Is it possible for three carbon double bonds to form next to each other in a carbon chain? In a typical carbon double bond $sp^2$ hybidrization is used to explain the resulting bond. When two carbon atoms A and B form a double bond one $sp^2$ orbital from A bonds to another $sp^2$ orbital from B and one $p$ orbital from A bonds to a $p$ orbital from B. This leaves two $sp$ orbitals sticking out the end of the ethene-like molecule. I think one could stick two ethene-like molecules end to end to stick together with two $sp^2$ orbitals from each end but then I think one would have an interfering $p$ orbital that would subtract and leave a bond order of 1. I think one might then have further hybridization and end up with three 1.67 bonds (instead of one single and two double bonds) but not three double bonds. The second possibility is sticking together two $sp^2$ orbitals and one $p$ orbital from A to two $sp^2$ orbitals and one $p$ orbital from B leaving behind one $sp^2$ bond sticking out on each end. I think one could stick together such pair with another pair rotated at an angle. After hybridization I think one ends up with three net 2 bonds together but this is confusing to me. A third possibility is sticking together two ethyne-like molecules together but I think that would be a hybrid of two 3-bonds and a -1 bond which would be 1.67 again.

There's two different ways to stick double bonds next to each other.

The first is to put the two double bonds right next to each other, the "cumulenes":

Image from Wikipedia Commons

In these compounds, the pi orbital of one double bond does NOT hybridize/delocalize with the pi orbital of the adjacent double bond. For example, in the structure on the far left, the three carbon propadiene, the two end carbons are hybridized $\ce{sp^2}$, and the center carbon is hybridized $\ce{sp}$. The remaining two $\ce{p}$ orbitals of the center carbon participate in the double bonds, one to each. But because the two $\ce{p}$ orbitals are perpendicular to each other, there's no communication between the two pi bonds - they're perpendicular to each other, too. One of them is in the plane of the screen, and the other is in the plane going in/out of the screen. (Or, it is in the simplest model of organic bonding that's taught. Once you get into things like molecular orbital theory things get more complicated, but the fact of two non-interacting orbitals remains.)

You can see this by looking at where the hydrogens are located in propadiene.

There's another way double bonds can be combined, and that with an interveneing single bond, the "polyenes":

Image from Wikipedia Commons

In a polyene, all the carbons are $\ce{sp^2}$ hybridized. Unlike a cumulene, the pi bonds of a polyene are all on the same "side" of the polyene. For example, in the polyene shown here, all the pi bonds are sticking in/out of the screen.

This means that you can draw resonance structures where the double bonds "travel" along the carbon chain, effectively swapping the double and single bonds. Like benzene, the distinction between double and single bond starts to break down, and it's more correct to think of a continous pi bond throughout the whole of the polyene chain.

But unlike benzene, the end effects for the bond-swap resonance form means it's less stable than the "normal" bond locations. This means it contributes less to the hybridized form. So while the polyene chain doesn't behave entirely like an alternating single/double bond form, it's not like it's really a "1.67" or "1.5" bond order, either.


How does initial rate of reaction imply rate of reaction at any time? My textbook begins the Chemical Kinetics section by describing the "Initial Rates Method" of determining the rate of reaction. I understood it as the following (for a first order reaction):
$$\ce{A -> B+C}$$
$$\text{rate}_\mathrm{i}=k[\ce{A}]_\mathrm{i}$$

The subscript $\text{i}$ shows that it is the initial rate. This I agree with, as it is suggested by experimental data. However, on the next page, the textbook makes a leap and (implicitly) states that this relationship holds at any time:
$$\text{rate}=-\frac{\mathrm{d}[\ce{A}]}{\mathrm{d}t}=k[\ce{A}]$$
This becomes a differential equation, the justification for which the textbook did not provide. How do we know that the change of concentration with respect to time depends on the concentration itself? I don't see how this can be assumed from the initial rates method.

First things first

When we talk about the rate of a reaction, we mean the rate at which the products appear. This means that it is also equal to the disappearance of the reactants. So, when we quantify the rate in terms of reactants, they must be in opposing magnitude. $$\text{rate }= \frac {\Delta [\text{product}]}{\Delta t}=-\frac{\Delta [\text{reactant}]}{\Delta t}$$

From A to B (and C)

Suppose that I am running the reaction $\ce{A ->B +C}$ in two separate vessels, flask 1 and flask 2, each with a different concentration of $\ce{A}$. We'll say that flask 1 is more concentrated. If you agree that the initial rate is given by $\text{rate}_\mathrm{i} = k[\ce{A}]_\mathrm{i}$, then it follows that for each the initial rate is $\text{rate}_{i,fl1} = k[\ce{A}]_{i, fl1}$ and $\text{rate}_{i,fl2} = k[\ce{A}]_{i, fl2},$ respectively.

If we were to add something to flask 2, would it change the reaction rate? As long as we didn't add any $\ce{A},$ the rate would be the same. The rate is dependent on $\ce{A}$ but nothing else. So let's add a little bit of $\ce{B}$ and a little bit oc $\ce{C}$. We'll add the same amount of both. Let's call that amount $x$ and set it equal to the difference in concentrations in $\ce{A}$ in the two flasks. $$x=[\ce{A}]_{i,f1}-[\ce{A}]_{i.f2}$$

Now let's let the reaction in flask 1 proceed until the concentration of $\ce{A}$ is equal to the amount in flask 2. To distinguish the conditions now from the initial conditions, we'll call this time $t=1$ (arbitrary units). By the stoichiometry of the reaction, we know the amount of $\ce{A}$ and $\ce{A}$ will be $\Delta[\ce{A}] = [\ce{A}]_{i,f1}-[\ce{A}]_{i,f2}=x$ There is no perceivable difference between the two flasks, and the reaction rate is thus the same in flask 1 as in flask 2. Because we have chosen arbitrary concentrations, we can abstract it to any concentration. $$\text{rate}_{t=1,f1}=\text{rate}_{i,f2}=k[a\ce{A}]_{i,f2}=k[\ce{A}]$$

Finally, we want the instant rate. We do this by decreasing the time interval to an infinitesimal amount. $$\lim_{\Delta t \to 0}\frac{-\Delta[\ce{A}]}{\Delta t}=-\frac{\mathrm{d} [\ce{A}]}{\mathrm{d} t}=k[\ce{A}]$$


Nuclear zero-point energy and quantum delocalization effects of water molecules What do we mean by "Nuclear zero-point energy" and "quantum delocalization effects" of water molecules ? It's said that without these effects "water" becomes somewhat unbounded, which is not very well understandable to me. Can you please help me to clear my doubts.

So, what was actually said in the paper1 with respect to AIMD simulations of water is the following:

  Nuclear zero-point energy and quantum delocalization effects are neglected making classical water overbound.

To understand the meaning of this statement we first need to know how AIMD works. Ab Initio Molecular Dynamics (AIMD) is a molecular dynamics simulation in which potential energy surfaces for nuclear motion are obtained from first principles (ab initio) rather than represented by a force field. The quantum chemical calculation of potential energy surfaces are done in the Born–Oppenheimer approximation in which nuclei are considered to be stationary classical particles.

As its very name signifies, it is, of course, just an approximation. It ignores that nuclei move and aren't perfectly localized in space. The nuclear motion can be approximately separated into vibrational and rotational the first of which energetically dominates the second. Taking the vibrational motion on the nuclei into account results in some additional so-called zero-point vibrational energy that rises the total energy. Other quantum effects (tunneling, delocalization) can be also taking into account for the nuclei which results in some further changes in energy.

Altogether, the more and more quantum effects are taking into account, the closer are the results of the simulations to the physical reality. AIMD simulations ignore few quantum effects, and thus, do not quite match the reality. For instance, water tends to be overbound in such simulations comparing to what we have in the real world. That is, essentially, the meaning of the statement quoted above.

1) A.D. Hammerich, V. Buch, F. Mohamed, Ab initio simulations of sulfuric acid solutions, Chemical Physics Letters, 2008, 460, 423-431, http://dx.doi.org/10.1016/j.cplett.2008.06.053.


Does the synthesis of beta-keto phosphonates from esters have a name? Consider the Horner–Wadsworth–Emmons reaction; one of its reactants is a phosphonate, usually stabilised by a β-carbonyl group which is then deprotonated in α position (figure 1). These phosphonate esters can be synthesised in a Michaelis–Arbuzov reaction from a phosphorous acid triester and an alkyl halide (figure 2). Another viable way is to start from diethyl ethylphosphonate, deprotonate that (e.g. with n-BuLi) and let it attack a carboxylic ester much like a Grignard reagent would (figure 3).

Figure 1: Horner–Wadsworth–Emmons reaction. Image taken from Wikipedia, originally created by Wikipedia user ~K.

Figure 2: Michaelis–Arbuzov reaction. Image taken from Wikipedia, originally created by Wikipedia user ~K.

Figure 3: Alternative synthesis of HWE phosphonates. Image self-drawn.

Since about every reaction in organic chemistry has a name of some kind, is the reaction in figure 3 also a named reaction, and if so, what is the reaction’s name?

My inclination is to call this reaction a "phosphono-Claisen condensation." A  Claisen condensation is the alpha-acylation of an ester with another ester. In this case, the typical carbon-centered ester is replaced with a phosphonate, so phosphono-Claisen condensation specifies that difference.  Avoiding the proper name, "alpha-acylation of a phosphonate" is a suitable description of the reaction.

Montchamp and coworkers published a paper describing the "phospha-Claisen condensation" reaction, which they limit to the condensation of two phosphonates to give a 1,1-diphosphono compound.

The website maintained by the venerable Hans Reich describes this reaction as "Claisen-like":

  (MeO)2P(O)CH2Li will also react with carboxylic acid derivatives like
  esters, lactones and amides in a Claisen-like reaction to form keto
  phosphonates.

The Organic Chemistry Portal describes the reaction as "the condensation of esters and phosphonates."

A brief survey of the literature featuring this reaction type (Thanks SciFinder!) doesn't show any convention for naming this reaction beyond calling it the acylation of a phosphonate. In my SciFinder searching, the earliest usage of this reaction is from 1970, and that paper cites a Corey paper from 1966. The latter paper doesn't have a proper experimental on the reaction but contains the note:

  Anions derived from 1 (X = H or alkyl) are of value in the synthesis
  of keto phosphonates 1 (X = RCO) by reaction with esters. The
  preparation of these intermediates, which are of value in olefin
  synthesis [see H. Takahashi, K. Fujiwara, and M. Ohta, Bull. Chem.
  Soc. Japan, 35, 1498 (1962)], is usually impractical by the
  Michaelis-Arbusov reaction.

It's a little nebulous who performed this chemistry first, but perhaps someone could have christened it the "Corey-Claisen condensation" or the "Corey variant of the Claisen condensation". Of course there is already a Corey-Claisen rearrangement so this would be really confusing.


How to pool fractions after chromatography? I used chromatography to purify my protein and now I have fractions with different concentrations ranging from 1-16 mg/ml. Should I pool all these fractions or just the ones with high concentration (where under the actual peak)?

Thanks in advice

If all fractions contain the same protein, I'd combined all and concentrated the combined solutions to desired concentration using biochemical techniques such as dialysis. If they contained more than one protein, then I'd combined all fraction collected under the correct peak. However, it is better ask your supervisor (or adviser) before you combined them. The person who wrote the protocol knows which one to collect. Yet, you can look at following reference to get an idea how to concentrate the combined protein solutions:

Sarah M. Andrew, Julie A. Titus, Louis Zumstein, "Dialysis and Concentration of Protein Solutions," Current Protocols in Immunology 1997, 21(1), A.3H.1-A.3H.5 (https://doi.org/10.1002/0471142735.ima03hs21).


Electrolysis of water for demonstration: ideal electrolyte for safety, stability, conductivity, and ease of maintenance I'm developing a system for educational purposes where electrolysis of water takes place. The aim is to practically show how electrical energy can be converted to mechanical, by electrolyzing water, igniting the mixture of $\ce{H2}$ and $\ce{O2}$ produced, and accelerating a mass upwards.

Currently I'm using citric acid in distilled water ($4\%$ in weight, $18~\mathrm L$ water), platinum-coated mesh electrodes, $500~\mathrm W$ max power supply ($0\!-\!65~\mathrm{V_{DC}}$, $0\!-\!10~\mathrm A$). I'm wondering if there's any better option that meets the following requirements:

Not harmful to maintainers (that have to prepare and replace the liquid) and bystanders
Stable (not producing precipitates, and in general that can minimize the need for replacing the liquid)
Low/no interaction with the materials in the tank (Plexiglas, brass, platinum, silicone tubes, ABS)
Maximizes conductivity (the current setup at $20~^\circ \mathrm C$ has a conductance of $250\!-\!300~\mathrm{mS}$)
Not suitable for microorganism development (in the current setup algae develops in about two weeks)

If the last point can't be improved with just the electrolyte, I'd kindly ask for a suggestion for an additive that is not going to spoil the points above.

I suggest $1\%$ w/w $\ce{NaOH}$. Point-by-point:

$1\%$ w/w is $\sim\!10~\mathrm{g\over L}$, which for $\ce{NaOH}$ at $40~\mathrm{g\over mol}$ is $\sim\!0.25~\mathrm{M}$, corresponding to a $\mathrm{pH}$ of $13.4$. Standard personal protective equipment (coats, gloves, goggles) should suffice for observers and maintainers.
$\ce{NaOH}$ is extraordinarily soluble, so you will have no difficulties with precipitation.  You will want to keep the system covered when not in use, if you're not already, to minimize evaporation.  You may get some riming on the interior, but rinsing with DI water should take care of it.
According to Schweitzer$^\dagger$, the materials in your system have the following resistances to $10\%$ (not $1\%$) $\ce{NaOH}$:

Plexiglas (acrylic): Resistant, to $80~^\circ\mathrm F$
Brass: Good (less than 0.02" penetration per year), to $200~^\circ\mathrm F$
Silicone elastomer: Resistant, to $80~^\circ\mathrm F$
ABS: Resistant, to $140~^\circ\mathrm F$
Platinum: ..... darn well better hold up! (Though, if there are any gaps in the coating, the substrate might be exposed to attack.)

Per the first chart here (PDF link), the conductivity of $1\%~\ce{NaOH}$ should be about $50~\mathrm{mS\over cm}$, which is over ten times the $\sim\! 4.5~\mathrm{mS\over cm}$ of your current $4\%$ citric acid recipe.
I salute any common microorganisms able to grow in a $\mathrm{pH}~13+$ solution. (See, e.g., Table 3-5 here, where the maximum survivable $\mathrm{pH}$ for any species listed is $11$.)

Obviously, if you should switch, monitor the health of the system materials to be sure the above literature chemical resistance data actually holds in reality.  There's always the chance some small component of the system is of a material you haven't listed, too.

(To note, I re-thought the vinegar-plus-sodium-bisulfate approach after discovering that both brass and acrylic/Plexiglas are poorly resistant to vinegar, and that the bisulfate is not likely to provide as much of a conductivity boost as I was thinking.)

$^\dagger$ P.A. Schweitzer. "Corrosion Resistance Tables, Part C, P-Z." 4th Ed. New York: Marcel Dekker, 1995, pp. 2685-2688. (Amazon link)


Why is it a good idea to put some salt in the bread dough? The salt that you can typically find in a kitchen is the $\ce{NaCl}$, isn't this salt supposed to act as an antiseptic?

I find a lot of recipes that suggest the addition of salt when preparing the mix for the leaven of the bread.

Isn't the salt supposed to act against the formation and the culture of bacteria that are carrying forward the leaven?

Common table salt is not necessarily an antiseptic in the conventional "sterilizing" sense. What gives it weakly antiseptic properties in biological context is its ability to absorb water from cells through osmosis, should the concentration of the salt outside the cell be too high, the cell would lose too much water through its membrane and die.

However this is strongly dependent on concentration. For example, according to this source, a concentration upwards of 20% is required to kill bacteria; at lower concentrations the salt will only hinder growth, which is most likely the reason it is added (in moderation) to bread.


General Oxidant - Reductant Battery In general, is it true that if I connect an aqueous oxidant to an aqueous reductant via a wire then a current flows along the wire? In particular, does current flow in the following scenario?

Imagine putting one end of a wire in hydrogen peroxide (a common oxidizer) and the other end inside water containing formic acid or pure carbon (reductants).

I understand that standard Energizer alkaline batteries are composed of separated $\ce{Zn}$ and $\ce{MnO2}$. Connecting the terminals allows electrons to flow from the $\ce{Zn}$ to the $\ce{MnO2}$, oxidizing/reducing them to $\ce{ZnO}$ and $\ce{MnOOH}$.

is it true that if I connect an aqueous oxidant to an aqueous
  reductant via a wire, will a current flow along the wire? In
  particular, does current flow in the following scenario?

No true. Nothing will happen as the circuit is incomplete. The oxidant or the reductant species have to see each other "face to face". Guess what is the most powerful reducing and oxidizing agent in the world? It is the cathode or the anode respectively. Consider an electrolytic cell which is calm and unstirred. The anode and the cathode are dipping in the solution. The moment you turn on the current, only the species which is next to the electrode gets oxidized or reduced. Nothing happens to the bulk right away, so physical promixity is important in electrochemistry.


The "rules" for LCAOs in Molecular Orbital Theory In our course on physical chemistry, which involves MOT, we have been taught that in the LCAO approach, the wave function for a molecule … say hydrogen ion ($\ce{H2+}$), can be approximated by a linear combination of two atomic orbitals.

From what I could gather from the very little I know about variational method is that this linear combination is going to be then optimised to give the minimum energy to find the ground state of the molecule in question. Is this correct?

Also, if we are simply using the linear combination as a test/guess function for a variational method, why should we restrict ourselves to only (say $+$ and $-$) two combinations?
That is, how does the rule "Number of molecular orbitals = Number of atomic orbitals constituting them" arise?

Also, why is there the restriction in form of the rule that "Only atomic orbitals of close energies and proper symmetries combine to give MOs"?

You are right to be questioning the validity of this method, and I congratulate you for doing so. This is actually an extremely important skill, that differentiates the best students from the rest.

The variational method, like many methods from physical chemistry, is a method of approximation (a model) to what really happens. There exist a whole hierarchy of methods for computing the molecular orbitals of molecules (which are themselves models, being the stationary states of the Schrödinger equation), including at the top end post-Hartree-Fock theory and density functional theory. These methods provide quantitative information about molecular orbitals, but also require serious computer power (which is generally unavailable to undergraduate students). However, all of the basic physics can be explored and understood with simpler models that can be solved on a few sheets of paper, such as LCAO theory, hence why we teach them to undergraduates.

As regards to your question, we do indeed optimise a linear combination. We start off by assuming that the total molecular orbital wavefunction can be approximated using a linear set of atomic orbitals:

$$|\Psi\rangle = c_1|\phi_1\rangle + c_2|\phi_2\rangle + \cdots + c_n|\phi_n\rangle$$

We then need to find the coefficients. The variational principle (also known as the Rayleigh-Ritz method) states that the coefficients that give the best approximation to the wavefunction will minimize the energy, given by

$$\mathcal{E} = \frac{\langle\Psi|\hat{H}|\Psi\rangle}{\langle\Psi|\Psi\rangle}$$

Now, the second part of your question involves how to compute the two terms in this fraction. Without going into massive mathematical detail, the LCAO method can be recast into a matrix problem rather than an integral problem using the atomic orbitals as a form of basis vector. In the simplest case (Hückel theory) we assume that they are normalized, such that the denominator is always 1 or 0 (see later). The problem now is mostly how to determine the numerator.

In brief, each element of the Hamiltonian matrix is given by

$$H_{ij} = \langle\phi_i|\hat{H}|\phi_j\rangle$$

The variational principle applied to this matrix implies that the optimized energies are the eigenvalues of the Hamiltonian matrix, and the coefficients are given by the eigenvectors. Since an $n$ x $n$ matrix has precisely $n$ eigenvalues, this implies the "number of molecular orbitals" rule. Notice that the reason for the +/- combinations is by our own design (LCAO). We could have picked more complex trial functions by assumption, but this would make computation far more difficult.

It most certainly is true that in order for there to be a significant interaction, two orbitals must be close in energy. The detailed reasons are complex, but essentially it comes down to the size of the $\langle\phi_i|\hat{H}|\phi_j\rangle$. Orbitals that are far apart in energy have small values of this term hence interact weakly. This doesn't necessarily mean they can't form molecular orbitals, however, but these effects are negligible and are not important in understanding the chemistry, so are neglected in simple models (although are often included in some of the most complex modern methods).

The reason that only atomic orbitals of the same symmetry give molecular orbitals is the overlap integral $\langle\phi|\phi\rangle$. This is the total sum (integral) of the product of the AOs. This is zero for different symmetries. This must be non-zero to give an MO. Consider a $\mathrm{2p_z}$-orbital and a $\mathrm{1s}$-orbital for instance (for illustration purposes only!). The s-orbital is spherically symmetric, with all points the same sign. The $\mathrm{p_z}$-orbital has a "dumbbell" shape, with equal areas of different signs above and below the $xy$-plane. Therefore, the total sum of the product is zero. No overlap = no interaction.


Building an average model from MD snapshots I am performing a molecular dynamics simulation, and I am taking "snapshots" of the system at regular intervals.

The aim is that of finding the preferred geometry of a molecule in solution.

Are you aware of any software capable of automating to some extent the superposition of different geometries, and maybe of building an "average" structure?

This is possible with the Python package MDTraj. Assuming you have your trajectory saved as trajectory.dcd and topology in topology.pdb:

import mdtraj as md
from mdtraj.geometry.alignment import compute_average_structure

traj = md.load('trajectory.dcd', top='topology.pdb')
average = compute_average_structure(traj.xyz)

I believe it will RMSD align the trajectory first, but you may need to specify that explicitly first.

You might also want to check out MDAnalysis as well, it may have similar functionality. I've used both packages and both are still actively maintained.


What is the effect of polar protic solvent on nucleophilicty? Question

Attempt:

I could figure out the answer correctly using simple rules governing nucleophilicity . But my doubt is how does polar protic solvent affect nucleophicity. And what would be the answer in case of polar aprotic solvent?. Please help.

Protic solvents affect nucleophilicity because they form a shell around the nucleophile via their hydrogen which has a positive partial charge and therefore interacts with it, thus the nucleophilicity will decrease. This effect is stronger for smaller nucleophiles as they can be "trapped" more easily. So taking the classical nucleophiles you obtain this order:

$$\ce{I^-} > \ce{Br^-} > \ce{Cl^-} > \ce{F^-}$$

Aprotic solvents affect nucleophilicity in a more complex way. You can sometimes read that the previous order of nucleophile strength is inversed and small nucleophiles are favored:

$$\ce{F^-} > \ce{Cl^-} > \ce{Br^-} > \ce{I^-}$$

But be careful, this is not always true. There is more consideration to take.

A "soft" anion is highly polarizable and has more electron density to give, making it a good nucleophile compared to "hard" ones which are tightly holding their electrons.
This order is not respected by some polar aprotic solvents, acetone is the only one coming in my mind.


Why are there no edge dislocations in a face-centered cubic (FCC) lattice? In crystallography, we know that the Burgers vector is parallel to the dislocation direction in a screw dislocation, and perpendicular in a edge dislocation.

Furthermore, I have in my lecture notes "FCC cannot have edge dislocations" but there is no explanation of why, aside from "in FCC, the Burger's vector belongs to the family of the dislocation direction vectors"

I don't understand what this means.

I have to stretch my memory to remember how this goes (and read a bit from here).

First, how do you define burgers vector? You take the end point of the dislocation and make a circle around it as it were a perfect lattice, the extra (or missing step) from the full loop is the burgers vector.

Now, is the burgers vector also a lattice vector? If yes, then it is (edge)dislocation. If no, then it is only a partial (edge)dislocation (partial dislocations have their own name).

Now, how does this work on a fcc-lattice case. On quick read for link I pasted, it seems that fcc-lattice cannot have edge dislocations, because the burgers vector is not a lattice vector.

You may want to go this through by yourself and check that I didn't make any mistakes.


Absorption Spectrum Using a Spectrophotometer After using a spectrophotometer, I have obtained absorbance values. To create an absorption spectrum, should I convert these values to percentage absorbance? (I've seen it in textbooks as such)

If I were to convert absorbance to percentage absorbance, would I simply multiply all values by 100%?

This is simple mathematical conversion. Firstly, you find the fraction of each absorbance value.

Let, the absorption spectrum has a three absorbance values say $A_1, A_2$ and $A_3$.

Now, find the fraction of each absorbance as per following:

fraction of $A_1 = A_1/(A_1+A_2+A_3)$

fraction of $A_2 = A_2/(A_1+A_2+A_3)$

fraction of $A_3 = A_3/(A_1+A_2+A_3)$

Now convert all the respective fraction values into percentage as per:

percentage of $A_1 = (\text{fraction of }A_1 \cdot 100)\%$

percentage of $A_2 = (\text{fraction of }A_2 \cdot 100)\%$

percentage of $A_3 = (\text{fraction of }A_3 \cdot 100)\%$


What is the reaction between piperine and concentrated sulfuric acid? Piperine was chemically characterized by reaction with acids. This was done by dropping concentrated $\ce{HNO3},$ $\ce{HCl},$ and $\ce{H2SO4}$ to a sample of piperine. Both $\ce{HNO3}$ and $\ce{HCl}$ produced yellow solutions while sulfuric acid produced a blood-red solution.

What happens to piperine as it reacted with $\ce{H2SO4}$ and why it turned blood-red? Also, what is the difference of that reaction with the reaction of piperine with $\ce{HNO3}$ and $\ce{HCl}?$

Edit: Late addition for viewer's convenience:

The piperine structure is shown below:

I think OP is confused with the colorimetric method of detection of piperine, which uses the concentrated sulfuric acid as one of reactant with chromotropic acid, 1,8-dihydroxynaphthalene-3,6-disulfonic acid (Chromotropic Acid Test; Ref.1 & 2). Accordingly, this colorimetric method, developed in 1965, relied on hydrolyzing the methylenedioxy group of piperine by chromotropic acid in the presence of concentrated sulfuric acid to formaldehyde, upon which a definite color was developed with chromotropic acid (Ref.3). Color intensity is proportional to the amount of piperine present. Water inhibits color development, but a small amount of methyl alcohol is necessary.

Note: It is good to know that chromotropic acid test is very sensitive to presence of formaldehyde and detect them in very low concentrations (Ref.4). It is also fact that chromotropic acid is specific to formaldehyde (Ref.2).

References:

Leila Gorgani, Maedeh Mohammadi, Ghasem D. Najafpour, Maryam Nikzad, “Piperine—The Bioactive Compound of Black Pepper: From Isolation to Medicinal Formulations,” Comprehensive Reviews in Food Science & Food Safety 2017, 16(1), 124-140 (https://doi.org/10.1111/1541-4337.12246).
Clark E. Bricker, Hilding R. Johnson, "Spectrophotometric Method for Determining Formaldehyde," Ind. Eng. Chem. Anal. Ed. 1945, 17(6), 400–402 (https://doi.org/10.1021/i560142a021).
Horace D. Graham, “Quantitative Determination of Piperine. I. The Komarowsky Reaction,” J. Food Sci. 1965, 30(4), 644–650 (https://doi.org/10.1111/j.1365-2621.1965.tb01818.x).
Jingping Zhang, David Thickett, Lorna Green, "Two Tests for the Detection of Volatile Organic Acids and Formaldehyde," Journal of the American Institute for Conservation 1994, 33(1), 47–53 (doi:10.2307/3179669).


What is this copper compound I made? I started with copper (II) sulfate solution left over from a recrystallization. I added sodium carbonate (definitely an excess) and about 200 of water and got a lot of bubbling. I heated it, for no particular reason. Few hours later I filtered the solution, which now looked dingy gray. I got a deep blue solution with a pH of around 11, and what was in the filter was black...

The filtrate is still deep blue with no precipitate, I'm washing and drying the residue from the filter, but I'm still curious what I made. I should have gotten basic copper carbonate, I've done this before to get copper carbonate, I can't remember if I've used sodium carbonate or bicarbonate last times though. I didn't think it made a difference.

There are two surprising phenomena in this experiment.

First : $\ce{CuSO_4}$ reacts in a unusual way with $\ce{Na_2CO_3}$. The equation is :  $$\ce{2 CuSO_4 + 2 Na_2CO_3 + H_2O -> Cu(OH)_2·CuCO_3 + CO_2 + 2 Na_2SO_4}$$ This forms some bubbles of $\ce{CO_2}$ as you have seen. And it produces a strange precipitate of the so-called basic copper carbonate, $\ce{Cu(OH)_2·CuCO_3}$, which is well known in geology as the mineral called malachite. This formula is sometimes written as $\ce{Cu_2CO_3(OH)_2}$. This is what you have obtained as your insoluble deposit.

Second : Basic Copper carbonate gets redissolved in a big excess of carbonate ions, producing a deep blue solution containing an anion which is probably $\ce{[Cu(CO_3)_2]^{2-}}$ This is what you have obtained in your blue filtrate.


Demonstrational smelting iron I'm planning an instructional demonstration of smelting. It seems you may be able to take rust (hematite, $\mathrm{Fe_2O_3}$) and carbon (perhaps from charcoal briquettes), place this in a crucible of ample size, light the carbon on fire, and the formed iron may melt and accumulate on the bottom of the crucible.

Yes, iron burns, everyone knows you can light steel wool.  I'm hoping that the interface above the iron will become encrusted with enough slag and ash that there won't be much of the formed iron burning, and what's specific about steel wool versus trying to light a spoon on fire is that the steel wool has a very high surface-to-mass ratio, enabling it to be heated well as well as provide ample mass transfer area.

What are the problems with this demonstration, and what kinds of solutions exist for this?  For example, I foresee that I need to use lots of extra carbon to make sure that any iron formed doesn't reoxidize.

The biggest problem you will have is getting a high enough temperature, not the oxidation of the iron.

Putting the crucible in a charcoal furnace won't produce a high enough temperature to cause the reaction or the liquefy the iron: you need to have a crucible that can be driven to a much higher temperature. Normally this is done by blowing air (or, if you have it, oxygen) into the furnace. In practice what is often done is to mix the iron oxide and an excess of carbon in a single vessel: the carbon burns to create the temperature and also acts as the reducing agent to create the iron. This still requires a source to force air into the mixture. This appears to be how ancient craftsmen did the reaction, kicking off the iron age.

For an account of how to do this see this blog describing a modern recreation of some of those ancient techniques.


Why Prussian blue 'blue' in colour? Why $\ce{K_4[Fe(CN)_6]}$ give blue ppt with $\ce{Fe^{3+}}$.

I know that potassium ferrocyanide is used for the qualitative analysis of ferric salts. It forms a Prussian blue ppt of $\ce{Fe_4[Fe(CN)_6]_3}$. However I couldn't justify the blue colour of Prussian blue.

Edit. In my view the colour may be due to charge transfer between ligand and central atom. However I have no reason to accept it.

Prussian blue contains both $\ce{Fe(II)}$ and $\ce{Fe(III)}$. Mixed-valence compounds, such as the one in question, absorb certain wavelengths of visible light resulting due to the phenomenon of inter valence charge transfer (IVCT).
General Scheme for IVCT is:
$$\ce{L_nM^+-bridge-M'L_n + h\nu ->L_nM-bridge-M'+L_n}$$

In this case, orange-red light around $680\ \mathrm{nm}$ in wavelength is absorbed, and the reflected light appears blue as a result.

Additionally, changes colour from blue to colourless upon reduction. This change is caused due reduction of the $\ce{Fe(III)}$ to $\ce{Fe(II)}$, and thereby eliminating the IVCT responsible for Prussian blue's namesake blue colour. This is also why Prussian Blue exhibitis electrochromic behaviour.

For a more detailed description read this and this. Alternatively, you can look it up in a textbook.


Determining stoichiometry of an unsaturated hydrocarbon via combustion $60\ \mathrm{cm^3}$ of oxygen was added to $10\ \mathrm{cm^3}$ of a gaseous unsaturated hydrocarbon. After explosion and cooling to room temperature, the residual gases occupied $45\ \mathrm{cm^3}$. What volume of carbon dioxide was formed in the reaction?

I’ve tried to make an unbalanced equation out of it:

$$\ce{(CH2)$_n$ + O2 -> CO2 + H2O}$$

My problem now is I don’t know how to proceed to the next step. How do I even find the moles? Is the equation correct? How do I balance this equation?

First step: balance your equation.

$\ce{(CH2)$_n$ + O2 -> CO2 + H2O}$ is not balanced, this is:

$$\ce{(CH2)_{\it n} + \frac{3 }{2}{\it n}~O2 -> {\it n}~ CO2 + {\it n}~ H2O}$$

Second step: perform mole balances on relevant species.

The balanced equation has $1 + \frac{3}{2}n$ moles of gas reacting, and $2 n$ moles of gas formed.  However, the question states that the products are cooled to room temperature, which would eliminate the water from the gas phase.  Thus the reaction were are really interested in is

$$\ce{(CH2)_{\it n}_{\rm(g)} + \frac{3 }{2}{\it n}~O2_{(g)} -> {\it n}~ CO2_{(g)} + {\it n}~ H2O}_{(l)}$$

This means that $1 + \frac{3}{2}n$ moles of gas react, and $n$ moles form for each mole of reaction $\xi$.  Thus, the reaction causes a loss in moles of gas of is $-(1 + \frac{1}{2}n)$ moles per mol $\xi$ of reaction.  If we assume the ideal gas law applies, then we could also say that the loss in volume from the reaction is $(1 + \frac{1}{2}n)~{\mathrm{{cm}^3}}$ per $\xi~{\mathrm{{cm}^3}}$ of reaction.

Thus a total gas balance is:

$$\rm GAS~IN - GAS~OUT = GAS~REACTED $$
$$(60~\mathrm{{cm}^3} + 10~\mathrm{{cm}^3}) - (45~\mathrm{{cm}^3})=(1 + \frac{1}{2}n)~\xi$$

A $\ce{CO2}$ balance is

$$\rm \ce{CO2}~IN - \ce{CO2}~OUT = \ce{CO2}~REACTED $$
$$ 0 - x = - n \xi $$

Third step: assess the resulting equations

These equations are not solvable for $x$ because they depend on both $\xi$ and $n$.  That is, there are two equations, and three unknowns.  Writing additional balances on oxygen or on fuel will not help because we will not know the gas remaining for either species, so each new equation would introduce new variables. Thus, we will need additional assumptions to solve the problem.

Fourth step: determine additional assumptions.

Probably we should assume that either (a) fuel was in excess or (b) oxygen was in excess, which is equivalent to assuming that (a) oxygen was completely consumed or (b) fuel was completely consumed.  At least one of these assumptions makes sense because combustion reactions are essentially irreversible.  In my answer I'll assume oxygen was in excess, meaning that the fuel was completely consumed.  You should make the alternate assumption and repeat the procedure.

Fifth step: apply additional assumptions.

Assuming fuel was completely consumed, a fuel balance is

$$\rm FUEL~IN - FUEL~OUT = FUEL~REACTED $$
$$\rm (10~\mathrm{{cm}^3}) - 0=\xi $$

This conveniently gives us a solution for $\xi$, the extent of reaction.  It is $10~\mathrm{{cm}^3}$.

Sixth step: use new information to solve existing equations

We can substitute that solution for $\xi$ into the total gas balance to get that $1 + \frac{1}{2}n = 3$, which implies that $n=4.$  That and our solution for $\xi$ allows us to solve the problem: if oxygen is in excess, then 40 $\mathrm{{cm}^3}$ of $\ce{CO2}$ were formed.


Is Nuclear Spin Isomer a kind of allotrope? In my book there's a question:

  Does $\ce{H2}$ shows allotropy ? Enlist its allotropes and their application?

I pondered  over different way Hydrogen can be arranged(I don't know but I have weak intuition that hydrogen bonding can play a role) but to no resolute

After awhile I googled to find different answers :Some are enlisting NSIs of hydrogen as allotrope

So can NSI(s) of an element can be considered allotrope ?

If yes then how Nuclear Spin Isomer  can affect their physical form/structures?

*Wikipedia also doesn't enlist any allotrope of $\ce{H2}$ as such.
https://en.wikipedia.org/wiki/Allotropy

Some peer review journal articles and an India high school textbook and various study guides do refer to the spin isomers as allotropes.

There are physical differences such as different heat capacities.  See Orthohydrogen, Parahydrogen and Heavy Hydrogen for comprehensive information.

However, aside from the spin isomers, there are genuine allotropes of hydrogen.  Solid hydrogen has two normal pressure allotropes, a face centered cubic and a hexagonal closest packed allotrope.

Also, at high pressure, there is metallic hydrogen.


Why is only one part of buffer in the reactants side? Given that buffers consist of either

a weak acid and its basic salt, or
a weak base and its acidic salt

and that the buffer consisting of CH3COONa and CH3COOH is added to water (H2O) ...

The reaction is:
$$\ce{CH3COOH + H2O <--> CH3COO- + H3O+}$$
(Equivalent reaction is $\ce{CH3COO- + H3O+ <--> CH3COOH + H2O}$)

Why aren't both parts of the buffer that is being added to the water represented in the reactants side of the reaction?

In other words, how is it that one part of the buffer is making another part of the buffer if both needed to exist in order to be a buffer to add in the water in the first place?

A buffer is a solution that is meant to prevent the pH from drifting away from its intended value in both directions (i.e. the solution should neither become more acidic nor more basic than desired). For the buffer to accomplish this, two components are added one of which is intended to counteract acidic additives the other to counteract basic additives.

Whether or not these two components are conjugate Brønsted acids and bases does not matter; a solution of e.g. ammonium acetate will also buffer. Every liquid inside the human body is buffered by multiple components to a very narrow pH range, so many different compounds together can also produce a buffer effect. However, the combination of an acid and its conjugate base is typically the most effective and narrow-range buffer system available in a typical lab context — and also the easiest to set up and use.

While in the buffered solution the two components of a buffer can indeed be said to react; this can be proven with isotope labelling. For example, if you had $\ce{^13C}$-labelled acetic acid and $\ce{^12C}$-labelled acetate, these two would react in the following way:

$$\ce{CH3^12COO- + CH3^13COOH <=> CH3^12COOH + CH3^13COO-}\tag{1}$$

This reaction does not, however, describe the buffering action.

Buffering itself can only take place if some acidic or basic compound is added to the solution as hinted above. Depending on what is added either equation $(2)$ or equation $(3)$ will take place.

$$\begin{align}\ce{CH3COO- + H3O+ &<=>> CH3COOH + H2O}\tag{2}\\
\ce{CH3COOH + OH- &<=>> CH3COO- + H2O}\tag{3}\end{align}$$

The first equation $(2)$ is the one if something acidic is added, equation $(3)$ takes place if a basic compound is added.

As you can see, one component of the buffer basically exists to quench basic additives (the acidic part) while the other only exists to quench acidic ones (the basic part). Therefore, in a buffering equation you will only see one of the two active at once.


Are there reactions operated at high temperature at industrial level? Temperature is an important parameter when you deal with a chemical reaction, first of all $\Delta G$ and $\Delta G^\circ$ depend on temperature, and as a consequence the spontaneity of a reaction depends on temperature. Second, also the equilibrium constant $K$ depends on temperature so, it is possible to modify the reaction yield by varying it.

Finally, the kinetic constant $K_{V}$, and so the reaction speed, are related to temperature. Said that, I was wondering if at industrial level, there are some important reactions which are operated at high temperature (I mean at least more than $\pu{40 °C})$ in order to obtain at least one, or more of the effects I've written (or some other that I don't know). Of course, I also include reactions such as electrolysis and electrochemistry reactions.

The cracking reactions transforming heavy fuel, tar, domestic or heavy oils into lighter molecules like gasoline, kerosene or white spirits, they are all carried out at temperatures higher than 200 °C. The electrolysis of alumina to produce metallic aluminium is done at about 960 °C. The production of iron from ore and charcoal is done in a blast furnace at 1400 °C


Halogenation of an alpha-halogen ketone in acidic conditions When a methyl ketone is submitted to halogenation in acidic conditions with two equivalents of a halogen, the first halogenation happens at the secondary alpha carbon atom , while the second halogenation happens at the primary alpha carbon atom. Why is it so? Why doesn't it happen on the same carbon atom (the secondary one)?

The same mechanism discussed in my answer to your earlier question on acid catalyzed halogenation alpha to a carbonyl still applies here.  For reference, here is the same link supplied earlier that shows the mechanism for this acid catalyzed halogenation.

In that earlier answer I said that the enol that "produces the more stable double bond (the one that is more highly substituted)" is the enol that will form and react nucleophilically with the halogen to produce the alpha-halo carbonyl product.  In light of your new question, the statement that more highly substituted double bonds are more stable needs some qualification.

The closer in electronegativity a substituent is to carbon, the more it will stabilize a double bond.  Consequently, the more alkyl groups attached to a double bond, the more stable the double bond (barring large steric effects).  On the other hand, fluorine destabilizes a double bond.  It is so electronegative that it prefers to bond with orbitals high in p character.  When forced to bond to a double bond $\ce{sp^2}$ orbital (that is enriched in s character) destabilization results (see this earlier fluorobullvalene example for more on why fluorine "hates" to be on a double bond).

Based on its electronegativity, chlorine should be somewhere between fluorine and an alkyl group in terms of its ability to stabilize a double bond.  If we compare the heats of formation of 1-butene (-0.1 kJ/mol) and trans-2-butene (-11.2 kJ/mol) we see that placing the methyl group on the double bond stabilizes the molecule by 11.1 kJ/mol.  If we compare the heats of formation of allyl chloride (-5.6 kJ/mol) and trans-1-chloropropene (-12 kJ/mol) we see that the placing the chlorine group on the double bond stabilizes the molecule less, only 6.4 kJ/mol.

These differences mean that we will have much more of the enol without the chlorine on the enol double bond, then we would have of the enol without the methyl group on the enol double bond.  So we have more of the unsubstituted enol in the chlorine case and the enol with the chlorine on the double bond is much less nucleophilic (less reactive) due to the electron withdrawing inductive effect of the chlorine substituent.  These two factors combine to make the second alpha-chlorination take place on the unsubstituted side of the molecule.

As to steric effects, they probably don't play a significant role.  One metric commonly used to compare "steric size" is the cyclohexane A value.  The A value for chlorine is 0.43, much smaller than the A value for methyl (1.7), suggesting that a chlorine substituent is smaller than methyl substituent.


Why does ring contraction take place in Wagner–Meerwein rearrangement? An example of Wagner–Meerwein rearrangement given in my textbook is substitution by $\ce{AgNO2}$ in iodocyclohexane. One of the product, cyclopentylnitromethane is through ring contraction.

I don't understand how this took place. Can anyone explain the mechanism?

Kornblum1 demonstrated that nitrocyclohexane 3 was the major isolable product from the reaction of iodocyclohexane 1 and silver nitrite. Earlier work by Rosanow2 (1915) claimed the additional formation of tertiary nitro compound 7, a result that Kornblum did not confirm. If indeed nitrocyclopentane 7 did form, a ring contraction is required. Ring contraction of cyclohexyl carbocation 2 sacrifices the greater stability of a secondary carbocation for a primary carbocation (4) and presumably some loss in stability through ring contraction in forming cyclopentyl carbocation 6. There is no evidence that primary nitrocyclopentane 5 is formed. However, a hydride shift in 4 would afford the more stable tertiary carbocation 6 that would lead to 7.

1) N. Kornblum and C. Teitelbaum, J. Am. Chem. Soc., 1952, 71, 3076.
2) See reference in Kornblum paper.


Which of the following carbanion is most stable? The carbanion which is most stable is

The answer is 1. But why won't the answer be 2 because it shows resonance, whereas 1 only shows I effect and resonance effect is more dominant?

Some points to check while comparing stability of a carbanion:

Inductive effect: preferentially $\ce{-I}$ effect
Resonance: preferential $\ce{-M}$ effect
Hybridization: preferentially attached to carbon atom having more s-character

... many more: stability
Now the problem is option 1 and 3 are in itself very 'nice' (w.r.t. stability) possible cases of Hybridization-stabilization and Resonance-stabilization respectively.
So the natural (what I thought) way out here is comparing $\ce{pKa}$ 's of acetylene and toluene respectively for option 1 and 3.
from here :

$\ce{pKa}$ of acetylene is about 25.
$\ce{pKa}$ of toluene is about 43.

Less $\ce{pKa}$ denotes how (much more) feasible is for the acid to dissociate by deprotonating itself.
So, acetylene carbanion is more stable than toluene carbanion


Elevated boiling points in alkanes with even number of carbons Why do alkanes with even number of carbon atoms have greater boiling points than those with odd number of carbon atoms?

There is not really any difference between the boiling points of even and odd carbon alkanes.  For single-chain alkanes the boiling point just goes up as a smooth curve versus chain length.

The melting points do show a small oscillating component with the even chains being higher.  If we imagine the chains in their idealized zig-zag configuration, which would be relatively favorable for the solid state, we find that the even-length chains have a center of inversion which makes them fit better than odd-length ones.  This effect fades, however, with increasing chain length.

Reference:

https://en.wikipedia.org/wiki/Alkane


Moisture Absorbancy & Odor Elimination What would be the typical DRY ingredients in moisture absorbing and odor eliminating materials that work especially well on perspiration/organic odors?

Silica gel is often used to absorb moisture; activated carbon is effective at odor removal; and zeolites (also called molecular sieves) come in different varieties and can be effective for both purposes.


Is it possible to separate/remove sugar from common drinks, such as juice? Is it possible to remove sugar from a drink (for example, juice) using common household equipment?

Why not add yeast to your juice and allow it to ferment?  This will depend on the juice having pH and sugar concentrations; look at commercial wine making kits which provide the proper equipment and recipes. If you conduct the fermentation properly without introduction of air, the sugar will be converted to alcohol, which later can be boiled off (see above) to give the sugar-free version of your juice. Otherwise, with air present, the fermentation will go on to form acetic acid and give the juice a vinegary taste.


Flammable lighter than air gases (safeish) I'm an artist currently using propane and butane in my show where I put the gas into a soap bubble and light it to produce a nice flame.

But I'd like to start working with a lighter than air flammable gas, that's relatively easy to get hold of, and as safe as it possibly can be.

The propane/butane I use comes in small canisters and the biggest risk, as I see it, is forgetting to close the tap.

Any ideas?

I would propose to use hydrogen, as it is pretty easy to obtain right there where you need it and it's the lightest gas ever.
You can even watch a  video where the whole setup for the bubble experiment is described  (at the end).
Besides hydrogen, methane looks good to me, and it is also in our everyday life, being the natural gas. But the setup for using household methane is imagined more complicated to me than the one with hydrogen.
Just be very careful with any flammable gas, don't make it another Hindenburg :)


The entropy change during micellization I was reviewing the thermodynamics of micelle formation from the Wikipedia page and this book. I understood most of the arguments but micellization is supposed to be an entropy driven process with a positive entropy change in the process of micellization. But isn't this a bit counter-intuitive?

  In the unassociated form, the surfactant molecules are free to roam
  about in the solution much like normal solutes. But in the form of a
  micelle, they get associated, grouped into an associated whole, which
  reduces the freedom of independent movement and should therefore
  decrease the entropy (randomness) of the system? The entropy change
  should be negative during micellization. Why is it positive?

I understood the mathematical arguments leading to the develpment of the approximate formula $\Delta G=RT\ln(\text{CMC})$.

The micellization releases water molecules which initially solvate the hydrophilic head groups. The release of these waters greatly increases the entropy.

Reference: Dong et al., Chem. Rev. 2010, 110, 4978-5022.


Anhydrous calcium sulfate as a drying agent I know that anhydrous $\ce{CaSO4}$ is used as drying agent.

I also know that anhydrous $\ce{CaSO4}$ (dead burnt plaster) is prepared by heating gypsum $\ce{CaSO4.2H2O}$ at a temperature above $\pu{393K}$, which is an irreversible process.

How does $\ce{CaSO4}$ perform the drying if it can't accept any $\ce{H2O}$ molecules? Is there some other mechanism here? Or do I have some wrong information fed in my brain?

I think your problem is with the idea of "irreversible" process.

The term irreversible here means that, at $\pu{393 K}$ and in the conditions where this is performed, the reaction

$$\ce{CaSO4.2H2O <=>  CaSO4 + 2 H2O}$$

is pushed towards products so much, that it can be considered "one-way" only:

$$\ce{CaSO4.2H2O -> CaSO4 + 2 H2O}$$

This is dependent on the temperature, and likely other conditions: for example, water exists as a gas at this temperature at a pressure of $\pu{1 atm}$ or lower. If you remove some of the water while heating (such as vacuum, or by circulating dry air around to get the water out), then you further push the equilibrium to the right (and make the backwards reaction impossible since water is no longer present, thus making the process irreversible).

Once you have your dry $\ce{CaSO4}$, it can be cooled down to room temperature (in a dessicator, to avoid it absorbing water). At temperatures below $\pu{393 K}$, the reaction will work backwards:

$$\ce{CaSO4 + 2 H2O -> CaSO4.2H2O}$$

Thus, your dry $\ce{CaSO4}$ will absorb water from the environment, becoming a drying agent. You may regenerate it by reheating it to $\pu{393 K}$, and so on and so forth.


What causes the "white water" smell? I can detect a distinct smell when I am near water going over a weir. I could tell that I was near "white water" if I was blindfold. Can anyone tell me what causes this odour?

I don't have any journal article citations to back me up, but I've always understood that this distinct smell is from water aerosol.  You can detect the same smell during a fine misting rain as well as near water sprinklers (if the sprinklers produce a very fine mist of water).  But you don't get this smell from water vapor (steam) or bulk water, so there is clearly a specific range of water droplet size that triggers the olfactory response.

I've personally always wanted to know what the specific size range of these aerosol droplets needs to be to trigger the olfactory response.  I'm confident someone has studied this, but I don't know which journals to search to find the answer.


What is the rate expression for decomposition of $\ce{H2O2}$ in the presence of a catalyst? What is the rate expression for decomposition of $\ce{H2O2}$ in the presence of a catalyst?

As far as I know for catalase, it is like this: $$\ce{Rate = k[H2O2][Catalase]}$$

I don't know what it is for the catalysts $\ce{MnO2}$ and $\ce{Fe2O3}$. Another question in my mind is that even if it's similar to that of catalase, since these are powdered and so do not have a concentration, how can I express it?

It is kind of emergency so any help would be greatly appreciated.

$$\ce{2H2O2 + Catalyst ->2H2O + O2}$$

I found on web that both reactions are probably (still not sure even though I saw it) first order reactions.

Edit: Is it possible that there shouldn't be anything to be added and it will only affect the $\mathrm{k}$?

For catalase, it is most likely Michaelis-Menten kinetics.
At typically low enzyme concentration, this would be first order in catalase. For hydrogen peroxide, it is first order at low concentrations (<< Km) and approaches zero order at high concentration (>> Km). This behavior is also called saturation kinetics.

I don't know what it is for the catalysts $\ce{MnO2}$ and $\ce{Fe2O3}$. Another question in my mind is that even if it's similar to that of catalase, since these are powdered and so do not have a concentration, how can I express it?

For catalysts in the solid phase, it is the surface area (compared to the volume of the solution) that matters. You might see similar saturation kinetics, indicating that at some concentration of reactant, all the surface area is used so that an increase in reactant concentration no longer translates into an increase in rate.


Nucleophilic attack at a carbonyl - angle of attack What is the specific angle between the carbonyl carbon and the nucleophile during the nucleophilic addition reaction?

Let's take the following approach to answer your interesting question:

Use the bent bond model to describe the carbonyl
Calculate the bond angles around the carbonyl carbon
Treat the reaction as an $\mathrm{SN2}$ reaction
Use trigonometry to deduce the angle of nucleophilic approach

The bent-bond model is widely used in chemistry to describe highly-strained systems such as cyclopropane and double bonds (ref_1, ref_2).  In the case of a double bond, we wind up with a two-membered ring (which nicely explains the high heat of formation of such compounds). Here we will use it to describe the carbonyl double bond.  Here is a picture of what the bent bond description of a carbonyl looks like (it's the top left diagram that's of interest, note too that it is a picture of an amide carbonyl).

(image source)

In a simple carbonyl compound like formaldehyde the $\ce{H-C-H}$ bond angle is around 118°.  Using Coulson's theorem (ref_2, ref_4) we can use this bond angle to estimate the hybridization of the carbonyl $\ce{C-H}$ bonds.

$$\ce{1+\lambda^2cos(118)=0}$$ where $\ce{\lambda^2}$ represents the hybridization index of the $\ce{C-H}$ bond. From this equation we find that $\ce{\lambda^2=}$ $\mathrm{2.13}$, or the carbonyl $\ce{C-H}$ bonds are $\ce{sp^{2.13}}$ hybridized.

We can now sum up the s-character in the 4 bonds around the carbonyl carbon (it must add up to 1) and deduce the hybridization of the 2 $\ce{C-O}$ bent bonds.

$$\ce{\frac{2}{1+\lambda^_{C-H}^2}+\frac{2}{1+\lambda_{C-O}^2}=1}$$

Solving this equation we find that $\ce{C-O}$ bent bonds are $\ce{sp^{4.54}}$ hybridized.

Substituting this information back into Coulson's theorem allows us to estimate the $\ce{O-C-O}$ bond angle as being equal to 102.7°.

In an $\mathrm{SN2}$ reaction the nucleophile approaches 180° from the leaving group ("backside attack").  In this case, one of the carbonyl $\ce{C-O}$ bonds is the leaving group.

Using simple trigonometry we can estimate what the angle is for 180° attack of the $\ce{C-O}$ bond relative to the plane of the carbon and oxygen nuclei. Half of the $\ce{O-C-O}$ bond angle is 51.4°, leaving 38.6° to reach 90°. So the angle for backside attack relative to the plane containing the $\ce{C}$ and $\ce{O}$ atoms is $\mathrm{90°+38.6°=128.6°}$

The actual angle for attack has been found to be around 107° (see @Philipp 's comment on the Bürgi-Dunitz angle).  This angle was first determined by extensive, detailed crystallographic investigations and later supported by computational studies. The approach used here is no more than a quick, "back of the envelope" method. Yet it still provides an answer within about 20° of Bürgi and Dunitz's work. For such a quick, crude method, I would consider this result to be in qualitative agreement. It supports the back side attack $\mathrm{SN2}$ model, illustrates how a $\ce{C-O}$ bond participates as a leaving group and how we wind up with a tetrahedral carbon.


organic chemistry of alkyl halides and alcohols What happens if a primary alcohol that also bears a halide substituent on the alcoholic carbon (the carbon to which the hydroxyl group is attached) when it is reacted with copper at 300 °C ?
   I know primary alcohols, on treatment with copper at 300 °C react to give  aldehydes.  But I am confused what happens if halides are present on the same alcoholic carbon.

You are correct that copper metal is a catalyst for alcohol dissociation to aldehyde + $\ce{H2}$.  The reaction of ethanol, for example, can be written as:

$$ \ce{CH3CH2OH  ->[\ce{Cu}~\textrm{metal}][300~^{\circ} \textrm{C}] CH3CHO + H2} $$

Your question about halo-substituted alcohols is very interesting, because those species are very unstable.  If I understand you correctly, you are not referring to halohydrins such as 2-chloroethanol, which are species that have a halo substituent beta to the alcoholic carbon, but rather to species such as 1-chloroethanol where the halo and hydroxyl substituents are on the same carbon atom.
alpha halo alcohols are very unstable, and tend to eliminate halo-acid and form the corresponding carbonyl compound.  For example, gaseous chloromethanol spontaneously eliminates hydrogen chloride to form formaldehyde at 296 K (~23 °C) with a half-life of about 300 seconds.
Thus I think the best answer to your question is that:

Halo alcohols like 1-chloroethanol or chloromethanol are already oxidized to the same degree as carbonyl compounds.
Even before reacting them with copper metal at 300 °C, they are likely to decompose to the carbonyl compound via elimination of hydrogen halide.
Unlike the reaction of simple alcohols on hot copper metal, this reaction is not a redox dissociation, but is a simple elimination reaction.$%edit$


Can gases be "immiscible"? Anyone know of any examples of gas mixtures that phase separate, analogously to the way immiscible liquids do?

They would have to be extremely non-ideal. I might guess one might have to be much higher molecular weight than the other and at least one if not both would have to have significant intermolecular interactions (although of different kinds).

Good question but there is no such thing as immiscible gases. Immiscibility is caused by surface energy/tension.  Gases do not have surfaces and therefore lack surface tension.


What's the type of bonding in La@C60? Endohedral fullerene is the name given to a brand new chemical species with the following attributes:

  A metal, usually a transition metal, is "trapped" inside the fullerene, like $\ce{C60}$ and $\ce{C82}$. These transition metals are often La, Sc, Ce, Y, or, Ba and Sr etc.

Without any "binding" forces, the atom couldn't be possibly trapped inside the fullerene.

Here it's been discussed that the bonding is the type of "donor-acceptor". Later within, it was explained that

  Though there isn't empirical proof, It's believed for these species to have metallic structure.

I have two questions:

Is there another simpler name for this "donor-acceptor" bonding? My thoughts lead to wierd names!
How "close" is this type of bonding to metallic bonding?

Edit: Citation: The second article:

Publisher: [AIP]; German journal of chemical physics. Volume 115, number 15, page 7215. (15 Oct 2001) (Juelich GMBH or something; Couldn't figure the name out, maybe German fellows would help.)

DOI: 10.1063/1.1406500

Authors: R. Klingeler, G. Kann, I. Wirth, S. Eisebitt, P. S. Bechthold, M. Neeb, and W. Eberhardt (Dr. Rudiger Klingeler is the main author)

Title: $\ce{La$@$C60}$: A metallic endohedral fullerene.

Unfortunately the other mirror links are behind paywalls. (But this article is available in Researchgate, in case you've got institutions to be able to create an account)

$%edit$

That the question says: "Without any 'binding' forces, the atom couldn't be possibly trapped inside the fullerene", shows you are misunderstanding fullerenes.

Atoms such as Helium and $H_2$ can be confined within the interior of the fullerene.

There are even reports of extraterrestrial helium trapped in fullerenes during astroid or comet impact.

The helium can move around within the fullerene.  It is sterically confined rather than bonded.

In the case of La, the metal ion does not remain neutral, but loses about 3 electron to the fullerene.  So, in addition to being sterically confined in the cage, there is the interaction between postively and negatively charged species.

As calculated in a later article Computed Structure and Energetics of
La@C60 International Journal of Quantum Chemistry, Vol 104, 272–277 (2005), the La has a charge of somewhere betwen 2+ and 3+.  In the lowest energy state, the La is off-center with respect to the fullerene, located between the center and the center of a hexagon face.  Also, the fullerene cage is distorted from icosahedral by Jahn-Teller distortion.

The La ion is located 2.62 Angstroms from the nearest carbon atom (4.6 Angstroms from the furthest), when at the potential miniumun, but is expected to move within the fullerene unless the temperature is very low.  The reference also says "there is obviously a relatively large negative charge on some carbon atoms of the cage", without specifying how this charge is distributed.

Overall I would say, whereas in the case of helium and dihydrogen being confined in the fullerene is completely because of van der Waal repulsive force of avoiding the walls, in the case of La, there is attractive electrostatic force (ionic interaction) between the La3+ and negatively charged carbon atoms.

See also Theoretical investigation of structural and thermodynamic properties
of lanthanum carbides LaCn (n = 2–6) which finds strong ionic interactions between La and Cn due to charge transfer.


Help with idea for an optical pH sensor I am experimenting at home with the idea of making an optical pH sensor that can read the colour of pH indicator using photodiodes.
I am wondering if there is some kind of ion permeable membrane that could be used to contain the indicator solution, so it can 'read' the pH of the solution being tested, without mixing with it.
I would like something that is fairly easily obtainable to me as a home experimenter.
The sensor could possibly be used make an automated pH corrector for my local aquaponics farm.
I looked into commercially available pH sensors. The glass bulb type and ISFET type suffer from drift and need regular calibration. There are optical types available, but they are too expensive for me.

I really like the concept but the problem is to find a membrane that is permeable to water and H3O+ ions but impermeable in respect to the indicator.
I understand you need something as in the picture below.
Dialysis tubing are commonly used to prevent bigger molecules from crossing the membrane, but unfortunately the typical lower limit is for molecules of around 1 KDa so much more than a pH indicator molecule.
 The only option available seems reverse osmosis membranes, for example used in water purification filters. However, the resulted filtered water has an altered pH (ref.1) therefore not suitable.
Finding a membrane that is permeable to water and H3O+ ions but not to small polar molecules (i.e.Phenolphthalein, blue bromothymol, Anthocyanins etc.) is something difficult to achieve especially if you have also the requirement of not altering the pH.
I think that is the reason why we still don't have a pH meter that works as you described.

References:
1)Nir, Oded & Bishop, Noga & Lahav, Ori & Freger, Viatcheslav. (2015). Modeling pH Variation in Reverse Osmosis. Water Research. 87. 328–335. 10.1016/j.watres.2015.09.038.


Explanation for behavior of acetic acid in aquaponic/hydroponic systems? Because aquaponics is hydroponics with fish & bacteria, options for adjusting pH are limited.

Primarily people use phosphoric acid, hydrochloric acid, and nitric acid (heavily diluted).  A less popular but easy to acquire material is vinegar.

I've tried using it in a very alkaline solution (100 ppm calcium carbonates, 8.4 pH) and noticed that the pH would drop by a small amount before returning to it's original state.  Anecdotally, other aquaponic hobbyists have noticed a similar effect that is not replicated by the other acids. What is happening that the pH would lower temporarily and then rise back up over several hours?

The source of the acid shouldn't matter much in adjusting the pH by a small amount.  What is the source of the calcium carbonate?

If it is dissolved calcium carbonate only, as you might find in hard water, then this is strange.  The solution is buffered by the combination of dissolved carbon dioxide and carbonate minerals, so its pH might be difficult to change, but once it has been changed it should stay where you put it, more or less.

If you have solid calcium carbonate (cement, limestone gravel, some pottery (notably not terracotta) or marble chips, perhaps), then what is happening makes a lot more sense.  You add acetic acid, lowering the pH and consuming some of the carbonate from the solution.  There is a large reservoir of carbonate available though, in the form of the carbonate minerals, and that eventually redissolves and the solution becomes more basic with time.


Current in a Lithium Ion Battery I am trying to find the current transferred to the anode and cathode of a Lithium Ion Battery at different voltages. I know the temperature (approximately) of the battery, the battery's starting voltage, the voltage at 30sec intervals of a constant power discharge (from which current can be deduced), the dimensions of the battery and the initial amount of charge in the battery. What could I do to find this current transfer?

Power is the energy transferred per unit time ($\frac{J}{s}$) and is determined by the potential drop (V) and the current (I) at a given point in time as $P = IV$ so to get the current, solve for $I=\frac{P}{V}$. This gives you the current at a given point in time.

Your question has a few points that might be worth addressing. As Dr Moishe Pippik commented, "current transferred" is a little confusing, either instantaneous current ($I$), or total charge transferred ($Q$), which is the current integrated over time: $\sum_{t} I(t)\Delta t$ in your case. So if what you really want is the total charge transferred, you would have $Q = \sum_{t}\frac{P\Delta t}{V(t)} $. Strictly speaking you should exclude t=0 point to avoid double counting.

You also asked about both the anode and the cathode. The current for both will be the same number but with opposite sign.


How does Resonance stabilize a molecule? How does resonance lower the potential energy of the molecule? Take $\ce{O3}$ as an example.

With the concept Resonance you can describe delocalized electrons in some molecules. To go with your example:

(source)

Imagine the ozone molecule to be frozen in one of those two states. Positive and negative charges are localized, which is generally connected to a higher energy, i.e. a more reactive species. This would be an extreme case for ozone - in reality those electrons will be delocalized all over the molecule. So, what is the reason that molecules with localized electrons/charges have a higher total energy and are, hence, less stable?

Maybe you remember the Particle in a box. I wont go into the quantum chemistry here, but one of the take-away-messages is, that the energy for particles in a infinite square well, which is a very rough approximation of the potential in which the outer electrons reside in, does decrease with $ L^{-2} $. $L$ is the size of the box in which the particle can exist. So, the more delocalized the electrons are, the less total energy in a system.

With this, to correct you question a little bit: Resonance is not the reason why molecules are more stable, it is a concept which describes the bonding in a molecule, in which electrons are delocalized, better than a single Lewis Structure.

I would like to have a feedback if this answer is somewhat understandable.


Why doesn't a halide anion react with primary or secondary alcohols using SN2? In the tenth edition of Organic Chemistry by Solomons and Fryhle, section 6.18A of the "Ionic Reactions" chapter, in the last paragraph, they have mentioned that:

  Use of a weakly basic anion such as a chloride ion ($\ce{Cl^-}$ ), or an acetate ion ($\ce{CH3CO2^-}$) or a weakly basic and a highly polarizable one such as $\ce{Br^-}$, $\ce{I^-}$ or $\ce{RS^-}$ increases the likelihood of substitution (SN2). (emphasis mine)

But when we react alcohols with concentrated hydrohaloic acids (is that the right term?) with alcohols, we see that the rate of reaction decreases from a tertiary to a primary alcohol, indicating that the reaction is SN1. For primary alcohols, it is shown in the textbook to proceed through (SN2) pathway. [There is no such reaction mentioned with hydrobromic or hydroiodic acid (atleast I haven't come across such a reaction)].

Elsewhere, I have read about the Lucas test, which involves reacting the alcohol with hydrochloric acid and zinc chloride, where nearly no reaction occurs with primary alcohols.

My question is, why does this reaction not occur using an SN2 pathway for secondary and primary alcohols in the Lucas test? Why is it biased towards an SN1 reaction?

The issue is that when carrying out the Lucas test you have a polar protic solution (due to the alcohol which you are testing), in which case Cl$^-$ is well solvated and thus not a very good nucleophile. As you are not using a strong nucleophile, you disfavor the S$_\text{N}$2 mechanism, but since you have a primary alcohol, you are not really favoring S$_\text{N}$1 either. Hence, why no reaction happens and the Lucas test works to identify primary alcohols. I think your question of Why is it biased towards an S$_\text{N}$1 reaction? might be a bit misleading because the reaction is neither biased towards S$_\text{N}$1 nor S$_\text{N}$2 for primary alcohols - it simply does not happen at all (at room temperature). In the case of secondary and tertiary alcohols, however, I would definitely say the reaction is biased towards S$_\text{N}$1 since you have polar protic conditions, a weak nucleophile, and a more stable carbocation intermediate.

Upon raising the temperature, I suspect that the reaction would then proceed via an S$_\text{N}$2 mechanism for a primary alcohol (as, despite not having a strong nucleophile, the barrier would be lower than via S$_\text{N}$1), but remain S$_\text{N}$1 for secondary and tertiary alcohols. Hope this helps!


How to write the last cell reaction for the following cell? Consider the following electrochemical cell:
$$\ce{Zn|Zn^{2+}}(0.1~\mathrm{M})||\ce{Zn^{2+}}(0.01~\mathrm{M})\ce{|Zn}$$ How should I write the overall cell reaction for this cell? I can identify the oxidation and reduction reactions but when they are added, all the species cancel out, leaving my with no overall equation.

Anode: $\ce{Zn -> Zn^{2+} + 2e}$

Cathode: $\ce{Zn^{2+} + 2e -> Zn}$

The species do cancel out. However, the concentrations of the species do not. It is a bit of a matter of preference as to how you express this. The way I would write it is:

$$\ce{Zn^{2+}_{(aq)} ($c = 0.01\text{ M}$) + Zn_{(s)} <=> Zn^{2+}_{(aq)} ($c = 0.1\text{ M}$) + Zn_{(s)}}$$

for which $E = E^0 - \frac{RT}{2F}\,\ln\!\left(\frac{0.1}{0.01}\right) = -0.030 \text{ V}$ at 298 K, since $E^0 = 0 \text{ V}$. A negative value of $E$ agrees with the reverse reaction above being spontaneous, which is of course what you would expect since $0.1 > 0.01$.


Is melting of NaCl a physical and chemical process? If I melt $\ce{NaCl}$, I get a molten salt. This means a liquid formed only by ions ($\ce{Cl^-}$ and  $\ce{ Na^+}$). That could mean that the crystal $\ce{NaCl}$ experimented a chemical change:
$$\ce{ NaCl->Cl- + Na+}$$ Like a chemical reaction... The ions have now more space to move and interact with much more ions than in the solid phase. I am trying to get an explanaition why the melting temeprature and the melting heat of molten salts show some erratic behaviour, I mean they do not follow a simple pattern such as that found for n-alkanes or n-alkanols.

In fact the lattice structure of $\ce{NaCl}$ is composed by $\ce{Na^+}$ and $\ce{Cl^-}$ even when is in the solid state we don't write it with the charge only for convention.

Ionic compounds are different from organic compound, they have complex lattice structure and are in fact compose by two species. Breaking a ionic bond required more energy than an organic compound where you deal only with intramolecular forces.


Why is the ionization energy of sulfur anion more than that of oxygen anion? Which of the following are in the correct order of their ionization energies? (multi-answer question)

  $\ce{O > S > S- >O-}$
  $\ce{F > F- > Cl- > Cl}$
  $\ce{O > O- > S- > S}$
  $\ce{F > Cl > Cl- > F-}$

The answers are 1 and 4.

Why is ionization energy of $\ce{S-}$ greater than $\ce{O-}$? The sulfur ion has an extra shell than an oxygen ion, so the atomic size of sulfur should be greater than oxygen. Therefore, the ionization energy of an oxygen ion should be more than a sulfur ion because the electrons in oxygen are closer to the nucleus when compared to sulfur. But why is the reverse true?

The ionisation energy of an anion is the same as the electron affinity of the element. $$\text{IE}(\ce{A-}) = \text{EA}(\ce{A})$$

Your reasoning is not wrong and one might expect exactly this behaviour. Another point for this reasoning is the substantial lower electronegativity of sulfur ($2.6$) versus oxygen ($3.4$).  However, experimental findings contradict this reasoning.
$$\begin{array}{lll}
&&/\mathrm{kJ\cdot{}mol^{-1}}\\\hline
\text{IE}\ce{(O)}  &        &=1314\\
\text{IE}\ce{(S)}  &        &=1000\\
\text{IE}\ce{(S− )} &=\text{EA}\ce{(S)} &=200\\
\text{IE}\ce{(O− )} &=\text{EA}\ce{(O)} &=141\\\hline
\end{array}$$

Sources: Wikipedia articles on ionisation energies, electron affinities

Basically it all boils down to electron-electron repulsion. Oxygen ($152~\mathrm{pm}$) is much smaller than sulfur ($180~\mathrm{pm}$) (van der Waals radius, source). This goes hand in hand with the smaller orbitals (lower main quantum number for the valence shell)  The electrons can therefore only occupy lesser space and repulse each other more. This is mainly the reason, why sulfur has a greater electron affinity than oxygen.

The same reasoning may be applied to fluorine and chlorine.


In the Lewis structure for barium fluoride, why does barium lose its valence electrons? I searched for $\ce{BaF2}$ Lewis structure, and found out that the Barium atom loses its 2 valence electrons to form $\ce{F- Ba^2+F-}$.
Before seeing it I thought each of $\ce{Ba}$'s valence electrons form a single bond with each of the fluoride atoms $\ce{F-B-F}$ which also completes the octet for F atoms,  and the formal charge in this case will be $7-6-\frac{1}{2}(2) =0$ for F atoms and $2-0-\frac{1}{2}[2\cdot 2] = 0$ for Ba, which works out.
I'm looking for an explanation on why Ba loses its electrons to form $\ce{F- Ba^2+F-}$ instead of $\ce{F-Ba-F}$.

First, a little bit of background. I assume you're already familiar with atomic orbitals — functions that describe the wave-like behavior of an electron in an atom. There's also molecular orbitals (MO), that describe its behavior in a molecule. You can ignore the odd-looking labels in images and such, since they're not entirely relevant to this question.
Atomic orbitals interact to form molecular orbitals (so called LCAO). The number of resulting molecular orbitals is the same as the number of the atomic orbitals that interact in the first place. To visualize this in simple interactions, you can use a molecular orbital diagram, in which orbitals of lower energy are drawn as lines placed lower than the others. There are three types of resulting MOs:

Bonding MO, which is usually much lower in energy than the starting atomic orbitals
Anti-bonding MO, which is usually much higher in energy
Nonbonding MO, which would have the same energy as the starting atomic orbitals

You can imagine that electrons are rearranged when a molecule is formed, and they fill the orbitals with the lowest energies first, so the order is bonding first and anti-bonding last. In a symmetric molecule, like the boilerplate example $\ce{H2}$, this is what the diagram looks like:

Image taken from this helpful page on Libretexts
_________________________________________
Back to your question, the energy difference between the atomic orbitals and the bonding orbital (presented as the space between the lower bonding MO and the higher atomic orbitals of each H atom) is the same, so they're said to 'contribute the same to the bonding MO'. Bonds between homonuclear diatomic molecules are therefore considered to be purely covalent (whoops, almost!). Covalent compounds are the ones you're taught to draw Lewis structures for, apply formal charges, etc., but it's only one extreme of the spectrum.
The next example would be the MO diagram of HCl.

Cl's 3p atomic orbitals are lower in energy (check the provided link for the reasoning); this time round, the atomic orbitals do not contribute equally to the molecular orbital, and Cl's orbital's contribution to the bonding orbital is greater. Electronegative atoms are generally said to have atomic orbitals of lower energy. This is probably more simply and familiarly stated as "the more electronegative atom pulls/attracts the bonding electrons". Note that, now, there's also an energy gap between the initial atomic orbitals. The $\ce{H-Cl}$ bond is, indeed, somewhere between covalent and the other extreme.
Now to the punchline, what happens when the gap is too large? This would be what an MO diagram of LiF would look like:

Thanks Geoff!
The gap is so large that, essentially, the energy difference between molecular orbitals filled with electrons, and fluorine's atomic orbitals would be very small compared to the energy difference between F's 2p orbitals and Li's 2s. Put in simpler terms, the very electronegative fluorine has such a stronger pull than Li that the electrons are almost fully 'localized' on the fluorine, and Li has approximately lost its electron. This is what we call an ionic bond, the other end of the spectrum.
Most bonds between very electronegative atoms and very electropositive ones are considered ionic, and there's the guideline that metal-nonmetal bonds are often ionic. Don't let it trip you up though, since it's a vague general rule, with plenty of exceptions. As you might have suspected, there's no truly, purely ionic bond either, and it's a spectrum.


Biodegradation time for biodegradable shopping bags? How long does it take for the plastic/polythene shopping bags marked biodegradable to biodegrade?

It depends on the environment and on the definition of degrade.
In an anoxic landfill, a very long time before any change is noticed. Even in a wet environment, open to air and sunlight, only the cornstarch additive decomposes.
Initially, it was thought that adding cornstarch would help speed decomposition (and provide an outlet for surplus farm products), but studies such as this show that though some speed in decomposition is gained (shown by water absorption), tensile strength of the plastic is decreased by adding starch, so yet more plastic is needed to make the material sufficiently strong (and the bags still rip).
Also, though the corn-starch additive can decompose rapidly, this study shows that the polyethylene, though now fragmented, does not further decompose. This leads to run-off of microparticles into lakes and oceans.
So if your definition of biodegrade means how long does it take for the bag to shred, according to yet another study performed in a heated environment with added oxygen, about a year. At the end of that time, though, perhaps 80% of the initial mass remains, mostly polyethylene fragments. Much of this eventually is washed into oceans, is eaten, and winds up in us, as well.


Direct connection in a zinc copper battery Why can't you just hook up a piece of wire between a piece of zinc and  copper and get an electric current? What is it about the electrolyte that makes zinc give up electrons?  Also, why will the lack of a salt bridge stop the electric current from moving?   Why  does balancing out the positivity and negativity at the cathode and anode matter?

A close circuit is needed for electrons to flow. The electrolyte and salt bridge serves to complete the circuit. Additionally, the salt bridge also serves to function of maintain electrical neutrality in each half-cell, preventing the accumulation of charges, aiding the flow of electrons.

If charges were to accumulate in each half-cell, the potential difference between the two half-cells, which is the key driver of electrons, would decrease over time to 0 V and no electrons would flow.

There have been many discussions on this site. To cite a few:
What is the real purpose of the salt bridge in an electrochemical cell? (look at the comments)

Why is it important to use a salt bridge in a voltaic cell? Can a wire be used?

electrochemical (Galvanic) cell solution purpose


Is chromium the shiniest metal? I have seen many pictures of chromium, and it's pretty lustrous as I must say. Here are some examples. Is it the most lustrous metal known?

Edit: By lustre, I mean suppose if we have a mirror smoothly polished made out of a material, then I consider lustre as the percentage of reflection of light within the wavelength of visible light (400 - 700 nm). Then according to this scheme, can we consider chromium as the most reflective for visible spectrum?

You likely refer to remission, especially if you would collect spectroscopic data with an integration sphere at hand.  More generally, you refer to reflectance, which depends on the material, the selected wavelength, and the roughness of the top surface -- assuming the selected wavelength is not able to penetrate all across the reflective layer.

(source)

Metals and non-metals (e.g. polymer lamp housings) may be covered by a layer of chromium, and then may be polished to obtain a very smooth surface; in contrast to other metals such a surface remains "bright", even when exposed to air.

Beside the silver mirror already known from sugar chemistry, mercury / almagam based mirrors were quite common till about 1800.  The liquid nature of mercury renders this material still useful in the niche of liquid mirror telescopes.

Glancing briefly over the (n,k) values of metals provided here provides access to calculate the reflectance $R$ as the materials in question

$$ R = \frac{(n-1)^2 + k^2}{(n+1)^2 + k^2}$$

Matching the experience to see much more often silver top-coated mirrors in the optics lab, than of other materials, silver is of higher reflectance than chromium:

(The value at 10 µm equally provided by the referenc intentionally was omitted by mine.)

So why spokes are chromium plated, and not with silver? I speculate the much higher hardness of chromium (Mohs 8.5) than the one of silver (Mohs 2.5) contributes to the reasons why the former is used, and not the later.  It remains a smooth surface.  Maybe electrochemical reasons (chromium likely already is part of the steel / alloy the rims are made of; rather than silver) and the ease Ag would be tarnished by the environment (thinking about $\ce{Ag2S}$) are other reasons, beside obvious cost in material and processing.


Difference in ozonolysis reagents What is the difference when following up ozonolysis with DMS (dimethyl sulfide), Zn/water, or just water? Or is there really no substantial difference at all?
In my textbook, the alkenes section uses DMS or Zn/water but then in the alkynes section it uses water with no explanation as to why that happens. My professor also uses them interchangeably, which gets confusing.

Dimethylsulfide (DMS) is commonly added to an ozonolysis reaction after it is complete (e.g. it is the first step in the reaction work-up).  I can see how zinc could serve the same purpose.  Both of these reagents can reduce oxidized materials.  When the ozonolysis is complete the reaction contains the raw ozonide along with numerous undesirable side-products such as peroxides, hydroperoxides and tetroxanes - all of these materials are potentially explosive.  The DMS or zinc will reduce them to carbonyls and alcohols and remove the explosion hazard.  I suppose that tap water is somewhat acidic or basic depending upon where you live, and it might, therefore, also serve to reduce these materials.

It is one thing to say or write "add water during workup" in a lecture or text, it's another thing to actually deal with explosive materials in the lab.  Personally, I wouldn't chance it. I'd use a real reducing agent such as DMS or zinc, instead of a water work-up.


Are Tellurium Fluoride or Antimony Fluoride Ionic? I would like to know if any of the tellurium fluorides or antimony fluorides are ionic.  I am aware that the difference in electronegativity is greater than the 1.665 cutoff indicating ionic bonds.  Melting point and reactivity seem to somewhat support that tellurium(iv) fluoride is ionic but this is not proof. Being an unusual case of an ionic bond between nonmetallic atoms only though, I would like to know if anything empirical (e.g. spectroscopic) has been observed to support this.

TeF4 is polymeric and low-melting.  Not ionic.  Covalent.
In general, the elements that are metalloids (on the "border" of metal and non-metal) like Te and Sb, tend to behave chemically like nonmetals (looking at their compounds).   But physically as metals (shiny, malleable, conductive).  So that's another good clue that these elements would be form covalent compounds (molecules or polymers ["network solids"]) with fluorine.
Finally even for hard core transition or p group metals, you will find that the lower oxidation count species tend to be ionic.  But as you get to tetravalent species a fair amount are molecular or polymeric (i.e. covalent) and few are ionic.  No penta or higher flourides are ionic.  And the few tetravalent ionic compounds tend to be pretty far to the left of the table (e.g. Zr)...pretty hard/electropositive atoms, not soft (i.e. polarizable) or less electropositive like those in the p group.  (To which Te and Sb are even more marginal metals.)
The Wikipedia article on Compounds of Fluorine has some good details on the concepts and good references to review articles (and Greenwood/Earnshaw Chemistry of the Elements).
https://en.wikipedia.org/wiki/Compounds_of_fluorine#Nonmetal_fluorides
P.s. Note also that it is important to specify which species of fluoride you mean (for elements below the second row).  Te forms a tetra and hexa fluoride (and some more complicated variants)--you did specific tetra, but that's not the only choice.  Also for Sb, you didn't specify which compound and both the tri and penta fluoride are well known.


Does BET theory apply to type H3 and H4 isotherm for calculating surface area? BET applies to type 2 and 4. That is basically unanimous.

The sample that shows loop of H3 type have (very likely) no micropores (small uptake in low p/p0 range). In this case you can apply BET theory in common p/p0 range (0.05-0.3). The sample that shows H4 hysteresis have micropores (large uptake in low p/p0 range). The BET theory can be applied here, but in a much lower p/p0 range. For activated carbon, for example, one would look for a linear(!) range between p/p0 = 0.02-0.1. Generally, the BET theory is not suitable for microporous materials.


Are highly nitrated cyclopentadiene compounds possible? Has anybody synthesized di- or higher nitrated cyclopentadienes possibly with methyl or other groups in the 1 position (analogs of TNT)?

Mono-nitrated cyclopentadienes and their benzo-analogues (indene, fluorene) have been prepared in the context of studies on nitro-aci-tautomerism.

As far as higher substitution is concerned, there seems to be at least a crystal structure of potassium dinitrotrichlorocyclopentadienide reported in Acta Cryst., B

In the abstract, the authors (Y. Otaka, F. Marumo, and Y. Saito) mention the preparation of this compound by Haga in 1969, (sorry, reference is behind paywall) via

  nitration of hexachlorocyclopentadiene with alkali nitrite in an aqueous or ethanolic solution

NOTE: Hexachlorocyclopentadiene once was a favoured starting material in the synthesis of insecticides via Diels-Alder reaction. However, it is a potent teratogen and the "mother" of least five members of the dirty dozen (persistent organic pollutants) nowadays banned for good reason.

You might want to read the MSDS and then decide that you'd rather prefer to work with something else.


Calcium carbonate toxicity I can't make heads or tails of the information on the net — although there seems to be plenty of it, it seems contradictory.

Supposedly, calcium carbonate is used in antacids—yet there are warnings about ingestion and contact with skin.

I need to handle it in pure powder form. I'm mainly concerned about skin contact, should it be avoided? (Am I getting confused between two or more forms of calcium carbonate?

Looking at the Material Safety Data Sheet (MSDS), most of the issues with it seem to stem from the physical consequences of it being in fine dry powder form. This means that if you get it on your skin, it can get into cracks and pores and easily dry out your skin to the point of cracking, and will cause irritation or worse if you get it in your eyes, for example. It's not toxic if you get it on you, though -- rubbing a lump of limestone against your skin isn't going to do anything except abrade it.

So, yes, if you're working with the powder, wear gloves. And safety glasses. And long sleeves. (See also: lab coat.)

If in doubt, follow the MSDS.


Basic Intro to Quantum Chemistry. Two Slit Experiment and Bohr Model We just learned about the 2 slit experiment in Quantum Chemistry today, where electrons behave as waves when nobody is looking and behave as particles when they are being observed.

So, what would happen If I were to observe an atom (such that all it's electrons now behave as particles instead of waves) for a long period of time?

I know that real atoms have wave electrons, but when you observe them, don't they "turn" into particle electrons and obey Newtonian law like the 2 slit experiment? So what keeps the electrons from falling into the nucleus when you observe them? I was considering how the earth doesn't "fall" into the sun because of a momentum perpendicular to the centripetal force, but the mass of an electron is so much smaller than the nucelus, I'm not sure it will work.

If anyone can give a simple explanation that would be great, I asked the professor and he only said the the Bohr model was inaccurate and it's difficult to observe electrons.

Thanks in advance.

As some commenters have stated, this is a complicated topic, and if you really want to understand it you would probably be best to study quantum mechanics itself. Although it's detailed it's also interesting!

However, as to your question, the first thing to ask is what does it mean for an electron to be "observed"? It can't mean just looking at them with your eyes, because then any matter we looked at would suddenly go crazy when all of its electrons change their behaviour. It also doesn't mean just looking in a normal light microscope, because electrons are too small to see in a light microscope. In fact to really properly "see" an electron, we would need to use very powerful high energy equipment. So high energy in fact, it would give the electron enough energy to fly out of the atom it was attached to. After that it perhaps could well fly off like a "normal particle" electron with a Newtonian path.

Now the obvious question everybody then asks is, is there a way to "look at" the electron bound to an atom without giving it so much energy? And the answer according to physics is no, it is actually impossible, to really "look at" a bound electron enough to make it act like a particle, it always takes a big amount of energy. You could "look at" it just a little bit, just enough to find what energy level it is in for example - but as you probably know this isn't enough to stop it behaving like a spread out wave (atomic orbital).

Now as to your question about why it doesn't fall in to the nucleus, you have actually already answered it. Why would it? The earth doesn't fall in to the sun. You mentioned that electrons are very light, but that doesn't really matter. If we hollowed out the earth to make it much lighter that wouldn't affect its orbit. Now the real reason is a kind of "quantum mechanical" version of this argument that I could explain if you like, but it is complicated and really this is the key reason.


Acid-Catalyzed Hydration Is there any particular reason why this mechanism is used in the solution manual rather than an acid-catalyzed hydration?

I was thinking of having the double bond attack a hydronium ion, which would create a carbocation. This would then create a place for water to attack. Finally, the water would be attacked by water to reform hydronium ion.

Why wouldn't this work?

Does this have anything to do with the fused ring?

Acid catalyzed hydration probably would work, but might well be accompanied by other materials derived from the carbocation intermediate, such as rearrangement and elimination products, which would make this route less desirable.  The mercuric acetate route does not involve free carbocations, so such side products will not contaminate the desired product.


Can air conditioning cause urine smell? On rainy days there is a strong urin smell on trains in Tokyo.
I suspect this smell comes from ammonia.
On Wikipedia I read:

  Ammonia and ammonium salts are also found in small quantities in rainwater […] Ammonium salts are found distributed through fertile soil and in seawater.

Both is valid for Tokyo. In rainy season especially there is a lot of rain. And Tokyo is located at the sea.

Air conditioners work by drying the air.
Tokyo trains make extensive use of air conditioning.
Could this process intensify or make the ammonia to fall out and thus create the strong urine smell?
What other possibility is there for the strong smell?

First, human sweat contains urea, so it doesn't require incontinent commuters. Despite constant cleaning, this effluvia accumulates in the rail cars.
Second, air conditioners precipitate water, entrained dust and volatile organic compounds, making an ideal environment for bacteria, fungi and protists to swim around, excreting their wastes. Microbial volatile organic compounds are none too pleasant, and may have the smell of urine or ammonia (though the waste products of yeast in baking bread can be quite appealing).
Third, there are probably birds in and around the train stations. Birds produce uric acid, which is rapidly converted to ammonia by fecal flora.
Fourth, in older drainage systems, rain water and waste water may be mixed. Given a high water level, wastes may be closer to the surface... or overflow.