How to read a chromatography calibration curve?

I assume your teacher explained the HPLC separation. If injected a mixture of four ions you will get four peaks. Each peak has an area, which is proportional to the concentration of the substance.

Imagine you wanted to determine the concentration of chloride ions in your tap water. What you would do is that you will prepare several known concentrations of chloride ions using $\ce{NaCl}$ solutions, say $0$ to $\pu{350 ppm}$ (look at your calibration chart). Inject them one by one, and measure the peak area, let us say the second peak is that of chloride ion. The peak shape is Gaussian in real work but you can "estimate" them as triangles. Recall triangle's area is very easy to calculate.
First step: Plot the peak area for concentration. What you get is a calibration curve.
The next step is to find a mathematical equation that fits this data.
Fortunately it is linear in your case of the form $y=mC$ where $y= \text{peak area}$, $m=\text{slope}$, $C=\text{concentration}$ in $\pu{ppm}$.
Now you would inject your tap water in the HPLC, you do not know the concentration ($C_1$) but you do know the peak area.
The last step is called interpolation. You go from the known area, and read the corresponding concentration axis. Viola! Now you know the chloride concentration in your tap water!
Alternatively, you use the equation, $y=0.1312C$, so I know the area of the tap water chloride injection, say, it was $\pu{40.0 units}$, so $C_1 = 40/0.1312 = \pu{304.9 ppm}$.
In typical tap water it is around $\pu{100 ppm}$, I just made up those numbers.
Bonus: Rationalize the units of the peak area on the $y$-axis! Why it is conductivity times minutes?


Which plastic type is used inside dishwashers?

One possible reason for the different behaviour of the drinking bottle, compared to the lining of the dishwasher may be indeed the different composition of the polypropylene deployed.  The part of poly indicates that many hundreds and thousands smaller propylene molecules are brought together to constitute the polypropylene.  Yet there are different ways this may happen, as this picture (source) depicts:

Along the polymer chain, there a methyl groups (marked as $\ce{CH3}$) occupying more volume than hydrogen atoms (marked by $\ce{H}$). It is imaginable that all of these methyl groups are attached on one side of the chain only (the upper picture), or consistently "once on one, the next time on the opposite side" (the lower picture).  Not shown here, there may be a randomized distribution, too, where you can't predict on which side the next methyl group will be attached (not shown here) as an intermediate situation.

Such alterations in the molecular structure may alter how individual polymer molecules interact with each other.  It influences if the polymer becomes more crystalline, or more glass-like; and by the same token, it influences the melting point of the material.  Indeed, the reference states melting points of 130 degree Celsius (266 Fahrenheit), as well as 171 degree Celsius (340 Fahrenheit).

Now think about the production of the bottle: it is done by (blow) molding.  A manufacture aims to produce a high number of bottles per unit of time, and preferentially at a low cost to spend, for example to heat the polymer that it becomes flexible enough to form it.  So it is better to select a polymer with a lower melting point, and by admixing / blending with other polymers to polypropylene, the melting point probably may be lowered further.

On contrast, a manufacturer of a linings of dishwashers equally takes into account the processes while the dishwasher is used: chemicals (detergent), hot water during the washing, and heating to eventually dry the tableware.  So the selection prefers a polymer that tends to be more crystalline and hence exhibits a higher melting point.

You may encounter situations showing you that this comes with an disadvantage, too:  The more crystalline, high melting polypropylene liner often is more brisk than the more glassy, low melting polypropylene of your bottle.


According to this cooking site, Crisco brand vegetable shortening melts between $\mathrm{47^oC-48^oC}$ (yes, I know $\mathrm{48^oC}$ is less than $\mathrm{50^oC}$, but it seems pretty close to me). The best way to use such a non-certified melting point "standard" like this is to use your thermometer side-by-side with a properly calibrated thermometer.

Vanillin may be your best bet (melting point of $\mathrm{81^oC - 83^oC}$).  Again, using a calibrated thermometer to check the actual melting point would be the ideal situation.

Although your question asks about calibrating based on melting point measurements of household items, household items just aren't usually of the purity required for accurate melting point based thermometer calibration. In the absence of a calibrated thermometer as discussed above, the best way to calibrate your thermometer is probably to use the freezing and boiling points of water (don't forget to adjust boiling point for altitude if you don't live at sea level), then check the melting points of Crisco and vanillin if you want. Good agreement with these stated melting points would be a good indicator of proper thermometer calibration, while poor agreement could either be due to improper thermometer calibration or impure material.


Kinetic aspects:
Low concentrations of interacting ions decrease rate of precipitation, leading to bigger crystals.
A high ionic strength of spectator salts leads to lower activity coeeficients , and  decreased rate of precipitation, leading to bigger crystals.
Higher viscosity leads to lower collision rates, decreased rate of precipitation, and to bigger crystals.
Low temperaturee leads to bigger crystals, as rate of collisions leading to crystallisation is lower due higher viscosity, lower molecular speed and lower rates of interactions leading to precipitation.
Note that kinetic aspects may be hard to involve in the proces control, aside of concentrations.
Thermodynamic aspect:
Crystals of low sizes have high surface Gibbs energy and therefore bigger crystals are thermodynamically preferred.
This is initially overrun by the kinetics of the precipitation, but in long term there is ongoing recrystallization in favour of bigger ones. This can be extremely slow for very little soluble salts.
This process is speeded up by high temperature, what increases rates of the opposite processes of dissolution and crystallization. AFAIK, the gravimetric procedure to determinate sulphates as barium sulphate involved heating $\ce{BaSO4}$ suspension near below boiling point for I guess 1-2 hours to speed up recrystallization. As bigger crystals are easier to manage, espacially in this particularly case, as $\ce{BaSO4}$ is known to form too fine crystals.


Why is high molar molar mass a requirement of good primary standards for volumetric analysis?

This practice is done in analytical chemistry in order to minimize the relative weighing error on the balance. Preferring a larger formula weight for a primary standard has nothing to do with impurity levels. We have to start with the highest purity standard.

For example, you wish to prepare a 0.010 M solution of oxalic acid dihydrate in 1 L flask. Its formula weight is  126.07 g/mol. You need to weigh out 1.2607 grams of the acid on the balance. Imagine the balance has a small constant error, the relative error in weighing is smaller for a larger formula weight compound as compared to a smaller formula weight standard for the same concentration. Imagine if the molecular weight of oxalic acid were 60 g/mol, how much do you need to weigh out?


Dipole moment of ferrocene conformations

To find out whether a compound can have a nonzero dipole moment, consider its symmetry. A dipole moment is a vector which must be transformed onto itself by every symmetry operation present in the molecule’s point group.

It should be easy to see that te eclipsed conformation has five planes of symmetry perpendicular to the cyclopentadienyl rings ($5\sigma_\mathrm v$). Another sixth plane of symmetry is sandwiched between the two rings and passes through the iron atom ($\sigma_\mathrm h$). The only vector that will be transformed onto itself by all these planes of symmetry is the null vector $\vec{0}$. Therefore, this conformation has zero dipole moment.

The staggered conformation also has five planes of symmetry perpendicular to the ring planes as the eclipsed conformation ($5\sigma_\mathrm v$). It also features a centre of symmetry in the iron atom ($i$) as well as improper rotation ($S_{10}$). The inversion centre alone is enough to determine that only the null vector $\vec 0$ will be transformed upon itself. Therefore, this form also has no dipole moment.

Both you and the book are correct.


Computational chemistry software to generate files of cartesian coordinates and zmatrix of molecules

This is not the answer to your question, but it is a comment about an approach to the aims, based on the comments. If coding is an option, RDKit is a good option, specifically adding constraints (restraints) to the ForceField and optimising. Given a molecule:
from rdkit import Chem
from rdkit.Chem import AllChem

acetate = Chem.MolFromSmiles('CC(=O)C')

Normally the 3D conformer is generated with:
AllChem.EmbedMolecule(acetate)

And it stops there. However, then you can manually control the force field —either UFF (more elements) or MMFF (classic choice).
ff = AllChem.UFFGetMoleculeForceField(mol)
ff.Initialize()
print(ff.Minimize()) # this tells you if it failed.

Within this, you can add constraints, such as:
ff.AddDistanceConstraint(0, 1, 2., 2.1, 1_000_000_000)

Note that the weight needs to be set absurdly high for absurd lengths. To check it work you could do:
dm = Chem.Get3DDistanceMatrix(acetate)
print(f'{dm[0,1]} Å. {ff.CalcEnergy()} kcal/mol')


Particle suspension and re-suspension of nano-particles

Try mercaptoacetate surface treatment to put a negative charge on their surfaces for repulsion around pH 8 or 9 basic solution.  Uncoordinated mercaptoacetate is oxidatively dimerized to the disulfide by air (oxygen).  The drug S-mercaptocysteine as its salt (basic pH) should be more aggressive toward binding.  3-Mercapto-1-propanesulfonic acid will bind at lower pH.

The dose must be low enough that you do not disaggregate your particles, especially over time.  You can permeable membrane desalt an excess.


Liquid nitrogen superheating with a layer of oil?

Thank you Nicolau Saker Neto.

Superheating is possible up to 300°C for water, yet disappointingly only -163°C for liquid nitrogen.

Source: http://www.nist.gov/data/PDFfiles/jpcrd276.pdf


The two rings in biphenyl are inclined at an angle of 44° to each other because of the large steric repulsion between the ortho hydrogens on each ring.

For Hückel's rule to be applied, the molecule must be able to be planar and it must be monocyclic.

Due to the molecule not being planar, the two rings are not strongly conjugated to each other and you cannot apply Hückel's rule to the whole system. Additionally, the whole system is not monocyclic. Hückel's rule can be applied to each ring individually and they are both aromatic.


What are the currently most accurate sources for lists of the values for elemental properties

You should teach your student not to use numerical values with as many significant figures as possible. Knowing how to make sense to decimals is part of the scientific culture. The distance between two cities cannot be given to the least micrometer : it has no meaning. The weight of an object has no use if it is given such as $123,45678910111213 g$. The least figures which can be given for sure is the milligram (here 6). Put on a balance, such an object will display fluctuating next figures. Already the figure $7 (0.1 mg$) will fluctuate at the least draft. To be sure of the next decimals you must put your balance in an insulated and closed room without nobody entering during the whole day, in order to avoid drafts. And even, the least dust grain falling from the air will change one or two of the following figures. So please do learn to use numbers with maybe $2, 3, 4$ or $5$ significant figures. But not with ten or twenty significant figures. It is not the way science works. So why looking for the newest published values which have one or two more significant figures ? The given weight $123.45678910111213 g$ is not better than $123.456 g$. It only gives an irrational feeling of superiority. That's all. It has no meaning for usual applications. Nobody will believe you if you say that you have calculated a weight of $123.4567891011121314 g$, or measured a temperature of $12.3456789 °C$.


Why is carbon monoxide a monodentate ligand

It is possible for carbon monoxide to coordinate through oxygen, as reported here.  However, there is indeed a strong tendency for carbon monoxide to coordinate through carbon, essentially due to the electronegativity difference noted by the OP.  Cyanide is more likely to coordinate through N than carbon monoxide through O, simply because the electronegativity difference is not so large.  Even cyanide requires special circumstances to get it to coordinate through nitrogen, often involving the carbon being "preoccupied" with another atom -- just like carbon monoxide in the referenced article.


Why does adding salt to water help distillation?

Good question. We frequently "salt out" semi-soluble organic compounds from water because the ionic nature of the salt is more compatible with the water and less compatible with the (less ionic) organic compound. This is used, for instance, to drive sodium stearate (soap) from water. Adding salt could help drive more lanolin out of the water, or what is probably the situation, making an emulsion of water and lanolin less stable so that you can get more lanolin and less "grunge". The lanolin is then purified by filtering, which would remove salt crystals that formed as the water evaporated. And filtering would remove other dirt.

One description of obtaining lanolin suggests driving off all the water, then purifying by adding olive oil and water to remove dirt. But then it raises the question: why not remove the lanolin before all the water is evaporated? The salt would remain in the water. https://permies.com/t/65837/fiber-arts/extract-lanolin-wool

It's also possible that the water pulled out by the wool extracts a great deal of the salt, leaving the lanolin with only a small portion of the original salt and water (depending on how much water you use in the first place).


What is the origin of the maximum in rotational heat capacity?

Most heat capacities go through a maximum as the temperature increases. $C_V = \left( \frac{dU}{dT} \right)_V$ so a maximum in $C_V$ corresponds to a minimum in $\left( \frac{dT}{dU} \right)_V$, i.e. the point where the temperature changes very little as energy is being supplied to the system.  At this point (most of) the energy is being used to excite molecules from the ground state to the excited state, rather than increasing the kinetic energy (i.e. the temperature) of the molecules.

I have made video about it here.

For a more quantitative explanation, consider a two-state system with a singly degenerate ground-state and a excited state with degeneracy $g$, that is $\varepsilon$ high in energy than the ground state.  The heat capacity can be written as
$$C_V = \frac{\varepsilon^2}{RT^2}p_0p_1 $$
where $p_0$ and $p_1$ is the probability of being in the ground and excited state, respectively
$$ p_0 = \frac{1}{1+ge^{-\varepsilon/kT}}$$
$$ p_1 = \frac{ge^{-\varepsilon/kT}}{1+ge^{-\varepsilon/kT}}$$

Now some plots for $N_A \varepsilon$ = 2.5 kJ/mol and $g$ = 100

The first plot shows $p_0p_1$ (green), $C_V$ (red) and  $\frac{\varepsilon^2}{RT^2}$ (blue, right $y$-axis).  This shows that the general shape of $C_V$ as a function of temperature comes from $p_0p_1$

The second plot shows $p_0$ (red), $p_1$ (green), $p_0p_1$ (yellow) and $\frac{\varepsilon^2}{RT^2}$ (blue, right $y$-axis) - right $y$-axis.  This shows that the shape of  $p_0p_1$ come from the shape of  $p_0$ and $p_1$, and that the peak occurs where  the ground and exited states are equally likely ($p_0 = p_1$).  The peak in $C_V$ occurs at slightly lower temperature due to the $\frac{\varepsilon^2}{RT^2}$ term, but ultimately the peak in $C_V$ derives from the peak in $p_0p_1$.


Reactions involved in Titanium Anodisation

I was able to locate what appears to be a useful summary of reactions and reduction potentials that are all titanium-specific here.

The reaction $\ce{ Ti + O2 -> TiO2}$ is illustrative of $\ce{Ti}$ being oxidized while oxygen is reduced. I agree with you that it is simplistic-looking and might not appear as traditional an example of redox chemistry as you require.

You might investigate the high-temperature reaction of titanium metal and water (steam), which produces $\ce {TiO2}$ and $\ce{H2}$ gas (titanium is oxidized, hydrogen is reduced), which might be a little clearer. This reaction (and several others) are given on the page I've linked above.


Energy capture photochemistry with ionizing radiation?

The energy scales of x-ray and gamma radiation are too large to be chemically helpful in the way that the electromagnetic around visible wavelengths is.

X-rays, if absorbed, typically excite core electrons of atoms. As we know, core electrons are (to a first approximation) not affected by the details of chemical bonding in the valence, so any attempt to use a chemical bonding motif to capture x-rays (e.g. conjugation of porphyrins) is of the wrong energy scale. You could use the Auger Effect to eject a valence electron with x-ray scale energy. This does constitute radiationless conversion of the light into electronic energy, but it doesn't meet your qualifications for storing and extracting work.

Gamma rays are typically produced by the decay of one nuclear state to another, and are vastly different in energy scale than electronic states, so there's no way to do chemistry with gamma radiation, at least directly. The Mössbauer effect does allow us to use gamma rays to understand local structure in a crystal, but even then most of the energy is absorbed into exciting the nuclear state, which we cannot control chemically.


You are correct in drawing this. The "answer" molecule depicted by your book is different from the one it first depicted, so it cannot be right.

You correctly depicted the high energy conformation of the molecule as a rotation of the low energy that shows the methyl groups aligned. This is high energy because of the intramolecular repulsion of the hydrogen atoms and the electron clouds.


Ludwig Gattermann's The Practical Methods of Organic Chemistry (translated into English, 1896) has two references to reduction of aromatics with zinc dust:

The first one refers to Liebigs Annalen, issue 140, page 295 (1866):

Ueber die Reduction aromatischer Verbindungen mittelst Zinkstaub, Adolf Baeyer

This roughly translates to "On the reduction of aromatic compounds using zinc dust".
The second one refers to Chemische Berichte, issue 1, page 43 (1868):

Ueber Alizarin und Anthracen, C. Graebe C. Liebermann

This translates to "On alizarin and anthracen". Wiley, who now owns the Berichte, cites the page numbers as 49-51.
Name of the reaction
The reaction is given without name, even in modern textbooks. Maybe it should be called Baeyer reduction of aromatic oxygen-containing compounds.
Mechanism of the reaction
Maybe this would be a good research project because it does not seem like the mechanism is known.


How can I separate these two alcohols from each other out of the aqueous solution?

I am tasked with separating these two compounds.

If you want to keep both compounds, degrading one of them is obviously not an option.

  They are both solvated in water.

This can be changed - reversibly ;-)

Your left compound is a diol. React the mixture with a ketone in the presence of an acidic catalyst.

Typically, these reactions are performed to protect ketones by heating them in with ethanediol in toluene in the presence of toluene sulfonic acid. There is no reason not to use it the other way around to protect the diol.

The reaction product is a cyclic acetal (1,3-dioxolane) which is much less  soluble in water that the parent diol.


Are my intuitions about how batteries work right?

I'll assume you are someone less than 16 years of age, since usually it's around 16-17 years (11th or 12th grade) that you'll probably learn how batteries work.

You're assumptions about how electrons are taken from the zinc cell and moved to the copper cell are correct. But considering your setup, as the charge moves, an electric field and a potential difference would build up, won't it? In such case a battery would eventually stop functioning.

While batteries do stop functioning after a while, it's due to the reaction being completed (all of the zinc converted to $\ce{Zn^{2+}}$ and/or all of the $\ce{Cu^{2+}}$ deposited as copper onto the copper electrode. Charge separation does not occur. Then how do the charges remain balanced, you might ask. And that's where the salt bridge comes into the picture.

A Salt Bridge is a tube that connects the two half-cells to maintain electrical neutrality. This tube makes sure that the ionic solutions on both sides are electrically connected, but at the same time prevents diffusion of both solutions to become one (see the link to know exactly how this works).

Feel free to comment if you need any help or clarifications.


What is entropy change in electrochemistry?

I think your diagram seems plausible; the physical process happening here is when charging an applied potential difference (voltage) causes lithium ions to leave the crystal lattice and migrate to the cathode. When the battery discharges, the reverse happens, and lithium ions rejoin the anode.

The definition of reversibility is the infinitesimal change in some thermodynamic quantity in the system with respect to its surroundings without increasing the entropy. The electrode is crystalline, so Li+ ions can rejoin the lattice without increasing the configurational entropy, whilst infinitesimally changing the chemical potential.

If the temperature is low enough for crystallisation to be entropically and energetically favourable with respect to dissolving, then the free energy of the system will be reduced by adding an ion to the crystal lattice. The chemical potential induced by charging causes the lithium ions to return to the anode when discharging.


Nernst equation for a fuel cell when gases are stored separately

Given that the standard potential for a fuel cell is 1.229 V what is causing the difference in our measurement?

Multiple sources (e.g. here, here and there) estimate the voltage to be much less than the standard potential, especially when there is a load.

Individual fuel cells produce relatively small electrical potentials, about 0.7 volts, so cells are "stacked", or placed in series, to create sufficient voltage to meet an application's requirements.

So it is surprising that the voltage is higher than the standard potential. Perhaps two cells were stacked to provide about 1.4 V, similar to the voltage of a standard alkaline battery, and about twice what you would expect from a single cell under load.

For [H2] I am using the molar concentration of the gas in its containing cylinder (I believe this accounts for water vapour).

For the standard hydrogen electrode, the standard state is defined as a partial pressure of 1 atm. This is much lower than a concentration of 1 M. In any case, the Nernst equation does not predict a voltage that is significantly higher than the standard potential.


Regioselectivity in Kharasch addition of bromotrichloromethane

This looks like a typical Kharasch (chain) reaction. The reaction is initiated by the formation of peroxide radicals (by homolytic cleavage of the weak O–O bond). These peroxide radicals then go on to abstract Br• from the starting material:

Note that the alternative generation of Br• is unfeasible, as radicals tend to attack at less hindered sites: hence the phenyl radical attacks the bromine which only has one attached atom, and not the carbon which has four flanking halogens. The weaker C–Br bond is also preferentially broken over the other three C–Cl bonds, so there is no chance of generating a CCl2Br• radical.

The trichloromethyl radical then adds to C-1 of the alkene, resulting in the more stable secondary radical. This can now abstract a bromine atom from unreacted bromotrichloromethane to give the product and a new Cl3C• radical.

Unfortunately, with additions of X–Y across a C=C double bond (as opposed to additions of H–X), the designations Markovnikov and anti-Markovnikov lose their meaning. A Markovnikov addition of H–X would be one where the hydrogen adds to the less substituted carbon. However, here it is not clear whether to define "Markovnikov" as the regioselectivity where X adds to the less substituted carbon or where Y adds to the less substituted carbon.


I am really glad that you are reviving these challenge questions NotEvans.
As for the synthesis, the amide disconnection leading to fragments A and B seems like the most simplifying first disconnection:

Synthesis of fragments A and B is outlined below:

Esterification of the readily available acid with isopropyl alcohol under Yamaguchi conditions;
Demethylation (this is why I've chosen to make the isopropyl ester, it shouldn't be destroyed with $\ce{BBr3}$);
Formation of a thiocarbonate crucial for the introduction of two fluorine atoms;
JACS 2010, 132(51), 18199;
Aldol condensation of the ester with formaldehyde;
Ester hydrolysis;
Conversion of the acid to acyl chloride.

p-Nitroaniline is brominated;
Sonogashira coupling;
Gold-catalyzed hydroamination;
Fmoc protection of the indole nitrogen turned out to be necessary at this point (so it doesn't interfere with the coupling with fragment A);
Reduction of the nitro group.

Now comes the endgame:

Amide formation;
Corey-Chaykovsky reaction of a stabilized sulfure ylide is selective for enone-like systems;
Piperidine removes Fmoc;
The indole nitrogen attacks glycidol thus opening the epoxide on the more accessible site;
Reduction of the ester with $\ce{LiBH4}$ should be selective. Several equivalents are needed because of the free hydroxyl groups. My only concern is the stability of the cyclopropane, but other hydride donors could be tried.

EDIT: Here is the synthesis of the alkynyl ester needed to make fragment B:

18. Methylacetoacetate is methylated twice;

Formation of enol triflate;
Base-induced elimination of the triflate gives the ester.

  Total step count: 20

  Longest linear sequence: 12


How to estimate the dielectric constant of a liquid?

Yes, such relations exist.

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

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

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

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


A strategy different from the one described by Yusuf Hasan would be to look at the neutral isoelectronic molecule. The oxygen atom is isoelectronic to the fluorine cation, so $\ce{OF+}$ is isoelectronic to $\ce{O2}$.

It is difficult to write a nice Lewis structure for $\ce{O2}$, see https://chemistry.stackexchange.com/a/15061. Experiments show that dioxygen is paramagnetic, i.e. it has unpaired electrons. If you try to write this as Lewis structure, you end up with 7 electrons on each oxygen (counting electrons in bonds twice), i.e. no octets either.

For $\ce{OF+}$, you should be able to write the same Lewis structure (and the same MO diagram). Because fluorine has a different number of protons than oxygen, however, the energy levels will be different and "the best" Lewis structure might be different as well.

As Yusuf Hasan mentioned, the struggles to write a "nice" Lewis structure for this species hints at high reactivity, so I think this is the main lesson of trying to write a Lewis structure. Which of the Lewis structure is best at approximating the real situation becomes secondary, and it makes sense to move on from a Lewis structure model to more comprehensive models of bonding for species such as this one.


No. Radioactivity means that the nucleus of a particular atom is unstable, and the nucleus will decay to a more stable state either by decomposing or emitting energy (as a photon). This instability will not make other nuclei unstable unless the decay of one of the radioactive atoms causes a fission process in one of the atoms in the desk. This is a secondary effect though and surely will not happen once the material is removed.

Another way of thinking about this is that there are materials in which people house radioactive substances. These materials do not become radioactive or else there would be no point in having them in the first place. Not just any material will work though because radioactive emission products might be transparent to many materials or are energetic enough to melt/decompose other materials. So, I would bet the desk would get destroyed, but from normal decomposition not from decay.

Finally, the reason that it seems like things become radioactive once there has been radioactive material nearby is because we usually think about the products of nuclear fission. These products will have quite long half-lives and thus the material will stay around for a long time decaying.


Will isopropyl alcohol dissolve the coating/paint of a notebook?

Unfortunately you won't really know until you try it because there doesn't appear to be any info about what they used as a coating. Having said that, isopropanol isn't a superbly strong organic solvent so while it may remove some of the coating, it's not likely to immediately eat through your laptop like how acetone might. Still, follow user137's advice and test it on a very small area before you try anything else, and give it time before you go using it everywhere else. If you don't see any damage and you decide to go through with whatever it is you want to do (I assume cleaning it), just make sure you wipe it off with a damp cloth (water) followed by a dry cloth as soon as you get the results you want to minimize any possible damage.


Why is H in OH group is more shielded than in H in CH2 group (ethanol NMR spectroscopy)?

It is a myth that electronegativity is the only things that effect chemical shifts. However, electronegativity is only one of many things that effect chemical shift of a proton. Other things effecting chemical shift are magnetic anisotropy (e.g., large deshielding effects on aromatics and alkenes), steric compression (e.g., measurable $\Delta \delta$ values of alkene protons on cis- and trans-isomers), anisotropy of double bonds (e.g., measurable positive $\Delta \delta$ values of methyl group of methylcyclohexene and  methylcyclohexane, $\delta_{\mathrm{sp^2}(\ce{CH3})}-\delta_{\mathrm{sp^3}(\ce{CH3})} \approx 1.60-0.92= 0.68$), solvent effects, etc.
All of above mentioned factors effect chemical shifts of proton attached to electronegative atom such as $\ce{OH}$ and $\ce{NH}$. In addition, solvent used and solute concentration are also fators changing chemical shifts of these protons. For example, let's consider alcohol $\ce{OH}$ protons. In dilute solution of alcohols in non-hydrogen-bonding solvents such as $\ce{CCl4, CDCl3,}$ and $\ce{C6D6,}$ the $\ce{OH}$ signal generally appears at $\delta \ 1\!- \!2$. For instance, residual peak in $\ce{CDCl3}$ is always appeared at around $\delta \ 1.6$. At higher concentrations in non-polar solvents,  the $\ce{OH}$ signal moves downfield ($+\delta$) because when concentration increase, the $\ce{H}$-bonding increases as a result. For instance, the $\ce{OH}$ signal of ethanol comes at $\delta \ 1.0$ in a 0.5% solution in $\ce{CCl4}$, and it appears at $\delta \ 5.13$  in the pure ethanol (5-HMR-2 Chemical Shift):

The solvent used to take the given spectrum is unknown. Based on the chemical shift of $\ce{OH}$ signal, one can assume solvent might be $\ce{CDCl3}$, which is the most common among NMR solvents, which does not promote $\ce{H}$-bonding. As you can judge by the image above, it might have taken in the concentration of 1-5% range.


Is a strontium-fluorine battery the highest voltage battery using pure elements?

The half-potential you've given for strontium is only for the first ionisation. The half-potential you'd actually get is -2.899V for the stable dication to give a cell potential of 5.769V. From the CRC Handbook[1], lithium has the lowest element->stable ion potential of -3.0401V, which is why it's common in batteries.

[1]: Lide, David R., ed. (2006). CRC Handbook of Chemistry and Physics (87th ed.). Boca Raton, FL: CRC Press. ISBN 0-8493-0487-3.


Painting corroded 9V battery contacts with vinegar

Using vinegar, petroleum jelly, or glycerol to clean the contacts would leave some residue.  Vinegar might dissolve it somewhat more quickly than water, but you have to be sure you get all the vinegar off the contact, or it could cause more corrosion.  Petroleum jelly - I think someone was kidding you.  Vinegar & petroleum jelly don't mix.

Much easier: use a small piece of emery paper or steel wool to grind the deposit off.  Keep sanding until the metal is shiny.  Use a damp rag to remove the powder.  Don't let the ground-off powder fall into the circuitry.

Edit:  added steel wool.


Calculating the strength of an ionic bond that contains poly-atomic ions

With enough effort, Born–Haber cycle can be extended to polyatomic ionic solids, however it's practically never done in practice due to the lack of experimental data or because it's impossible to obtain any.

From [1, p. 117–118] (emphasis mine):

  Lattice energies cannot be measured experimentally since they represent hypothetical processes:

  $$\ce{M^n+(g) + X^n-(g) → MX(s)}$$

  However, the following reaction sequence, relating the heat of formation, $ΔH_\mathrm{f}$ of a crystal $[\ce{M(s) + 1/2 X2(g) → MX(s)}]$ to $U[\ce{M+(g) → MX(s)}]$ is thermochemically equivalent (and $ΔH_\mathrm{f}$ can be measured).

  $$
\begin{array}{ccc}
\ce{&M^0(g) &+ &X^0(g) &→ &M+(g) &+ X-(g)} \\
&↑\small ΔH_\mathrm{s} &&↑\small D &&↓ & \\
\ce{&M(s) &+ &1/2 X2(g) &→ &MX(s)}
\end{array}
$$

  In this diagram, $ΔH_\mathrm{s}^\circ$ gives the enthalpy of sublimation of the metal $[\ce{M(s) → M^0(g)}],$ $D$ gives the dissociation energy, or bond energy of the diatomic gas $[\ce{1/2 X2(g) → X^0(g)}],$ $IE$ gives the ionization energy of the gaseous metal $[\ce{M^0(g) → M+(g)}],$ and $EA$ gives
  the electron afﬁnity for the formation of the gaseous anion $[\ce{X^0(g) → X-(g)}].$
  The lattice energy is obtained through the relation:

  $$U = ΔH_\mathrm{f} - \left(ΔH_\mathrm{s} + \frac{1}{2}D + IE + EA\right)$$

  One difﬁculty with using a Born–Haber cycle to ﬁnd values for $U$ is that heats of formation data are often unavailable.
  Perhaps the greatest limitation, however, is that electron afﬁnities for multiply-charged anions (e.g. $\ce{O^2-}$) or polyanions (e.g. $\ce{SiO4^4-}$) cannot be experimentally obtained.
  Such anions simply do not exist as gaseous species.
  No atom has a positive second electron affinity; energy must be added to a negatively charged gaseous species in order for it to accommodate additional electrons.
  In some cases, thermochemical estimates for second and third electron afﬁnities are available from ab initio calculations.
  Even so, if there are large covalent forces in the crystal, poor agreement between the values of $U$ obtained from a Born–Haber cycle and Madelung calculations can be expected.

Already for some simple species such as oxoanion it is impossible to determine their electron affinity $EA$ in gaseous phase:

$$
\begin{align}
\ce{O^0 (g) + e- &→ O^- (g)}\label{rxn:R1}\tag{R1} \\
\ce{O- (g) + e- &→ O^2- (g)}\label{rxn:R2}\tag{R2} \\
\hline
\ce{O^0 (g) + 2 e- &→ O^2- (g)}\label{rxn:R3}\tag{R3}
\end{align}
$$

Enthalpy of reaction \eqref{rxn:R2} cannot be determined experimentally and can only be derived from the lattice energy $U$, which causes Catch 22 situation[2, p. 61].

References

Lalena, J. N.; Cleary, D. A. Principles of Inorganic Materials Design, 2nd ed.; John Wiley: Hoboken, N.J, 2010. ISBN 978-0-470-40403-4.
Smart, L.; Moore, E. Solid State Chemistry: An Introduction, 4th ed.; CRC Press: Boca Raton, FL, 2012. ISBN 978-1-4398-4792-3.


XRF is "non-destructive," but what actually happens to the sample?

The title of the question is about sample heating by ED-XRF instrument. Yet, OP is wondering its limitations and like to understand what kind of compounds analysts might lose to some sample heating if they switch to ED-XRFs (bench-top?). I hope this may be addressed by MaxW, when OP has said whay kind of samples they are analyzing. However, general knowledge on these instruments would predict both WD- and ED-XRF instruments would heat samples regardless of what kind of instrument is in use, but I'd suggest ED-XRF would heat less than that of WD-XRFs since they are low energy instruments.

Actually, to me, the bigger question is which kind is more sensitive to particular elemental analysis, regardless of sample kind (e.g., environmental or archaeological). My literature survey on the subject indicate that ED-XRF instruments is more qualitative than quantitative toward analysis, specifically towards analysis on low-$Z$ and medium-$Z$ elements (References 1 & 2).

The authors of ref.2 pointed out that the energy dispersive X-ray fluorescence analysis of low-Z-elements can generally be seen as a challenge. They have adjusted the instrument in several ways to get better data. The abstract of the article has well reflected the point:

  Photoninduced energy dispersive X-ray fluorescence analysis in total reflection geometry has been applied to low-$Z$ elements. To avoid the absorption in the $\ce{Be}$-window a windowless $\ce{Si(Li)}$-detector has been used. Icing of the detector crystal is prevented by a thin $\ce{C}$-foil. Physical reasons of the low energy background are reduced using a high energy cut-off reflector. Measurements of low-$Z$ elements from $\ce{O}$ ($Z= 8$) onwards are possible and detection limits in the $\pu{ng}$ region are obtained.

Anybody interested in these adjustments should get the article, which has explained them with some depicted diagrams.

On low-$Z$ element analysis, authors of Ref.2 claimed that:

  WD-XRF is the best XRF approach for measuring and quantifying low-$Z$ elements. ED-XRF spectrometers are subject to limitations, such as Bremsstrahlung radiation and escape peaks, which affect detection and quantification of low-$Z$ elements in particular. Although we will demonstrate in Section 2.3 that, under the right conditions, pXRF spectrometers can simulate the performance of bench-top ED-XRF instruments, there are two notable exceptions: $\ce{P}$ and $\ce{Na}$ (Note: the notation pXRF spectrometers are for portable-ED-XRFs).

The following diagram from Ref.2 well explained point, which shows exceptional correlation ($R^2=0.995$) of expected to measured concentrations of $\ce{P2O5}$ when WD-XRF is used:

References:

Alice M. W. Hunt, Robert J. Speakman, “Portable XRF analysis of archaeological sediments and ceramics,” Journal of Archaeological Science 2015, 53, 626-638 (https://doi.org/10.1016/j.jas.2014.11.031).
C. Streli, H. Aiginoer, P. Wobrauschek, “Total reflection X-ray fluorescence analysis of low-Z elements,” Spectrochimica Acta Part B: Atomic Spectroscopy 1989, 44(5), 491-497 (https://doi.org/10.1016/0584-8547(89)80055-2).


Using molecular orbital for determining leaving group trend and nucleophilicity

MO theory can certainly be used to explain substitution reactions from a mechanistic perspective, and to help rationalize trends of relative reactivity. However, I think that the best way to build a mental model that gives you practical predictive power is to also incorporate the simpler ideas that mostly reduce to Lewis acid/base theory that most introductory textbooks teach.

Remember that "nucleophile" and "leaving group" are roles that molecules take when they participate in a specific type of reaction. They are not intrinsic molecular properties. MO theory can fill in part of the picture, but other factors (such as steric hindrance in the nucleophile or electrophile), solvent selection, heat, concentration, etc. can all affect reaction rates and how well a given compound takes up the role of being a nucleophile or electrophile in a given reaction context.

So, to answer your questions:

1) The energy of the HOMO does play a role in determining nucleophilicity since, from a mechanistic perspective, a substitution reaction occurs when the HOMO of a nucleophile overlaps with (and populates electrons into) the LUMO of the electrophile. However, HOMO energy levels are difficult to predict off-hand. A better explanation for the hypothetical difference in nucleophilicity between H2O and H2O+ is that H2O+ is strongly electron-deficient, with the oxygen lacking an octet. Therefore, it is extremely unlikely that the oxygen will act as a nucleophile. This is just Lewis acid/base theory.

2) The LUMO is, by definition, unoccupied. There are no electrons in the LUMO that could act in a nucleophilic manner towards another orbital. On the flip side, if the HOMO is occupied then that orbital is "full" and there is no way that it could act as an electrophile by accepting more electrons.

3) No - see the second paragraph of my answer. The energy of the electrophile's LUMO in question -- the sigma-antibonding orbital -- is certainly relevant, but many other factors are at play: sterics, the bond length between the leaving group and the electrophilic carbon, solvent effects, etc. The most useful, generalizable approach to predicting LG ability is Lewis basicity; the weaker the base, the better the leaving group. Thus, H2O is a better leaving group than OH-, Br- is better than Cl-, etc.

4) From a Lewis/acid base standpoint, one would predict I- to be a good leaving group since it is a very weak base. MO theory predicts that I- will be a better nucleophile than the other halides because its valence electrons reside in 5p orbitals, which are much higher in energy and more close in energy to the sigma-antibonding LUMO of the electrophile.


How to detect nM levels of Copper(I) Oxide in blood?

The main analytical problem in your case of detecting Cu(I) oxide in blood is that you do not know the fate of the particles once they enter blood stream. As I said earlier, Cu(I) is thermodynamically not stable in any aqueous system. It oxidizes to Cu(II), which is water soluble. The non-destructive technique to detect species is X-ray photoelectron spectroscopy, but its detection limit is very poor. It may take a day to collect weak signals from such a low concentration of Cu(I) oxide. The sample has to be bone-dry as well! This is a species specific analysis, i.e. one can distinguish oxidation states of metals by deconvolution.

A practical suggestion is that you do a control test, i.e. measure copper levels before treatment in blood. As a result, all copper originates from your copper (I) oxide dispersion. You would not know the specific species of copper but you will get the total content. For that one can use common techniques such as graphite furnace atomic absorption analysis (only a few microliter of sample is required).


How to calculate the number of waters of hydration?

All of the questions involve the same reasoning, so let's look at question #2.

What is the % oxygen in $\ce{MgSO4}$ when there is no water of hydration (x=0)?  It would be given by $$\mathrm{\%~ oxygen~ =~ \frac{(4 \cdot atomic~ wgt~ O)}{(atomic~wgt~ Mg + atomic~ wgt~S + (4 \cdot atomic~ wgt~ O))}}\\\mathrm{=~\frac{(4 \cdot 16)}{(24_{.}31+32+(4 \cdot 16))}}\\=53.2\%$$  How would the equation change if we added one $\ce{H2O}$ of hydration to the molecular formula? How about if we added 2 or 3?


Can nitric acid be used to create an acidic medium for redox reactions?

Yes, you can.

However, nitric acid is an oxidizer, so it can interfere in some cases. If a strong reducer (like magnesium) is involved, nitric acid may act as an oxidizer even in diluted solutions. Obiously, it acts as an oxidizer in concentrated solutions, producing rather nasty nitrogen oxides, and if used without prior knowledge, it may produce unstable explosives.

For this reason, one generally use hydrochloric acid or sulfuric acid. Mind you, in case oxidation is not a problem, nitric acid may be used with no problems. Another reason to use sulfuric acid is that it is dirt cheap. It is the cheapest alternative.

Still, in concentrated solutions sulfuric acid does show oxidating properties, so in some cases sulfuric acid is not suitable either. Unfortunately, there is no common and reasonable alternative in the case, except significanlty weaker phosphoric acid.


Phenol or ethanol as a nucleophile for ether formation?

You are adding excess sodium ethanolate so there is the possibility of deprotonation of phenol: $$\ce{C6H5OH + C2H5O- <=> C6H5O- + C2H5OH}$$

If we look at the $\text{p}K_\text{a}$ of ethanol and phenol we have 16 and 9.95 respectively and so the equilibrium above will strongly favour the right hand side. Therefore, the phenoxide ion will dominate as a nucleophile in the mixture, and the major product will be ethyl phenyl ether.


What is the most rare element in periodic table

Well, the first question to answer would be: where?  The rarest element in the universe? the sun? in meteorites? on earth? in the ocean? in humans?  We can tap in to the curated datasets provided by Mathematica to get the answer:

Elements up to atomic number 98 have been found in trace quantities; however, the dataset does not contain any information about the trans uranium elements which were once thought to be synthetic.
Here's a plot of abundance as a function of atomic number:

We can see that the elements beyond about Z = 82 (lead) have a marked decrease in abundance, and this is consistent with all of the elements with atomic number above 82 being radioactive.
Where do we come from?
We can compare the elemental abundance of these individual "spaces" to speculate on where humans come from.  Below are plots showing the correlation of the elemental abundances.  The plot labels give the best fit equation:

The best slope is observed when comparing our abundance to that of the Earth's crust; however it does appear as if the scatter is smallest for the Us vs. the Ocean plot.


Nomenclature of MnO4^-: is it an oxyanion?

As far as I know, an oxoanion just refers to an anion which has oxygen covalently bonded to the central atom; the central atom can be pretty much anything apart from the Group 1 and 2 metals, since the only oxygen compounds those form are ionic oxides/peroxides/superoxides. For the 3d transition metals, you have $\ce{VO3^-}$ (vanadate(V)), $\ce{CrO4^{2-}}$ (chromate(VI)), $\ce{MnO4-}$ (manganate(VII)), and $\ce{FeO4^{2-}}$ (ferrate(VI)), amongst others.

I would not recommend reading too much into the nomenclature, because a lot of it is just what chemists are comfortable with after using for a long time - it's like a language that you just get more familiar with over time. But there are some general rules. The 'per...ate' name, such as perchlorate $\ce{ClO4-}$, implies a higher oxidation state than the normal '...ate' compound, which is chlorate $\ce{ClO3-}$ in this case. For chlorine, only certain oxidation states are available, namely the -1, 0, +1, +3, +5 and +7 oxidation states. Apart from 0, which occurs in diatomic chlorine $\ce{Cl2}$, you will notice that these are in increments of 2 - which means that you can generate the whole list of chlorine anions by adding one more oxygen, but keeping the charge of the anion constant: $\ce{Cl-}$, $\ce{ClO-}$, $\ce{ClO2-}$, $\ce{ClO3-}$, $\ce{ClO4-}$. This tendency to form compounds 2 oxidation states apart is very pronounced for the p-block elements, i.e. Group 15, 16 and 17.

However, for d-block elements such as $\ce{Mn}$, this isn't the case because single d electrons are easily lost and gained. So after permanganate, $\ce{MnO4-}$, which features $\ce{Mn}$ in the +7 oxidation state, the oxoanion with the next highest oxidation state is the manganate ion $\ce{MnO4^2-}$ which has $\ce{Mn}$ in the +6 oxidation state. (As far as I know the ion $\ce{MnO2-}$ is not called "manganate".)


How can I tell if a certain carbon/hydrogen molecule can actually exist?

When drawing organic structures Carbon should have 4 "lines" coming from/to it while Hydrogen can have only 1 covalent bond. So if you could draw your molecule following these rules, it probably can exist. Here is an example:

H       H
 \     /
  C=C=C
 /     \
H       H

Alternatively there could be this structure with the same number of "lines" coming from/to Carbon:

  H   H
   \ /
    C     H
   / \   /
  H   C=C
     /   \
    H     H

As for the naming.. It's more complicated since it would depend on whether there is only one = bond or more than that. You could watch CrashCourse Organic Chemistry intro that talks about some naming conventions. But if you're stumbled, you could count the number of Carbons and Hydrogens and google for it: C3H4 or C3H6.


How can I control the oxygen in an erlenmeyer flask in a bioprocess?

Controlling oxygen in a flask can be achieved by bubbling a purge gas of some description through it. This will entrain and remove oxygen from the solution as well as creating an oxygen-depleted headspace.

If you have a regulated nitrogen supply, a rubber stopper with a hole in it, an open-ended pasteur pipette and a length of hose, you can bubble nitrogen into the solution to purge oxygen. This requires the vessel to be open somewhere else to equalise pressure, which is achievable with a multi-neck flask but not with a standard Erlenmeyer flask. You can get around this with a stopper with two holes. Continual positive pressure should keep the external atmosphere out.

Depending on how vigorously you want to shake your flask, this may or may not be reasonable. If you are using a stirrer bar you will obviously want to give the pipette sufficient clearance.

As far as controlling addition of oxygen, this would seem to be a bit trickier as you are probably not going to be providing continual positive pressure. You might be able to get around this, depending on the specifics of what you are doing, by pumping in oxygen in the same way and using a balloon or a gas syringe or something to deal with the overpressure.


What's the future of an organic and/or analytical chemist?

The answers to this question are going to be as varied as the people answering them. Let me give you the advice I give my undergraduate advisees.

The future is what you make of it. Don't let random people on the internet tell you which field to specialize in or which research to do. The decision need to be personal, because you need to care about what you are doing so that you are motivated to excel.

  Instead of worrying about which type of chemistry to study, find the problems you want to solve.

Research in chemistry is pretty interdisciplinary nowadays, both across the old subfields of organic, analytical, physical, etc. and between chemistry and the other sciences. Rather than become obsessed about organic vs. analytical or whatever, read about current research in chemistry. Think about the types of research problems you want to solve. Usually that will help you figure out which pathway to follow (i.e. which courses you need to take, which specializations to pick up, which lab to join, etc.). Research, both in academia and elsewhere is going to be more problem- than field-focused.

As an undergraduate, I was torn between organic and physical chemistry. I chose organic chemistry as a focus in graduate school, joined a research group focusing on polymer synthesis, and did a lot of analytical work as part of my research. We collaborated with a theoretical group to understand how monomer sequence would affect optoelectronic properties of conjugated oligomers and polymers, so I also had to learn a lot about the physical models of my conjugated systems. Another part of my group was working on polyesters with controllable biodegredation pathways for applications in medicine. The team on that project included grad students focusing on organic, inorganic, and analytical chemistry collaborating with engineering students in another group.


The mechanism of the reaction between 2-chloroethanol and hydroxyl ions in aqueous solution

Yes it is. Because alcohol is a weak acid and it is reasonable to consider first step is irreversible because of the second step almost occuring simultaneously. And also because that the second step is much faster than the first.


Will the total equilibrium pressure increase in gas phase reaction PCl5(g) = PCl3(g) + Cl2(g) after chlorine is removed?

First it is necessary to understand that due to the stoichiometry of the reaction, if you move the equilibrium to the products side, pressure will be increased. This is based on the fact that 1 mole of gas produces 2 moles of gas. In other words the number of molecules in the system is increased. By removing all of the $\ce{Cl2}$ you force the reaction to move to the products side (Le Chatelier's Principle). Thus more $\ce{PCl3}$ will be produced along with $\ce{Cl2}$. Eventually you will end up with more atoms inside the system and that will lead to increase of the pressure.


Why do we not see silicon in a structure like graphite?

It's possible, and it's already been done! The compound is called Silicene.

However, silicene generally isn't perfectly flat. (Then again, as  Karl Ratzsch points out in the comments, graphene isn't perfectly flat either). The Wikipedia article notes that it usually takes a puckered shape.

I don't know the reasons for the puckered shape, but I recall being quite surprised that silicene could be synthesized at all. The reason why is that I was taught in high school that silicon very rarely forms double bonds due to its large size. For a visualization, please refer to the absolutely awful image below (if anyone could provide an actual graphic I'd be grateful).

The black atoms on the left are carbon. The $p$ orbitals overlap to a significant degree, which means that pi-bonds can easily form. The green atoms to the right are silicon. You can see that, because of the larger atomic radius, the p-orbitals don't overlap as much, meaning that double bonds are generally not favorable in silicon. This explains why compounds like silenes are relatively unstable.

Pi systems are crucial to the chemistry of carbon--they allow carbon to form multiple bonds, flat rings (like in graphene), and generally be stable in a lot of configurations. The lack of pi-bonding in silicon is sometimes given as a rationalization for why silicon-based lifeforms can't exist, and explains why it's so hard to get silicon into the flat sheets that form silicene.


On a periodic table it shows that Zn and Cd and Hg all have 0 electron affinity. Why is that?

They all have 2 electrons in their highest occupied s sublevel and 10 electrons in their highest occupied d sublevel (both sublevels are full).  As a result, they can not complete any orbitals by adding an electron and they can not put an electron in a sublevel that is occupied.  All of the other transition metals will either add an electron to complete the highest s sublevel, or into an already occupied d sublevel, or (in the case of palladium) add one to an s sublevel that is empty.

Cd, Hg and Zn thus are unique in that they will all have to put an unpaired electron alone into a p sublevel.


The analyte spends the same time in the mobile phase as it tooks for the mobile phase to pass through the column.

  An analyte's time in the different phases is dependent on it's affinity.

The time $t_\mathrm{m}$ is given.

The ratio of an analyte's time in the different phases is dependent on it's affinity and is equal:

$$\frac{t_\mathrm{s}}{t_\mathrm{m}}=K_\mathrm{D} \cdot \frac{V_\mathrm{s}}{ V_\mathrm{m}} $$.

where

$t_\mathrm{s}, t_\mathrm{m}$ are time is stationary and mobile phase
$K_\mathrm{D}$ is the phase distribution constant
$V_\mathrm{s}, V_\mathrm{m}$ are volumes of stationary and mobile phase.


Is polylactic acid (PLA) soluble in ethylacetate?

D- and L- polylactic acids come from both enantiomers of the monomeric lactic acids.
There will be no difference in solubility between both polymers.
Besides, lactic acid is normally produced by fermentation and the L- form is obtained, so the standard PLA used in 3D printing will be L- PLA.

I have tested the solubility of a piece of PLA and it is not soluble in ethylacetate in one hour.
It does even not seem to be attacked by it.

However, when you rub a 3D printed piece with a paper soaked in ethylacetate, some polishing is attained.

I make basically structural pieces, so I normally do not polish them and therefore I do not know if this is enough for your purpose.


Yes, temperature and heat do affect the spontaneity and the equilibrium of reactions. The term heat can be referred to enthalpy, which can be quite confusing in context, as I believe you are asking about raising the temperature.

Let's first take a look at the equation for the Gibb's Free Energy and raising temperature:

$$ \Delta G^\circ  = \Delta H^\circ - T\Delta S^\circ$$

The Gibb's Free energy does not only relate to the spontaneity of a reaction, the magnitude of Gibb's Free energy describes how much "free energy" is available, or how much work is possible.

Since, entropy, enthalpy are state functions, we can calculate them by using Hess's Law and using tabulated values for standard thermodynamic data:

$$\Delta H^\circ_{rxn} = \Sigma \Delta H^\circ_{products} - \Sigma \Delta H^\circ_{reactants} $$

$$\Delta S^\circ_{rxn} = \Sigma \Delta S^\circ_{products} - \Sigma \Delta S^\circ_{reactants} $$

For simplicity's sake, I will use the combustion of methane that mannaia has so kindly provided and run the reaction at STP using heats of formation:

$\Delta H^\circ_{\text{comb of} \ \ce{CH_4}} = -890 \text{kJ} \cdot \text{mol}^{-1}$

$\Delta H^\circ_{\text{f of} \ \ce{H_2O} \ \text{(l)}} = -285.83 \text{kJ} \cdot \text{mol}^{-1}$

$\Delta H^\circ_{\text{f of} \ \ce{CO_2} \ \text{(g)}} = -285.83 \text{kJ} \cdot \text{mol}^{-1}$

$$\ce{CH4(g) + 2O2(g) -> CO2(g) + 2H2O(l)}$$

$$ \Delta G^\circ  = \Delta H^\circ - T\Delta S^\circ$$
$$ \Delta G^\circ  = -604.53 \text{kJ} \cdot \text{mol}^{-1} - (298.15 \text{K}) *97.39 \text{J} \cdot \text{mol}^{-1} = -633.7 \text{kJ} \cdot \text{mol}^{-1} $$

If we use the enthalpy of combustion for methane:

$$ \Delta G^\circ  = \Delta H^\circ - T\Delta S^\circ$$
$$ \Delta G^\circ  = -1569.34 \text{kJ} \cdot \text{mol}^{-1} - (298.15 \text{K}) *97.39 \text{J} \cdot \text{mol}^{-1} = -1598.38 \text{kJ} \cdot \text{mol}^{-1} $$

What would occur if we raise the T term? Obviously, the Gibbs Free Energy would become larger in magnitude and more negative, thus being more spontaneous. Inversely, if we lower the temperature to such a degree that $T\Delta S^\circ = \Delta H^\circ$, the reaction would not occur at all, and if $T\Delta S^\circ > \Delta H^\circ$, we would need to input energy in order for the reaction to proceed forward.

In general:

If $\Delta H<0$ and (-) and $\Delta S>0$ (+), the reaction is spontaneous at any temperature.

If $\Delta H>0$ and (+) and $\Delta S<0$ (-), the reaction is NOT spontaneous at any temperature.

If $\Delta H<0$ and (-) and $\Delta S<0$ (-), the reaction is spontaneous at low enough temperatures.

If $\Delta H>0$ and (+) and $\Delta S>0$ (+), the reaction is spontaneous at high enough temperatures.

It is possible to think of heat as a product or reactant. Combustion reactions are highly exothermic. Le Chatlier's principle includes raising the temperature as a stress. Alternatively, we can think about the equilibrium of this reaction to see how much product will be formed by looking at the magnitude of the equilibrium constant, $K$.

Accordingly the van't Hoff isochore equation is written as:

$$ln \frac{K_2}{K_1} = \frac{\Delta H}{R}(\frac{1}{T_1} - \frac{1}{T_2}) $$

In summary, an increase in temperature favors the formation of products if the reaction is endothermic. Inversely, a decrease in temperature favors the formation of products if the reaction is exothermic.

As, Uncle Al has stated:

  Ambient oxygen makes no difference

The atmosphere is composed of about 70% nitrogen and 20% oxygen. If this reaction was to be observed in nature, there would be almost a limitless amount of oxygen for the reaction to rpoceed. Temperature does play a role in the stoichiometry of the reaction and thus whether or not the combustion is complete or incomplete. However, it is possible to limit one of the reactants as mannaia has done, specifically oxygen in a laboratory setting.

As related to the kinetic molecular theory of matter, increased temperature will increase the frequency of collisions of particles with sufficient energy to cleave bonds and reform them, driving the reaction further to the right. This is commonly done with the commercial production of ammonia using Van Haber process. Although this is thermodynamically unfavorable, when run at high temperatures, the rate of formation of ammonia is much faster.

In regards to say, an open environment, the temperature would increase the kinetic energy of the oxygen molecules, and so the particles would be spaced further apart. Describing this quantitatively... well, an experiment should be carried out. I would assume if one was so inclined to find if that would increase the equilibrium, rate, or completeness of combustion I'm sure they would find differences.


Determining the orbitals that overlap to form a sigma bond

The basic answer is go by what is available.

Hydrogen does not have any other occupied orbitals than s-orbitals. Hence it must use an s-orbital for bonding. This goes both directions.

Carbon requires four bonds. In order to achieve this, it must hybridise in some way, otherwise, an electron pair would be in an s-orbital and an empty p-orbital would be unavailable for bonding. Hence carbon uses hybrid orbitals in both directions.

Chlorine could hybridise, but it doesn’t have to fulfill its bonding needs. In general, an electron pair in an s-orbital is more stable than one in a p-orbital because it is ‘closer to the nucleus’. (That layman’s argument is bad but gets us a long way.) So chlorine will happily use its p-orbitals for bonding.
This goes both ways, by the way, so a carbon-chlorine bond ($\ce{C-Cl}$) is a hybrid-p sigma bond.

Finally, let’s consider oxygen and nitrogen. They would love to just bond with their p-orbitals, but even if hydrogen is the bonding partner that leads to too much steric strain. Therefore, they hybridise just enough to alleviate the steric stress while still maintaining a maximum s-character for a/the lone pair. So those would love to form bonds with their p-orbitals but typically can’t (so they must use hybrid orbitals). Phosphorus and sulfur are large enough to generally allow bonding only with p-orbitals.


Am I using these spectra to identify the compound correctly?

Structure elucidation from spectrocopic analysis needs some experience. I have feeling that you are novice to this field by your explanation of the structure. You cannot determine the molecular formula by the data given and, hence you cannot determine the degree of unsaturation of the molecule. Yet, based on $\ce{^1H}$-$\mathrm{NMR}$, you can propose there are at least two aromatic nuclei (probably mono-substituted) in the molecule, be case there is a 10-proton multiplet in the aromatic region ($\pu{7-8 ppm}$). This assignment is supported by the 4 resonances in the region of $\pu{110-160 ppm}$ in $\ce{^{13}C}$-$\mathrm{NMR}$ spectrum. That also suggest these two aromatic nuclei are identical. Two aromatic rings are accounted for 8 degree of unsaturation.

The given FT-IR spectrum suggest there is a carbonyl group in the molecule, which is confirmed by the resonance at about $\pu{205 ppm}$ in $\ce{^{13}C}$-$\mathrm{NMR}$ spectrum. The peak at about $\pu{205 ppm}$ would also suggest that the carbonyl is not an aldehyde, but a ketone by its chemical shift ($\gt \pu{200 ppm}$), which would be confirmed by the lack of aldehydic peak at about $\pu{10 ppm}$ in $\ce{^1H}$-$\mathrm{NMR}$. The carbonyl group  is also accounted for another 1 degree of unsaturation.

Now, let's analyze $\ce{^1H}$-$\mathrm{NMR}$ spectrum. There is a 3-proton singlet at around $\pu{2.2 ppm}$ for isolated methyl group (without a neighboring hydrogen), indicating it is a benzylic methyl or next to a carbonyl group. Also, there is another one-proton singlet at about $\pu{5.1 ppm}$ for a methine group (-$\ce{CH}$-), indicating its highly deshielding nature. That deshielding must come from two aromatic ring and carbonyl group together. Thus, based on these data, I can deduct the structure as 1,2-diphenylpropn-2-one as depicted in the diagram below (blue box):

The highest peak observed in mass spectrum of the compound at $\pu{210 m/z}$ with law abundance and the base peak ($100\%$ intensity) observed at $\pu{167 m/z}$ confirmed the structure. As depicted in the red box (vide supra), the fragmented cationic radical at $\pu{167 m/z}$ is very stable, which explained the absence of any other fragmentations (as stated by OP, it is the only other significant peak in MS).


Changing the spectroscopy type to infrared in GC-MS

GC-MS is just one common method of what is known as 'hyphenated' analysis. GC-IR instruments do exist (see examples), although usually involve some pretty specialised cells to allow for gas flow through the sensor. Hyphenated instruments are limited pretty well by your imagination (and budget) only. LC-NMR-MS is a great example of extended hyphenation, and is a powerful emerging tool for automated analysis of complex sample mixtures.


From the comments:

  What do you want to do with the hot water? Swim in it. I was thinking
  of thousands of liters

That's an interesting idea, but unfortunately, I don't think adding chemicals to a pool in order to heat it is a good idea (especially yellowish chemicals). The water temperature would drop in a few hours, tops, and you'd have to be constantly adding more chemicals to it. At some point, you'd have a pool of chemicals instead of a pool of water. Not to mention that the chemicals would probably end up being more expensive than the electricity or the gas bill to heat up the pool to the same temperature the conventional way.

A nice way of heating up pools is using solar water heating systems, and coupling them with the conventional heating systems, if necessary.

So... Is it feasible? I'd say it isn't. But since you said this was more of a thought experiment... The only scenario I can imagine where the best solution to heat up a pool is to add chemicals to it is: You have a small pool, very little time to heat it up, you want to heat it up to sauna-like temperatures, and although you'll only use it for a couple of hours you are willing to spend a lot of money to do so. In this case, I'd add some water to the pool, and then add the following reagents to it, making the following reaction:

$$\ce{NaOH + HCl -> NaCl + H2O} \\ \Delta H = -56\,200\ \mathrm{\frac{J}{mol}}$$

You can do the math, but I assure you it heats up pretty quickly. We used to neutralize concentrated acid in ~10 L plastic buckets, and if we "accidentally" neutralized too much at a time the bucket would start to heat up to a point you couldn't even touch it.

NaCl is kitchen salt, so no issues with toxicity if you do the math correctly. Salt water also makes you float better, which I think is a nice quality in a pool. There would be salt on the bottom, but I think that's nice too, makes it feel like the sea.


Comparing Electron affinity and electron gain enthalpy at 0 K

In my book it is given that first electron gain enthalpy is greater than second for elements. Should we compare the magnitudes in such cases or the actual numbers with signs?

Successive Electron Gain Enthalpies
After the addition of one electron atom becomes negatively charged and second electron is to be added to a negatively charged ion. Thus, due to electrostatic repulsion energy is always required for addition of second electron. First electron gain enthalpy is always more negative than second electron gain enthalpy (which is essentially positive). Author made an ambiguous statement, as we can't just compare the magnitudes. For more clarity see:
$$\ce{S_{(g)} + e- -> S- _{(g)}}             \Delta_{eg} H^- _1 = \pu{-200kJ mol^{-1}}$$
$$\ce{S- _{(g)} + e- -> S^{2-}_{(g)}}             \Delta_{eg} H^- _2 = \pu{600kJ mol^{-1}}$$.

Does the same comparision hold with Electron Affinity at 0K?

Generally, if energy is released when an electron is added to an atom, electron affinity is taken as positive, contrary to thermodynamic convention. That is, if $\Delta_{eg} H^-$ is negative EA is positive (since temperature mentioned is absolute zero).
At absolute zero, $\Delta_{eg} H^- = \pu{-EA}$. Thus if $\Delta_{eg} H^- _1$ is negative, EA is positive and successive EAs will be increasingly negative. Here, you can say first electron affinity is greater than second (actual numbers with sign).

For example, Be has positive Electron Affinity and F has negative Electron Affinity. So should I consider signs and say Be has more Electron Affinity than F or consider magnitude and say F has more Electron Affinity than Be?

First, $\ce{Be}$ has negative electron affinity ($\pu{-0.5 eV atom^{-1}}$) and $\ce{F}$ has positive electron affinity ($\pu{+3.4 eV atom^{-1}}$). So, fluorine has more electron affinity than beryllium, consider the actual numbers with signs).

How to compare their Electron gain enthalpies also at 0K?

Since, at 0K, $\Delta_{eg} H^- = \pu{-EA}$. Thus, first electron enthalpy of fluorine is more negative than that of beryllium. Strictly speaking, it's value is less than that of beryllium.


Reducing temperature on dissolved Pb(NO3)2

Assuming that the solution is saturated, meaning that if you add more salt to the solution it will not dissolve, and neglecting eventual evaporation of water, we have a mass of water of about 228g for 171g of salt (assuming solubility of 75g in 100g at 40°C).

Now, when cooling down to 18°C the solubility drops to ~55g / 100g water. This means that in our solution we still have
$$228~\text{g}~\ce{H2O} \times \frac{55~\text{g}~\ce{Pb(NO3)2}}{100~\text{g}~\ce{H2O}}=125.4~\text{g}~\ce{Pb(NO3)2}\; . $$

The remaining 45.6g of the compound will have precipitated. Whether they form nice crystals or not depends on various other things, see this.

Now, if the solution is not saturated (meaning your salt can still be fully dissolved, but there may be room for more in the solution) the calculation is impossible if you don't know the mass of the water at the beginning.


Does using same calibration curve for chemical analysis introduce correlation in sample error/uncertainty values?

Please note that calibrating an instrument once a month is another story as compared to constructing the calibration curve from that instrument. The former is meant to check the accuracy of the instrument's output and the latter serves to determine the concentration of a substance by noting the response of the instrument to 6-7 standards.
Also note that all instrumental analysis techniques are relative with very very few exceptions (gravimetry, microwave spectroscopy, coulometry which are absolute etc.). The rest all need standards for comparison.
In analytical chemistry, it will be considered a bad practice to make a calibration curve one day and use it for the entire month. The good practice is to make a fresh calibration curve for every batch of samples and there should be a "reference" check which keeps on validating the calibration curve after a certain number of samples have been analyzed.
The main reason is that most instruments will show what is called as the drift i.e., a low frequency and hence long term noise in the baseline. The baseline response can rise and decline slowly.
Also the word correlation is perhaps not suitable in your situation. Correlation has a very specific meaning in signal processing/instrumental analysis. What you imply is the influence of interferences and something called as the matric effect. Suppose, you have b, along with c,d,e,...,f. Yes, the presence of the additional components can indeed influence the signal of b. In most cases, the signal of b is decreased causing a lower result than expected.
Beyond your hypothetical situation, one can say nothing. Which instrument you had mind?


When did the realistic typesetting of chemical structures start?

this isn't a definite answer but it might show the direction:

I have an old copy of Namensreaktionen in der Organischen Chemie, 2nd ed. by Helmut Krauch and Werner Kunz, published in 1962 at Alfred Hüthig. First edition must have been published in 1960 or 1961.

Most of the arenes in the sketches have the proper shape. Once in a while they seem to be deliberately stretched to align centres that react.

All through the book there's a consistent notation for saturated centres, hydrogen atoms are mostly not omitted.


The wikipedia page on Phosphoresence has a red pigment. Calcium Sulfide. I can imagine that the lower energy (higher wavelength, lower energy) is more stable because the triplet is not as high in state energy as lower wavelength emitters. Because of the stability the time to emit is longer and the intensity lower.


What is the role of hydrochloric acid in a copper electroplating bath?

As already mentioned by @GM it indeed is used to take out $\ce{Cu^{2+}}$, but it is also used to take out $\ce{Cu+}$.

I know this because I once used a mixture of $\ce{HCl}$ and $\ce{H2O2}$ to etch copper boards, in that process both $\ce{Cu^{2+}}$ and $\ce{Cu+}$ were dissolved by the formation of $\ce{CuCl2}$ and $\ce{CuCl}$ with $\ce{Cl-}$ from $\ce{HCl}$.


Do isotopic groups cause optical activity?

From a 2007 paper::

  Here we show that instrumental advances in Raman optical activity, combined with quantum chemical computations, make it possible to determine the absolute configuration of (R)-[$\ce{^2H1}$, $\ce{^2H2}$, $\ce{^2H3}$]-neopentane. This saturated hydrocarbon represents the archetype of all molecules that are chiral as a result of a dissymmetric mass distribution. It is chemically inert and cannot be derivatized to yield molecules that would reveal the absolute configuration of the parent compound. Diastereomeric interactions with other molecules, optical rotation, and electronic circular dichroism are, in contrast to the well-known case of bromochlorofluoromethane, not expected to be measurable. Vibronic effects in the vacuum ultraviolet circular dichroism might reveal that the molecule is chiral, but the presence of nine rotamers would make it extremely difficult to interpret the spectra, because the spatial arrangement of the rotamers’ nuclei resembles that of enantiomers. The unequivocal spectroscopic determination of the absolute configuration of (R)-[$\ce{^2H1}$, $\ce{^2H2}$, $\ce{^2H3}$]-neopentane therefore presented a major challenge, one that was at the very limit of what is possible.

Reading between the lines:

The described molecule looks like this:

Optical rotation is not expected to be measurable.  Probably some fantastically minute but nonzero specific rotation could be computed with advanced quantum mechanical methods, but it would be no fun to try to measure it.
The difference in the electronic configuration of the isotopically substituted enantiomers in the ground state is negligible.
But the electronic configuration of rotationally or vibrationally excited states is not negligible.  Therefore, some technique that can measure the optical properties of rovibrational excited states might be able to discern the difference.  (I'll leave it to experts in this area to explain why Raman optical activity worked better than vUV-CD.)


I have a sodium silicate sealant in my car and need to find a way to soften it

I do not know much about engines, but from a chemical point of view, the clog that formed in your engine is most likely a hydrated silica. This compound is very insoluble and unreactive - there are only few things that can dissolve silica, e.g., hydrofluoric acid or a hot concentrated solution of sodium/potassium hydroxide, but I wouldn't pour any of these to an engine. Therefore, it seems that your only option is to dismantle the engine and try to remove the clog mechanically.


Do all metal nitrates give nitrogen dioxide on decomposition?

TL;DR: Decomposition of nitrites always afford $\ce{NO2}$ if you apply the correct reaction condition to it.
Long answer
To generalize the decomposition of metal nitrates (I concluded from the article above, and hereby using a generic monovalent metal M for the equations), we can say this:

The first stage of decomposition is always the decomposition to nitrites: $\ce{MNO3 -> MNO2 + 1/2 O2}$

Note that nitrites are almost always (if any) less thermally stable than the corresponding nitrates. Thus, the formed nitrites from the reaction above quickly decomposes. I think that (and from the article), nitrites decompose to form the oxide and $\ce{N2O3}$, the latter disproportionates into $\ce{NO}$ and $\ce{NO2}$: $\ce{2 MNO2 -> M2O + NO + NO2}$.

Everything becomes interesting from here. It is due to the fact that the nitrite formed from the first step and not decomposed (kinetics matter) can be oxidized by either the $\ce{NO2}$ or $\ce{O2}$ back to the nitrate. This is especially significant when the nitrite is stable enough ($\ce{KNO2, NaNO2}$ etc). In fact, the decomposition of $\ce{KNO3}$ is reversible (equilibrium) at about 650 - 750 °C in the atmosphere. In the other hand, nitrites of transition metals (which are mostly covalent in character) are so unstable that they decompose rapidly and thus only appear as the so-called reactive intermediates.

Finally, the oxide can decompose into the metal and $\ce{O2}$. This is a characteristic of metals that lie in the less-reactive end of the reactivity series. If the decomposition temperatures of the oxide and the nitrate are different enough, you can control the temperature to get the oxide or the metal ($\ce{HgO}$ from $\ce{Hg(NO3)2}$ is an example.)

The re-oxidation of stable nitrites can be avoided by two ways: either do the reaction under a flow of inert gas to displace the gaseous products away, or crank up the heat more to increase the reaction temperature, which means faster decomposition of nitrites. For example, decomposition of $\ce{KNO3}$ goes to completion at about 800 °C.
Reference

High Temperature Properties and Decomposition of Inorganic Salts Part 3, Nitrates and Nitrites, Kurt H. Stern, Journal of Physical and Chemical Reference Data 1, 747 (1972); DOI: 10.1063/1.3253104


As correctly pointed out in the comments, the compound you present belongs to the class of sydnones. According to Hückel's rule, it should be aromatic, as the molecule is planar and contains $4n+2$ π-electrons.

However, calculations performed in 1998[1] that mainly looked at chemical shifts and inferred aromaticity from the magnitude of observed diamagnetic shielding, show that sydnones in fact are not fully delocalized, i.e. they are not aromatic. To quote from the linked publication:

  Mesoionic compounds  are  planar  five-membered  heterocyclic  betaines
  with at least one side chain whose α-atom is also in the ring plane and with dipole moments of the order of 5 D. Electrons are delocalized over two regions separated by what are essentially single bonds. One region, which includes the α-atom of the side chain, is associated with the HOMO and negative π-charge whereas the other is associated with the LUMO and positive π-charge.

The publication is also riddled with images that depict syndones like this:

It ends with another quote from Gilchrist's book "Heterocyclic Chemistry", 2nd edition, 1992, p. 329:

  It does not therefore seem useful or appropriate to classify sydnones and other mesoionic compounds as aromatic, despite the existence of a cyclic array of p-orbitals.

Reference

Simas, A. M.; Miller, J.; de Athayade Filho, P. F. Are mesoionic compounds aromatic? Can. J. Chem. 1998, 76 (6), 869–872. DOI: 10.1139/v98-065.


The compound is called copper(I) chloride dihydrate - carbon monoxide adduct. It is formed during the Gatterman-Koch reaction.

Gatterman-Koch reaction is a reaction that produces aromatic aldehyde from aromatic hydrocarbons. The reagent used are carbon monoxide, hydrochloric acid and a Friedel-Craft catalyst like copper(I) chloride ($\ce{CuCl}$) or aluminium chloride ($\ce{AlCl3}$). So, the adduct is formed by the reaction of copper chloride and carbon monoxide during the course of reaction. If aluminium chloride is used, then a complex would be formed i.e aluminum trichloride complex, $\ce{[HCO(AlCl4)]}$.

To know more about the mechanism and role of reagents, see this previous question on chem.SE..

Other References

https://en.wikipedia.org/wiki/Gattermann_reaction
https://www3.aiche.org/proceedings/Abstract.aspx?PaperID=358705


Why does the oxidizing strength of the chlorates decrease with increasing oxidation number?

The chlorate system is very complex. I doubt that the open-circuit potentials can be actually measured in solution, but rather are calculated from heats of formation. If the term "oxidizing strength" is used to compare systems, it should be rigorously defined; in this case, it seems that "oxidizing strength" is defined as "open-circuit voltage". But this could be a poor definition - even woefully poor.

In particular, the comparison of the open-circuit voltages does not include the potential current derivable from the reaction of a mole of each oxychloride. The output energy of the electrolytic cell (if you could just test each reaction, one at a time) would be the output voltage times the number of electrons transferred. By that standard, perchlorate will consume 7 times as many hydrogens as hypochlorite, and thus give 7 times as much current, with only a slightly lower voltage.

A comparison of the heats of reaction (just for the acidic equations) proves to be much more dramatic:

2 moles hypochlorous acid (total heat of formation = -56 kcal (CRC Handbook) produces   2 moles of H2O (total heat of formation = -136 kcal)

2 moles chlorous acid (-28 kcal) --> 4 moles H2O (-272 kcal)

2 moles chloric acid (-46 kcal) --> 6 moles H2O (-362 kcal)

and 2 moles perchloric acid (-62 kcal) --> 8 moles H2O (-544 kcal).

The heats of formation of the chlorine acids are similar, but the heat of the reactions is much higher for the more highly oxidized acids.

A mental picture using water: imagine 4 dams each 200 feet high and 200 feet wide. The lake behind the first is filled with water 163 feet deep and is 2 miles long. The second lake is 164 feet deep and 6 miles long. The third is only 147 feet deep, but 10 miles long. The fourth is 142 feet deep but 14 miles long. A pressure gauge at the dam would indicate the height (similar to the open circuit potentials above), but if the dam is used to provide power, the fourth dam/lake will outshine the other three.

Assuming that all the electrons don't react simultaneously, the open circuit potentials are a measure of the activation energy of the first step of the overall redox reaction: in order to get zero current, you just have to stop the first step. This activation energy will be sensitive to many factors, and we occasionally oversimplify, ignoring the succeeding steps by using non-quantitative words like "strength". The total oxidation strength will be a combination of all these steps, and the open circuit potential may not reflect this combination.


Why does carbon dioxide not form addition compounds?

A fairly simplistic argument is that in CO, the lone pair on oxygen is of a much lower energy than the lone pair on carbon (because oxygen has a greater effective nuclear charge and therefore pulls electrons closer to itself). Therefore it is quite rare (very rare indeed) to find any compound that has the oxygen acting as a donor atom; CO only really reacts via its lone pair on C.

Carbon dioxide only has low-energy lone pairs on O, and is consequently a lousy Lewis base.

One could easily refine this by bringing in MO schemes, but at 3 a.m. in the morning, this is good enough to convince me.


Is there a simple assay to determine the elemental carbon content of biochar?

Assuming your charcoal is mostly:

water (evaporates)
carbon (burns)
ashes (does neither)

I would :

weigh a sample of your charcoal to get the mass $\ce m_0$
dry it in an oven at low-ish temperature (50°-80° or more, whatever doesn't burn your charcoal) for a few hours.
weigh it again to get $\ce m_1$
burn the sample in air as thoroughly as possible
weigh what's left to get $\ce m_2$

Now you can calculate :

mass of water (+ other volatiles) = $\ce m_0$ - $\ce m_1$
mass of carbon (+ other non-volatile burnables) = $\ce m_1$ - $\ce m_2$
mass of ashes = $\ce m_2$


When does ipso attack occur in electrophilic aromatic substitution?

In ipso substitution first step is usually attack of the nucleophilic species at the ipso carbon of the aromatic
ring (the carbon bearing the leaving group in this case).

Scheme 1

The first step is followed by elimination of the leaving group and regeneration of the aromatic ring.

Scheme 2

This mechanism greatly resembles the tetrahedral mechanism . The attacking species forms a bond
with the substrate, giving an intermediate, such as 1, and then the leaving group
departs. We refer to this mechanism as the SNAr mechanism.The reaction between 2,4,6-trinitrophenetole and methoxide ion involves ipso substitution.Intermediates of this type are stable salts, called Meisenheimer or Meisenheimer–Jackson salts,7 and many more have been isolated .

Scheme 3

In the reaction of dinitro compound  with piperidine,same ipso substitution operates.

Scheme 4

In the reaction between picryl chloride and hydroxyl ions same ipso substitution operates.

Ref

1.MARCH’S ADVANCED ORGANIC CHEMISTRY REACTIONS, MECHANISMS,AND STRUCTURE
SIXTH EDITION  ,Michael B. Smith ,Jerry March

2.Organic Chemistry, Volume 1, 6/E By Finar


Why does cobalt have a higher atomic mass than nickel?

Protons in each nucleus repel each other electrostatically.

As atomic number increases, there is more repulsion and an increasing advantage of spreading out the protons with neutrons.

Calcium-40 is the last (greatest atomic number) observationally-stable isotope with an equal or greater number of protons than neutrons.

Cobalt-55 and Nickel-56 are both unstable to beta decay, because it is favorable to emit a positron converting a proton into a neutron.

The natural abundance of isotopes are not necessarily in proportion to the stability, but instead stellar and cosmic processes.  As explained in the 1963 article Computer Guides Professor Hoyle to New Hypotheses:

  Working with Professor W. A. Fowler of the California Institute of Technology, he had begun with an analysis of the distribution of isotopes of the common metals, and found that nickel presented an unusual case; in the natural metal on Earth, nickel-58 is some three times more abundant than nickel-60, which is contrary to what one would expect from considerations of nuclear stability. He found the solution to this puzzle in the effect of neutrino emission from highly evolved stars, leading to extremely short life times, in the range of 20 minutes to 10 hours, and also in the phenomenon of electron-positron pair production. Nickel-58 could be made in an explosion in a large star which occurred so rapidly that the process of synthesis was cut short.

So, while the details of that article may be out of date, the reason cobalt has a higher natural abundance atomic mass than nickel is a combination of nuclear stability and stellar processes.

See On the e-process - Its components and their neutron excesses for further discussion of natural abundance ratios of Co and Ni isotopes.


The one with the lowest set of locants is correct. I.e. the one with lower first number in the locant set; if they are same, then the second one, etc. But, it is done on the ordered sets of locants, (3,6) vs (2,5) not (5,2). The substituent prefixes are then sorted alphabetically. So, the correct name is 5-ethyl-2-fluoroheptane.


Acid-base titration: Unknown base in burette or flask?

The base should be in the burette, and the acid in the flask. With this technique, the point of equivalence is obtained with a rather high degree of accuracy.


Why do you need to convert copper (II) hydroxide to copper (II) oxide before converting to copper (II) sulfate?

I would like to add to what Jason has already mentioned:

1.Copper Hydroxide is comparatively tougher to handle since it contains a lot more impurities ( when we are converting Cu(II)nitrate to hydroxide using NaOH, the impurities could be sodium nitrate, excess sodium hydroxide. )

Copper Hydroxide itself is blue, so the equivalence point (when exact amount of copper hydroxide has reacted with concentrated sulphuric acid) of the reaction would be hard to detect, and so yield would drop (since you would again have to neutralise the excess sulphuric acid before purification or conversion to copper metal).
In contrast to this CuO is a black precipitate and can be easily isolated (comparatively) by a Hirsch funnel (or any such equivalent apparatus). This improves the overall yield of further steps and eases the process of purification.

Hope this helps.


Like the comments have said, black residue is usually from organic material that has scorched.

If the pan somehow got to $840 ^{\circ}$C, then the calcium carbonate would decompose into calcium oxide and carbon dioxide, neither of which would lead to a black residue.

$$\ce{CaCO3 -> CaO + CO2}$$

Organic materials are mostly made of hydrocarbons. When you burn the organic matter, you're left with pretty much all carbon. This occurs when there is not enough oxygen to completely burn all of the carbon - when there is enough oxygen, some of the carbon combines with it to form $\ce{CO2}$ and $\ce{CO}$ and the hydrogen combines with oxygen to form $\ce{H2O}$. When there is a lack of oxygen, some carbon is left over, and this is what we see as the black substance.


The chemical reaction between lemon juice and polystyrene foam cups

It is not any acid in the juice, but limonene, the natural, fragrant, oil found in citrus fruit, and similar terpenes that are good organic solvents.

In fact, these oils are sold for the purpose of cleaning. Note, though limonene itself is considered comparatively nontoxic, after it has dissolved the polystyrene, it might be a slightly more dangerous, since any unreacted styrene monomer in solution is somewhat more toxic.


As long as everything is clamped and balanced - so that there isn't excessive weight on the joint connecting the flask to the condenser - the system should be OK.  After all, many rotary evaporators have the flask plus condenser tilted at an angle.


Which side of the C-O-C bond breaks in esters during hydrolysis in basic medium?

To answer this,Think about how an ester is formed.

In the formation of an ester, wherin you react an alcohol with an acid in presence of conc.$\ce{H2SO4}$

$\ce{RCOOH + R'OH -> RCOOR' + H2O}$

Now what we have found by replacing the oxygen with an isotope of oxygen is that

$\ce{RCOO'H + R''OH -> RCOOR'' + H2O'}$

What this reveals is that the acid loses an $\ce{OH- group}$ and the alcohol loses an $\ce{H+}$

So summing up, O atom stays with the carbon which is not attached to the $C=O$ group.


Online values of the boiling points indicate a greater bond strength in $\ce{H2O2}$.

The boiling point of water is $\pu{100 ^\circ C}$ while that of hydrogen peroxide is about $\pu{150 ^\circ C}$. Here is a list of factors that might contribute to this difference (adapted from source)

Stronger dispersion forces in $\ce{H2O2}$ (larger molar mass)
Higher dipole moment in $\ce{H2O2}$
More hydrogen bond acceptors in $\ce{H2O2}$

As for the number of hydrogen bonds, if we look at the structure of ice, each hydrogen atom of water is involved in a hydrogen bond (each oxygen is involved in two). If we assume that each hydrogen can make one bond at maximum, we would not expect more hydrogen bonds per molecule for  $\ce{H2O2}$. The crystal structure of solid  $\ce{H2O2}$ confirms that.

  But peroxide has a larger number of hydrogen bonds.

We can't infer that from the molecular structure alone. As stated above, in the solid state, the number of hydrogen bonds per molecule is the same. However, the density of solid $\ce{H2O2}$ (1.7 g/mL) is much higher than that of ice. This is true of the liquid forms as well (1.45 g/mL vs about 1 g/mL). This might allow for more dispersion forces (better contacts).

In liquid water, there are 3.4 hydrogen bonds per molecule on average (source). The higher density of $\ce{H2O2}$ compared to water suggests that atoms are closer in $\ce{H2O2}$, but you have to consider that the composition of the two substances are different. On a per-atom basis, the available volume is comparable.

  Is the strength of hydrogen bonding greater in hydrogen peroxide or water?

The question is a bit vague. If you compare one $\ce{H2O2}$:$\ce{H2O2}$ hydrogen bond to one water:water hydrogen bond, the hydrogen bond length is very similar (at least in the solid phase - source). Because the bond partners are the same atoms and the geometry around the oxygen is not that different in the two molecules, this would suggest that the strength of the bond is similar as well.


Explain isotopic uranium/perchloric acid digestion

If that is a perchlorate digestion, the technician needs more training.  The hood door must be lower despite the Official marker.  His cotton lab coat must NOT be exposed to vapors!  Sleeves should be inside long gloves.  He has a bottle of perchloric acid next to a running hot plate!  I'd have a 5-gallon bucket of water nearby if the gloves ignite.  As perchlorate digestions occasionally explode, and he's doing a lot of them, I'd wear goggles under the face shield and wear Dyneema or Spectra body armor under the lab coat.  One might have some concern for hands (longer pair of tongs) and neck, too.

He's apparently cooking off a sample of organics-contaminated uranium whatever for trace contamination or isotopic analysis.  Everything oxidizable will be gone, leaving uranium oxides.

That is NOT marked as a perchlorate fume hood!  Everything organic downwind of those fumes is a potential high explosive, and certainly a violently flammable material including its own oxidizer.  There will be massive corrosion of all metals.  You don't want to be standing, living, downwind of the exhaust stack.  OTOH, the operation is low cost/sample, and anybody off the street is a potential lab tech.


How does optimizing the molecular orbital coefficients in CASSCF improve its multi-reference capabilties?

I think you are maybe confusing how dynamical and static correlations are treated with different method. Also CASSCF by itself is not a multi-reference method.
CI in general is able to describe both dynamical and static correlation (FCI does at least). What is treating dynamical correlations (but not static) is the truncation scheme using different degrees of excitation, or example CISD, or in a similar way CCSD. Those are called single-reference methods, because they generate configurations based a single reference wave function.
To deal with static correlations you can for example use the Active Space approach (CASCI), or select configurations by hand (since you might not need to many here).
The term multi-reference refers to doing both, dynamic and static correlation, by generating excitations starting from multiple configurations. One first does a calculation for static correlation (usually MCSCF or CASSCF). Then a second calculation adds dynamical correlations by using the configurations from the first calculation (all or only the most important ones) as multiple reference points (e.g. MRCI-SD or MRCC-SD).
Optimizing the MO coefficients in a MCSCF calculation (e.g. CASSCF) is about optimizing the one-electron basis for your specific CI problem. Since you include more than just one configuration, the HF orbitals are not optimal anymore. In principle you could combine MCSCF with CISD as well, but usually CISD generates to many configurations making such an approach unfeasible.
TL;DR
The optimization of the orbitals in MCSCF does not directly improve the multi-reference capabilities. But it improves the description of static correlation, so you get a better starting point for your multi-reference calculation.
Maybe this explains better the need to optimize both sets of coefficients:

[...] The presence of several important configurations poses a difficult challenge for ab initio electronic-structure theory. The single-configuration Hartree-Fock approximation, by its very construction, is incapable of representing systems dominated by several configurations. By the same argument, methods designed to improve on the Hartree-Fock description by taking into account the effects of dynamical correlation, such as the coupled-cluster and Moller-Plesset methods, are also not suitable for such systems. Furthermore, to carry out a CI calculation 'on top of the Hartree-Fock calculation'is problematic as well since the Hartree-Fock model is inappropriate as an orbital generator: the orbitals generated self-consistently in the field of a single electronic configuration may have little or no relevance to a mutliconfigurational system.
An obvious solution to the multiconfigurational problem is to carry out a CI calculation where the orbitals are variationally optimized simultaneously with the coefficients of the electronic configurations, thereby ensuring that the orbitals employed in the wave function are optimal for the problem at hand and do not introduce a bias (towards a particular configuration) in the calculations. This approach is referred to as the multiconfigurational self-consistent field (MCSCF) method. [...] it may be used either as a wave function in its own right (for a qualitative description of the electronic system) or as an orbital generator for more elaborate treatments of the electronic structure.

Taken from the introduction to the MCSCF chapter in the book T. Helgaker, P. Jorgensen, and J. Olsen. Molecular electronic-structure theory. New York: Wiley, 2000.