Obtaining ammonium perchlorate via double displacement reaction

The Wikipedia article for ammonium perchlorate gives a hint at how this can be done. The hint is that ammonium perchlorate is curiously less soluble than sodium perchlorate, and thus with the right concentrations, you would get ammonium perchlorate to precipitate by mixing solutions of sodium perchlorate and and, say, ammonium chloride.

$$\ce{NH4Cl(aq) + NaClO4(aq) -> NH4ClO4(s) + NaCl(aq)}$$

That being said, the best advice that can be given on how to make ammonium perchlorate is

  Don't. Especially do not use heat!

Ammonium perchlorate is fairly dangerous explosive that can decompose with not much provocation into a whole heck of a lot of gas for just a little bit of solid.

$$\ce{2NH4ClO4(s) -> N2(g) + Cl2(g) + O2(g) + 4H2O(g)}$$

That this decomposition produces toxic chlorine gas should only make you want to prepare ammonium perchlorate less.

Safely producing, storing, and transporting this compound is going to require a lot of specialized equipment and training (which is expensive). The danger here is not of the type that may lead to funny smells and dizziness in poorly ventilated rooms. The hazards associated with the compound involve loss of life or limb and serious structural damage if you look at it funny. Don't. For more information into the sort of destructive power ammonium perchlorate has, consider that this is the rocket fuel that got the space shuttle into low earth orbit. This stuff has the power to level industrial complexes with the forces of an earthquake.


What is deuterium-depleted water actually used for?

When you perform 1H-NMR spectroscopy in solution, there are many cases where you want to use deuterated solvent, so that signals coming from solvent hydrogen nuclei don’t interfere with the signal from your target molecule. Hence the frequent use of deuterated water, deuterated acetone, deuterated methanol, and deuterated chloroform. Aprotic solvents are also in common use, including carbon tetrachloride.

The situation is reversed when you want to perform 2D-NMR: you then need deuterium-free solvents, so their signal doesn’t mess with the molecule you're interested in. Hence, you will be interested in buying deuterium-depleted water.

(If you're wondering about real use cases for solution deuterium NMR, you can browse the literature for articles, there are plenty. One recent exemple is this one.)


As $K$ is a thermodynamic constant, and if you define your reaction as

$A + B \rightarrow{} C + D$

leading to

$K = \frac{c_C \cdot{} c_D}{c_A \cdot{}c_B}$

then -- thermodynamically -- yes.  Especially for chelate complexes (like with EDTA), entropically favoured. This answer excludes (macroscopic) kinetics


First, when it comes to periodic boundary conditions, you should make sure that your system does not interact with itself through the boundaries.
A way to check it is to see if the electron density is low in vacuum (in case your system has some void separating surfaces).

Energies should also be checked to be converged with respect to cutoff energies, $k$ point selection, etc..
This means that you have attained the "maximum performance" you could; e.g. a well-converged result means that improving parameters do not significantly changes results.

Furthermore, convergence with respect to slab thickness is also important: as you increase slab thickness, the properties of the inner layer should approach the properties of the bulk.
How fast this happens depends on your system.

A good reference for all this is this one.


What is the world's finest powder substance?

Clusters of just a few atoms, such as gold atomic clusters, and quantum dot nanoparticles could be considered very fine powders, though they usually are suspended in another substance to keep the clusters from agglomerating. Particles this tiny have different properties from the bulk material, both because of quantum effects (there are few particles, so there is not a continuous valence band) and because more atoms are at the surface interface than are surrounded by similar atoms in the interior.

See Quantum dot for more information.

I hope this fits your definition of a "fine powder". You might also be interested in pyrophoric iron.

BTW, ruby glass is made using colloidal, i.e. super-finely divided, gold.


Which gas is easier to compress, the ideal gas or a real gas?

Let's use the van Der Waals equation for this. (Real gasses also described elegantly by the Viral equation)

You were false that "pressure of a real gas is less than that of the ideal gas". It is greater. Also the volume is smaller.

Thus if it has a higher pressure and smaller volume, it typically is harder to compress (i.e. the compression factor increases) but it also depends on temperature. Here is a nice plot of compression factors vs temp for some common gasses from the cambridge chem guide.
http://www.chemguide.co.uk/physical/kt/realgases.html


Color of precipitate formed on reaction of acetylene with Tollen's Reagent

You are correct that ammoniacal silver nitrate is Tollen's reagent. Tollen's reagent is most notable for its reaction with aldehydes to produce a silver mirror, but it also reacts with several other types of compounds, notably alpha-hydroxy ketones and terminal alkynes.

With terminal alkynes, the alkyne reacts as a acid to form a silver acetylide which precipitates. In the case of acetylene, both hydrogens can react: $$\ce{HC#CH + 2AgNO3 + 2NH4OH -> AgC#CAg + 2NH4NO3 + 2H2O}$$

The colour of the precipitate is variously described as yellow, gray-white, or white. The actual colour you observe is probably strongly dependent on the purity of the sample and the size of the precipitated particles.


I found it quite hard to follow both the question and the answer, so I am going to state a few relevant things and paraphrase the answers of @LDC3 to add some clarity.

Chemical shift (ppm) and peak area (integral)

These are independent quantities. One tells you about the environment (shift) the protons are in, and the other tells you how many protons are in that environment, compared to the number of protons in all the other environments.

That the areas need to be considered as a ratio instead of absolute values is important: if the molecule was symmetric, with an ethyl group either side of a ketone e.g. C₅H₁₀O you would see a peak ratio of 3:2 (two environments), but there would actually be 6 protons in the first environment and 4 in the second. This is a bit like empirical vs molecular formulae.

Some tips for deducing structures

1) Start with 'characteristic' shifts. These are usually the lowest and highest shifts.

Here the 4.1 ppm shift is characteristic of ester protons, i.e. -(C=O)OC H₂R (R can also be a proton too!) where, much like the data sheet above, bold is to tell you which protons are being referred to in a more complicated structure. The shift at 0.9 ppm is characteristic of protons which are attached to a standard saturated alkyl chain. This tells you that those protons aren't very near any oxygen, nitrogen or other electron withdrawing or electron donating groups.

2) Try and explain the other shifts using functional groups that you have identified from your characteristic shifts

The shift at 2.0 ppm is likely to be alkyl chain protons which are next to an electron withdrawing group, which conveniently you have already identified as an ester group by looking at your characteristic shifts! In this case, that means you can put the 0.9 ppm shifted protons at one end of your structure, because there is nowhere else to put them.

3) Check your peak assignment makes sense when you look at your peak areas

The last thing to check is which end of the molecule the 0.9 ppm shifted protons go on. i.e. is the molecule methyl propanoate or ethyl ethanoate? To do this you can look at the peak areas. For this example it is quite easy to check because there are 3 protons at 4.1 ppm, so you know that you have a methyl ester group -(C=O)OC H₃. Our R group from earlier is an H after all! The resulting structure is the one shown in the example: methyl propanoate.


Angular momentum - Question on Zare's use of the Vector Model to Estimate Probability

As mentioned in the comment by Weijun Zhou, the length of the vectors $\mathbf{j}_1$ and $\mathbf{j}_2$ does not change as $\mathbf{j}_2$ precesses about $\mathbf{j}_1$, only the total vector $\mathbf{j}$ changes its length. Knowing this just requires application of the sum and chain rule of derivations that is

$
-\frac{1}{2}\frac{d}{dt}|\mathbf{j}|^2=-2|\mathbf{j}|\frac{1}{2}\frac{d}{dt}|\mathbf{j}|
$

and the time derivatives of $\mathbf{j}_1$ and $\mathbf{j}_2$ are zero.


Is there a database which give reaction constant?

The information you are looking for is out there, but sometimes you might have to work a little for it.

The example reaction you post is a redox reaction. If we split the redox reaction into its two half-reactions, we can look up their reduction potentials (easier to find sometimes than equilibrium constants) and use the Nernst equation to calculate the equilibrium constant. I use Wikipedia for my list of standard potentials, but only because the values are identical to those in textbooks that I own.

  Oxidation half-reaction

$$\ce{Fe^{2+} -> Fe^{3+} + e-} \ \ \ E^\circ_{ox}= -0.77 \ \text{V}$$

  Reduction half-reaction

$$\ce{O2 + 4H+ + 4e- -> 2H2O} \ \ \ E^\circ_{red} = +1.229 \ \text{V}$$

  Full balanced redox reaction

$$\ce{4Fe^{2+} + O2 +4H+ -> 4Fe^{3+} + 2H2O}$$

  Cell potential

$$E^\circ_{cell}=E^\circ_{ox} + E^\circ_{red} = -0.77 \ \text{V} + 1.229 \ \text{V} = 0.459 \ \text{V}$$

  Nernst Equation

$$E^\circ_{cell}=\frac{RT}{zF}\ln{K_{eq}}$$

$$k_{eq}=e^{\dfrac{E^\circ_{cell}zF}{RT}}$$

In this equation, $z$ is the number of electrons transferred, $T$ is the temperature in kelvins, $R$ is the ideal gas constant, and $F$ is the Faraday constant. At 0 Celsius (273.15 K):

$$K_{eq}=e^{\dfrac{0.459 \ \text{V} \cdot 4\text{ mol}\cdot 96.49\times 10^3 \frac{\text{C}}{\text{mol}}}{ 8.314\ \frac{\text{J}}{\text{K}\cdot \text{mol}}\cdot 273.15\ \text{K} }}=7.565\times 10^{33}$$

If you can look up the enthalpy change or free energy change for the reaction (many of which are available at webbook.nist.gov/chemistry, then you can use a variant of the van 't Hoff equation

$$\Delta_r G^\circ = -RT\ln{K_{eq}}$$
$$K_{eq}=e^{\dfrac{-\Delta_R G^\circ}{RT}}$$


Why is 2-ethylpentane is not a valid IUPAC name, but 3-ethylpentane is?

Due to IUPAC conventions, the main alkane name is based on its longest chain. Here is an image showing different chain lengths for both alkanes. The numbers denote the longest alkane chain:

As you can see, 3-ethylpentane is symmetrical; no matter what chain you count along, the longest chain will always be 5 carbons long (pentane), and the ethyl group will always be on the third carbon (3-ethyl). Together, this is 3-ethylpentane.

However, if you look at '2-ethylpentane', there's a new longest chain which isn't 5 carbons long anymore. The image shows a new chain which is actually 6 carbons long (hexane), and there's now a methyl group on the third carbon (3-methyl). Together, this is 3-methylhexane. Because of this, '2-ethylpentane' isn't a real structure, since drawing it will leave you with 3-methylhexane as shown in the image.


In which direction does imidazole affect the pH?

Imidazole is an amphoteric molecule and can therefore act as an acid and a base.
 (courtesy of wikipedia)
It's acidity is given with $\mathrm{p}K_a=14.5$ for the proton as $\ce{N~1}$. The basic site is the lone pair at $\ce{N~3}$. Its basicity is given through the following equilibrium: $$\ce{Im + H+ <=> IMH+}$$ with a $\mathrm{p}K_a=7.05$.
(source)
You can obtain more information about this nice little molecule on its wikipedia page.

So in conclusion, it very much depends on the solution with which you are working, which effect of imidazole takes over.


Sonication to purify solid product- the crude product just disappeared

So I think we have established the problem (thank you @user6376297) . The material still contained enough iPrOH to cause it to dissolve in ether rather than give a filtrable solid because it was not adequately dried. What you need to do with your ethereal solution of product is remove the solvent under reduced pressure then dry the resulting solid under vacuum for at least 5hrs and preferably overnight. Then briefly (~1 min) sonicate the dried solid with ether and collect by filtration.
There is a lesson here: if you have a literature procedure that delivers your desired product in good yield then you follow it - without shortcuts.


Calculating the Presence of Anthocyanin in a Solution

It is not a trivial exercise because it is not a single compound in red cabbage which causes this color (anthocyanin is a class not a single compound). There is a whole master's thesis dedicated for isolating red cabbage pigments from the University of Ohio. https://etd.ohiolink.edu/!etd.send_file?accession=osu1437655932&disposition=inline

One can determine monomeric anthocyanin using UV-Vis spectrophotometry by measuring absorbances at two wavelengths. It is always good to mention what apparatus your school has.

Giusti MM, Wrolstad RE. 1996. Characterization of red radish anthocyanins. J Food Sci 61(2):322-6.

Giusti MM and Wrolstad RE. 2005. Characterization and measurement of anthocyanins by UVvisible spectroscopy. In: Handbook of Food Analytical Chemistry. RE Wrolstad, SJ Schwartz (eds). John Wiley & Sons Inc. New York .p: F1.2.1-F1.2.13


The molarity of the $\ce{HCl}$ and the $\ce{Na2HPO4}$ is the same. So 50 ml of $\ce{HCl}$ converts the $\ce{Na2HPO4}$ to $\ce{NaH2PO4}$. The remaining 25 ml of $\ce{HCl}$ essentially converts half the $\ce{NaH2PO4}$ to $\ce{H3PO4}$. So the solution is as if 6.875 millimoles of $\ce{NaH2PO4}$ were mixed with 6.875 millimoles of $\ce{H3PO4}$ in a total of 100 ml of solution.

The Henderson-Hasselbalch equation is:

$$\text{pH} = \text{pKa} + \log{\dfrac{\ce{[H2PO4^-]}}{\ce{[H3PO4]}}} = 2.1192 + \log{\dfrac{0.06875}{0.06875}} = 2.1192 \ce{ ->[Rounding] } 2.12$$

Let's check....

For the phosphate species let $C_\mathrm{P}$ be the concentration of all the phosphate species. The following equations give the fraction of each species as a function of $\ce{[H^+]}$. The equations were most easily calculated using a spreadsheet.

$$[C_\mathrm{P}] = \ce{[H3PO4] + [H2PO4^-] + [HPO4^{2-}] + [PO4^{3-}]}$$

$$\dfrac{\ce{[H3PO4]}}{[C_\mathrm{P}]} = \dfrac{\ce{[H^+]^3}}{\ce{[H^+]^3} + K_\mathrm{a1}\ce{[H^+]^2} + K_\mathrm{a1}K_\mathrm{a2}\ce{[H^+]} + K_\mathrm{a1}K_\mathrm{a2}K_\mathrm{a3}} = 0.49999$$

$$\dfrac{\ce{[H2PO4^-]}}{[C_\mathrm{P}]} = \dfrac{\ce{K_\mathrm{a1}[H^+]^2}}{\ce{[H^+]^3} + K_\mathrm{a1}\ce{[H^+]^2} + K_\mathrm{a1}K_\mathrm{a2}\ce{[H^+]} + K_\mathrm{a1}K_\mathrm{a2}K_\mathrm{a3}} = 0.50001 $$

$$\dfrac{\ce{[HPO4^{2-}]}}{[C_\mathrm{P}]} = \dfrac{\ce{K_\mathrm{a1}K_\mathrm{a2}[H^+]}}{\ce{[H^+]^3} + K_\mathrm{a1}\ce{[H^+]^2} + K_\mathrm{a1}K_\mathrm{a2}\ce{[H^+]} + K_\mathrm{a1}K_\mathrm{a2}K_\mathrm{a3}} = 4.0791\times10^{-06}$$

$$\dfrac{\ce{[PO4^{3-}]}}{[C_\mathrm{P}]} = \dfrac{K_\mathrm{a1}K_\mathrm{a2}K_\mathrm{a3}}{\ce{[H^+]^3} + K_\mathrm{a1}\ce{[H^+]^2} + K_\mathrm{a1}K_\mathrm{a2}\ce{[H^+]} + K_\mathrm{a1}K_\mathrm{a2}K_\mathrm{a3}} = 1.1272\times10^{-16} $$

Thus the assumption that the H_H equation can be used is valid.

The textbook answer is wrong.


It looks like you're correct.

This molecule has been given 2-ethyl-1-hexene or 2-ethylhex-1-ene

Another name for your molecule would be 3-methylidenedecane.

Added from Wikipedia:

The trivial name carbene is the preferred IUPAC name.

The systematic names methylidene and dihydridocarbon are constructed according to the substitutive and additive nomenclatures, respectively.


Hybridization refers to the orbitals that each atom is using to make bonds.  In $\ce{HNO}$, the $\ce{N}$ and $\ce{O}$ are $sp^2$ hybridized.  This means they form a sigma bond with an $sp^2$ orbital from each and the pi bond from the interaction of the non-hybridized $p$ orbital on each.  The $\ce{N}$ also has a $sp^2$ orbital overlap with the $1s$ of $\ce{H}$, which forms a sigma bond.  Each of the lone pairs on the $\ce{N}$ and $\ce{O}$ is in an $sp^2$ orbital.

In $\ce{HCN}$ the carbon is $sp$ hybridized, as is the $\ce{N}$.  $\ce{C}$ and $\ce{N}$ form a sigma bond with $sp$ overlap, and the pi bond from the interaction of both of their non-hybridized $p$ orbitals (to form a triple bond).  The lone pair of $\ce{N}$ is in an $sp$ orbital, and the $\ce{C-H}$ bond is from an $sp$ overlap with the $1s$ of $\ce{H}$.

None of these molecules have $sp^3$ hybridization.


How to rationalise the drop in pH when excess aluminium nitrate is added into an aqueous ammonium solution?

As andselisk has already pointed out in a comment, the hexaquo complex of aluminium is a known "cation acid"
$$\ce{[Al(H2O)6]^{3+} + H2O<=> [Al(H2O)5(OH^-)]^{2+} +H3O+}$$

A practical application is the use of solid $\ce{KAl(SO4)2·12 H2O}$ (alaun) for blood clotting during wet shave.


Stability comparison on the basis of violation of Bredts rule between 2 bicyclo carbocations

The carbocation's energy would be effectively minimised when it adopts an $\ce {sp^2}$ hybridisation scheme, that is, when it adopts a trigonal planar molecular shape. This would ensure that all the precious s orbital character of the carbon atom would be used for bonding with the other three surrounding atoms, leaving a vacant p orbital. However, this bicyclic structure prevents the carbocation from adopting this scheme, destabilising the carbocation formed. This is the basis of Bredt's rule.

On p. 435, Carey & Sundberg (2007) wrote the following:

  ... bridgehead carbocations become accessible in more flexible structures. The relative solvolysis rates of the bridgehead bromides 1-bromoadamantane, 1-bromobicyclo[2.2.2]octane, and 1-bromobicyclo[2.2.1]heptane illustrate this trend. The relative rates for solvolysis in 80% ethanol at 25 degrees Celsius are shown.

When there is greater flexibility in the structure, the carbocation is better able to adopt a structure that is closer to the preferred planar structure, thus it is better able to be minimise its energy. This results in a lower activation energy and an increased rate of reaction, explaining the data above.

Reference

Carey, F. A., & Sundberg, R. J. (2007). Advanced Organic Chemistry Part A. Structure and Mechanisms (5th ed.). Springer.


The 'crystals' are polyformaldehyde, more commonly called paraformaldehyde.

Heat it up, with some base catalyst.


Let us say that height of HCP is h then what is the distance of tetrahedral voids

Actually $h/8$ and $(7h)/8$ are correct ... but not the whole truth.  TL, DR the correct set is $\{h/8,(3h)/8, (5h)/8, (7h)/8\}$.

Here is how to get $h/8$ and $(7h)/8$:

Think of the unit cell as consisting of a "base" having parts of seven atoms, then three middle atoms entirely inside the cell, then the top layer having parts of seven more atoms.  Define $h$ as the height from the middle of the base atoms to the middle of the top atoms.  Then each of the three atoms at height $h/2$ above the base is the common vertex of two tetrahedea.  One has its base at the base of the larger cell, and since the center of the tetrahedron is one fourth of the way from the base to the apex it's height is $(h/2)/4=h/8$.  The second of the two tetrahedea is upside down, with its base at the top of the cell, and the height of its center is $h-(h/8)=(7h)/8$.

Now for the curveball: these cannot be all solutions.  We forgot to include the possibility of voids involving atoms from surrounding cells!  If we include these surrounding atoms we discover that there is a layer of tetrahedral voids at $h/8$ above the mid-level, matching those at $h/8$ above the base, and another layer at $h/8$ below the midlevel matching those just below the top.  These additional voids are hard to "see" because they involve atoms in surrounding cells.  Nevertheless they are there, the complete solution set is $\{h/8,(3h)/8, (5h)/8, (7h)/8\}$, and an equal number of tetrahedral voids (three per cell) lie at each of these heights.


Why H2O is a liquid (water) and H2S is a gas?

The reason is hydrogen bonding.

The relatively large electronegativity difference between oxygen and hydrogen (about 2.3) is the reason why the partially positively charged (try to think why they are partially positively charged) hydrogens are attracted to the lone pairs on the oxygen atoms of other water molecules.

The relatively smaller electronegativity difference between sulphur and hydrogen (about 0.5) is too small to polarise the covalent bond to a required extent. Secondly, the lone pairs on the sulphur atom are delocalised in the sp3 orbitals and the d orbitals - an option unavailable to oxygen. So, the “negative charge” of the lone pairs is also distributed and the interaction is lowered.

So, the hydrogen bonding in water causes the water molecules to be associated with each other due to intermolecular forces of attraction, whereas in the case of hydrogen sulphide, these forces of attraction area much weaker.

So, lesser energy is required to overcome the forces of interaction between the hydrogen sulphide molecules than those between water molecules. This energy is available at room temperature and hence, hydrogen sulphide is a gas, while water is still a liquid.


Why does green ferrous sulfate solution change color to yellow upon addition of hydrochloric acid?

Dry ferrous sulfate heptahydrate is green(ish) and is expected to make a green(ish) aqueous solution.

Anhydrous ferrous chloride is described as green to yellow, and the dehydrate and tetra hydrate are green to blue-green (CRC Handbook).

Now color is broad and continuous (see the visible spectrum),

not digital, like how many fingers am I holding up. This set of curves relating ferrous to ferric comes from https://www.sciencedirect.com/science/article/abs/pii/S0143720816309263 There's a continuity, not a change that can be described as this color to that color.

There is also an effect called dichromatism (not dichroism) in which the perceived color of a material that has some depth will change depending on the depth/thickness or concentration observed (https://en.wikipedia.org/wiki/Dichromatism).
In addition, ferrous ion picks up oxygen easily and becomes ferric (ferric chloride is deep red in concentrated solution (~20% or more), but quite definitely yellow in more dilute solutions (~5%)). The reaction 2Fe$^2$$^+$ + 0.5 O$_2$ + 2H$^+$ --> 2Fe$^3$$^+$ + H$_2$O is favored with lower pH. So your supposition that the (greenish) ferrous ion in solution is picking up oxygen and producing some yellowish ferric ion is almost certainly true. As ferric is formed green will diminish and yellow will increase; the spectral curve will just slide over from the green to the yellow.
However, because there is no quantification mentioned, everything is ishy. Greenish, yellowish, pHish, and yellow and green are both in the green and in the yellow solutions. You also don't mention specifically what other acids you tried.
To explain this to a class, you need some numbers. What you could do is get the visible absorption curve of fresh ferrous sulfate of known concentration (do it at two concentrations, at least), and then get the spectra after addition of (at least) two levels of known concentration HCl. Versus time, because the oxidation will probably take a few minutes in order to get enough exposure to air/oxygen. That will suggest how much of the color change is due to oxidation.
But you should also take the known concentrations of ferrous sulfate and add sodium chloride at the same molar concentrations as the HCl, to see how much color difference occurs because of chloride complexation while keeping the pH essentially constant. These solutions will yellow in time also, but slower than the acidified solutions.
These are some of the variables in your experiment, but how to express this is a short time will be up to you.


Clarification of textbook concepts relating to "perfect", "ideal", and "real" gases

The authors attempt to explain that in many sources the term "ideal gas" is used in place of "perfect gas" to indicate a gas following the ideal gas law and which has the property that the molecules do not interact. They equate the absence of an intermolecular interaction with the statement that the "interactions are zero". In other words, there is no intermolecular potential between the idealized gas molecules in a "perfect gas". These idealized molecules lack a volume, so that even steric (excluded volume) interactions expected for hard spheres are absent. They are "point particles".

The authors further explain that it is preferable to use the word "perfect" rather than "ideal" because the word "ideal" is used in other contexts where the condition of no interaction is not observed, ie during ideal mixing. Ideal mixing occurs when different molecules have the same volume and intermolecular potential in the pure and in the mixed states.

Opinion-based aside

BTW I've always been partial to the term "ideal gas". IMHO the term "perfect" seems to imply that the molecules excel in some way, whereas "ideal" simply means that it is the product of human minds, an abstraction, and not necessarily a real thing.

IUPAC does not to have any qualms about continued use of the term "ideal gas", so I don't see reason to stop using this term.


In an absorption experiment a collimated beam of light is passed through the sample. The intensity of this is measured before and after the sample. The absorption is measured as optical density which is $D=[c]\epsilon _{\lambda} l$, where $[c]$ is the concentration, $\epsilon_\lambda$ the extinction coefficient at wavelength $\lambda$ (a measure of the strength of absorption), and l the path length through the sample. At each wavelength the measured light intensities are $I=I_0\exp(-D)$ where $I_0$ is the initial intensity. This is called the Beer–Lambert law.

Now as you mention the sample is placed into its excited state where it remains for some period of time. This depends on the molecule being studies but may be from picoseconds (relatively rare) to several nanoseconds to microseconds (also rare). In an absorption spectrophotometer the light intensity used is low, so the excited state population is always very small. As any fluorescence produced is emitted over $4\pi $ radians (in solution) the amount captured by the detector will be small, just on geometric grounds as the detector is placed some distance away from the sample and the light passes through slits before entering the detector.(The slits & grating give spectral resolution). Also note that the fluorescence is mainly at a longer wavelength than absorption but the zero zero line in absorption and emission can overlap. The geometric reason just given suffices to make absorption accurate even in this case.

In laser experiments the situation can be different, particularly when pulsed lasers are used. It is quite easy to observe 'gain' in  the sample, this means that transient signals that are more intense than the measuring (probe) beam are observed. This is due to stimulated emission caused by the probe beam. It is also quite easy to see amplified stimulated emission (ASE) from a cell containing a fluorescent dye if the laser beam is focussed.

EDIT:

An explanation as to why stimulated emission is not normally observed in an absorption experiment and why fluorescence is observed under similar conditions is given below.

In the simplest case of two levels $1$ and $2$ , where $1$ is the lowest level the rate of absorption from $1$ is $N_1B_{12}\rho(\nu)$ where $N_1$ is the population of level $1$, $B_{12}$ the Einstein coefficient for (stimulated) absorption from level $1$ to $2$ and $\rho (\nu)$ the energy density per unit frequency interval ($\pu{Jm^{-3}/s^{-1}}$). The rate of transitions from $2$ to $1$ is $N_2[B_{21}\rho(\nu) +A_{21}]$ where $A_{21}$ is the rate of spontaneous emission from level $2$ and $B_{21}$ the coefficient of induced (stimulated) emission. Note that $A$ does not depend on the number of photons present. In is convenient to assume that the molecule is bathed in thermal radiation that follows  Planck's blackbody law at temperature $T$,

$$\rho(\nu)= \frac{8\pi h\nu^3}{c^3}[e^{h\nu/(k_BT}-1]^{-1}$$

At equilibrium the rate of change $dN_1/dt=dN_2/dt=0$ then

$$ \frac{N_1}{N_2}= \frac{B_{12}\rho(\nu)}{B_{21}\rho(\nu) +A_{21}}$$

The ratio of populations is also given by the Boltzmann distribution, also at temperature $T$

$$ \frac{N_2}{N_1} = \frac{g_2}{g_1}e^{-h\nu/(k_BT)} $$

where $g_1,\,g_2$ are the degeneracies of the two levels.

Combining these last two equations and solving for $\rho$ gives

$$\rho(\nu) = \frac{A_{21}/\,B_{21} }{ \frac{g_1B_{12}}{g_2B_{21}}e^{h\nu/(k_BT)}-1}$$

For this relationship to be true $g_1B_{12}=g_2B_{21}$ and $A_{21} = n(\nu)B_{21}h\nu$ where

$$n(\nu)= 8\pi\nu^3/c^3$$

and is the number of radiation modes per unit volume per unit frequency.

This relationship means that the ratio of induced to the spontaneous emission rate (leading to fluorescence) is equal to the number of photons in this mode.  The figure shows that in the visible part of the spectrum,  $\approx 400$ to $700$ nm at at thermal emission at room temperature the number of photons mode is somewhere between  $10^{-10}$ and $10^{-100}$, the exact number does not matter much as it is so small. Thus this means that the spontaneous emission rate (i.e. fluorescence) far exceeds that of stimulated emission. Under normal circumstances where the radiation is weak, as in a spectrophometer, and $N_2 \ll N_1$, the linear absorption spectrum is obtained.  If many photons can be concentrated into a few modes,  then for these modes, stimulated emission may exceed spontaneous emission; the result is a laser amplifier.

The figure shows the average number of photons/ mode with temperature, assuming Planck black body distribution vs. the photon frequency. The vertical lines are approximately the range of visible light. The horizontal dashed line is at $300$ K.


There are a number of vegetable gums, such as carageenan, pectin and gum arabic, and man-made methyl cellulose which might be compatible with low concentrations of $\ce{H2O2}$, which you can determine easily. Any impurities, particularly transition metals, may make $\ce{H2O2}$ decompose, so it might be best to keep the gelling agent and peroxide separate until just before use.

If for oral use, this site suggests 0.75% to 1.5% solutions, one quarter and one half the strength of the common antiseptic N.B. $\ce{H2O2}$ bleach is much stronger and is not for oral use. Note that there are dangers with use of peroxide as a mouthwash.


Ring contraction in a carbocation due to ring strain and back bonding

The rearrangement has little to do with ring strain and a lot to do with resonance stabilization.

The ring strain in a cyclobutane ring is ~26.3 kcal/mol, while the ring strain in a cyclopropane ring is ~27.5 kcal.mol (reference).  This isn't much of a difference, certainly not enough to drive a reaction one way or the other.

The reason that cyclobutyl carbocations generally rearrange to cyclopropylcarbinyl carbocations is due to resonance stabilization.  Here are some drawings that may help, if you build a model and look at the model that will help even more. The $\ce{C-C}$ bonds in a cyclopropane ring are approximately $\ce{sp^{3.74}}$ hybridized (see here), said differently, they have a lot of p-character in the bond.  In fact, the p-character is so high that a cyclopropane compound will absorb bromine much like an olefin. If that cyclopropane high p-character bond (orbital) can align itself with the p-orbital on the carbocation center, it will stabilize the carbocation, just like a p-orbital on an adjacent double bond (the allyl system) stabilizes a carbocation.

here's another representation

Typically we discuss two conformations of the cyclopropane ring with the carbocation p-orbital, a bisected and a perpendicular conformation.  In the bisected conformation the cyclopropane $\ce{C-C}$ bond lies in the same plane as the carbocation p-orbital.  The carbocation p-orbital and the $\ce{C-C}$ orbital which resembles a p-orbital (because it is high in p-character) overlap!  In the perpendicular conformation, the cyclopropane $\ce{C-H}$ bond is in the same plane as the carbocation p-orbital, the $\ce{C-C}$ bond is not - there is no carbocation stabilization by the cyclopropane ring in this conformation.  Molecules where the bisected conformation can be attained show stabilization due to the stabilizing overlap between the carbocation p-orbital and the p-like $\ce{C-C}$ cyclopropane ring bond.  Sometimes we use resonance structures such as the following to denote this stabilization.


Enthalpy change defined at constant pressure only?

Is change in enthalpy defined only at constant pressure?

No. The definition of $\Delta H$ is $\Delta H = \Delta U + \Delta (PV)$  Simple as that.

  I know Q-W=del(U) And Q at constant pressure equals del(H) (enthalpy
  change).

  Can Q (heat given to the system) be used interchangeably with del(H) in the first equation?

No.  Only if the applied pressure is held constant during the change.

  Is del(H) =nCp del(T)? How?

No.  Only for an ideal gas.  For real gases, this equation is not correct outside the limit of ideal gas behavior.  For an ideal gas, $\Delta U=nCv\Delta T$ and $\Delta (PV)=nR\Delta T$, so $\Delta H=n(C_V+R)\Delta T$.  And then $C_p=(\partial H/\partial T)_P=C_v+R$, so $\Delta H=nC_P\Delta T$.

For a real gas, liquid, or solid,
$$dH=nC_PdT+\left[V-T\left(\frac{\partial V}{dT}\right)_P\right]dP$$
Note that the term in brackets is zero for an ideal gas.


What would be a stable way to hold lithium hydride powder?

If the LiH needs to be exposed to ambient conditions, it should be enclosed is a moisture-impermeable container, e.g., a glass ampoule which is closed off. Glass can be blown quite thin - but then you need to be quite careful. As DrMoishe suggested, PE or PP would be good, but I'm thinking of it as a closed tube, not as a binder.


What are typical runtimes for CASSCF calculations?

There is nothing trivial about MCSCF calculations because it is hard to predict a priori how long a calculation will take. There are well-defined equations for calculating how many determinants

$$
D(n,N,S) = \binom{n}{N/2+S} \binom{n}{N/2-S}
$$

or configuration state functions (CSFs)

$$
D(n,N,S) = \frac{2S+1}{n+1} \binom{n+1}{N/2-S} \binom{n+1}{N/2+S+1}
$$

will be required for your problem; $n$ is the number of active orbitals, $N$ is the number of active electrons, and $S$ the total spin. The calculation cost will scale with this number and can be further reduced with the use of point group symmetry. (Actually, depending on your system and the property of interest, calculations without symmetry can be meaningless, like for excited states and their energies.)

These equations should make it clear that the cost grows very rapidly, considering that for CASSCF, a full CI is being performed within your chosen active space. This is not what makes MCSCF calculations tricky in practice. Here is a plot of the total energy after the first 30 iterations.

This is because the occupation numbers are very close to 0/1/2, and they aren't changing with each iteration. Notice that the occupation numbers have inverted themselves, and repeated at every step is Rot=23,19. Orbital 19 is within the active space, but 23 is external. This is an indication that the orbitals in the active space are not the correct ones. From the paper:

  the 12 or 13 valence electrons of the neutral or anion, respectively, were distributed among 14 valence active orbitals: 3p of sulphur plus 3d, 3d' and 4s orbitals of iron. With this active space we found that all important correlating orbitals are included in the active space.

MCSCF is not a black box; it's one of the few quantum chemical methods left that requires chemical intuition to even get started. This a (12,14) active space; the one in your calculations is (12,8). The procedure I use for running MCSCF calculations is generally:

Converge a single-reference calculation. This could be HF, DFT, MP2, etc. The usual recommendation is to form MP2 natural orbitals, then check the occupations numbers. If the occupation deviates more than 0.02 from an integer value, then it should probably be in the active space. This is in addition to considering the electrons and orbitals you wanted to correlate to begin with, along with antibonding partners. For example, in ethylene this is probably the pi bonding/antibonding pair, leading to a (2,2), and for 3d metal complexes this is whatever coordination (anti)bonds are present, with potentially the double d shell. I've had some luck with using DFT orbitals for metal complexes, where perturbation theory might be a bad idea due to near-degeneracies or cost. The goals here are to confirm your active space and provide the best possible starting orbitals for MCSCF.
Look at your orbitals. Of course, read the Mulliken or Lowdin population analysis, but from personal experience, I've caught many potential mistakes by doing this. It also helps when visualizing how the single-state, single-reference orbitals transform to the natural orbitals (potentially state-average) that result from MCSCF calculations.
Reorder your orbitals if necessary before running the MCSCF. Except for the most trivial cases, the orbitals that belong in the active space are not the orbitals that come straight out of your single-reference calculation. If they were, then you might not be running MCSCF to begin with. I also tend to look at the reordered orbitals one more time before doing MCSCF to confirm I haven't made any mistakes.
Read in the previous set of (reordered) orbitals into a separate MCSCF calculation and cross your fingers. Monitoring the occupation numbers of your active space is a good guide for seeing how the calculation is converging.
Do whatever post-processing you need for your project. For me, this is looking at final occupation numbers, the orbitals themselves, and then doing MR-CISD or some other multireference correlated calculation.

Doing an auto-occupation procedure without starting from previously-converged orbitals means that 1. you don't have any guarantees about the contents of your active space, 2. you start from orbitals that are close to Hartree-Fock in quality, and 3. the calculation has to do far more work in optimizing both sets of variational parameters (the MO coefficients $\{C\}$ and the CI coefficients $\{c\}$) since the internal and external orbitals (where the CI coefficients are frozen) are still far from convergence using a single-reference method.

To walk through part of this workflow, I ran a PBE0/aug-cc-pVTZ calculation (converges much faster than the SCF before MP2) to look at the Lowdin MO populations:

! pbe0 aug-cc-pvtz cc-pvtz/jk ri rijk tightscf usesym zora

%output
 print[p_orbpopmo_l] 1
 end

* xyz 0 5
Fe      0.000000      0.000000      0.000000
S       1.960000      0.000000      0.000000
*

The MO indices are S 3s: 14, Fe 4s: 20, Fe 3d: 15,16,17,18,19, and Fe 3d': 33,34,36-40. Automatic occupation would have gotten the occupieds right, but the virtuals wrong. I'm not sure which of the 7 virtual MOs that might be the Fe 3d' are the correct ones; there is probably some trial-and-error here, but these should be visualized. Another point is that I've only looked at the ordering for alpha-spin orbitals; for difficult cases such as transition metals, the spatial ordering for beta-spin orbitals can be very different. The reordering for alpha- and beta-spin orbitals might be different. Again, this is speaking from experience.

The final exhibit is the RI-MP2/aug-cc-pVTZ natural orbitals that might enter an MCSCF calculation based on their occupation numbers:

N[ 15]( A2) =   1.97761944
N[ 16]( A1) =   1.97753198
N[ 17]( B2) =   1.97009578
N[ 18]( B1) =   1.97009578
N[ 19]( A1) =   1.03828526
N[ 20]( A1) =   0.99215102
N[ 21]( B1) =   0.98317529
N[ 22]( B2) =   0.98317529
N[ 23]( A2) =   0.02013554
N[ 24]( B2) =   0.01947713
N[ 25]( B1) =   0.01947713
N[ 26]( A1) =   0.01816184

To be thorough, this should be combined with looking at the Lowdin MO populations of these, which might require reading them into another (SCF) calculation and not performing any iterations.

However, it is well-known that converging MCSCF equations is generally difficult. In my limited experience, DALTON, GAMESS, and Molcas have less trouble than ORCA. I haven't used Molpro for MCSCF. Which program you choose dictates what post-MCSCF calculations can be performed. The DIIS algorithm in ORCA, in particular, can stall even when it shouldn't. If you can afford it, the Newton-Raphson algorithm (switchstep nr), where the active space orbital Hessian is calculated at each step, works very well to force convergence once SuperCI is done. ORCA makes it easy to get started with CASSCF, and it has plenty of knobs, but it has a very slow integral engine, doesn't prevent you from doing stupid things like GAMESS, and more advanced features are poorly documented. The workflow described above (namely the plotting orbitals and rotating them) becomes very tedious as well, but that has more to do with MCSCF in general.

Here are some good resources that describe both practical and theoretical aspects of performing MCSCF calculations: 1, 2, 3, 4.


What are good lubricants for Bunsen burners?

We've been running into a lot of problems the last several years with our laboratory Bunsen burners becoming jammed and/or corroded. I've just come to find out that our lab tech has been lubricating the burners with glycerol.

Does the issue itself seem to have been caused by the glycerol?

Admittedly, we don't use many Bunsen burners nowadays as they're seen as being fairly unsafe in the kind of organic lab I work in where we handle large amounts of solvent, however we do have a well vented hood with several burners in.

The ones we do have are very old indeed (pre 50's) and still work fine without the need for any maintenance (i.e. greasing).

Bunsen burners themselves are incredibly simple pieces of kit, if you do have issues with corrosion (i.e. because of an overly humid environment or harsh chemical vapours), the most sensible thing would be to disassemble, remove the corrosion and re-assemble.  The use of any kind of grease should be un-neccesary, especially as very few of them (other than very specialised products) can withstand long term exposure to heat without degradation.


Can I safely use a separatory funnel to extract the precipitate from a solution?

Good thought, but Waylander is correct.

However, you might consider centrifugation to consolidate the ppt, then pour off the supernatant. If you don't have a centrifuge, and the quantity of liquid is not too large (100 mL or less), you could make your own, since it seems that your ppt settles fairly fast.

Put the solution in a tough plastic bottle, cap it, and get a pretty strong string or light rope, tie the bottle SECURELY and twirl it over your head. Maybe outdoors is better for this, unless you are very careful. A few gentle swings (~20) should be enough, if it is going to work at all. I have had some success with this method a few times.

Secure tying is key, since the extra Gs will put a strain on your knots. I did something like (vertical swing, about 4 foot circle) this inside a room with 12 foot ceilings, and the plastic bottle hit the ceiling and splashed (I did not have the cap on). Fortunately, the stain was not very noticeable, but I did lose some of the material.


Clarification on absorption spectra and crystal field theory

You would expect a sharp peak for a well-defined structure which has bond lengths and angles that do not significantly change. In a crystal structure at low temperatures you may be able to observe sharp peaks rather than broad bands.

In solution, however, there is a number of effects that broaden the peaks, making the entire thing less sharp:

aqua ligands can dissociate and reassociate; if the photon hits the complex while the structure is different, the wavelength will be, too
typically, the ligands of a complex vibrate both symmetrically and antisymmetrically. Both vibrations lead to structure changes, therefore electronic changes and thus changes in the absorption wavelength.

Different solvents will give different peak broadening (I saw a graph once, but I can’t remember where it was).

The titanium(III) spectrum is also more complex, by the way. If you look closely, you can see a shoulder on the low-energy side of the main peak. This is because the excited state is Jahn-Teller distorted, and there is not one but two transitions possible that slightly differ in energy. (The ground state is very, very weakly Jahn-Teller distorted, because there is not much energy to be gained.)

After a molecule has absorbed the energy of a photon and went into an excited state, there is not only the simple rerelaxation that would emit a photon of the same wavelength to return to the ground state. Rather, by a set of complicated mechanisms, the energy can be transformed into vibration (more and more vivid than before) and thus slowly released in form of heat. That is why a solution that absorbs light will warm slightly.


I just looked in the literature, and it has been done even without catalyst in a "reactive distillation column" at high temperatures (patent US201113194873). If you can remove the water formed, you can technically drive the reaction forward, even if it is slow. A catalyst will just make that process easier, requiring less heat and time.

Phosphoric acid is a much weaker acid than sulfuric acid, and it does not have the same dehydrating properties. However, if you have a proper setup, it should do the trick. A big problem with phosphoric acid is that the most concentrated you can normally get is 85%, which means you still have to remove 15% of water somehow. If you don't, you'll likely hydrolyse any aspirin formed and get your starting material back.


Does glucose react with Brady's reagent?

From Wikipedia:

  The reaction of $\ce{NH2OH}$ with an aldehyde or ketone produces an oxime.
  $$\ce {R2C=O + NH2OH·HCl + NaOH → R2C=NOH + NaCl + H2O}$$

For the penta-acetate ester of glucose not to react with hydroxylamine would indicate that the ester is more stable, or there is steric hindrance to the reaction. Since hydroxylamine is heat sensitive, I am guessing that you did not heat the reaction. Given time, you should see the oxime form.

From Wikipedia:

  A positive test is signaled by a yellow, orange or red precipitate (known as a dinitrophenylhydrazone).
  $$\ce {RR'C=O + C6H3(NO2)2NHNH2 → C6H3(NO2)2NHNCRR' + H2O}$$

2,4-DNP also reacts with an aldehyde (or ketone), so both glucose and the penta-acetate ester of glucose should react. If you get a different color for the penta-acetate ester, it is due to the substituents on the glucose.


Was nitroglycerin invented as an explosive?

Probably ... but he doesn't actually say

$Nitroglycerin^1$ was discovered by Dr. Ascanio Sobrero in 1847. Both Sobrero and Alfred Nobel were students of Prof. Théophile-Jules_Pelouze.

It is rather unfair to describe chemistry of this era as "random". 1847, of course, was mid-1800s rather than early 1800s, and it was some two decades after Wöhler's famous demonstration that organic chemicals could be synthesized from inorganic ones. Organic chemistry of this era was more in its adolescence than its infancy and was being pursued in a systematic scientific manner with publications in peer reviewed journals.

Some of these journals have now been digitised and are searchable, so we have some idea of Pelouze's work, and his students. Much of it involved preparation and study of nitro-esters of simple carbohydrates. They don't seem to say why, but in the context of the era, it is very likely they were looking for practical explosives. Braconnot had discovered nitrostarch in 1832; Pelouze apparently first prepared nitrocellulose in 1838, but didn't recognise it as a new substance. Then in 1846 -- just one year before Sobrero nitrated glycerin -- no fewer than 3 chemists prepared and described nitrocellulose in the same year, with Otto being first to publish.

For a long time, it appeared that nitrocellulose was not a practical material as it is very difficult to purify, and extremely unstable when impure. So it is not surprising that a year later, we find one of the early nitrocellulose investigators now conducting a program of testing nitrates of a range of simple organic compounds, especially carbohydrates. (Cellulose itself is a carbohydrate.) They don't say exactly why in anything I have read. It is possible that they are using nitrations to investigate fundamental chemistry; for example, such a process does help to identify the structure of simple sugars. However, they test them all for explosive properties and refer to them all as "fulminant" (i.e. explosive.)

Sobrero's paper on his discovery of nitroglycerin has been digitised. (Note: article is in French.) He mentions earlier work with sugars and notes that glycerin is not a carbohydrate, but doesn't seem to say what inspired him to try to nitrate it except that it is "analogous" with other materials that his group had nitrated. Incidentally, in this paper he describes purification methods at some length, and also discusses the toxicity:

  ... for it is sufficient to take a very small quantity (which can be taken
  by slightly wetting the tip of the little finger) on the tongue to
  feel a powerful migraine for many hours. This action on the human body has
  been observed by many people in my laboratory ...

However, he doesn't mention its extreme shock sensitivity. Later, Sobrero came to the conclusion that nitroglycerin was too dangerous to use, and it was, of course, his colleague Alfred Nobel who did the considerably longer and more complicated studies to find a way to render it practicable. (The same situation occurred with nitrocellulose; discovered by Schönbein, Böttger and Otto, but it was Abel who did the real spade work to make a practical industry for it.)

Note that whilst many of the substances in this answer were historically called nitroX where X is the original moiety, none of them are nitro compounds in the modern sense. Rather, all explosives discussed in this answer are nitrate esters. So nitroglycerin, for example, should really be called glyceryl trinitrate.


TKN: Why are they called Missouri tablets?

As suggested in the comments, the predecessor of the method that uses "Missouri" tablets was developed at the University of Missouri (Columbia, Missouri, USA). I believe the first report was from 1972 and named the method as the "Missouri Automated Nitrogen Method (MANM)". They also refer to it as the "Missouri Technicon" method because it uses a Technicon AutoAnalyzer instrument. Many variations of this method have been described, and the specific formulation of Missouri tablets is for one of them.

An automated spectrophotometric method, utilizing Technicon AutoAnalyzer modules, has been developed to determine total nitrogen in fertilizers containing only ammoniacal, nitrate, and urea nitrogen. This colorimetric system employs the Berthelot ammoniaphenate-hypochlorite reaction. A homogeneous chromous/titanous reduction system for the automatic reduction of nitrates was interfaced with the digestion unit and color development manifold, resulting in a total nitrogen system for fertilizers with an effective analysis rate of $\pu{20-25 samples/hr}$. A new sample retrieval system removes digested samples from the helix. The totally automated system was optimized to obtain the highest sensitivity for nitrogen. Data were compared for $\pu{458 samples}$ analyzed by the automated method and the comprehensive nitrogen method (CNM), 2.053–2.054. The average difference between the 2 methods (MANM – CNM) was +0.04% nitrogen. The average relative difference was 1.12%. The average value for $\pu{140 samples}$ of $\ce{KNO3}$ primary standard analyzed by the MANM was 13.84% nitrogen (theoretical content 13.85%). The results of these standards ranged from 13.58 to 14.14% nitrogen and the standard deviation was 0.11%. The relative standard deviation was 0.80%. Of fertilizer samples received for analysis, less than 1% contained insoluble organic nitrogenous material which was unsuitable for analysis by the MANM. These few samples were not included in the statistical evaluation. The MANM is rapid, accurate, and precise and is a convenient and dependable automated method for the determination of total nitrogen in fertilizers containing only ammoniacal, nitrate, and urea nitrogen.
 Source: Charles W. Gehrke, John S. Killingley, Larry L. Wall, Sr. (1972) Missouri Automated Nitrogen Method (MANM) for Fertilizers. Journal of Association of Official Analytical Chemists 55:467-480. https://doi.org/10.1093/jaoac/55.3.467

A rapid automated method for total protein nitrogen has been developed, using the Technicon AutoAnalyzer with block digestor. Ammonia is determined by the automated ammoniasalicylate reaction at a rate of $\pu{40 samples/hr}$. A number of variations in digestion parameters have been evaluated. Hydrogen peroxide was used as a digestion accelerator. A salt-acid ratio of 1:1 and a block temperature of $\pu{425 ^\circ C}$ were chosen. The catalysts evaluated included $\ce{HgO, CuSO4, SeO2,}$ and $\ce{TiO2/CuSO4}$. Maximum nitrogen recovery in the shortest time (30 mill) was achieved with HgO as the catalyst. The results from multiple analyses of 10 experimental samples of different refractoriness with the block digestor method, using $\ce{HgO}$ or $\ce{CuSO4}$ as a catalyst, compared well with the results by the official AOAC Kjeldahl method. The accuracy, precision, economy, and saving of space offered by the Missouri-Technicon block digestor method make this an attractive alternative to classical Kjeldahl analysis for large numbers of total protein determinations in samples of different refractoriness. The AutoAnalyzer cartridge manifold is simple and reproducible in its performance, and a large dilution of the sample is made on dialysis which eliminates the matrix effects from the sample digests. At least $\pu{250 samples}$ can be analyzed in $\pu{9 hr}$ with 1 instrumentation setup.
Source: Larry L. Wall, Sr., Charles W. Gehrke (1975) An Automated Total Protein Nitrogen Method. Journal of Association of Official Analytical Chemists 58:1221-1226. https://doi.org/10.1093/jaoac/58.6.1221


Mixing and matching different anions and cations can easily produce different compounds.  For instance, sodium and potassium chloride, and sodium and potassium nitrate, are all soluble in water and do not react to form a precipitate. When the aqueous solution dries, all four compounds are possible.  So, mixing sodium chloride and potassium nitrate will result in 2 anions and 2 cations in solution, allowing all four pairs to possible precipitate.

Now, will the precipitate as pure compounds or not is another question.  That requires looking at the relevant phase diagrams and seeing what is thermodynamically favored (and consideration of kinetic factors like how fast you evaporate the liquid).

As an example, the phase diagram for KCl and NaCl exhibits a miscibility gap starting at 505C.  This means that below that temperature they will phase separate into a Na-rich chloride and a K-rich chloride.  At 100C the concentration of the 'impurity' phase is less than 1%.

The nitrate diagram also shows a two phase region below 100C.

So, mixing (Na,K) (Cl,NO3) will result in all four precipitating out.


Like what @GeoffHutchison said,

$\ce{SiO2}$ is not an ion. It is a network solid. Its net charge is zero.

From charge balance, since oxygen holds a -2 formal charge, then silicon must hold a +4 formal charge in order to balance out. $(+4) + (+2 \times -2) = 0$


How to calculate ∆rGº from entropy and ∆fGº in different temperatures?

For a small temperature change like that, you can make use of dG=-SdT.  Determine $\Delta rG^0$ and $\Delta r S^0$ at 298, multiply the latter by 10 , and subtract it from the former to get $\Delta rG^0$ at 308K.


I think this may be an overreaction.  First, MSDS are designed for workers working with industrial quantities in industrial conditions.  Second, the danger comes from concentration of the peroxide from evaporation or, more typically, distillation.  Small quantities of peroxides in solution are not usually an explosion hazard.  Third, I have never seen isopropanol listed in lists of hazardous peroxide prone chemicals as used by lab safety departments (ethers are the typical risk here).  The closest I have seen is sec-butanol, which is the secondary alcohol from butanol, and I've only seen this listed once, in a list that seemed much more cautious than some other lists.  Isopropanol is the secondary alcohol from propanol.  Given your storage, cool and dark, I think you are fine.  Remember that 91% isopropanol is sold in drugstores, though these may contain stabilizers and you larger bottles may not.

But having said all this, Lighthart's brief answer is correct though probably unlikely.  If you DO see any crystals, don't even try to open the bottle (crystals in the threads are the most common source of explosion).  But again, seems unlikely.

If you are still concerned, there are peroxide test strips you can buy that are very easy to use and can tell you what concentration of peroxides are present.


What happens to air at 1000 degrees Celsius

At $\pu{1000^oC}$, oxygen and nitrogen will not react substantially or probably even noticeably if at all. Besides that, whatever organic material you are using for a hot air balloon will most definitely ignite immediately. The average temperature that most organic materials char and eventually ignite is $\pu{270^oC}$. At almost double the kinetic energy at $\pu{1000^oC}$, most things will almost instantaneously ignite. Note that in air, temperature diffuses rather quickly so even if the fire or whatever you are using to get to $\pu{1000^oC}$ will probably be significantly colder when it reaches anything flammable. In fact, most fuel used in normal hot air balloons burn at over $\pu{1000^oC}$, but it diffuses very quickly that the average temperature of the air in the balloon is around $\pu{110^oC}$


What is the mass fraction of C2HCl3 in the liquid phase?

I'm sorry. I mislead you. Temperature changes, and the boiling points of the liquid have nothing to do with this problem as I erroneously commented.

The correct approach is to simply apply Raoult's law. For some temperature T, let

    $p^*_i =$  partial pressure of some pure compound $i$
    $x_i =$ mole fraction of the compound $i$ in solution.

then let $p_i$ be the partial pressure of compound i above a liquid mixture. For an ideal system then:

    $p_i = p^*_ix_i$

The problem states "The mole fraction of $\ce{CHCl3}$ in the vapor phase is 1."

As you deduced, since there is no $\ce{C2HCl3}$ in the vapor phase there can't be any $\ce{C2HCl3}$ in the liquid phase either.


Thermochemistry questions regarding enthalpy

You should not separate the species on the left and assign enthalpy variations to each of them. On Scenario 1, you say that 1 mol of methane reacts with 2 mols of oxygen, and the variation of enthalpy for this is −889 kJ (since it is 1 mol of methane). If you had 2 mols of methane, and 4 mols of oxygen, $\Delta$H would be 2*(-889) kJ, and so on. As you mentioned, if only 1 mol of oxygen were available, only 0.5 mol of methane would react and $\Delta$H would be 0.5*(-889) kJ
So, for Scenario 2, you said it was given that

the oxygen produces 400kJ of energy

This probably means that, for each mol of oxygen that reacts with ethanol, 400 kJ of energy is released. So yes, if you wanted the $\Delta$H for the reaction, you should multiply by 3 and get that $\Delta$H = -1200 kJ mol-1.
This is in agreement with the measured molar enthalpy of combustion of ethanol (-1211 kJ mol-1)


They were writing it fancy in old times.
Sodium acetate would have been "essigsaures Natrium", potassium sulfate "schwefelsaures Kalium" and so on.

Therefor, "ölsaures Alkali" is an alkali salt of oleic acid (Ölsäure).


What library/program can draw molecules in 3D programmatically?

The "jmoldata" command (using JmolData.jar from Jmol) converts xyz/pdb and many other molecule formats to nicely rendered 3D images based on the user-provided Jmol scripts.

(answer based on the hint from the user @porphyrin)


Separating two sugars according to their heavy atoms - Steered Molecular Dynamics

I have realized the problem. The force constant was too low. Increasing it to 25.0 from 10.0 sorted things out.


Are there natural or artificial enzymes that break down metals?

Silly answer

There is no enzyme that grinds down an Airstream, but an alligator might do the job!

Serious (but disappointing) answer

In nature, most metals seem to exist as oxides, sulfides, carbonates, halides, etc. and not as bare metals. Consequently, there was/is probably little evolutionary benefit in tackling the bare metals.

Breaking down something to into very small particles, as you described it, is a mechanical process that leaves the oxidation state of the metal unchanged. It is not a chemical transformation.

Nothing cool with enzymes an metals?

There's a lot! Two topics come to my mind:

Chemosynthesis

There's a bunch of bacteria that can oxidize $\ce{Fe(II)}$ to $\ce{Fe(III)}$

at different pH levels
in the presence and absence of oxygen
in the presence and absence of light

Acidithiobacillus ferrooxidans and Leptospirillum ferrooxidans - look at their family names ;-) - operate at $pH\ < 4$, use the oxygen from the air and munch away pyrite.

Metal hyperaccumulation

Some plants and bacteria harvest enormous amounts of metal ions even from highly contaminates soils and store them. The metals are held by by the action of proteins. If you're interested, here is a freely available article on PLOS One Pathogens with lots of further references.


The type of H-NMR spectra observed for S=PF2H and other inorganic molecules similar to this

The everything OP and  orthocresol discussed in the comment session is in this reference book (Ref.1). I include its PDF file (Ref.1) for benefits of OP and future readers. An example of coupling is depicted in the following figure (from Ref.2):

This spectrum is qualified as other inorganic molecules similar to $\ce{S=PF2H}$ molecule. What I want to show is alike different coupling constants in $\mathrm{^1H \ NMR}$ due to the molecular structure, bond distance, etc., the coupling constants in $\mathrm{^{31}P }$ and $\mathrm{^{19}F \ NMR}$ are also changed as well. For instance, $^1J_\ce{HH}$ is $17$-$\pu{20 Hz}$ in $\mathrm{^1H \ NMR}$ organic compound. However, $^1J_\ce{PH}$ is $\pu{836 Hz}$ for this particular molecule ($\ce{PF2H(^{15}NH2)2}$) while $^2J_\ce{PH}$ is $\pu{11.8 Hz}$ for same molecule (c.f., avarage $^2J_\ce{HH}$ is $\approx \pu{7 Hz}$ for a hydrocarbon with more than 2 carbons).
This particular spectrum consists of 90 peaks due to additional coupling of $\ce{P-^{15}N}$ coupling. If the compound is not labelled with $\ce{^{15}N}$, the number of peaks would reduced to 30 peaks, a relatively simplified spectrum, but still complicated coupling pattern of doublet of triplet of quintet.
References:

Michael O’Neill, In Assigning Inorganic NMR Spectra: a programmed approach to solving NMR problems in main group chemistry; Copyright © 2019 Michael O’Neill (ISBN-13: 978-1-0864-12017) (PDF).
David W. H. Rankin, Norbert W. Mitzel, Carole A. Morrison, In Structural Methods in Molecular Inorganic Chemistry; First Edition, John Wiley & Sons: Chichester, United Kingdom, 2013 (ISBN-13: 978-0-470-97278-6).


Isothermal process gives V = A(T)exp(-KP)?

$A$ is already an implicit function of $T$, so your answer is essentially correct.

Starting from the definition of $\kappa$ and integrating both sides with respect to $P$, we obtain $$-\int\kappa\,\mathrm{d}P = \int\frac{1}{V}\frac{\partial V}{\partial P}\,\mathrm{d}P,$$ which yields as solution $-\kappa P = \ln V + f(T)$, where $f(T)$ is our "constant" of integration. Rearranging and exponentiating yields $V(P,T) = e^{-f(T)}e^{-\kappa P} = A(T)e^{-\kappa P}$, $A(T) := e^{-f(T)}$, and we are done.


Standard solution and extracted solution : same concentration?

I found the answer to my question, and I am going to share it just in case.
Apparently, when extracting the pollutants from the foam, I am also extracting some foam-specific materials that influence the pollutants' behaviour during the SPME part of the extraction.
I tested that by extracting the blank foam, then adding a standard amount of pollutants to the extraction solvent. This gave the same result as extracting a spiked foam, meaning higher peak areas than a standard solution.

In short : matrix effect from the foam. I don't yet know why or what, but it's a matrix effect.


In Reimer-Tiemann reaction why does phenol attack the carbene from ortho position?

First of all, in the phenoxide ion, the lone pair on the oxygen atom doesn't remain fully available, it remains in continuous conjugation with the double bond of the benzene ring. In the resonance hybrid, there is also electron density on the Carbon atom in the ortho and para position,and also definitely on the oxygen atom itself. But, Carbon being less electronegative acts as a better nucleophile in this case rather than Oxygen.  So, attack from the Carbon atom rather than Oxygen atom is more probable.  More over, attack from the ortho position, though creates a chance of Steric crowding, but the intermidiate generated from the attack by ortho-position creates an ordered transition state.The metal ion(say, $\ce{Na+}$) can easily interact with the electronegative $\ce{O}$ and $\ce{Cl}$ atoms, and can increase the stability of the transition state. Here is a picture of the intermidiate and a 3D view of it,

This ordered transition state will not appear in case of attack by the Oxygen atom. This is also another reason why that transition state is more favoured, and the reaction rate through that step is higher than any other possible steps.


How to compare the acidity of this multicyclic hydrocarbon to the acidity of cyclopentadiene?

The method presented by Yusuf is a simplified model, which works best for well-behaved molecules. Unfortunately in this case it presents the 'correct' result based on incorrect assumptions. (Even though I did a calculation, I am not 100% convinced this is actually the correct result in the first place.)

In order for the resonance to stabilise a negative charge, the π-systems must be aligned, which is the case for the deprotonation of cyclopentadiene, but it certainly is not true for the other molecule F (Loong said it's called 3a2H-benzo[3,4]pentaleno[2,1,6,5-jklm]fluorene). Here is the calculated structure at the ri-BP86/def2-SVP level of theory:

As you can see, it is domed, and there are three (almost) independent aromatic systems. Through deprotonation this will not change much.

A reason for its relatively high acidity can be found in applying Coulson's Theorem, where the bond angle between two $\mathrm{sp}^x$ hybrid orbitals can be calculated from their angle $\omega_{ij}$ and their hybridisation indexes $\lambda_x$:
$$\cos(\omega_{ij}) = -\frac{1}{\sqrt{\lambda_i\lambda_j}}$$
Since the molecule is C3v symmetric, therefore $\lambda_i = \lambda_j \mathop{:=} x$, this simplifies to:
$$ x = -\sec(x)$$

The inner angle is $\approx\pu{102^\circ}$, therefore $x \approx 5$. The term for hybrid orbitals is actually just an abbreviation:
$$\mathrm{sp}^{x} = \mathrm{s}^{\frac{1}{x+1}}\mathrm{p}^{\frac{x}{x+1}}$$

Given that we just form linear combinations, we can see:
\begin{align}
 1\times\mathrm{s}, 3\times\mathrm{p}
   &\leadsto 3\times\mathrm{sp}^5, 1\times\mathrm{sp}
\end{align}
Given the rounding I have done, the remaining orbital is very roughly speaking an $\mathrm{sp}$ orbital; in the same ballpark as acetylene, from which we know that it has a very acidic proton.

The reason for the high acidity is therefore not the stabilisation of the conjugate base, but the destabilisation of the acid. This is the completely reverse reasoning as cor cyclopentadiene, in which the corresponding base is massively stabilised by resonance.

Given the above, how are you supposed to know, and how are you supposed to judge these compounds?

In my opinion, the exercise is something more on the level of quantum chemistry course and belongs into the canon of a masters student. While simply models will accidentally produce the correct result, it is important to see, that they completely fail for F. Structure-property relationships cannot be inferred from a 2D-graph of the molecule. To be honest, the reasoning above was also rather post-hoc, after I have calculated the molecule and analysed its structure.

I believe, (as long as you are not a human calculator), that there is no simple way to tell which of those molecules is more acidic; there is no easy way to weigh the stabilising effect of resonance against the destabilising effect of pyramidalization.

On the ri-BP86/def2-SVP level of theory (which is rather crude, but I don't have a supercomputer at home) it really comes out that F is more acidic. The following isodesmic reaction yields a difference in electronic energy of about $\Delta E_\mathrm{el} = \pu{-92.8 kJ mol-1}$.

Important: Even though the above sound reasonable, it still neglects plenty of things and has other shortcomings; to name the most important: thermal corrections, solvent effects, level of theory.

Addendum:

Upon deprotonation the structure does not change very much (see larger image), it remains very domed, see the overlay of the structures below.

Judging from the three highest occupied molecular orbitals (HOMO, HOMO -1, HOMO -2) we can still presume some delocalisation of the negative charge into the adjacent aromatic rings.

I'm talking more about hybridisation, Bent's rule, and Coulson's theorem in the following: Are the lone pairs in water equivalent?


Which of the following atoms has the largest first ionization energy?

I actually think this is an unfair question, unless you were given some additional information, or you were specifically instructed to memorize specific ionization energies, rather than learning the trends.  The trend for ionization energies is that they increase from left to right across rows, and from bottom to top in columns.  Based on only that information, I can't see how one should distinguish between $O$ and $Br$ ... yes, it happens to be true that $O$ has a higher IP than $Br$, but how are you supposed to tell? Suppose the example had used $S$ instead of $O$, and $I$ instead of $Br$?  Those elements have the same relative positions in the table, but in that case, it would be $I$ that has the larger ionization energy.

So, in this case I think your confusion was quite understandable .. I would never construct a test question like this unless the student had some additional information.  It's not even true that you can always use the rule about them increasing from left to right along rows .. in the first row, $N$ actually has a slightly higher IP than $O$, because there is a penalty for disrupting the half-filled p-shell in $N$, which is anomalously stable.  You can see the periodic trends here.


How explosive is benzene compared to TNT?

All flammable organic liquids could, in principle, create large fuel-air explosions but the conditions are very hard to achieve accidentally
The first important point to note is that benzene should not be a particular worry. There are plenty of other, widely used, chemicals and mixtures that could, in principle, cause large explosions. Standard automotive petrol (gasoline) for example is present in most vehicles (non-diesel ones anyway) and is transported in 40 tonne tankers on most roads. Fires and explosions happen, but rarely, and they are never known to be of the thermobaric type you are worried about. Most motor vehicles have 40kg or so of fuel that is just as flammable as benzene but after accidents they rarely catch fire or explode and never do so with the power of a fuel air thermobaric explosion (the hollywood trope that all crashed cars explode is pure nonsense).
But big fuel air explosions are known to be possible and have been developed into some very effective military weapons. These are known as thermobaric bombs and the largest are just about the biggest non-nuclear explosions possible in existing military arsenals. But a quick read of how the weapons work (see this wikipedia entry) suggests that they require some fairly notable technology to get right. Largely this is because the fuel has to be dispersed at just the right concentration in the air to get a detonation or deflagration rather than a fiery but unimpressive "phut". It isn't easy to achieve this by accident with a large volume of fuel.
Bad maintenance and safety procedures in chemical and oil plants can lead to accidental fuel-air explosions. The Buncefield explosion in a UK oil storage depot was caused by one. It was a huge explosion probably involving about 300 tons of fuel (but that is only about 0.1% of the fuel stored in the depot a lot of which caught fire but didn't explode). Another famous UK example is the Flixborough disaster caused by a fuel-air explosion caused by a leaky pipe in a cyclohexane plant. In both incidents poor management and engineering contrived to make a dangerous event possible and standards were rewritten to prevent future mistakes. That such event are rare should be of significant comfort if you are worried about the dangers of transporting tankers full of benzene. It is worth noting that none of these (very bad0 accidents came close to involving the majority of the flammable compounds stored nearby. It is hard to get the right fuel-air mixture with very large volumes of fuel, thank goodness.
A more common form of fuel air explosion might also put the rare chemical plant incidents in perspective. This is explosions caused in flour mills and grain elevators. a mixture of finely divided flour or grain dust is explosive and plants that handle flour and grain have to be specially designed to avoid dangerous dust explosions. These are far more common that explosions in chemical plants. While this may not be much comfort, it does suggest that worrying about benzene is not that big a real world risk.


Identifying the C3, C4, S4, and S6 symmetry operations in the Oh point group

At first I thought the character table was counting both positive and negative rotations, but then the order for the $C_2'$ and $C_2$ rotations should be doubled if that were the case ($12C_2'$ and $6C_2$).

You were very close. The missing symmetry operations are indeed ‘positive and negative’ but only those that lead to non-identical transformations. In a $C_2$ axis, it does not matter whether you turn clockwise or counter-clockwise. Consider the following mock-up of the central planar square of an octahedron with the ligands labelled A, B, C and D:

 A
BMD                      (1)
 C

If you apply $C_2$, you will arrive at the same orientation (2), regardless of which direction you rotate in:

 C
DMB                      (2)
 A

However, when rotating by $C_4$ clockwise and counterclockwise actually lead to two different results (3) and (4):

 B
CMA                      (3)
 D

 D
AMC                      (4)
 B

These two are not identical transformations. In fact, they have distinct symbols; depending on your direction of rotation one is $C_4$ and the other is $C_4^3$. You can also interpret them as rotation by $90^\circ$ and rotation by $3 \times 90^\circ = 270^\circ$. $C_2$ and $C_4^2$ correspond to the same transformation as well.

The same case can be made for $C_3$, the reverse rotation is $C_3^2$.

The $S_n$ cases are not quite as simple. Well, actually $S_4$ is ($S_4$ and $S_4^3$) but that’s a coincidence. $S_6$ shows this more clearly. Remember that $$S_6 = C_6 \oplus \sigma_\mathrm{h}$$

Therefore, performing $S_6$ twice is equivalent to $C_3$:

$$S_6 \oplus S_6 = S_6^2 = C_6 \oplus C_6 \oplus \sigma_\mathrm{h} \oplus \sigma_\mathrm{h} = C_3 \oplus E = C_3$$

Thus, $S_6$ applied twice and $S_6^5$ applied twice give $C_3$ and $C_3^2$, respectively, so $S_6^2$ and $S_6^4$ are already taken care of. A last case needs to be made for $S_6^3$: This is identical to $i$ as can be seen below:

$$S_6^3 = S_6 + S_6 + S_6 = C_6 \oplus C_6 \oplus C_6 \oplus \sigma_\mathrm{h} \oplus \sigma_\mathrm{h} \oplus \sigma_\mathrm{h} = C_2 \oplus \sigma_\mathrm{h} = i$$

Thus, while you could imagine up to six different $S_6$ transformations, only two of those ($S_6$ and $S_6^5$) give new transformations not covered elsewhere.


Dividing different units of measurement?

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

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


What is the maximum extent of a hydride shift to form the most stable carbocation?

Advances in Physical Organic Chemistry, volume 19, discusses the relative transition state barriers for 1,2-, 1,3-,1-4-, and 1,5-hydride shifts.

The specific cations considered are 2,3 dimethyl-2-butyl, 2,4 dimethyl-2-pentyl, 2,5 dimethyl-2-hexyl and 2,5 dimethyl-2-heptyl.

Interestingly, they report that whereas the barrier for 1,4 shift in 2,5 dimethyl-2-hexyl is 12 kcal/mole, this decreases dramatically to less than 6 kcal/mole for 1,5 shift in 2,6 dimethyl-2-heptyl

The reference draws the transition states as rings, with the H being transferred as a bridging member of the ring, a five membered ring for the 1,4 shift and a six member ring for the 1,5 shift.

The reference also makes a comparison to intermolecular hydride shift between isobutane and t-butyl cation, which has a barrier of 3.6 kcal/mole.

While the reference does not discuss the possibility of longer distance than 1,5 shifts, it seems that because the ends of a long ion can meet each other, long distance (in terms of number of bonds) shifts are possible because through space the distance is short.


Can molecular structure be determined without modern characterization methods?

Structure elucidation via degradative studies and decomposition chemistry has sadly all but passed from the chemists' toolbox, but this was the method once used to determine molecular structures up until about half a century ago. One of my very favourite examples that I use to teach structure determination by NMR is the story of strychnine. Strychnine (below) is by no means a trivial structure; it is hard enough to solve by 'modern' spectroscopic means. The story of how Robinson and Woodward, standing upon the shoulders of those before them, unravelled the the structure of this alkaloid over a decade and a half is an intriguing story, and one well worth investigating further. (AMBIX, Vol. 48, Part 3, November 2001)

Essentially, the structure is slowly pieced together through a series of degradation reactions. It is a methodology that relies on chemical processes, reactivities and properties only. For instance, it was worked out in the late 19th century that strychnine contained an amide as a lactam through a hydrolysis reaction to give strychnic acid. Were it not a lactam, two separate products would have been formed, rather than the one. Suggestions of the quinoline skeleton came from the reactivity of a dimethyl derivative in a series of chemical tests. Over the next 20 or so years, this was revised and superceded by more likely candidates. So on and so forth over a number of years, each little piece of the puzzle is fitted together, all through the lengthy process of chemical reactions to give smaller easier to identify products. A number of complicated framework structures were proposed, mulled over and discarded for a number of years, and final structure was tested through total synthesis reproduction. This whole process requires an incredibly talented chemist, and it simply makes my mind boggle at the thought of how these guys operated.

And though, it can be argued that UV spectroscopy did play a role in the final jigsaw piece in 1947, the correct structure was proposed by Robinson in 1942, and total synthesis confirmed this in 1954.

This is just one example of determination of molecular structure through degradative or decomposition chemistry, and many complicated structures were determined this way. Thank goodness for high field NMR.


When does Hartree-Fock fundamentally fail?

Restricted HF (RHF) fails anytime you have partial occupancy due to (near) degeneracy of the HOMO. For example, $\ce{H2}$ dissociates to a mix of $\ce{2H}$ and $\ce{H- + H+}$ and the dissociation energy is way off. Similarly, singlet $\ce{O2}$ is not well described by RHF if you can get it to converge at all.

The solution is to use more than one MO wavefunction (determinant) and this is sometimes referred to as "non-dynamical" (rather than "strong") correlation.

Unrestricted HF (UHF) can usually treat these systems qualitatively correct.


Where to find pre-optimized files for Quantum Chemistry Softwares?

Short answer: we're building one for molecules

In your case, you don't actually want molecules. You're mentioning crystals. In that case, I suggest the Crystallography Open Database or if you have access, try the commercial Cambridge Crystal Structure Database. The COD is planning a theoretical crystal database, but I don't think it's online yet. These are experimental x-ray structures, but should be good starting structures for solid-state simulations.

For molecules, my group is building one. It's the Pitt Quantum Repository There are a few related efforts that I'm aware of.

With the PQR, you can either search through the website (even on mobile). The database is keyed by InChI keys, so if you have an existing structure you can generate the InChIKey and then get the structure, e.g.:

D-(-)-Morphine -> .mol2 file at http://pqr.pitt.edu/api/mol2/BQJCRHHNABKAKU-KBQPJGBKSA-N

The long answer is that some commercial software vendors consider such libraries as violations of the license. No joke, I've planned this for years to augment Avogadro, but the company behind a leading program told me that they would void my license to distribute output files in this fashion.

With the rise of community and open source quantum chemistry software, this is moot. Use programs that let you distribute output files. What remained was:

Gathering the 3D coordinates (i.e., a low-energy conformer) of X thousand/million molecules
Obtaining the computer time to optimize all of them
Devoting a server and disk space to serving them
Having a great website to help view, browse, and search them


Mechanism of acid-catalyzed addition of water across an alkene

No, the person on your ship is incorrect.

An electron pair can only be transferred to electrophiles, which are species having a vacant orbital present on them. $\ce{H3O+}$ doesn't have a vacant orbital, so that can't happen. Here is a description as to why:

However,  $\ce{H+}$ ions, which have vacant orbitals can accept the pair and have it shift, while the other carbon sharing the double bond loses its electron, making it electrophilic, and can be approached by a nucleophile, in this case, $\ce{H2O}$:


How can I eliminate the harmful gas produced from the heating/manipulation of galvanized steel?

Galvanizing by definition is zinc coating. The zinc boils off ( as 1700 F)  is approached and immediately oxidizes to ZnO , a white powder. ZnO has the unusual characteristic of causing zinc chills = brass founders ague = oxide shakes,  IF inhaled while less than one hour old . After that is is pretty harmless and used in various lotions and creams and rubber products.  So avoid the fumes when welding. However , one recovers from "chills" in about 24 hours so not toxic ( I can verify no lasting affects as I had "chills" many decades ago). There is another potential problem ; zinc can cause liquid metal embrittlement of steel . This occurs where the zinc is not exposed to air as if two surfaces are placed together and the edges sealed by welding. I don't know where you think chromium is coming from other than steel made with scrap ( the majority) typically contains about 0.04 % tramp chrome ( and Ni ,and Cu , and Mo). So essentially all steel  welders in the world are exposed to this tramp chrome.


Is methanol even a strong enough reagent for such a reaction?

Methanol is a very poor nucleophile, so the rate of the $\mathrm{S_N2}$ pathway will be extremely slow. This is much likelier to be a $\mathrm{S_N1}$ reaction. The starting material is a benzylic bromide from the point of view of the alkyl halide, both pathways are quite possible, but the lack of a good nucleophile suggests the predominance of the $\mathrm{S_N1}$ route.

So, you go through a carbocation intermediate, which undergoes an alkyl shift to generate a more stable carbocation.

Both carbocations are benzylic and secondary, so hyperconjugation isn't the driving force for the rearrangement, as your comment hints at. Instead, it's simply related to the relief of ring strain as you expand a four-membered ring to a five-membered ring. Four-membered rings are pretty unstable to begin with and fusing one to a benzene ring increases the strain even further since you impose more geometrical constraints on the ring.

Interception by methanol and deprotonation gives the desired product.


The octet rule only applies firmly to elements below main energy level 3 (in fact, there are rare instances in which even that isn't entirely accurate, but molecular orbtal theory needs to be invoked to explain those situations). At main energy level 3, which Sulfur occupies, unfilled d orbitals become available, and those d orbitals serve to accommodate the additional electrons in Sulfur's bonds. In reality, Sulfur is a much larger atom than Oxygen, and the π orbital overlap between the two elements is fairly poor, giving the bonds significant single-bond character. You might therefore see resonance structures in which single bonds are drawn instead, with formal positive charges assigned to Sulfur and formal negatives assigned to Oxygen.


To rotate on a single bond in a Fischer Projection, all 3 groups move in the same direction. $\frac {1} {2\ \ \ 3} \rightarrow \frac {2} {3\ \ \ 1}$. See this page for more information.


Defining enantiomer of morphine by R/S designation

The structure that you are showing is neither R nor S.

As it has five chirality centers you need to look at all of them.

$\hskip3in$

You can see in the above figure that there are three R and two S centers in the configuration of morphine that you posted.


Determination of (R) and (S) configurations from a Haworth projection

In a Haworth projection, the hexagonic shape of the ring already implies the direction of two bonds. The two bonds on the ring are always ‘bent ring-inwards’ when viewed from the outside. Thus, the two perpendicular bonds (typically vertical) must ‘point away’ from the ring: in a forward direction for the front carbons of a ring and in a backward direction for the rear side of the ring.

However, I notice that the question uses (R) and (S) a lot and attempts to argue a lot with it, so a note must be dropped on those stereodescriptors, too. It is important to realise that the (R) and (S) stereodescriptors can change even if the conformation of the carbon in question did not invert or otherwise change conformation. This is due to the way priorities are determined in the CIP system. Let me exemplify this.

First, remember that α-D-glucopyranose, β-D-glucopyranose and the aldehyde form of D-glucose are three different compounds with different physical properties. The α and β forms are diastereomers, more specifically epimers, more specifically anomers of each other and e.g. display different optical rotation. The chain-form of glucose is different altogether. I will construct all further discussion of this answer in a way that it does not matter whether we are dealing with α or β-D-glucopyranose, so you may always assume a mixture.

Now let’s consider the absolute configuration of carbon 3 in both the chain and pyranose forms.

Pyranose

Obviously, the oxygen atom has highest priority.
Obviously again, hydrogen has lowest priority.
We have two carbon atoms of the (b) and (c) priorities; initially undistinguishable. Thus, we must go down one level.
Both carbons are attached to the atoms (in order of priority) $\ce{O, C, H}$. Again, this is a tie; we must go down one level.
Both oxygens are attached to a hydrogen; this is identical and the end of the sub-chain. We must go back to the next, lower priority of the former atom.
The two carbon atoms differ: one is the acetal group which expands to $\ce{O, O, H}$. The other is connected to $\ce{O, C, H}$ (carbon 5). Thus, we have resolved the issue.
Going towards lower carbon numbers is the higher priority.

Therefore, carbon $\ce{C{3}}$ has an (S) configuration.
Chain form

Again, oxygen obviously has highest priority.
Equally again, hydrogen obviously has lowest priority.
The two carbons are initially equal.
Both are again attached to $\ce{O(H), C, H}$.
The two subsequent carbons differ again. This time, one is the aldehyde group which expands to $\ce{O, (O), H}$, the other is $\ce{C{5}}$ which expands to $\ce{O, C, H}$. Since a ghost oxygen atom has a higher priority, going up the Fisher projection has the higher priority.

We arrive at the same (S) conclusion as before.

Now, we will construct the same discussion for carbon $\ce{C{4}}$.

Pyranose

Obviously, the oxygen atom has highest priority.
Obviously again, hydrogen has lowest priority.
We have two carbons, initially indistinguishable again. We go down a level.
Both carbons ($\ce{C{3}}$ and $\ce{C{5}}$) expand to $\ce{O, C, H}$. We must go down a level.
One of the two oxygens is connected to a carbon atom. Thus, $\ce{O(C), C, H}$ wins over $\ce{O(H), C, H}$.
Going towards $\ce{C{5}}$ has a higher priority.

Thus, carbon $\ce{C{4}}$ is (S) configured.
Chain form

Bla, oxygen.
Bla, hydrogen.
Blabla, two carbons. Again, we descend a level to $\ce{C{3}}$ and $\ce{C{5}}$.
Both carbons expand to $\ce{O, C, H}$. We descend a level.
Both oxygens are attached to a hydrogen and nothing else. They are indistinguishable.
The two carbons we are now examining are $\ce{C{2}}$ and $\ce{C{6}}$. The former expands to $\ce{O, C, H}$, the latter to $\ce{O, H, H}$. Therefore, Going towards $\ce{C{3}}$ has a higher priority.

Thus, carbon $\ce{C{4}}$ is (R) configured.

Note that while the configuration of $\ce{C{3}}$ did not change, $\ce{C{4}}$’s absolute configuration changed from (R) to (S) upon formation of the pyranose ring. Thus, you cannot rely on absolute configurations to verify a correct Fischer to Haworth transformation for carbohydrates!


Percentage ionic character when electronegativity is given

Linus Pauling proposed an empirical relationship which relates the percent ionic character in a bond to the electronegativity difference $\Delta \chi$.

Percent ionic character $= (1-e^{-(\Delta \chi/2)^2} )\times 100$

But I'd like to correct the definition of percent ionic character in your question using dipole moment  $\mu$ (not Observed value of ionic character):

Percent ionic character = $\Large\frac{\mu_{\text{observed}}} {\mu_{\text{calculated}  }}$ $\times 100 \%$

Where $\mu_{\text{calculated}}$ is calculated assuming a 100% ionic bond.

For more details please see this page.


How does sugammadex (a modified cyclodextrin) draw rocuronium into its lipophilic cavity?

After consulting with a professor, the answer seems obvious.

The two negative charges repel each other when the free cyclodextrin molecule is in solution, resulting in an "open" conformation that has the carboxylate groups facing "out". The thioether chains interact with the rocuronium as it is drawn into the large open face of the molecule. As the positively charged nitrogen moves into place, the carboxylate groups "close", and lock the drug into place.


From Cubic crystal system on Wikipedia:

  In the rock-salt or sodium chloride (halite) structure, each of the two atom types forms a separate face-centered cubic lattice, with the two lattices interpenetrating so as to form a 3D checkerboard pattern.

If LiX crystallizes like NaCl, then the unit cell length is twice the distance of the shortest Li - X distance. The magnitude of the answer given in the book looks like it could be an interatomic distance, but the calculation does not make any sense to me (it does say $a = 2a$, as @Ivan_Neretin already pointed out in the comments. Also, when a calculation drops the units midway and sticks them back in the result, it demonstrates a certain nonchalance that does not instill confidence in the source.

The question asks about the unit cell parameter a (in a cubic system, this is all you have to specify to describe the dimensions of the cell because the other two lengths are the same and the angles are all right angles). If that's all you want, the internal structure of the unit cell has no bearing on the calculation (there might be systematic absences in a particular space group, but the question says that the 111 reflection was observed, so there is no issue there either). Why not just plug in the angle, get d, and calculate the unit cell length from that?


Radioactive Isotopes form Compounds? The Hershey-Chase Experiment

Radioactive isotopes have similar chemical properties to that of nonradioactive element of same atom. They actually do from chemical bond with compound. You even heard the name of radio-pharmaceutical that involved the complex of radioactive metal with ligand or the radioactive isotope is attached to the pharmaceutical through covalent bond.

The simplest example is FDG is 18F-labled glucose analogue.

Carbon-11 is another radioactive isotope that form covalent bond similar to Carbon-12 and used in Nuclear medicine as PET imaging agent. C11-Methionine is an example used in the field of Nuclear Medicine for the diagnosis of brain tumour

Iodine-131 is another sample of radioactive isotope that is used for the detection of hypothyroidism and also form bond with compound. Like in metaiodobenzylguanidine is used in neuroblastoma

.

I can gave you lots of others examples but i think these are enough for your understanding. I hope, this will be helpful.


A "dry" electron is an electron released from the event of an ionization. Depend on source the electron can either be defined as:

  "The $e^-$ is a 'dry electron' that still retains some of the excess
  kinetic energy acquired from the ionizing event."$\ce{^{[1]}}$

or,

  "The 'dry-electron' (defined as an electron before solvation)
  ..."$\ce{^{[2]}}$

A dry electron is thus, an electron that is not assiciated with any atoms, molecules or solutions.

Reference

$\ce{[1]}$ Thormod Henriksen, David H. Maillie; Radiation and Health; page 157

$\ce{[2]}$ J.D. Simon; Ultrafast Dynamics of Chemical Systems; page 144


Are there general rules for Base + Metal Compound reactions like there are for Acids?

On the same level of generalization, the rule is "no reaction", but this rule has quite a few exceptions (probably more than the rules you quoted above).

Water-soluble bases would react with amphoteric metals to produce hydrogen and the corresponding salt:
$$2\ce{Al + 2KOH + 6H2O \to 2KAl(OH)4 + 3H2}$$


Draw a diagram to show the shape of Chloromethane and explain why it has this shape.

Carbon has in his outer shell 4 electrons and needs 4 more electrons to get the full shell like the noble gas Ne.

Chlorine has in his outer shell 7 electrons and needs one electron to get the full shell like the noble gas Ar.

I give you an answer because the question about the tetrahedral structure of Chloromethane contradicts the perceptions of physics about a s- orbital with 2 electrons and 3 p-orbitals with 2 electrons in each p-orbital. To avoid the contradiction between the claimed s- and p-orbitals and the empirical founded tetrahedral structure the sp-hybridizations were postulated. The carbon in Chloromethane has four sp3-hybridizations with all atoms around.

If you have a look around you will see that the hybridizations are the norm. Let the s- and p-orbitals be a toy for physicists.

What you have to check is, always to get pairs of electrons. That electrons like to be pairwise was expressed by the Pauli exclusion principle. It is a principle and not an explanation why this happens this way in atoms. But if you need an imagination of how the electrons form pairs you could think for yourself of the electrons as tiny bar magnets (they indeed have a magnetic dipole moment) and two of them form a magnetic pair. This pairs around the carbon will be in some equilibration to each other. Since chlorine has a lot of other electron pairs it pushes the 3 hydrogen atoms. So you get this shape:

All images from Wikipedia


Can Gay-Lussac's law be observed in the reverse direction?

When you decrease the temperature of an ideal gas held at constant volume, what you are doing is transferring energy as heat from the gas into the surroundings. You do this by keeping the surroundings at a targeted final temperature and placing the gas in thermal contact with the surroundings, allowing heat to dissipate. When you reach the final state, you measure the temperature and pressure of the gas and find, in agreement with Gay-Lussac's law, that both T and P have decreased. Then if you ask yourself, which did I lower first, you cannot say. They decreased simultaneously.

The pressure and temperature are coupled (dependent) properties of the ideal gas when the volume is held constant. This is a consequence of the thermodynamic state postulate that for a simple compressible system, the properties of the system are fixed once two independent ones have been specified. The fallacy in the question here is the assumption that one might be able to change the temperature and that a pressure change somehow follows after. The cooling experiment changes both simultaneously. You never actually choose to alter one variable rather than the other.