Is there a relation between transition density and density differences?

By definition, the transition density is not the same as the difference density. In the following derivation, the transition density is $\mathbf{T}^{\mathrm{CIS}}$ and the difference density is $\mathbf{P}^{\Delta}$. Although it may seem like they can be formed from the same quantities, the true difference density contains orbital relaxation terms and can be used to build a relaxed density, which is the correct excited state density. Neglecting those terms (the $P_{ia}^{\Delta}$) will give an unrelaxed density. I will also try to show that the unrelaxed difference density is still not equivalent to the transition density.

Derivation of individual CIS densities

Here is the derivation of the transition density for doubly-occupied spin-restricted MOs, with quoted blocks and equations directly taken from the original CIS paper.

The CIS approximation is identical to the TDA approximation in TD-DFT. Full TD-HF or TD-DFT follows the RPA equations, which include deexcitation terms in addition to excitation terms. The equations would be different for RPA, but conclusion would be identical.

The standard indexing convention is followed: $\mu, \nu, \lambda, \sigma$ for AOs, $i,j,k,l$ for occupied MOs, $a,b,c,d$ for virtual MOs, and $p,q,r,s$ for any MO. Define the Hartree-Fock (or SCF-type, valid for DFT as well) density as a sum over occupied MOs:

$$
P_{\mu\nu}^{\mathrm{HF}} = \sum_{i=1}^{n} c_{\mu i} c_{\nu i}
$$

  The two-particle CI-singles density matrix, $\mathbf{\Gamma}^{\mathrm{CIS}}$, can be written in terms of the HF ground-state density matrix and the ground-to-excited-state transition density matrix, $\mathbf{T}^{\mathrm{CIS}}$:

  $$
\Gamma_{\mu\nu\lambda\sigma}^{\mathrm{CIS}} = \frac{1}{2} \left[ P_{\mu\nu}^{\mathrm{HF}} P_{\lambda\sigma}^{\mathrm{HF}} + 2 T_{\mu\nu}^{\mathrm{CIS}} T_{\lambda\sigma}^{\mathrm{CIS}} - P_{\mu\sigma}^{\mathrm{HF}} P_{\lambda\nu}^{\mathrm{HF}} - 2 T_{\mu\sigma}^{\mathrm{CIS}} T_{\lambda\nu}^{\mathrm{CIS}} \right]
$$

  where $\mathbf{P}^{\mathrm{HF}}$ is given (above) while $\mathbf{T}^{\mathrm{CIS}}$

  $$
T_{\mu\nu}^{\mathrm{CIS}} = \sum_{ia} a_{ia} c_{\mu i} c_{\nu a}
$$

I can't find the definition of $\{a\}$ in the paper, but it is the set of coefficients from a converged CIS calculation that describe the single-particle excitation from every occupied orbital to every unoccupied orbital. As an aside, they are the elements of the $\mathbf{X}$ vector in the RPA eigenvalue problem:

$$
\begin{pmatrix}
\mathbf{A} & \mathbf{B} \\
\mathbf{B} & \mathbf{A}
\end{pmatrix}
\begin{pmatrix}
\mathbf{X} \\
\mathbf{Y}
\end{pmatrix}
=
\omega
\begin{pmatrix}
\mathbf{1} & \mathbf{0} \\
\mathbf{0} & \mathbf{-1}
\end{pmatrix}
\begin{pmatrix}
\mathbf{X} \\
\mathbf{Y}
\end{pmatrix}
$$

$\mathbf{A}$ and $\mathbf{B}$ are parts of the orbital Hessian. If you diagonalize the whole thing, you get the set $\{\omega\}$ of all excitation energies back as eigenvalues. As mentioned above, setting $\mathbf{B} = \mathbf{0}$ gives the TDA or CIS approximation, so there is no longer a deexcitation vector $\mathbf{Y}$.

Importantly, we always start from the set of ground-state MO coefficients $\{c\}$ and describe excitations on top of them.

  The CI-singles excited-state density matrix, $\mathbf{P}^{\mathrm{CIS}}$, is also constructed as a sum of HF and excited-state terms:

  $$
P_{\mu\nu}^{\mathrm{CIS}} = P_{\mu\nu}^{\mathrm{HF}} + P_{\mu\nu}^{\Delta}
$$

  Here we have introduced $\mathbf{P}^{\Delta}$, the CI-singles $\Delta$ density matrix. This can also be called a "difference density matrix", since it represents the changes in electronic distribution upon excitation. It is not, however, the same as the transition density matrix $\mathbf{T}^{\mathrm{CIS}}$ defined above. As we shall demonstrate, it is the use of the true CI-singles density matrix required by (the CIS gradient expression that isn't important here) and not the simple one-particle density matrix (1PDM) which allows the realistic computation of charge distributions, orbital populations, and electronic moments of the excited state. To see this distinction, first consider the $\Delta$ density matrix which would be added to the HF density matrix in order to generate the 1PDM for an excited state. In the MO basis, it is a symmetric matrix with both occupied-occupied (OO) and virtual-virtual contributions

  $$
\begin{align}
P_{ij}^{\Delta} &= -\sum_{ab} a_{ia} a_{jb} \\
P_{ab}^{\Delta} &= +\sum_{ij} a_{ia} a_{jb}
\end{align}
$$

  with the occupied-virtual (OV) elements all zero. The true CI-singles density matrix required in (the CIS gradient expression) will have exactly the same OO and VV contributions, but the OV terms are not all zero. The appearance of these off-diagonal block elements in the excited-state density matrix can be interpreted as orbital relaxation following the initial gross charge rearrangement due to excitation. That is to say, the CI coefficients will by themselves describe some of the gross features of charge redistribution in the excited state, but the response of the wave function to an external perturbation will account for further refinement in electronic properties. These OV terms can be found by solving a single set of CPHF equations:

  $$
L_{ai} = \sum_{bj} \left[ (ij||ab) - (ib||ja) \right] P_{bj}^{\Delta} + (\epsilon_a - \epsilon_i) P_{ai}^{\Delta}
$$

  where the $\mathbf{L}$ vector is the CI-singles Lagrangian:

  $$
\begin{align}
L_{ai} &= C1_{ai} - C2_{ai} + \sum_{kl} P_{kl}^{\Delta} (al||ik) + \sum_{bc} P_{bc}^{\Delta} (ab||ic) \\
C1_{ci} &= -2 \sum_{jab} a_{ia} a_{jb} (cb||ja) \\
C2_{bk} &= -2 \sum_{ija} a_{ia} a_{jb} (ik||ja)
\end{align}
$$

  The total CI-singles $\Delta$ density matrix required for $\mathbf{P}^{\mathrm{CIS}}$ can be generated by backtransforming the entire MO basis $\Delta$ density matrix defined by $P_{ij}^{\Delta}$, $P_{ab}^{\Delta}$, and $P_{ia}^{\Delta}$:

  $$
P_{\mu\nu}^{\Delta} = \sum_{pq} P_{pq}^{\Delta} c_{\mu p} c_{\nu q}
$$

Show that the transition density and unrelaxed difference density are not equivalent

We want to compare $T_{\mu\mu}^\mathrm{CIS}$ and $P_{pq}^{\Delta}$, but they are not in the same basis, so one must be transformed. Because I don't want to mess with the block-diagonal structure of $\mathbf{P}_{\mathrm{MO}}^{\Delta}$, I will transform $\mathbf{T}$ to the MO basis:

$$
\begin{align}
T_{pq}^{\mathrm{CIS}} &= \sum_{\mu\nu} c_{\mu p} T_{\mu\nu}^{\mathrm{CIS}} c_{\nu q} \\
&= \sum_{\mu\nu} \sum_{ia} c_{\mu p} c_{\mu i} a_{ia} c_{\nu a} c_{\nu q}.
\end{align}
$$

Now consider the unrelaxed difference density, meaning the ov/vo blocks are zero due to neglecting orbital relaxation effects:

$$
P_{pq}^{\Delta} =
\left(
\begin{array}{c|c}
-\sum_{ab} a_{ia} a_{jb} & \mathbf{0} \\
\hline
\mathbf{0} & +\sum_{ij} a_{ia} a_{jb}
\end{array}
\right)_{pq}.
$$

Here is a slightly hand-waving proof. I am not sure if the MO coefficients in the double sum can be simplified, but it doesn't matter; assume they are unity. $T_{pq}^{\mathrm{CIS}}$ contains terms with $a$, while $P_{pq}^{\Delta}$ contains terms with $|a|^2$. Unless there is an extra transformation that can be done, I don't see how they can be equivalent.

If you want a little test, generate some fake matrices and see for yourself. $\texttt{x} \equiv X_{ia} \equiv a_{ia}$.

#!/usr/bin/env python

from __future__ import print_function

import numpy as np

np.random.seed(42)

nocc = 2
nvirt = 2
norb = nocc + nvirt
nov = nocc * nvirt
nbasis = norb

x = np.random.rand(nocc, nvirt)
c = np.random.rand(nbasis, norb)
cocc = c[:, :nocc]
cvirt = c[:, nocc:]

p_delta_oo = -np.einsum('ia,jb->ij', x, x)
p_delta_vv = np.einsum('ia,jb->ab', x, x)

p_delta_ov = np.zeros(shape=(nocc, nvirt))
p_delta_vo = p_delta_ov.T

p_delta = np.asarray(np.bmat([[p_delta_oo, p_delta_ov],
                              [p_delta_vo, p_delta_vv]]))

# same as np.dot(cocc, np.dot(x, cvirt.T))
t_ao = np.einsum('ia,mi,na->mn', x, cocc, cvirt)
# same as np.dot(c.T, np.dot(t_ao, c))
t_mo = np.einsum('mp,mi,ia,na,nq->pq', c, cocc, x, cvirt, c)

print(t_mo)
print(p_delta)

which gives

[[ 1.82856579  1.89892758  0.59008502  3.09931192]
 [ 1.48134568  1.53535055  0.49193325  2.47904162]
 [ 0.54242862  0.56250364  0.17874155  0.9109361 ]
 [ 1.79242654  1.85791848  0.59456093  3.00118592]]
[[-1.75629929 -1.76345302  0.          0.        ]
 [-1.76345302 -1.77063588  0.          0.        ]
 [ 0.          0.          1.22441763  1.71443377]
 [ 0.          0.          1.71443377  2.40055604]]

Final comment

My original understanding of the term "difference density" comes from a coworker who was taking the real-space representation (cube files) of two state densities, say the ground state $0$ and an excited state $n$, and subtracting them.

$$
\Delta \mathbf{P}^{0\rightarrow n} = \mathbf{P}^{(n)} - \mathbf{P}^{(0)}
$$

Is this equivalent to the definition of the difference density from above?

$$
\begin{align}
\mathbf{P}^{(n)} - \mathbf{P}^{(0)} &= \mathbf{P}^{\mathrm{CIS}(n)} - \mathbf{P}^{\mathrm{HF}} \\
&= \left( \mathbf{P}^{\mathrm{HF}} + \mathbf{P}^{\Delta(n)} \right) - \mathbf{P}^{\mathrm{HF}} \\
&= \mathbf{P}^{\Delta(n)}
\end{align}
$$

Looks good!

References

James B. Foresman, Martin Head-Gordon, John A. Pople, and Michael J. Frisch. Toward a systematic molecular orbital theory for excited states. The Journal of Physical Chemistry 1992 96 (1), 135-149, DOI: 10.1021/j100180a030
Frank Neese. Prediction of molecular properties and molecular spectroscopy with density functional theory: From fundamental theory to exchange-coupling. Coordination Chemistry Reviews 2009 253 (5-6), 526-563, DOI: 10.1016/j.ccr.2008.05.014


Band MO of Sodium Filling of electrons confusion

From left to right, it shows 1, 2, 3, and 4 atoms of sodium, with a total of 1, 2, 3, and 4 valence electrons. So this diagram is different from others which have the uncombined AO's on the left and right, and the combined MO's in the middle. Here, we are adding more and more atoms to go from discreet energy levels (left 4 columns, up to four atoms) to a band structure (rightmost column, many atoms in bulk metal).

The only column that shows the AO is the left-most because for that one, there is only one atom, which means that there can't be any MOs.

Maybe this diagram makes more sense to you (the full and empty circles on the right show how the AO's are combined to make the bonding, non-bonding and antibonding MO's):

Source: http://xbeams.chem.yale.edu/~batista/vaa/nabands.gif


Platinum group metals: Why are platinum and palladium great catalysts and not the others?

I think some of the other transition elements have also catalytic properties. For example, rhodium is generally used in addition to platinum or palladium in catalytic converters to oxidize carbon monoxide and hydrocarbons on one hand, and reduce $\ce{NO_x}$ to nitrogen on the other hand.

So, the choice is often made on the basis of relative cost.

But of course, not all transition elements are adequate for these catalytic properties:

Metals such as silver and copper have a high affinity for sulfur and form metal sulfides and sulfates. As this happens, there will be progressively less metal available to function as a catalyst. While platinum tends not to become totally or irreversibly poisoned, i.e It can be efficiently recycled.
Another point is linked to "Tammann temperatures" of these metallic elements:  Actually, in a catalytic converter, metal is used in the form of nanoparticles dispersed on the surface of a porous support material. When  temperature increases the particles starts to becomes mobile and coalesce (i.e. the catalytic activity is lost). This process is well-known under the name "sintering" and it becomes tangible as the metal is close to its Tammann temperature. This temperature is often defined as the half of the metal's melting point on the absolute temperature scale. In fact, metals such as gold, silver and copper have a Tammann temperature that is well below the average temperature of exhaust-gas ($600-700^0\mathrm{C}$).
Tammann Temperature is $750^0\mathrm{C}$ for platinum, and $640^0\mathrm{C}$ for palladium. While it is only $405^0\mathrm{C}$ for copper, $345^0\mathrm{C}$ for silver and $395^0\mathrm{C}$ for gold.

To sum up, the candidate transition element for use as a catalyst in a catalytic converter:
1) has a low affinity to poisons  such as sulfur compounds;
2) can be efficiently recycled;
3)  has a relatively high melting point.


is it possible to break the bonds of diatomic elements such as fluorine or iodine and create positive and negative ions by electron bombardment?

Sure, why not. It's just the question of your bombarding electrons having the right energy. At low energies, a bombarding electron will likely be captured by one of the iodine atoms:
$$
\ce{I2 + e- -> I- + I} \, .
$$
But somewhere around $8.62 \pm 0.06 \, \mathrm{eV}$ according to this study, you'll indeed have the ion pair formation:
$$
\ce{I2 + e- -> I- + I+ + e-} \, .
$$
There are some other processes which contribute to the ion pair formation. For details, read the paper quoted above.

With respect to the formation of $\ce{I2+}$ note that to form it a bombarding electron have to, formally speaking,

Either first break the $\ce{I-I}$ bond homolyticly and then remove two electrons from one of the $\ce{I}$ atom which requires it to carry about 30 eV of energy (the sum of the homolytic bond dissociation energy and of the first and the second ionization energies).
Or break the $\ce{I-I}$ bond heteroliticly and then remove the second electron from $\ce{I-}$ ion which requires it to carry also about 30 eV of energy (the sum of the heterolytic bond dissociation energy and of the second ionization energy).

$%edit$


Is it appropriate to call this material a hybrid composite?

There is a slight deviation in the literature as to what the hybrid composites are.
Originally, materials were considered hybrid composites if both inorganic (e.g. clay/silicates as the hydrophilic component) and organic (e.g. organic polymer as the hydrophobic component) constituents were involved in building up the matrix (see e.g. a review [1]).

Another review [2] and the more recent definition from Encyclopedic Dictionary of Polymers [3, p. 373]:

  Hybrid Composite n Advanced composite with a combination of different high-strength continuous filaments in the matrix.
  Also, composite in which continuous and staple fibers are used in the same matrix.

provide a more permissive definition, according to which hybrid composite materials are the materials in which a combination of reinforcing layers from two or more types of fibers is utilized.

With this in mind, in order to provide a definitive answer for yourself, you probably want to do surface imaging (e.g. with AFM or SEM) to prove that really are more than one type of fibers present.
If you are planning to publish this data, be prepared that you will be asked to provide rheological data as well.

References

Wahab, M. A.; Kim, I.; Ha, C.-S. Silica- and Silsesquioxane-Containing Polymer Nanohybrids. In Macromolecules Containing Metal and Metal-Like Elements; Abd-El-Aziz, A. S., Carraher, C. E., Pittman, C. U., Zeldin, M., Eds.; John Wiley & Sons, Inc.: Hoboken, NJ, USA, 2005; pp 133–160. DOI: 10.1002/0471712566.ch6.
Perov, B. V.; Khoroshilova, I. P. Hybrid Composite Materials. In Polymer Matrix Composites; Shalin, R. E., Ed.; Soviet Advanced Composites Technology Series; Springer Netherlands: Dordrecht, 1995; pp 269–304. DOI: 10.1007/978-94-011-0515-6_6.
Encyclopedic Dictionary of Polymers; Gooch, J. W., Ed.; Springer New York: New York, NY, 2011. DOI: 10.1007/978-1-4419-6247-8.


Yes, aqueous ammonia is likely damaging in time to metal Aluminum in the presence of air and any salts (from outside dust,...) that may further serve as an electrolyte.

Ammonium salts, themselves are good in transporting solvated electrons, and here you are, unintentionally, likely creating a metal-air battery scenario.

Possible half-cell reactions:

$\ce{ Al -> Al(3+) + 3 e- }$

$\ce{ O2 + 2 H2O + 4 e- -> 4 OH- }$

Resulting in a net cell reaction of:

$\ce{ Al(3+) + 3 OH- -> Al(OH)3 }$

The aqueous ammonia may also assist here in the removal of the Aluminum hydroxide by forming a "dissolved/peptised/supersaturated hydrated alumina" (per a paper 'The Precipitation of Aluminium Hydrous Oxide and Its Solubility in Ammonia' by Prideaux and Henness).

In essence, Aluminum is now a more rapidly corroding anode.

Nice experiment, albeit, unintended.


Until this post, $\mathrm{sp^1,~sp^2~\text{and}~sp^3}$ meant for me, that the hybrid orbitals would consist from one s orbital and one, two or three p orbitals. I thought of it being so, as this seems to be what the most textbooks suggest, as can be seen in the following image:

One 2s orbital and three 2p orbitals hybridize to four $\mathrm{sp^3}$ orbitals.

This scheme seems to work but it also seems to be a working simplification as can be seen by my question about $\mathrm{sp^5}$ hybrid orbitals, which simply cannot consist from one s orbital and five p orbitals, as there are only three p orbitals in reach.

Then what does this nomenclature actually mean?

As Martin commented, it is more like the following:
$$\mathrm{sp^n \equiv s^\frac{1}{n+1}p^\frac{n}{n+1}}$$ or to say it slightly different, a $\mathrm{sp^n}$-bond has
$\frac{100}{n+1}~\%$ s-amount and $\frac{100\,n}{n+1}~\%$ p-amount.

\begin{array}{lll}
\hline
\text{bond} & \text{s-amount} & \text{p-amount}\\
\hline
\mathrm{sp} & 50~\% & 50~\%\\
\mathrm{sp^2} & 33~\% & 67~\%\\
\mathrm{sp^3} & 25~\% & 75~\%\\
\mathrm{sp^4} & 20~\% & 80~\%\\
\mathrm{sp^5} & 17~\% & 83~\%\\
\ldots & \ldots & \ldots \\
\mathrm{sp^n} & \frac{100}{n+1}~\% & \frac{100\;n}{n+1}~\%\\
\hline
\end{array}


What did actually happen when Staudinger tried to synthesise diamond?

This is not really an answer, but more like a historical and literature review. Anyway, I hope it can help to further understand the reaction the the ideas behind.

First of all, thaditional Wurtz reaction suggests formation of 2D bonds such as $\ce{R^1 - R^2}$, whereas the idea in Staudinger's experiment is to obtain hree-dimensional network, ideally the the $\mathrm{sp^3}$ diamond hybrid state. Reaction of alkali metals with organics with high halogen content most likely will always lead to an explosion. For this reason qualitative Lassaigne's test should always be performed carefully, and one must not use fire extinguishers with tetrachloromethane for burning sodium.

According to the great review in [1, p. 8], the ﬁrst documented record of a combustion reaction between a metal and halocarbon compound refers to the ways of obtaining aluminium carbide for technical methane production. In 1907 Camille Matignon observed a very exothermic reaction between hexachlorobenzene and aluminium [2]:

$$\ce{10 Al + C6Cl6 ->[\pu{225^\circ C}] 2 Al4C3 + 2 AlCl3}$$

In 1913 H. Staudinger published his attempts to synthesize ethylenedione $\ce{O=C=C=O}$ [3]. He suggested  abstracting the halogen atoms ($\ce{X}$) from either oxalyl chloride and -bromide $\ce{(COX)2}$ by a reaction with potassium or its sodium alloy ($\ce{NaK}$). This is one of the experiments performed by Staudinger (doing science was much more fun back then):

  In einer starken Bombe wurde aus 5 g Kalium und 5 g Natrium die flüssige Kalium-Natrium-Legierung hergestellt, dazu ein Röhrchen mit 8 g Oxalylbromid gegeben, und die Bombe nach dem vollständigen Evakuieren zugeschmolzen. Wenige Sekunden nach dem Zertrümmern des Röhrchens trat äußerst heftige Explosion unter gewaltiger Detonation ein, die nur durch glücklichen Zufall ohne erhebliche Beschädigung des einen von uns verlief.
  Liquid potassium-sodium alloy was prepared from 5 g of potassium and 5 g of sodium in a strong bomb, then a tube of 8 g of oxalyl bromide was added, and the bomb was melted after complete evacuation. A few seconds after the tube had been trimmed, and an extremely violent explosion ensued, with a great detonation, which proceeded only by chance without considerable damage to one of us.

Formation of an unknown highly reactive intermediate such as Dioxyacetylenkalium was suggested, but it was not confirmed experimentally.

A decade later, in 1922, Staudinger reported explosive reactions of alkali and alkaline earth metals with partially and perhalogenated solvents ($\ce{CH2Cl2}$, $\ce{CCl4}$). He again assumed the formation of different very instable species on contact that would be very sensitive to mechanical impact and thus trigger an explosive reaction. [4, part "5. Explosion mit Alkalimetallen"].

At the same time he also proposed usage of reaction between various alkali and alkaline earth metals ($\ce{Na, K, Ba}$) with halogenated organics with short $\ce{R}$ ("Rest des Moleküls") ($\ce{CH2Cl2}$, $\ce{CHCl3}$, $\ce{CCl4}$, $\ce{CH2Cl-CH2Cl}$, $\ce{CHCl2-CHCl2}$, $\ce{CHCl2-CCl3}$, $\ce{C2Cl6}$, $\ce{CHCl=CHCl}$, $\ce{CHCl=CCl2}$, $\ce{CCl2=CCl2}$, also corresponding $\ce{Br, I}$-derivatives) as an ammo detonating charges:

DE Patent 396,209 1922 "Verfahren zur Darstellung von Sprengmitteln" (DPMA Publikationsnummer CH 100 199 A);
DE Patent 391,346 1922 "Verfahren zur Initialzündung von Sprengstoffen" (DPMA Publikationsnummer CH 100 200 A).

He also underlined that the more halogenated the molecule is and the less other groups are present, the more vigorous the explosion will be. In addition, potassium shows better overall performance due to electropositivity ("Vergleicht man die Reaktionsfähigkeit der ver
schiedenen Alkalimetalle, so ist das elektropositivste, das Kalium, das reaktionsfähigste, Lithium das reaktionsträgste"), but various amalgams or alkali metal mixtures will outperform potassium (flux).

Based on this, later its been proposed that the explosive reaction between sodium and tetrachloromethane to make diamond. His idea was implemented in late 1980s for obtaining diamonds in TNT/RDX detonation soot [5].

Finally, an autoclave synthesis based on original Staudinger work (funny enough, they didn't cite it) implementing sodium and tetrachloromethane [4] resulted in successful nanodiamond synthesis (in a 2% yield), and, subsequently, a well-recieved report in Nature in 1998 [6]:

$$\ce{CCl4 + 4Na ->[\pu{700^\circ C, 48 h};  Ni-Co (cat)] C(diamond) + 4 NaCl}$$

  We call it the reduction-pyrolysis-catalysis (RPC) route. According to the free energy calculations $\Delta G^\circ (\text{diamond}) = \pu{-416.7 kcal mol^{-1}}$ and $\Delta G^\circ (\text{graphite}) = \pu{-417.4 kcal mol^{-1}}$,
  $$\ce{CCl4 + 4 Na -> C(diamond) + 4 NaCl}$$
  and
  $$\ce{CCl4 + 4 Na -> C(graphite) + 4 NaCl}$$
  are thermodynamically spontaneous. Formation of graphite (amorphous carbon) and diamond is possible, and the yield of graphite (amorphous carbon) and diamond may be determined by kinetics.

The role of $\ce{Ni-Co}$ catalyst is not clear yet, and athors suggested than experimentation with other transition metals can potentially improve the process.

Review [1] points out that idea of diamond synthesing is mentioned in Staudinger's monography [7], though I failed to find this book, so unfortunately I cannot provide any substantial details from there.

Bibliography

Koch, E.-C. Metal-fluorocarbon based energetic materials; Wiley-VCH: Weinheim, 2012. ISBN 978-3-527-32920-5.
Matignon, C. Comptes rendus hebdomadaires des séances de l’Académie des sciences 1907, 145, 676–679.
Staudinger, H.; Anthes, E. Ber. Dtsch. Chem. Ges. 1913, 46 (2), 1426–1437. DOI 10.1002/cber.19130460222.
Staudinger, H. Angewandte Chemie 1922, 35 (93), 657–659. DOI 10.1002/ange.19220359302.
Greiner, N. R.; Phillips, D. S.; Johnson, J. D.; Volk, F. Nature 1988, 333 (6172), 440–442. DOI 10.1038/333440a0.
Li, Y.; Qian, Y.; Liao, H.; Ding, Y.; Yang, L.; Xu, C.; Li, F.; Zhou, G. Science 1998, 281 (5374), 246–247. DOI 10.1126/science.281.5374.246.
Staudinger, H. Arbeitserinnerungen; Hüthig Verlag, 1961.


How do I figure out the amount of water (mL) and amount of Aluminum (g)?

The first step you need to take is to determine the moles of your copper sulphate.

This is simple arithmetic. $$moles = \frac{mass}{molar.weight}$$
The mass of the substance is your measured 5 grams. The molar mass of your compound can be obtained from Wikipedia. In this case, anhydrous copper sulphate has a molar mass of $159.61\frac{g}{mol}$.

Plugging into the equation: 0.0031 $moles = \frac{5}{159.61}$

Now that you know the moles of your copper sulphate, you should be able to work out the the needed amounts of your other compounds using stoichometry.


What is CIF (a PCB developer) and what does it do?

Developers are usually solutions of $\ce{NaOH}$ or $\ce{KOH}$ (perhaps with some additives). Negative photoresists uses organic developers/solvents. From your description, I assume, that you use a positive photoresist. Positive photoresists are more soluble after exposition to the UV light - polymer bonds are disturbed and therefore more soluble (in areas where exposed). Please see illustration. For more detailed information look for Photoresists and Photolithography.


If you add AlCl3 to water, why does the pH decrease?

A more clarified reaction in addition to Berry Holmes' answer:

$$\ce{AlCl3 + 4H2O <=> [Al(OH)4]- + 3Cl- + 4H+}$$

The reason this happens is the Lewis-Acidity of $\ce{Al^3+}$ as well as the higher affinity for $\ce{OH-}$ ions when compared with $\ce{Cl-}$.


NCl₃ + H₂O → NH₃ + HOCl , the resulting solution is acidic, basic or neutral?

Here's our initial reaction, $$\ce{NCl3 + 3H2O -> NH3 + 3HOCl}$$ $\ce{NH3}$ is a weak base and $\ce{HOCl}$ is a weak acid, but there is more of the acid, so at this point the solution would be mildly acidic. However $\ce{NH3}$ and $\ce{HOCl}$ are known to react as follows $$\ce{NH3 + HOCl->NH2Cl + H2O}$$ and in aqueous solution the resultant chloramine can decompose as shown in the following equation $$\ce{3 NH2Cl ⟶ N2 + NH4Cl + 2 HCl}$$  Clearly, an acid is generated in this last reaction.  So although the initial products are neutral, eventually the solution will be strongly acidic.


Why does the SCF not converge in a structure optimization with GAMESS of the thiourea molecule at the PM6 level of theory?

TL;DR: PM6 only works for elements up to fluoride in GAMESS.

I have played around with the GAMESS version installed at our cluster, the ancient 5 DEC 2014 (R1) and I came to the conclusion that there must be a 'bug' in the implementation of PM6 with GAMESS. (I reproduced the problem with 20 APR 2017 (R1); I have also downloaded 30 Jun 2019 (R1), but I won't try it as I think the result is the same.) When I tried it with Gaussian 16 Rev. B.01 on the PM6 level of theory it converged fine.

Actually, I have looked at the documentation, and it seems to be officially not supported (from INTRO.DOC):

Capabilities

   [...]
   3. Calculates semi-empirical MNDO, AM1, or PM3 models
      using RHF, UHF, ROHF, or GVB wavefunctions.
   [...]

Or from INPUT.DOC:

        * * * semiempirical basis sets * * *

GBASIS = MNDO - selects MNDO model Hamiltonian
       = AM1  - selects AM1 model Hamiltonian
       = PM3  - selects PM3 model Hamiltonian
       = RM1  - selects RM1 model Hamiltonian
       = DFTB - selects tight binding Hamiltonian

I am pretty sure the error is associated with sulfur. I have tried a few molecules, among them $\ce{H3C-NH2}$, $\ce{(H3C)2NH}$, $\ce{(H3C)2CO}$ (see input below), urea, which converged fine. Whenever I tried a molecule containing sulfur the SCF seemed to be oscillating. I have tested $\ce{CS2}$, $\ce{SO2}$, $\ce{(H3C)2CS}$, and (obviously) thiourea. (With PM3 everything worked fine.)

My best guess is that they simply didn't finish implementing it. You could issue a support ticket on GitHub: issues. I found an article by (the here known) Jan Jensen from 2013: Project status: PM6 and PCM in GAMESS. And then from 2014: PM6, DFTB, GAMESS, enzyme design, protein-ligand binding, and protein structure validation. I stopped there.

I'd recommend using a different program, the obvious choice would be MOPAC (http://openmopac.net/). Gaussian has it implemented and a couple others probably too. Or you switch to a completely different method like xtb, HF-3c (in ORCA), DFTB (e.g. DFTB+), etc.. Depending on your system size, you can probably also get DFT based calculations done.

Here is also something related: GAMESS crash course?

 $CONTRL SCFTYP=RHF RUNTYP=OPTIMIZE MAXIT=100 MULT=1 COORD=UNIQUE
    NOSYM=1 $END
 $BASIS GBASIS=PM6 $END
 $DATA
 C3H6O
C1
CARBON      6.0      0.156332162        -0.059231571         0.417752882
OXYGEN      8.0      1.200569512        -0.535604279         0.831319560
CARBON      6.0     -0.037914597         1.466530485         0.340822109
HYDROGEN    1.0      0.279389785         1.860652999        -0.631337255
HYDROGEN    1.0     -1.089141584         1.742544046         0.481697509
HYDROGEN    1.0      0.545018653         1.982387227         1.112307247
CARBON      6.0     -1.053925162        -0.882111362        -0.061565447
HYDROGEN    1.0     -0.987243509        -1.102851552        -1.132983096
HYDROGEN    1.0     -1.121341433        -1.838731138         0.468944868
HYDROGEN    1.0     -1.993938204        -0.344384630         0.106761952
 $END


Sulfuric acid from elemental sulfur and hydrogen peroxide

You will likedly find it faster to perform the first oxidation by burning the elemental sulphur in a little tinfoil boat to produce gaseous sulphur dioxide. then set up some aperatus sucking the gasses produced to bubble up through the hydrogen peroxide to produce sulphuric acid. The peroxide does not oxidize sulphur dioxide to sulphur trioxide because it forms sulphuric acid directly.

Does require some aperatus but will be much faster than waiting for the peroxide. Calculate the gibs free energy if you are not sure if a reaction will be spontaneous.


Computing i0 in the Butler-Volmer equation: How to interpret Wikipedia and Wolfram Alpha

k0 has units of frequency (1/time)
All concentrations are surface concentrations (mol/area)

Sometimes people use k0 with units length/time and volume concentrations (mol/volume)

On Wikipedia: Coxy stands for Ox concentration * at equilibrium *. Cred stands for Red concentration * at equilibrium *. It's for 1 electron. Multiply by n for n electrons.

On Wolfram: [Ox] stands for Ox concentration at electrode, [Red] stands for Red concentration at the electrode. These concentrations are not necessarily at equilibrium.


Weak acids compared to strong ones in pH > 7 solution

When the ratio of conjugate base concenteration to conjugate acid concenteration is more than 1000 , we  say that the  dissociation of acid is complete.
According to Henderson–Hasselbalch equation the ratio of conjugate base concenteration to conjugate acid concenteration gets more than 1000 , when PH-PKa is more than 3.
For strong acids, this ratio is more than 1000 in PH>2 ((  actually it is true for most strong acids )) , but for weak acids , there is no specefic PH . Actually, higher PH , higher dissociation ratio.
So ,when the PH gets higher, the dissiciation ratio gets greater and more poroton will be given up.
Actually, when PH-PKa gets more than 3 for weak acids , they behave like strong acids and allmost all of the acid will dissociate.


What is the relation between the following two optical isomers?

you can check this out it is not exactly the same but Cl is replaced by OH


Can two metals combine to form a compound?

In short, yes they can. What exactly forms however is quite complicated.
Certain rules have been created to estimate structures but in general it's not as easiy as predicting simple inorganic compounds.

It's part of a topic that we call intermetallic compounds. Sometimes also called alloys, although this term should not be used. We destinguish four different type of compounds that might arise from two metals.

Intermetallic Compound: This has a clear stoichiometry (Sodium thallide)
Intermetallic Phase: There is a certain range
Solid Solution: Two metals can be mixed in all ratios
Alloys: A compound that has more than one of these phases present

Why these compounds form however is often attributed to certain abilities like radii, sometimes electronegativity and something we call valence electron concentration.

In general we seperate the periodic table into four groups:

Group I + II
Group III - XI
Group XII - XIV (parts)
The rest (if metallic)

Now we can consider each combination of two, either from the same group or from different groups. For example group I + group I will often yield simply a solid solution are both compounds simply mix. Group I + say group XIII however have quite some difference in electronegativity. Here we get stronly polar or perhaps even more like ionic bonds, as compounds like NaTl form.

In between stuff like Laves-phases can form, compounds that mostly form deltahedra only, or Hume-Rothery-Phases like brass. At some point you get a lot of possible phases and some means to explain why they form and not something else depending on size, electronegativity, valence electron concentration, etc.

I had a few lectures on these topics and every new lecture would basically introduce a new set of compounds with new rules. You could relate them back to this simple scheme I gave you above but it would always have it's own limitations and set of rules.


RobChem has already pointed out in his comment that your assumption is not quite correct.

Take a look at $\mathrm{p}K_\mathrm{a}$ values for $\ce{ROH}$ and $\ce{RSH}$ in water from the CRC Handbook of Chemistry and Physics and/or other online sources, such as this or that.

\begin{array}{lrr}
\mathbf{R} & {\ce{\mathbf{OH}}} &\ce{\mathbf{SH}}\\
\hline
\ce{H} & 15.7 &7.0\\
\ce{Et} & 15.9 & 10.6\\
\ce{(H3C)3C} & 18.0 & 11.7\\
\ce{C6H5} & 9.9 & 6.6\\
\end{array}

In all the cases above, the $\mathrm{p}K_\mathrm{a}$ value for $\ce{RSH}$ is smaller than that of $\ce{ROH}$.


Can I synthesize iron acetate like this?

Red iron oxide, Fe2O3, from a paint or art supply store, or as polishing rouge (no additives).  dissolve in aqueous acetic acid, filter off any solids, and go on from there.

Dissolving iron in acetic acid will give you Fe(II)acetate.  Air or peroxide oxidation may (probably will) give you Fenton chemistry and muck up your organics.


How to find the molar mass of an unknown metal carbonate through a gas evolution reaction?

$$\ce{HCl + M2CO3 -> MCl + CO2 + H2O}$$

Since you already figured out, that there is a one-to-one ratio of carbonate to carbon dioxide, you can simply write
$$n(\ce{CO2}) = n(\ce{M2CO3}).$$

With the formula for the Molar mass,
$$M(\ce{M2CO3}) = \frac{m(\ce{M2CO3})}{n(\ce{M2CO3})},$$
you can simply substitute one for the other, hence
$$M(\ce{M2CO3}) = \frac{m(\ce{M2CO3})}{n(\ce{CO2})}.$$


1.0 g of AlCl3 is added to 100.0 ml of water. What would be the resulting pH?

Not being given any other information assume the reaction goes to completion and that it is:

$$\ce{AlCl3 + 6H2O ->[excess H2O] Al(H2O)5(OH)^{2+} + H+ +3Cl-}$$

molecular mass $\ce{AlCl3 = \pu{133.34 g/mol}}$. So:

$$\ce{\pu{mol AlCl_{3}} = mol [Al(H2O)_5OH^{2+}] = mol [H+]} = \frac{\pu{1 g}}{\pu{133.34 g/mol}} = \pu{7.496\cdot 10^{-3} mol}$$

$$\pu{Molarity} = \dfrac{\pu{moles}}{\pu{volume}} = \dfrac{\pu{7.496\cdot 10^{-2} mol}}{\pu{0.100 L}} = \pu{0.07496 M}$$

pH = - log(0.07496) = 1.12

Now the truth is a bit more complicated. You can assume the reaction happens in two steps. The first step will go to completion.

$$\ce{AlCl3 + 6H2O ->[excess H2O] Al(H2O)6^{3+}  +3Cl-}$$

The second step won't go to completion since $\ce{Al(H2O)6^{3+}}$ is a weak acid. There will be an equilibrium favoring the left side. But with no value for $K_\mathrm{eq}$ you are stymied.

$$\ce{Al(H2O)6^{3+} ->[excess H2O] Al(H2O)5(OH)^{2+} + H+}$$

$$K_\mathrm{eq} =\dfrac{\ce{[H+][Al(H2O)5(OH)^{2+}]}}{\ce{[Al(H2O)6^{3+}]}}$$


Dissolving Metal Reductions of Organic Compounds

There are plenty of examples where ethanol is present at over 10ml/g for this reaction. The key thing is that this is transfer of a single electron from Na initially. Where would this go onto ethanol? Clearly, it is faster to transfer an electron to an unsaturated compound i. e. The carbonyl compound. Once this occurs, protons are transferred. The movement of electrons and protons is fast compared to getting the two requisite H's together to form hydrogen and drive the alternative reaction.


What happens to a quaternary ammonium cation at a pH above its pKa?

Amine compounds in which the nitrogen is bound to four carbon atoms through covalent bonds are known as quaternary ammonium salts. They are stable compounds that will not converted back to their relevant amines by treatment with base. Generally, they are very soluble in water because of their ionic nature, and also soluble in commonly used organic solvents due to (long chain) organic substituents (e.g.: phase-transfer catalysts). Because they do not contain a free electron pair on nitrogen, in general, the quaternary ammonium ions are inert towards chemical reactions. Thus, pH changes course only them to change their counter ions.

In fact, not only quaternary ammonium salts are relatively unreactive, but also, they do not diffuse readily across biological membranes when used in medicine, because they exist as permanent cations.


Temperature dependency of enthalpy H and entropy S

Three step process to evaluate temperature dependence

There is a change in enthalpy and in entropy when changing the temperature of a sample. It follows that the enthalpy of reaction and the entropy of reaction can also change when changing the temperature. The easiest way to get values for a temperature $T_2$ if you have values for a temperature $T_1$ is to do a thought experiment with three steps, starting with reactants at the temperature $T_2$:

Change temperature of reaction mixture from $T_2$ to $T_1$
Let the reaction proceed at a temperature of $T_1$
Change temperature of product mixture from $T_1$ to $T_2$

The net process is the reaction at the temperature $T_2$, where you don't know the entropy and enthalpy change yet.

Changes depend on differences in heat capacity

If we are working at constant pressure, the enthalpy change is (with $C_P$ the heat capacity of the reaction mixture):

$$\Delta H = C_P \Delta T$$ and the entropy change is:

$$\Delta S = C_P \ln \frac{T_\text{final}}{T_\text{initial}}.$$

Because steps 1. and 3. change the temperature in opposite directions, a lot cancels out. If we define  $\Delta C_P$ as the difference in heat capacity between reactant and product mixture, we get:

$$\Delta(\Delta_r H) = \Delta C_P \Delta T$$

$$\Delta(\Delta_r S) = \Delta C_P \ln \frac{T_2}{T_1}$$

Compared to other changes with temperature (e.g. Gibbs energy or equilibrium constant), these changes tend to be modest. To give an idea of the magnitude, the molar heat capacity of a diatomic ideal gas is $c_P = 3.5 R$, or about 30 J / (mol K). Given that values for $\Delta_r H$ are typically given in kJ/mol, the change in enthalpy when changing the temperature is relatively small. For the entropy, the "correction term" is not proportional to $\Delta T$ but to the logarithm of the ratio of temperatures. If you start at room temperature and increase the temperature by ten Kelvin, the value of the logarithm will be about 0.03, so the effect is also relatively small.


DNA contamination

DNA is water-soluble and non-volatile (large molecules). You would not expect it to be present in a butane extract. However, traces of DNA are difficult to remove. If you look for DNA (using PCR, for example), you will probably find it. After all, it is a forensic molecule of choice.

Contamination with living microbes

I don't think living cell would survive the extraction. Fungal spores might, though.

Human consumption

There are lots of legal foods where the presence of microbes is desired - yoghurt, beer, cheese are just some examples. All fresh food has its share of microbes, and healthy humans have them too - on the skin and in our digestive tract. Many of these microbes have a role in keeping us healthy, but some make us sick.

As the laws concerning your project differ with location, and the opinions on benefits and risks of your product range widely, I won't comment on those except to say that residual butane would concern me more than traces of microbial DNA.


The following is just a mind game:

Oxidize at the benzylic position to get an arylketone using oxone
Convert the ketone to an ester by Baeyer-Villiger reaction
Cleave the ester
Run a Barton-McCombie deoxygenation on the resulting alkanol

Please note that I havent't done it and I'm not planning to. I'd rather try to avoid the benzyl group in the starting material if possible.


Why does magnesium form covalent bonds with carbon?

The crystal structure of calcium carbide gives a subtle hint of covalent bonding.  It looks like a rock salt structure, but note that the carbon atoms in the anions have their lone pairs directed towards the calcium atoms and so are favorably oriented for covalent overlap. The structure might be considered either a rock salt structure or a series of parallel $\ce{-Ca-C#C -}$ polymeric chains.

Source


Quoting Wikipedia,

  Plasticizers (UK: plasticisers) or dispersants are additives that increase the plasticity or decrease the viscosity of a material.

As far as most research papers say, water is not plasticized, but rather it acts as a plasticizing agent for other substances such as starch films as mentioned in this article. Water makes these films permanently deformed upon external stress and resists reverting back to its original shape.


Salt Water Metal etching 50 (Euro-)Cent Coin

ECM (electrochemical machining) is a safe (besides risk of getting electrocuted: keep your hands dry) way to carve metal. However, only electrically conductive materials can be machined. Both aluminum and copper are conductive, therefore you can carve it safely.


What is the reason for isotropy in amorphous solids?

Assume that the spheres in the drawing are atoms.

In the second picture atoms are not orderly arranged. No matter where you go in crystal of a amorphous solid, the disarrangement will be same. Since in every direction, disarrangement is same, physical properties along every direction is also same. Therefore, amorphous solids are isotropic.

In the first picture, the arrangement is orderly. So, when you go in any direction, the arrangement will be different. It is also clear in the rectangles shown in the picture. Try doing it yourself on a piece of paper. Try finding any four different directions along which the arrangement is same.

Ps: Ignore my poor drawing skills.


What is it in the composition of natural fabrics that cause them to retain smoke odors versus synthetics?

Adsorption is a very complicated phenomenon which involves so many interactions. Smoke has thousands of chemicals too, therefore one cannot pin-point a single cause. It is just like the concept of stains. Some stains on cotton never come off easily, which means that molecular interaction is very strong between the dye in the stain and cotton fiber. As a rough estimate, the -OH groups on cellulose may be providing the hydrogen bonding sites to nitrogen containing aromatic (in a chemical sense) compounds, hence stick for a long time. I am drawing an crude analogy because silica which has plenty of silanols (Si-OH) groups bind amines too strongly.


Which ring expansion in cyclobutyl(cyclopropyl)methanol is favourable?

It is clear to me that no matter which ring you open the first, you'd get the same product as A. Michael Lautman gave the correct product , first opening the cyclopropyl ring. Michael's mechanism is fair, but I'm reluctant to say it is the most preferable path because it gives you bicyclobutyl $2^\circ$ carbocation as an intermediate, which needed extra 1,2-hydride shift to get stabilized as $3^\circ$ carbocation (total of 5 steps). However, if you have opened the cyclobutyl ring first, you'd get the same final product A, through a relatively low energy cyclopropylcyclopentane carbocation (total of 5 steps) even though it is also needed a 1,2-hydride shift (in relatively more stable cyclopentyl $2^\circ$ carbocation when compared to $2^\circ$ carbocation in cyclobytyl ring in Michael's mechanism) to get stabilized as $3^\circ$ carbocation. Complete mechanism is depicted below:

This mechanism is supported by following reference:

G. K. Surya Prakash, V. Prakash Reddy, G. Rasul, J. Casanova, G. A. Olah, “The Search for Persistent Cyclobutylmethyl Cations in Superacidic Media and Observation of the Cyclobutyldicyclopropylmethyl Cation,” J. Am. Chem. Soc. 1998, 120(51), 13362–13365 (https://doi.org/10.1021/ja9828962).


How to determine which of two similar compounds has the larger dipole moment?

The dipole is the vector sum of each atoms's partial charge times its distance from the origin. Lacking this information individual bond dipole values can be used

In the first case it is that of $\ce{CH3}$ vs. the halogen, and then which halogen has the greater electronegativity. Looking up some bond values it seems that a $\ce{C-F}$ bond has the slightly larger dipole ($1.6$ vs $1.5$ Debye ) so the difference is small and not easy to predict as the C-F bond is shorter than the C-Cl.

In the second molecule, geometry can be used as the two groups are the same on each molecule but positioned differently. The para molecule will have the smaller dipole vs. the meta if both groups act in the same direction, either both to withdraw or add electron density with respect to the ring, as then they act to cancel one another to some extent.

If they act in a different direction then the para molecule will have the larger dipole vs the meta. Looking for dipole moments it seems that phenol has a dipole of $1.45$D pointing into the ring and nitrophenol $3.95$D out of the ring so act in different directions and so the para should have the larger dipole.

(To add two vectors of length b and c at an angle $\theta$ the resultant c is found using the cosine rule $c^2 = a^2 + b^2 -2ab\cos(\theta)$ . If the dipoles point in the opposite direction make either a or b negative)


Which is most reliable excitation energy: LUMO-HOMO energy or TD-DFT excitation energies?

When you can, use TDDFT, in part because of the oscillator strengths

I like to point out to my group that optical excitations consist of two things consistently (e.g., in a UV/Vis spectra). There are the excitation energies (the x-axis) and the intensity or extinction coefficient (the y-axis). To some degree, the peak widths give a bit more information (e.g., into vibrational structure).

The example you give above is a great example. Most of the excitations are forbidden and have no oscillator strength. I'd pick the same excitation, although I'm curious what the molecule is - usually there are much higher oscillator strengths.

You're right that there are papers and groups that use the HOMO and LUMO eigenvalues. While some DFT methods (e.g. optimally tuned range-separated hybrids) do provide HOMO-LUMO energies that correspond nicely with experiments, there's nothing physical about Kohn-Sham orbital eigenvalues. Moreover, in general, unfilled orbitals are not handled well by DFT methods.

It's been a while since I looked for reviews, but general TDDFT excitation energies are ~0.2-0.3 eV from experimental absorption energies after correcting for solvent effects, etc.

In my group, we also do what I call a "consensus" model. That is, when you run a TDDFT calculation, you get the HOMO and LUMO energies anyway. Since there are slightly different random and systematic errors from each approach, you can calibrate both against experiment and use a multivariate linear regression. The result decreases the average error significantly.


You burn the sulfur and lead the fumes through a solution of $\ce{H2O2}$. Afterwards you boil that solution for a moment to remove the rest of the peroxide.

Use a vertical tube reactor or something so you can lead a soft stream of oxygen gas (or pressurised air, in a pinch) over the sulphur.


Would adding acetic acid from household vinegar to washing machine make it too dilute to be effective?

As a weak acid, acetic acid will act as a buffer in the washing machine solution.

Following similar calculations from the first answer of the question,What's the pH of Vinegar when it generally contains 5% Acetic acid?, we can calculate the concentration and pH of acetic acid in the washing machine solution. 1 Cup is ~ .237 L and 40 gal is ~ 152 L. Using the concentration of acetic acid in vinegar from the post above (0.847 mol/L):

(0.847 mol/L)(0.237 L)=(X mol/L)(152 L), X=0.00136 mol/L

pH of the washing machine solution = 0.5[4.76-log(0.00136)] = 3.80

So, although the acetic acid is very dilute, there is still a significant pH drop from the addition of 1 cup of white vinegar. Changes in pH can strongly affect the solubility of many compounds, which would seem to be helpful in the cleaning process. Furthermore, the denaturing of proteins is promoted at lower pH, so lowering the pH could be particularly useful in the removal of protein-based stains (grass, bodily fluid, egg, etc.).


Why are silicon and germanium used as semiconductors but not other metalloids?

The statement, "metalloids like [a]ntimony and [t]ellurium are not used as semiconductors", is untrue:

Tellurium thin-film transistors have been fabricated.
Bismuth nanowires have been used as electronic gates.
Carbon, in both diamond and graphene allotropes, is used.
Phosphorus, in the form phosphorene, structurally similar to graphene, has been used as a semiconducting photodetector.

That said, there are various reasons why these semiconductors have not replaced silicon, which is plentiful, usable to beyond 400 K, resistant to degradation by air, and has an established fabrication technology, for the majority of electronic uses. GaAs, GaN, SiC and other semiconducting compounds are being used increasingly, though, for high-speed electronics, high-temperature devices and lasers.

See the Wikipedia list of semiconducting elements and compounds... it's longer than you might expect.


Generally speaking: knowing the history of your sample (e.g. earlier steps of synthesis allowing / excluding elements that could numerically fit into the list of suggested fragments) and additional spectroscopic characterisation (UV-Vis, IR, Raman, NMR, etc.) may be helpful.

The problem you describe may be attenuated by using a mass spectrometer of higher resolving power.  Instead of submitting the sample to the detector behind a smaller LC-MS coupled analysis, providing a purified sample to a larger, tuneable sector mass spectrometer, for example.

An important alternative is coupling of several mass spectrometers along one line, where after initial ionisation primary fragments are separated from each other, and their corresponding subsequently generated daughter fragments are eventually characterised, like in a tandem MS/MS:

(source)

An overview about the later provides this wikipedia entry.


Potassium vs sodium bicarbonate in fire fighting

To answer this question, firstly we must understand both how fires are started, and how they are perpetuated. It is common knowledge that in order for fire to exist there must be heat, fuel, and an oxidizing agent (usually oxygen). Once started, a fire is then self perpetuated by a radical chain reaction. Removing any of these three ingredients or interrupting the radical chain reaction can be effective ways of extinguishing a fire. Both sodium and potassium bicarbonate are effective at performing several of these actions, though there are slight differences between them.

Differences between $\ce{NaHCO3}$ and $\ce{KHCO3}$

  It is widely believed that fireextinguishing powders can function as both energy-absorbing materials and solid surfaces on which free radicals can be destroyed. Heat may be absorbed by the heat capacity of the solid [$C_p(\ce{KHCO3})=90.05\:\mathrm{J\:mol^{-1}\:K^{-1}}$ vs. $C_p(\ce{NaHCO3})=87.61\:\mathrm{J\:mol^{-1}\:K^{-1}}$], the heat of fusion, the heat capacity of the liquid, heat of dissociation from breaking of chemical bonds [$\Delta H_\mathrm{decomp}(\ce{KHCO3})=1063.28\:\mathrm{kJ\:mol^{-1}}$ vs. $\Delta H_\mathrm{decomp}(\ce{NaHCO3})=995.72\:\mathrm{kJ\:mol^{-1}}$], and heat of vaporization. All of these contribute to the total endothermicity of the fireextinguishing powder.

  From a chemical aspect, it has been found that potassium salts are more effective than sodium salts, and iodide anions are more effective than chloride anions. Presumably, there is a catalytic path for destruction of free radicals, such as $\ce{H}$, $\ce{O}$, and $\ce{OH}$, utilizing the potassium in the salts. It must be remembered that any powder that has a chemical fireextinguishing capability will also have a heat-absorbing (endothermic) capability.$^{[1]}$

The Chain Radical Mechanism

  Combustion of hydrocarbons is thought to be initiated by hydrogen atom abstraction (not proton abstraction) from the fuel to oxygen, to give a hydroperoxide radical ($\ce{HOO}$). This reacts further to give hydroperoxides, which break up to give hydroxyl radicals. There are a great variety of these processes that produce fuel radicals and oxidizing radicals. Oxidizing species include singlet oxygen, hydroxyl, monatomic oxygen, and hydroperoxyl. Such intermediates are short-lived and cannot be isolated.$^{[2]}$

The Chain Radical Breaking Mechanism

  The mechanism by which chemical inhibition occurs when certain powders are added to fuel-air flames is not completely understood. Solid particles may scavenge chain propagating species by surface adsorption or reaction; or alternatively, the solids may vaporize in the combustion zone to produce gaseous products that react with active species via homogeneous gas reactions.

  Most investigators seem to favor the latter mechanism, although direct experimental verification is very lacking. If homogeneous gas reactions are, in fact, responsible for the inhibition, then for cases in which potassium salts are the inhibiting agents, the chemical reactions

  $$\ce{KOH(g) + H->K(g) + H2O}\:\:\:\:\:\Delta H = -33.2\:\mathrm{kcal\:mol^{-1}}$$
  $\hspace{78 mm}$and/or

  $$\ce{KOH(g) + OH -> KO(g) + H2O}\:\:\:\:\:\Delta H = -1.7\:\mathrm{kcal\:mol^{-1}}^{[3]}$$

[1] Fire Extinguishing Powders

[2] Oxygen Fueled Combustion Reaction Mechanism

[3] Flame Inhibition by Potassium Compounds


Why is dG=dH-TdS?

It isn't. That would only be true at constant temperature. In general:

$$\mathrm{d}G = \mathrm{d}H - T\,\mathrm{d}S - S\,\mathrm{d}T$$


Ok, back to the start on all of this....

If you were looking for pure compounds or azeotropes what I'd want is a setup where a fractional distillation column was used with a very precise thermometer and a balance measuring the weight of the distillate. You could then nicely plot temperature vs weight.

Assuming that you just want to detect the pure compounds and/or azeotropes and not collect them.

Part of the problem here is that it takes some time to get the whole fractional distillation column up to temperature. So the temperature of the first drops is suspect.

If you're using a heating mantle you might be willing to boil the pot dry. But you wouldn't want to do that with a flame. So you might lose a bit on the back end of the distillation too.

The fractional distillation column doesn't have an infinite number of plates so the transitions from one compound/azeotrope to the next will be a bit "fuzzy." So if you were trying to collect "pure" compound/azeotrope fractions you'd avoid collecting the distillate from the transitions in those "pure" compound/azeotrope fractions.

Does this help?

EDIT

I guess I'd like two thermometers. One over the pot and one at top of the fractional distillation column.

Some sort of computer controlled temperature setting for the pot would be nice too.


Boiling temperature of salt water at low pressure

According to this 1927 paper, "THE VAPOR PRESSURE OF THE SATURATED AQUEOUS SOLUTIONS OF CERTAIN SALTS" at 25 °C, 6 mol per kg salt solution gives a pressure of 18.5 mmHg instead of 24.5 mmHg for pure water.
So yes there should be a measurable difference (unless salt destroys the diaphragm pump).

Edit: This recent paper "Water vapour pressure above saturated salt solutions at low temperatures", gives the same results


Major products of the reaction of 2-bromo-2-methylpropane and propanoic acid

The commenters are correct; this reaction, based on the steric hindrance of the tertiary alkyl halide and the propionic acid, I would expect a mixture of SN1 and E1, and minimal if any second-order reactions to the mechanism.

Specifically, the propionic acid would slowly approach the bromide substituent, and the acidic carboxyl proton (pKa ~ 5) would begin to associate with the bromide, which is a good leaving group. This slow formation of a tertiary carbocation (stabilized by hyperconjugation of the central empty p orbital through the C-H sigma bonds) then leads to the propionate binding to the carbocation with no rearrangements occurring.

Even though this reaction is at room temperature, E1 should not be ruled out, as there is no reason why a carboxylic acid cannot grab a proton from the 9 available ones on the alkyl halide. Due to the necessity for the proper orientation for beta-elimination though, it is plausible that E1 is less likely.

Regardless, I would project both SN1 and E1 to be the most likely reactions to occur. However, if you were to pick only one, I would pick SN1 based on the reaction temperature (and call E1 a "secondary" major product if that makes sense).


Why +M effect of -NH2 is more than that of -NHR and -NR2?

It is because the bulky alkyl groups provide repulsion to the lone pair such that their orientations change. Due to the change in orientation the %s character changes and the lone pair becomes less available. This is not the case with hydrogen.


pH affects the shape of proteins. In the case of fermentation a collection of enzymes is responsible for the metabolic processes that occur. - an enzyme is a protein which performs a metabolic process. For example sucrase is an enzyme which breaks sucrose down into fructose and glucose. They are kind of like organic catalysts.

Proteins are made of amino acids strung together in a long line, based off DNA. The amino acids bond up into a long polymer. Then cross linking occurs between the R groups on the amino acids. This makes the unique shape of the protein, because the cross-linking turns the protein into a 3-d shape.

Amino acids may be acidic amino acids, or basic amino acids, due to the R group on the amino acids. If the pH is increased, this affects the shape of proteins, by disrupting the bonds in the protein. In the case of fermentation, you say the rate increases when it get's more acidic - when the pH is lower. This is because the organisms - the yeast - producing the enzymes to ferment glucose, have adapted to acidic conditions. This means that through natural selection the enzymes today have been selected because they work best in acidic conditions, which bend the protein into the correct shape to allow fermentation to occur.

Read: http://www.icr.org/article/glycolysis-alcoholic-fermentation/

In this article it says that 10 enzymes turn glucose in pyruvic acid:http://en.wikipedia.org/wiki/File:Pyruvic-acid-2D-skeletal.png

And then two enzymes turn this into ethanol and CO2. The last two enzymes function best in acidic conditions, presumably because they have evolved to work best in the acidic conditions made by the pyruvic cid. However all fermentation stops after the pH drops below about 4.2. The optimum pH is about 4.8 - 5.0. After this we can assume that the pH bends the proteins out of shape too much, and the protein is said to have been denatured.

So to conclude... The rate increases because enzymes are all suited to specific conditions, dictated by natural selection - the conditions they have evolved around - and in the case of fermentation the last 2 enzymes - turning pyruvic acid to ethanol and CO2 - work best in acidic conditions, as they have evolved to those conditions. But increase the pH too much... and the protein denatures, and fermentation stops completely.


Why would CuSO4-EDTA drops be used in Pb titrations?

I'm not an expert on coordination chemistry, and in particularly, metal-indicator complexes in EDTA titrations for heavy metals. But, I assume, the addition of $\ce{[Cu(EDTA)]^2-}$ solution to your unknown $\ce{Pb}$ sample is to get a sharp end-point (example of Replacement or Substitution Titration). Since, the extra metal ion added ($\ce{Cu^2+}$) has already complexed with EDTA, it won't interfere with the calculations. Meantime, both $K_{\mathrm{f}}$s of $\ce{[Cu(EDTA)]^2-}$ and $\ce{[Pb(EDTA)]^2-}$ are so closed ($5\times 10^{18}$ and $2\times 10^{18}$, respectively) that following equilibrium exists:
$$\ce{[Cu(EDTA)]^2- + Pb^2+ <=> [Pb(EDTA)]^2- + Cu^2+}$$
Where, $$K_{\mathrm{eq}} = \frac {[\ce{(Pb(EDTA))^2-}][\ce{Cu^2+]}}{[\ce{(Cu(EDTA))^2-}][\ce{Pb^2+]}} = \frac{K_{\mathrm{f}(\ce{(Pb(EDTA))^2-})}}{K_{\mathrm{f}(\ce{(Cu(EDTA))^2-})}} = \frac{2}{5} = 0.4$$

Thus, released $\ce{Cu^2+}$ cpmplexed with Xylenol Orange to give stable red color at operating pH (e.g., Figure 1A, Ref.2). My assumption is based on EDTA titration of $\ce{Pb^2+}$ and $\ce{Bi^3+}$  ion mixtures in the presence of Xylenol Orange as an indicator. In that titration, pH of a test sample solution containing $\ce{Pb^2+}$ and $\ce{Bi^3+}$ is adjusted to $3$  and add the indicator (Xylenol Orange). The test solution becomes red due to the complex formation of $\ce{Bi^3+}$ and Xylenol Orange (Figure 1B). When titrate with EDTA, at the first end point, the solution becomes yellow, the color of the free indicator at that pH. To titrate for $\ce{Pb^2+}$ at the same solution, it is needed to adjust pH to 5 - 5.5 until the appearance of red color again (indicating the complexation of $\ce{Pb^2+}$ with Xylenol Orange). Now, the slution can be titrated with EDTA until the turning of red to yellow. It is important to note that solution of $\ce{Pb^2+}$ ions does not become red with Xylenol Orange until pH is above 5. That means, the [$\ce{Pb^2+}$ + Xylenol Orange] complex is unstable below pH 5.

Also note that use of ammonium acetate ($\ce{NH4OAc}$) as a neutral buffer is misleading. $\ce{NH4+}$  and $\ce{AcO-}$  are not a conjugate acid/base pair, which means that they do not constitute a buffer at pH 7. However, dissolution of ammonium acetate salt in water results in pH 7, but this pH is highly labile (Ref.1). Ammonium acetate does provide buffering around pH 4.75 (the $\mathrm{p}K_{\mathrm{a}}$ of acetic acid) and around pH 9.25 (the $\mathrm{p}K_{\mathrm{a}}$ of ammonium). Thus, advantage for using $\ce{NH4OAc}$ is a drop from pH 7 to around pH 4.75, which is considerably less dramatic (e.g., addition of $\pu{1 mM} \: \ce{H+}$ to a $\pu{10 mM}$ neutral pH solution will decrease the pH to 5.8) than doing the same to pure water. Remember, during the titration, with out "real" buffer, pH of the solution changed due to release of $\ce{H+}$ ions by complexation.

References:

L. Konermann, “Addressing a Common Misconception: Ammonium Acetate as Neutral pH “Buffer” for Native Electrospray Mass Spectrometry,” J. Am. Soc. Mass Spectrom. 2017, 28(9), 1827–1835 (https://doi.org/10.1007/s13361-017-1739-3).
Interesting reading: O. J. V. Belleza, A. J. L. Villaraza, “Ion charge density governs selectivity in the formation of metal–Xylenol Orange (M–XO) complexes,” Inorganic Chemistry Communications 2014, 47(1), 87–92 (https://doi.org/10.1016/j.inoche.2014.07.024).
Relevant: S. Murakami, “Semi-Xylenol Orange Complexes with Bivalent Metal Ions,” J. inorg, Nucl, Chem. 1981, 43(2), 335–343 (https://doi.org/10.1016/0022-1902(81)90020-8).


TL;DR  Yes, the reaction is feasible at $\pu{60 ^\circ C}$. Just extending my comment into an answer.

Goethite and ferrihydrite(not ferrihydrate) are two different things. They are polymorphs of iron oxyhydroxide ($\ce{FeOOH}$). Goethite is $\ce{α-FeO(OH)}$ while ferrihydrite is amorphous/nanocrystalline hydrated iron oxide, officially formulated as $\ce{FeOOH•1.8H2O}$ but also exist in many hydrated states and so formula is theoretically indeterminate as its water content is variable.

Reaction between iron(III) nitrate hydrate in alkaline solution has been studied by Atkinson et al.1 at different concentrations and amounts. One case is: 50g of $\ce{Fe(NO3)3}$ salt dissolved in 825 ml of water is added to $\ce{200 ml 2.5 N}$ potassium hydroxide solution. At pH 12 and maintaining the temperature at $\pu{60 ^\circ C}$, it took 24 hours to complete the reaction. They also noted that in this goethite synthesis, ferrihydrite is the initial precipitate that results from the rapid hydrolysis of Fe(III) solutions, before aging to goethite at higher temperature.
Pure ferrihydrite synthesis is difficult to isolate because it requires careful control of pH and temperature and its composition can changes if any of this parameter changes. Also, ferrihydrite is the only polymorph which is called nanomineral because its crystal size is in the range of 2 nm to 5nm. Nonetheless, its synthesis has also been discussed in the same paper where ferric nitrate is dissolved in potassium hydroxide but the pH is maintained at 7-8. Vigorous stirring and centrifugation yields the ferrihydrite.
An alternative reaction where the pH is not needed to maintain strictly is by adding $\ce{Fe(NO3)3.9H2O}$ and $\ce{NH4HCO3}$ in a 1:3 mole ratio at $\pu{80-100 °C}$ 2 or iron(II) salt or iron(III) salt with silicon or germanium at pH 5-11 3.
Notes and References

Laboratory synthesis of goethite and ferrihydrite of controlled particle sizes by
Milton Villacís-García, Mariana Ugalde-Arzate, Katherine Vaca-Escobar, Mario Villalobos,
Rodolfo Zanella, Nadia Martínez-Villegas, Boletín de la Sociedad Geológica Mexicana
Volumen 67, núm. 3, 2015, p. 433-446 (PDF)
Novel Synthesis and Structural Analysis of Ferrihydrite
Stacey J. Smith, Katharine Page, Hyunjeong Kim, Branton J. Campbell, Juliana Boerio-Goates, and Brian F. Woodfield
Inorganic Chemistry, 2012 51 (11), 6421-6424
DOI: 10.1021/ic300937f
SYNTHESIS OF FERRIHYDRITE AND FEROXYHYTE, NAGANORI YOSHINAGA and NOBORU KANASAKI
Faculty of Agriculture, Ehime University, Matsuyama 790, Japan, Clay Science 9, 43-51 (1993), (PDF)
Simple hydrolysis of iron(III) nitrate yields $\ce{\alpha-Fe2O3}$. Fusion of potassium hydroxide and iron(III) oxide ($\ce{Fe2O3}$) in air produces potassium(VI) ferrate but in laboratory, $\ce{K2FeO4}$ is prepared by oxidizing an alkaline solution of an iron(III) salt with concentrated chlorine bleach.


Calculating partial pressure of carbon monoxide

For simplicity, let's assign numerical indices to the compounds of interest — all gaseous products participating in equilibrium:
$$\ce{ZnO(s) + \underset{1}{CO(g)} <=> \underset{2}{Zn(g)} + \underset{3}{CO2(g)}}$$
Partial pressure of carbon monoxide can be found via its mole fraction $x_1$ and given total pressure $p$:
$$p_1 = x_1p\tag{1}$$
To find unknown mole fraction, two facts can be taken into consideration:

As Karsten also pointed out, the molar ratio between the products is $1:1$ so that $x_2 = x_3$. Which gives
$$x_1 + x_2 + x_3 = 1 \quad\implies\quad x_2 = x_3 = \frac{1 - x_1}{2}\label{eqn:2}\tag{2}$$

All three mole fractions are linked together in the expression for the average molar mass:
$$\bar{M} = x_1M_1 + x_2M_2 + x_3M_3\label{eqn:3}\tag{3}$$

The average molar mass can be found via average density of the gaseous mix $\bar{ρ}$ by applying ideal gas law:
$$pV = nRT$$
$$p = \frac{nRT}{V} = \frac{mRT}{\bar{M}V} = \frac{\bar{ρ}RT}{\bar{M}}$$
$$
\begin{align}
\bar{M} &= \frac{\bar{ρ}RT}{p} \\
  &= \frac{\pu{0.344 g L-1} × \pu{0.082 L atm K-1 mol-1} × \pu{1300 K}}{\pu{1 atm}} \\
  &\approx \pu{36.67 g mol-1}\tag{4}
\end{align}
$$
Now, knowing numerical value for $\bar{M},$ we can combine \eqref{eqn:2} and \eqref{eqn:3}:
$$\bar{M} = x_1M_1 + \frac{1 - x_1}{2}M_2 + \frac{1 - x_1}{2}M_3$$
Grouping the like terms and expressing $x_1$:
$$
\begin{align}
x_1 &= \frac{M_2 + M_3 - 2\bar{M}}{M_2 + M_3 - 2M_1} \\
  &= \frac{\pu{44.01 g mol-1} + \pu{65.38 g mol-1} -2×\pu{36.67 g mol-1}}{\pu{44.01 g mol-1} + \pu{65.38 g mol-1} -2×\pu{28.01 g mol-1}} \\
  &\approx 0.67\tag{5}
\end{align}
$$
Finally, the partial pressure of carbon monoxide:
$$p_1 = 0.67×\pu{1 atm} = \pu{0.67 atm}$$


Temperature and moles of gas given volume, pressure, enthalpy and constant pressure molar heat capacity

Hint: A general outline of what I think is the idea behind this problem/ This is a quick back of the envelope estimate:
$$\ce{NaN3(s) -> Na(s) + 3/2N2(g)}$$
The reaction given above takes place, and releases:

$$\Delta H = -23.1\ \mathrm{kJ} $$

60% of this heat is transferred to gaseous nitrogen, and since the process is adiabatic no additional heat enters/leaves the system. All of this energy goes into expansion work

$$Q = \frac{60}{100} \times 23.1 \ \mathrm{kJ} = 13.86\ \mathrm{kJ} $$

Temperature of $\ce{N2(g)}$ (assuming initially it is at $298\ \mathrm K$) can be found using (first, assuming one mole of azide decomposes releasing 3/2 moles of nitrogen)

$$Q = nC_{\mathrm m,p} \Delta T$$

$$T_{\text{final}} = \frac{Q}{nC_{\mathrm m,p}} + T_{\text{initial}} = 615.3\ \mathrm{K}$$

Volume occupied by 3/2 moles of nitrogen at the above temperature and given pressure is

$$V = \frac{nRT}{P} = 30.3\ \mathrm{L}$$

You need another $9.7 \ \mathrm{L}$ of gas to completely fill the airbag, I am sure you can complete the calculation. I'll come back and post a full, detailed answer later.

Complete Answer

So, we need a volume of $ 40 \ \mathrm{L}$, at an unspecified temperature.

$$\frac{PV}{n_{\ce{N2}}R} = T_\text{final}$$

Additionally, $$T_{\text{final}} = \frac{Q}{n_{\ce{N2}}C_{\mathrm m,p}} + T_{\text{initial}}$$

I am assuming since the process is adiabatic no additionally heat enters/leaves the system. The way I see it, the energy generated during this decomposition would heat the nitrogen produced to some (as of yet unknown temperature), and would be used for doing expansion work. I assume everything is initially at $298 \ \mathrm{K}$

$$\frac{PV}{n_{\ce{N2}}R} = \frac{Q}{n_{\ce{N2}}C_{\mathrm m,p}} + T_{\text{initial}}$$

From the reaction stoichiometry, $n_{\ce{N2}} = \frac{2}{3}n_{\ce{NaN3}}$

Thus, the expression we are interested in is

$$\frac{PV}{(2/3)n_{\ce{NaN3}}R} = \frac{Q}{(2/3)n_{\ce{NaN3}}C_{\mathrm m,p}} + T_{\text{initial}}$$

Now, it is just a matter of plugging in appropriate values, and solving for $n$

I used,

$ P = 2.5 \ \mathrm{atm}$
$ V = 40 \  \mathrm{L}$
$ R = 0.0821 \ \mathrm{L \ atm \ mol^{-1} \ K^{-1}}$
$ Q = 0.6 \times 23.1 \times 1000 \times n_{\ce{NaN3}} \ \mathrm{kJ}$
$C_{\mathrm m,p} = 29.12 \ \mathrm{J\ mol^{−1}\ K^{−1}} $

Answer given below:

   Solving all of this I get no. of mol of azide $\approx 1.33401 \ \mathrm{mol}$ and, no. of mol of nitrogen produced $ \approx 2.001015 \ \mathrm{mol}$. Additionally, $T_{\text{final}} = 615.307 \ \mathrm{K}$ Plugging all of this back into the ideal gas equation and solving for volume at the given pressure gives me $40.4\ \mathrm{L}$


Is there anything that can react with molecular gaseous nitrogen?

From Wikipedia

  Elemental lithium reacts ("burns") in an atmosphere of $\ce{N2}$ to give lithium nitride:
  $$\ce{6Li + N2 → 2 Li3N}$$

  Magnesium also burns in nitrogen, forming magnesium nitride.
  $$\ce{3 Mg + N2 → Mg3N2}$$


Making a non-CN dissolved gold solution ideal for electroplating

The problem is your amount of gold that you have. Normal chemical baths contain several grams per liter. So you will need to operate in small volumes. You can use this article - Chemical Papers
June 2012, Volume 66, Issue 6, pp 617-620
Date: 23 Dec 2011
Etching and recovery of gold from aluminum substrate in thiourea solution, that is using thiurea. Be aware of the low stability of this bath. The bath will degrade quite fast, but that is common problem of all non-cyanide gold baths.

As the article deals with a gold dissolution, I would suggest to reduce the gold back to a solid state. Use the article or this Hydrometallurgy
Volume 55, Issue 2, March 2000, Pages 137–152, Gold electrowinning from aqueous–alcoholic thiourea solutions and then do the electrodeposition.


Dipole and size of a thioether relative to hydrocarbon

I just realised that I got the dipole business utterly wrong as sulfur has two lone pairs.
I checked dimethyl sulfide, propane and dimethyl ether and was surprised how differently they behaved. The former has a total dipole moment of 1.5 D ref, while the second near zero (0.1 D) as expected from a non-polar molecule, while the last has 1.3 D.
Whereas there is only a 0.03 difference in electronegativity between C and S (vs. .89), the geometry is different, which I did not know about. Wikipedia on says:

thioether article: "C–S–C angle approaching 90°. The C–S bonds are about 180 pm."
ether article: "C–O–C linkage defined by a bond angle of about 110° and C–O distances of about 140 pm".
alkane article: "154 [pm] for a C–C bond [and] with an angle of 109.47° between them."

So the side chain of methionine should actually be able to hydrogen bond with stuff (not electrostatically, but dispersively). So I was surprised to find papers I had not found prior (ex1, ex2).

S-π are probably the strongest force nevertheless (the bond has an enthalpy of 2.6, 1.1 or 0.7 kcal mol-1 depending on the angleref, which is comparable to N−H···:O hydrogen bond (1.9 kcal mol-1))


Basis sets - negative contraction coefficients?

Diagonalization of a matrix means to find the eigenvalues of the matrix and put them into a diagonal matrix:

$$\mathbf{S}^\text{diag}=\mathrm{diag}\left( \lambda_1,...,\lambda_n \right)\tag{1}\label{diagmat}$$

for $\lambda_i$ being the eigenvalues of the matrix $\mathbf{S}$.
When we want to find the eigenvalues of the matrix $\mathbf{S}$ we solve an eigenvalue problem:

$$\mathbf{S}\mathbf{c}=\bar{\lambda}\mathbf{c}\tag{2}$$

for $\mathbf{c}$ being a matrix of the eigenvector, and $\bar{\lambda}$ being a vector of the eigenvalues.
This equation can also be written out in sum-form for the $i$the eigenvalue:

$$\sum_kS_{jk}c_{ik}=\lambda_ic_{ij}\tag{3}\label{eigproblem}$$

Now as you have stated we can define our atomic orbitals as a linear combination of primitive basis functions:

$$\chi_i=\sum_k c_{ik}\phi_k\tag{4}\label{AO}$$

This can be used to evaluate the self-overlap of the atomic orbitals:

$$S_{ii}=\int_\Omega\chi_i^*(\bar{r})\chi_i(\bar{r})\mathrm{d}\bar{r}=\int_\Omega\left|\chi_i(\bar{r})\right|^2\mathrm{d}\bar{r}=\left< \chi_i \left| \chi_i\right.\right>\tag{5}\label{selfoverlap}$$

This can now be expanded out using Eq. (\ref{AO}):

$$S_{ii} = \left< \sum_k c_{ik}\phi_k \left| \sum_j c_{ij}\phi_j\right.\right>\tag{5}$$

Now using the definition of the overlap-integral $S_{jk}=\left< \chi_j \left| \chi_k\right.\right>$ the above equation can be formulated as:

$$S_{ii}=\sum_{j,k}c_{ij}^*S_{jk}c_{ik}\tag{6}$$

Now using Eq. (\ref{eigproblem}) this can be expressed as:

$$S_{ii}=\sum_{j,k}c_{ij}^*\lambda_ic_{ij}=\lambda_i\sum_j\left|c_{ij}\right|^2\tag{7}$$

Now since $S_{ii}$ can be seen to be larger than or equal to zero from Eq. (\ref{selfoverlap}) and $\left|c_{ij}\right|^2$ must also be larger than or equal to zero, it follows that the eigenvalues $\lambda_i$ must all also be larger then or equal to zero.
I.e. from Eq. (\ref{diagmat}) the diagonalized overlap matrix cannot have any negative values.


Why are reversible reactions spontaneous?

A negative $\Delta G^0$ implying that a reaction will proceed spontaneously or a positive $\Delta G^0$ implying that a reaction will not proceed spontaneously is just a "rule of thumb" to give you a feel for what to expect.  But all a negative $\Delta G^0$ really means is that the equilibrium constant for the reaction is large (and formation of the products is highly favored), and a positive $\Delta G^0$ means that the equilibrium constant for the reaction is small (and formation of the products is not highly favored).  However, if you start out with only reactants present initially, irrespective of the sign of $\Delta G^0$, the reaction will proceed spontaneously.  However, if $\Delta G^0$ is negative, at final equilibrium, the extent of the reaction will be greater.


Why is HNO3 a stronger oxidising agent than H3PO4?

Let me create background first. Oxidizing agents are the chemical that helps something else oxidize and itself gets reduced.(reduction in charge)

$\ce{N}$ in $\ce{HNO3}$ is in the +5 oxidation state - how do we know that? $\ce{H}$ is +1, $\ce{O}$ is -2 and the overall $\ce{HNO3}$ has a zero net charge. The same goes for $\ce{H3PO4}$ resulting in +5 oxidation state of $\ce{P}$.

That +5 is the highest common oxidation state for $\ce{N}$. If $\ce{N}$ is reduced to say +4 as in $\ce{NO2}$ or +2 in $\ce{NO}$, it's charge is reduced and $\ce{N}$ is therefore an oxidizing agent for that particular reaction

So, $\ce{HNO3}$ is an strong oxidizing agent because the $\ce{N}$ can be readily reduced.

But $\ce{H3PO4}$ is a poor oxidizing agent. Here are two basic reasons:

Nitrogen does not possess $d$-orbitals in valence shell and so its covalency is limited to 4. $\ce{N}$ can however achieve a formal oxidation state of +5 as  in the $\ce{NO3-}$ ion. The inability of $\ce{N}$ to unpair and promote its $2s$ electron results in that $\ce{N(+5)}$ is less stable than $\ce{N(+3)}$. However, $\ce{P(+5)}$ is more stable than $\ce{P(+3)}$ and phosphoric acid shows less oxidising properties.
The affinity of phosphorus for oxygen is greater than that of nitrogen; as a result, phosphonic acid ($\ce{H3PO3}$) is good reducing agent.

$\ce{H3PO4 + 2H+ + 2e-> H3PO3 + H2O}$; $E=-0.276 \ce{V}$


Why do we take the compressibility factor equal to one when graph is plotted?

I think that at $p = 0$ the compressibility factor ($Z = pV_m/RT$) itself is not particularly well-defined since $p = 0$, $V_m = \infty$ and $T = 0$. Zero pressure is not a state that we can reach.

However, if you use any equation of state and take the limit as $p \to 0$ (or $V_m \to \infty$, which is equivalent), you will find that $Z \to 1$. Try it with the virial equation of state:

$$pV_m = RT\left( 1 + \frac{B}{V_m} + \frac{C}{V_m^2} + \cdots \right)$$

This happens with all gases because the individual gases are identified by their virial coefficients $B, C, \ldots$ and the terms involving these constants vanish as $V_m \to \infty$.

A simple qualitative explanation is that in the limit of $p = 0$ all gases behave ideally; for an ideal gas, $Z$ is of course equal to $1$ because of the ideal gas law $pV_m = RT$.


Material with refraction index similar to Si

Over the range of the visible spectrum, ~1.8 - 3.1 eV, or 700-400 nm, the refractive index (nD) of Si varies from about 3.5 - 5, though it is also absorbent and somewhat reflective over that range. Gallium(III) phosphide, $\ce{GaP}$ is one material that comes close to that. It should be fairly stable in air at STP.

Of course, it all depends on your intended use. At what wavelength should nD match that of Si? At what temperature and in what atmosphere will the material be? Does the crystal structure of the material need to match that of Si, so that a regular layer can be deposited? More information is needed for the answer to be of any use.


Regular black ink is pretty viscous it has a dynamic viscosity ($\eta$) of 4.177.

In comparison a 50% mixture of methanol and glycerol has a viscosity of 5.8

Some other solvents to consider are

Hexanol which has a viscosity of about 5.4
Heptanol which has as viscosity of
about 7.3

These long-chain alcohols, however, will not dissolve significantly in water.

Some more viscosity data can be found here (it's not an ideal place for viscosities but it is free)

Linseed oil is also not water soluble. In any case here is another link

Ultimately it seems like you'd need to try a few different mixtures to get desired viscosity, solubility and evaporation speed.


I was under the impression that acetoacetic ester has more resonance structures so it more acidic.

That's actually wrong. Deprotonated acetoacetic ester has structures where electrons are donated from ester oxygen into conjugated system. This, while stabilizing system as a whole, lessens effect of electron donation from deprotonated carbon. Think about it this way: carbonyl oxygen wants electrons. While in acetylacetone electrons may be harvested from deprotonated carbon only and in acetylacetic ester they may be harvested from both deprotonated carbon and ester oxygen, stabilization effect in the first case is higher.

It is well known, that stabilization effect on acidity of nearby $CH$ grows in row
$COO^- < -CO(NH_2) < -COOR < -COR < -CHO < -NO_2$

The more electron deficient the conjugated system is, the more stabilized carbanion becomes, thus the less resonance structures with transfer of the charge onto carbonyl oxygen the group has, the more it stabilize carbanion. Thus, $\alpha-CH$-acidity in carbonyl compounds correlates with electrophilic activity.

The reason is that negative charge from deprotonated carbon must be delocalized as much as possible, which is especially easy with strongly electron-accepting groups. In case the compound already has an electron pair conjugated with carbonyl group,  this is more problematic.


What exactly is Q in the Nerst equation for a ¨lemon battery¨ of Cu and Zn in varying concentrations of acid?

The 'Q' in Nerst equation is the same Q which you might've used while dealing with chemical equilibria, otherwise known as the Reaction Quotient. The expression is same as the equilibrium constant (K) and yes, they both include partial pressures and concentrations.

The units of partial pressure is the same as the unit of pressure (atm or equivalent), which can pinned down by knowing what partial pressure is in the first place. If I give you a container to begin with, filled with hydrogen and neon, and ask you about the partial pressure of hydrogen, I'd expect that you'd say:

  "Had the container been filled with only  hydrogen and then the pressure exerted by hydrogen would've been measured, then I'd say that partial pressure of hydrogen (in the first case) would have been equal to this measured amount."

Now, to write down Q we need to have the equilibrium reactions with us:

At the cathode (copper in this case), hydrogen ions are forming molecular oxygen due to their reduction, so the situation appears to be like:
$$ \ce{2H+ \text{(aq.)} + 2e- -> H2 (g)}$$
At the anode, zinc is getting oxidized:
$$ \ce{Zn (s) -> Zn^2+ \text{(aq.)} + 2e-}$$

Recall that solid species are omitted from the equilibrium expression, so Q of the reaction would be written as:

$$\displaystyle\pu{\frac{P_{H_2} \cdot [Zn^{2+}]}{[H^+]^2}}$$


Short answer:

Non-bonding MOs are usually lone pairs, but a lone pair does not necessarily mean there is a non-bonding MO.

Long answer:

You are trying to match what you learned from Lewis structures to the orbital picture. In general, that does not work (there may be exceptions).

Let's look at the MO diagram:

It is tempting to assign the bonding $\sigma_{2p}$ to the first (sigma) bond in $\rm O_2$. The two $\pi_{2p}$ bonding orbitals would then results in 2 $\pi$ bonds, hence we get a O-O triple bond. The further two electrons in the anti-bonding $\pi_{2p}^*$ orbitals could be assigned to one oxygen each, thus we have a biradical. The remaining two electron lone pairs might be assigned to the two $\sigma_{2s}$ orbitals. This is actually one of the two valid Lewis structures for $\rm O_2$.
(The other possible Lewis structure has only a double bond and two lone pairs at each oxygen atom.)

The problem is that the lone pairs are ascribed to only one atom, but the MOs are delocalized over the molecule. Thus there is no direct relationship between them. A non-bonding orbital is usually just an atomic orbital, which you may thus ascribe to a lone pair. But that does not mean each lone pair leads to a non-bonding orbital.

It is however possible, to localize the molecular orbitals (by making appropriate linear combinations of MOs) in such a way you get atomic orbitals, which you might match again with the lone pairs. The thing here is, that there are many different sets of orbitals which are all valid descriptions of the bonding situations.

Also note, that MO diagrams are just a simplification as well.

Source of the MO Diagram


Nomenclature of an alkyne with bromo and hydroxy substituents

In this case, the relevant criteria for the numbering are:

lower locants for suffixes (here: ‘ol’)
lower locants for multiple bonds (here: ‘yne’)
lower locants for prefixes (here: ‘bromo’)

The corresponding subsection of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) reads as follows:

  P-14.4 NUMBERING

  When several structural features appear in cyclic and acyclic compounds, low locants are assigned to them in the following decreasing order of seniority:

  (…)

  (c) principal characteristic groups and free valences (suffixes);

  (…)

  (e) saturation/unsaturation:

    (i) low locants are given to hydro/dehydro prefixes (…) and ‘ene’ and ‘yne’ endings;

    (ii) low locants are given first to multiple bonds as a set and then to double bonds (…);

  (f) detachable alphabetized prefixes, all considered together in a series of increasing numerical order;

  (…)

Therefore, the correct name for the compound given in the question is ‘7,7-dibromooct-5-yn-4-ol’ (not ‘2,2-dibromooct-3-yn-5-ol’) since the locant ‘4’ for the suffix ‘ol’ is lower than ‘5’.


Change of first order rate constant with temperature

First of all, decomposition of $\ce{N2O5}$ is a first order reaction, not a zero order reaction. So, your question headline needs to be corrected.

You can justify the given assertion by simple mathematics. For a first order reaction, we know that, $$t_{1/2} = \frac{\ln \space 2}{k}$$ which is independent of initial concentration. It only depends on the rate constant which will change if you change the temperature.

From Arrhenius equation also, we know the relation,
$$\ln\left(\frac{k_2}{k_1}\right) = \frac{E_a}{R}\left(\frac{1}{T_1} -\frac{1}{T_2}\right)$$

So here we have three rate constants for three different temperature. Thus we will have at least two sets of above equations. Dividing one by another we will have,
$$\frac{ln\left(\frac{k_2}{k_1}\right)}{ln\left(\frac{k_3}{k_2}\right)} = \frac{\left(\frac{1}{T_1} -\frac{1}{T_2}\right)}{\left(\frac{1}{T_2} -\frac{1}{T_3}\right)}$$

Now, from the relation between half life and rate constants we can have

$$\frac{\ln\left(\frac{\left(t_{1/2}\right)_1}{\left(t_{1/2}\right)_2}\right)}{\ln\left(\frac{\left(t_{1/2}\right)_2}{\left(t_{1/2}\right)_3}\right)} = \frac{\left(\frac{1}{T_1} -\frac{1}{T_2}\right)}{\left(\frac{1}{T_2} -\frac{1}{T_3}\right)}$$

Now from the given assertion by putting the values you have to just verify whether both the sides are giving the same value or not. In this case both sides, gives the value ≈ 1.2. So, I guess this assertion is thus correct.


Can weathering change the composition of something so that it appears different?

Yes, weather can and probably will change the composition of materials and make them different.

However, forensic analysis usually takes that into account. There are some properties that might change because of weathering (you shouldn't compare water content, for example), but others won't. According to the sample article, the difference was found in the chemical composition of the paints.  Since details were not disclosed, one can only guess what these differences are, but some possibilites are:

Presence or absence of metals in the paints
Substances that wouldn't appear or disappear because of wheatering
Different isotopic ratios of elements
Traces of the solvent indicating they were not the same
Probably others

Assuming the analysis were performed correctly, it is usually not hard to say two samples are different. On the other side, when the results are similar, you can hardly say they are equal, but only that they match and that no differences were found.


Your values for $x$ are perfectly correct and lead to the correct concentrations, though I can understand why you are perturbed by negative numbers. Let's see what's going on.

First, remember that $x$ is not the final concentration. It has to do with the change in concentration. If $x$ is negative, then that means the concentration is changing in the opposite direction than what you thought it was. We'll come back to that.

Let's take one of those values of $x$ and plug it into your equilibrium concentration expressions. Let's start with $x=-0.167$.

For $\ce{N2O4}$:

$$[\ce{N2O4}]=\dfrac{1.00-x}{4.00}=\dfrac{1.00--0.167}{4.00}=\dfrac{1.167}{4.00}=0.292$$

For $\ce{NO2}$

$$[\ce{NO2}]=\dfrac{0.500+2x}{4.00}=\dfrac{0.500+2(-0.167)}{4.00}=\dfrac{0.164}{4.00}=0.041$$

Notice anything? The first concentartion is spot on with one of the answers, and the second is probably just a rounding error somewhere (yours or mine).

  Now about that negative value for $x$.

Notice that you have the concentration of $\ce{N2O4}$ decreasing in your ICE table, but the final concentration is higher than the initial (0.292 vs 0.250). Your assumption about the direction of the shift in the reaction toward was not correct. Not all reactions shift toward the right. How could we have known that the reaction would have shifted toward the left? We could have calculated the reaction quotient $Q_c$ at the initial conditions. If $Q_c > K_c $, then the products are in excess and the reaction shifts to the left. If $Q_c < K_c$, then the reactants are in excess and the reaction shifts to the right. We calculate $Q_c$ by putting the initial concentrations into the law of mass action.

$$Q_c =\dfrac{[\ce{NO2}]^2}{[\ce{N2O4}]}=\dfrac{\left(\dfrac{0.500}{4.00}\right)}{\dfrac{1.00}{4.00}}=\dfrac{0.125^2}{0.250}=6.25\times 10^{-2}>5.9\times 10^{-3}$$

Thus, your ICE table should have the signs of $x$ reversed: the concentration of $\ce{N2O4}$ is increasing and the concentration of $\ce{NO2}$ is decreasing.

$$\begin{array}{|c|c|c|} \hline
\ & \ce{N2O4} & \ce{NO2} \\ \hline
I & \dfrac{1.00}{4.00} & \dfrac{0.500}{4}\\ \hline
C & \dfrac{+x}{4.00} & \dfrac{-2x}{4.00} \\ \hline
E & \dfrac{1.00+x}{4.00} & \dfrac{0.500-2x}{4}\\ \hline
\end{array}$$

$$5.9\times 10^{-3}=\dfrac{\left(\dfrac{0.5.00-2x}{4.00}\right)^2}{\dfrac{1.00+x}{4.00}}$$

Now $x=+0.167$ and $x=+0.338$. $x=+0.167$ leads to the same answer since the signs are switched everywhere.

The other value $x=\pm 0.338$, whether negative or positive, is not valid since $0.500-2(.338)<0$.


Improving Elephant Toothpaste experiment

You might try horseradish as a catalyst, rather than yeast. Use the fresh root, not the prepared sauce, and test a few different plants - some are more effective than others.
Cosmetic supply stores stock 6% (20 volume) $\ce{H2O2}$, but that concentration is more likely to cause skin irritation or bleach fabrics.
The most effective catalyst for decomposing  $\ce{H2O2}$ that I've used is potassium permanganate, $\ce{KMnO4}$, but it has a few liabilities:

It's a strong oxidant and can start a fire when mixed with organic chemicals.
It stains skin and fabric anything from pinkish-purple to black. The stain can be removed with sodium bisulfite, $\ce{NaHSO3}$, sold as an iron-stain remover. That can be a fun demonstration but is not suitable for young students to use.


Covalent atomic radii: oxygen vs nitrogen

The van der Waals radius of nitrogen is larger than that of oxygen, and has been calculated as such for quite a long time.
$$
\begin{array}{lll}
\hline
\text{Reference} & R_\ce{O} & R_\ce{N} \\
\hline
\text{Pauling, 1939} & 1.40 & 1.5 \\
\text{Bondi, 1964} & 1.52 & 1.55 \\
\text{Zefirov, 1974} & 1.29 & 1.50 \\
\text{Gavezzotti, 1983–1999} & 1.40 & 1.50 \\
\text{Batsanov, 1995} & 1.51 & \\
\text{Wieberg, 1995} & 1.5 & 1.6 \\
\text{Rowland, 1996} & 1.58 & 1.64 \\
\hline
\end{array}
$$
As is evident from the table, $R_\ce{N} > R_\ce{O}.$ An interactive webpage of this reference can be seen here, which has the van der Waals radii of quite a few other elements as well.
Edit:
After OP mentioned the abnormality in Covalent Radii, I checked the radii in Concise Inorganic Chemistry, by J.D. Lee. The Covalent radius of Nitrogen $R_N = \pu{0.77 Å}$, whereas that of Oxygen is the same; $R_O = \pu{0.77 Å}$. Lee mentions that the radii are taken from Tom Moeller's Inorganic Chemistry book (which is from 1952)
OP should also be aware of the method used for calculating covalent radii: The Covalent Radius is defined as half the length of a single bond between two uncharged atoms. According to a more recent reference2 (which is used by Wikipedia), $R_N = \pu{73.4 pm}$ and $R_O = \pu{70.2 pm}$
Even more recent references, such as (3) use a different method of calculating covalent radius:

The covalent radius for nitrogen was obtained from N–N bond
distances in substituted hydrazines (only H or C atoms bonded to
N) with three-coordinated nitrogen atoms, excluding all structures
with an R factor larger than 10% or presenting disorder or
errors. The resulting value, $\pu{0.706(13) Å}$, is the average of 2200
crystallographically independent data and was used throughout
with three decimal figures for calculating other radii from N–
element bond distances to minimize rounding-off errors.
The covalent radius for $\ce{O}$ was derived from a
sample of 10,000 acyclic $\ce{C–O}$ bond distances, in a search limited to
purely organic compounds with two-coordinate oxygen and four-coordinate carbon atoms with R ≤ 5%, and the value obtained was $\pu{0.676(28) Å}$

A comparision of covalent radii has also been provided by the authors:
$$
\begin{array}{lll}
\hline
\text{Author} & R_O\ (Å) & R_N\ (Å) \\
\hline
\text{Cordero et al.} & 0.706(13) & 0.661(19) \\
\text{Alcock} & 0.702 & 0.659 \\
\text{Mingos} & 0.74 & 0.72 \\
\text{Butler and Harrod} & 0.75 & 0.73 \\
\text{Wells} & 0.74 & 0.74 \\
\hline
\end{array}
$$
References

Batsanov, S. S. Van Der Waals Radii of Elements. Inorganic Materials 2001, 37 (9), 871–885. DOI: 10.1023/A:1011625728803. (PDF)
Sanderson, R. T. “Electronegativity and Bond Energy.” Journal of the American Chemical Society, vol. 105, no. 8, Apr. 1983, pp. 2259–61.  doi:10.1021/ja00346a026.
Cordero, Beatriz, et al. “Covalent Radii Revisited.” Dalton Transactions, no. 21, 2008, p. 2832. doi:10.1039/b801115j.


General methods of Analysis for FTIR Results

Let me start by saying that the XeF$_4$ is a standard example of group theory applied to IR, which is generally treated in Uni courses, for which you can find plenty of notes on the web. For these reasons, I will not explain it, but you can rather put a bit of effort and read it yourself.

In general, the level of information you can extract depends on the nature of the molecules you analyse. For example, for organic molecules, the analysis is qualitative and limited to functional group identification. This is because usually they have a high number of atoms, thus there will be many vibrations. Since the chemical composition is quite homogeneous, all these vibrations have similar frequencies, and they overlap. Thus a quantitative analysis is difficult, and the only information you extract concerns some particular bond whose signal is isolated from the mess (usually carbonylic and carboxylic bonds, triple bonds, X-H bonds). An additional problem lies in conformational flexibility.

Now, for molecules like yours, which are small, have few vibrations, are symmetric etc, you can achieve a very detailed level of analysis. But, as I said, I will not discuss it, since you can read it yourself with a quick google search.
The basic thing to keep in mind is the frequency of oscillation of a classical harmonic oscillator, given by

$$ \omega = \sqrt{\frac{k}{\mu}} \\
\mu = \frac{m_1 m_2}{m_1 + m_2}$$

where $k$ is the force constant of the bond, and $\mu$ is its reduced mass.
So, what this tells you is that stronger bonds vibrate at a higher frequency. It also tells you that bonds involving lighter atoms vibrate at a higher frequency. Another thing that you can deduce is that bending frequencies are lower than stretching frequencies.
So, to give you a hint, what you can say immediately by looking at those frequencies you reported is that the 586 cm$^{-1}$ vibration is a stretching.


Explanation for a basic decomposition of water experiment

If the hydrogen and oxygen molecules take up the same amount of space, then why did the tube over the negative terminal have much more than twice the gas as the other one?

Good questions but keep in mind that you did not mention the concentration of sodium chloride. If you have too much salt in water, chlorine gas is produced (eventually bleach) instead of oxygen. This is why you did not see the expected 1:2 vol. ratios.
If possible, use the lowest possible salt concentration for the experiment or even better, can you use a small amount of Epsom salt? Then there is no issue of chlorine formation and only oxygen gas will form. You will certainly see 1:2 vol. ratio.

The atomic mass of oxygen is nearly 16 times that of hydrogen, so then even if there was twice as much hydrogen in the product, wouldn't the oxygen still take up more space?

This issue was tackled by scientists long time ago. The key point is that if you have 1 billion molecules of $\ce{H2},$ and 1 billion molecules of $\ce{O2},$ they will occupy the same volume at a given temperature and pressure. There is a named law behind it. The volume of molecules way too small as compared to the volume they occupy.
The radius of oxygen molecule is 152 picometer (symbol: $\pu{pm}$) and that of hydrogen is 120 picometer $(\pu{1 pm} = \pu{10^{-12} m}),$ so masses and volumes do not correlate exactly.


How are the van der waal constants 'a' and 'b' related to molecular size?

The real gas equation or van der Waals’ equation of state is
$$\left(P + \frac{an^2}{V^2}\right)(V - nb) = nRT $$

where the constants a and b are known as van der Waals’ constants.

$a$ is a measure of van der waals forces of attraction existing between the molecules of a given gas. Greater the value of $a$ greater is the strength of the forces. It doesn’t affect molecular size.

On the other hand, $b$ is the incompressible volume per mole for a gas. It is directly proportional to the size of gas molecules.