Potential energy of electron density at the nucleus Consider the hydrogen atom for simplicity.

The electronic density at the nucleus is not null.

The attractive potential between a small volume of electronic density $\mathrm dV$ (at position $\vec r$) and the nucleus (at position $\vec R$) is:

$$\frac{-Z\rho(\vec r)}{|\vec R - \vec r|}\mathrm dV$$

How is the energy not going to (negative) infinity near the nucleus then? How is this accounted for to give the traditional $\pu{-0.5 Ha}$ energy value of the hydrogen atom?

I think it's useful to expand slightly on the comments because I think the language here should be as clear as possible.

You're concerned about the convergence of the integral of a function when some piece of the function is unbounded, i.e., for $f(x)$ on domain $[a, b]$, $\exists c \in [a,b]$ such that $f(c) = \infty$. Frequently, integrals of this type do not converge.

Your function certainly looks like that but the resolution of this problem is actually hidden in the $dV$. Typically, we treat the integration of the wave function as a separable problem, i.e, we integrate the radial and spherical components separately. When you look at the integral over the distance from the nucleus, you get a factor $4\pi r^{2}$ because you are integrating shells and $dV = 4\pi r^{2} dr$. So even though $\frac{1}{r}$ diverges at zero, $\frac{r^{2}}{r}$, while not defined at zero, does converge.

$$\lim_{r\rightarrow 0}\frac{r^{2}}{r} = 0$$

From this perspective, no piece of the integral is infinite, and the integral converges.


How are pure substances (elements, compounds, etc) produced? How are large quantities of the substances we use in the lab generally produced? (As in the process from getting the stuff out of the ground to use in the lab) How 'pure' are they? Also: is it possible to create relatively pure substances with household goods?

No lab reagent is 100% pure. But you can get really close to that.

How 'pure' are they?

Read the label or do the analysis. It varies A LOT from one reagent to another.

Is it possible to create relatively pure substances with household goods?

Depends on the substance, depends how pure you want it. For instance, you can very easily distill water at home, but pure water is hard to get.

How are large quantities of the substances we use in the lab generally produced? (As in the process from getting the stuff out of the ground to use in the lab)

You'll need to be more specific. There are entire classes you can take at the uni on the large scale production and purification methods of this and that type of reagents. I see you used the ionic-compounds tag, but that's far from specific. A more answerable question would be "How are compounds X and Y produced and purified?"


Rate of change of reaction rate of 0th, 1st, 2nd order reaction? Does the rate of change of the reaction rate for a 1st order reaction change linearly over time? My teacher says it does, but my textbook seems to contradict this. Could anyone clarify?

The logarithm of the concentration of one of the components will vary linearly. However, the concentration does not vary linearly with time, only its logarithm does.

Reaction Rate $= -\frac{d[A]}{dt} = kA = $ constant * [concentration] for a first order reaction.

If you integrate this using separation of variables you end up with

$\ln \frac{[A]_t}{[A]_0} = -kt$

if you plot $\ln[A]_t$ verse time the slope will be constant.

If you plot concentration as a function of time it will not be constant because we can rearrange the above equation to get

$[A]_t = [A]_0 \exp(-kt)$

which is clearly exponential and not linear.

So yes, for a first order reaction, the reaction rate is constant because it depends linearly on the reaction rate constant which is... constant because it is the slope of the logarithm of concentration verse time. Here is a useful link:

www.chem.libretexts.org/reaction_rates

Here is a short answer to a similar question that is useful: chemistry.stackexchange.com


Why are a calico strainer and chloroform used in a British Pharmacopoeia assay of milk of magnesia? The following is an assay for milk of magnesia in the British Pharmacopoeia (emphasis added):

  Dissolve the sodium hydroxide in $\pu{150 cm^3}$ of purified water, add the light magnesium oxide, mix to form a smooth cream and then add sufficient purified water to produce $\pu{2500 cm^3}$. Pour this suspension in a thin stream into a solution of the magnesium sulfate in $\pu{2500 cm^3}$ of purified water, stirring continuously during the mixing. Allow the precipitate to subside, remove the clear liquid, transfer the residue to a calico strainer, allow to drain and wash the precipitate with purified water until the washings give only a slight reaction for sulfate. Mix the washed precipitate with purified water, dissolve the chloroform in the mixture.

What is a calico strainer, and why is it used here? Is it better than using, say, Büchner filtration?

As well, why is chloroform used (see emphasized text, above)?

Ok I understand where you got chloroform from.

This is not the assay of milk of magnesia, but the extemporaneous preparation.

In extemporaneous preparations chloroform is used as:

a general anaesthetic
a carminative
flavouraing agent
preservative

So in the case of mik of magnesia it has two main roles that is a carminative and a preservative.

Hope this helps


Prove that idempotency is a necessary and sufficient condition for a densiy matrix to be N-representable As I understand, a necessary and sufficient condition for a density matrix $P$ to be represented by a wavefunction $|\Psi\rangle$ is that it is idempotent, i.e. $P^2=P$. It is easy to see that if $P=|\Psi\rangle\langle \Psi|$ then $P$ is idempotent:
$$
P^2=|\Psi\rangle\langle \Psi|\Psi\rangle\langle \Psi|=|\Psi\rangle\langle \Psi|
$$
since by normalization $\langle \Psi|\Psi\rangle=1$. However, I have had trouble with the other direction - how can I show that if $P^2=P$ then $P=|\Psi\rangle\langle\Psi|$ for some $\Psi$?

I have been able to demonstrate that if $P^2=P$ then
$$
P=\sum_{j} |\Psi_j\rangle\langle\Psi_j|
$$
where the sum over $j$ has no more terms than the dimension of the density matrix $P$. My argument was analogous to the one found here. I suspect that I'm almost there - I think it's just a change of basis that I haven't seen yet. How can I complete this proof?

A density matrix $D$ is positive semidefinite, hermitian, and has trace one.
Because of hermiticity we may assume that it is diagonal.
Let's denote the eigenvalues with $\lambda_i$ .
Because it is positive semidefinite, we have $0 \leq \lambda_i \leq 1$.

The matrix $D^2$ has eigenvalues $\lambda_i^2$. Since it is idempotent ($D = D^2$) and has trace one, we may write:
$$\sum_i \lambda_i = \sum_i \lambda_i^2 = 1$$
The only possibility for this to be true is, if there is one and only one $j$
for which $\lambda_j = 1$.

So $D$ has one $1$ in the diagonal at the $j$-th column and all other values in the matrix are zero.
You can easily represent this matrix with:
$ e_j e_j^T$. ($e_j$ being the $j$-th unit vector.)

If you transform back to you original nondiagonal form, you obtain your $\Psi$.


Effect of optical activity I know that optical activity means polarisation of light but what is so special in that that it changes the properties of the substances significantly

TL;DR version: Chirality is a basic property of a molecule, while optical activity is a bulk physical property possessed by samples of chiral compounds. Specific rotation is an intrinsic property of chiral compounds and a measure of optical activity under standardized conditions. Optical rotation is not responsible for the disparate behavior of enantiomers of biologically active compounds.

Chirality is an intrinsic property of a molecule (or, indeed, of certain geometric structures, more generally). A chiral molecule is one that is not superposable on its mirror image molecule (its enantiomer), meaning that no series of rotations or translations will make the two geometrically identical in three-dimensional space. One of the consequences of chirality is that chiral molecules in bulk will measurably rotate plane-polarized light by a specific amount and in a specific direction. (The distinction here is crucial, since any molecule[s] will in general cause some rotation, but the random distribution and spatial orientation of the large number of molecules in any given sample will produce no observable net rotation if the molecules are achiral. The rotation caused by one molecule is likely to be negated by another identical molecule in an opposite spatial orientation.) The key point is that chirality is an intrinsic property of a molecule, while optical activity is a bulk physical property of said molecules gathered en masse. Specific rotation, on the other hand, can be viewed as an intrinsic property, since it is defined for a specific sample under an exact set of conditions (light wavelength, path length, concentration [for samples in solution], density [for pure liquid samples], temperature).

Optical activity is not the cause of any changes in the properties of chiral compounds, nor is it responsible for the sometimes wildly different biological effects of certain enantiomers (which I presume may be what you're actually referring to, forgive me if I'm wrong). The enantiomers of chiral molecules can often differ widely in their biological effects due to the basic nature of proteins, which are the constituent biological molecules that most if not all biologically active substances target in one way or another. Proteins are constructed from amino acids, which often themselves contain multiple chiral centers, and these proteins go on to fold into very specific and complex conformations. Enzyme active sites and cell receptors (among various other biological structures) are mostly comprised of proteins with very specific three-dimensional structures. As such, they interact significantly only with molecules that have the appropriate shape. Hence, enantiomers of chiral substances, which differ in their three-dimensional shape, will exhibit variable degrees and types of interactions with proteins, and can hence impact biological targets in disparate ways. This is largely what makes chirality so important to chemists, and partly explains why so much of the most respected (not to mention economically important) cutting-edge work recently (at least in organic chemistry) has revolved around enantioselective or enantiospecific synthesis techniques.


Is there an open Chemical database that includes GHS/CLP classifications? I'm looking for a Chemical database that

is up to date (via a firm, experts or crowdsourcing)
is open and free to use (best via an API)
contains the CLP/GHS classifications for chemicals

I have not found any database that has these three points (e.g. ChemSpider is actually not really open and does not contain CLP/GHS classifications)

I was really hoping any of you might know of one?

Since you're a Dutch neighbour, the European Chemicals Agency might be an option.


Why are electrons transferred in an ionic bond? I've been taught at school that in an ionic compound like $\ce{NaCl}$, electrons are transferred from the $\ce{Na}$ to the $\ce{Cl}$ to form $\ce{Na+}$ and $\ce{Cl-}$ which both have full outer shells. But what causes these electrons to be transferred?

The modern definition of electronegativity is due to Linus Pauling. It is the power of an atom in a molecule to attract electrons to itself.

When the electronegativity difference between the two atoms  is greater than 2.0, the bond is considered ionic; i.e. there is complete transfer (between the two atoms) of the electrons in the bond.
The electronegativity (according to Pauling scale) of chlorine is 3.0, while the electronegativity  of sodium is 0.9.


Why does the light catalyzed radical addition of tetrachloromethane to an alkene produce only a single regioisomer? This reaction always occurs in the following, regiospecific way:

$$\ce{CCl4 + R-CH=CH2 ->[$h\nu$]  R-CHCl-CH2-CCl3}$$

The same should apparently also happen when there is a carbon that has two substituents (e.g. methyl groups) and yield only one major product:

$$\ce{CCl4 + (CH3)2C=CH2 ->[$h\nu$]  (CH3)2CCl-CH2-CCl3}$$

I believe the initiation step goes like this:

$$\ce{CCl4 ->[$h\nu$] CCl3^. + Cl^.}$$

Can't both of these radicals, that form in the initiation step, react with the double bond and form a secondary or tertiary radical carbon? Why does only $\ce{CCl3^.}$ apparently attack the double bond?

While this reaction is highly regioselective, it is not 100% regioselective. Consider the nearly identical reactions:

$$\begin{align}
\ce{CCl4 + H2C=CH-(CH2)5CH3 ->[(PhCO2)2][90\text{-}105ºC] Cl3C-CH2-CHCl-(CH2)5CH3} \hspace{.63cm} \text{75%}^{[1]}\\
 \ce{CCl4 + H2C=CH-COH-(CH3)2 ->[(PhCO2)2][80ºC] Cl3C-CH2-CHCl-CH(OH)-(CH2)5CH3} \hspace{.63cm} \text{70%}^{[2]}\\
\end{align}
$$

Granted these reactions use radical initiators instead of light, but that wouldn't make the difference between 100% and 70-75% regioselectivity. The reality is that at the start of the reaction, $\ce{.CCl3}$ and $\ce{.Cl}$ are both competing to attack the alkene, and $\ce{.Cl}$ does in fact attack faster. Looking at the activation energies for the abstraction of hydrogen atoms by different radicals, we gain insight into their relative stability:

$$ \small
\begin{array}{lcc}
\hline
\text{Radical} & \ce{CH3-H} & \ce{CH3CH2-H} & \ce{(CH3)2CH-H} & \ce{(CH3)C-H}^{[3]} \\
\hline
\ce{.F} & \text{1-1.5} & \text{<1} & & \text{<1} \\
\ce{.Cl} & \text{3.4} & \text{1.1} \\
\ce{.Br} & \text{17.5} & \text{13.0} & \text{9.5} & \text{6.9} \\
\ce{.CH3} & \text{14.0} & \text{11.6} & \text{9.6} & \text{8.1} \\
\ce{.CF3} & \text{10.9} & \text{8.0} & \text{6.5} & \text{4.9} \\
\ce{.CCl3} & \text{17.9} & \text{14.2} & \text{10.6} & \text{7.7*} \\
\hline
_{^* \text{units are } \mathrm{kcal\ mol^{-1}}}
\end{array}
$$

As this reaction continues, however, chlorine atoms are abstracted from $\ce{CCl4}$, creating the product along with $\ce{.CCl3}$ radicals. Abstraction of $\ce{.CCl3}$ is highly disfavored due to steric effects. Thus, the radical chain reaction is as follows:

$$\ce{CCl4 ->[$h\nu$] .CCl3 + Cl.}$$

$$\ce{.CCl3 + CH2=CH-R ->  Cl3C-CH2-\stackrel{\displaystyle .}{\ce{C}}H-R}$$

$$\ce{Cl3C-CH2-\stackrel{\displaystyle .}{\ce{C}}H-R + CCl4 ->  Cl3C-CH2-CHCl-R + .CCl3}$$

The less than ideal yields of the radical addition of $\ce{CCl4}$ can be attributed to the strength of the $\ce{C-Cl}$ bond (higher activation energies correspond to slower rates of reactions). Because $\ce{CCl4}$ is in competition with the polymerization of the $\ce{.CCl3}$-alkene adduct, $\ce{BrCCl3}$ and $\ce{CBr4}$ are generally preferred for their higher selectivity, owed to the lower strength of the $\ce{C-Br}$ bond. One such example is shown below.

$$\ce{BrCCl3 + H2C=C(C2H5)2 ->[$h \nu$] Cl3C-CH2-CBr(C2H5)2} \hspace{1cm} {\text{91%}^{[4]}}$$

$^{[1]}$M. S. Kharasch, E. W. Jensen, and W. H. Urry, J. Am. Chem. Soc., 69, 1100 (1947).
$^{[2]}$P. D. Klemmensen, H. Kolind-Andersen, H. B. Madsen, and A. Svendsen, J. Org. Chem., 44, 416 (1979).
$^{[3]}$F. A. Carey, R. J. Sundberg, Advanced Organic Chemistry Part A: Structure and Mechanisms, Springer Science, New York, 2007, pg. 1034.
$^{[4]}$M. S. Kharasch and M. Sage, J. Org. Chem., 14, 537 (1949).


Determining the boiling point of a substance given the triple point and critical point If I have the pressure and temperature of a substance in its triple point, as well as its pressure and temperature in the critical point, how can I determine the boiling point of the substance at 1 atm of pressure?

What you need is the Clausius–Clapeyron equation, which is used to relate vapor pressure and temperature (or external pressure and boiling point).
$$\ln{\left(\dfrac{P_1}{P_2}\right)}=\dfrac{\Delta H_{\text{vap}}}{R}\left(\dfrac{1}{T_2}-\dfrac{1}{T_1}\right)$$
Step 1: Find the enthalpy of vaporizations $\Delta H_{\text{vap}}$ by plugging in the two sets of temperatures and pressures that you have.
Step 2: Use your $\Delta H_{\text{vap}}$ and the temperature and pressure of the triple point to determine the boiling temperature at 1 atm.


Cleaning an overhead water tank using Alum or other chemicals? Alum is generally used in water tanks to settle down solid impurities. Since it contains aluminium I suspect it to be toxic to humans. So are there any safe chemicals to clean large overhead tanks without removing the water or  doing much effort. I am asking about solid impurities and not disinfection .

Aluminum is remarkable safe for human consumption, also considering that it is present in high concentration essentially everywhere, and it is not known to have any significant specific biological roles. Wikipedia happens to have a nice well researcher section on aluminum's possible effects on human and animal health: Aluminium: health concerns. Alum is also routinely added to water in low concentrations for elimination of solid particulate, as you mentioned.

My main concern for your application is the fact that typical methods of water purification from solid particulate, just like addition of alum, will tend to form flocs in your water tank that will then need to be removed! Depending whether this scum will float on top of the water, or sink to the bottom of the tank, the removal may be more or less easy, also depending on what kind of access you have to this tank. I doubt if any of those solutions would count as not "doing much effort"...


Searching for a solvent with a low carbon dioxide solubility I would like to use a liquid solvent for some experimentation with a carbon dioxide solubility as low as possible under usual pressure and temperature conditions (say, $\pu{1 atm}$ and $\pu{293 K}$).

Until then, I was unable to find much information on the topic:

A study of $\ce{CO2}$ solubility in water and seawater by Weiss [1], indicating that $\ce{CO2}$ is less soluble in saline water than in pure water.
Publications concerning several organic solvent (acetone, cyclohexane, ethanol, methanol, etc.), a sum-up in IUPAC SDS Vol 50 [2]. To the naked eye, it seems that all of them can dissolve much more $\ce{CO2}$ than water, being poor candidates to answer my needs.
Publications concerning olive, linseed or other origin oils (see Simon and Gutknecht [3] for instance). However such fats seems to also have a higher solubility coefficient than water.

I do not know if such liquid does exist, but it would be great. Up to now, my best candidate is salty water.

References

Weiss, R. F. Carbon Dioxide in Water and Seawater: The Solubility of a Non-Ideal Gas. Marine Chemistry 1974, 2 (3), 203–215. https://doi.org/10/cb6n2j.
Carbon Dioxide in Non-Aqueous Solvents at Pressures Less than 200 kPa; Fogg, P. G., Clever, H. L., Eds.; Solubility data series; Pergamon Press: Oxford, 1992. ISBN 978-0-08-040495-0. PDF
Simon, S. A.; Gutknecht, J. Solubility of Carbon Dioxide in Lipid Bilayer Membranes and Organic Solvents. Biochimica et Biophysica Acta (BBA) - Biomembranes 1980, 596 (3), 352–358. https://doi.org/10/bj647m.

Update

I understand the "like dissolves the like"[1] rule of thumb so that a highly polar solvent would be the worst for a non-polar molecule such as $\ce{CO2}$. Water as a solvent would then be the best choice (possibly salted to further reduce its $\ce{CO2}$ solubility) for my needs.

But as Rolf pointed out, the dissociation of $\ce{CO2}$ into carbonate ions also happens and might increase its solubility. For such reasons an aprotic solvent could be useful.

However, when comparing between water (molar fraction 0.00061) which is a strongly polar (dielectric constant: 80 and dipole moment 1.85$~$D) but protic solvent and dimethyl sulfoxide (molar fraction 0.0091[2]) which is also polar (dielectric constant: 47 and dipole moment 3.96$~$D) but aprotic solvent, dimethyl sulfoxide as a higher solubility for carbon dioxide than water.

What can be the reasons for such unexpected outcome? Are my reasonings and assumptions correct or am I flawed in my thinkings?

References

Polar Protic? Polar Aprotic? Nonpolar? All About Solvents. - Master Organic Chemistry. https://www.masterorganicchemistry.com/2012/04/27/polar-protic-polar-aprotic-nonpolar-all-about-solvents/
Li Hua & Chen Wanren, 2005, Solubility of dilute SO2 and CO2 in dimethyl sulfoxide, Physics and Chemistry of Liquids, 43:3, 289-298, DOI: 10.1080/00319100500083583

To find a solvent with low solubility for non polar gases like CO2 you have to compare the polarity of the solvents. A very polar solvent like water is the best choice.
There's a related question here: Dissolving of non polar gases in water (liquid)

At the end of this answer I have added two links. The first one gives you the solubility of CO2 in water. According to this source, at room temperature and around atmospheric pressure of Co2, the molar fraction in water is 0.00061.
The second link compares many other solvents. At room temperature and 1atm of CO2 the molar fraction is 0.012 in hexane that is the lowest for straight chain alkenes(Fig.1). Overall, among alkanes and cyclic alkanes, cyclohexane is the only one that at room temperature has a CO2 solubility below 0.01(fig.2).
Among aromatic hydrocarbons, you can reach a molar fraction as low as 0.006 if you use 1-Methylnaphthalene(fig.3) and this value is interestingly similar to what you can reach using methanol(you can find a graph comparing alcohols there as well(pag.129). Apparently the CO2 solubility with aromatic compounds is comparable to alcohols(polar).

With all this in mind, water is the best option and if you add salts the solubility of CO2 decreases further as you mentioned.

Solubility of Carbon Dioxide in Water:
https://sites.chem.colostate.edu/diverdi/all_courses/CRC%20reference%20data/solubility%20of%20carbon%20dioxide%20in%20water.pdf

CARBON DIOXIDE IN NON-AQUEOUS SOLVENTS:
https://srdata.nist.gov/solubility/IUPAC/SDS-50/SDS-50.pdf


Why does the dipole moment of molecules go from positive to negative? In physics the dipole moment is said to go from the negative to the positive pole, check for instance https://en.wikipedia.org/wiki/Electric_dipole_moment (Elementary definition).

This makes sense to me since in an electric field (which is directed from positive to negative) the dipole moment aligns with the electric field, such that in the aforementioned definition the positive pole points towards the negative pole of the electric field and vice versa.
(also see: https://physics.stackexchange.com/questions/89157/motion-of-a-dipole-in-an-electric-field)

Now, in all representations of molecular dipole moments I could find online the dipole moment is directed from the positive to the negative pole.

Check for instance: https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Physical_Properties_of_Matter/Atomic_and_Molecular_Properties/Dipole_Moments

or just have a look at google images for the search term 'dipole moment molecule'.

This doesn't make any sense to me, is there any good reason for this discrepancy?

You have assessed the problem correctly. Physicists and chemists use just the opposite directions. Keep in mind that both are arbitrary choices, however the physicist's convention is consistent with electrostatics. All you have to do is to be aware of this direction in the chemistry literature. I have rarely seen any student ever thinking about it which is rather sad.

If you want more details read this article Misconceptions in Sign Conventions: Flipping the Electric Dipole Moment

  "Most chemists draw the direction of an electric dipole backwards. This inconsistency in sign convention is more than a trivial oversight or matter of convenience. We are introducing an incorrect convention that is central to our student’s understanding of fundamental thermodynamics; which way is up? As a student matures from general chemistry to organic through analytical, physical, inorganic, and biochemistry they are reintroduced to similar, powerful concepts that hold our discipline together. However, this reconnection of chemical principles comes with each subdiscipline’s own idiosyncrasies and a highly contextual framework. Reexamination of a central concept from the perspective of a new subdiscipline should not introduce misconceptions about that concept. When misconceptions introduced through chemical language can be avoided, we should change the way we speak."


How do point mutations affect complementary base-pairing and widths of the nucleotide pairs? Point mutations are divided into two categories: transitions and transversions. Since point mutations only occur during DNA replication (an integral process of both mitosis and meiosis), the mutated nitrogenous bases only appear on the newly synthesized DNA strands, rather than on the parental strands that serve as templates during DNA replication.
A transition occurs when a would-be purine is replaced by another purine, or a would-be pyrimidine is replaced by another pyrimidine. In contrast, a transversion occurs when a would-be purine is replaced by a pyrimidine, or vice versa. Because the two strands of a DNA molecule always follow the strict rule of complementary base-pairing, there must be some impact on chemical structures of the DNA molecule when either a transition or a transversion occurs.
Specifically, under normal circumstances, an adenine forms two hydrogen bonds with its complementary thymine, and a cytosine forms three hydrogen bonds with its complementary guanine. When a transition occurs, such as when a would-be thymine is replaced by a cytosine, would hydrogen bonds still form between the two bases? And how many hydrogen bonds will there be? Same questions for when a transversion occurs, such as when a would-be thymine is replaced by a guanine.
Another relevant point to ask is the distance between the two DNA strands that constitute a DNA molecule. The rule of complementary base-pairing helps to maintain equal widths for each complementary base pair. Thus, an A-T pair and a C-G pair are of the same width. But when a transition occurs, would the new A-C combination have a different width? Same question for when a tranversion occurs.
Thanks to anyone who answers my questions.

When Point mutations occur in one strand of DNA, irregular base-pair emerge. One of the main features of this irregular state is that the width of the double-strand differs from the standard form(usually larger width as the new, irregular base pairings are weak) in that specific area. In other words, the portion of DNA which has the defective base-pair bulges out or protrudes leading to the asymmetry of DNA.(analogous but not the same as Thymine dimers) Most of the times the DNA repair system finds this and deals with it in the correct way. I do not know about the exact number of hydrogen bonds that appear in these situations but perhaps it obeys either Watson-Crick, Hoogsteen, or Wobble rules.


What are the most extreme chemicals? Some chemicals certainly have extraordinary properties. One can consider them extreme in a sense, that they are the prime example for a concept, or they achieve the highest or lowest values in a certain area.

This Q&A attempts to compile a list of questions asked on this network, which have a background related to such extreme values or properties. we decided to add such a list because it is a popular search term.

It is a community effort, hence everybody with more than 100 reputation points can edit the answer. If you think something is missing in this list, just go ahead and add it. This post will be locked to prevent adding new answers, though. The corresponding discussion can be found on meta.chemistry.se. If you think there are some obvious points missing here, please go there and open a discussion.

Densest gas
Densest element
Greatest range between melting and boiling temperature
Highest melting point
Lowest solubility product constant
Most dangerous: Is there any substance that's a 4-4-4 on the NFPA diamond?. But also, don't forget that the NPFA diamond doesn't even classify ClF3 as flammable (though everything else is flammable in it) and by any criterion it is far more dangerous than the other compounds in those answers)
Strongest acid
Strongest organic acid
Strongest reducing agent

Good external lists:

Most extreme compounds known to man.


Confused about the direction in which the bond will break Why does the bond break in different fashion in the two cases?

For case X, a 3° carbocation is formed, bit why isn't this the scenario for case Y?

A 3° carbocation is anyday more stable than a 1° carbocation, isn't it?

The reactions happen on different centres because they follow different mechanisms.

Epoxides are actually surprisingly stable. They don’t undergo spontaneous ring-opening reactions by themselves unless the oxygen atom has been protonated previously. That is what the $\ce{H+}$ in reaction path X is there for. After having protonated the oxygen, the ring-opening reaction is charge-neutral and thus releiving the Baeyer strain to generate a tertiary (more stable) carbocation is favourable. This carbocation can be captured by a nucleophile in an $\mathrm{S_N1}$-type mechanism.

This does not happen under basic conditions as removing a negative charge would not help the epoxide opening. However, under these conditions solvent molecules such as methanol are easily deprotonated creating much better nucleophiles. These nucleophiles are then able to attack the epoxide under $\mathrm{S_N2}$ conditions. Hereby, an anion attacks a neutral molecule to generate an anion, so the ring-opening is again charge-neutral. However, $\mathrm{S_N2}$ attacks are more likely to happen on less hindered, primary carbons.

Taking this into account, path X that follows an $\mathrm{S_N1}$ path, generates the most stable carbocation hence the nucleophilic attack happens on the tertiary carbon. Path Y that follows an $\mathrm{S_N2}$ mechanism includes an attack on the less hindered, i.e. primary carbon.


Reaction of sulfate ion in copper sulfate electrolysis I am planning to try copper plating a piece of metal by performing electrolysis on an aqueous solution of copper sulfate. I plan run an electrical current with the metal I want to plate as the cathode and a platinum anode. I understand that the copper ions will become reduced when they plate the metal, but I am curious what will happen to the sulfate ions. Will it oxidize to form sulfur trioxide with the produced oxygen from the electrolysis?

A typical solution for copper plating contains about $0.5\ \mathrm{mol\ l^{-1}}$ $\ce{CuSO4}$ and $0.5\ \mathrm{mol\ l^{-1}}$ $\ce{H2SO4}$:

$$\begin{align}
c(\ce{Cu^2+})&=0.5\ \mathrm{mol\ l^{-1}}\\
c(\ce{H+})&=1\ \mathrm{mol\ l^{-1}}\\
c(\ce{SO4^2-})&=1\ \mathrm{mol\ l^{-1}}\\
\end{align}$$

The positive ions $\ce{Cu^2+}$ and $\ce{H3O+}$ are attracted to the cathode. The redox potentials $E$ for $\mathrm{pH} = 0$ show that cathodic reduction of $\ce{Cu^2+}$ to $\ce{Cu}$ is preferred over reduction of $\ce{H+}$ to $\ce{H2}$:

$$\begin{alignat}{2}
\ce{Cu^2+ + 2e- \;&<=> Cu}\quad\quad &&E^\circ = +0.340\ \mathrm{V}\quad\quad &&&E=E^\circ+\frac{0.059\ \mathrm{V}}{2}\log 0.5 = 0.331\ \mathrm{V}\\
\ce{2H+ + 2e- \;&<=> H2}\quad\quad &&E^\circ = +0.000
\end{alignat}$$

At the anode, the redox potentials $E$ imply that oxidation of $\ce{H2O}$ to $\ce{O2}$ is preferred over oxidation of $\ce{SO4^2-}$:

$$\begin{alignat}{3}
\ce{O2 + 4H+ + 4e- \;&<=> 2H2O}\quad\quad &&E^\circ = +1.229\ \mathrm{V}\\
\ce{S2O8^2- + 2e- \;&<=> 2SO4^2-}\quad\quad &&E^\circ = +2.01\ \mathrm{V}
\end{alignat}$$

Nevertheless, $\ce{SO4^2-}$ actually is oxidised at the anode:
$$\ce{SO4^2- -> SO4- + e-}$$
The reason for this reaction is that negative ions are attracted to the anode. Under neutral conditions, $\ce{OH-}$ ions would migrate to the anode where they would be oxidised to $\ce{O2}$. However, under the given acidic conditions, the concentration of $\ce{OH-}$ is very low. The predominant negative ions are $\ce{SO4^2-}$, which migrate to the anode.
Since the produced concentration of $\ce{SO4-}$ is low, the $\ce{SO4-}$ ions do not combine to $\ce{S2O8^2-}$. Instead $\ce{SO4-}$ reacts with water:
$$\ce{4SO4- + 2H2O -> 4SO4^2- + O2 + 4H+}$$

Therefore, the net reaction is the above-mentioned oxidation of $\ce{H2O}$ to $\ce{O2}$. The $\ce{SO4-}$ ions act as catalyst:

$$\begin{align}
\ce{4SO4^2- &-> 4SO4- + 4e-}\\
\ce{4SO4- + 2H2O &-> 4SO4^2- + O2 + 4H+}\\
\hline\ce{2H2O &-> O2 + 4H+ + 4e-}
\end{align}
$$


What is the definition of of 'compound', 'mixture', 'element' and 'molecule'? I am looking for the precise definitions, as I am very confused as to what they are exactly because although I mostly understand what they mean, I have encountered some conflicting definitions that confused me.

As it stands, this is what I understand them to mean:

  Compound: two or more different atoms bonded together.
  Mixture: two or more different atoms together but not joined.
  Molecule: two particles (same or different) bonded together.
  Element: only 1 type of atom; this definition is applied to things both bonded and not to itself.

I don't know whether these terms apply microscopically, macroscopically, or both. For example, I think 1 $\ce{H2O}$ molecule is a compound but is a bathtub of them called a mixture (as it contains more than 1 atom type), a compound, or both? That is, to be a compound, do all the atoms in the compound have to be bonded physically together? Generally, to what extent can these four terms overlap?

I would very much like a systematic way of thinking about these terms.

I can't really provide a systematic approach, but I can attempt to clarify (as a student myself).

Elements are classes of atoms. Atoms of the same element are similar (if not identical) in their physical and chemical properties (but be aware of Isotopes which are physical variations among atoms of the same element).
A definite (I suppose, systematic) way to distinguish elements is that their corresponding atoms have different numbers of electrons orbiting the nucleus.
The term is also used to describe a collective of the same atom (element). A bar of gold (Gold being an 'element'; a chemical class) is said to be an element itself.
Molecules are a group of atoms covalently bonded to each other (which can be considered a 'direct connection', if you will). The molecule can consist of atoms of the same element, or atoms of different elements.
If you joined to molecules via covalent bonding, you'd have created a new molecule (it's nothing special)
Compounds are (in the chemical sense of the term) made of two or more elements. This means a lot of molecules (like H2O) are considered compounds (though H2 is not). 'Compound' is often used interchangeably with the word substance, which describes a collective of molecules (water is a substance. It is a large quantity of neighboring H2O molecules).
A substance (and a compound, when a compound is used to describe a quantity of molecules) only consists of the same molecule. In water, all the molecules are H2O. If there were also some H2 molecules among them, it would no longer be considered a compound/substance (but instead; a mixture).
These molecules are not directly bonded to each other (or they'd be one big molecule), they are just in close* proximity to one another and are attracted to one another, keeping them in proximity. Even when the substance is a solid, like ice, they are not directly connected, the molecules are just slower (less energetic) and closer together (and in some special cases, like ice, stronger, such as Hydrogen Bonding. Don't be confused, though it is referred to as 'bonding', there is no direct covalent bond, just an attraction).
Mixtures are collectives of different molecules or atoms. The molecules and atoms are like that in a substance; they're not bonded to each other, but are attracted, and in a mixture specifically, the molecules/atoms are not all the same. Juice, for instance, is a mixture. It contains some Vitamin C molecules, some water molecules, some sugar molecules, etc.

Here is a horribly drawn diagram to help clarify the distinctions.

Reflecting on this, here is what in your understanding needs specific attention.

Compound: two or more different atoms bonded together A molecule or a group of identical molecules.
Mixture: two or more different atoms or molecules together but not joined (they're not covalently bonded).
Molecule: two or more atoms(same or different) bonded together.


It is possible to make an homogeneous solution of water and oil? I wondered if in certain condition like very high pressure(so that water and oil remain liquids) and very high temperature(so that the variation of the Gibbs energy is negative--> ∆G = ∆H-T*∆S) we can make an homogeneous solution of water and oil.

Or, alternatively, mixing water and oil meanwhile applying an intense electric field so that the oil molecules get a charge such that they become soluble in water.

Do you think these methods could work? Is there any other way to mix them?

Generally polar and non polar liquids do not dissolve in one another to any great extent. Practically, you would have to find a temperature composition phase diagram for the oil you want to use. This will tell you if there is a region in which the water and oil are mixed as a homogeneous solution. Alternatively if you add surfactant then a ternary phase diagram (triangle plot) is what you need to look for since than micellar or liquid crystal phases may form.

As you mention the free energy I have tried to summarise what happens on dissolving a non-polar molecule in water.
The enthalpy $\Delta H$ change on placing, alkane, into water is small, slightly positive or negative depending on the compound, -9kJ/mol for ethane, +1.6 for toluene for example.  This can be thought of as the cost  making a void in the water to hold the molecule.

Water is unusual in having a strong and highly directional hydrogen bonding nature. Thus adding a non-polar molecule into water is determined by entropic effects. This generally means that $T\Delta S$ is negative and so $\Delta G$ is positive, for cyclohexane into water $\Delta G$ = +28 kJ/mol.

The entropy of water decreases on contact with non-polar molecules since these molecules cannot form hydrogen bonds. The water has to bridge around the non-polar molecules, as this is a lowering of entropy compared to not bridging. This bridging partly restricts the water molecules compared to bulk water. This option is preferable as the alternative of breaking hydrogen bonds, and leaving H atoms pointing at the non-polar molecules,is too energetically costly. Bridging makes it possible to form some H bonds at the expense of entropy loss. The figure shows this (from Southall et al. J. Phys. Chem. p521, 106, 2002 )

As the temperature increases the enthalpy increases as does T$\Delta S$, but $\Delta G$ remains largely constant and remains positive, see sketch below. This suggest that  water and oil will not easily mix.


VSEPR theory and hybridization in determining the shape of a molecule Our chemistry teacher told us that both VSEPR theory (which says that the electron pairs in the valence shell of an atom arrange themselves in such a way that repulsions among them are minimized and this arrangement of the electron pairs determines the shape of a particular molecule)  and hybridization (which is the intermixing of a particular number of atomic orbitals to form equal number of new orbitals which have same shape and energy)  can be used to determine the shape of a molecule and that they are independent of each other. I understand how the shape of a molecule can be predicted using the VSEPR theory but i cannot understand how hybridization helps in determining or rather predicting the shape of a particular molecule. I have searched the net but could not find any useful resource.

How does hybridization help in determining the shape of a molecule?

The VSEPR theory can indeed be used to predict geometry a priori. You would count the number of substituents around a central atom, counting any lone pairs as substituents also, and then find an arrangement that maximises the distance between all substituents. This theory works well for elements of the first period and compounds in which the central element is in the highest oxidation state, but tends to fail very rapidly for all others, most notably because it cannot account for anything outside of the traditional 2-electron-2-centre bond and because it considers all lone pairs equal even though they are typically not.

The hybridisation theory can only be used to predict geometry in a very limited number of cases that involve carbon atoms, and it only works because carbon is extremely regular. However, using hybridisation to predict carbon geometry is basically circling back onto VSEPR: you check how many atoms are bound to a certain carbon and thereby deduce whether this carbon atom will be $\mathrm{sp,sp^2}$ or $\mathrm{sp^3}$. Carbocations and carbanions can also be accomodated for (by either counting the anionic lone pair or ignoring it — as in VSEPR).

In general, however, it is not possible to predict the geometry following a certain hybridisation. Rather, one determines the geometry by an independent mean (e.g. quantum chemical calculations, gas phase diffraction), deduces the bond angles (and bond lengths) from the geometry and then derives a hybridisation. In short: hybridisation follows geometry, not the other way around.


What is the simplest formula for the mineral beryl which contains 5.03% Be, 10.4% Al, 35.5% Si and 53.58% O? Beryl is a mineral which contains 5.03% $\ce{Be}$, 10.4% $\ce{Al}$, 35.5% $\ce{Si}$ and 53.58% $\ce{O}$. What is the simplest formula for beryl?

So I tried to follow a 3 step rule to solving this equation:

Divide each % by it's own atomic mass
Divide each of those by whichever is smallest
Find the lowest whole #

Amount of $\ce{Be}$ = 5.03% / 9 = .558 moles
Amount of $\ce{Al}$ = 10.04 / 26.98 = .372 moles (smallest number)
Amount of $\ce{Si}$ = 31.35% / 32 = .979 moles
Amount of $\ce{O}$ = 54.58% / 16 = 3.3 moles
Proportionate amount of $\ce{Be}$ = .558 / .372 = 1.5
Proportionate amount of $\ce{Al}$ = .372 / .372 = 1
Proportionate amount of $\ce{Si}$ = .979 / .372 = 2.63
Proportionate amount of $\ce{O}$ = 53.58 / 16 = 3.3
From here I don't know what I have done wrong, but these numbers are all whack silly and I can't see to get a number that leads me to the answer.
The correct answer is $\ce{Be4Al(SiO3)8}$

Firstly the "correct" answer seems to be wrong, which is not an uncommon occurrence, so please do not always take answers at the back of books to be correct. Wikipedia lists the formula of beryl as $\ce{Be3Al2(SiO3)6}$ which is much more in accordance with what you have done.

That aside, there are two things you need to correct.

The percentages that you have in your first step are different from those quoted in the title. For example, for silicon, you divided 31.35 by 32, but the title says that there is 35.5% silicon. Please, be more careful when you are pressing numbers into a calculator or the computer. The same issue exists for aluminium and oxygen.
In your second step, oxygen, you seem to have done a calculation pertaining to the first step? You meant to do 3.3/0.558, correct?

Fix those and you should obtain the correct answer - and not the one in the back of whatever text you are using.


Which element has the highest binding energy per nucleon The graph of binding energy per nucleon (Hyperphysics) seems to peak at Iron-56. However, Wikipedia says that Nickel-62 has the highest binding energy per nucleon of any known nuclide (8.7945 MeV). Isn't this a direct contradiction? Why does the binding energy per nucleon peak at Iron-56, but then Wikipedia says Nickel-62 has the highest binding energy per nucleon?

Can't be expertly comment on the plot without knowing exactly how the plot was constructed. But I can make the following observations...

User Mithoron makes a good point that it depends on how the Y-axis of the plot is taken. You can plot average mass per nucleon in which case iron-56 is the heaviest atom per nucleon. However if you plot binding energy per nucleon, then nickel-62 has the highest binding energy per nucleon.
For elements with stable isotopes I'd guess only the stable ones are shown. For elements with unstable isotopes than some selection process is necessary so as not to clutter up the plot too much.
I'll point out that the "peak" is marked as "Iron Group." It isn't entirely clear what isotopes that means. Wikipedia notes that in astrophysics The iron group in astrophysics is the group of elements from chromium to nickel which are substantially more abundant in the universe than those that come after them – or immediately before them – in order of atomic number.
In terms of stellar production much more iron than nickel is produced. The iron  produced in stars is about about 92 % iron-56. Nickel is only about 3.6% Nickel-62.
I'd guess that the plot is a "best fit line." Here is a different plot showing more detail. You can see that not all isotopes fall on "the line."

Here is another plot at link pointed out by user IanBush in a comment. The "best fit" curve doesn't pass through all the data points.


Dividing different units of measurement? 1) $$\frac{35 \; \text{cm}^2}{0.62 \; \text{cm}} = 56.4516$$

For the units in the answer, do you put $56.4516$ cm$^2$ or just $56.4516$ cm?

1) $$\frac{0.075 \; \text{g}}{0.0003 \; \text{cm}^3} = 250$$
Same here, do you put $250$ g, $250$ cm$^3,$ or $250$ g/cm$^3$?

What does $g/cm^3$ mean? It means you have divided a mass in grams by a volume in $cm^3$ (the slash character"/" is really a division). So yes, your second value is 250 $g/cm^3$.

In the first case, you can simplify $cm^2/cm$, just like in any equation, to $cm$. So your value is 56.4516 cm.


How to give an evidence in support of the arenium ion mechanism? Outline the mechanistic steps involved in the arenium ion mechanism for
  aromatic electrophilic substitution. Draw the energy profile diagram for the proposed mechanism. Give one evidence in support of the proposed mechanism.

I can solve the first two questions but I got confused about what the last question exactly asking for. Any help will be appreciated.

You say you've already drawn the energy profile diagram for your proposed mechanism , now you want to make sure that your reaction mechanism proceeds through the most stable transition state , ( Least activation energy ) and therefore , on the energy profile diagram , the peak occuring at the transition state should not be too high .Also you wanna make sure that your final products have energies lesser than the  reactants , because in that case , your $\Delta H$ would be $-ve$ and from the equation $\Delta G=\Delta H-T\Delta S $, there would be a greater chance your reaction moves forward.

You also should have a look at this


Inductive effect of hydrogen isotopes Why does the electron-donating inductive effect (+I) of the isotopes of hydrogen decrease in the order $\ce{T} > \ce{D} > \ce{H}$?
(where T is Tritium and D is Deuterium)

Google has nothing to offer. Does it have to do anything with mass, as the order implies?

Yes, it has a lot to do with mass.  Since deuterium has a higher mass than protium, simple Bohr theory tells us that the deuterium 1s electron will have a smaller orbital radius than the 1s electron orbiting the protium nucleus (see "Note" below for more detail on this point). The smaller orbital radius for the deuterium electron translates into a shorter (and stronger) $\ce{C-D}$ bond length.

A shorter bond has less volume to spread the electron density (of the 1 electron contributed by $\ce{H}$ or $\ce{D}$) over resulting in a higher electron density throughout the bond, and, consequently, more electron density at the carbon end of the bond. Therefore, the shorter $\ce{C-D}$ bond will have more electron density around the carbon end of the bond, than the longer $\ce{C-H}$ bond.

The net effect is that the shorter bond with deuterium increases the electron density at carbon, e.g. deuterium is inductively more electron donating than protium towards carbon.

Similar arguments can be applied to tritium and it's even shorter $\ce{C-T}$ bond should be even more inductively electron donating towards carbon than deuterium.

Note: Bohr Radius Detail

Most introductory physics texts show the radius of the $n^\text{th}$ Bohr orbit to be given by

$$r_{n} = {n^2\hbar^2\over Zk_\mathrm{c} e^2 m_\mathrm{e}}$$

where $Z$ is the atom's atomic number, $k_\mathrm{c}$ is Coulomb's constant, $e$ is the electron charge, and $m_\mathrm{e}$ is the mass of the electron.  However, in this derivation it is assumed that the electron orbits the nucleus and the nucleus remains stationary.  Given the mass difference between the electron and nucleus, this is generally a reasonable assumption.  However, in reality the nucleus does move too.  It is relatively straightforward to remove this assumption and make the equation more accurate by replacing $m_\mathrm{e}$ with the electron's reduced mass, $\mu_\mathrm{e}$

$$\mu_\mathrm{e} = \frac{m_\mathrm{e}\times m_\text{nucleus}}{m_\mathrm{e} + m_\text{nucleus}}$$

Now the equation for the Bohr radius becomes

$$r_{n} = {n^2\hbar^2\over Zk_\mathrm{c} e^2 \mu_\mathrm{e}}$$

Since the reduced mass of an electron orbiting a heavy nucleus is always larger than the reduced mass of an electron orbiting a lighter nucleus

$$r_\text{heavy} \lt r_\text{light}$$

and consequently an electron will orbit closer to a deuterium nucleus than it will orbit a protium nucleus.


Calculating the charges of anions and cations I am new to chemistry. Could someone explain me how to calculate the charges on anions and cations. I've heard about some crossing method (the left subscript becomes the right superscript and the right subscript becomes the left super script.)

Example: $\ce{Rb2S -> 2Rb+ + S^2-}$

But what about:
$\ce{K2MnO4}$ and $\ce{Zn(ClO)2}$

This trick doesn't seem to apply here, so how can I find out the charges?

Some basic knowledge of the atoms in the periodic table is necessary to extract the ionic charges. In the case of $\ce{K2MnO4}$, for example, we know that the potassium ion is positive, and that the manganate ion is negative. We see that the compound is neutral: has zero net charge. So how can we place charges to get the molecule to be neutral, given the subscripts?

Potassium is a Group 1 element, meaning it only has 1 valence electron, placed in the  4s orbital. Potassium is eager to lose this electron to fulfil the octet rule. Therefore potassium ions have one less electron than protons, and the charge is +1. Since we have two $\ce{K^+}$ ions, the manganate ion must have a charge of -2 to make sure the molecule has zero net charge.

$$\ce{K2MnO4 <=>[\ce{H2O}] 2K+ + MnO4^2-}$$

Zinc is one of two transition metals that only have one (natural) oxidation state: +2. The electron configuration is [Ar]3d$^{10}$4s$^2$. When zinc becomes a cation, it loses the two 4s electrons. Knowing this, we see that the hypochlorite must have a charge of -1: the two hypochlorite -1 charges counter the 2+ charge of zinc.

$$\ce{Zn(ClO)2 <=>[\ce{H2O}] Zn^2+ + 2ClO^-}$$

I recommend not relying on simple rules, such as the one you mentioned. Rather, try to understand why the rule came about, and understand when the rule fails, and why. It really helps to be able to "read" the periodic table, as a lot of basic information is incorporated into it.


Difference between reversible and irreversible processes On a quiz, I was asked for the difference between reversible and irreversible processes.

I said:

For a reversible process, you can change direction at any time without disturbing equilibrium.
For an irreversible process, the macroscopic variables may not always be defined.

I was given 1/4 point for this answer. I asked the professor what the issue is with the answer, and she could not articulate a clear reason, and eventually said she'll re-think it.

Do you see a problem with my answer? If so, please explain.

In a reversible process, at each point along the process path, the system is only slightly removed from being in thermodynamic equilibrium with its surroundings.  So the path can be considered as a continuous sequence of thermodynamic equilibrium states. For an irreversible process, the system is not close to thermodynamic equilibrium with its surroundings at each point along the path.

At any point along the path of a reversible process, both the system and the surroundings can be returned to their original states without significantly affecting anything else.  This cannot be accomplished if the process is irreversible.


Calculate hardness of water A sample of water has its hardness due to only $\ce{CaSO4}$. When this water is passed through on anion exchange resin, $\ce{SO4^2-}$, the $\ce{SO}$ ions are replaced by $\ce{OH-}$. A 25.0 mL sample of water so treated requires 21.58 mL of 0.001 M sulfuric acid for its titration. What is the hardness of water expressed in terms of $\ce{CaCO3}$ in ppm? Assume the density of water is 1.0 g/mL.

I do not understand where does $\ce{CaCO3}$ come from if we started with $\ce{CaSO4}$at the beginning.

The problem isn't going through gyrations just to drive you crazy. It is really the way that water analysis works. The "overall" hardness is expressed in terms of the "equivalent ppm of $\ce{CaCO3}$".

The two biggest contributors to water hardness are typically $\ce{CaCO3}$ and $\ce{MgCO3}$. Now let's say that you want to buy a water softener to replace cations $\ce{Ca^{2+}}$ and $\ce{Mg^{2+}}$ with $\ce{Na^+}$ and the $\ce{CO3^{2-}}$ with $\ce{Cl^-}$. Figuring how many gallons of water you'd use in a week and knowing your water's hardness as "equivalent ppm of $\ce{CaCO3}$" you could figure out how much ion exchange resin you'd need. So there is a functional reason that "equivalent ppm of $\ce{CaCO3}$" is helpful.


Would Al ions still be reducd to Al metal if it was aqueous soltion? To my knowledge, Aluminium is mostly mined in the form of Aluminium oxide or Al2O3 and then the Bayer process is used to extract and refine the pure Aluminium.

Now, The Al2O3 must be in a molten state when the process occurs. Why is that? Why must it be molten?

Also, Would the Al(3+) ions still be reduced to the Al metal if it was in an aqueous solution instead of being molten?

Thanks,

The Bayer process does not produce aluminium. It produces sodium aluminate, and finally aluminum oxide, which is later used in the Hall–Héroult process (without water) to produce aluminium metal.

As @Mithoron implied, aluminium will displace the hydrogen from water, so cannot be made in an aqueous environment.


Is hydroboration a pericyclic reaction? According to Ahn.N.T "Frontier Orbitals" hydroboration isn't a pericyclic reaction because boron uses two AOs and not one. The same applies for cheletropic reactions, which aren't pericyclic either. But the IUPAC goldbook says different: http://goldbook.iupac.org/P04491.html - that cheletropic reactions are pericyclic. So is hydroboration, too?

Based on the IUPAC Goldbook definition, the only two requirements for a pericyclic reaction are:

Concerted mechanism
Cyclic transition state

Hydroboration reactions (and similar reactions) are pericylic reactions by these criteria, as are all cheletropic reactions.

The language about fully conjugated transition states can be used to explain some pericyclic reactions, but not all, and thus IUPAC uses the phrasing may be viewed as.... Hydride shifts, (and all sigmatropic rearrangements ene reactions) involve non-$\pi$ orbitals (even so, the transition states do tend to follow the Hückel rule).


Is there a way to search for adducts by exact mass? We found an unknown of exact mass 217.082. Is there a tool that one can search by an exact mass get a list of possible compounds. We have considered desmethylnaproxen as a possibility, but found its mass as 217.086 for the proton adduct.

I guess that would be what you are looking for:
http://www.chemcalc.org/mf_finder/mfFinder_em_new
This one gives you theoretical combinations of atoms, that fit the mass.

You could also check the sdbs database for organic compounds.
https://sdbs.db.aist.go.jp
This one gives you existing compounds but it is rounding to 1 decimal so it gives you everything with 217.1 which is not perfect.

I just used the frist calc and then checking all close results (green) at SigmaAldrich to see, which formula matches a compound they sell. There are just two from the List:
$\ce{C10H16ClNO2}$ and $\ce{C11H11N3O2}$

So I would focus on these.


Conversion of grams to milliliters using kg/m3 density Convert 100 grams of a solution to milliliters with a density of $681.9\rm~\frac{kg}{m3}$. It also has a density of 5.69 pounds per gallon.

Are any or all of these correct?

$$\rm100~g \times \frac{1~mL}{681.9 \frac{g}{cm^3}} = 0.146649~mL\quad(\approx 0.15~mL)$$

$$\rm1,000~g \times \frac{1,000~mL}{681.9~\frac{g}{cm^3}} = 1,466.4906~mL\quad(\approx 1,500~mL)$$

$$\rm1~kg \times \frac{1~L}{681.9~\frac{kg}{m^3}} = 1.4664906~L (\approx1.5~L)$$

Or, can it be as simple as: 100 g x 681.9 = 68.190 mL? The unit kg/m3 is what's confusing me. I'm used to g/cm3 or g/mL for densities. Does 681.9 g/cm3 = 681.9 kg/m3, while scaled-up a thousand-fold? Can I consider using the same density with the math involved by substituting kg/m3 for g/cm3 in the equation, or do I have to utilize a conversion within the metric system?

Concerning the units in your equation none of them is correct (As all your formulations would end up in a square volume unit).

Definition of volumetric mass density

The density $ \rho $ is given as

$$ \rho = \frac{m}{V} $$

resulting in a mass and density dependent Volume as

$$ V = \frac{m}{\rho} $$

Direct way

Although you could calculate this by just substituting the physical quantities by the given values and units, and then take care of the unit conversion afterward:

$$ V = \frac{m}{\rho} = \frac{100\,\mathrm{g}}{681.9\,\mathrm{kg}\,\mathrm{m^{-3}}} =\\= \frac{100\,\mathrm{g}\,\mathrm{m^{3}}}{681.9\,\mathrm{kg}} = \frac{100\,\mathrm{g} \cdot \left(10^2\,\mathrm{cm}\right)^{3}}{681.9 \cdot (10^3\,\mathrm{g})} =\\= \frac{100 \cdot 10^6\,\mathrm{cm}^{3}}{681.9 \cdot 10^3} = \frac{100 \cdot 10^3}{681.9}\,\mathrm{cm}^{3} =\\\approx 146.65\,\mathrm{cm}^{3} \equiv 146.65\,\mathrm{ml} $$

I would suggest to convert the density from $ \mathrm{kg}\,\mathrm{m^{-3}}  $ to $ \mathrm{g}\,\mathrm{cm^{-3}} $ as a first step:

Easier way

Of course easiness is something subjective, but I believe this is easier to use.

Calculate conversion-factor for density

First the combined conversion-factor for the units-fraction is calculated

$$ \frac{\mathrm{kg}}{\mathrm{m}^3} \stackrel{\left[1\right]}{=} \frac{\left(10^3\,\mathrm{g}\right)}{\left(10^2\,\mathrm{cm}\right)^3} \stackrel{\left[2\right]}{=} \frac{10^3\,\mathrm{g}}{10^6\,\mathrm{cm}^3} \stackrel{\left[3\right]}{=} 10^{-3}\,\frac{\mathrm{g}}{\mathrm{cm}^3} $$

Start with substituting each units SI prefix:
Care has to be taken with the conversion of $ \mathrm{m}^3 $, since the power applies to the whole substituted expression ($ \mathrm{m}^3 = \left(10^2\,\mathrm{cm}\right)^3 $).
In the last step the exponential expressions are summarized ($ \frac{10^3}{10^6} = 10^{3-6} = 10^{-3} $).

Then calculate the new numeric value of the density in the target unit:

$$ \rho = 681.9\,\mathrm{kg}\,\mathrm{m^{-3}} \stackrel{\left[4\right]}{\equiv} 681.9 \cdot \left(10^{-3}\,\mathrm{g}\,\mathrm{cm}^{-3}\right) \stackrel{\left[5\right]}{=} 0.6819\,\mathrm{g}\,\mathrm{cm}^{-3} $$

Substitute the unit with the previously calculated expression
Multiply the numeric terms

Calculate the volume

Now the unit-adjusted density is used:

$$ V = \frac{m}{\rho} = \frac{100\,\mathrm{g}}{0.6819\,\mathrm{g}\,\mathrm{cm^{-3}}} = \frac{100}{0.6819}\,\mathrm{cm^{3}} \approx 146.65\,\mathrm{cm}^{3} \equiv 146.65\,\mathrm{ml}  $$

Conversion-factors

According to SI Prefixes the following applies:

$ 1\,\mathrm{m} \equiv 10^2\,\mathrm{cm} \\
1\,\mathrm{kg} \equiv 10^3\,\mathrm{g} $

And of course:

$ 1\,\mathrm{cm}^3 \equiv 1\,\mathrm{ml} $


Knowing that CI is variational for any basis set, why is coupled-cluster variational in the full correlation limit? This is the question asked in Martin's comment to my answer to: Why is Coupled Cluster not variational?

In the comment I was also asked to provide a publication as a reference:

The CI (configuration interaction) energy, given by Eqs. 3.9-3.10 of this, is an upper bound to the exact ground state energy (i.e. CI is variational). Coupled cluster energies are not generally variational, as explained in the above linked question, except in the case of CC($N$) for an $N$-electron system. The reason is because CC($N$) for an $N$-electron system is equivalent to FCI (full configuration interaction) which is an example of CI, and all CI energies are variational.

Since I was asked for a publication as a reference, a good one is this, in which it is shown that CC and CI become the exact same in the full-correlation limit.


Lowering the melting point of water with salt I was trying to calculate amount of salt needed to lower the melting point of water by 5 °C. According to my solutions, I need to make sure $$n\mu(\ce{H2O(s)}) =n\mu(\ce{H2O(l)})+ RT\log(x)$$where $ x=\frac{n}{n+2m}$

I, however, thought that I need to make sure that

$$n\mu(\ce{H2O(s)})+m\mu(\ce{NaCl(s)}) =n\mu(\ce{H2O(l)})+m\mu(\ce{NaCl(aq)})$$

After all, the solid ice still has a certain contribution. And I would have assumed that water can be seen as the solved according to the first convention? So why don't we need to consider the solid ice, and what is wrong with my way of thinking?

You should write independent equilibrium expressions for the chemical potential of each species.

In a freezing point depression problem the solvent in frozen form is at equilibrium with the solvent in liquid form, so you can write that
$$\mu(\ce{H2O(s))}=\mu(\ce{H2O(sol))}$$
Next, by Raoult's law (assuming it holds), the chemical potential of solvent in the solution (the right-hand term) differs from that of the pure liquid by the term $RT\log(\chi_1)$ where $\chi_1$ is the mole fraction of solvent, so that
$$\mu(\ce{H2O(s))}=\mu(\ce{H2O(l))} + RT\log(\chi_1)$$
This is just the beginning. Next you need to recognize that the standard Gibbs free energy change for melting is just $\Delta_m G^\circ=\mu(\ce{H2O(l))}-\mu(\ce{H2O(s))}$ (assuming the solution is at $\pu{1 bar}$ or approx normal pressure) and invoke the Gibbs-Helmholtz equation in differential form:

$$\left(\frac{\partial(\Delta G /T)}{\partial (1/T)}\right)_p=\Delta H$$

so that

$$\left(\frac{\partial\log(\chi_1)}{\partial (1/T)}\right)_p=-\frac{\Delta_m H^\circ}{R}$$

Now you integrate the van't Hoff equation from $\chi_1 = 1$ (pure solvent) to $\chi_1 = 1-\chi_2<1$, while assuming $\Delta_m H^\circ$ to be constant over this range of $T$. Finally, assuming $T\approx T_m$, expressing $\chi_1$ in terms of solute molality, and assuming the solute concentration is low, results in the freezing point equation
$$\Delta T = K_m m_2 $$
where $K_m$ is the cryoscopic constant
$$K_m=\frac{RT_m^2M_2}{1000 \Delta_m H^\circ}$$
Here $M_2$ is the molar mass of the solute.
You can use this last expression to compute the molality of solute $m_2$ required to depress the freezing temperature by $\Delta T$.

The process of dissolving the solute is not important (whatever happens before it is fully dissolved doesn't matter as long as it is soluble). Also, it is assumed an ideal solution forms, which means interactions of the solvent with solute, and between solvent molecules are equally strong. And the salt remains in solution (none is incorporated into the solid phase). Note also that liquid solvent remains in equilibrium with solid solvent at the melting point $T_m$, and that we assume that the solute concentration is dilute (this is important to derive the last expression, but Raoult's law may hold even for substantial solute concentrations).


Is the photoelectric effect possible for semiconductors? Is the photoelectric effect possible for semiconductors?

I imagine that this might be, if possible, a two-photon process: excitation of an electron from a filled to conductance band, and then on to ejection. Is anyone familiar with such materials, or even whether this is possible?

I can't readily think of a theoretical reason against it. Two-photon events are often less likely than one-photon, however I could imagine irradiation by a laser of bandwith covering both the band-gap wavelength and that necessary for ejection would do it.

If it does indeed happen, I'd be glad to hear a short summary of how it differs from the photoelectric effect of metals.

Given that Jaques Pankove's excellent book on 'Optical Processes in Semiconductors' devotes chapter 13 to 'Photoelectric Emission', you can rest assured that the photoelectric effect is alive and well in semiconductors.  While somewhat more complex than emission from simple metals, there is no need for two-photon processes. The 'tricky' part is that you have to look at where the electron states are, and then how much energy is needed to take it out of the semiconductor.

The work function is the energy difference from the Fermi energy to the vacuum level.  There may not be any states at the Fermi energy though.

The electron affinity is the energy difference from the bottom of the conduction band to the vacuum level.

One has to take the band structure into account, since direct and indirect transitions will likely have different energies required.
Also, surface band bending may make interpretation of experimental results more difficult. Again, perusing Pankove may prove useful.


IUPAC Naming of CH3C(p-ClC6H4)2CH(Br)CH3 According to the locant rule, the numbering should be done from left end, as from left end locant positions are 2,2,3 and from right it is 2,3,3.

So, the IUPAC name should be 3-bromo-2,2-bis(4-chlorophenyl)butane. But according to most textbooks, its name is 2-bromo-3,3-bis(4-chlorophenyl)butane. Is alphabetical prefrence given to bromine, and if so, why?

According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), a structure with two or more identical units that are linked by di- or polyvalent groups can be named in two ways:

by ordinary substitutive nomenclature, in which one of the parent structures is chosen as the senior parent structure and the remainder of the structure is expressed by substituent prefixes
by multiplicative nomenclature, in which two or more parent structures are connected by symmetrical or unsymmetrical single or concatenated substituent groups

Both methods yield correct names, which may be used in general nomenclature. However, the IUPAC recommendations define a preferred IUPAC name (PIN), which is the name preferred among two or more names generated from two or more IUPAC recommendations.
For the compound that is given in the question, the preferred IUPAC name is actually generated using multiplicative nomenclature (see also What is the rationale of the IUPAC name for DDT?), which yields the name 1,1'-(3-bromobutane-2,2-diyl)bis(4-chlorobenzene).
In general nomenclature, nevertheless, you may choose the butane chain as the senior parent structure and generate the complete name using substitutive nomenclature.
The most important criteria for the numbering in substitutive nomenclature according to the current version of the IUPAC Recommendations are:

P-14.4 NUMBERING
When several structural features appear in cyclic and acyclic compounds, low locants are assigned to them in the following decreasing order of seniority:
(…)
(c) principal characteristic groups and free valences (suffixes);
(…)
(e) saturation/unsaturation:
  (i) low locants are given to hydro/dehydro prefixes (…) and ‘ene’ and ‘yne’ endings;
  (ii) low locants are given first to multiple bonds as a set and then to double bonds (…);
(f) detachable alphabetized prefixes, all considered together in a series of increasing numerical order;
(g) lowest locants for the substituent cited first as a prefix in the name;
(…)

Since the compound that is given in the question has no principal characteristic group and the butane chain, which is taken as the parent structure, has no multiple bonds, the rules (c) and (e) are not relevant in this case.
Thus, all substituents are considered together according to rule (f). Therefore, the correct name is 3-bromo-2,2-bis(4-chlorophenyl)butane and not 2-bromo-3,3-bis(4-chlorophenyl)butane since the locant set ‘2,2,3’ is lower than ‘2,3,3’.


Name of a proton bonded to a hydroxyl carbon What do you call the proton in a CH-OH configuration?

When discussing an NMR spectrum, I can talk about the hydroxyl proton, but it seems clumsy to refer to "the proton on the hydroxyl carbon".

I would call it the $\alpha$-proton to the $\ce{-OH}$. The alpha proton is the proton bonded to the alpha carbon (i.e., the hydroxyl carbon).


Chemicals in Urinal Odor Eaters What chemicals are used in so-called urinal "odor eaters"? Do they actually destroy odors, or do they just mask them? Also, are these chemicals subject to any environmental laws or regulations?

What chemicals are used in so-called urinal "odor eaters"?

There are 1,4-dichlorobenzene-based and non-PDCB products. In the latter, the chemical mixture from the linked MSDS is:
SODIUM SULPHATE, SODIUM DODECYL SULPHATE, COCAMIDE MEA, DIPENTENE, CITRAL, LINALOOL,BENZYL ALCOHOL

  Do they actually destroy odors, or do they just mask them?

They mask odors.

  Also, are these chemicals subject to any environmental laws or regulations?

Yes, as detailed in the MSDS sheets linked above. Please note that my knowledge of urinal cakes and related products is not exhaustive.


Criterion for gravimetric analysis of silver If the only criterion for gravimetric analysis of silver is solubility of  precipitate, which one of the following two compound are we supposed to choose? $K_\mathrm{s}(\ce{AgCl}) = \pu{1.78E-10}$ and $K_\mathrm{s}(\ce{Ag2CO3}) = \pu{8.13E-12}.$
a) $\ce{AgCl}$
b) $\ce{Ag2CO3}$
c) both of them are suitable
d) none of them

As I learned from textbook, one of the criteria for choosing the right one is: $K_\mathrm{s} < 10^{-7}$—thus I eliminate option d.
I also read that it is meaningless to compare the solubilities of two salts having different formulas on the basis of their $K_\mathrm{s}$ values. Therefore, I would choose option c—both have $K_\mathrm{s} < 10^{-7}.$ But from experience I found out that $\ce{AgCl}$ is used way more often—but that is probability because of other criteria—which I do not have to bear in mind in this question.
Is my reasoning correct?

The solubility of $\ce{AgCl}$ is equal to $\sqrt{K_\mathrm{s}} = \pu{1.3E-5 M}.$
The solubility $s$ of $\ce{Ag2CO3}$ is such that  $K_\mathrm{s} = 4s^3.$ So that its solubility $s$ is equal to $s = \pu{1.2E-4 M}.$ This is ten times more than the solubility of $\ce{AgCl}.$
For gravimetric purposes, $\ce{AgCl}$ is a better choice.


Kinetics question regarding Rate Law The acid catalysed hydrolysis of an compound $\ce{A}$ at $\pu{303K}$:
$$\ce {A ->[$k$] B}$$ has a half-life of $\pu{100 min}$ when carried out in a buffer solution of $\mathrm{pH=5}$ and $\pu{10 min}$ when carried out at $\mathrm{pH=4}$. Both the times the half-life are independent of the initial concentration of $\ce{A}$. Find its Rate Law.

I am planning something like this:
Let $$\text{Rate} = k [\ce{A}][\ce{H+}]^x\tag1$$
Now let:
$$k'=k[\ce{H+}]^x\tag{constant}$$
And then:
$$k' t_\frac 12 = \ln2$$

Is it conceptually correct?

Note:

I have found tons of solutions over the internet all with different answers.

The answer given in my book is x=1 (x is same as is used in eq(1)) and I found the same myself but some sites give answer of the same question as x=2.

We can do this question in two steps, finding the rate law with respect to both A and $\ce{H+}$ individually.
$$\text{Rate}=[A]^y[\ce{H+}]^x$$

The first part is simple, they've given that the half life is independent of $[A]$, therefore it is first order with respect to $\ce{A}$.
Therefore as you've found,
$$\text{Rate}=[A][\ce{H+}]^x$$

Now, onto finding the order with respect to $\ce{H+}$.
From what I've seen on the contradictory answers on the net that OP has mentioned, they state that since the rate is inversely proportional to concentration of $\ce{H+}$, it is second order with respect to $\ce{H+}$ by the relation:
$$t_{1/2}\, \alpha\, [A]^{1-n}$$
where $n$ is the order with respect to A in a reaction.
Is this true? Yes. Is it correct in this context? No. This is because this is true only when the reactant in question actually decreases in the mixture, i.e.
$$-\frac{\mathrm d[A]}{\mathrm dt}=k[A]^n$$
Here, it is specifically given that it is in a buffer of $\mathrm{pH} = 4$. This means that there is no change in the actual concentration of $\ce{H+}$.
Therefore, what you've done is conceptually sound and is the correct method.
$$k'=k[\ce{H+}]^x = \text{constant}$$
\begin{align}
\frac{t_{1/2}(\mathrm{pH}=5)}{t_{1/2}(\mathrm{pH}=4)} &= \frac{k'(\mathrm{pH}=4)}{k'(\mathrm{pH}=5)} = \frac{k[\ce{H+}]^x(\mathrm{pH}=4)}{k[\ce{H+}]^x(\mathrm{pH}=5)} \\
\implies \frac{100}{10}&=\left(\frac{10^{-4}}{10^{-5}}\right)^x = 10^x \\
\implies x &=1
\end{align}
Therefore, the rate would be:
$$\text{Rate}=k[A][\ce{H+}]$$


Hydrogenation side product: potential mechanisms for C-C bond making In the analysis of the product mixture of a new catalytic hydrogenation  reaction I'm trying out I am getting this side product molecule (in low quantities; less than 1% w/w).

Is this a likely / explainable side product? What would be a potential mechanism?

I'm skeptical because the C8 increases to a C9. I assumed C-C bond making wasn't a very common occurrence in this sort of reaction? Or is it?

The reaction conditions are fairly mild: 10 barg & 80 C. Somehow I associated this sort of hydrogenation with functional group identity changes & not Carbon-backbone changes.

Or am I wrong?

More info. Solvent is Methanol. pH is alkaline by small qty. of NaHCO3 initially added. Catalyst is Pd on C.

Edit: Supposing the MeOH solvent reacts with the starting material, would this be the reaction to expect?

Any mechanism to go from this alcohol-ether to the impurity I'm seeing?

First of all, I would check for impurities in styreneoxide. There is a good chance it was prepared from styrene having some 2-phenylpropylene in it, and this could account for your impurity.

Other than that, I would think among lines that hydrogenation is usually reversible, so above strong hydrogenation catalys phenylethanol is in equilibrium with corresponding aldehyde and methanol is in equilibrium with formaldehyde. Crotone condensation of said carbonyl compounds with subsequent hydrogenation may account for the impurity you see here.


Change in Gibbs free energy of vaporization reaction In my chemistry book, I see the equation

$$\Delta G_{rxn} = \Delta G_{rxn}^0 + RT \ln(Q)$$

where $\Delta G_{rxn}^0$ is the change in Gibbs free energy under standard conditions (1 atm pressure in this case), R is the gas constant, T is the absolute temperature, and Q is the reaction quotient. Consider the reaction

$$\ce{H2O (l) -> H2O (g)}$$

Clearly in this example, $Q = P_{\ce{H_2 O}}$, the pressure of water vapor.

Here is my confusion. The earlier expression for $\Delta G_{rxn}$ implies that anytime $P_{\ce{H_2 O}} = 1\ \mathrm{atm}$, we have $\Delta G_{rxn} = \Delta G_{rxn}^0$, which is $8.59 \mathrm{kJ/mol}$ for this reaction. We should see this regardless of temperature, according to this expression.

However, when $T = 100\ \mathrm{^\circ C}$, liquid water and water vapor are in equilibrium at $P_{\ce{H_2 O}} = 1\ \mathrm{atm}$! And if they are in equilibrium, it stands to reason that we should see $\Delta G_{rxn}=0$, which clearly we do not.

Could anyone help me understand this apparent contradiction? My only guess at the moment is that it arises because $\Delta G_{rxn}^0$ is not temperature-independent and at the higher temperatures, perhaps the value of $8.59\ \mathrm{kJ/mol}$ becomes inapplicable. If anyone could explain this to me I would be very grateful!

Yes. You are correct. $\Delta G^0_{rxn}$ is a function of temperature, and is equal to zero at 100 C for this reaction.


Can we dissolve salts with higher lattice enthalpy by providing heat? It is well-known that

A salt with a higher lattice energy will not dissolve at all, because then $\Delta_{\text{sol}}H^\ominus$ will be negative.

So, the reaction should be endothermic, right? My question is:
If then we provide heat to the reaction, then we should be able to dissolve such salts. Is this true?

Heat itself it just indirect factor. It is temperature what determines the shift of solubility products. If the dissolution is endothermic process, the salt solubility product would increase with higher temperature.
But if the solubility product is still small even at high temperature, like e.g. $\pu{e-16}$ instead of $\pu{e-20}$, then only a tiny extra salt amount would dissolve when heated.
The effective result depends on the particular case. The typical cases where significant solubility increase is observed are  potassium salts  $\ce{KNO3, KClO3, KClO4}$. But they have limited solubility even at low temperature - see the solubility table.
Generally, the dependence of an equlibrium constant of a reaction, including dissolving/precipitation process, is described by van't Hoff equation

$$\frac{\mathrm{d}}{\mathrm{d}T} \ln K_\mathrm{eq} = \frac{\Delta H^\ominus}{RT^2}$$

For solubility products, being equlibrium constants, is valid the following reaction:
$$\Delta G^{\circ}_\mathrm{dis} = \Delta H^{\circ}_\mathrm{dis}  - T \Delta S^{\circ}_\mathrm{dis} = RT \ln K_\mathrm{sp}$$


Confused about carbocation stability For which element 'X' is the carbocation $\ce{CH3-X-CH2+}$ the most stable?

A. X = S

B. X = N

C. X = O

I thought since oxygen is the most electronegative among the options, it would create the greatest electron deficiency, which would in turn facilitate a lone pair donation from the O atom (which happens with boron halides), so the answer should be oxygen. However the answer given is nitrogen. I really can't understand why. Is my logic wrong? Or is there more to this question?

A carbocation is electron deficient. In an electron-neutral state, carbon would have four valence electrons that form four bonds resulting in an overall octet structure. To form a carbocation, one of these bonds needs to be removed, resulting in a sextet on the carbon atom; and one could also consider carbon now only having three valence electrons that form three bonds. Whichever picture we choose, carbon is lacking electrons.

(A more true picture would, of course, involve orbitals and their interactions, but still leave us with the conclusion that there are too few electons around the carbon atom.)

The three possible neighbours differ, amoung others, in their electronegativity — their ability to draw electrons towards them, away from what is next to them. Oxygen is the strongest, nitrogen second strongest and sulphur is the weakest.

In the case of the carbocation, I have already shown that it is electron-deficient. If we place a more electronegative element next to it, that element will draw the electrons away from the cation even more; this makes the carbon even more electron-deficient and thus destabilises the system. On the other hand, if a more electropositive element such as silicon were neighbouring the carbon, it would push electrons towards the carbon — exactly what the cation wants since it’s lacking electrons.

By this reasoning, sulphur, being the least electronegative element on the list should best stabilise the carbocation in question. But the answer gives nitrogen; so there must be a second effect. This would be mesomeric stabilisation.

For all three elements, one can draw a resonance structure where instead of a single $\ce{C-X}$ bond and a lone pair on $\ce{X}$ one has a $\ce{C=X}$ double bond shifting the positive charge to $\ce{X}$. This resonance structure is considerably less prevalent with sulphur (a third period element) than with oxygen or nitrogen; the reason being the spacial distribution of sulphur’s p-orbitals not fitting too well with carbon’s. Oxygen and nitrogen, on the other hand, have p-orbitals of a similar size and distribution so mixing works well. (Yes, this is also more complex on a more realistic level, but this visualisation works well enough for the simple case here.)

The question is now, whether the resonance stability provided by oxygen is stronger or weaker than that of nitrogen and if the difference is larger or smaller than the energy difference caused by the different electronegativities discussed earlier.

Again arguing that oxygen is more electronegative and thus draws electrons towards it, it would equally be less happy to share its unbonded electron pair with carbon to relieve carbon’s electron deficiency. Thus, one may expect nitrogen to provide a greater stabilisation to a neighbouring carbocation.


Why must a buffer solution contain both a weak acid and a salt solution of its conjugate base? Doesn’t the weak acid dissociate to produce the conjugate base on its own?

I understand that when a strong base is added to the buffer solution, the hydroxide ions will react with the hydrogen ions formed via the dissociation of the weak acid. The equilibrium position will shift rightwards to replenish these hydrogen ions, ensuring that few hydroxide ions remain in the solution.

Conversely, when a strong acid is added to the buffer solution, the hydrogen ions will react with the conjugate base to form the weak acid molecule, thereby ensuring that most of the additional hydrogen ions are removed from the solution, and the pH of the solution remains stable. In the literature I’ve been reading so far, it is mentioned that a salt solution containing the conjugate base ensures that there is a continuous supply of the ions, but can’t the ions simply be replenished by the dissociation of more weak acid molecules, as in the previous case?

You are correct that a weak acid will only partially dissociate in water. Taking for example the weakly acidic salt $\ce{NH4Cl}$ of which the ammonium cation is acidic, in water we get reaction (1).

$$\ce{NH4+ + H2O <=> NH3 + H3O+}\tag{1}$$

To analyse this reaction we need the equilibrium constant which in the case of acids is defined as an acidity constant $K_\mathrm a$ as shown in (2). (Wether we use $\ce{H+}$ or $\ce{H3O+}$ is irrelevant.)

$$K_\mathrm a = \frac{[\ce{NH4+}][\ce{H+}]}{[\ce{NH3}]}\tag{2}$$

Using the tabulated $\mathrm pK_\mathrm{a}$ value of 9.25, we can calculate the amount of $\ce{NH3}$ in a $\pu{1M}$ solution of $\ce{NH4Cl}$:

$$\begin{align}K_\mathrm a &= \frac{[\ce{NH4+}]^2}{[\ce{NH3}]}\\[0.3em]
K_\mathrm a (c_0 - [\ce{NH4+}]) &= [\ce{NH4+}]^2\\[0.5em]
[\ce{NH4+}]^2 + K_\mathrm a [\ce{NH4+}] - K_\mathrm a c_0 &= 0\\
[\ce{NH4+}] &= \frac{-K_\mathrm a \pm \sqrt{K_\mathrm a^2 + 4 K_\mathrm a c_0}}{2}\\
[\ce{NH4+}] &= \pu{2.37e-5 M}\tag{3}\end{align}$$

Whoops. We may have a good concentration of ammmonium chloride but only a very minor amount of it is still ammonium, $99.998~\%$ are conjugate base. That doesn’t look like a good buffer system. If you were to add an acid, that would be fine since the excess base can capture that. But if you add more base the pH will go up rapidly, because there are no ammonium ions in solution to buffer it.

Only if you add both a significant amount of an acid and its conjugate base will you have significant amounts of both compounds in solution which are then able to buffer by capturing excess protons or hydroxides.


How does acidity affect electrode potentials? What is the role of acidity in the decrease of formal potential of $\ce{Fe^{3+}$/$Fe^{2+}}$ in a sulfuric acid and phosphoric acid mixture?

I've observed that the formal potential of the $\ce{Fe^{3+}$/$Fe^{2+}}$ electrode in a mixture of sulfuric acid and phosphoric acid, $E_0 = \pu{+0.61V}$ is much lower than the standard electrode potential, $E_0 = \pu{+0.76V}$.

I came up with a plausible explanation that phosphoric acid would cause $\ce{Fe^{3+}}$ to get precipitated as  $\ce{FePO4}$ and the presence of excess acid will help in the formation of a complex $\ce{[FeHPO4]+}$. Am I correct in my observation? How would the situation change if the medium was basic?

There are numerous reactions that involve iron and many involve either $\ce{H+}$ and $\ce{OH-}$. Take a look at this table and search for "$\ce{Fe}$". The non-standard potential for each one depends on the concentration of the species involved, and for the ones with hydrogen or hydroxide ions this means it includes the pH. This is described by the Nernst Equation. The potential varies as the natural log of the concentration, so you would add or subtract the pH or pOH, depending on the half reaction in question.


Can Ca and K exist as an ionic compound? Recently in the Chemistry.SE chatroom a user posted this to celebrate the New Year:

Can Ca and K exist as an ionic compound? If not, why is it impossible?

  Ca with atomic number 20, K 19
  so 2019 ->Happy New Year->2020

As a creator of this memetic illustration, I feel obliged to post a brief answer.

Potassium is known for forming several metal alloys, such as a widely used Sodium-potassium alloy NaK.
However, CaK doesn't exist [1]:

  The Ca-K system was found by Klemm and Kunze [67Kle] to exhibit virtually complete immiscibility in both solid and liquid states. No compounds were found by X-ray diffraction.

  Cited Reference
  *67Kle: W. Klemm and D. Kunze, "Systems of Alkali and Alkaline Earth Metals," Proc. Int'l. Symp. on Alkali Metals, London Chem. Soc., Special Publ. No. 22, 3-22 (1967). (Equi Diagram; Experimental; #)

References

Pelton, A. D. The Ca−K (Calcium-Potassium) System. Bulletin of Alloy Phase Diagrams 1985, 6 (1), 34–34. DOI: 10.1007/BF02871178.


Ambiguity in definition of primary valency in Werner's theory In Concise Inorganic Chemistry (5th Edition) on page 196 , the primary valency of $\ce{[Co(NH3)5Cl]Cl2}$ is given to be 2.

In other textbooks and in the question here, we get the idea that:

The primary valency of a compound is equal to its oxidation state.

In 'Inorganic Chemistry(4th Edition)' by Catherine E. Housecroft and Alan G. Sharpe':

Werner referred to the oxidation state of the metal ion as its ‘primary valence’ and to what we now call the coordination number  as  its  ‘secondary  valence’.

This further suggests that that they are in fact exactly identical and just a matter of how they are referred to.
The oxidation state of cobalt in $\ce{[Co(NH3)5Cl]Cl2}$ is +3 and hence would suggest a primary valency of 3.
What is the correct primary valency of the compound? I am aware that these concepts are not in modern use yet it would be useful to know how Werner originally put forward this idea.

George B. Kauffman has edited an excellent book, "Coordination chemistry: A century of progress." It was published as an ACS Symposium Series. In the chapter, he writes, "If history teaches us anything, it teaches us that the latest view is not always the best and that change is not always progress." I may add that there is no authority in science. I am glad that you challenged textbooks.
Basically, in Werner's time, the prevalent theory of the complexes was that the complexes are chained compounds, which was not consistent with the solution conductivity and optical isomerism of the complexes. Werner proposed the primary and secondary valence concept.
Contrary to most modern chemistry books, including that respectable Alan G. Sharpe's texts, Werner's primary valence matches the "oxidation" state of the metal in simple cases. The oxidation state matches Werner's primary valence when there is no anion in the coordination sphere. The textbook version of Werner's story fails for the type of compounds you just asked, such as $\ce{[Co(NH3)5Cl]Cl2}$.
Here the oxidation state is indeed 3. However, Werner's primary valency of this compound is 2.
Kauffman mentions

Werner postulated two types of valence—Hauptvalenz, primary or ionizable valence, and Nebenvalenz, secondary or nonionizable valence. According to Werner, every metal in a particular oxidation state, that is, with a particular primary valence, also has a definite coordination number, that is, a fixed number of secondary valences that must be satisfied. Now, whereas primary or ionizable valences can be satisfied only by anions, secondary or nonionizable valences can be satisfied not only by anions but also by neutral molecules containing donor atoms such as nitrogen, oxygen, sulfur, and phosphorus.

The keyword is that primary valence is the ionizable valence. The ionizable valence of the complex in question is 2. $\ce{[Co(NH3)5Cl]Cl2}$ would generate three ions in solution $\ce{[Co(NH3)5Cl]^{2+}}$ and two $\ce{Cl-}$. These experiments were easily doable in Werner's time by conductivity measurements. Secondly, as discussed in the comments, argentometric titrations would show only two chloride ions rather than three.


Chelation and its relation to transition metal ligands Can someone explain chelation and its relation to monodentate and tridentate ligands? I dont understand what they mean by chelate effect and chelating rings.

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

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


Effect of solutes on surface tension What happens to the surface tension of a solvent, say, water when we mix a solute in it?
For the sake of concrete discussion, the examples that I encountered in a test were sodium chloride $(\ce{NaCl}),$ methanol $(\ce{MeOH}),$ and sodium alkyl sulphates $(\ce{ROSO3Na}).$

My guess is that since the surface tension can be defined as the force per unit length on a line drawn on the surface of the liquid, if the solute attracts $\ce{H2O}$ more than another $\ce{H2O}$ molecule, then the surface tension should increase.

Using this logic, I predict the following:

Addition of methanol should decrease the surface tension because $\ce{MeOH}$ forms lower number of hydrogen bonds than pure water.
Sodium chloride addition should increase the surface tension because ion dipole interactions are stronger than dipole dipole interactions or H-bonds.
The $\ce{ROSO3Na}$ kind of confused me. I assume they intended for me to perceive it as detergent and I know that decreases the surface tension. They actually asked to recognize the appropriate plot of variation in surface tension with concentration and the graph I guessed attained a constant value after a certain concentration. I thought it might be due to micelle formation.

If anyone can confirm or deny my intuition, or point me in the direction of any data that would be helpful.

Davies and Rideal1 published the following graph. This graph nicely describes the action of various solutes on a solution.

It's clearly visible that:

Addition of $\ce{MeOH}$ reduces the surface tension nonlinearly
Addition of $\ce{NaCl}$ increases the surface tension linearly, but by a small amount
Addition of a surfactant such as $\ce{RSO3Na}$ decreases the surface tension drastically, up until Critical Micelle Concentration is reached.

Smith and Gillham2 present a more detailed graph of the effect 1-butanol has on the surface tension of water.

References:

Hansen, Robert S. “Interfacial Phenomena (Davies, J. T.; Rideal, E. K.).” Journal of Chemical Education, vol. 39, no. 7, 1962, p. A552. doi:10.1021/ed039pA552.
Smith, James E., and Robert W. Gillham. “Effects of Solute Concentration-Dependent Surface Tension on Unsaturated Flow: Laboratory Sand Column Experiments.” Water Resources Research, vol. 35, no. 4, 1999, pp. 973–82. doi:10.1029/1998WR900106.


How to convert wedge–dash structure into Fischer projection? I am very confused when a compound with more than one chiral centre is given. How to find whether the two compounds are enantiomers or diasteromers?

Please also explain how to convert more than one chiral compound with wedge–dash structure to Fischer projection.

Method 1

Method 2

Quoting from

http://oscar.iitb.ac.in/OSCARPP/Chemistry/UploadedStoryboards/SCH023_3P0431Wedge-dash%20to%20Fischer.ppt

The following is conversion of molecules with two chiral centres from wedge to fischer projection yhrough a series of operations as shown below.

Reference :

http://oscar.iitb.ac.in/OSCARPP/Chemistry/UploadedStoryboards/SCH023_3P0431Wedge-dash%20to%20Fischer.ppt


Why does Ammonium sulfate cause proteins to precipitate? I'm trying to understand an observed reaction from a chem practical:

You have a solution of ammonium sulfate, which you add to a bovine serum albumin solution. The proteins precipitate, and the solution becomes cloudy. w/v % of $\ce{(NH4)2SO4}$ at this point is 57%.

When you add water (in a ratio of 2 parts water: 1 part serum solution)to the mix, the process reverses, and the solution becomes clear again.

This can be explained by the ammonium sulfate solution 'stealing' water from the proteins, which causes them to precipitate, and then the process reversing once more water is added. (Right?)

Question: Does the ammonium sulfate 'steal' the water because it wants to reach equilibrium, or because it's supersaturated, or what?

Well, I wouldn't know how to sum it up better than chemwiki.ucdavis:

  To achieve this conformation the molecule folds in such a way that all of the hydrophobic parts of a molecule are aggregated together and the hydrophillic groups are left to interact with the water. In the case of proteins it is the charged amino acids that allow selective salting out to occur. Charged and polar amino acids such as glutamate, lysine, and tyrosine require water molecules to surround them to remain dissolved. In an aqueous environment with a high ionic strength, the water molecules surround the charges of the ions and proteins. At a certain ionic strength, the water molecules are no longer able to support the charges of both the ions and the proteins. The result is the precipitation of the least soluble solute, such as proteins and large organic molecules.

If you ask what the aminoacid content of bovine serum albumin is, I'll cite:

  Glycine,
  illanine,
  Valine,
  Leucine,
  Isoleucineh,
  Proline,
  Phenylalanine,
  Tyrosine,
  Tryptophan,
  Serine,
  Threonine,
  Half-cystinee,
  Methionine,
  Arginine,
  Histidine,
  Lysine,
  Aspartic acid and
  Glutamic acid. Source


Is p-bromophenol more acidic than p-chlorophenol? p-chlorophenol is more acidic than p-fluorophenol due to presence of vacant d orbitals. But then is p-bromophenol more acidic than p-chlorophenol since it has vacant f orbitals also? But I guess that’s wrong, isn’t it? Please give your suggestions.

Disclaimer: This is not necessarily the "correct" answer as many factors play roles and it would be difficult to describe the observation so simply.

I'm not sure I can provide an easily "correct" answer, but I can try to explain some of the many factors that influence the acidity.

Most notably, the factors we should consider are inductive effects and delocalization of charge.

The inductive effects, based on electronegativity, would dismiss the experimental observation, so I'll look at delocalization of charge.

The delocalization of charge, which introduces a negative charge by the halogen is greatest by F>Cl>Br>I. This would reduce acidity, as a more negative benzene ring would not desire to gain another negative charge. Thus, this is one factor that supports the experimental observation.

Thus, we may be able to explain the observation, if only partially, by the delocalization of charge.

I adapted the answer to a similar question on http://www.quora.com/Organic-Chemistry/Which-is-more-acidic-ortho-fluorophenol-ortho-chlorophenol-ortho-bromophenol-or-ortho-iodophenol-Why . I would have preferred more sources but I could not find a very good source.


Why could it be assumed that the standard enthalpy change is independent of temperature down to 25 °C? For the reversible reaction of pyruvic acid to 2,2-dihydroxypropanoic acid, we calculated enthalpy using NMR and extrapolated from area (integration) and peaks etc.

We then plotted $\ln K$ vs $1/T.$ I am confused by this questions because I thought enthalpy was temperature dependent.

I guess my question was unclear. The exact question we are asked is:

  With the assumption that the standard enthalpy change is independent of $T$ down to 25 °C, calculate an extrapolated value for $K$ (25 °C) and compare this to your experimental $K$ value determined at 25 °C. Is this a fair assumption? Why?

I do not understand if this assumption is fair and why?

Generally speaking you can write that
$$\Delta H^\circ(T) = \Delta H^\circ(T_\mathrm{ref}) + \int_{T_\mathrm{ref}}^T \Delta C_p \,\mathrm dT \tag{1}$$
where $T_\mathrm{ref}$ might be $\pu{25 ^\circ C}$ and the pressure is 1 bar (in your case, approximately 1 bar).
If your temperature window is not too large you may get away with assuming $\Delta H^\circ$ is constant. This amounts to assuming that the second term in the equation, dependent on $\Delta C_p$, is small compared to the first. If it isn't then rather than using the integrated van't Hoff equation

$$\ln\left(\frac{K_2}{K_1}\right)=\frac{-\Delta H^\circ}{R}\left(\frac{1}{T_2}-\frac{1}{T_1}\right) \label{eqn:2}\tag{2}$$
maybe you should attempt a polynomial fit to $\ln K_\mathrm{eq}$ and taking the derivative of the polynomial wrt $T^{-1}$ to evaluate $\Delta H^\circ$ as a function of $T$.

To test whether \eqref{eqn:2} is appropriate you might evaluate the quality of a fit to that equation or check the linearity of a plot of $\ln K$ versus $T^{-1}$.