How exactly does a battery work (specifically Pb/H2SO4/H2SO4/PbO2 lead acid battery)? Specifically talking about the lead-acid battery, I understand (or think I understand) this much:
The anode, in the presence of the electrolyte, $\ce{H2SO4}$ will be oxidised:
$$\ce{Pb (s) + SO4^2- (aq) -> PbSO4 (s) + 2e-}$$

At the same time at the $\ce{PbO2}$ electrode, two electrons from the $\ce{PbO2}$ solid are lost to the electrolyte:
$$\ce{PbO2 (s) + 2e- + 4H+ (aq) + SO4^2-(aq) -> PbSO4 (s) + 2H2O}$$

This all occurs without a solid conductive path (say a wire) between the anode and cathode, the cathode will build a positive charge while the anode builds a negative charge and ions will diffuse appropriately during this process.

When a wire is connected between the anode and cathode, electrons that have built up on the surface of the $\ce{Pb}$ will migrate to the $\ce{PbO2}$ which is of lower potential and if a load is placed along the way work can be done.
So batteries will discharge by themselves without even contributing to any work? Is this not true for all batteries?

The part that confuses me is that most people say that batteries store chemical energy.

I understand that the redox reactions release energy which can be used as electrical energy but if batteries discharge without contributing to any meaningful work then why are they still considered to be storing energy? Aren't they converting chemical energy to heat continually unless they are put to use?

In order for oxidation-reduction chemical reactions to occur, the electrons must somehow go from the anode electrode where the oxidizing (electron donor) takes place to the cathode electrode where the reducing (electron acceptance) occurs.

Ideally, the electrolyte, in this case sulfuric acid, should allow the movement of positively charged ions but prevent the movement of electrons through it.  This forces the electrons to go through a circuit, external to the cell, to go from anode to cathode thus enabling the redox to occur.  Without electron flow, said redox reaction will not occur.

In the real world, there will be some leakage current through the electrolyte and other parts of battery.  We can eliminate the leakage with a more insulating electrolyte, but this would hurt the charging process and make the battery less rechargeable.  So, what we have is an economical compromise.


How well do d-block electrons actually shield? I've been using 'Chemical Structure and Reactivity: An integrated approach' by James Keeler and Peter Wothers to study periodicity, among other sources. However, there seem to be some commonly accepted contradictions used, and I don't understand how to reconcile these ideas:

1) D-block electrons shield poorly (s>p>d>f)

2) Using Slater's scale, d block electrons would, in fact, shield as well as most other (n-1) orbitals. Even though this is a rough scale, I take it to mean that d block electrons do have a notable shielding effect.

3) D-block contraction is the effect of increased Zeff due to the poor shielding of d-block electrons, and the addition of an equal number of protons.

Therefore, for example, down Group 13: the filling of the d-block orbital corresponds to an extra increase in Zeff with respect to the 4s/4p compared to the 3s/3p orbital. The 4s/4p orbitals will therefore be lower in energy than expected: shielding does not completely offset the extra protons.

4) Across the first row d-block elements, the energy of the 4s orbital does not fall very much. This is because the increase in nuclear charge is offset almost entirely by the addition of a shielding d-block electron, implying almost perfect shielding.

These four points seem, to me, to be essentially using d-block electrons at both ends of the spectrum when convenient. Is there a way to reconcile these ideas, or where have I gone wrong in my logic?

Let us read Keeler and Wothers's Chemical Structure and Reactivity carefully. (It's really a great book. I wish I had used it in the past.) We excerpt the relevant paragraphs that detail the points you mention, and highlight particularly important statements regarding screening. All page numbers are taken from the second edition.

Point 2, on Slater's rules.

[Energy ordering:] (1s) (2s, 2p) (3s, 3p) (3d) (4s, 4p) (4d) (4f) ...

If the electron being considered is in an ns or np orbital, then electrons in the next lowest shell (i.e. that with (n-1)) each contribute 0.85 to $S$. Those electrons in lower shells (i.e. (n-2) and lower) contribute 1.00 to $S$.

If the electron being considered is in an nd or nf orbital, all electrons below it in energy contribute 1.00 to $S$ [the shielding constant]. (p. 262)

Point 3, on Z$_\text{eff}$ and d-block contraction.

...there is a downwards kink in the [s-block] orbital energy on moving from the 3s to the 4s [orbitals]. [...] This difference is due to the filling of the 3d orbitals between calcium and gallium. The effects of all the extra protons in the nucleus is to cause an additional lowering in the energy of the valence orbitals for the p-block elements in Period 4. However, the effect is not really drastic: the 4s and 4p orbitals in gallium do not seem to have experienced the full effect of the ten extra protons. We interpret this by saying that the effect of the extra protons on the orbital energies of the 4s and 4p has, to a large extent, been cancelled out by the electrons that have been added in the 3d orbitals.
We could rephrase all of this by looking at the problem from the perspective of effective nuclear charges rather than orbital energies. The increase in Z$_\text{eff}$ for gallium is slightly greater than might have been expected by comparison with boron and aluminium because of the filling of the d-block. (p. 266)

Point 4, on orbital energies.

...between scandium (Sc) and copper (Cu) electrons are being added to 3d orbitals, and such electrons form an effective screen for the 4s electrons. To put it another way, as far as the 4s electrons are concerned, on moving from one element to the next between scandium and copper, the effect of the extra proton is largely cancelled out by the addition of the electron to the lower 3d orbital. Consequently the energy of the 4s falls rather slowly. (p. 261)
The outer electron in potassium occupies the 4s AO, rather than the 3d. This is because the 4s orbital penetrates to the nucleus more effectively than does the 3d, resulting in the energy of the 4s being lower than that of the 3d. [...] ...both the 3s and the 3p have subsidiary maxima close in to the nucleus, but the first maximum for the 3s is very much closer in. The 3d has no such subsidiary maxima, and therefore is much less penetrating than 3s or 3p.
...the electrons in the 4s do not screen the 3d particularly well since much of the electron density from the 4s is further out from the nucleus than that from the 3d.
From scandium onwards the energies of both the 4s and 3d AOs drop steadily: this is simply the result of the increase in nuclear charge not being quite offset by the increase in electron-electron repulsion. Put another way, the electrons in the 3d sub-shell do not screen one another particularly well. Although both AOs fall in energy, the 4s falls less steeply, which can be explained by noting that this AO is quite well screened by the 3d electrons, not least as these are in a lower shell. (p. 627)

We see that Keeler and Wothers are at least self-consistent regarding points 2, 3, and 4: 3d electrons are effective at shielding 4s and 4p electrons and not effective at shielding 3d electrons. I believe, therefore, that the contradiction lies in point 1, which should be corrected as follows:

d-block electrons shield poorly s or p electrons in the same shell

I suspect the statement that d-block electrons shield poorly is taught in reference to the concept of penetration. If we look at plots of radial probability density for the 3s, 3p, and 3d orbitals, we might naively conclude, on the basis of mean distance from nucleus, that 3d electrons are closest to the nucleus, which means that they should shield 3s and 3p electrons well. In reality, because of penetration, this is not the case. Thus we conclude that 3d electrons shield (s or p electrons in the same shell) poorly, and the misconception arises if we try to extend this poor shielding to 4s electrons and the like.


Calculate Work Done for Reversible and Irreversible Adiabatic process How to calculate it.

Work done for adiabatic process is (P1V1 - P2V2)/(γ-1) but this is also the work done for reversible process. How?

And How do we calculate for irreversible?

For an irreversible process, the pressure of the gas within the system is typically non-uniform spatially, and so there is no unique value that can be used in conjunction with the ideal gas law to calculate the work.  But, at the interface with the surroundings, the force per unit area exerted by the gas on the interface always exactly matches the pressure of the surroundings, which is usually referred to as $P_{ext}$.  So, if $P_{ext}$ can be controlled externally as a function of the volume, one can calculate the amount of work done by the gas on the surroundings.  In many homework problems encountered in thermodynamics, the value of $P_{ext}$ is often specified as being constant during an irreversible process.

In addition to the effects of pressure and temperature nonuniformities in a gas experiencing a rapid irreversible expansion or compression, there is also another physical mechanism acting.  This is related to the viscous behavior of a gas that is deforming.  In the case of a rapid irreversible deformation, the gas pressure depends not only on the gas volume, but also on the rate of change of volume.  The behavior is analogous to that of an ideal spring and damper in parallel (as in the front suspension of a car).  The force exerted on a spring/damper combination during a rapid stretching or compression depends both on (a) how much the combination is stretched, but also on (b) how rapidly it is stretched (because of the effect of the damper).  A very crude approximation to the behavior of an ideal gas that is experiencing a rapid (irreversible) rate of deformation is provided by the equation:  $$P=\frac{nRT}{V}-\frac{k}{V}\frac{dV}{dt}=P_{ext}$$ where k is a positive quantity related to the viscosity of the gas.  So the gas pressure on the piston face depends both on the amount of volume change and on the rate of volume change.  The equation indicates that, during rapid expansion (i.e., dV/dt > 0), the pressure exerted by the gas on the surroundings is less than that given by the ideal gas law, while, during rapid compression (i.e., dV/dt < 0), the pressure exerted by the gas on the surroundings is greater than that given by the ideal gas law.


High or low spin complex of Zn2+ and Ti3+ The zinc ion in its +2 state can show no more pairing, since all the electrons are already paired - so in any complex it forms it will have the same magnetic moment as it does right now. $\ce{Ti}$ in its +3 state will show both outer and inner orbital complexes without any electrons pairing.

So while calling these complexes high spin or low spin complexes is not useful, since there is no possibility of pairing involved in either - but all the same, won't they be called high spin complexes, since no lowering of magnetic moment occurs?

My teacher said, however, that since $\ce{Zn}$ in +2 state forms only outer orbital complexes, its complexes will be called high spin - and $\ce{Ti}$ in its +2 state similarly when it forms inner orbital complexes (which it does without pairing), will be called low spin because the complex is inner orbital. But shouldn't such a complex of $\ce{Ti}$ be a high spin complex? (I'm saying this in reference to VBT).

Are you familiar with ligand field stabilisation energy? This approach answers your question. Interaction of the central metal ion with the surrounding ligands causes splitting of the $d$ orbitals into an essentially upper and lower energy level, the form of the splitting is rather complex and depends directly upon the symmetry point group that the ligand complex belongs to.

Either way, high spin complexes only form if the reduction in inter electron repulsion energy is greater than the energy splitting of the orbitals, with the definition of high spin being a none aufbau filling of the split orbitals. It does not make sense to describe the electronic structure of a $\ce{Zn^2+}$ ion in a complex as high or low spin, there is only one configuration available as there are 10 $d$ electrons present, and the 5 $d$ orbitals split to form 5 new orbitals of varying degeneracies (dependent upon the point group of the complex and the symmetry of the coordination environment).

$\ce{Ti^2+}$ has 2 valence d electrons available to be involved in bonding with the ligand environment, however there is no coordination environment that I am aware of (The splitting would have to be extremely small and other thermodynamic factors such as the ionic contribution to the Gibbs energy of complex formation would mean that the complex would never be observed anyway - such a low splitting would only be the result of very poor overlap which would be caused by large separation between the central ion and the ligands, resulting in a lower ionic term) that causes splitting where the magnitude of the splitting is less than the repulsion term invoked when both electrons occupy the same molecular orbital, be it degenerate or not.

The $\ce{Ti^3+}$ ion has only 1 $d$ electron, it will definitely reside in the lowest available molecular orbital. However, the first excited states, ignoring hyperfine splitting, of both titanium ions will be high spin - the magnitude of the splitting usually corresponds to wavelengths of light in the visible region of the spectrum, hence many transition metal complexes are coloured.


How to explain the periodic trends in boiling points in groups? Observing the trend of boiling points of the compounds listed, choose the appropriate terms to fit into the blanks:
\begin{array}{lr}
\hline
\text{Compound} & \text{b.p.}/\pu{°C}\\
\hline
\ce{H2Te} & −2 \\
\ce{H2Se} & −41.5 \\
\ce{H2S} & −60.7 \\
\ce{H2O} & 100 \\
\hline
\end{array}
The ___ the compound, the stronger the ___ present between molecules. The stronger the ___, the higher the boiling point of the compound?
More polar, smaller, London dispersion forces, larger, less polar, dipole-dipole and ion-dipole forces

I noticed that the electronegativity for each compound increases from top to bottom, so does higher electronegativity implies higher boiling point?
I've also tried putting more polar the compound, the stronger the dipole-dipole and ion forces. No luck, I still got it wrong.

I think it is wrong to say there should be a trend as we study the trends of individual atoms, not molecule and compound itself. Boiling points depend on the intermolecular force which depends on the polarity of the whole molecule, which in turn depends on the constituting species, so there should not be specific trend.
It is only the matter of comparison between the species in a molecule. Moreover, there is not a single type of intermolecular forces which bonds molecules such as dispersion, metallic or hydrogen: all have different strength.


What are the conditions for liquid metal to be corrosive to another, different solid metal? I've had already been aware of the fact that metallic aluminum can be degraded by coming in contact with mercury, but also came across another question more recently here explains a similar reaction with liquid gallium.

So my question is more general regarding corrosive action of any liquid metal on other metals.

What are the more general conditions for a liquid metal to be corrosive to another, different solid metal?

Perhaps the word corrosive is not the most suitable choice. Corrosion refers to a chemical reaction. If I understand it right what's happening, at least in the case of $\ce{Ga}$ and $\ce{Hg}$ on $\ce{Al}$, is a weakening of the grain boundaries by displacement and amalgamation.

Another interesting thought occurred to me. Liquid sodium is often used as a coolant in nuclear power plants. It's pumped through metal pipe (material unknown). So here is a good example of a combination that doesn't lead to corrosion. Are there examples other than I've mentioned that lead to corrosion? So what makes the difference?

There are no general conditions- indeed there no general conditions for the any of the forms of corrosion. Solubility is a factor and many other factors but it just comes down the actual interactions between a given liquid metal and another metal and the service conditions such as temperature, velocity, impurities.

Liquid-metal attack has the following categories:

simple solution,
alloying between liquid metal and solid metal,
intergranular penetration,
impurity reactions,
temperature-gradient mass transfer,
concentration-gradient mass transfer, or dissimilar-metal
mass transfer.

With that many means- determining what liquid metal damages what solid metal  is very difficult.

http://web.ornl.gov/info/reports/1956/3445603500959.pdf


Why so much color change in synthesis of triphenylmethanol? In lab today, I prepared a solution of Grignard reagent (phenylmagnesiumbromide) in anhydrous ether. I then added a solution of diphenylmethanone in anhydrous ether (AKA benzophenone) to the Grignard reagent and observed a brilliant array of color change.

Immedietely when the first drops of the benzophenone were added to the cloudy grayish grignard mixture (grey color due to impurities in starting magnesium), the reaction turned a dark purple before quickly changing to a crimson red, before once again changing to a bright pink. As more benzophenone was added, the reaction mixture changed, once more, to a milky white color and promptly hardened.

I understand the white final color and hardness is due to formation of my intended product: triphenylmethoxide magnesium salt...

But WHAT in the world was happening in that reaction with all those fantastic color changes?

It sounds like you might have generated a species with this structural moiety:

This can happen if your solution was concentrated enough or the boiling of your ether was vigorous, so the local overheating caused the formation of $\ce{MgO}$, the guy above, and $\ce{Br-}$, which gives a very distinct bright red-pink color, which is VERY bright and only requires a bit of the side product to form. Though this carbocation is kinda stable, it will find the $\ce{Br-}$ in the solution and form triphenylmethylbromide as a minute impurity.


What purpose does mercury dichloride have in fireworks? My local newspaper ran an article covering the local rules about fireworks.  One of those rules specifically calls out "snakes and fountains containing mercury bichloride" as forbidden.

Neither Wikipedia's article on mercury(II) chloride nor its article on mercury lists any fireworks-related use.  What purpose does the mercury dichloride serve?  Is it just a way to color the sparks, or does it have some more fundamental purpose?

Most likely, mercury was used as a colorant, specifically red.  This list shows mercury providing a red color in flame tests, and the same would presumably be seen in fireworks.

Several other, less dangerous elements also give red or reddish colors.  This site identifies lithium as providing the color in red fireworks, and also mentions lithium carbonate specifically.

Be safe this 4th, handle fireworks with care if you do so at all!


Why doesn't TEMPO react with itself? TEMPO ((2,2,6,6-tetramethylpiperidin-1-yl)oxyl, or (2,2,6,6-tetramethylpiperidin-1-yl)oxidanyl) has the following structure:

Given that it's a free radical, why doesn't it react with itself, despite being capable of reacting with other radicals in the system itself)?

In addition, according to Wikipedia, it's a red-orange, sublimable solid. How can it be a solid despite being a radical?

I believe the first part of your question is asking why the oxygen with the unpaired electron doesn't find a second one to form a peroxide.

The answer is actually quite simple. It is by design. Next to the nitrogen, you see the 4 methyl groups that give TEMPO the "TEM" part of it's name. These methyl groups give TEMPO a lot of steric bulk around the oxygen, and this bulk makes it very difficult for two TEMPO radicals to form an oxygen-oxygen bond with each other. There is just not enough space. You can build a model to convince yourself of this.


Selective bond breaking Can a specific bond be broken by an engineered energy wave corresponding to the bond length?

For example, can we break the $\ce{C-N}$ bond and not the $\ce{C-H}$ in a hydrocarbon chain?

In general, to break one bond vs another on demand in the presence of many others  is a very difficult thing to achieve. Not possible at all by heating molecules thus there has been much effort to do so using lasers.

But this becomes very difficult also. This is  because on exciting a molecule into an excited electronic state, such that it has enough energy to break a bond, and into the type of vibration involving the bond you wish to break,  the energy then leaks out to other different vibrational normal modes and can then end up anywhere. So it does not just stay where the energy was initially placed.

This 'leaking' occurs due to vibrational anharmonicity which provides a 'coupling' with which one normal mode can swap energy with another. The process can be very fast and even if femtosecond laser pulses are used energy still moves all over the place and competes with accumulating energy into the 'victim' bond effectively defeating the process. The area is, however, one of very active study so progress in this area is to be anticipated.


Crossed Cannizzaro reaction What is the major product of the following reaction:

I think methanol will be formed. Although formaldehyde is more reactive to nucleophilic addition, the rate determining step of the Cannizzaro reaction is the hydride transfer. So instead of a nucleophilic attack of $\ce{OH-}$ on formaldehyde, nucleophilic attack takes place on the aromatic aldehyde.
Is that correct?

Background

It seems that you understand the mechanism of the Canizzarro and Crossed Cannizzaro reaction, but let me supply a link to an earlier answer on the Cannizzaro reaction itself for others who might want to review the reaction. Also, here is the mechanism for the Cannizzaro reaction from the same link (but note, as we'll discuss below, the first step is really an equilibrium).

In the Cannizzaro reaction an aldehyde disproportionates into the corresponding  alcohol and acid. The maximum yield of the alcohol or acid is therefore only 50%.  If you've worked in the lab for 2 weeks synthesizing 50 mg of the aldehyde, and if your next step is the Cannizzaro reaction to prepare the alcohol from it, then a 50% yield is fairly distressing!

Here is where the Crossed Cannizzaro reaction can come in handy.  If it's the alcohol that your after, then using formaldehyde as a sacrificial reagent allows you to achieve a much higher yield of the desired alcohol.  In the Crossed Cannizzaro the formaldehyde is always oxidized to the acid and the other aldehyde is always reduced to the alcohol - you could get a 100% yield of alcohol!

Your Question - Why is that?

The first step in either the Cannizzaro or Crossed Cannizzaro reaction is an equilibrium between the aldehyde(s) and hydroxide ion to form a tetrahedral intermediate as pictured below. In the case of Crossed Cannizzaro with formaldehyde, the formaldehyde carbonyl is extremely reactive and the equilibrium will lie strongly on the side of the formaldehyde tetrahedral intermediate, and consequently most of the other aldehyde will remain present as the aldehyde.

You are correct to note that the rate determining step is the second step where hydride is transferred from the tetrahedral intermediate to any remaining carbonyl compound. But look at the 2 possible transition states for the Crossed Cannizzaro with formaldehyde (pictured above) - they are virtually the same, just some minor differences.  Hence the 2 possible rate determining steps should have similar activation energies.  This means that the energetics of the rate determining step will not control the reaction outcome (e.g product distribution), rather the initial equilibrium will be controlling.

If most of the formaldehyde is converted to the tetrahedral intermediate, then most of the other aldehyde remains as the aldehyde; and if the 2 possible rate determining steps from the tetrahedral intermediates have similar rates (they do as explained above), then the most likely reaction will involve the formaldehyde tetrahedral intermediate (major tetrahedral intermediate present at equilibrium) and the other aldehyde (major remaining carbonyl compound present at equilibrium).

Hence, in a Crossed Cannizzaro the formaldehyde is always oxidized to formic acid and the other aldehyde is always reduced to the corresponding alcohol.


Help with uncertainties When diluting a mixture of two reactants with lets say a 100ml of H2O, how do you add this to the uncertainties of the concentrations of the reactants? Should the relative uncertainty of the added H2O be added to the relative uncertainty of the volume in both reactants?

When adding values together, you should add together their absolute uncertainties. For example, if you started with $500\pm5\mathrm{mL}$ and add $100\pm1\mathrm{mL}$, your new volume value is $600\pm6\mathrm{mL}$.

You would add the relative uncertainties if you are multiplying values. For example, if you started with $1\pm.06\text{ moles}$ and want to find the molarity in a solution with a volume of $500\pm5\mathrm{mL}$, your molarity will be $2\pm.14\text{M}$. The uncertainty was obtained by finding the sum of the relative uncertainties and multiplying it by the concentration to obtain an absolute uncertainty, as shown below.  $$(\frac{.06}{1}+\frac{5}{500})(2\text{M})=\frac{7}{100}(2\text{M})=.14\text{M}$$

For more details on uncertainty rules, this page from the University of Victoria gives some of the basics.


Can I convert elemental nitrogen to nitrous oxide? Can I convert nitrogen to $\ce{N2O}$, to conduct barking dog reaction, where it's needed as oxidizer?

There are several ways to fairly safely make $\ce{N2O}$ in the laboratory that don't involve the difficulties associated with trying to react molecular nitrogen.  The following reactions all are mentioned in this Wikipedia article which I suggest reading, along with the references.  Depending on the quantity you are making, I'd be very cautious about using the thermal decomposition of ammonium nitrate as this can kind of blow up on you (literally). Of course every reaction has it's own risks and it's up to you to decide what you are most comfortable with.

I think the direct oxidation of ammonia over a manganese dioxide-bismuth oxide catalystis an intriguing method$\mathrm{^{(1)}}$

$$\ce{2 NH3 + 2O2 -> N2O + 3 H2O}$$

This also seems like a fairly straight-forward route:

$$\ce{2 HNO3 + 8 HCl + 4 SnCl2 → 5 H2O + 4 SnCl4 + N2O}$$

And this seems to me a bit of a more complex synthyses:

$$\ce{2 (NH2)2CO + 2 HNO3 + H2SO4 → 2 N2O + 2 CO2 + (NH4)2SO4 + 2H2O}$$

Again, I'll say molecular nitrogen makes a lousy starting material for most syntheses, including $\ce{N2O}$.

1) Suwa T, Matsushima A, Suziki Y, Namina Y (1961). "Synthesis of Nitrous Oxide by Oxidation of Ammonia". Kohyo Kagaku Zasshi, Showa Denka Ltd. 64: 1879–1888.


Factors to consider to predict EPR What are the factors to consider when predicting the EPR spectrum of $\ce{Cu(Ox)P(CH3)3}$ where $\ce{OX}$ is the oxalate ion?

I understand that there are several factors to consider for EPR (Electron Paramagnetic Resonance) when we have transition metal complexes like: coupling, anisotropy, high spin/low spin, Jahn Teller distortion, Zero Field Splitting, or Kramer's rule. My question is which of these are important for this special copper complex and why?

The EPR experiment measures the interaction of unpaired spin density ($\rho(\alpha)$) at the nuclei.  This is quantified by an equation called the Fermi contact interaction.

If you have some prior knowledge of "where to put the dot" in your structure (highly localized), then generally you can assume that the EPR pattern is split most by that nucleus.  Splitting patterns are determined by standard rules in accord with the Zeeman effect.

The nucleus must have a non-zero nuclear spin to be observable. Certain nuclei such as hydrogen-1 and many transition metals have spin. It is very common that isotopic impurities (e.g. carbon-13) appear though the most prevalent isotope is EPR-silent.

(There are other complications, such as nuclear quadrupole moment and broadening due to exchange of chemical sites.)

An additional issue is that the spin density is generally spread over the molecule, and interacts with multiple nuclei.  It turns out to be easier to measure the spectrum, and then rationalize it given the particular structure and nuclear composition.

Accurate prediction of EPR spectra is a rather difficult problem, as quantum chemical calculations generally don't emphasize basis function placement at the nucleus, as they tend to focus on the "valence" and not the cusp-like nature of the radial distribution.  (Here, Slater-type orbitals would be better in principle than Gaussian-type, but STOs of sufficient number and of high-enough quality of Hamiltonian are virtually impossible).  Finally, solvent and powder effects can render the gas-phase ab initio calculations virtually irrelevant.

EPR simulation/interpretation of fine structure is much more difficult than NMR spectra, because in NMR spectra, one has the chemical shift to conveniently move splitting patterns out of the way.  In EPR, the splittings are often "piled" on top of each other, and make it necessary to use computer simulation to obtain precise fits.

In your specific example, I would identify the nuclei with spin (probably Cu, certainly phosphorus-31, and then hydrogen-1).  Then, identify where you think the unpaired density will be, and then take the nuclear spin + 1 and that will be the "guesstimated" isotropic spectrum.  The pattern will follow some variant Pascal's triangle if I=1/2, otherwise, you have to figure it out by writing out the number of possible configurations.

Edit: I have not even started on the characteristics of triplet spectra, as here the electrons interact.  One thing you can hope for is that all of your electrons are paired (singlet), and so it's EPR silent.


Find whether a gas cools or heats up on Joule-Thomson expansion A gas that follows $P(V - nb)= nRT$ is subjected to Joule-Thomson expansion. Tell whether it cools or heats up.

$$\mu = {\partial T \over \partial P} = {\partial P(V - nb)/nR \over \partial P} = \frac1{nR}\left(V - nb + {\partial V \over \partial P}\right) = \frac1{nR}\left(V - nb - {nRT\over P^2}\right)$$

Now how do I determine whether $\mu >0 $ or $\mu < 0$ without knowing anything  about temperature or anything else ?

The equation for dH is:  $$dH=C_pdT+\left[V-T\left(\frac{\partial V}{\partial T}\right)_P\right]dP$$


Long term composition of matter If I have an igneous rock shown by geologic study to be 10,000 years old, approximately what percentage of the atoms (or protons, etc) in the rock might be the same ones that were in it 10,000 years ago?  Is it reasonable to even think of a "particular", individual atom?  Is it the "formation" of particles that persists over time, or is it the actual "matter"?

To answer your question, we can use the idea of radioactive decay;

According to the IB Physics Book from Oxford University, radioactive decay is "a naturally occurring process in which the nucleus of an unstable atom will spontaneously change into a different nuclear configuration by the emission of combinations of alpha particles, beta particles, and gamma radiation. Alpha particles are simply a Helium atom while beta particles consist of either electrons or positrons (the antiparticle of the electron).

Radioactive decay is a random process and therefore it is impossible to predict which exact nucleus may decay next. However, with a large sample of nuclei, say of your igneous rock, it is highly probable that in a given amount of time, a predictable amount of nuclei will decay, even if we cannot predict which ones exactly. Hence we have the idea of half-life, which is the amount of time it takes for the half of the initial nuclei of a sample to decay.

  approximately what percentage of the atoms (or protons, etc) in the rock might be the same ones that were in it 10,000 years ago?

It depends on what the igneous rock consists of; if it is made out of ununoctium-294, an infinitesimal amount of it will remain, as the half-life of it is 5 milliseconds. On the other hand, if the rock consists of Uranium-238, after 10 thousand years the composition of the atomic composition of the rock will have barely changed, as Uranium-238 has a half-life of 4500000000 years. So to answer your question, you would need to look at the half-life of your igneous rock. For example, if your igneous rock has a half life of 10 thousand years, then only 50% of the nuclei would be what they were 10 thousand years ago. However, it would also depend on how much of the rock is radioactive, as MaxW mentioned.

  Is it reasonable to even think of a "particular", individual atom?

As already mentioned, it is difficult to predict which particular nucleus will decay next, therefore, looking at the substance as whole facilitates predicting how much will decay in a certain period of time.

  Is it the "formation" of particles that persists over time, or is it the actual "matter"?

I do not quite understand this question. Nonetheless, I would say that it is the matter that persists over time, as the matter consists of these particles. In the photo provided below, you can see that the nuclei are decaying, but they are not decaying row for row, yet more in a random manner. Hence, I would not consider it to be the formation of particles that consists over time, but the matter as a whole.


Increasing temperature in system of dynamic equilibrium According to the article Summary of Le Chatelier's Principle on dynamicscience.com.au, for the process

  $$\ce{N2(g) + 3 H2(g) <=> 2 NH3(g)}\quad ΔH = \pu{-92 kJ/mol}$$

the forward reaction is exothermic, while the reverse reaction is endothermic.

If the forward reaction is exothermic, increasing temperature decreases $K,$ because it drives the equilibrium backwards, in order to absorb the heat energy.

However, in the graph, there is an abrupt change in the reaction rate of forward and reverse reactions. Why does the forward (exothermic) reaction rate rise up so abruptly, while the reverse (endothermic) reaction does not go up so high?

Shouldn't it clearly be the opposite, as increasing temperature favors the reverse reaction?

Shouldn't it clearly be the opposite, as increasing temperature favors the reverse reaction?

Yes, they made a mistake. For all the other scenarios, they paired a figure of the rate changes with a figure of the matching concentration changes. For this scenario (increase in temperature) they matched it with a correct figure of concentration changes when the temperature is decreased:

If you change the label in the figure posted by the OP, it would reflect what happens when increasing the temperature: Both rates increase, but the reverse rate increases by a larger factor, resulting in a net reverse reaction until equilibrium is reached again.

Here are sketches for the rate changes both for increase in temperature (low to high, upper panel) and decrease in temperature (back to low temperature, lower panel). As you can see, the reverse rate is more sensitive to temperature (higher activation energy because the product is lower in energy than the reactant for this exothermic reaction):


How can I protect my skin from KMnO4? We have national elections coming up in two days, and like every time the election officials will mark people's thumbs with a dab of some purple solution that leaves a brown stain on the skin (and thumbnail) that takes a while to grow out. I'm pretty sure that's a potassium permanganate solution - that stuff behaves exactly the same.

Now I'm not into electoral fraud (there are other controls in any case) but I dislike having that stain on my fingernail for a month or more afterward. Previously I've had partial success by rubbing butter into my skin, but the stain still gets through. I've also tried making some plain old sugar or even glucose available, hoping that it would get oxidized first, but I can't conclusively say that that worked.

Any other tricks to avoid, or at least strongly mitigate, these stains? There's a window of about a minute or two where the purple solution sits on the cuticle while I'm unable to do anything about it.

If it is permanganate, the brown $\ce{MnO2}$ stain will decolorize and solubilize in acidic reductant - lemon juice, perhaps boosted with a little crushed and dissolved Vitamin C.   I'd go for crystal/gentian violet or Disperse Red 9 as a marker, or a reactive Cibacron or Procion dye.

Cthulhu for Fearless Leader!  Why vote for the lesser evil?


What are the differences between diesel and unleaded gasoline? I am curious about the flammability of diesel fuel and its properties, as I don't deal with it on a regular basis (at all).

In particular, I want to know how the fuel works when powering a diesel engine.

Things I know about gasoline engines:

Compresses gasoline and air
Ignites compressed gasoline and air with spark from spark plug
The ignition pushes the piston back turning the crank shaft

I have heard that diesel is less flammable than gasoline and that diesel engines don't require spark plugs. How does this work?

Diesel engines vs. gasoline engines

Diesel engines don't rely on spark plugs, but they still work by igniting the fuel to generate a force that moves a cylinder in the engine.  The air/fuel mixture in a diesel engine is ignited by the compression of the same cylinder during the "compression" stroke.  That is, like most gasoline engines in cars, diesel engines are four-stroke engines.

The compression stroke, in a gasoline engine or in a diesel engine, causes heating of the cylinder contents as the compression happens.  Diesel engines are designed to have a more severe compression of the cylinder contents during the compression stroke.  This results in much higher heating of the contents; in fact the temperature becomes so high that the ignition of the fuel and air becomes spontaneous.

You may ask, where does the energy to power the compression stroke come from?  It comes from inertia in the engine as it is running.  Another way to say it is that the previous power stroke provides the energy for the compression stroke.  (The power stroke is when the exploding air fuel mixture pushes the cylinder out, doing useful work.)  There is so much work released in the power stroke that there is enough to move the car forward as well as provide inertia for the next compression stroke.

Here are some more web links with helpful explanations:

Wikipedia:

  The diesel engine (also known as a compression-ignition or 'CI' engine) is an internal combustion engine in which ignition of the fuel that has been injected into the combustion chamber is initiated by the high temperature which a gas achieves when greatly compressed (adiabatic compression). This contrasts with spark-ignition engines such as a petrol engine (gasoline engine) or gas engine (using a gaseous fuel as opposed to gasoline), which use a spark plug to ignite an air-fuel mixture.
  The diesel engine has the highest thermal efficiency of any standard internal or external combustion engine due to its very high compression ratio and inherent lean burn which enables heat dissipation by the excess air.

How stuff works:

  In theory, diesel engines and gasoline engines are quite similar. They are both internal combustion engines designed to convert the chemical energy available in fuel into mechanical energy. This mechanical energy moves pistons up and down inside cylinders. The pistons are connected to a crankshaft, and the up-and-down motion of the pistons, known as linear motion, creates the rotary motion needed to turn the wheels of a car forward.
  Both diesel engines and gasoline engines covert fuel into energy through a series of small explosions or combustions. The major difference between diesel and gasoline is the way these explosions happen. In a gasoline engine, fuel is mixed with air, compressed by pistons and ignited by sparks from spark plugs. In a diesel engine, however, the air is compressed first, and then the fuel is injected. Because air heats up when it's compressed, the fuel ignites.

Be sure to watch the animation at How Stuff Works.  It's quite helpful!

Diesel fuel vs. gasoline

In diesel engines, premature combustion can be a problem.  This happens when the fuel ignites "too early" in the compression phase, before maximum temperature and thus maximum efficiency can be obtained.  This means that by design, diesel fuel is a bit harder to ignite than gasoline.  It's made of slightly less volatile hydrocarbons.

As noted by @steveverrill in the comments, the above paragraph was incorrect.  A better explanation (straight from his comment) is

  Diesel is harder to ignite at ambient temp due to its lower volatility / higher flash point (approx 52 °C vs -43 °C). This means it's difficult to form an explosive mixture which could be ignited by a spark (which is a local area at much higher than ambient temp.) But inside an engine both diesel & gasoline are fully evaporated. Under these conditions diesel is easier to ignite due to its lower autoignition temperature (approx 265 °C vs 280 °C). Autoignition temperature is tuned by additives to help ensure diesel can be ignited just by compression, whereas gasoline cannot ignite without a spark.

Wikipedia again

  Petroleum-derived diesel is composed of about 75% saturated hydrocarbons (primarily paraffins including $n$, $iso$, and cycloparaffins), and 25% aromatic hydrocarbons (including naphthalenes and alkylbenzenes). The average chemical formula for common diesel fuel is $\ce{C12H23}$, ranging approximately from $\ce{C10H20}$ to $\ce{C15H28}$.

Wikipedia on gasoline

  The bulk of a typical gasoline consists of hydrocarbons with between 4 and 12 carbon atoms per molecule (commonly referred to as C4-C12). It is a mixture of paraffins (alkanes), cycloalkanes (naphthenes), and olefins (alkenes).

Thus, diesel fuel contains mostly hydrocarbon molecules that have on average 12 or so carbon atoms, while gasoline contains hydrocarbons that contain on average 8 or so carbon atoms.  The heavier, bigger molecules in diesel fuel means it evaporates more slowly at ambient conditions, requires more heating (i.e. more compression) before it spontaneously ignites.


Is there a clear distinction between the various intra- and intermolecular interactions? Wikipedia lists dipole-dipole, hydrogen bonding and Van der Waals under intermolecular interactions, and hydrophobic, ionic and covalent under intramolecular interactions.

Is there a clear distinction between each and every one of these interactions? For example: isn't hydrogen bond a private case of VdW? aren't ionic bonds and hydrophobic interactions private cases of dipole-dipole interaction?
Is there one single chart that differentiates all of these types of interaction?

No, there aren't clear distinctions, particularly between intramolecular and intermolecular interactions.

First, a caveat that we generally divide molecules along covalent bonds. So those are strictly intramolecular, in my opinion. (Network solids are a bit tricky to classify.)

Let's go through the rest one at a time.

Dipole-dipole: These are usually intermolecular, but certainly in larger, more flexible molecules, intramolecular polar-polar interactions can change the molecular shape and preferred conformation. (Think about tying together two bar magnets on one piece of string. Certainly they'll interact.)
Hydrogen bonds: Yes, many of these are intermolecular. However, there definitely are intramolecular hydrogen bonds:

Van der Waals: Again, these can be intermolecular. In even a moderate molecule, intramolecular non-bonded van der Waals interactions can change conformations and geometries. Consider axial/equitorial preferences in cyclohexanes or other types of "steric clashes" because of big bulky groups. These are the effects of van der Waals interactions.

OK, so all the intermolecular interactions are also intramolecular ones too.

Hydrophobic interactions: I usually think of these as intermolecular but similar considerations make sense as intramolecular for the reasons I outline under van der Waals interactions above.
Ionic: I'll just think of these as different partial charges in a molecule. Clearly electrostatic interactions occur both inside a molecule (e.g., electrostatic non-bonded interactions changing conformations and geometries) and as intermolecular forces, e.g., water-cation interactions.

So besides covalent interactions, which we use to divide "inside" and "outside" a molecule, all of these are both intermolecular and intramolecular interactions.

I'm not sure I've seen a great chart. Maybe someone else can comment.


Tests on sodium peroxoborate When I prepare sodium perborate by mixing sodium hydroxide with metaborate ($\ce{NaBO}$) and hydrogen peroxide and cooling it I applied a test to it by acidifying it with $\ce{HCl}$ and adding $\ce{KMnO4}$. A precipitate appeared and it turned black. What is the reason?

I tested another sample of sodium perborate by acidifying it with $\ce{NH4Cl}$ and after that KI and this turned white. Again, why?

In your first test, the $\ce{KMnO4}$ probably reacted with the $\ce{H2O2}$ excess remaining after your preparation. The ionic equation for the reaction is

$$\ce{2MnO4- + 3H2O2 -> 2MnO2 + 2H2O +3O2 + 2OH-}$$

The $\ce{MnO2}$ formed is an insoluble black precipitate. Although there is an $\ce{OH-}$, the reaction apparently occurs in neutral and faintly acidic medium as well.

The result of your second test doesn't seem to be very clear. Sodium perborate apparently reacts very quickly with any water so if you added any water during the second test, it must have reacted to form oxygen gas bubbles (which appear white).


How to dissolve chalk (CaCO3)? I've read that vinegar can dissolve $\ce{CaCO3}$, I know that extreme heat can decompose it, but how to dissolve $\ce{CaCO3}$ at home? I need to dissolve chalk in my home because of repairing something. Can someone help me do it without heating $\ce{CaCO3}$? Any replacement for that?

$\ce{CaCO3}$ is sedimenting when mixed with water, and therefore is not easy to remove with hot water. Nevertheless several methods exist that actually dissolve it that are better than vinegar:

You could add hydrochloric acid again to convert it into soluble Calcium chloride, according to equilibrium:
$$\ce{CaCO3 (s) + 2 HCl (aq) <=> CaCl2 (aq) + H2CO3(aq)}$$
You could use carbonic acid instead
$$\ce{CaCO3 (s) + H2CO3 <=> Ca(HCO3)2}$$
this is the reason shells are getting dissolved by an increase of the amount of carbon dioxyde in solution:
$$\ce{CaCO3 (s) + CO2 (aq) + H2O <=> Ca(HCO3)2}$$
But I recommend Sulfamic acid ($\ce{H3NSO3}$), is frequently used for removing rust and limescale, replacing the more volatile and irritating hydrochloric acid, it may be a little more expensive but is certainly not toxic.


How would you know if the light emitted is of one wavelength or if it is a mixture of two or more wavelengths? I am just learning some chemistry after a long time without studying it and I have thought about something.
Some copper compounds emit green light when heated to flame. So, I would like to know if it is known if an emitted light is of one wavelength or if it is a mixture of two or more wavelengths.
Thanks in advance

The green light from copper flame is sadly not a single wavelength. It is very good question though because not many students even think about it. In order to "see" what wavelengths are present in a light source, you either need a spectroscope (old school way) or a monochromator with a detector (modern way). Of course you can search about Bunsen-Kirchoff spectroscope. It uses a prism to disperse the light falling on the prism.
Depending on how much interest and money you have you can easily construct a spectroscope in a shoebox (several published articles out there). Ten years ago or so I used to make CD diffraction spectroscopes in shoeboxes or cardboard boxes for seeking answers to queries like yours.
If you want spent some money ($400-500) you can buy a high quality pocket spectroscope with a wavelength scale . Then you can enjoy all light sources (except the Sun). It is fascinating to see and explore so many phenomena. You will find out by looking at the moon that it has some dark lines (never look at the Sun from a spectroscope). I am not suggesting to buy third rate products from Amazon because twice I purchased from them and the wavelengths were all wrong!
The monochromator is a fancier product. I never had the courage to buy one. They are very expensive. The most common design is a Czerny-Turner design. I am sure Prof. EdV had even fancier products which we have never fathomed :-)
This picture shows white light entering a slit, it falls on a mirror which makes all the rays parallel. The mirror sends the light to a high quality diffraction grating, the reflection diffraction grating separates wavelengths in space. The light from reflection grating falls on a focusing mirror, and the mirror focuses lies on the exit slit. The exit slit has a detector. You would slowly rotate the grating. If there are multiple wavelengths in the light source (say a flame) the detector will record a spike every time a different wavelength reaches the detector.


What chemicals have been named after Star Wars characters? The molecule below is whimsically named Yoda1 because it "uses the force" to open channels in a class of animal proteins known as piezo ion channels (which appear to be part of their pressure sensing nervous systems). See this Nat. Commun. 2018, 9 (1) for a description and other links.

This raised the question for me of how many and what other interesting chemicals have been named after Star Wars characters?

I heard about one other Star Wars compound named after a character: Kinobeon A.[1]

There is also a supramolecular structure named after a Tie fighter.[2]

Apparently there is also a Yoda1 analogue (note the additional sulfur atom) called Dooku1.[3]

References:

Kanehira, T.; Takekoshi, S.; Nagata, H.; Matsuzaki, K.; Kambayashi, Y.; Osamura, R. Y.; Homma, T. A novel and potent biological antioxidant, Kinobeon A, from cell culture of safflower. Life Sci. 2003, 74 (1), 87–97 DOI: 10.1016/j.lfs.2003.06.033. Mirror: www.chm.bris.ac.uk
D'souza, F.; Gadde, S.; El-Khouly, M. E.; Zandler, M. E.; Araki, Y.; Ito, O. A supramolecular Star Wars Tie Fighter Ship: electron transfer in a self-assembled triad composed of two zinc naphthalocyanines and a fullerene. J. Porphyrins Phthalocyanines 2005, 09 (10), 698–705 DOI: 10.1142/S1088424605000812.
Evans, E. L.; Cuthbertson, K.; Endesh, N.; Rode, B.; Blythe, N. M.; Hyman, A. J.; Hall, S. J.; Gaunt, H. J.; Ludlow, M. J.; Foster, R.; Beech, D. J. Yoda1 analogue (Dooku1) which antagonizes Yoda1-evoked activation of Piezo1 and aortic relaxation. Br. J. Pharmacol. 2018, 175 (10), 1744–1759 DOI: 10.1111/bph.14188.


Does partially/completely removing solid shift the equilibrium? For the reaction: $$\ce{P4(s) + 5O2(g) <=> P4O10(s)}$$
Let's say the reaction is at equilibrium, and SOME $\ce{P4O10(s)}$ is removed from the reaction. Will that affect equilibrium? My assumption is no, since $\ce{P4O10(s)}$ is not included in the equilibrium expression.
However, for me, it gets confusing when let's say $\ce{P4O10(s)}$ is COMPLETELY removed. In this case, I would assume that the reaction shifts to the products (since no products are present). I'm not sure if this is right.
Help is appreciated. Thank you!

I am using a saturated sodium chloride solution as an example here because it is more common and easier to reproduce at home. This system is at equilibrium, with salt dissolving and crystallizing at the same rate.
If you take away the solid NaCl from a saturated solution, both the forward and the reverse reactions stop. Obviously, salt crystals are no longer dissolving because there aren't any. Less obviously, there is no crystallization process anymore (there are no crystals that solvated ions could attach to). So this is not a dynamic equilibrium, i.e. the reaction or process no longer is observed at all.
This is different from a system out of equilibrium where one direction of the process is faster than the other direction. In that case, the system will approach equilibrium, and the process (in either direction) will continue to happen when equilibrium is reached (which is why it is called a dynamic equilibrium).
You might ask how we can ever get solid sodium chloride again. If the solution is sufficiently supersaturated, nuclei will form (in solution, or on the container or on some impurity), and then crystal growth will be observed again.


Does oxygen catalyze the decomposition of sucrose? When a bottle of sports drink that contained sucrose is opened, is it expected that the sucrose contained therein will split into its constituents (fructose and glucose) more rapidly than it will via acid hydrolysis?

I understand that in acid hydrolysis there is nucleophilic substitution (provided the acid is protic). Does a similar phenomenon occur when the contents of the bottle are exposed to oxygen?

I am trying to interpret some unexpected results I obtained from an experiment, and I assume what I have detailed here is what happened. However, I cannot find any relevant information to confirm or deny my hypothesis.

RESULTS:

The experiment showed 0.72 g of glucose was present per 100 mL of solution. The bottle label states that it is 0.5 g per 100 mL.

If any sucrose hydrolysis were occurring due to oxygen as a factor, it's more likely due to acids formed as it oxidizes species in the solution (for example, alcohol groups to carboxyls), which conceivably could include sucrose itself. But, it seems more likely that the $\ce{CO2}$ in air would be a factor because $\ce{CO2}$ dissolution increases the acidity of the solution, in addition to any other acids present (citric, for example).

However, sucrose hydrolysis is really slow unless the acidity and/or heat is significant ($\mathrm{pH} < 0$, or heat near boiling), so I doubt either of these explanations, unless these are really old, and/or had been heated.  More likely the glucose was there in the first place.

The way to properly answer this question would be to do control studies in which known amounts of sucrose and acid are initially present and tested with or without exposure to air having varying amounts of oxygen and carbon dioxide.


What is the mole fraction of dihydrogen in a mixture, given the relative rates of effusion of its components? A mixture of $\ce{H2}$ and $\ce{He}$ at $\pu{300 K}$ effuses from a small hole in a gas chamber. What is the mole fraction of $\ce{H_2}$ in the original mixture, if $3.0$ times as many $\ce{He}$ atoms as $\ce{H2}$  molecules escape from the hole in one minute?

The answer apparently relies on the fact that $\ce{H2}$ collides with the walls $\displaystyle\frac{1}{\sqrt2}$ as often as $\ce{He}$, and we therefore need $\sqrt{2}$ as many $\ce{He}$ atoms as $\ce{H2}$  molecules to yield an equal amount of substance escaping. Considering that the ratio is in fact 3, the final answer is then that the mole fraction of $\ce{H2}$ is

$$x(\ce{H2}) = \frac{1}{3\sqrt{2} + 1}.$$

Can anybody please explain how they got the $\displaystyle\frac{1}{\sqrt2}$ part?

If you are familiar with Graham's Law, you will know that the ratio of the rates of effusion of two gases is inversely proportional to the square root of the ratio of their molar masses.  That is:

$$\frac{R_1}{R_2}=\sqrt{\frac{M_2}{M_1}}$$

Here, $M_1 = M(\ce{H2}) = \pu{2 g//mol}$ and $M_2 = M(\ce{He}) = \pu{4 g//mol}$, so the lighter of the two, hydrogen, effuses at $\sqrt{2}$ the rate of helium, or conversely, the heavier, helium, effuses at $\frac{1}{\sqrt{2}}$ the rate of hydrogen.  That's where the fraction came from.

Since we know that helium is escaping three times faster than hydrogen, there must be three times more helium than would product an equal rate of flow for the gases.  An equal flow would occur when there is $\sqrt{2}$ times as much helium, and triple flow would occur with $3\sqrt{2}$ times as much helium.

The total amount of gas present is then:
$$\ce{H2} + \ce{He} = 1 + 3\sqrt{2}$$
and the mole fraction of hydrogen is:
$$x(\ce{H2}) = \frac{n(\ce{H2})}{\text{total amount of substance}}
= \frac{1}{3\sqrt{2} + 1}$$


Calculating the ph of an unknown solution by measuring the precipitate You have an unknown metal hydroxide, but no meters or indicators. You have lead (II) nitrate, which is soluble in it, and lead (II) hydroxide that is insoluble in it. You take one liter of the unknown substance and add lead (II) nitrate, causing the formation of a precipitate. You add more lead (II) nitrate until the precipitate stops forming, and then add a little more. After you filter and dry the precipitate, you have 3.81 grams of it. What was the approximate ph of the original solution?

I know that the metal hydroxide would be a base, but I’m having a hard time figuring out anything else about this problem. My first thought was to figure out the molar mass of the precipitate to try and determine how much lead nitrate was added to the unknown substance, but I can’t figure that out if I don’t know what the precipitate is composed of. I thought about figuring out the molarity of the total solution to figure out how much lead nitrate was added, but again I couldn’t see a way to do it without more information.

I also thought, since we’re working with pH, we’re supposed to do something with dilution and changing powers, but I don’t see any way to do it.

This problem has me completely stuck. I just don’t see where to start. Any help would be greatly appreciated.

You have to assume your unknown hydroxide is completely dissociate in aqueous solution. If so, then you can calculate $\mathrm{pOH}$ of the solution as $\mathrm{pOH} = -\log \ce{[OH-]}$. Then, $\mathrm{pH} = \mathrm{p}K_\mathrm{w} - \mathrm{pOH} = \mathrm{p}K_\mathrm{w} +\log \ce{[OH-]}$. Now, you see, since $\mathrm{p}K_\mathrm{w} = 14$, only unknown is $\ce{[OH-]}$.

Now look at the reaction of your unknown hydroxide with lead (II) nitrate:
$$\ce{2 OH- (aq) + Pb(NO3)2 (aq) -> Pb(OH)2 (s) + 2 NO3 (aq)}$$

Since you know the precipitated amount of $\ce{Pb(OH)2}$ isolated from $\pu{1.0 L}$ of unknown hydroxide solution, you can calculate original $\ce{[OH-]}$:

$$\text{Amount of } \ce{OH-}\text{ in original solution} \\
= \pu {3.81 g}\text{ of } \ce{Pb(OH)2} \times \frac{\pu {1 mol}\text{ of } \ce{Pb(OH)2}}{\pu {241.21 g}\text{ of } \ce{Pb(OH)2}} \times \frac{\pu {2 mol}\text{ of } \ce{OH-}}{\pu {1 mol}\text{ of } \ce{Pb(OH)2}} \\
= \pu {0.0316 mol}\text{ of } \ce{OH-} $$
Thus, $ \ce{[OH-]} = \pu{0.0316 M}$, and hence,
$$\mathrm{pH} = \mathrm{p}K_\mathrm{w} +\log \ce{[OH-]}= 14 + \log(0.0316) = 12.5 $$

Asking is approximate $\mathrm{pH}$ of orignal solution. Thus, this answer is acceptable.  However, keep in mind that trace of $\ce{OH-}$ will remain in solution due to the solubility of  $\ce{Pb(OH)2}$ in water $\left(K_\mathrm{sp} = 1.42 \times 10^{-20}\right)$.


Why does one of Dalton's compound elements with oxygen and phosphor consists of four atoms instead of two? I stumbled across a peculiar detail while reading the chapter about the  Chemical Revolution in Bowler and Morus' book Making Modern Science.
According to this book, Dalton argued that compound elements are created in the simplest possible way. For instance a compound consisting of equal number of X- and Y-atoms would then consist of one X-atom and one Y-atom. Still, in this illustration from Dalton's New System of Chemical Philosophy, oxygen with phosphor (compound 51) consists of two oxygen and two phosphor atoms. What is the reason for this?

The plate is from Dalton's book A New System of Chemical Philosophy.

The explanation of the plate is on pages 546-547 of the book.

The explanation says that "51" is "phosphorous acid" with weight 32.

(keep in mind that Dalton is basing this on O=7 and P=9)

"52" is "phosphoric acid" with weight 23.

The discussion of phosphorous acid is mostly on pages 408-410.

Initially, he contradicts the plate saying "I am inclined to believe" that phosphorous acid is one O and one P with a weight of 7 + 9 = 16.

But then he says:

  If this be the case, it may appear singular that none of the other elements exhibit acid properties when combined with 1 atom of oxygen ; but it should be observed, that the phosphoric oxide is in a liquid form, and disposed to separate into phosphorus and phosphoric acid, circumstances that do not combine in regard to the other oxides.  In fact, phosphorous acid may be considered as phosphoric acid holding phosphorous in solution, rather than a distinct acid.

So, if phosphoric acid (one phosphorous atom and two oxygen atoms accord to Dalton) "holds" another phosphorous atom, the result is what is show on the plate.


Decomposition of alkene in acidified potassium permanganate A low boiling point liquid A, molecular formula $\ce{C5H10},$ reacts with chlorine to form B, molecular formula $\ce{C5H10Cl2},$ and with $\ce{HCl}$ (gaseous) to form substance C, molecular formula $\ce{C5H11Cl}.$ Compound A decomposes in acidifed potassium permanganate solution to form two acids D (molecular formula $\ce{C4H8O2}$) and E (molecular formula $\ce{CH2O2}$).
Find the systematic names of A, B, C, D, E.

I'd like to verify that my answers are correct and I would also like a good explanation of why the answers are what they are. My answers are:
A - pent-1-ene
B - 1,2-dichloropentane
C - 1-chloropentane (Minor), 2-chloropentane (Major)
D - butanoic acid
E - methanoic/formic acid
Currently, I doubt my answer because of C, where I got two possible compounds from applying Markovnikov's rule. Any help would be great.

gaseous hydrochloric acid or anhydrous,

we have an alkene where each side is attached to the same number of hydrogens —> both “equally substituted”.

Products in this case, 3-chloropentane and 2-chloropentane.


Could I get a hint on the structure of this molecule based on HNMR spectra? Chemical formula is $\ce{C7H6O3}$. The IR spectrum has a peak at $\pu{3238 cm-1}$, which I believe could be due to an alcohol, phenol, or carboxylic acid. There's an additional peak at  $\pu{3015 cm-1}$, which I believe is due to either an alkene or aromatic group. The presence of aromatic and phenol peaks makes me think there's a phenol group in this compound.

What I'm stuck on are the four protons in the 7-8 ppm range. Especially the one split into a quartet. If I'm right about the IR, I just can't think of a way to get one proton with three neighbors to split with in a phenol.

That is not a quartet, but probably two overlapping signals. You always have to pay attention to the signal ratios and not just the number of split signals. The ratio here is 2:2:2:1, that is not a quartet ratio. A quartet would be 1:3:3:1.

There is a certain distortion in strongly coupled systems, the "roof effect" you can also see in this spectrum. This distorts the ratios a bit from the ideal ones, but not so much as to explain the ratio in the pseudo-quartet here.


Does Mie Scattering occur in Liquid or is Rayleigh scattering? Rayleigh scattering occurs when the dimensions of the scatter is much smaller than the wavelength of the incident electromagnetic radiation.

Mie scattering occurs when the dimensions of the scattered is much larger than the wavelength of the incident electromagnetic radiation. An example is when light is scattered by small water droplets in clouds.

So does Rayleigh or Mie scattering occur in liquid or bulk water?

It seems the answer is Mie scattering. But in Raman spectrometry. The incident light say 532nm is rayleigh backscattered in liquid. This is why the equipment needs to have notch filter, etc. So why does Rayleigh scattering (?) occur in liquid/bulk water when the particles are much larger than the wavelength of light? Or is it Mie Scattering that the raman device is filtering in the incident light backscattered?

So does Rayleigh or Mie scattering occur in liquid or bulk water? So why does Rayleigh scattering (?) occur in liquid/bulk water when the particles are much larger than the wavelength of light?

What is the approximate size of water molecule? It is on the order of 2.7 Angstroms or 0.27 nanometers. When we are talking about visible light, the wavelength is 500 nm/0.27 times larger than the scatterer! In case of pure water and with visible light, Rayleigh scattering is observed along with Raman effect. If you disperse larger "particles" in water such as milk, one sees Mie scattering.


Chemical Reaction for making Acetic Acid from Glycerol inside Fermenter I have been trying to find a chemical reaction for the formation of acetic acid from glycerol. I have been searching different literature, but apparently the reaction can not be found. I was hoping maybe anyone here would know the reaction.

$$\ce{C3H8O3 + X -> C2H4O2 + X}$$

Some Background

I am working on producing Succinic Acid (SA) from Glycerol (GLA) inside a fermenter. I know that inside the fermenter, the GLA is converted into succinic acid, water and acetic acid. I have been able to find a reaction for the formation of succinic acid, which is given by the following,

$$\ce{C3H8O3 + CO2 -> C4H6O4 + H2O}$$

But I can not seem to find a reaction for the acetic acid. I need an equation because I want to know the stoichiometric relation between the GLA and AA.

Additional Question

Another thing is the buffer solution reaction required to keep the $\mathrm{pH}$ constant inside the fermenter. I have been studying that either ammonia or $\ce{Ca(OH)2}$ can be used. I know the chemical reaction,

Calcium Hydroxide + Succinate -> Calcium Succinate + Water.

The calcium succinate can be reacted with conc. sulfuric acid in order to crystallize the SA crystals, since sulfuric acid can release the free succinic acids. I understand the theory behind these reactions, but I can not seem to find chemical reactions, from where I can see the stoichiometric coefficients.

Short answer: under anaerobic conditions, it is impossible to produce acetate as the sole product from glycerol with wild-type E. coli.

Explanation: Under anaerobic conditions (and assuming no substrates for anaerobic respiration have been added), a significant concern is redox balancing. Since electrons are not transferred to oxygen (as they are under aerobic conditions), the number of electrons in the substrates must equal the number of electrons in the products.

Glycerol as a substrate presents problems in this regard, because it is slightly more reduced than typical hexose substrates (eg glucose). Furthermore, the desired product acetic acid is more oxidized than glycerol (or hexoses for that matter).

Looking at the core metabolic pathways of E. coli, we immediately see that 1 molecule of glycerol is converted to 1 molecule of acetate. E coli is unable to utilize all three carbon atoms of glycerol make 3 molecules of acetate from 2 molecules of glycerol. The third carbon atom of glycerol is lost as either formic acid or $\ce{CO2}$ (with concomitant production of hydrogen gas). So we can write an initial stoichiometry of

glycerol ($\ce{C3H8O3}$) + $\ce{H2O ->}$ acetic acid ($\ce{C2H4O2}$) + formic acid ($\ce{CH2O2}$) + 4 electrons + 4 $\ce{H+}$.

[Note that the formic acid can be converted to $\ce{H2}$ and $\ce{CO2}$. The exact ratio of hydrogen to formate will depend on conditions.]

Unfortunately, there is no pathway for disposing of the excess electrons. You would need to add a respiration substrate such as fumarate (converted to succinate) or nitrate or nitrite (both can be reduced to ammonia).

One alternative approach would be to ferment the glycerol to a mixture of ethanol and formic acid or hydrogen gas and then chemically oxidize the ethanol to acetic acid. The balanced stoichiometry of that fermentation is

glycerol ($\ce{C3H8O3}$) $\ce{->}$ ethanol ($\ce{C2H6O}$) + formic acid ($\ce{CH2O2}$)

or

glycerol ($\ce{C3H8O3}$) $\ce{->}$ ethanol ($\ce{C2H6O}$) + $\ce{H2 + CO2}$


How to attach a liebig condenser to a flask for reflux heating? I know it's probably a stupid question, but I can not figure this out. I have a liebig condenser that I usually use for distillation with a Kjeldahl bulb, which makes it easy to attach to a flask because the bulb has a stopper on both ends?

Now I have an experiment that calls for reflux heating, where the condenser needs to be held vertically over the flask. One end of the condenser is a cork-fit opening, and the other end is a spout. The flask also has a cork-fit opening.

Maybe I don't understand your question well, but often in the lab when we do reflux heating, we don't really attach the liebig condenser.

You fix the flask using a clamp. And gravity holds the liebig condenser into place. You can put a clamp around it to hold it straight but you'd better not tighten it too much because these things are fragile.

I hope it helps you :)

If it's not clear maybe it would be easier to answer if you posted pictures of your glassware!


On the periodic table: Why are groups of elements organized by 'letter' Why are the groups of elements on the periodic table organized into areas represented by the letters s,p,d,f,g, and h?
What does this mean?

The letters are related to the electron orbitals, which were originally observed through spectroscopy. The lines shown in the spectroscope were named sharp, principal, diffuse and fine (or fundamental).

With a strong magnetic or electrostatic field, these separate into one, three, five or seven lines, or energy levels. There can be up to two electrons (with opposite spin) with the same energy level, according to the Pauli exclusion principle, so this implies there can be double the numbers above, or 2 s electrons, 6 p electrons, 10 d electrons and 14 f electrons. Thus, the Periodic Table is based on the number of electrons in each elements orbitals.


Why can't nitrate ion in aluminium nitrate solution oxidise copper when a copper spoon is used to stir the solution? I have been recently studying electrochemistry and I came across the electrochemical series. A question was asked saying what will happen if a copper spoon is used to stir an aluminium nitrate solution,the answer was no reaction will take place as Aluminium is more stable in its ionic form and copper cannot reduce aluminium ion, but can't nitrate ion which is above copper in the spectrochemical series oxidize copper and be transformed into NO which is more stable than Cu. Why doesn't it happen?

Minor correction: It is not the spectrochemical series but rather the electrode potential tables. Those tabulated values are given under standard conditions
You are indeed correct in thinking that nitrate ion should oxidize copper. It is a game of pH. If you have very low pH and nitrate ions, the following reaction takes place (in the absence of other ions)...
$$\ce{Cu(s) + 4 HNO3(aq) -> Cu(NO3)2(aq) + 2 NO2(g) + 2 H2O(l)}$$


pH of a solution of acetic acid and ammonium acetate After studying ionic equilibrium, I was just making some wild questions and I was unable to calculate the $\ce{pH}$ of this solution. Assume, 1 mole of each was added to $1L$ solution of water.

My attempt :

$\ce{CH3COOH}+\ce{NH4OH<=>CH3COONH4}+\ce{H2O}$

This reaction will happen till all $\ce{CH3COOH}$ is consumed by $\ce{NH4OH}$. More of $\ce{NH4OH} $ will be formed by Le Chatelier's principle. So, the answer must be same as that of a weak acid and weak base salt : $pH=\dfrac{1}{2}(pK_w+pK_a-pK_b)$.
But how will we calculate if there was $2$ mole of acetic acid?

You have some mistaken assumptions.

$\ce{NH4OH}$ and $\ce{CH3COONH4}$ are not molecules that exist.

The assuption "all $\ce{CH3COOH}$ is consumed" is incorrect.

Why do you say more $\ce{NH4OH}$ will be formed? How is this possibe? What can it be formed from?

Review that Henderson–Hasselbalch equation


Reduction potential as a function of pH I have read that Latimer diagrams can be represented both in an acidic medium (pH = 0) and in a basic medium (pH = 14).

I would like to know how to calculate the reduction potentials as a function of pH. That is, if I know the reduction potential for the pair $X^{+}/X$ at pH = 0, calculate it for another pH.

I thought the Nerst equation could be applied, but I'm not sure about that.

Nernst equation is the only way of solving this problem. Let's consider the example of the reduction of permanganate. The half-reaction is:
$$\ce{MnO4-  +  8 H+  +  5 e-  ->  Mn^2+  +  4 H2O}$$

The potential can be written as:

$$E  =  E^\circ  +  \frac{0.059}{5}\log \left(\frac{\ce{[MnO4-][H+]^8}}{\ce{[Mn^2+]}} \right)$$

If you want to see explicitly the $\mathrm{pH}$, you can rewrite the preceding expression, which gives:

$$E  =  E^\circ  +  0.012 \log \left(\frac{\ce{[MnO4^-]}}{\ce{[Mn^2+]}}\right) - \left(\frac{8\times 0.059}{5} \right) \mathrm{pH}$$

or :
$$E  =  E^\circ  +  0.012 \log \left(\frac{\ce{[MnO4^-]}}{\ce{[Mn^2+]}}\right) - 0.094 \: \mathrm{pH}$$


What does "stoichiometric amounts" mean? Can anyone inform me what the term stoichiometric amounts means? I know what stoichiometry means. It is the:

  The relationship between the relative quantities of substance taking part in a reaction or forming a compound.

But what is "stoichiometric amount" supposed to be?

Sometimes, practical chemistry does not work like it's supposed to on paper. A perfectly balanced reaction equation may dictate that two reactants should be in a 1:1 stoichiometric ratio, but in practice, you get much better yields or reaction times if you use an excess of one.

When that happens, you are no longer using a stoichiometric amount, as specified by the balanced equation, but a super- or substoichiometric amount.

Welcome to chemistry in the real world.


Finding Kp at another temp with /\Hrxn Took a pic of the paper I tried it on it on, has the exact problem description written at top, and what I tried. Figured that would be easier then typing all that. The answer is supposed to be 2.0.

You used the wrong version of $R$.  Try using $R = 8.314 J mol^{-1} \ K^{-1}$.


Why is ortho-hydroxybenzoic acid more acidic than its para-isomer? Why is ortho-hydroxybenzoic acid $(\mathrm{p}K_\mathrm{a} = 2.98)$ more acidic than its para-isomer $(\mathrm{p}K_\mathrm{a} = 4.58)$?

March's Advanced Organic Chemistry (7th ed) gives the reason to be intramolecular hydrogen bonding between the $\ce{OH}$ and $\ce{COO-}$ groups of the conjugate base.

However, note that intermolecular hydrogen bonding (between the $\ce{OH}$ and $\ce{COO-}$ of two different molecules) is possible in the para-isomer. Interestingly, this question also states that intermolecular hydrogen bonding is relatively stronger than intramolecular one. So, shouldn't the para-isomer be more acidic?

I don't think that in the first link it is stated that the intermolecular $\ce{H}$-bonding is stronger than the intramolecular one. On the contrary, the accepted answer starts as:

"I don't think there is much of a difference between the strengths of intramolecular or intermolecular hydrogen bonds."  With which I agree.

In that post they discuss the effect of intermolecular and intramolecular $\ce{H}$-bonding on the melting and boiling points of the compounds.

The answer by @AllYouCanEat86 that follows in the same post I think is more relevant to your question.

Therefore in our case there are intermolecular $\ce{H}$-bonds in p-hydroxybenzoic acid but they don't last for long due to the thermal motion and higher degrees of freedom of the $\ce{COO-}$ and $\ce{OH}$ groups. In contrast, in the ortho-isomer the $\ce{COO-}$ and $\ce{OH}$ groups are locked in close proximity due to the aromatic ring and therefore are forced to maintain the $\ce{H}$-bond for much longer. This could be thought as an equivalent to the chelation effect in metal complexes. Lastly, one could also argue that in the case of o-hydroxybenzoinc acid, there is both the option of intermolecular and intramolecular $\ce{H}$-bonding compared to the para where there is only intermolecular $\ce{H}$-bonding.


Is there a standard scholarly reference for lattice constants of crystals of the elements? I need to discuss the lattice constants of bulk crystals of several metal and semiconductor elements. I can find plenty of tables and numbers that are probably "close enough" but for a paper I'd like to cite a standard source.
From  https://periodictable.com/Properties/A/LatticeConstants.html I have the following numbers, but I don't want to use a "dot com" as a scholarly source, and I can not figure out how to use Wolfram Alpha (the source for this website) or understand where WA gets its numbers. They list several references here https://reference.wolfram.com/language/note/ElementDataSourceInformation.html , but it will be a challenge to track them all down one by one. I'm hoping someone will recognize one of them or simply be able to mention it.
I only need (at a minimum) three decimal places for the lattice constants in Angstroms (though four is better if they are known), but what I can't find is a citable, schollarly source that covers all of these and is likely to cover other common elemental crystals when I need them in the future.
element.   lattice constant
             (a, b and c)
              Angstroms

   Au          4.0782
   Ag          4.0853
   Pb          4.9508
   Ge          5.6575
   Si          5.4309

The CRC Handbook of Chemistry and Physics contains a dedicated compilation by H. W. King, titled «Crystal Structures and Lattice Parameters of Allotropes of the Elements».  In case your research library is closed, you may access some of its editions freely or borrow them with the library card of archive.org.
In case of the 97th edition (by 2016), the section starts by page 12-16.  (Elements liquid or gaseous at ambient conditions are included.)

source


What's the most appropriate name for this reaction? I recently wrote a test in which I was asked:

  What is this reaction $\ce{CH3CH=CH2 ->[CO + H2O][H+]CH3-CH(COOH)-CH3}$ known as?

  Wurtz Reaction
  Koch Reaction
  Clemmensen's Reduction
  Kolbe's Reaction

My thoughts on this:

It's definitely not Wurtz Reaction, because that reaction eliminates terminal halogens and attaches together two alkyl groups. Also, that reaction needs sodium in dry ethereal medium.
It's definitely not Clemmensen's reduction, because that needs zinc amalgam and acid. (Although acid is present, the reduction won't take place without the zinc, right?).
Now, my options are Koch and Kolbe's reactions. As Koch's reaction (Gatterman-Koch Formylation) adds a carbonyl group, and not a carboxylic acid group, I decide to go with Kolbe's reaction. However, as per Wikipedia, this reaction requires carbon dioxide and sodium phenoxide. The carbon dioxide can be taken care of, by the water gas shift reaction that converts $\ce{CO + H2O->CO2 + H2}$. But now, there's no sodium phenoxide. What to do now?

Basically, my doubt is whether there's a mistake in my procedure of eliminating the options, or if there's something I'm missing out somewhere which may lead to make a different choice?

As Koch's reaction (Gatterman-Koch Formylation) adds a carbonyl group, and not a carboxylic acid group, I decide to go with Kolbe's reaction.

You started right, and then took the wrong turn ;-)

Under acidic conditions

the proton adds to the $\ce{C=C}$ double bond and gives a secondary cation
the cation reacts with carbon monoxide and yields an acylium cation
the latter reacts with water to result in the formation of a carboxylic acid

The whole process is known as the Koch reaction (without the Gattermann).


Which metals hydroxides dissolve in excess NH4OH? Some metal hydroxides like $\ce{Ag(OH)}$ and $\ce{Cu(OH)2}$ dissolve in excess $\ce{NH4OH}$ to form metal ammonia complexes.

I want to know which other common metal hydroxides dissolve in excess $\ce{NH4OH}$. Also, can it be somehow predicted ? I couldn't find much relevant information on the Internet.

The Wikipedia article for "Metal amine complex" gives many examples.  According to the article:

  Almost all metal ions bind ammonia as a ligand, but the most prevalent
  examples of ammine complexes are for Cr(III), Co(III), Ni(II), Cu(II)
  as well as several platinum group metals.

Specific examples include:

$\ce{[Pt(NH3)4][PtCl4]}$
$\ce{[RhCl(NH3)5]Cl2}$
$\ce{[Cr(NH3)4Cl2]Cl}$
$\ce{[Co(NH3)6]Cl3}$
$\ce{NH4[Cr(NCS)4(NH3)2]}$
$\ce{[Zn(NH3)4]^2+}$
$\ce{[Ag(NH3)2]+}$
$\ce{[Ni(NH3)6]^2+}$
$\ce{[Cu(NH3)4(H2O)2]^2+}$

Several others are listed just in that one Wikipedia article, along with a fair list of references.


How many dichlorinated products including stereoisomers can be isolated when (S)-2-chlorobutane reacts with Cl2 /hv? This is my attempt at solving this problem. I get 6 products. However my book says the answer is 5 products.

Elaborating on what Waylander said, The C-2 carbon chlorine should be in S configuration so both (III) and (IV) structures can't be correct and only one with S configuration(IV) will be considered and rotating that will result in the other structure.


What is the fire point of sucrose with an average particle diameter of 60 micrometres? What is the fire point of sucrose with an average particle diameter of 60 micrometres? I can't seem to find any documentation on this, online.

I managed to find the answer in the MSDS of sugar given by the US Department of Labour, after some searching.

In the MSDS found here there is an ignition temperature (which I imagine to be the same as the fire point) of 350 C, with a minimum explosion concentration of 0.045 g/l in air. There also seems to be a possible ignition from a dramatic increase in pressure (mentioned in the MSDS to be 5,000 psi/sec), though no further dangers are expressed.


Bond rotation in buta-1,3-diene Why don't all the atoms in s-cis-1,3-butadiene lie in one plane?

All carbon atoms are $\mathrm{sp^2}$ hybridised and I feel that there would be no $\ce{C-C}$ bond rotation due to the conjugation that leads to a double bond on $\ce{C2-C3}$ making them all lie on a plane.

Is it any different for s-trans-1,3-butadiene?

1,3-Butadiene (Preferred IUPAC name: Buta-1,3-diene) is the organic compound, which is industrially important as a monomer in the production of synthetic rubber. It is also the simplest conjugated diene among organic compounds. Since all carbon atoms are $\mathrm{sp^2}$ hybridized, orbital overlap between double bonds is possible and, theoretically, two different conformer shows maximum overlapping, which are s-trans conformers with $\ce{C2-C3}$ dihedral angle of $180^\mathrm{o}$ and s-cis conformer with $\ce{C2-C3}$ dihedral angle of $0^\mathrm{o}$. However, due to steric hindrance, this s-cis conformer is approximately $\pu{16.5 kJ/mol}$ higher in energy than its s-trans counterpart, making this geometry a local energy maximum. Thus, in contrast to the s-trans geometry, which is global minimum, s-cis geometry is not a conformer (Wikipedia).

The double bonds of the s-cis geometry are twisted to give a dihedral angle of around $38^\mathrm{o}$, is considered to be a second conformer (called gauche-rotamer) that is around $\pu{12.0 kJ/mol}$ higher in energy than the s-trans conformer. The high level ab Initio calculations, utilizing coupled cluster theory with quasi-perturbative triple excitations and augmented quadruple $\zeta$ level basis sets, done by Feller and Craig in 2009 showed that the transition state separating the trans- and gauche-rotamers lies $\pu{26.8 kJ/mol}$ above the trans global minimum (Ref.1). These calculations further showed that the gauche rotamer lies $\pu{12.6 kJ/mol}$ above the trans-rotamer and the s-cis-form is a transition state $\pu{2.0 kJ/mol}$ higher than the gauche-rotamer. Wikipedia state that:

  This increased rotational barrier and strong overall preference for a near-planar geometry is evidence for a delocalized π system and a small degree of partial double bond character in the C-C single bond, in accord with resonance theory.

This is illustrated in the following scheme with the most stable conformer of 1,3-butadiene, which has the s-trans conformation (Ref.2):

Quantum chemical calculations on Π-Electron Delocalization in Butadiene has also done and published in 2006. The resultant bond lengths were illustrated in the following diagram and the total abstract of the paper was reproduced here for readers with interest:

  Abstract: Equilibrium structures have been determined for s-trans-1,3-butadiene and ethylene after adjusting the rotational constants obtained from rotational spectroscopy by vibration-rotation constants calculated from the results of quantum chemical calculations. For butadiene, the formal $\ce{C=C}$ bond length is $\pu{1.338 Å}$, and the formal $\ce{C−C}$ bond length is $\pu{1.454 Å}$. For ethylene, the $\ce{C=C}$ bond length is $\pu{1.3305 Å}$. These values appear to be good to $\pu{0.001 Å}$. It is shown for the first time that π-electron delocalization has the structural consequences of increasing the length of the formal double bond by $\pu{0.007 Å}$ and decreasing the length of the formal single bond by $\pu{0.016 Å}$. Comparisons are made with structures computed with several quantum chemical models. The MP2/cc-pVTZ results agree best with the new $r_e$ structure.

References:

D. Feller, N. C. Craig, “High Level ab Initio Energies and Structures for the Rotamers of 1,3-Butadiene,” J. Phys. Chem. A 2009, 113(8),  1601–1607 (DOI: 10.1021/jp8095709).
N. C. Craig, P. Groner, D. C. McKean, “Equilibrium Structures for Butadiene and Ethylene:  Compelling Evidence for Π-Electron Delocalization in Butadiene,” J. Phys. Chem. A 2006, 110(23), 7461–7469 (DOI: 10.1021/jp060695b).


Periodic acid oxidation of carbonyls? The following reaction seem to be valid;
$\ce{CH2OH-CHOH-CH2OH ->[HIO4]CH2O + HCOOH + CH2O + 2H2O}$
Apparently, there are similar reactions by 2,3-dihydroxypropanal, and also by the straight chain form of fructose.
The first two reactions involve doing the Malaprade reaction on molecules with adjacent alcohol and aldehyde groups, and the last one involves doing it on a ketone. Both times, it seems sufficient to just cleave the carbon-carbon bonds and oxidise each resulting molecule one step up the ladder.
As far as I know, the Malaprade reaction is for vicinal diols, and I can't stretch the mechanism to work for carbonyls.
Can anyone help me with how these reactions work?

In 1928, Malaprade demonstrated that periodic acid reacted with ethylene glycol to produce iodic acid and formaldehyde (Ref.1). Hence, the oxidation of adjacent diols with periodic acid or its salt in aqueous solution is now generally known as the Malaprade reaction, the mechanism of which is depicted below:

The reaction proceeds faster under acidic conditions and can be applied to higher polyhydric alcohols, which behave similar to ethylene glycol. The necessity for the reaction is to have 1,2-dihydroxy functionality $\left(\ce{>C^1(OH)-C^2(OH)<}\right)$, and upon exposer to periodic acid, the middle $\ce{C^1\!-C^2}$ bond would oxidatively cleave to give two compounds with gem-diol functionality $\left(\ce{>C^1(OH)2 \ and \ (OH)2C^2<}\right)$, which are essentially carbonyl groups in aqueous solutions. The reaction has also been further extended to the cleavage of α‐hydroxy carbonyl compounds, 1,2‐dicarbonyl compounds, α‐amino alcohols, α‐amino acids, and polyhydroxy alcohols since then, and successfully applied for structural analysis specifically of sugars:

(source, Ref.2)
As shown in the figure, each $\ce{C-C}$ oxidative cleavage of $\ce{C-C}$ bond gives $\ce{OH}$ group to each carbon. For example, 1,2-bond chevage between $\ce{-CH(OH)-CHO}$ gives aldehyde carbon another $\ce{OH}$ group, making it formic acid as shown in the diagram. Meanwhile, 1,2- and 2,3-bond cleavages gives $\ce{C}$2 carbon two extra $\ce{OH}$ groups so that resultant molecule is formic acid again. Thus, complete oxidation of $\pu{1 mol}$ of D-glucose gives $\pu{5 mol}$ of formic acid and $\pu{1 mol}$ of formaldehyde.
The requirement is to have 1,2-diol functionality. So, how ketone and aldehyde with $\alpha$-hydroxy groups get oxidized? In aqueous solutions, ketones and aldehydes are in equilibrium with their corresponding gem-diols, thus providing 1,2-diol feature.
The 1,2-diol does not have to be cis-orientation. The trans-diol will also get oxidized but with slower rate (see here and Ref.3).
References:

Zerong Wang, "Malaprade Reaction (Malaprade Oxidation)," In Comprehensive Organic Name Reactions and Reagents; John Wiley & Sons, Inc.: New York, NY, 2010 (https://doi.org/10.1002/9780470638859.conrr406). ISBN: 9780471704508.
Fathia Mohammed Ibrahim,  Mubark Elsayed Osman, "Elucidation of Sugars Structure through Periodic Acid Oxidation Cleavage," International Journal of Science and Research (IJSR) 2018, 7(1), 1152-1155 (https://www.ijsr.net/search_index_results_paperid.php?id=6121702)(PDF)
G. J. Buist, C. A. Bunton, J. H. Miles, “149. The mechanism of oxidation of α-glycols by periodic acid. Part V. Cyclohexane-1 : 2-diols,” J. Chem. Soc. 1959, 743-748 (https://doi.org/10.1039/JR9590000743).


Why is PbCl₄ covalent? My answer: The inert pair effect in $\ce{Pb}$ causes it to pull back the electrons, resulting in polarisation.

My teacher's answer: An ionisation state of $+4$ is too difficult to achieve, and it is not gained even when 4 chlorine atoms get reduced, as the later ionisation enthalpies of lead (or any metal) are high.

My point is, if we don't think of the inert pair effect, we are generalising this, and moreover, high ionisation energies of lead should be explained by the inert pair effect.

Which explanation do you think is better?

As a side note, I would like to add that my teacher pointed out that the inert pair effect is an outdated concept. Is it so? If yes, why is it rejected?

The high covalent characted is just explained by the fact of $\ce{Pb^{4+}}$ being very polarizing and $\ce{Cl-}$ fairly polarizable.

The inert pair effect would give an explanation about why it is difficult to reach that oxidation state, and why it has high tendency to get reduced, but there is no need to require to this concept to explain the covalent character of the compound.

Regarding your last question, I do not see why the inner pair concept should be rejected. Is just another model that can explain in a simple way properties or behaviors of certain elements. There are better models and orbital calcualtions that can also do the job? Of course, but that is no reason for avoiding the use of the model. Models that are still useful to explain stuff easily, can be employed even if more accurate ones do exist. One good example is ligand field theory, which is still useful even when molecular orbitals theory is more accurate and can still explain the same (and many more) things than the former.
Even though, a theorem or model HAS to be rejected if a new one is found that explains properties better and is incompatible with the old one.


HDEHP and complexes formed in extraction I did an solvent extraction of $\ce{Eu(III)}$ using $\ce{HDEHP}$ from nitric and hydrochloric acid solutions to determine which acid, and which concentration, will yield the highest distribution coefficient.

My results were interesting, I got that the hydrochloric acid extractions had much lower distribution coefficients, and less dependence on acid concentration (the max distribution coefficient was at 0.3 M).  The nitric acid extractions were far more dependent on acid concentration with the global max at 0.01 M and another local max at 0.6 M, as well as being higher in general than the $\ce{HCl}$ distribution coefficients.

I did a literature review, and based on this paper, I think I know the structure of the $\ce{HDEPE/Eu}$ complex.  My original complexes were $\ce{EuCl3}$ and $\ce{Eu(NO3)3}$ dissolved in the acid solution, and extracted with 0.8 M HDEHP in dodecane.  I know the melting point of solid $\ce{EuCl3}$ is about $\sim\pu{600°C}$ which is far higher than $\ce{Eu(NO3)3}$ ($\sim\pu{85°C}$) so I'm wondering if the stronger bonds in the $\ce{EuCl3}$ complex are preventing the $\ce{HDEHP}$ complex from forming, or if the conjugate bases are somehow participating in the complex formation between $\ce{HDEHP}$ and $\ce{Eu(III)}$.

Does anyone have any insight into my results?

We need some more details such as what diluent you used. Maybe you were using pure DEHPA as the organic phase which is something which I would suggest that you do not do.

I would like to know if you did triplicate experiments, sorry if I offend you but I have no idea how good you are at the measurement of D values by solvent extraction. I like to think I am good at it, but I always use triplicate experiments. I will shake three different vials for each set of conditions. This allows me to see how reproduceable my work is, it also guards against me doing something stupid with one of the vials.

If you only do single extractions then it is possible that a random error causes the experiment to give misleading results. I hold the view that a beginner needs triplicate shaking experiments more than anyone. When you are new to solvent extraction you are more likely to make errors, also when you have done it for a while you might get careless and then you would need the triplicate experiments again to guard against mistakes.

The triplicate experiments will not prevent you making mistakes, but they will help to alert you to the fact that something has gone wrong.

I also worry about the use of very low acid concentrations, the Eu will tend to form hydroxides.

I think that you need to pay attention to the activity coefficents. I would be much more happy if you had used sodium nitrate to keep the ionic strength of the aqueous phase constant. I would want to do the experiment with a constant nitrate concentration to allow you to separate the effects of changing the nitrate concentration from that of the solvated protons.

I would like to point out that at very high nitrate concentrations the extraction mechanism for lanthanides with DEHPA can change from the normal ion exchange mechanism to a solvation mechanism which is similar to the extraction of plutonium from nitric acid by tributyl phospahte (PUREX process),


Stability of Sulfides - backbonding? A stability model premised on Lewis acid-base covalent bonding can be proposed to rationalize the stability shown by sulfides such as CuS and Ag2S. For instance, although the oxide anion is a stronger Bronsted base than sulfide anion, sulfide anion can be (and usually is) a stronger Lewis base than oxide anion since S has a lower EN than O. Furthermore, recall that Lewis base strength depends on the Lewis acid with which the Lewis base interacts. For the sulfide anion, since sulfide anion possess empty valence 3d orbitals, Lewis acid-base interactions are often very strong if the Lewis acid is a low oxidation state metal ion with a fairly large number of valence d electrons. for such cases, sulfide-to-cation sigma coordinate covalent bond formation couples with pi-type coordinate covalent bond formation because metal ion d-electron density "drifts" back to empty valence 3d orbitals on sulfide anion. This results in covalent bond interactions that have substantial, strong multiple ond character and formation of a metal-sulfide lattice which is rather macromolecular in character. In sum, such as lattice must be very stable ...

1) Is the above excerpt describing pi-backbonding?

2) Is pi-backbonding more favorable for metals in a low rather than a high oxidation state because such metals still have substantial electron density that needs to be stabilized - and preferably stabilized by something more electronegative than a metal?

3) Metal d-orbitals overlap with p or d-orbitals of the non-metal. How does this work? I always see diagrams such as these:

Is the overlap between the metal d-orbital and non-metal p-orbital as poor as depicted? Are the d-orbitals really at a 45 degree angle relative to the p-orbital?

4) How well is the above excerpt written? I feel that two improvements could be made:

A) The passage seems to imply that Bronsted acidity and basicity do not depend on the corresponding base or acid. Wrong implication. HCl in HBr solvent won't be a strong acid.

B) Why electron density might just "drift" to the non-metal could be explained explicitly. I.e. electronegativity.

1) Yes and no ... the d-d orbital overlap will almost certainly have pi-orbital character (angular node along bond axis), and it could reasonably be considered "backbonding" because the electron density is going in the canonically "unexpected" direction (i.e. away from positively charged metal center).  On the other hand, chemists (particularly inorganic chemists) usually restrict the phrase "pi-backbonding" to refer to back donation from the d-orbitals of a neutral or ionic metal center into anti-bonding orbitals of a ligand, typically an olefin or a carbonyl ligand.

2) Yes .. that's essentially correct ... d-orbitals tend to be rather diffuse, and so can overlap with lobes of diffuse orbitals on ligands (or counter-ions, as with the metal sulfides).  In traditional metal-ligand pi-backbonding, lobes the d-orbitals overlap with lobes of an unoccupied anti-bonding orbital of the ligand.  In the example given above, they overlap with lobes of unoccupied d-orbitals on the sulfur.  For a given system, unoccupied, or virtual, orbitals tend to be much more diffuse than occupied orbitals.

3) Those diagrams are just intended to capture the qualitative aspects of the bonding interaction, such as the symmetries of the orbitals involved.  Remember, the orbitals are 3-dimensional shapes that describe the probability of finding an electron with a particular set of quantum numbers in a given region of space.  Depictions like the one in your post are helping to visualize the orbital by assuming a fairly tight "tolerance" for the probability. For example, those images might depict the surfaces that enclose 75% of the total probability density.  If you expand that to 90%, or even 99%, the lobes maintain the same basic shapes but get much larger, which might make the overlap seem more significant, but could also be confusing for other reasons.

4) I think it is quite well-written, but I suppose it would depend on the broader context in which it was being presented.  I guess it is presented at an inappropriately high level for an introductory chemistry textbook, for example.