How do you connect a battery to a galvanic cell in order to reverse the flow of electrons? Assume you have a galvanic (voltaic) cell with E = +2V. If you wanted to reverse the current by inserting a 9V battery, which end of the battery should connect to the cathode of the galvanic cell?
The Kaplan answer key states that the "positive end of the battery should connect to the cathode."
Since electrons flow from anode to cathode in the galvanic cell, we want them to flow from cathode to anode to reverse the current. In a battery, electrons move from the negative to positive end. Therefore, why wouldn't the positive end of the battery be on the galvanic cell's anode to attract electrons in order to reverse the current?
Could someone point out the flaw in my reasoning?
Thanks :)

The electrochemical signs and labels have confused several generations of students. You have to remember a couple of things about the labels for the electrolytic and galvanic cells.
Anode: The electrode at which oxidation takes place.
Cathode: The electrode at which reduction takes places.
The signs of the electrodes of a galvanic cell are electrostatic signs (positive and negative) of the electrodes. This is an important point to keep in mind, which is often not emphasized during teaching. So (-) sign on the battery terminal indicates excess of electrons and (+) sign indicates deficiency in electrons.
In a galvanic cell, the cathode is positively charged and the anode is negatively charged. Work this out.
In order to reverse the flow of electronic current in the 2V galvanic cell, you need to connect the electron rich electrode of the 9V battery to the electron rich electrode of the galvanic cell i.e., you will connect the negative terminal of the 9 V battery with the negative terminal of the 2V galvanic cell and connect the positive terminal of the battery with the positive end of the galvanic cell.
Kaplan's wordings is quite bad!
Consider the analogy of water supply in your own house: When you open a tap, water spontaneously flows out because the actual source of water is at a higher gravitational potential than your tap. What do we need to do to reverse the direction of water flow in the tap. We need to connect the tap to a powerful water pump which overcomes the gravitational potential energy of the water source and pumps the water against the gravitational potential. This battery connection is exactly the same because the battery generates 9V and the galvanic cell produces 2V.


Does a tertiary carbocation rearrange to another tertiary carbocation? I strongly believe that a carbocation should not rearrange to another if there are no immediate benefits(like a Greedy Algorithm).

The doubt hit me while solving this question. The answer mentioned in the book answers (b) but my answer comes out to be (a).

Here is the mechanism:

Option B is the correct one. The process of dehydration of alcohols follow E1 elimination mechanism which involves two steps out of which the first one is slow ionisation of C-X bond( where X is any hetro atom) and the second step is  fast removal of $H^+$ ion .Since the first step is slow it provides enough time for the carbocation to rearrange and form a stable carbocation . The mechanism itself is self explanatory. Firstly as Oxygen possess lone pairs of electrons so the $H^+$ ion attacks the OH group as a result the $H_2O $ molecular gets eliminated resulting in a positive charge there.  Now hydride shift takes place as the so formed $ \textbf{Carbocation is more stable as it has 7 alpha hydrogens compared to the 5 alpha hydrogens of the previous carbocation}$. Now the C-C bond undergoes ionisation to form a three degree Carbocation and a carbanion . Now this carbanion forms a bond with the Carbocation formed earlier to form a six member ring . Now the resulting carbocation undergoes methyl shift which produces more stable three degree Carbocation. Now a $H^+$ ion from the neighbouring carbon atom is ejected as a result the pie bond is formed.


Will silicon dioxide react with hydrochloric acid? I have the following equation, based on the reaction of $\ce{SiO2}$ with $\ce{HF}:$

$$\ce{SiO2 + HCl -> SiCl4 + 2 H2O}$$

I want to know if this reaction is indeed feasible, what conditions may be needed to induce it, or if it isn't possible why that's the case.

According to the paper by Chung and Sturm [1], $\ce{HCl}$ will do a very small amount of etching at high enough temperatures (525 °C will give you about 1 nm/min), but the presence of hydrogen limits the reaction because the formation of $\ce{HCl}$ is preferred to the formation of $\ce{SiCl4}.$

References

Chung, K. H.; Sturm, J. C. Chlorine Etching for In-Situ Low-Temperature Silicon Surface Cleaning for Epitaxy Applications. In ECS Transactions; ECS: Chicago, Illinois, 2007; Vol. 6, pp 401–407. DOI: 10.1149/1.2727426.


How to estimate the dielectric constant of a liquid? Are there any approximate relationships between the (static) dielectric constant of a liquid and properties of the individual molecules?

For example, suppose I know the polarizability and dipole moment of a molecule, as well as it's density in the liquid phase. Can I estimate the dielectric constant of the liquid from these quantities (or other similar ones)?

Yes, such relations exist.

Notably there are two general theories for polar and non-polar fluids that connect the polarizability $P$ with the dielectric constant $\epsilon$ as reviewed in this article by Harvey and Lennon, International Journal of Thermophysics, Vol. 26, No. 1, January 2005.

The Clausius-Mosotti equation for non-polar liquids is:
$$P = \frac{\epsilon-1}{\epsilon +2}$$

Whereas an equation developed by Kirkwood for polar liquids is:
$$ P = \frac{(\epsilon-1)(2\epsilon+1)}{9 \epsilon}$$

The article then continues to explain that you can estimate the polarizability from theoretical relations involving the molar density $\rho$ and the dipole moment $\mu$, but in your case the two equations above are sufficient since you assume that you know $P$.


How many ligands in Heme? How many ligands does Fe has in heme ?
I am not sure how many are in heme and how many in cytochrome C.
Its about iron complexes.

Heme: The Iron molecule is bound to 4 Nitrogen atoms, but take a look at porphyrin. All the nitrogens are part of one actual structure. You can visualize the iron molecule in the center of the larger compound's structure, replacing the two Hydrogen atoms in N-H:
The large structure surrounding and coordinating with Fe is also known as a polydentate ligand since it bonds to Iron at more than one site. So there is just one ligand on Fe.

Cytochrome C: is a heme protein, so you may expect the same answer: A single polydentate ligand binds to Iron. This page has a PDB Model of it in a plugin.
As one ligand coordinates to four sites in this case, other ligands typically attach (e.g. oxygen binding in hemoglobin) to fill the octahedral group (up to 6 ligands attached to Fe the coordinate center).


How to calculate wavenumbers of normal modes from the eigenvalues of the Cartesian force constant matrix? I would like to get the wavenumbers (Frequencies -- in Gaussian) from the eigenvalues of the force constant matrix (Force constants in Cartesian coordinates: in Gaussian).

Could someone please explain how to perform this calculation? The Cartesian force constants are given in Hartrees per Bohr squared, while the wavenumbers are in $\pu{cm^-1}$.

Short answer: (1) mass weight each force constant (see below). (2) Diagonalize the matrix. (3) Convert the eigenvalues ($\{ k_1^{\prime\prime} \}$) to wavenumbers $\tilde{\nu}=\sqrt{k_i^{\prime\prime}}/(2\pi \times 137 \times 5.29\times 10^{-9})$.  137 is the speed of light in atomic units and $5.29\times 10^{-9}$ converts bohr to cm.

More detail

Harmonic vibrational frequencies come from solving $F=ma$ assuming $E(x)=\frac{1}{2}k(x-x_0)^2 = \frac{1}{2}k^\prime l^2$ where $l=\sqrt{m}(x-x_0)$ and $k^\prime = k/m$. $F=-(\partial E/\partial x)$ so $F=ma$ becomes
$$ \frac{\partial^2 l}{\partial t^2}+k^\prime l = 0 $$
The solution to this differential equation is
$$ l = K \cos(\sqrt{k^\prime}t) +\varepsilon$$
where $K$ and $\varepsilon$ are arbitrary constants.

The solution tells us that the displacement coordinate will increase and decrease in a repeating fashion every $\sqrt{k^\prime}/2\pi$ times, which is the definition of frequency ($\nu$). The units of $k^\prime$ are energy/(mass length$^2)$ or 1/time$^2$, because energy = (mass)(length/time)$^2$.  So the units of $\nu$ is 1/time.  This can be converted to a wavenumber ($\tilde{\nu}$, 1/length) by dividing $\nu$ by the speed of light.

Diatomic molecule

You have 6 Cartesian coordinates $\{q_i\}$ and 6x6 Cartesian force constants (some have the same value),
$$ k_{ij} = \left( \frac{\partial^2 E}{\partial q_i \partial q_j} \right) $$
and it's not clear what to do with the cross terms.
So you diagonalize the matrix to get rid of the cross terms, but you need to mass-weight first for the reasons described above.
$$k^\prime_{ij} = \frac{1}{\sqrt{m_i m_j}}k_{ij}$$
After diagonalization of $\mathbf{k^\prime}$ you have 3N-5(6) non-zero force constants $\{ k_i^{\prime \prime}\}$ , which you can convert to wavenumbers by $\sqrt{k_i^{\prime \prime}}/2\pi c$


What product will be produced when a sucrose and potassium nitrate mix burns? What product will be obtained get when a sucrose and potassium nitrate mix, which are the ingredients for smoke bomb, are burned in a combustion container?

$$\ce{KNO3 + C12H22O11 -> ?}$$

You will get potassium carbonate ($\ce{K2CO3}$), carbon dioxide ($\ce{CO2}$), water ($\ce{H2O}$), and nitrogen gas ($\ce{N2}$):
$$\ce{48KNO3 + 5C12H22O11 -> 24K2CO3 + 36CO2 + 55H2O + 24N2}$$

See this yahoo question and this science forums thread.


Do dyed-alcohol thermometers tolerate cold/hot stems? How do the red organic solvent thermometers respond to having the tip submerged while the stem is at a different temperature? I understand mercury thermometers could easily handle this, and bimetallic strip probe thermometers fail utterly.

To put it simply, if I'm heating just 2-3 cm of liquid, can an organic solvent thermometer still give accurate readings?

The very shortest answer is no. Since asking this question, I've learned that thermometers have an insertion depth they are calibrated to, presumably assuming the rest of the thermometer is somewhere between room temperature and the bath temperature. Some thermometers are "total immersion", meaning the conductivity of the fluid doesn't matter if used as intended, because it's expected to conduct through the sides as well.

Given what I've learned, the first half of the answer to this question is that if a thermometer is calibrate to be immersed 2 cm, then immersing it deeper will give a wrong reading.

This article explains further: https://www.coleparmer.com/tech-article/glass-thermometer-selection-guide. In particular, correction factors are given for when a thermometer is not inserted at its calibrated depth:

  Emergent Stem Corrections. Avoid inaccurate readings when a total
  immersion thermometer cannot be properly immersed. Determine the
  approximate stem correction with these formulas:

For mercury Fahrenheit thermometers:
°Correction = 0.00009°F x n x (T-t)
For mercury Celsius thermometers:
°Correction = 0.00016°C x n x (T-t)
For spirit-filled Fahrenheit thermometers:
°Correction = 0.0006°F x n x (T-t)
For spirit-filled Celsius thermometers:
°Correction = 0.001°C x n x (T-t)

  where T is the bath temperature (the temperature indicated on the
  thermometer), t is the average temperature of the emergent stem, and n
  is the number of emergent thermometer degrees. Hold an auxiliary
  thermometer next to emergent stem to determine t.

While these corrections aren't meant to be applied backwards (for a partial immersion thermometer that's immersed too deep), the numbers speak for themselves: a colored organic thermometer is not nearly as accurate when it's either immersed too far or too shallowly. I plugged in some numbers for a very nearly worst case scenario (cold room with a strong draft where the thermometer is calibrated to be fully inserted but is actually only inserted up to the 0⁰C mark):

T=100⁰C
t=20⁰C
n=100
mercury °C correction = 0.00016°C x n x (T-t) = 1.28⁰C
spirit °C correction = 0.001°C x n x (T-t) = 8⁰C

The final answer is that no, dyed-spirit thermometers do not tolerate being at the wrong insertion depth, at least not if your experiment's parameters are too far off from what the thermometer manufacturer intended.


How do self-relighting (trick) birthday candles work? You may have seen those birthday candles, where, upon blowing them out they smoulder for a bit and then (magically) relight.

How do they work? Why can't regular candles do this (is it possible for a regular candle to relight itself, anyway?)

In normal candles, as you blow them out, you will see burning embers in the candles, which vaporises the wax and thus causes a ribbon of paraffin wax to rise up (this looks like a wisp of smoke). These embers are often not hot enough to light the ribbon of wax, so the candle goes out.

However, in trick candles there is magnesium put in the wick. When normally burning, the liquid wax shields it from oxygen so it cannot ignite. However, when the candle is blown out, this magnesium is able to react with oxygen (ignite) and light up the paraffin wax ribbon rising, reigniting the candle (or rather, the volatile materials above the candle). The magnesium is often powdered or in thin strips for low ignition temperature and easy ignition.

Sometimes regular candles do reignite, in these cases, the embers are hot enough to light up the paraffin wax vapour


Why does copper sulphate (CuSO4) in water increase surface tension? So, recently for school, I did an experiment on surface-tension. We tested Water, Salt, Sugar, Dissolvable Panadol, Coffee Powder, Vinegar, Detergent, Flour, Methylated Spirits, Oil & Detergent, Copper Sulphate, Iron Sulphate, and
Ammonium Chloride.

Our theory at the time was:

Copper sulphate has a high surface tension when mixed with water because it has a lot of hydrogen bond acceptors that are not used. As more hydrogen bonds increase the surface tension, the increased amount of possible hydrogen bonds thus increasing the surface tension. What the copper sulphate is doing is being a whole bunch of extra links connecting the hydrogen atoms of water molecules together.

My Current Theory:

As Hydrogen likes Oxygen, the large amounts of Oxygen bonds in CuSO4 strongly attract to the H2O, which increases the pull force. Also, both H2O & CuSO4 are polar substances, helping them stick together.

I have no idea if any of this is true; our chemistry teacher had no idea either. Does anyone have a better theory/proof as to why this happens?

Note; We did not use capillary tubes as the order time would exceed the time our assignment due date. Also, our data might just all be noise too, as our method was not the most error-free design possible.

Our results are here.

(Relevant Results:)

                | Average | Test1 | Test2 | Test3 | Test4 | Test5 |
Control         |  3.21   |  3.25 |  3.4  |  3.1  |  3.1  |  3.2  |
Copper Sulphate |  3.94   |   3   |  4.7  |  3.5  |  4.5  |   4   |

Method:

Fill beaker with 40 mL of water (4th line)
Mix in approximately 3 mL amount of solute.
Using a pipette, drop water into a bottle cap until liquid is flat with the top.
Fill a pipette to 3 mL. Slowly drop in the water until it overflows, taking note of how much is used (i.e. emptied once & 2 mL used = 5 mL on top)
Record the results in a table. Repeat 4 more times.
Repeat the procedure with different solutes in different beakers.

From Sayok#6589 on Discord:

  When dissolving CuSO4 in H2O, it does not remain as CuSO4. It's a soluble salt, so you make [Cu(H2O)6]2+ and [SO4]2-. The first coordination shell of either ions is filled with water, forming electrostatic/polar bonds between the water molecules and the ions.
  Because the water molecule are forming bonds with charged ions, their polarity is slightly increased compared to a regular water molecule.
  Surface tension of H2O is formed because of the polarity of the water molecules. They like each other and want to stick with each other. In general, the more polar a solution is, the more surface tension it can have.
  Increasing the polarity of the water molecules because of the presence of the charged ions in solution thus results in a small increase in the surface tension.


Why is caustic soda (NaOH) used to clean beer bottles? I see in many blogs in the internet that cleaning bottles with $\ce{NaOH}$ is very effective. I also realize that industries use caustic soda to clean bottles. Why is caustic soda very effective at cleaning bottles? What makes it so special from other solvents which made it preferable for cleaning?

Caustics such as $\ce{NaOH}$ and $\ce{KOH}$ are effective in converting insoluble fatty acids in organic fats and oils into soluble soaps in a proccess called saponification. The caustics also react with cell membranes, DNA and RNA, destroying undesired yeasts, bacteria and other microorganisms.

That said, caustics react with the glass itself, removing the surface and etching the glass. Use will weaken and damage containers.

Use caution! Caustics also destroy skin and clothing, and are particularly damaging to the eye.


What is a polyfuranoside? I came across this molecule on wikipedia, and all that was written about it was two words: Docosanasaccharide and polyfuranoside.

https://en.wikipedia.org/wiki/Chemical_glycosylation#Notable_synthetic_products

The term polyfuranoside is a combination of

the Greek prefix poly, which means many and
the term furanoside, which indicates that the structure is somewhat related to furan, a five-membered (aromatic) heterocycle that contains an oxygen atom. Forget about the aromaticity here, what counts is

five-membered ring
contains an oxygen atom

If you take a look at the large diagonal of the structure, you'll recognize a large number (12) of these structure elements connected by glycosidic bonds, hence the other description, docosanasaccharide.


Why does ethanol have a health rating of [2] in NFPA 704 whereas methanol, which is far more toxic, has a rating of only [1]? Why Ethanol has the blue health hazard rating of 2 in NFPA 704 where as both Acetone and Methanol which are far more toxic than Ethanol have a rating of only 1?    (Click on each link to see the NFPA diamond)

Ethanol
Methanol
Acetone

I believe that the rating is dependent on toxicity tests.

The acute oral toxicity LD50 value of acetone is 5800 mg/kg

The acute oral toxicity LD50 value of methanol is 5628 mg/kg

The acute oral toxicity LD50 value of ethanol is 3450 mg/kg

Low LD50 values means that only a small amount of chemical is needed to kill someone. According to these values, you would need to ingest a larger amount of acetone or methanol to die.I am not sure how valid these toxicology test are. Fun fact: fructose is deadlier than methanol according to LD50 values.

Furthermore, there appears to be some variation in the values as well. Wikipedia lists the LD50 value of ethanol as 7,060 mg/kg.
Maybe this question can be addressed at Biology SE.


Acidic cleavage of an ether In the following question;

Apparently, an SN2 reaction ensues, where the iodine atom's supposed to cleave off with the less sterically hindered alkyl branch from both the oxygens. This is the solution:

That's what happens with the lower oxygen, but what's going on with the top one? In the second step, why doesn't the methyl attached to it cleave off with the iodine atom?

The Wikipedia page on ether cleavage says:

  Depending on the specific ether, cleavage can follow either $\mathrm{S_N1}$ or $\mathrm{S_N2}$ mechanisms. Distinguishing between both mechanisms requires consideration of inductive and mesomeric effects that could stabilize or destabilize a potential carbocation in the $\mathrm{S_N1}$ pathway. Usage of hydrohalic acids takes advantage of the fact that these agents are able to protonate the ether oxygen atom and also provide a halide anion as a suitable nucleophile.

So in essence, the mechanisms for such reactions are usually a trade-off between $\mathrm{S_N1}$ and $\mathrm{S_N2}$ pathways, with adequate carbocation stability playing a major role in the determination of the pathway (as intermediate cases like secondary cations often lead to a "confusing" case where it is difficult to say qualitatively whether the path is majorly $\mathrm{S_N1}$ or $\mathrm{S_N2}$).

In particular, the Wikipedia page cited above has pretty much answered your question under the $\mathrm{S_N2}$ ether cleavage tab:

  If the potential carbocation can not be stabilized, ether cleavage follows a bimolecular, concerted $\mathrm{S_N2}$ mechanism. In the example, the ether oxygen is reversibly protonated. The halide ion (here bromide) then nucleophilically attacks the less sterically hindered carbon atom, thereby forming methyl bromide and 1-propanol.

So for your case, the upper oxygen after protonation can cleave the $\ce{C-O}$ via a $\mathrm{S_N1}$ pathway to generate a reasonably stable potential carbocation which is delocalized over the benzene ring.
Hence,it can simply remove a molecule of methanol(which is a good leaving group by itself) on it's own without needing an iodide to assist in the removal.

The strongly acidic solution of $\ce{HI}$ will anyway ensure it that the methanol undergoes $\mathrm{S_N2}$ to form methyl iodide later. Also, from the point of view of a nucleophile,it would be much easier to perform an $\mathrm{S_N2}$ reaction on a separated methanol molecule than on the entire substrate molecule initially given in the question due to steric considerations,so that also becomes a contributing factor.

But as for the lower oxygen,the poor fellow does not have such a choice. The potential carbocation that can be generated by cleaving the lower $\ce{C-O}$ bond is a phenyl cation,which is pretty unstable as explained here. So, like the Wikipedia page said and illustrated in the example above, the pathway has to be a concerted $\mathrm{S_N2}$ mechanism now so as to take the reaction forward.

Note: Even if the said $\mathrm{S_N2}$ starts out slow, Le Chatlier's principle will ensure that ultimately, most of  the reactant goes into the product as equilibrium is attained.


What causes the yellow colour in the Briggs-Rauscher oscillating reaction? The color changes are due to different concentrations of supposedly $\ce{I2}$ (which causes the yellow/amber color), $\ce{I-}$ (colorless) and $\ce{I3-}$ (dark blue).

Since iodine ($\ce{I2}$) is barely soluble in water, I do not understand why in this case it would be.

If one was to put just iodine in water and mix it thoroughly, the color change of said water would not be as visible to the extent of the Briggs-Rauscher reaction.

The color is in fact caused by the elemental iodine. At the 'yellow' stage of  reaction the concentration of iodide is low, so the triodide anion (yellow at low concentration and red-brown at high) does not form, hence there is no formation of the blue triodide-amylose complex.

The OP wonders how the iodine concetration can be high enough to give the solution its yellow coloration, given the low solubility of elemental iodine in water?

First let me make a remark that crystalline iodine is very sluggish to dissolve in water to reach its maximal concentration. Even when making potassium iodide - iodine solutions - it takes a lot of stirring, anyone who worked in an analytical lab knows that.

But this is not the answer to the question. You have to consider that the iodine in solution does not come from a solid phase going to the liquid phase - it is generated in a homogenous chemical reaction. Therefore, it can reach concentrations far beyond the maximal solubility, forming a metastable supersaturated solution. This stage of the oscillation does not last long enough to allow nucleation and precipitation of the elemental iodine. The subsequent increase of iodide concentration causes formation of triodide, which along with starch gives the blue color and so on.


Heats of combustion of cycloalkane homologues Heat of combustion increases as the number of carbons increases, but on comparing ethyl cyclopentane, ethyl cyclohexane, andethyl cycloheptane, shouldn't you consider ring stability too?

In that context, it should be 6-membered ring < 7-membered ring < 5-membered ring.

Why isn't this so?

Have a look at this table from Organic Chemistry by Wade and Simek [1, p. 134]:

Every time a $\ce{-CH2}$ group is added, the heat of combustion is increased by almost 650 kJ/mol; whereas the maximum relief in ring strain attained by adding that $\ce{-CH2}$ group is not more than 25 kJ/mol.

So there is no competiton between increase in heat of combustion due to more number of carbon atoms and decrease in it caused by ring stability. Everytime you see two hydrocarbons with different number of carbon atoms, be assured that the one with more carbon will realease more heat on combustion.

P.S. If in case you are unfamiliar with how ring strain is quantitatively measured [1, p. 133]:

  The molar heat of combustion of cyclohexane is nearly twice that of cyclopropane, simply because cyclohexane contains twice as many methylene ($\ce{-CH2}$) groups per mole. To compare the relative stabilities of cycloalkanes, we divide the heat of combustion by the number of methylene ($\ce{-CH2}$) groups. The result is the energy per $\ce{-CH2}$ group. These normalized energies allow us to compare the relative amounts of ring strain (per methylene group) in the cycloalkanes. The reference value of 658.6 kJ (157.4 kcal) per mole of $\ce{-CH2}$ groups comes from an unstrained long-chain alkane.

References

Wade, L. G.; Simek, J. W. Organic Chemistry, 9th ed.; Pearson: Glenview, IL, 2017. ISBN 978-0-321-97137-1.


Misleading features of Lewis Dot Representation Today I had an embarrassing experience when I sat to help my nephew learn Lewis' Dot Representation and realized that I actually don't get it at all. I have a physics background and always thought that the Lewis Structure in chemistry mimics the physical picture, but soon realized that it doesn't.

First of all, the "sockets" where the dots go on Lewis Diagrams do not correspond to actual physical orbitals. They are neither 4-fold symmetric, nor are equivalent (s-orbitals are spherically symmetric, p-orbitals are lobed, etc), so I couldn't give any explanation to my nephew what their physical interpretation is.

Also, I noticed that the rule for filling these "sockets" do not follow the Aufbau rule. For example, in the ground state of carbon, Nature first fills the $\ce{2s}$-orbital and just after that inserts an electron in two of the $\ce{2p}$-orbitals. In contrast, the rules of Lewis' Dot Representation say that C should have a single dot in each of its 4 "sockets", and just after they get exhausted, one starts pairing them. (I wonder how this principle explains compounds like methylene, in which carbon has its $\ce{2s}$-shell pair present... but anyway.)

All of this prompted me to ask, what is the story behind Lewis' Dot Structure and what goals Gilbert Lewis pursued when he introduced it? I guess if I learn this, I'll be able to explain its misleading features and not get embarrassed in front of a 14-year-old.

Lewis dot structures predate the advent of modern quantum theory and quantum chemistry. There is no Schroedinger equation associated with them -- in fact placing a dot on an atom (seems to me that it) violates the concept that electrons must be indistinguishable.

There are no slots, there are no "octet rules".  There are only particles, and kinetic and potential energy that depend on their masses and charges. They take a configuration that minimizes the total energy with respect to their location in space.

So my advice is to learn Lewis dot structures as a heuristic device that can be figured out for the purposes of passing General Chemistry and speaking with the vast majority of chemists who have continued to suffer with learning about octet rules, resonance structures, etc. and the "astonishment" when these rules fail. They are hard to learn, because one can not draw on experiences learned about other physical systems that depend on e.g. QM, electrodynamics, etc.

There is a subject known as Valence Bond theory, that uses (spin-adapted) Lewis structures as a basis. But if you knew what went into constructing a proper quantum mechanical valence bond wave function, you might as well say "sure, anything so long as it forms a complete basis."  i.e. we might as well talk about molecules being made up of millions of plane waves.

You did mention orbitals: "Orbitals" only arise due to the fact that they are complete bases of Spherical Harmonics and Laguerre Polynomials that were used by Schroedinger to solve the case of the hydrogen-like atom -- a spherically symmetrical system.  They do not really apply to molecules, and it is common to use other bases to describe molecules such as Gaussian-Type orbitals, plane waves, etc.  Atomic orbitals, molecular orbitals (constructed from linear combinations of AO's) and plane waves have never been observed.  Mathematically, orbitals may be complex, and leaves me wondering why we place so much stock in them except as useful complete bases with convenient boundary conditions, e.g. periodic with a unit cell or decaying in the limit of infinite distance.  (In contrast, the electron density is real and observable.)


Fluoride Color Reaction I am trying to find visual ways to demonstrate water contaminants. For example, chlorine reacts to N,N-diethyl-p-phenyldiamine (DPD), forming a magenta color. The chemical doesn't need to be accurate, as this is more of pass-fail test of the water. What reacts with fluoride to provide similar results?

Fluoride has a high affinity for $Fe^{3+}$, and $Fe^{3+}$ forms deeply coloured complexes with phenols.  The decomposition of the latter by fluoride ion is the basis for colorimetric determination of fluoride.

Ligands used in this kind of assay include 5-phenylsalicylaldehyde and resacetophenone.  For occasional testing of well water I'd recommend any analytical lab.  I'd be very surprised if Merck didn't offer any fluoride test strips, at their price of course, although whether they go down to the 5 ppm range is a different question.  For those that really like to spend money, there's fluoride-sensitive electrodes.


Fraction of concentration of benzene in air water We have a 120 ml water sample with an unknown concentration of benzene. If the bottle size is 160 ml (with 40 ml of air) and benzene has reached equilibrium, what fraction of benzene is left in the water?

We are only given Hcc (Henry constant - Cair/Cwater) 0.225 and Log Kow (to which I believe is needed for a later question) 2.53.

I'm not sure where to start. We do not know mole fraction so we cannot find partial pressure using Henry's law. Also this rules out Raoult's Law.

Also we can't use the Ideal gas law since we do not have a lot of the variables. I'm stumped. But we can make assumptions such as T for example.

Any suggestions?

You don't need the original concentration or any of the variables you are missing. Henry's Law gives you everything you need. The usual formulation of Henry's Law at equilibrium:

$$c_{i,s}=k_{H,i,s}c_{i,s}$$

Where $c_{i,air}$ is the concentration of solute $i$ in the vapor above the solution, $k_{H,i,s}$ is the Henry's Law constant for solute $i$ in solvent $s$, and $c_{i,sol}$ is the concentration of $i$ remaining in solution.

We can rearrange Henry's Law to give you what you need - a ratio of the amount of benzene in the air to the amount in the water:

$$k_{H,i,s}=\frac{c_{i,air}}{c_{i,sol}}$$

The other snippet of information you need is the relationship between moles and concentration. There is a reason you were given volumes.

$$n_{air}=c_{air}*V_{air}$$
$$n_{sol}=c_{sol}*V_{sol}$$

You only need the ratio of $\dfrac{n_{air}}{n_{sol}}$, so you really only have one unknown. It's algebra time!

$$K_{H,i,s}=\frac{c_{i,air}}{c_{i,sol}}=\dfrac{\dfrac{n_{air}}{V_{air}}}{\dfrac{n_{sol}}{V_{sol}}}$$

$$\dfrac{K_{H,i,s}V_{air}}{V_{sol}}=\dfrac{n_{air}}{n_{sol}}$$


Migratory aptitude in Baeyer-Villiger reaction In short, I do not understand how or why the migratory aptitude is as listed in every textbook I've read.

Firstly, they talk about a positive charge being built up that is stabilised by the movement of the most substituted group.

I cannot fathom such a build up in my mechanisms.

Please help me understand this in a mechanistic fashion.

The migratory aptitude list for the Baeyer-Villager oxidation is as it is because that is how the molecules undergoing the reaction behave. When we are learning about new reactions from a textbook, we often read about the mechanism and theoretical explanations of chemo-, regio-, and stereoselectivity. It is easy to forget that the experimental observations came first.
The migratory aptitude list $(\ce{H>R3C>R2HC}\approx\ce{Ph(>H)>RCH2>Me})$ is the result of many experiments. Nor is it a linear list as is usually presented. $\ce{H}$ migrates more readily than tertiary groups, which migrate more readily than secondary groups, which migrate more readily than phenyl groups (which migrate more readily than $\ce{H}$).
The mechanism and theoretical explanations came later. The partially complete mechanism from the Wikipedia page is shown below. The key step (center) is a concerted collapse of the tetrahedral intermediate with a [1,2]-sigmatropic shift of $\ce{R}^1$ and ejection of the carboxylate leaving group. Note that this mechanism shows none of the proton transfer steps.

Since the whole thing is supposedly concerted, I can understand your objection to the build-up of $\delta^+$ on $\ce{R}^1$. However, with the exceptions of the anomalous behavior of $\ce{H}$ and $\ce{Ph}$, the migratory aptitude list is the same as the carbocation stability list. Thus, researchers looked for a link between the two phenomena and a way to relate one to the other. The link may not be real, but the trends are there. To a student learning organic chemistry, it is much friendlier at first to apply an existing trend in reactivity to a new reaction (with an exception), than to learn a whole new trend that is unrelated but cosmetically similar.


Python command line tool to get chemical structure In computational chemistry, we need to provide an starting molecular structure to start our calculations. Is there any way to get that from a chemical database? Say, I want to run H2, it would call something like

url="www.chemicaldatabase.com"
H2=(url,H2)

or it can use smile/other standard symbols for chemical representation.

There are multiple approaches in Python.

My suggestion would be to use a cheminformatics library like Open Babel or RDKit to convert from SMILES (for example) to 3D coordinates.

If you want to grab from chemical databases, I can suggest two approaches:

CIRPy - Uses the NIH chemical resolver to convert from names, SMILES, etc. into 3D structures.
Webel - This is a web-based cheminformatics tool. I've used it in the past, but I'm not sure if it's still maintained.

There are other databases, including my PQR and PubChemQC that offer QM-optimized geometries.

The catch with databases is that you might want the geometry of a molecule that's not in the database. In that case, Open Babel or RDKit is a better solution.

One other caveat. Nothing I've indicated above does a very good job with metal-containing species. If you want ferrocene, that's a trickier problem with current solutions.


Why are amines more soluble than ethers in water? For example, diethyl ether ($\ce{C2H5OC2H5}$) has a limited solubility of $6.05\rm~\frac{g}{100~mL}$ water at $25\rm~^\circ C$. However, diethylamine ($\ce{(C2H5)2NH}$) is a lot more soluble in water. Why are ethers less soluble than amines? Both can form 2 hydrogen bonds with water (ethers accept two H-bonds, and amines donate an H-bond and accept an H-bond). Oxygen is more electronegative than nitrogen, so shouldn't ethers be more soluble?

Diethyl Ether can accept hydrogen bonds readily, and you are correct in pointing out it would be better than diethylamine at doing such.  However, diethylamine can also donate a hydrogen bond, giving it two modes of solvation, possibly both simultaneously.

Also, there is some minor acid-base reactions that may occur forming diethylammonium hydroxide, which being an ion, would certainly be more soluble than diethyl ether.


Pb-based pigments and rhodizonate salts In a previous question, I learned that we can detect lead with sodium / potassium rhodizonate, giving a nice purple color when reacting with pure (elemental?) lead.

But in the paint, lead is primarily used in the form of salts such as lead oxide, lead carbonate, lead chromate...

What if the lead in my paint is not pure, but a salt? Would sodium rhodizonate react with such salts? (I insist on sodium vs potassium, since it's cheaper to get). And if not, how can I detect these salts?

According to Ashley et al, "a rhodizonate spot test kit [for] in-situ testing of lead in paint... can be used for screening (i.e., qualitative) purposes." The test is sensitive not only to metallic (elemental) lead, but also to lead compounds used as pigments. However, the reference cited also states that the sulfide and rhodizonate tests have a high incidence of false positives.

Sodium, potassium or even cesium salts of rhodizonic acid should be equally effective, since the active species is the rhodizonate ion, but be aware that the sodium salt is less stable in water solution and should be used within a few days.


What is the composition of diet coke? I've been trying to find the values of any sort of percentage composition of diet coke and I haven't found any luck other than one interviewee (Dr. MAUREEN STOREY [Senior Vice President, Science Policy, American Beverage Association]) who states

Soda is comprised mostly of water. A full-calorie soft drink has 90 percent water, and a diet soft drink is 99 percent water. Water is the most important nutrient that we have...

And that was all the information I have.

The major ingredient in Diet Coke is water (H$_2$O). The second ingredient is carbon dioxide (CO$_2$), about 30 psig, or three times atmospheric pressure. (I measured it with an automobile tire gauge.) Since CO$_2$ is soluble about 1.7 g/Kg at 20$^o$C at atmospheric pressure, Ref 1, this is about 5 g/L, or ~0.5%, using a density for Diet Coke of 1 g/L Ref 2.

So, in the list of ingredients for Diet Coke, Ref 3, everything will be between a maximum of 0.5%, down to about 0.01295% ("46 mg caffeine/12 fl oz" on the label of my 2L bottle).
And, as Matt Black noted, the recipe varies from country to country, and from time to time (and is noticeably different in cans vs fountain delivery, according to those with highly developed preferences), but the big impacts are provided by
#1. water
#2. carbon dioxide
#3. artificial sweetener
#4. caffeine and
#5. flavoring, which might include a little more acid, like citric or phosphoric.
Seltzer is just #1 plus #2; many caffeine-free beverages just leave out #4.
Ref 1. https://www.engineeringtoolbox.com/gases-solubility-water-d_1148.html
Ref 2. http://web.physics.ucsb.edu/~lecturedemonstrations/Composer/Pages/36.34.html
Ref 3. https://www.coca-cola.ca/brands/diet-coke; see also https://www.huffpost.com/entry/coke-zero-sugar-taste-difference_n_597f71d7e4b00bb8ff3875d9 for varieties of Diet Coke


Definition of major, minor, and trace elements in analytical chemistry What is the definition of major, minor, and trace elements in a sample? I am unable to find this anywhere on the internet. Is it something like:

Major elements are those > 0.1 %

Minor elements are those > 100 ppm < 0.1 %

Trace elements are those < 100 ppm

There is no definite classification of what are the major, minor and trace elements are. While the numbers you mentioned are indeed the rule of thumb, they are rather loosely applied. It depends on field (geochemistry, solid state physics, analytical chemistry). There are however several guidelines.

Major elements are the elements that define the material in question. Some examples are:

Gold and silver in electrum,
Yttrium, aluminum and oxygen in YAG,
Hydrogen and oxygen in the ocean.

If you change the major elements, you are essentially changing the material. If you replace the hydrogen with potassium then instead of water you will get potash, which is not the same material and has vastly different properties (although this example might be a bit extreme).
Major elements are usually measured in percentage (either mass or molar), and are commonly above 1% of the chemical composition of the material.
Trace elements are elements that occur in such small concentrations that they do not change the essence of what a material is. Some examples:

Copper and platinum in electrum. It's still a gold-silver alloy, just with some other metals. The properties (such as colour) may change slightly.
Neodymium, erbium, etc. in YAG. YAG is used to generate lasers, and doping it with trace amounts of other rare earth elements may change certain properties of the laser (wavelength). However, it's still a YAG.
Almost every other element other than H, C, Cl and Na in ocean water.

Trace elements usually have concentrations of below 0.1%. In this case it's easier to measure them in units of ppm (parts per million, this mostly refers to mass). Remember: 1% = 10000 ppm. In extreme cases ppb (parts per billion) may be the appropriate unit.
Minor elements are everything in between. Technically, this means things between 1% and 0.1%. You might think of sodium and chlorine in ocean water as an example of minor elements (even though they're both slightly above 1%). To be honest, other than a mention in introductory textbooks about analytical chemistry and geochemistry, I rare see this term used in the professional literature.


What does "atoms make no compromises" mean? I was watching The Ring of Truth - 5 Atoms and in the last part of the movie (the link starts at this part), Dr. Morrison said

  "When it comes to energy, atoms make no compromises. It is all or nothing for them." - Dr. Philip Morrison (MIT)

What does this quote mean? Why can't atoms make compromises in Energy? When he says it is all or nothing for them, does that just mean that atoms can only absorb certain energy levels?

Please help me as I'm a very confused Chemistry student who is even more confused after watching a video my teacher told me would help me understand atoms better.

The subject we're looking at here is the quantization of energy absorbed and released by atoms, which is a fancy word describing the fact that atoms can only absorb or emit very specific amounts of energy (in the form of light) and nothing in between.  This curious fact has everything to do with how electrons orbit the nucleus.

The electrons whirling around the nucleus do so only at specific distances and in specific patterns.  Some orbit in spheres, others in dumbell-shapes, and still others in even odder patterns (a search for "orbital shapes" turns up a few relevant images).  On top of that, smaller spheres (and other shapes) are nested within larger ones, in a manner Niels Bohr compared to the orbits of planets in the solar system (https://en.wikipedia.org/wiki/Bohr_model).

Now here's where the energy comes in.  Electrons can "jump" between orbitals (unlike the planets, thank goodness!) when they absorb energy and move farther from the nucleus.  When electrons jump orbitals, they gain potential energy, just like when we move farther from the Earth, we gain potential energy.  The farther up we go, the more energy we have to expend on our way down: think jumping out of a 1st story window or a 2nd story window, or even a 3rd story window.  Note that the electrons can only jump between the first, second, third, etc. "floors" (orbitals) of an atom.  No mezzanines or mid-levels.

Obviously, the more electrons an atom has, the harder it is to map all of the specific energies it absorbs and emits.  Hydrogen atoms are easy, however, as this equation depicts.  Just plug in the starting shell #, the final shell #, and you've got your specific wavelength (energy) of light.
http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch6/graphics/6_2.gif

The logical question to ask now is "why do electrons orbit only in those specific levels?"
And that's a tough one, involving a crazy (interesting) amount of mathematics.  Equations we've worked out from quantum mechanical theory describe "wavefunctions" for electrons that predict the probability of them being in a certain area (orbital) around the nucleus.  They're convoluted, but brilliant.  Take a gander here if you're curious: https://en.wikipedia.org/wiki/Wave_function.


Comparing the stabilities of allylic, benzylic, and tertiary radicals Which is the most stable free radical among the given species?
  (1) $\ce{CH2=CH-CH2^.}$ (allyl radical)
  (2) $\ce{C6H5-CH2^.}$ (benzyl radical)
  (3) $\ce{(H3C)3C^.}$ (t-butyl radical)
  (4) $\ce{C2H5^.}$ (ethyl radical)

The book claims the answer to be the allyl radical, option 1. But how can we definitely say that the allylic free radical is more stable, compared to the benzyl radical (option 2) which is also resonance-stabilised, or the tertiary radical (option 3) which has more hyperconjugative donation from α-hydrogen atoms?

This is a tough question.  I think it might even be unfair to ask such a question on a test in non-advanced classes.  In advanced classes it could make an interesting topic of discussion, but I'm still not sure that the "real" answer is known.

What can be said is that due to resonance, both the allylic and benzylic radicals are more stable than the t-butyl or ethyl radicals which are not resonance stabilized.  You can see from the following diagram that you can draw two resonance structures for the allyl radical.

For the benzyl radical you can draw even more resonance structures.

Superficially, this might suggest that the benzylic radical is more stable than the allyl radical. This was the "answer" and reasoning provided here.

However, if you do some thermochemical calculations (as done here), you arrive at the opposite conclusion that the allyl radical is roughly 2 kcal/mol more stable than the benzyl radical. The number of resonance structures alone is not a perfect indicator of stability.

For me, that is a pretty small difference in energy, and I think it would be fair to answer that the allyl and benzyl radicals have comparable stabilities.


Does concentration or pKa define acid strength? For example, is 3M HCl a stronger acid than 1M HCl?

I would reason that the concentration of an acid/base does not influence its strength. Strength is determined by the pKa, and, as per Le Chatelier's Principle, the initial concentration does not influence the equilibrium constant.

It may shift the equilibrium concentrations (meaning that the pH is higher for the 3M HCl), but it will not change $K_a=\frac{[H_3O^+][Cl^-]}{[HCl]}$ at equilibrium.

Is this right?

The term ‘strong acid’ is sometimes used in a rather fuzzy way and you ran into problems doing so. I prefer to use the term ‘strong acid’ only with respect to an acid’s $\mathrm pK_\mathrm{a}$ value and disregard all other influences. This gives a clearly defined measure of acid strength and we can easily sort various acids by their strength into stronger or weaker acids.

However, this is looking at the acid as a molecule. In real-life applications you are typically more interested in the property of a solution. To give a real-world example, imagine a $\pu{1M}\ \ce{HBr}$ solution and a $\pu{12M}\ \ce{HCl}$. solution. Obviously, $\ce{HBr}$ is the stronger acid, but the concentration of $\ce{HCl}$—also a strong acid and thus fully deprotonated—is higher. Therefore, the $\ce{HCl}$ solution is more concentrated or, as some would say, more acidic. It can do greater harm and it is able to protonate more Brønsted base molecules than its $\ce{HBr}$ counterpart.

If instead of examining a $\pu{1M}$ and a $\pu{12M}$ solution I had examined a $\pu{e-3M}$ and a $\pu{1M}$ solution, we could even base the discussion around the solution’s resulting pH: the stronger acid is much more diluted and will result in a solution of pH 3 while the weaker acid results in a solution of pH 0.


How to calculate the concentrations of conjugate species present in the solution of diprotic acid? Calculate the concentration of $[\ce{SO_3^-^2}]$ present in 0.010M solution of $\ce{H_2SO_3}$ in pure water?

($Ka_1=0.017)$, ($Ka_2=10^{-7.19}$)

I found this approach in reference text for solving the problem :

But I confused with the derivation of Eq.2, so I do my calculation according to Eq.1 to find the concentration of sulfite ion equal
$2.20×10^{-7}$ M.

SO :

I choose another approach to find the concentration of sulfite ion depending on substitution in this formula :
$$Ka_2 = \dfrac{\ce{[H+][SO3^{2-}]}}{\ce{[HSO3^-]}}\ $$
Thus

$$\mathrm{10^{-7.19}} = \frac{(0.00706 + {[SO_3^{-2}]}){[SO_3^{-2}]}}{(0.00706 - {[SO_3^{-2}]})}$$

From which I obtain
$$[\ce{SO_3^{-2}}] = 6.46 \times 10^{-8}$$
Which different from the answer obtained from the first approach.

I appreciate any help.


Le Châtelier’s Principle and Addition of Water With regards to gas volume, Le Châtelier’s principle predicts that an increase in the volume of the container causes the system to favor the side of the reaction that produces more moles of gas. However, would this principle apply equally well to an aqueous system, at equilibrium, that has now had water added to it? That would be analogous to expanding the volume of a gaseous system, wouldn’t it? Would adding water to an aqueous system at equilibrium affect the position of equilibrium at all?

Yes adding more water will affect the position of the equilibrium simply because you are adding more volume to the solution by adding more water. So adding volume is inversely proportional to pressure where the system needs to reduce the amount of water. Then the equilibrium will shift to the side with increased amount of pressure.


How does changing volume affect concentration of solutions? My teacher performed an experiment and made us watch.

He prepared 3 conical flask containing 50 mL mixture of sodium thiosulphate, potassium iodide and acetic acid.
He also had 3 measuring cylinders containing 12.5 mL, 25 mL and 50 mL of hydrogen peroxide of the same concentration.
He added few drops of universal indicator to each of the flask. Then he added the $\ce{H2O2}$ to them as well and started timing.

The general idea is that the indicator is going to change colour from orange to green, which is the colour when he stopped timing. The flask added with 50 mL $\ce{H2O2}$ had the shortest time recorded.

It is intuitive that higher volume of $\ce{H2O2}$ added will increase the speed of reaction but I don't know how to explain it. In kinetics we learn the effect of concentration on rate of reaction, but i cannot use this idea to explain because the concentration does not change! All three solutions of $\ce{H2O2}$ added has the same concentration because they are undiluted (He did not add distilled water to the $\ce{H2O2}$ solutions to make them the same volume).

So how can I explain why the use of higher volume (50 mL) will make the reaction go faster?

An example of the experiment can be viewed through youtube :
https://www.youtube.com/watch?v=Tlj5RvCSpLw

The concentration of $\ce{H2O2}$ doesn't change but when you have more of it in the reaction (50 ml vs. 12.5 ml) there are more moles of $\ce{H2O2}$ meaning more collisions speeding up the reaction.


Where to find a comprehensive list of chemical/biochemical etymologies? I've found lists like this, which explain what different root words mean. But I'm looking for explanations of why those roots mean those things.
For example, why does carbonyl refer to a carbon double bonded to an oxygen? What is the historical rational behind why the -yl implies the oxygen double bond?
I'm hoping this gives me a little more intuition to remember what different chemical terminology means without brute-force memorization.

It is a good question and many comprehensive works are available specifically on chemical etymologies. Many scholars have spent on portion of their life on chemical etymologies.  Ignore these useless down votes or close votes. The answer essentially boils down to how money are you willing to spend? I would say, if you are really keen, go for OED.
After going through so many etymological dictionaries of science (mainly chemistry & mathematics), I still feel that the Unabridged Oxford Dictionary  (> 20 volumes) is the best (https://www.oed.com/). It is not available for public use but one can buy subscription for a year at an affordable price. Most universities have it. There you can find the etymology for sure, the best part is that you can see the earliest use with original reference.
I will give you a taste of it. Say you want to find out where does the word alcohol come from?  You can extend this that the suffix -ol indicates alcohol group. Perhaps it is too much for you but you can see the extent of depth of analysis. It is sometimes surprising that there might be no connection of older meanings with current meanings.

post-classical Latin alcohol, alcool, alcol, alcofol kohl (galena
(lead sulphide) or stibnite (antimony sulphide)) (from 13th cent. in
British sources), spirit, essence obtained by distillation (a1527 in
Paracelsus in alcool vini , also alcohol vini ) (perhaps via Spanish
alcohol: see below) < Arabic al-kuḥl , Spanish Arabic al-kuḥul < al
the + kuḥl , Spanish Arabic kuḥul eye cosmetic, also denoting various
specific substances used as eye cosmetics or eye unguents (compare
kohl n.1) < the same base as Arabic kaḥala, Hebrew kāḥal (in the Bible
only in an isolated attestation in Ezekiel 23:40), both in sense ‘to
stain, to paint’, Akkadian guḫlu antimony (used as eye paint).
Compare Middle French alcohol, (rare) alcofol, French alcool,
†alcohol, †alcol, †alkol, †alkool, etc., the chief senses of which
are: ‘kohl, very fine powder’ (c1370 in a translation of Chauliac; the
precise sense is often difficult to determine in early quots.),
‘essence obtained by distillation’ (1620, originally and chiefly with
reference to spirit of wine), ‘ethanol’ (1792), ‘any of the members of
a similar class of chemical compounds’ (1834: see note below). Compare
also Spanish alcohol (c1200 as †alcofor; also †alcofol, †alcool,
†alcol, etc.), the chief senses of which are: ‘powder used as eye
cosmetic’ (c1200 as †alcofor), ‘galena, sulphide of lead’ (1541),
‘antimony, also any of various minerals containing antimony,
especially stibnite’ (a1555 or earlier; in a number of early instances
denoting minerals (from the second half of the 12th cent.) it is
impossible to tell whether the word denotes galena (sulphide of lead)
or antimony sulphide), ‘spirit of wine, ethanol’ (1730), ‘any of the
members of a similar class of chemical compounds’ (1865).

On your specific query
Carbonyl

< carbon n. + -yl suffix. Compare French carbonyle (1856 or earlier),
German Carbonyl (1854 or earlier).

Further probing of -yl

Formerly occasionally -ule, a terminal element of chemical terms, <
German -yl, < Greek ὕλη wood, matter, substance (see hyle n.), used
for ‘chemical principle, radical’. It was introduced by Wöhler and
Liebig ( Ann. der Pharm. (1832) III. 262), and first used by them in
the term benzoyl; other early names were éthyle (éthule), élayle
(Berzelius), dadyle, peucyle, citronyle, citryle (Blanchet and Sell).
Some fifteen in anglicized form, including acetyl, amyl, cinnamyl,
glyceryl, salicyl, appear in the Elements of Chemistry by T. Graham,
1842,

If you find OED too much to digest, the next best etymological dictionary of chemistry is

Elsevier's Dictionary of Chemoetymology (1st Edition) The Whys and Whences of Chemical Nomenclature and Terminology by Alexander Senning.
It has 400 pages. For example, the entry aniline shows

aniline C6H7N, derived from al-nil (Arabic: indigo plant), ultimately
from nila (Sanskrit: dark blue) – referring to the fact that aniline
was first obtained by degradation of indigo

The next one, which focussed on organic chemistry is

Organic Chemistry: The Name Game Modern Coined Terms and Their Origins (1st Edition)
by Alex Nickon and Ernest F. Silversmith,
This has plenty of fancy organic molecules from modern organic chemistry but at the end, they do list traditional names for example for benzene

benzene from Styrax benzoin, a tree native to Sumatra and Java. The
bark yields a resin, gum benzoin, from which "benzoic acid" was
obtained. Peligot (1833) and E. Mitscherlich (1834) heated benzoic
acid with lime to form C6H6; Mitscherlich named it "benzine." Liebig
preferred "benzol," but Laurent (1835) proposed "benzene."

Coming to free rides. Dr. John Andraos made a series of webpages by the name Modern Coined Glossary of Coined Names & Terms Used in Science to which I had heavily contributed more than a decade ago.
Here is a link: http://www.careerchem.com/NAMED/Glossary-Coined-Names.pdf
Another free source is Dr. William B. Jensen's lovely book "Ask the Historian". Simply Google "Ask the historian Jensen" and the first result is his book. He explain a lot of names, and even history of concepts. It was kind of him to make this book free of all of us. He has retired and he is no longer active.


Trying to understand the chemistry in an iron carbon battery I am trying to understand the chemistry that occurs in an iron carbon battery during charging.

The negative electrode is iron, the positive electrode is carbon. The electrolyte is iron(II) chloride. I understand that ferrous chloride dissociates in water to $\ce{Fe^2+}$ and $\ce{Cl-}$ ions. During charge the $\ce{Fe^2+}$ ions are attracted to to the negative iron electrode and the $\ce{Cl-}$ ions are attracted to the positive carbon electrode. The $\ce{Fe^2+}$ ions combine with electrons from the power supply to form metallic iron that plates on to the iron electrode.

My problem is understanding what happens at the positive carbon electrode. I'm told that iron(II) chloride loses electrons to the positive electrode and becomes iron(III) chloride. Fine, but how does this happen if all the $\ce{Fe^2+}$ ions are at the negative electrode? How does this happen if the iron(II) chloride has dissociated in to ions, there is no iron(II) chloride anymore?

If the power supply drives all the $\ce{Fe^2+}$ ions to the negative electrode and $\ce{Cl-}$ ions to the positive electrode how can any $\ce{Fe^2+}$ ions even be at the positive electrode to somehow recombine with $\ce{Cl-}$ ions back into iron(II) chloride that then loses electrons to the positive electrode and becomes iron(III) chloride?

Ions in solution

  [OP] I understand that ferrous chloride dissociates in water to $\ce{Fe^2+}$ and $\ce{Cl-}$ ions.

Yes. The same goes for the iron(III) species.

  [OP] I'm told that iron(II) chloride loses electrons to the positive electrode and becomes iron(III) chloride.

No, iron(II) loses electrons and becomes iron(III). The chloride is less relevant, and could be exchanged for a different ion.

  [OP]  How does this happen if the iron(II) chloride has dissociated in to ions, there is no iron(II) chloride anymore?

The reactant is $\ce{Fe^2+(aq)}$, not $\ce{FeCl2(s)}$.

Ion movement

  [OP] During charge the $\ce{Fe^2+}$ ions are attracted to to the negative iron electrode and the $\ce{Cl-}$ ions are attracted to the positive carbon electrode.

Yes, but a small fraction, only. Otherwise the repulsion between like ions would be too great.

  [OP] My problem is understanding what happens at the positive carbon electrode. I'm told that iron(II) chloride loses electrons to the positive electrode and becomes iron(III) chloride.

Just $\ce{Fe^3+(aq)}$. There might be a double layer, but no solid iron(III) chloride.

  [OP] Fine, but how does this happen if all the $\ce{Fe^2+}$ ions are at the negative electrode?

There are plenty of cations in the bulk of the solution, and plenty near either electrode.


No reaction in net ionic equation for lithium nitrate and sodium chloride? I've heard that nothing in the following net ionic equation reacts:

$$
\ce{Li+ + NO3- + Na+ + Cl- -> \text{"nothing"}}
$$

Is this true? Could someone explain why there would be no reaction?

Assuming all these ions are in an aqueous solution, according to solubility rules nitrates are soluble, chlorides (with the exception of AgCl, PbCl₂, and Hg₂Cl₂) are soluble, and salts containing group 1 elements are soluble so they all remain as individual ions as nothing can come together to form an insoluble compound and precipitate out of the solution.


Suitable coating material for protection of sensors I'm building a moisture/temperature sensor for outdoor use in soil. It uses capacitive sensing for moisture and an NTC thermistor for the temperature. It's going to be used for vegetables so food safety is a concern.

I don't think the PCB is food safe, or safe from corrosion/degradation due to the soil and moisture would creep into the FR4 material in the PCB and cause my measurements to drift so I would like to coat the entire thing to protect it from the elements, keep my vegetables free from toxins, and keep the measurements from drifting.

So these are the requirements in order of importance for the coating:

Not electrically conductive
Food safe
Doesn't degrade/react in/to soil/sun light
Hydrophobic
Resistant to abrasion when insertion/removing from the soil
Affordable
Low relative permeability (electromagnetic)
Decent thermal conductivity
Not a host for bacteria

I first considered a food safe epoxy but I had difficulties finding a reputable supplier of food safe epoxies in Germany (my German sucks). I did find a supply of food safe silicone which is nice in that it has all the properties above but I could only find it in large quantities and it's quite expensive. I also considered L/HDPE which seems to have all the criteria apart from the bacterial aspect (I'm not sure about that) but it seems like melting it to have a low enough viscosity to be able to form a solid but thin coating seems hard/impossible.

My knowledge on the topic of epoxies, plastics, rubbers and chemistry in general is limited, so I'm turning to chemistry@sx for recommendations to what I can use for a coating and what process to use for coating. And maybe I've misunderstood something of the materials I've considered so far... Also I'm sorry if this is the wrong place for this question, I'm not sure where else to post.

Various DIY capacitive moisture meters use a PCB for the sensor:

While not directly related to your question I thought I'd mention that you can buy these sensors for several dollars:

Purchasing a readymade kit and relying on the discount of mass production (along with a manufacturer's warranty) is probably easier than designing your own, and etching your own PCB.

I think how you plan to make the probe plays a part in how you plan on coating it; for example you could fit the parts in a test tube, that would meet all the requirements in your list. Such a capacitive sensor would operate similar to the touchscreen of a cellphone.


What determines the density of an element? I used to be under the (wrong) assumption that the density of an element correlates with it's atomic number $\mathrm{Z}$, I thought that since having more protons meant the atom weighed more; but of course that's wrong.

So why does make osmium $\mathrm{Os}$, with an atomic number of $\mathrm{Z}_{\mathrm{Os}}=76$ more dense than say uranium $\mathrm{U}$, with an atomic number of $\mathrm{Z}_{\mathrm{U}}=92$?

There are a couple of factors here that you need to consider.

First of all, increasing Z does mean that there are more particles in the nucleus; however, you're also adding electrons to the systems as well. Electrons dictate the size (i.e. volumen) of the atom, so the size of the atom could increase with increasing Z, which would mean that fewer atoms could fit into the same volume. To use your example, the covalent radius  of Os is 144 pm; the covalent radius of U is 196 pm. So just from this property alone, you would expect more Os than U atoms to pack into the same volume.

There is another factor to consider as well: the crystal system of the solid. At standard temperature and pressure, atoms for different elements in the solid state pack into different arrangements, which are called crystal systems. Some of these are much more efficient than others. For example, there are three variants of the cubic crystal system: primative, body-centered, and face-centered. From Wikipedia:

  Assuming one atom per lattice point, in a primitive cubic lattice with
  cube side length a, the sphere radius would be a⁄2 and the atomic
  packing factor turns out to be about 0.524 (which is quite low).
  Similarly, in a bcc lattice, the atomic packing factor is 0.680, and
  in fcc it is 0.740. The fcc value is the highest theoretically
  possible value for any lattice, although there are other lattices
  which also achieve the same value, such as hexagonal close packed and
  one version of tetrahedral bcc.

Now, depending on which crystal system the atoms in the solid take, you could get more or less atoms per unit volume. Unfortunately, determining which crystal system is favored for different elements is not straightforward, and requires some knowledge of the bonding between the atoms and other factors.


How to transform an aliphatic primary amine to the corresponding nitro compound? Could you suggest a reagent or procedure to transform an aliphatic primary amine to the corresponding nitro compound.

I know this is a bit backwards of a request because the usual procedure is the reduction of nitro groups to their corresponding amine, but there is a need of its electron-withdrawing properties for a subsequent condensation to form the intended intermediate. If this approach could be solidified, then it would make things a lot easier for this recent graduate, as the starting material is a lot less expensive.

I have found references to ozonolysis, but am unsure of it, as I have no previous experience with these sorts of reactions. I have also seen in abstracts where it is discussed formation of the diazonium compound, which was the used to form the nitro compound. There are a couple of others I have come across, but I am not sure of their efficacy or plausibility.

Using silica gel, primary amines can be oxidized to their corresponding nitro compounds using ozone ($60$-$70 \%$ yield). In this paper, dry silica gel (dried for $24\ \mathrm{h}$ at $450^\circ \mathrm{C}$) was mixed with an amine. The silica gel (ca. $\mathrm{30\ \mathrm{g}}$) containing the amine ($0.1$-$0.2\ \mathrm{wt/wt \%}$) was cooled to $-78 ^\circ \mathrm{C}$, and a stream of $3 \%$ ozone in oxygen was passed through it.

Aliphatic amines and aromatic primary amines are rapidly and effectively  oxidized to nitro compounds by dimethyldioxirane ($80$-$90 \%$ yield). Dimethyldioxirane is prepared by reaction of OXONE ($\ce{2KHSO5 \cdot KHSO4 \cdot K2SO4}$) with buffered aqueous acetone. In a typical reaction, $0.7\ \mathrm{mmol}$ of amine in $5\ \mathrm{mL}$ acetone is treated with $95\ \mathrm{mL}$ of dimethyldioxirane in acetone ($0.05\ \mathrm{M}$). The solution is kept at room temperature for $30\ \mathrm{min}$ with the exclusion of light.

Oxidation of amines to nitro compounds in high yield has also been carried out using oxyacids like peracetic acid, peroxytrifluoroacetic acid, and meta-chlroperoxybenzoic acid, though the anhydrous forms of these acids are hazardous and difficult to use.

Ono, N. The Nitro Group In Organic Synthesis; Wiley-VCH: New York, 2001.


What is the formula of potassium ammonium sulfate? I tried to get the formula of potassium ammonium sulfate. What I got is $\ce{K(NH4)2SO4}$ but I don't think this the right formula. I searched on the internet and found an answer ($\ce{KNH4SO4}$), but I don't think it is correct.

So what is the correct formula of potassium ammonium sulfate?

$\ce{K(NH_4)SO_4}$ is the correct formula. You have two unipositive cations and a single dinegative anion. The total charge must be zero.


Isn't the dual behaviour of subatomic particles contradictory? I recently came across Molecular Orbital Theory. In this, if two electrons are bonding, their waves merge and they form a proper bond. But this is directly opposite to the particle theory, according to which they should repel each other as they have like charges.

Take a nice simple system like a hydrogen molecule. It's true that the two electrons repel each other, but at the same time they are attracted by the positive charge of the two protons.

If we start with an $\ce{H2+}$ ion, with just a single electron, it should be obvious that the electron is tightly bound because it feels an attractive force from both the protons.

Now imagine adding the second electron to form a neutral hydrogen molecule. The second electron will be repelled by the first electron that is already present but it will be attracted by both the protons, and the attraction from the two protons wins. The binding energy of the second electron in a hydrogen molecule is about 15.4 eV, so it is even more strongly bound than the electron in a hydrogen atom.

Note that this is not related to the wave/particle duality. Wave particle duality is in any case a misguided concept since particles are quantum fields not waves or particles – they simply behave approximately like waves in some circumstances and approximately like particles in others. See this post on the Physics Stack Exchange for more on this.


Magnesium air-fuel cell or battery? I'm a bit confused.
I've got this "toy" ...
with the help of this site I was abled to find out that the reaction stops with a lack of air. Link to the old question.

The battery/fuel cell or however you call it consists out of following components:

magnesium plate
isolator which consists out of something like paper, the surface feels more like cotton
the black plate is called "black cathode" in the manual, my hypothesis is that it might be rubbered graphite or a polymer, because it has a very low electrical resistance.
the electrolyte is a $\ce{NaCl}$ dispension- I think it just need to lower $\ce{H2O}$'s electrical resistance
plastic stuff is only the case of the "cell"

I did some research about this and the results are very confusing to me.

In the manual they call it a fuel cell.
My chemistry teacher says that in his hypothesis it is impossible to be a fuel cell and has to be a battery. -Because of the great answer to my last question I doubt it or at least I'm too confused to agree.

My research results:

►Battery:

news about a battery for cars
an article about the upper link and about the toy

►Fuel cell:

Wikipedia-MAFT
a kind of datasheet of a "battery" which is in development

►My hypothesis for beeing called fuel cell

→ because fuel cells require according to their definition a constant source of fuel, it might be called fuel cell because of the fact that the "thing" requires air to react.

►My hypothesis for beeing called battery

→ oxygen as fuel?- some people might see it different and call it battery because of that (then there's no constant fuel source →battery) .

►My idea about the chemical equation is:

$\ce{Mg + H2O → Mg(OH)2 + H2}$  (but I doubt it, because I can't imagine that it is responsible to sell a toy which emitts $\ce{H2}$ - or what else happens with it?-maybe the black cathode has to do something with it?)

If you've got a better idea about the chemical equation- let me know.

I would like to know if there's a mistake in my thoughts- and if it's actually a battery or a fuel cell.(the main problem).

And if you've got a better idea about the black cathode component's material- let me know please.

please don't mark it as duplicate to my older question-there's a difference because I added information about the theories and reasearched a lot about it and found the reasons/arguments I've posted here out- the last question was doing it by its components.

What you have is classified as a Mg-air battery. You can read more about it here : https://en.wikipedia.org/wiki/Category:Metal-air_batteries

These are a class of primary(i.e. not rechargeable) batteries, where the anodic reaction is a metal being oxidised and the cathodic reaction is oxygen reduction.

The reactions are:

$\ce{Mg → Mg^{2+} + 2 e^{-}}$ on the anode

$\ce{O2 + 4 e^{-} + 4 H+ → 2 H2O }$ on the cathode, if the reduction goes to completion. Most of the time, it is only reduced to the peroxide or the superoxide which is considered incomplete reduction.

For it to be classified as a fuel cell, the fuel for the anodic reaction would need be brought from the outside(source: anectodal, personal thought).


Photosynthetic rate calculation What are the steps in calculating the photosynthetic rate ($\pu{\mu mole}~ \ce{O2} \pu{m2/ min}$) for a leaf that has an area of $\pu{0.0025 m2}$ and a $\ce{O2}$ evolution
rate of $0.096~\ce{O2}/\pu{min}$?

To do so , the first step is to convert it to $\mu L$ i.e. $\pu{960 \mu L} ~\ce{O2}/\pu{min}$.
The second step should be to convert it to $\pu{\mu moles}$. So, how do we do that ?
EDIT : In my textbook, the following calculation has been done but I am not able to understand it...

$\pu{960 \mu L}~\ce{O2}~\pu{/L} / [(273+23)/273) \times 22.423]$

I know that $22.423$ is the volume of a mole of gas and $\pu{273K}$ is $\pu{0^\circ C}$ but I am not able to understand how that fits here.

Using the general case equation
$$\frac{P_1V_1 }{T_1} = \frac{P_2 V_2} {T_2}$$
and given that pressure is constant at $\pu{1  atm}$ in your experiment, the equation simplifies to
$$\frac{V_1 }{T_1} = \frac{ V_2} {T_2}$$
We know $V_1=\pu{960 \mu L}$ and it appears that the experiment is being run at $\pu{(273+23)K}$, so we need to calculate what the volume ($V_2$) would be at standard T ($\pu{273 K}$).  This is given by
$$V_2 = \frac{V_1 T_2}{T_1}$$
if we next divide $V_2$ by $\pu{22.423E6 \mu~L}$, we now know how many moles of $\ce{O2}$ we have.


Relating collision theory to rate of reaction I want to relate the macroscopic advancement of a reaction, the rate of the reaction, with the microscopic collisions of the molecules. This is the type I have developed so far:

$$u = A^2 \times \frac{n_1}{m^2} \times\frac{n_2}{m^2}\times \frac{(m_1u_1^2+m_2u_2^2)}{E_\mathrm{act}}\times 1/t $$

where $u$ is the rate of reaction in $\pu{mol2 sec-1}$.

$A =$ the common area of the substances

$n_1/m^2 =$ the moles per square metre of the first substance

$n_2/m^2 =$ the moles per square meter of the second substance

$u_1$ and $u_2 =$ average velocity of molecules of each substance

$E_\mathrm{act} =$ activation energy of the reaction

Now the molecules of the first substance must bump to the molecules of the second substance with the appropriate orientation. How can I add this to my type?

It is interesting to see you creating your own approach. The comments above provide good responses. I would recommend looking at Buttonwoods suggestion regarding collision theory and cross-sections. However, it appears none have answered your question directly, so I will provide my response to it here.

Question: "Now the molecules of the first substance must bump to the molecules of the second substance with the appropriate orientation. How can I add this to my type?"

My Response: You recognize that orientation is important when it comes to chemical reactions. There are many ways to handle this, I will provide three short suggestions and you can consider them or use them to come up with something else.

Arrhenius (constant) approach: The pre-exponential factor, $\alpha$, in the following Arrhenius equation absorbs two important concepts of chemical reactivity - 1) the frequency of collisions, 2) the orientation of the molecules.

$$u = \alpha e^{-E_a/RT}$$

Note, I changed from the conventional notation in order to match and not confuse with your own. In fact, $\alpha$ is defined as the frequency of properly oriented collisions. So a simple constant multiplier that accounts for orientation can help modify your equation.

Boolean logic: You can possibly simplify things further by using a simple selection function that tells you whether the molecules are in proper orientation. Here is the simplest you could use:

$$f(x) =
\begin{cases}
1 \text{ if proper orientation}\\
0 \text{ if not proper orientation}
\end{cases}$$

Functional dependence: Last mentioned here would be the most creative approach to come up with a function that describes the orientation of your molecules. You will have to come up with the functional dependencies you need (i.e. a reaction coordinate) to help solve your problem. Here is an example of a reaction between atomic hydrogen with molecular hydrogen to make the same species using sinusoidal function where the independent variable (reaction coordinate) is the angle of the H-H bond relative to the hydrogen atom.

$$\text{Reaction:    } H-H + H^{\boldsymbol{\cdot}} \to H^{\boldsymbol{\cdot}} + H-H$$
$$\text{Function:    } f(\theta) = \frac{1}{2}\left[\cos\left(\frac{\theta}{2}\right)+1\right]$$

The interpretation of such a function as this would be 100% likely to react if the $H$ atom is in the plane of the bond with $H_2$ while 0% likely to react if $H_2$ is vertical. This isn't an exact reality, but just an example.


Are Lewis Structures the same throughout solids, liquids, and gases? For example, is the lewis structure for water constant even when it is in a solid, liquid, or gas phase?

Lewis structures are a representation of the atomic connectivities and electronic lone pairs present in an individual molecule of a substance. The molecule of the substance does not change in these states, so the Lewis structure would remain the same.

What does change in these states is the way individual molecules interact with each other. If you were to draw a diagram of how multiple Lewis dot structures of the molecule interact with each other, that would be different for these different states.


Why do atomic radii decrease across a period? It's said that atomic size/radius decreases across a period, in spite of addition of electrons. But how does this actually work?

Atomic radius decreases across a period because  electrons are being added to the same energy level and at the same time the nucleus is increasing in protons. The increase in nuclear charge $$Z_{eff}$$ attracts the electrons more strongly, (addition of one proton has a greater effect than addition of 1 electron) thus pulling  electron closer to the nucleus and making atomic radius decrease.


How to write equations for dissociation of ionic compounds in water? Please write a dissociation-in-water equation for the compounds $\ce{BeI2}$ and $\ce{LiI}$. Make sure to add the states of matter after each compound.

Currently, for $\ce{BeI2}$ I have the equation $$\ce{BeI2 (s) -> Be^2+ (s) + I2^2- (g)}.$$ I have yet to attempt the second one.

Your ionic charges are not correct for iodine. Looking at your attempt.

As $\ce{Be}$ is in group 2, the ionic charge for beryllium ion is fine, but iodine is in group 17, so its ion is $\ce{I-}$.  When the ions dissociate, they become aqueous or (aq) as the state of matter.

Then the ionic charges need to balance, thus:

$$\ce{BeI2 (s) -> Be^2+ (aq) + 2I- (aq)}$$

To balance the ionic charges in this example, you need 2 $\ce{I-}$ (iodine ions) to balance the $\ce{Be^{2+}}$ (beryllium ion).

A similar example (and further explanations) are provided on the UC Davis ChemWiki page Unique Features of Aqueous Solutions (including an example of the dissolution of $\ce{MgCl2}$ - another compound with group 2 and 17 elements).

So,

determine the group, hence ionic charge of each dissociated ion
balance these charges
state that the dissociated ions are aqueous

Now, use the process to determine the dissociation of $\ce{LiI}$


Non-foaming surfactant to add to water I'm looking for a surfactant that I can add to water to decrease the surface tension by 20 to 40%.  I'm working on a new kind of mixing device that is so designed such that it mixes relatively large volumes of two liquids and cannot be made small enough for laboratory scale.  So flow on the order of 1.5 to 2.5 liters/min.  I would like to be able to run my experiments for a number of minutes, so the estimate is 10 gallons for each hypothetical liquid (total amount would be 20 gallons for an experiment that would give me about 20 minutes to adjust parameters, take photographs, etc.).  Adding something like isopropyl alcohol can get expensive very fast (10% by weight in water lowers the surface tension to about 41 where pure water is 72).
Very important:  The surfactant needs to be non-foaming as release of gas will be problematic for these tests.
Thanks for any thoughts.

Conceptual thoughts about mixing two immiscible liquids (not on surface tension):

A reflux condenser used in the ochem lab equally has a circulation of about 2 L/min water (reference from an advertisement, which doesn't look too much off-range)

After the two liquids are mixed, what application will follow?  Do you target cleaning an engine, or deployment for food processing?  What are intended storage conditions once the two are mixed (e.g., time of storage / transport, temperature of storage / use)?  Altogether with the other liquid you want to mix with water, this may narrow your choice of the surfactant.

Mixing the two continuously may be achieved with a tube reactor with two inlets, one outlet, and a screw-shaped inner geometry

(reference, section «Liquid-Liquid & Liquid-Gas Mixers»)
Depending on parameters of your liquids (e.g., density, viscosity, flow rate, key word Reynolds number) you may enter the regime of turbulent flow until you divide the liquid into tiny droplets which increases the surface accessible for mixing.  The addition of surfactant may help to achieve the intended mixing, too.

(loc. cit.)

The formation of foam however is the result of an entrapment of a gas.  Whipping cream with a whisk for example intentionally aerates the cream.  Liquids like water may boil (i.e., generation of gas) because of sudden pressure changes while passing a propeller blade at a flow rate too high (cavitation), too.
Thus, mixing water and your other liquid in a tube reactor like the one mentioned above, in absence of air and at the right hydrodynamic state may prevent the formation of foam.  (You mau fight foam by ultrasound (example), though.)


Sensitivity vs. Limit of Detection of rapid antigen tests I'm comparing a bunch of SARS-CoV2 rapid antigen tests:

Columns 4 and 6 list the values for sensitivity and limit of detection (LOD). How come that a test with a several times lower limit of detection can have a worse sensitivity?
As an example, consider tests #2 and #4:
#2 - Sensitivity: 97.7% - Limit of detection: 2.0*10^2.4 = 502 TCID50/ml
#4 - Sensitivity: 91.4% - Limit of detection: 2.5*10^1.8 = 158 TCID50/ml
Shouldn't these values correlate, i.e. the more sensitive a test is the smaller the amount it is still able to detect?

In this situation, "sensitivity" refers to how often the test is able to correctly detect the presence of analyte. A test with high sensitivity means that there are few false negatives.
Limit of detection refers to how much of the analyte must be present before it is considered positive.
Imagine two different tests: test A uses an antibody with very high affinity for a specific spike protein sequence. Because of the high affinity, it can detect the spike protein even when there's not much around (good limit of detection). However, the antibody is less robust against other sequences, so it sometimes reports a false negative in the presence of spike protein variants (poor sensitivity).
Conversely, test B has an antibody with medium affinity for several different spike protein variants. It relies on high levels of the protein being present, but it works with many different variants and rarely gives false negatives. This test would have a high (aka poor) limit of detection and high (aka good) sensitivity.


Why does a photon have linear momentum, although it doesn't possess mass? We know that photon doesn't possess mass. But it has a tendency of colliding with surfaces, as we see in the photoelectric effect. As it doesn't possess mass, then how does it have linear momentum?

$E=mc^2$ and $E=hc/\lambda$

$mc = h/\lambda =p$

Here $m=0 \Rightarrow p=0$.
But photons can collide.

The formula for the momentum, $p = mv$ is only valid in the context of classical mechanics. For photons, energy and momentum are related by $E = pc$, which, because $E = hv = \frac{hc}{\lambda}$ leads to the following formula: $p = \frac{h}{\lambda}$. So the momentum is always non-zero and depends only on the wavelength.

See this Wikipedia page for more details.


If sulphuric acid barely evaporates, why we can smell it? If sulfuric acid barely evaporates, why we can smell it? Or does it give off sulfur dioxide gas?

There are at least two plausible explanations for why an odorless compound with very low vapor pressure could nevertheless generate an odor:

Sulfuric acid is highly reactive.  If you are pouring or dispensing sulfuric acid, you are probably generating some mist. This is a concern in industrial settings where sulfuric acid is used:

https://www.cdc.gov/niosh/docs/81-123/pdfs/0577.pdf

Thus, even if the sulfuric acid itself has neither a significant vapor pressure nor an odor, as the mist enters your nose it could react with substances in your nasal passages to create compounds that have an odor.  Some studies have reported the perception of odor from sulfuric acid mists:
https://www.epa.gov/sites/production/files/2014-09/documents/sulfuric_acid_interim_dec_2008_v1.pdf

And even without misting, particulates in the air, as they fall into an open container of sulfuric acid, could react with sulfuric acid to generate compounds that create a smell.  This effect is seen with metals -- the metallic "smell" is not from the metals themselves (which have an extremely small vapor pressure at room temperature), but from reaction products of the metals:

Why can we smell copper?


Equivalent conductivity of barium chloride at infinite dilution Find out the equivalent conductivity of barium chloride at infinite dilution given that the ionic conductivities of $\ce{Ba^{2+}, Cl-}$ are $\pu{127 ohm^-1 cm^2 equivalent^-1}, \pu{76 ohm^-1 cm^2 equivalent^{-1}}$ respectively.

Equivalent conductivity is defined as.

  The equivalent conductance of an electrolyte is defined as the conductance of a volume of solution containing one equivalent weight of dissolved substance when placed between two parallel electrodes 1 cm apart, and large enough to contain between them all of the solution.

One gram equivalent weight would be simply half of the entire compound so
$$\text{Equivalent Conductivity} = \dfrac{1}{2} \times 127 + \dfrac{1}{2} \times 2 \times 76$$

This comes out to be 139.5 however the answer is 203.

One must be really careful with the units while dealing with conductivity, conductance problems.

If you read the question carefully, the equivalent conductivity of $\ce{Ba^{2+}}$ and $\ce{Cl^{-1}}$ are provided to you.

So the molar conductivity of $\ce{Ba^2+}$ is: $2 \times 127 ~\pu{ohm^-1 mole^-1} = 254 ~\pu{ohm^-1 mole^-1}$

and that of $\ce{Cl^{-1}} = 76~\pu{ohm^-1 cm^2 mole^-1} $

Now apply Kohlrausch's law of molar conductivity of solution at infinite dilution:

$\lambda^o_{\ce{BaCl2}}= \lambda^o _{\ce{Ba^{2+}}} + 2\lambda^o_{\ce{Cl^{-1}}} \\ \implies  \lambda^o_{\ce{BaCl2}} = 254 + 2\times 76 = 406 ~\pu{ohm^-1 cm^2 mole^-1}$

Now, as you say, equivalent conductivity is $\dfrac{1}{2}$ times the molar conductivity for $\ce{BaCl_2}$ so equivalent conductivity of $\ce{BaCl_2} = \dfrac{1}{2}\times 406 = 203 ~ \pu{ohm^-1 cm^2 equivalent^-1}$ which is the correct answer.

Alternatively, observe that $203 = 127 + 76$ so Kohlrausch's law for equivalent conductivity of strong electrolyte at infinite dilution may be stated as:

  The equivalent conductivity of a strong electrolyte at infinite dilution is equal to the sum of the equivalent conductivities of the anions and cations.


Why are the secondary alpha hydrogens more acidic than primary alpha hydrogens (keto-enol tautomerization)? I learned in organic chemistry that secondary/tertiary carbanions are less stable than primary carbanions and therefore primary hydrogens are MORE acidic than secondary/tertiary hydrogens.

But lets take 3-methyl-2-butanone as an example. It seems that the more substituted alpha carbon is more acidic than the nearby primary carbon. In other words, it seems that the keto-enol tautomerization prefers to form a C=C with the secondary carbon over the primary carbon.

Why would a base prefer to deprotonate the secondary carbon over the primary carbon to trigger keto-enol tautomerization over the theoretically more acidic primary carbon?

You're actually asking two questions. There is both kinetic and thermodynamic acidity. There are times when performing deprotonation at low temperature will yield the kinetically favored conjugate base, i.e., the proton that is most exposed. If you increase the temperature of this intermediate, you can get to the point where proton transfers happen in reaction vessel to create the more stable thermodynamically favored conjugate base. Usually, this is the more substituted one.


What is the relation between chemical thermodynamics and chemical kinetics? Can one be understood on the basis of the other or are they not interrelated at all?

The first thing my kinetics textbook demonstrated was how thermodynamics ignores time taken for a process whereas kinetics considers the time and rate of the process and hence kinetics is a more effective way to view change.

Also, I saw examples of highly thermodynamically feasible reactions which were not kinetically favourable in the sense that they were so slow that the entire process could be neglected as not happening.

Can one understand the thermodynamics of a reaction by studying it kinetics?

Or can one figure out the kinetics of a reaction by thermodynamical consideration?

Also, would it be a wise observation to say that thermodynamics is a largely theoretical subject while kinetics is a more experimental field?

No. Both I think are equally theoritical or experimental. You must study both seperately. Thermodynamics doesn't speak about rates. While kinetics doesn't tell you stability.
If a reaction produces more than one product, thermodynamics will tell you which is more stable, while that doesn't necessarily mean it is the major product. It may happen that rate of formation of other product is more under certain conditionds. Read about thermodynamic vs kinetic product and you will get it.


Which of the following possesses the highest energy electron? Which of the following possesses the highest energy electron?

  $\ce{Br-}$
  $\ce{Ca^2+}$
  $\ce{Cr+}$
  $\ce{As}$

Don't bromine and arsenic have equally high energy electrons (i.e. in the $\ce{4p}$ subshell)? My book says that arsenic is the correct answer. But I don't see why $\ce{Br^{–}}$ is wrong.

My best guess is that your book wants you to argue that, since the bromine nucleus has a higher charge than arsenic, electrons are more strongly attracted to it, lowering their energy. And indeed this would be true - if we only considered interactions between the electrons and the nucleus, and ignored electron-electron interactions. Unfortunately, however, the situation in real life is more complicated than this, because electrons repel one another, “shielding” each other from being attracted to the nucleus (which, of course, is the reason why subshells have different energies in the first place). And the bromide ion has three more electrons than the arsenic atom, which actually turns out to be quite significant.

In fact Webelements gives the electron affinity of $\ce{Br}$ (i.e., the first ionisation energy of $\ce{Br^-}$) as 324.8 kJ mol$^{-1}$ while the first ionisation energy of As is 947.0 kJ mol$^{-1}$. So in fact it seems that, experimentally, the bromide ion has the highest energy electron (which therefore requires the lowest energy to remove it).

This is a more difficult question than it looks - the problem of how to deal with interactions between lots of electrons is still an active research field! A fairer question to do in your head would be to compare, say, $\ce{Br^{2+}}$ with As: in this case the number of electrons of each species is the same, so that the argument I put forward at the beginning of this response is actually correct here.


Does the shared electron in Covalent bonds revolve around nucleus? We know that electrons are charges that revolve around the nucleus. Then, when in covalent bonds the electron is shared; does the electron obey the rule?

I am guessing that you have an understanding of the atomic orbitals of an atom such as 1s and 3p atomic orbitals. Atomic orbitals are the probability distribution of where an electron is going to be $90$% of the time. Technically, electrons don't revolve around the nucleus but are rather quantum particles that are superimposed in several positions at the same time. However the description of electrons revolving randomly around the nucleus is suffice at this level.

When a covalent bond is formed, the electrons do still continue to revolve around the nucleus in orbitals, however a new kind of orbital is formed which are called molecular orbitals (rather than atomic orbitals). Here is a simple explanation of what molecular orbitals are.

Molecular orbitals are formed by constructively and destructively overlapping the atomic orbitals together. The simplest example of this is when $\ce{H2}$ is formed. Here, each H atom has a 1s atomic orbital. When they form, these atomic orbitals form 2 molecular orbitals. Where one molecular orbital is due to constructively overlapping the 2 atomic orbitals and the other molecular orbital is due to destructively overlapping the 2 atomic orbitals. This shown in the diagram below:

As you see that these 2 molecular orbitals that are formed are quite different to each other. The MO that is formed due to constructive overlap has greater electron density between the 2 positive nucleus. Meanwhile the MO that is formed due to destructive overlapping has decreases electron density between the 2 positive nucleus and in fact have a node between them. Therefore the first MO is lower in energy than the second MO.

Like AOs, the same rules apply to MOs. Each MO is able to hold 2 electrons and the electrons fill the lowest energy orbital first. So in $\ce{H2}$ the 2 electrons will fill the first MO and will revolve around the nucleus in that orbital (well technically only for $90$% of the time).

This theory can be applied to any compound, however it gets more complicated as the number of AOs increases and the number of atoms increases. However essentially, when atoms form covalent bonds, the electrons continue to revolve around the nucleus in MOs rather than AOs.


What can we say about the Gibbs free energy if it decreases or not? [Sorry if the question looks trivial but I have thought about it for days and red a number of books all with no success.]

In almost every thermodynamics book that has something to say about chemical reactions there is a proof for the fact that at constant pressure and temperature the process is spontaneous in the direction in which the Gibbs free energy decreases. In some of the books it is also remarked that this conclusion is for when there is no "additional work" (beside the work of pressure in expanding the system), if there is an additional work then except for $dG=0$ the reversible process will yield into that additional work gaining its maximum value:

\begin{align}
\text{The only work is expansive:}\qquad & dG\le 0\\
\text{There are some additional works present:}\qquad & dG\le \delta W_{\text{additional}}
\end{align}

In this regard I have a question:
 When we have some additional works there in the problem (like electric work in an electrochemical cell), so that the second inequality holds and the maximum work is assigned to $dG$ as $dG=\delta W_{Add,Max}$, why still we feel free to use the first inequality and claim $dG$ should be negative?

In the case of electrochemical cells we have: $\delta W_{E,Max}=-n\,F\,E_{cell}\,d\zeta$ wherein $\zeta$ measure the advancement of the reaction. So we would obtain in a reversible (i.e. infinitely slow) reaction $\Delta_rG=\frac{dG_{p,t}}{d\zeta}=-n\,F\,E_{cell}$. But all the books then state that since the reaction is spontaneous in the direction in which $\Delta_rG\le 0$ so we would obtain $E_{cell}\ge 0$, while we know the inequality $\Delta_rG\le0$ is obtained from $dG\le0$ which doesn't hold here.

My question in other wording reads like this. Does the electrochemical cell problem involve an electric work or not? If "yes" then from where do we know the Gibbs free energy should decrease? And if "no" then from where do we obtain the equality $\Delta_rG=-n\,F\,E_{cell}$?

I think the confusion may be because you are conflating two ideas about Gibbs free energy.

The first is that Gibbs energy is a thermodynamic potential - this means that on a given surface in phase-space (the NPT surface), it is the "driving" force for, well, everything. Including chemical reactions.

The second idea is that the net change in Gibbs energy when going from reactants to products indicates the direction that the reaction will (eventually) go, and that its magnitude tells you the maximum amount of work that could be done by that reaction.

The second idea follows from the first - this becomes easy to understand if you just picture the thermodynamic potential as a gravitational potential, since we all have a good intuitive understanding of gravitational potentials.

Imagine a hill representing the thermodynamic potential (in this case Gibbs energy):

The rock represents a reaction coordinate - say, the concentration of a reactant. What will happen to the rock? It will go downhill. What will its maximum energy be when it reaches the bottom? Whatever the difference in potential energies is. What is the minimum energy it would need to get back the top? Again, whatever the net difference is in potential energies.

When you ask about $dG$ being negative, you are asking about the differential change - in other words, what is the slope? Obviously, if the reaction proceeds, the slope must be negative - "things roll downhill"

When you ask about $\Delta G$ being negative, you are asking whether the net change is positive or negative. The system can never do more work than the net change in $\Delta G$, because that is the maximum change in thermodynamic potential energy.

For your problem, the reaction involves moving an electron against a potential - the cell potential. The reaction can provide $\Delta G$ amount of energy with which to move that electron.