable the intermediate must be. The IUPAC Gold Book defines intermediate as:

A molecular entity with a lifetime appreciably longer than a molecular vibration (corresponding to a local potential energy minimum of depth greater than $RT$) that is formed (directly or indirectly) from the reactants and reacts further to give (either directly or indirectly) the products of a chemical reaction; also the corresponding chemical species.

I am not aware of a corresponding "official" definition for a reactive intermediate, but in my experience it simply refers to, well, intermediates that display exceptionally high reactivity. Typically these correspond to high-energy species such as carbocations, radicals, carbanions, or carbenes, which exist only transiently, and are difficult or impossible to isolate. A simple example is the carbocation intermediate in an SN1 reaction.
Obviously all intermediates are reactive to one degree or another – which is why they are intermediates and not simply thermodynamic products – but the term reactive intermediate is usually reserved for those which are particularly reactive. So, reactive intermediates are a subset of reaction intermediates.
As for an example of an intermediate which is not particularly reactive, consider the Staudinger reduction of an azide. The initial steps leading to loss of nitrogen gas are extremely fast and (very vigorous) effervescence is observed immediately upon addition of PPh3:

but the iminophosphorane, i.e. RN=PPh3, reacts with water only very slowly to produce triphenylphosphine oxide and the primary amine. Typically you need to heat the reaction mixture for a number of hours, and if you didn't add water it would probably be quite happy on its own. I would therefore consider the iminophosphorane to be an intermediate (or reaction intermediate if you prefer), but I would not refer to it as a reactive intermediate.


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Why don't soluble gases that evolve in a reaction remain in solution?
Take for instance the following process:
$$\text{2NaCl}_{\ (s)} + \text{H}_2\text{SO}_{4 \ (aq)} \to \text{2HCL}_{\ (g)} + \text{Na}_2\text{SO}_{4 \ (aq)} $$
Why does the evolved gas leave the system instead of reprotonating water?

There is no water in the system, to begin with. You normally use concentrated acid for this reaction. With diluted acid, gas would remain in the solution indeed.

Given the reaction:
$$\ce{2NaCl(s) + H2SO4(aq) -> HCl(?) + Na2SO4(aq)}$$
It depends...
If the system is open, like a beaker sitting out on the bench, then HCl tends to escape from solution, esspecially if there is a large excess of $\ce{H2SO4}$.
If the system is closed, then Henry's Law relates the partial pressure of the compound in the gas phase to the concentration of the dissolved gas. The relative amounts of the compound in each phase of course will depend on the relative volumes of the gas and liquid.
$$\ce{HCl(g) <=> HCl(aq)}$$
$$H^{cp} = c/p$$
where:

$H^{cp}$ = Henry's constant
c = concentration of compound in aqueous phase
p = partial pressure of compound in gas phase

There is another consideration for HCl since it readily ionizes in water too. So the HCl which is dissolved in solution is dependent on two different equilibria.
$$\ce{HCl(aq) <=> H^+ + Cl-}$$


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Why +M effect of -NH2 is more than that of -NHR and -NR2?
Why does $\ce{-NH_2}$ shows more +M effect that $\ce{-NHR}$ and $\ce{-NR_2}$? My first conclusion was totally opposite of the reality. I thought, as the alkyl groups are attached to the Nitrogen the lone lone pair on nitrogen should be easily available. But it doesn't seem to be happening. What is the actual cause?

It is because the bulky alkyl groups provide repulsion to the lone pair such that their orientations change. Due to the change in orientation the %s character changes and the lone pair becomes less available. This is not the case with hydrogen.

The main reason for NH2 showing more +M than NHR and NHR more than NR2 is electronegativity
As in NR2 two bulky groups are present so bond angle between the two groups will increase and we know that bond angle is directly proportional to electronegativity due to which the electronegativity of nitrogen atom of NR2 increases due to which NR2 becomes a better group to take electron density rather than donating its electrons which in turn reduces this groups tendency to donate it's lone pair of electron to other groups easily
Same reason can be applied for NHR and NH2
Also because of the above reason the reverse order of -I effect is seen as NR2 most electronegative so more -I .
Hence order of +M NH2 >NHR> NR2
Order of -I effect NH2< NHR< NR2
Thanks .


text:
What is oil smoke point?
I understand that the smoke point is the temperature at which oil starts to produce smoke. But how exactly is that smoke produced? I've read that it is related to hydrolysis and oxidation, but I'd like to know a more precise link.

How "precise" an answer you want? Most oils are mixtures of many different compounds so discussions tend to be general rather than showing a specific chemical mechanism, or a particular overall chemical reaction.

@Max Well, at least some examples of reactions or mentions of a family of compounds.

Typically smoke point has to do with the composition and amount of free fatty acids in the fat/oil. Most fats and oils that you see on the market are mostly composed of triacylglycerols, with only small amounts of free fatty acids. The more free fatty acids you have the lower the smoke point. Similarly unrefined oils have a low smoke point. If you have fats/oils that contain high levels of unsaturated fatty acids, then that product will also have a lower smoke point than say a fat that contains significant amounts of saturated fat. In regards to the chemical mechanism, upon heating the triacylglcyerol structure gets broken down into free fatty acids and glycerol via a hydrolysis reaction. The free fatty acids volatilize and cause the visible smoke. Unsaturated fatty acid are more sensitive to heat and will be hydrolyzed quicker than saturated fatty acids. For more info read this article: https://www.ncbi.nlm.nih.gov/pmc/articles/PMC4424769/

The smoke point also known as
burning point of an oil or fat is the
temperature at which, under specific
and defined conditions, it begins to
produce a continuous bluish smoke
that becomes clearly visible.


text:
Why do we use the word phase and state of matter so interchangeably?
I would be happy if someone explains me why the word "phase" and "state of matter" is used so interchangeably.Lets say we have HBr and another compound in the system both gases.Then am I to take two phases on a single phase when I plug in the Gibbs formula in Phase rule?

We don't.$;!$

Ok In Gibbs Phase rule I have C-P+2.Now for a reaction which has HBr and say a gasesous organic compound giving a single liquid product then why don't I use 3 in place of C during the reaction?Hence HBr and the gaseous compounds are treated as just one phase in the formula.Explain this problem.

Because the third component is determined by the first two via the equilibrium constant. As for HBr and the gaseous compounds, they are treated as one phase because they are one phase.

Aren't they having the same state of matter?And please elaborate on the equilibrium constant.

@IvanNeretin Gibbs rule is not applicable except for equilibrium.Why then do we apply it even before the reaction starts/during the reaction/after the reaction?Secondly u said HBr and gaseous compounds are in same phase will that be the case if one of them would have been liquid?

Well, all different states of matter are different phases but all different phases are not different states of matter.  For Example the solid-solid transition of steel from FCC martensite to BCC Austenite at high temperature.

@A.K Why is it so?

We don't apply Gibbs rule except for equilibrium. As for being in the same phase, no, gases and liquids are not.

From the wikipedia page for phase (matter)

a phase is a region of space (a thermodynamic system), throughout which all physical properties of a material are essentially uniform [...] The term phase is sometimes used as a synonym for state of matter, but there can be several immiscible phases of the same state of matter.
A simple description is that a phase is a region of material that is chemically uniform, physically distinct, and (often) mechanically separable. In a system consisting of ice and water in a glass jar, the ice cubes are one phase, the water is a second phase, and the humid air is a third phase over the ice and water. The glass of the jar is another separate phase.

In the above system you can see that the ice and the glass jar are the same state of matter, but are two distinct phases. It's the same with oil and water. So, people who are being precise don't use the two terms interchangeably.

We don't.  Phase is more specific.  A couple examples of the differences:
1)  A slab of steel is heated in a furnace in preparation for being hot rolled to a strip.  The iron that predominates forms a face-centered cubic phase at the temperature in the furnace,  about 1250°C. By the time the slab has been rolled to a strip and cooled to a temperature, usually somewhere between 500 and 800°C, the steel has switched to the other low-pressure solid phase of iron, which is body-centered cubic.  The phases are different, but the slab/strip never melts in this process, they are both in the solid state of matter.  Often a single substance, like iron or water, forms more than one solid phase.
2)  Water and oil are poured together into a glass.  Only a very limited amount of water can dissolve into the oil and vice versa, so they remain separate phases -- indeed physically separated by Earth's gravity, as the oil phase in its liquid state of matter is less dense than the water phase in its liquid state of matter.  When we have two or more substances the liquid state of matter, despite its freedom of molecular motion, may split into more than two phases.

What about two gases?


text:
How to calculate the concentrations of conjugate species present in the solution of diprotic acid?
Calculate the concentration of $[\ce{SO_3^-^2}]$ present in 0.010M solution of $\ce{H_2SO_3}$ in pure water?
($Ka_1=0.017)$, ($Ka_2=10^{-7.19}$)
I found this approach in reference text for solving the problem :

But I confused with the derivation of Eq.2, so I do my calculation according to Eq.1 to find the concentration of sulfite ion equal
$2.20×10^{-7}$ M.
SO :
I choose another approach to find the concentration of sulfite ion depending on substitution in this formula :
$$Ka_2 = \dfrac{\ce{[H+][SO3^{2-}]}}{\ce{[HSO3^-]}}\ $$
Thus
$$\mathrm{10^{-7.19}} = \frac{(0.00706 + {[SO_3^{-2}]}){[SO_3^{-2}]}}{(0.00706 - {[SO_3^{-2}]})}$$
From which I obtain
$$[\ce{SO_3^{-2}}] = 6.46 \times 10^{-8}$$
Which different from the answer obtained from the first approach.
I appreciate any help.

You messed up calc in eq 2 in multiple ways...
$$K_1 = 0.017$$
$$K_2 = 10^{-7.19} = 6.5\times10^{-8}$$
$$K_1K_2 = 1.11\times10^{-9}$$
Also
$$K_1 = \dfrac{\ce{[H^+][HSO3^-]}}{\ce{[H2SO3]}}\tag{1}$$
$$K_2 = \dfrac{\ce{[H^+][SO3^{2-}]}}{\ce{[HSO3^-]}}\tag{2}$$
$$K_1K_2 = \dfrac{\ce{[H^+][HSO3^-]}}{\ce{[H2SO3]}}\times\dfrac{\ce{[H^+][SO3^{2-}]}}{\ce{[HSO3^-]}} =\dfrac{\ce{[H^+]^2[SO3^{2-}]}}{\ce{[H2SO3]}}\tag{3}$$
rearranging (3)
$$ \ce{[SO3^{2-}]}= \dfrac{K_1K_2\ce{[H2SO3]}}{\ce{[H^+]^2}}= \dfrac{K_1K_2(0.010-0.0076)}{0.0076^2} = 4.6\times10^{-8}\tag{4}$$

MaxC  According to the last solution I think  the solution in the reference text incorrect .thanks  for the support


text:
Stereogenic Center Question
In cholesterol, I don't understand why the two points I circled are stereogenic centers. Isn't the left carbon attached to an OH, H, CH2, and CH2? And isn't the right carbon attached to CH, CH2, CH, and H? And according to my understanding, a stereogenic center must be connected to four different groups. I must be doing something wrong here, but I cannot figure out what. Thanks for your help

In your first example, when you found two CH2 groups you must move to the next carbon to break the tie. In one case you get another CH2 but the other one leads to the carbon of the double bond which is treated as a quaternary carbon C3 is of the S-configuration. Use the same method at the other site. Read about Cahn-Ingold-Prelog rules.

While we see the strereocenters we just don't see the adjacent carbon but the adjacent group  this seeing the adjacent group we can follow that the Marked Carbon is connected by a sp3 and then sp2 carbon from bellow
And the Marked Carbon is connected by a sp3 and then also sp3 carbon from above
Hope you got it !]1

The sp3 arrow should point to the CH2 (C1) to determine C3. It should be compared to the designated sp2 carbon. Follow CIP rules.

Oh yes I did it by mistake


text:
Twisting stereoisomers with rings to determine R/S

I know that if the smallest atomic number group is not in the back, we can twist the molecule so that it is in the back and then determine if it is R/S (Left image). But if the chiral center is attached to a ring, can we still simply twist the molecule (right image)? If so, how? Or is there another way to make the smallest atomic number group go to the back for molecules with rings?

If you mean "rotate" (with the center in the desired chiral atom), then you're right. You can rotate freely.

Rotating structures is a recipe for disaster because errors in transcription will invariable occur. Use the best chirality tool you have---your hands. Leave the structure where it is! Use either hand to make a "thumbs up" gesture. Point your thumb in the direction of the lowest priority group. The remaining fingers should point in the direction of the highest to next to lowest priority group (OH>C2H5 >CH3). If not, you are using the wrong hand. Left hand = S. Right hand = R. Try to rotate structures 1-3 and see how it compares with the "hand method". Structures 1-3 are of the S-configuration. Cyclopentene 4 is Jon's structure with substituents. It is of the R-configuration (=CH>CH2>CH3), a conclusion reached without redrawing the structure.


text:
IUPAC naming of alkyl halide
Why is

4-Bromopent-2-ene

correct but

2-Bromopent-3-ene

not correct by IUPAC nomenclature? I don't see anything wrong in the second case.
Which rule does it violate?
Note that I am naming the same compound by the above mntioned names.

Which rule does it violate? E.g. P-15.1.7.2.3 (from section Types of Nomenclature – Substitutive nomenclature), which might not fit perfectly for our example, but it refers to much more detailed P-14.4 rule.

P-15.1.7.2.3
When several nomenclatural features are present in a structure,
  lowest locants are assigned in accordance With P-14.4. For example,
  in a cyclic unsaturated alcohol having one substituent group,
  lowest locants are assigned in the order:
  (a) characteristic group cited as suffix (-ol);
  (b) unsaturation (‘ene’ ending, for example);
  (c) detachable alphabetized prefxes (a methyl group, for instance).

Bromo substituent is considered detachable prefix, not a characteristic group, so only b) and c) rules do apply.
Therefore,
$\ce{CH3-CH=CH-CH(Br)-CH3}$ is 4-bromopent-2-ene
but
$\ce{CH3-CH=CH-CH(OH)-CH3}$ is pent-3-en-2-ol.


text:
What is the "coefficient of ionization" or "ionization coefficient"?
The question got bumped and I looked at it. The OP seems to use the two terms coefficient of ionization and ionization coefficient interchangeably and assigns the variable $\alpha$. The problem is in part asking about the pH of a 0.1 molar ammonia solution.
Am I correct to assume that the ionization coefficient is the fraction of the ammonia that has ionized?
My google foo is off today. I couldn't find a formula anywhere where the context made sense. The problem in question has nothing to do with a plasma.

Am I correct to assume that the ionization coefficient is the fraction of the ammonia that has ionized?  Yes.

It's degree not coefficient.

@Mithoron _ Thanks! I now suspect that the OP in the original problem may have translated it from another language and used an uncommon term.


text:
Calculating the free energy of separation in AMBER
I am currently busy with my master's research project and I'm trying to calculate the free energy of separation of two cellobiose molecules in ionic liquid, that is, calculate the free energy of separating two molecules from a distance of 35 nm to 215 nm. I have tried steered molecular dynamics with the Jarzynski relation, taking the logarithm of the radial distribution function and various umbrella sampling methods, but none of them seem to give me good FE profiles.
I think my system (1-butyl-3-methylimidaozlium acetate) just isn't going to give good FE profiles, but my supervisor is convinced that it will. I am currently trying thermodynamic cycles out so I hope that works.
So, my question is, what other methods for calculating the FE of separating the cellobiose molecules are there? And can you provide tutorials for me for those? I am aware of slow growth, ABMD, NFE, metadynamics, etc. but either these do not have any useful tutorials or don't work in the AMBER force field.
I have been struggling with this for more than two months now and my supervisor does not want me to stop my analysis. What other method can I use? Is there other computational methods available for determining the interaction between two systems?
Thank you in advance.

What does it mean to give 'good FE  profiles'? What are you actually trying to investigate? What do you need that free energy profile for? Your system seems small enough to not rely on force fields, maybe something more sophisticated would work better, e.g. DFT with a solvent model. Are you starting from the contact ion-pair? What structures are you using?

I'm trying to simulate the dissolution of cellulose in ionic liquid. So, what my supervisor asked me to do is determine the free energy of separating two cellobiose molecules from each other in ionic liquid as to model the spontaneity of two cellulose strands dissolving and separating in my solvent.

My systems are cellobiose and 800 1-butyl-3-methylimidszolium acetate molecules. The whole project is based on using force fields, so I cannot use DFT, QM, etc.

Please edit your question to make it clearer what you are doing; your recent comments suggest something a lot different than what your question asks.


text:
Reduction of hydrogen sulfide by silver
Why can $\ce{H2S}$ be reduced by $\ce{Ag}$ when left in air:
$$\ce{2 Ag + H2S -> Ag2S + H2}$$
If silver is a noble metal that should not reduce hydrogen?

Hydrogen sulfide won't tarnish gold or platinum but it will tarnish silver.

I believe silver is oxidized by oxygen to form $\ce{Ag2S}$, not displacing hydrogen from $\ce{H2S}$.
$$\ce{2Ag + 1/2 O2 + H2S -> Ag2S + H2O}$$

Well, if silver can form hydrogen during   dissolving in a cyanide solution ( gold needs oxygen, silver does not ), it could as well form hydrogen from $\ce{H2S}$. Just thought. $$\ce{2 Ag + 4 CN-  + 2 H2O -> H2 + 2 [Ag(CN)2]- + 2 OH-}$$

Check the electronegativity of hydrogen, silver and noble metals. In addition, though, consider that the product, $\ce{Ag2S}$, is a solid. The reaction is driven in one direction by precipitation, continually removing sulfur from the air.

If you consider the (probably incorrect) net reaction given by the OP, the simplest driving force argument is shaky because a solid and a gas reacts to a solid and a gas, see https://chemistry.stackexchange.com/a/175826


text:
Specifically how dangerous is a 10g chemical sample of pure arsenic and how do you properly dispose of it?
I'm not a chemist.  I'm an electrical engineering student and I recently purchased an element collection.  The collection came with a 10g arsenic sample.  I assumed it wasn't much of a thing because there were no warnings or anything but afterward when I started reading more I became concerned.  How dangerous is the really, what precautions must I take in handling the sample, and most importantly how do I properly dispose of this sample?
Thank you

I would suggest rewording this from "how dangerous is" to "what hazards can be anticipated with" to reduce the risk of a close vote.

This may assist you  http://www.sciencelab.com/msds.php?msdsId=9922970   short version - don't swallow it or inhale the dust

Nothing under the Waste Disposal section there :-( .

Not sure what you have, but most of the elements would need to be sealed (in glass for example) to remain "bright and shiny." A lot of the elements will react with oxygen, water and/or carbon dioxide from the air to at least tarnish if not completely.

A sealed bottle of arsenic isn't harmful; ingesting it would be. To get elemental arsenic into your body you would need to create vapour or dust. So don't handle it without gloves, don't heat it and don't eat it. So a sealed bottle in an element collection is low risk. Especially compared to arsenic pigments (once used in wallpaper) which released vapours that really were toxic to random bystanders.

What you have is a sample of arsenic in its 'metallic' form (it is actually a metalloid). This makes it not as dangerous as you might be imagining. The reason is that As(0) (what you have) is not absorbed easily by the body. What is actually extremely dangerous are soluble forms of arsenic eg. arsenate salts etc. Those are readily absorbed by the body and cause all sort of problems. Arsine ($\ce{ArH3}$) is pretty nasty for example since being a gas is readily absorbed in the lungs.
The above applies for many other compounds, a characteristic example being mercury. We have all heard about how toxic mercury is. However, what most people are not  aware is that it is the soluble mercury salts that is the issue here (eg. $\ce{HgCl2}$) and not as much metallic mercury. Actually metallic mercury is very hard to absorb by ingestion or through the skin so if ingested one will most likely survive (according to studies on animals). However, if one ingests the same amount in a soluble salt form (or any other absorbable form, eg $\ce{HgMe2}$)  he is in serious trouble (see here: Karen Wetterhahn). The main issue with metallic Hg is its vapors which can again be absorbed so it is not safe to have it in a non confined from (eg, mercury spillage).
Same applies for tin! The metal is (obviously) safe, as well as the salts which are not absorbed by the body. Organotin compounds on the other hand are readily absorbed and extremely dangerous.
Lastly, having been a mineral collector myself many years ago, i used to have a sample too (which i handled with bare hands sometimes something i dont recommend doing) and never had a health issue 20 years now.
In conclusion, i dont think you need to dispose your sample. Keep it in its box and avoid handling it just to be on the safe side as Waylander mentions in his comment. Definitely dont inhale any dust from it although that would be unlikely to happen unless you grind your sample.

Treat this with very great care; don't let it come into contact with you or anyone else as it is readily absorbed through the skin, lungs and gastrointestinal tract.
Arsenic is the classic poison known also to the ancients. Ten grammes is a huge quantity as it is deadly in quantities of approximately  0.1 g in the blood stream. The oxide and compounds such as the  arsenate, are far more poisonous than the metal as they replace phosphate in the body (which is harder for the pure metal to do) so someone can be poisoned directly by taking a large dose or by gradual exposure over a period of time.
For a fascinating read on this and other poisons look up 'A is for Arsenic: The Poisons of Agatha Christie' by K. Harkup, publ. Bloomsbury. '

As pointed out by ANM, it's elemental As not oxide.

@Mithoron Yes he does but there should be oxide on its surface or in the container. It is true that it is possible to 'eat' metallic arsenic as it acts as an antibacterial but eventually poor health results. In some  societies, in the past, this has been common. The corpses of such people also do not decompose rapidly for the same reason.


text:
Do individual crystals resulting from evaporating a mixture of two ionic compounds contain both cations/anions within the crystal lattice?
I have wondered this for awhile. Let's say I dissolve, without loss of generality, potassium chloride and sodium chloride in water, then evaporate the water. Would any of the resulting crystals consist of randomly dispersed potassium and sodium cations within the lattice or would each resulting crystal be purely potassium chloride/sodium chloride?

Crystals with randomly intermixed ions only occur when the ions have very similar diameter (really the same spatial charge distribution in its exterior, but alkali ions are in very good approximation spherical) and same charge. That would be called a single phase solid solution.
However, $\ce{Na+}$ and $\ce{K+}$ have quite different diameters ($r=102$ resp. $\pu{138 pm}$). Your mixed solution will crystallise as NaCl with a low amount of potassium ions and KCl with some potassium replaced by sodium ions. In other words, your isotropic solution decomposes into a solid two-phase mixture.
The two substances will surely have a different tendency (also temperature dependent) to include such defect ions, and the two compounds have different solubility, so one will precipitate first, so the concentrations in the residual solution changes greatly during the process, and consequently also the defect concentration in the single crystals.

To add to Karl's point, there is a process called zone refining. Think of a compound packed in a glass tube. You melt a small zone of the tube going from one end to the other. You can repeat the process. A purer compound results at one end and the contaminates are concentrated at the other end.

I did a lot of x-ray fluorescence. Every sample of NaCl which I had also had detectable amounts of $\ce{K^+}$ as well. It is really really hard (impossible?) to get any salt pure enough that XRF doesn't detect other elements.
Look at the Wikipedia article for Sea salt. It is far from being pure sodium chloride.

Interesting. However, sea salt is not the best example here, evaporation of all of the solvent can be barely called "crystallization".


text:
Disinfection solution ( Trichloroisocyanuric acid 96%)
I'm going to make a 100 liter solution to desinfect various tools used in Shiitake production so it's important the solution kills mold and mold spores.
I've been looking into various methods and chemicals. At first I thought about going with H2O2 since it's safe and does not generate harmful by-products. Since I do not have peroxide stronger than 3% readly available I discarded this route.
I thought about a solution of calcium hypochlorite taking the PH below 7 but since that would probably be quite dangerous since I'm not sure about the concentrations / ph levels used to generate minimum ammounts of chlorine gas. I surely don't want to be dealing with the release of chlorine gas if I can avoid it.
I looked into chlorine dioxide and althought extremely effective it's also very toxic and even has explosive properties. I discaded it immediately.
Lastly I saw that my supermarket carries Trichloroisocyanuric acid (pool shock) at 96%.
I was reading some info here and it's dangerous to mix with small ammounts of water and also I need to be careful about the PH.
These are the recommended concentrations for each use:

swimming pool disinfection: calculate the usage according to 2-3g/m3; take effect for 30-60 minutes.
Public places and ground disinfection: use 1500-3000-fold diluted water solution for spraying and wiping for 30 minutes. Infectious
  disease epidemic area, use 150-200-fold aqueous diluted water solution
  for spray disinfection.
disinfection of white clothing and blankets: apply 2000-3000 fold diluted solution for clean items; soak for 10 minutes; for seriously
  polluted items, diluted 500 times, soak for 20 minutes.
disinfection of bathroom and toilet scrub; use 300-fold diluted solution for reaction of 30 minutes.
sterilization of non-metallic medical equipment: use 500-1000 times diluted solution for soak of 30 minutes; disinfection of apparatus
  with pus and blood; apply 100-fold diluted water solution for reaction
  of 30 minutes.
disinfection of non-metallic tableware: use 500-1000 times diluted water solution for reaction of 30 minutes.

My questios are:
1. How many grams should I had to 100 liters of water for sterilization / desinfection of equipement? (I'm looking into a strong solution but also safe to use)
[EDIT] I found this patent  that states

with a sterilizing solution of a compound having an active-chlorine
  concentration of from 500 to 20,000 mg/l and being selected from the
  group consisting of sodium hypochlorite, calcium hypochlorite,
  chlorinated trisodium phosphate, chlorine dioxide, sodium
  p-toluenesulfochloroamide, p-toluene-sulfonsulfochloroamide,
  N-chlorosuccinimide, 1, 3-dichloro-5, 5-dimethylhydantoin,
  trichlororoisocyanuric acid, salts of trichloroisocyanuric acid,
  trichloromelamine and dichloroglycoluril, the pH of said sterilizing
  solution being from 8 to 10, and treating the thus sterilized article
  with an aqueous solution of hydrogen peroxide at a temperature below
  90° C to deactivate any remaining active chlorine, said solution of
  hydrogen peroxide having a concentration of lower than 30%.

Also fromhere

using a solution of concentration of 200-400 mg/kg for environmental
  disinfection and utensils disinfection.

Since my solution will be made with water I would assume mg/kg = mg/l
So for a 100 liter solution I would need to dissolve 20-40 grams, correct me if I'm wrong.
2. What would be the recommended PH value for maximum effectiveness but also safe to use?
On a side note the desinfection will be carried at a open area using adequate PPE.
[edit: I would also need to know an approximate time the equipement needs to be submerged and how long the solution would be effective]

Not a full answer, but I suspect that once the solution is made up, it shouldn't be stored for too long (or at least, it will become useless quite quickly)

Thanks for your input! I just need to know the time, so I can have an idea of how long it is effective. The solution will be made-used-discarded so not really worried about storage.


text:
Do you assume undrawn hydrogens are pointing away from you when determining R/S?
When determining R/S of stereoisomers, the group with the smallest atomic number needs to be in the back (dashed line). But if no group attached to a chiral center is drawn in dashed line, do we assume that the undrawn hydogren is dashed? I have attached an example below, the chiral center is circled in red. Thanks

We don't assume anything. The configuration of this center is unknown.

.. btw, it's not true that "the group with the smallest atomic number needs to be in the back (dashed line)".

so how would the R/S be determined in this molecule?

Jon: oh, there might be some confusion, I meant only for the structure representation. When determining the stereochemistry.. feel free to use your 3D spatial imagination and rotate your molecule in your head, you don't have to redraw the formula. But you cannot determine stereochemistry for chiral centers with unknown stereochemistry.

When no bond from the chiral center has up or down graphical indication, it can mean any or unknown stereochemical configuration. Which might be confusing, because it can also mean, that the stereochemical configuration on this atom was mistakenly omitted.
To explicitly indicate that any or unknown configuration, wavy line bond can be used:

(Hydrogen atom with the dashed bond on the (other) chiral center can be can be omitted at all, the stereochemical meaning is unambiguous, much more if the bond wedge was filled with black.)

Better practice is to indicate a specific stereochemistry explicitly, if you know or specifically presume it.  In the molecule presented you could, of course, do this by drawing the methyl group with a wedge pointing into the chiral center or, for the opposite configuration, a hashed line.
In some situations where specific stereochemistry is unimportant, you can get away with generic lines.  You could render the reaction of 2-butanol with acetic acid, to form an ester, quite well without fussing over the stereochremistry of the chiral center in the alcohol.  But in a case where you have multiple chiral centers and their relative stereochemistries are important (like d-l versus meso tartaric acid), or if you are tracking the configuration around a single chiral center (like $S_N2$ substitution on 2-butanol), putting the stereochemistry explicitly into the structure is best whenever it it possible.