Why is there only one ionization isomer of [Cr(NH3)5(NO2)]Cl2?

Because $\ce{NH3}$ is not a counter ion and thus would not be bound to the complex if not bonded to the center ion.  In $\ce{[Cr(NH3)5(NO2)]Cl2}$  the $\ce{Cl-}$ ions do not form dative bonds with the $\ce{Cr^{3+}}$ ion but are attracted by the positive charge of the $\ce{[Cr(NH3)5(NO2)]^2+}$ complex.  If you were to form$\ce{[Cr(NH3)4C(NO2)]Cl(NH3)}$, the $\ce{NH3}$ would not be datively bonded to the $\ce{Cr^3+}$ center it would have no bond or attraction to the complex and thus would be a free ion, not part of the complex anymore. Thus the compound would now be $\ce{[Cr(NH3)4(NO2)Cl]Cl}$ as opposed to $\ce{[Cr(NH3)5(NO2)]Cl2}$.


Amount of substance expressed in yoctomole units

Well, the closer you get to atomic detection limits, the less sense bulk units make, that is correct.

However, coming from lab scale units it is much easier to scale down directly with SI prefixes — the exact reason what they were designed for. Therefore, it makes sense that these papers might say ‘Our predecessors could only detect $\pu{10zmol}$, we have lowered the detection limit by a factor of $100$ to $\pu{0.1zmol}= \pu{100ymol}$.’ And that is basically what they say. I note that both abstracts immediately place a molecule count behind the amount in ymol, so they are well aware of the scale of their unit with respect to molecules.

Once you get to the full molecular scale, you will be dealing with a type of quantisation this way or that; whether you are reporting energy, charge, mass or amount. In none of the other cases is anything special described for atomic scales so why should we make an exception for moles? Just make sure that the value itself corresponds to a whole number of molecules within the boundaries of rounding errors.


What is an alloy called when it melts at the same temperature as something else?

Isotectic substances are those which melt at the same temperature.

Ref.: Heycock, C. T.; Neville, F. H. "On the freezing points of triple alloys of gold, cadmium, and tin". J. Chem. Soc., Trans., 1891, 59, 936-966

But be aware that this term is very rarely used.


Difference in the acidity of ethanoic and methanoic acid

What an interesting way of looking at acidity! How I would look at it would be the more standard way which is to compare the stability of their conjugate bases...

The ethanoate ion has an methyl group, which is electron-donating, donating electron density towards the oxygens, concentrating the negative charge. This makes it a more unstable conjugate base.

In stark contrast, the methanoate ion, does not have the electron donating group and thus, does not have such a high concentration of negative charge. This makes it the more stable conjugate base.

The explanation proposed can also explain the differential solvation of the two carboxylic acids...

With a higher charge density, ethanoic acid would form a more "ordered" hydration shell as the electrostatic attraction between the water molecules and the negative charge is greater.

With a lower charge density, methanoic acid would form a less "ordered" hydration shell as the electrostatic attraction between the water molecules and the negative charge is not as great.

Consequently, the decrease in entropy for the dissociation of ethanoic acid is much greater than that for the dissociation of methanoic acid.

Therefore, the dissociation of ethanoic acid would be less favoured than the dissociation of methanoic acid, based on both considerations of entropy and conjugate base stability, which are actually linked.


What is a relationship between Volumetric Analysis and Titration Theory?

I think you mean titration, not “tetration”. “Volumetric analysis” is another (somewhat dated) name for titration, as explained the first paragraph of the Wikipedia page.


What is the product when ferric chloride reacts with propan-1-ol?

There is no reaction between propan-1-ol and iron(III) chloride because propan-1-ol isn't acidic enough to react with it. Instead, phenol reacts with iron(III) chloride because it is acidic enough to form either the phenoxide or iron-phenol complex. On this basis, aliphatic and aromatic alcohol are distinguished. The mechanism of reaction of phenol and $\ce{FeCl3}$ can be found here.

According to this quora post:

  Ferric chloride test is specific for determination of phenol presence.
  So as mentioned above, phenol is more acidic and can form phenoxide
  anions. This can form complex with Fe(III) in $\ce{FeCl3}$, which has a
  blue/green/red color depending on the nature of the phenol. Alcohol is
  too weakly acidic to even form the alkoxide or the complex.

So what's happening?

It actually forms a sol-gel. It is a part of iron-oxide based gels.

One noble application of this is non noble metals electrocatalysts for high temperature PEM fuel cells. For this purpose, Fe-containing carbon aerogels were prepared in resorcinol-furfuraldehyde resins. The resins are made in premixed solution of $\ce{FeCl3}$ and propanol.


How many hydrogen bonds can methanol make?

Although this was an MD study (and might be said to be somewhat dated), quoting from Matsumoto and Gubbins (Ref. 1):

  Most of the molecules are in the two-bonding state, but a
  small fraction of molecules are in the one- or three-bonding
  state, as already pointed out in previous studies. 7
  ,8

According to this study most of the methanol molecules form 2 H-bonds but ~4-7% form 3.

References

Matsumoto, M., Gubbins, KE. The Journal of Chemical Physics 93(3):1981-1994 (doi 10.1063/1.459076).


I don't think this is an either this or that situation. As with most of the equilibria we know it depends very much on the circumstances we look at it. Many examples of keto-enol-tautomerism are heavily influenced by the solvent. Due to time and resource restrictions I won't be able to go in as much detail as I did in explaining whether enolates get protonated at the carbon or oxygen atom, but I ran a few calculations showing some of the controversy.
The calculations were carried out using Gaussian 16 Rev. B.01 at the DF-B97D3(BJ)/def2-TZVPP level of theory, and thermal corrections were obtained from normal coordinate analysis (in the harmonic approximation) at the optimised structures at the same level of theory using $T=\pu{273.15 K}$ and $p=\pu{1 atm}$ (because that is the G16 standard, sorry for the non-SI). The polarisable continuum model was employed except for the gas phase. All energies are in $\pu{kJ mol^-1}$ for reaction \eqref{tauto}, which means that negative values indicate that 4-pyridinol is more stable than 4-pyridone and vice versa for positive values:*
$$\ce{4-pyridinol <=> 4-pyridone}\tag1\label{tauto}$$
\begin{array}{lrrr}
\text{solvent} & \Delta E_\mathrm{el} & \Delta E_\mathrm{o} & \Delta G \\
\hline
\text{gas}     &  -3.4  &  -4.2  &  -3.7  \\
\text{water}   &  21.0  &  18.7  &  18.9  \\
\text{benzene} &   6.9  &   5.2  &   5.5  \\
\hline
\end{array}
From this we can imply (although there are too few points) that in condensed phase 4-pyridone is the predominant species, while in vacuum it would be 4-pyridinol. Unfortunately, at this time I cannot offer a satisfactory (and easy) explanation for this. The values above have not been calibrated, they are a mere snapshot. I don't want to wander deep into hand-wavy territory and propose a rule of thumb based on bond energies and/or similarities to other molecules.
It is probably true that water is able to stabilise a possible negative charge at the oxygen and at the same time stabilising the positive charge of the proton bonded to nitrogen, but that's about as much of an educated guess I would like to give up.
I want to caution everybody with expressions like the aromatic form when referring to a resonance contributor. Aromaticity is a concept which applies to a (sub-) structure and as such to all resonance contributors. This is why both tautomers are aromatic regardless of the depiction chosen.
In a similar way I want to caution against the argumentation with bond energies. Whenever resonance comes into play, we will have fractional bond orders, which deviate significantly from the conditions under which the tabulated values have been measured (or calculated).
The above notes apply directly to the two reasons your friend offers:

My friend gave two plausible reasons:

$\ce{C=O}$ is more stable than $\ce{C=N}.$

$\ce{-N=}$ will have greater repulsion between the lone-pair and the double bond than the same in $\ce{-O=}.$

There is neither an isolated $\ce{C=O}$, nor a $\ce{C=N}$ bond, as they are both part of the same π-system. The lone pairs cannot (directly) interfere with the π-system as they are orthogonal.
I wish I could offer you a better explanation, but with most chemistry there is nothing easy about simple systems.

Notes:

There is a more extensive theoretical study available, which compares THF, water, and methanol and focuses more on the solvent model. However, obtained values are in a similar region:
Nagy, P. I.; Alagona, G.; Ghio, C. Theoretical Investigation of Tautomeric Equilibria for Isonicotinic Acid, 4-Pyridone, and Acetylacetone in Vacuo and in Solution. J. Chem. Theory Comput. 2007, 3 (4), 1249–1266. DOI: 10.1021/ct6002252.

* For those unfamiliar with the notations: $\Delta E_\mathrm{el}$ refers to the electronic energy in the Born-Oppenheimer approximation; $\Delta E_\mathrm{o}$ refers to the observed energy, i.e. it is the former including the zero-point energy correction; $\Delta G$ refers to the Gibbs energy.


The real structure of difluoro(dioxo)xenon (XeO2F2)

Xenon is a noble gas and thus does not need to form any bonds to complete its octet. When such a bond is formed, this is known as a hypercoordinate structure: it contains more bonding partners than the most simple of simple Lewis structure theories would allow. Thanks to Linus Pauling, for many years chemists have been trying to invoke d orbitals to explain this bonding situation when the actual case is far more trivial and can be explained with simple octet considerations.

The two oxygen atoms which are depicted as double-bonded in the d orbital model are actually connected by a single bond. If you want, this bond can be formed by oxygen keeping its six valence electrons in three orbitals and using the forth, empty orbital to bond with a single orbital of xenon. Both electrons can be described as originally donated by xenon and a good way to draw it in the Lewis formalism is to use charge separation:

$$\ce{\overset{-}{O}-\overset{2+}{Xe}-\overset{-}{O}}$$

This leaves the two fluorine atoms. They actually form a slightly more complex bond known as a 4-electron-3-centre bond (thus fluorine only bonds to xenon in neutral compounds in multiples of 2). For a deeper explanation of the 4-electron-3-centre bond, see this previous answer of mine. In a Lewis formalism, this can only described with resonance structures as shown below:

$$\ce{F-\overset{+}{Xe}\bond{...}\overset{-}{F} <-> \overset{-}{F}\bond{...}\overset{+}{Xe}-F}$$

This bonding situation means that the fluoride atoms must occupy opposite positions with $\angle(\ce{F-Xe-F}) \approx 180°$. Thus, A is the correct depiction of  $\ce{XeO2F2}$ which was confirmed in a neutron diffraction experiment by Peterson, Willett and Huston.[1]

The paper determines the $\ce{O-Xe-O}$ angle as approximately $105°$. This is somewhat in-between a pure p orbital and an $\mathrm{sp^2}$ description of xenon, $15°$ from either. At first approximation it would, however, be fine to assume unhybridised xenon and derive the bonding situation from there with each oxygen bonded to one of the p orbitals and the two fluorines bonding to the third.

Reference:

[1]: S. W. Peterson, R. D. Willett, J. L. Huston, J. Chem. Phys. 1973, 59, 453–459. DOI: 10.1063/1.1679827.


That is not the Gibbs Free Energy at constant pressure and temperature.

Alphabetic soup definitions:

$G$ - Gibbs free energy
$H$ - enthalpy
$T$ - temperature
$S$ - entropy
$P$ - pressure
$V$ - volume
$w$ - work
$q$ - heat
$U$ - internal energy

Definition of $G$ is $$G = H - TS$$

Definition of $H$ is $$H = U + PV$$

The total differential of $G$ is

$$dG = dU + PdV + VdP - TdS - SdT$$

The definition for the change in $U$ is $$dU = dq + dw$$

There are two kinds of work: pressure-volume work (pv) and non-pressure-volume work (non-pv)

$$dw = dw_{\mathrm{pv}} + dw_\mathrm{non-pv}$$

Thus, substituting we get

$$dG = dq - PdV + dw_\mathrm{non-pv} + PdV + VdP - TdS - SdT$$

If the process is (approximately) reversible:
$$T =\dfrac{dq}{dS}$$

Substituting $q = TdS$ and simplifying, we get:

$$dG = VdP - SdT + dw_\mathrm{non-pv}$$

At constant pressure, $dP=0$, and at constant temperature $dT = 0$.

Therefore:
$$dG = dw_\mathrm{non-pv}$$

Hence, Gibbs free energy is sort of a measure of useful work we can get out of a system.

$$w_\mathrm{non-pv\ by\ system} = - w_\mathrm{non-pv\ on\ system}$$


Is there a way to quantify the 'surface area' or simply the shape of organic compounds?

It is possible to compute the surface area of a molecules, and to compute the surface area of each individual atom in a molecule. These areas depend on the conformation of the molecule. An example of a software that does this is MSMS (http://mgltools.scripps.edu/packages/MSMS).

With respect to the shape, it is possible to categorize molecules in a triangular diagram describing similarity with "rods", "disks" and "spheres" (e.g. http://pubs.acs.org/doi/10.1021/ci300415d). Details about this analysis were published here: http://pubs.acs.org/doi/10.1021/ci025599w.


Measuring bioanode potential against Ag/AgCl reference electrode

Coming to your experiment, there is a nice description along with sufficient details which shows you how to make connections.

Microbial Fuel Cells & Bacterial Power
https://agsci.oregonstate.edu/sites/agscid7/files/bioenergy/fuel_cell_interactive.pdf

Coming to the confusion: Electrochemists do not measure electrode potentials by using a multimeter as shown in general chemistry textbooks. The numbers shown on a multimeter display only qualitatively showing us that there is a potential difference between these two electrodes. Secondly, carbon cloth is the least reproducible electrodes in terms of its potential (heavily dependent on the history of how it was made). You won't find its values in standard tables. Therefore we cannot compare your experimental values with standard conditions (in the tables, the values are calculated under controlled conditions) because your system is nowhere close to standard conditions. It is like comparing apples and oranges.

An interesting part for you is, how to decide which one is cathode and which one is anode of your cell? Connect the electrodes in such a way that you get a positive potential difference on the multimeter. If you happen to get it, you have made the right connections i.e., negative terminal of your cell is connected to the negative terminal of your multimeter and so on.


Since we have $2.0 \times 10^{-4}\text{ mol}$ per $50\text{ mL}$ of solution, to get the amount of substance per liter, you multiply by:

$$ \frac{1000\text{ mL}}{50\text{ mL}}=20$$

So we get $4.0 \times 10^{-3}\text{ mol L}^{-1}$ as Klaus says.

You're being asked to find a rate of dye consumption. This should be a positive number, if the dye is being consumed - which it is. If dye was being consumed at a negative rate, more dye would appear over time!

Imagine we begin with $c = 1.0\text{ mol L}^{-1}$. We end with $c = 0.5\text{ mol L}^{-1}$. Thus we might say that $\Delta c = -0.5\text{ mol L}^{-1}$. We see that the change is negative, signifying that dye has been consumed. If this occured over the course of 180 seconds, we might write:

$$\text{average rate of consumption} = -\frac{\Delta c}{\Delta t} = -\frac{-0.5\text{ mol L}^{-1}}{180\text{ s}} = 2.77\times 10^{-3}\text{ mol L}^{-1}\text{ s}^{-1}$$

So even though the change in dye concentration is negative, the rate of dye consumption should be a positive number.


So their rate is equal to the constant k.

With other words, the change of concentration over time is linear:

\[ \frac{d[\ce{A}]}{dt} = k\]


You might try different solvents on a spot that does not have writing, to see if it removes the glue, and on glue with non-critical writing, to see if it removes or smears the writing as well as the glue.

Since it described as a clear synthetic resin, you might try the following sequence, from milder solvent to stronger:

Comparatively mild non-polar organic solvents, such as paint thinner or kerosene. If that does not dissolve the glue, try
Some Polar solvents such as acetone, methyl ethyl ketone or nail-polish remover. If that does not remove the glue, try
PVC pipe solvent (tetrahydrofuran).

N.B. All solvents listed are flammable, and some are fairly toxic. Use them in small quantities, closing the containers after getting a bit out, to prevent flash-back of flames, and use in well-ventilated areas.


Solute almost always changes the volume of final solution. The change (volume decrease or volume increase of final solution) of volume depends of solvent and solute chemicophysical properties, e.g. the full charge of each or the polar properties of each, or their molecular orbital structure and electron bond donors or acceptors. Two very non-interactive materials would not even mix, e.g. oil and water (H2O), though tiny tiny amounts of oil will "dissolve" in water as micelles, and tiny tiny amount of water will infiltrate oil, humidity of oil, just like humidity of breathable air.
If solute is not charged and does not have hydrogen-bonding-ready open orbitals or lone pairs of electrons, then the volume increases, both solute and solvent usually do not mix well.
If the solute is a ionic compound (e.g. salt, NaCl), then up to a certain amount of table salt the volume will decrease, the density will increase and reach the maximal amount of salt to reach the minimum amount of solvent, H2O, at a specific temperature and pressure.
If you mix a polar compound ready to accept or donate hydrogen bonds to H2O, then you get a similar picture as for NaCl in H2O, but not so drastic http://butane.chem.uiuc.edu/pshapley/genchem1/l21/vol.png , http://butane.chem.uiuc.edu/pshapley/genchem1/l21/1.html.


Oxidation number for nitrogen in ammonium sulfate

Ammonium sulfate is an ionic compound, composed of two ammonium ions ($\ce{NH_{4}^{+}}$) and one sulfate ion ($\ce{SO_{4}^{2-}}$). As such, you can neglect the sulfate entirely from your consideration of the oxidation state of the nitrogen, as well as treating each ammonium ion as a separate entity. Given the total charge on the ammonium cation, it should be easy to see why the nitrogen has the specified oxidation state.


Horizontal symmetry plane of cis-dichloroosmiumtetracarbonyl

Your compound has only one proper axis of rotation and it is $C_2$ as shown in the diagram. Therefore, it is the principle axis of rotation. By definition, $\sigma_\mathrm{h}$ is a plane of symmetry perpendicular to the principle axis of rotation. However, the plain you were talking about is paralleled to the principle axis of rotation. Thus, it is not a $\sigma_\mathrm{h}$ plane. Nonetheless, it is a $\sigma_\mathrm{v}$ plane, by definition.


Is it right to compare the molarity of a substance with the probability of its molecules reacting?

As Alchimista points out, two different principles are at work here. Concentration, or how many of the species are available, and are these species able to react when they do meet.

The transcript of the video at 3:32 states:

'some constant that takes into account things like temperature and how the molecules are actually configured. Because it's not dependent on them just being there. You have to worry about their kinetic energies...'

So, Molarity will not affect the probability that any two molecules that meet will react, but it does absolutely affect rate of reaction. If the reaction is possible (there is sufficient activation energy) then the more concentrated the substance, the more likelihood of reactable species meeting, and therefore a greater rate of reaction.


Is glycinate ligand a monoanion or a zwitterion?

,Assuming glycinate is in its anionic form ($\ce{H2N-CH2-CO2^-}$), one thing to consider is it may not act just as a simple anionic ligand like chloride or hydroxide.  You can count off four atoms from the nitrogen to either carboxyl-ate  oxygen and those atoms both have available electron pairs.  So you add the metal atom and you get a five-membered ring similar to that formed with ethylenediamine.  Thus, glycinate anion may act as a chelating agent.


Battery acid solution preparation from 98% ACS Sulfuric Acid

The answer begins with the general calculation of mixing solutions with various densities. If you are interested only in the particular numeric results, or if you are scared by algebra, , skip the theoretical part between the horizontal lines.

There is the mixing cross rule, deriving the ratio of mass of 2 mixed liquids from their mass percentage.
$$\frac { m_2 }{m_1} = \frac{|w_1 - w_0|}{|w_2 - w_0|}=k\tag{1}$$
respectively
$$m_2 = k\cdot m_1 \tag{2}$$
and
$$m_0 = m_1 + m_2 = m_1 + k \cdot m_1 =  m_1 \cdot \left(1 + k \right) \tag{3}$$
where
$m_1,m_2, m_0$ are respective masses of mixed solution 1 and 2 and  ofthe final solution 0
$w_1,w_2,w_0$ are respective mass percentages of mixed solutions 1 and 2 and the final solution 0
As we are interested rather in volumes, we involve solution densities:
$$V_1 = \frac{m_1 }{\rho_1}\tag{4}$$
$$V_2 = \frac{m_2 }{\rho_2} = \frac{m_1 }{\rho_2} \cdot k\tag{5}$$
$$V_0 = \frac{m_0 }{\rho_0} = \frac{m_1 \cdot \left(1 + k \right) }{\rho_0}\tag{6}$$
We have the given final volume $V_0$ of the final solution of the density $\rho_0$.
Then, for the solution 1:
$$m_1  = V_0 \cdot \rho_0 \cdot \frac{1}{\left(1 + k \right)} \tag{7}$$
respectively the volume:
$$V_1  = V_0 \cdot \frac{ \rho_0}{\rho_1 } \cdot \frac{1}{1 + k}\tag{8}$$
For the solution 2:
$$m_2  = m_1  \cdot k = V_0 \cdot \rho_0 \cdot \frac {k}{1+k} \tag{9} $$
respectively the volume:
$$V_2  = V_0 \cdot \frac{\rho_0}{\rho_2 } \cdot \frac {k}{1+k} \tag{10} $$

Applying the particular data:
Solution 0: 37% $\ce{H2SO4}$,The density  $\pu{1.267 g/mL}$,  target volume $V_0  = \pu{100 mL}$
Solution 1: 98% $\ce{H2SO4}$.The density  $\pu{1.8361 g/mL}$
Solution 2: water = 0% $\ce{H2SO4}$, the density $\pu{1.000 g/mL}$
From the equation (1):
$k = \frac{|98-37|}{|0-37|}=61/37 \simeq 1.649$
We plug in the data into equations (8) and (10)
For  98% $\ce{H2SO4}$:
$$V_1  = \pu{100 mL} \cdot \frac{ \pu{1.267 g/mL}}{\pu{1.8361 g/mL} } \cdot \frac{1}{1 + 1.649} \simeq \pu{26 mL}\tag{8a}$$
For water:
$$V_2  = \pu{100 mL} \cdot \frac{ \pu{1.267 g/mL}}{ \pu{1.000 g/mL}} \cdot \frac {1.649}{1+1.649} \simeq \pu{79 mL} \tag{10a} $$
Be aware volumes are not additive. After cooling down, the total initial volume shrinks from $\pu{105 mL}$ to $\pu{100 mL}$.
For subsequent dilutions, it is an advantage to keep some volume of the residual 37% solution and to do the dissolution in 2 steps.
First, dilute the calculated portion of 98% H2SO4 in this rest of 37% H2SO4. After cooling, pour this mixture to the calculated volume of water.
The amount of heat per dilution would be divided into these 2 steps.
DO NOT forget to pour acid into water, as not vice versa !! Pouring water into acid would very propably end
by a hot acid splash into your face, as mixing is very exothermic. Mix them carefully, slowly, continually mixing and cool it by a water bath.
Pour the acid along the glass rod and not directly. Wear a lab coat or at least an apron, gloves and shield if possible.
Be aware even tiny drops can cause after some days/weeks holes in your clothes.


This is an example of the Claisen rearrangement.  It is basically the oxa-analogue of the Cope rearrangement.  The Claisen rearrangement can occur with aromatic or non-aromatic allyl vinyl ethers.  The reaction mechanism involves a concerted, intramolecular, [3,3]-sigmatropic rearrangement.  In the aromatic case, the rearrangement is followed by a proton shift (keto-enol tautomerization) to yield the final product with the aromaticity restored.

Here's a diagram of the aromatic Claisen rearrangement

Here's a diagram for the non-aromatic counterpart.


Recrystallizing Sodium Acetate from solution

The Purification of Laboratory Chemicals by Perrin and Armarego states about sodium acetate this procedure:

  Cryst[alise]d from acetic acid and pumped under vacuum for 10h at 120$^\circ$.
  Alternatively, cryst[alise]d from aqueous EtOH, as the trihydrate. This material can be converted to the anhydrous salt by heating slowly in a porcelain, nickel or iron dish, so that the salt liquefies. Steam is evolved
  and the mass again solidifies. Heating is now increased so that the salt melts again. (NB: if it is heated too strongly, the salt chars.) After several  minutes, the salt is allowed to solidify and cooled to a convenient
  temperature before being powdered and bottled (water content should now
  less than 0.02%).


The bulky phenyl group will preferentially occupy an equatorial position. Therefore, the trans acid has the carboxyl group equatorial and the cis acid has it axial, as you have rightly pointed out.

The difference in acidity is to do with the resulting solvation of the carboxylate group. The negatively charged carboxylate group is more strongly solvated by water molecules, and has a larger effective steric bulk, than the neutral, protonated form.

So, there is an increase in the steric bulk of the group upon deprotonation. Such a process would be more favoured if the group is equatorial, i.e. the trans isomer is more acidic.

Of course, in the absence of pKa values, such an explanation is only theoretical in nature. Solvation is a very difficult concept to quantify objectively.


Is there a light cured resin (UV or another) that I could make in my house with easy to find chemicals?

There are simple light-curing chemicals, such as the gelatin-dichromate system, that you could make at home. However, the specifications for stereolithography are rather rigid, and it is doubtful that you could synthesize a monomer that doesn't corrode the system, that has the correct viscosity, has long-term stability and is not toxic.

You might do better taking up gum dichromate photography, which is 2D printing.


Shouldn't MSDS data on products be publicly disclosed?

There is a bit of a misconception here.

Of course, the material safety data sheet (MSDS) is publicly available. Just go to your favourite chemicals supplier, type in your substance and download the corresponding MSDS in your language for your jurisdiction. At least in Europe that is mandatory for all chemicals a company sells; I would be very surprised if there is any developped country where making MSDS publicly available be not mandatory.

On the other hand there is the question of how a value was determined. That is not necessarily public information. Of course, there is a set of standardised tests that companies would perform, but they don’t have to exactly mention whether they used method A or method G. They also have a reason to be nondisclosive about this: imagine an animal rights activist calling you pretending to just be an interested scientist and then using the information they get on the phone to start a name-ruining anti-ChemicalCompany campaign. You don’t want that, do you? And then, of course, there is the general secrecy issue. If you just invented a compound you want to put on the market as a drug, you don’t want anybody knowing anything about it until you’ve started selling it.

Of course, government bodies have a certain reason to know how the tests were perfomed and what the raw results were (otherwise the company could just print a number on their MSDS which the responsible person invented on the toilet). But you’re not government. You’ll have to somewhat trust the government to properly monitor the companies in question.

Of course, the chemist who works safely doesn’t need to know how dangerous something is. They will work with it in a way that they are not affected.


Ok.
Now you have 3 variables and 2 equations.
To solve the problem you need 3 equations.

Third equation you'll get from the partial pressure law:

  The total pressure of an ideal gas mixture is the sum of the partial pressures of each individual gas in the mixture.

Tip: suppose total pressure is 1.


Mustard gas hydrolysis (at a very high rate)

Luckily I got the mechanism via Google Image Search.

Internal attack by sulphur atom takes place during the reaction. This is called anchimeric assistance, neighbouring group effect or $\mathrm{S_N{NGP}}$. Wikipedia describes it well.


Abstraction of $\ce{H+}$ by strong base $\ce{NH2^-}$
$$\ce{CH3-COO- + NH2^- -> ^-CH2-COO- + NH3}$$

Carboxylation to form malonate
$$\ce{^-CH2-COO- + CO2 -> ^-OOC-CH2-COO-}$$

Free malonic acid
$$\ce{^-OOC-CH2-COO- + 2 H+ -> HOOC-CH2-COOH}$$

Dehydration of malonic acid with $\ce{P4O10}$ yields carbon suboxide (unbalanced)
$$\ce{HOOC-CH2-COOH + P4O10-> O=C=C=C=O + H3PO4}$$


Is [PdCl2(PMe3)2] diamagnetic or paramagnetic?

You have not considered the d orbital splitting which has occurred due to the presence of the bonded ligands. Yes, the $\ce{Pd^2+}$ ion adopts the $\mathrm{4d^8}$ electronic configuration with two unpaired electrons, as shown below. However, when it forms the square planar complex, the d orbitals split in energy levels and the electrons now occupy the new energy levels differently, still abiding by Hund's rule and the Aufbau principle. These d orbitals no longer possess any unpaired electrons and thus, the complex is not paramagnetic, but diamagnetic.

Image source: Wikimedia Commons

Image source:Wikimedia Commons


Why Gaussian soft (STO-3G) causes error for Helium anion?

An orbital of a closed-shell Hartree-Fock calculation can only contain two elctrons, one with alpha ("up") and one with beta ("down") spin.

You are calculating the Helium-anion. In a minimal STO-3G basis, it comes with one(!) basis function (or, loosly speaking, "orbital"), as it has two electrons in its neutral form. The anion however has three electrons. So, one closed-shell orbital is certainly not enough to host three electrons. The calculation works for Lithium a.) because gaussian automatically assumes that you want to perform an unrestricted-calculation as the number of electrons is uneven (this is also the case for the helium-anion, but I am not sure if you are aware of this). Furthermore,  Lithium has three electrons "by nature" (in its neutral state) and therefore, two basis-functions are enough to host four electrons.

You are using the STO-3G minimal basis, which adds one basis-function per two electrons of the system in the closed-shell case and one per electron in the open-shell case. Of importance is not the actual number of electrons in your system but the number of electrons the neutral atoms composing your system have. For a quick fix, substitute "hf/sto-3g" with "uhf/cc-pVDZ" and your calculation will run without issues.

Your system has an uneven number of electrons, so you cannot run a restricted (closed-shell) HF calculation. Either perform an unrestricted Hartree-Fock (UHF) or a restricted open-shell Hartree-Fock (ROHF) calculation. Both are available in gaussian09. As said, gaussian is already automatically doing this as a restricted HF calculation of a system with an uneven number of electrons is not possible.

Furthermore, as the calculation of the Helium-anion is quite fast, you should considering using a larger basis. You can start with "cc-pVDZ", a Dunning correlation-consistent basis-set, and move on from there.

For a comprehensive introduction to electronic-structure theory I suggest you read the book by Szabo and Ostlund. It assumes no previous mathematical or chemical knowledge above high-school-calculus und explains all the conceptual issues I have just mentioned in a detailed and understandible manner. Also, this book is quite affordable.
http://store.doverpublications.com/0486691861.html

If this is too much, I suggest you at least look up the Roothan-Hall and the Pople-Nesbet equations for running an electronic structure calculation in a basis, as Gaussian builds on them.


Allred-Rochow vs Pauling Electronegativity scale

Pauling's electronegativity scale is determined by considering the strengths of A-A, A-B, B-B bonds.

Allred-Rochow's electronegativity is determined by considering the charge experienced by an electron on the "surface" of the atom.

While the correlation between the two scales is not perfect, it is quite good, since the scales are calculated with different methods:

  The correlation between Allred–Rochow electronegativities (x-axis, in Å−2) and Pauling electronegativities (y-axis).
  By Physchim62 GFDL, CC-BY-SA-3.0, from Wikimedia Commons

Also the partial charge is a model, useful in order to make rapid evaluations on real things. Sometimes, when molecules with almost covalent bonds (such as nitrogen trichloride) are modeled with different electronegativity scales, the partial charge may be different, but the actual electron cloud may not agree with the model.

The real electron density can be then calculated with quantomechanical methods, which are much more accurate and yet complicated than the simple model of electronegativity.


Can a supersaturated solution get cooler during crystallization?

The system sodium sulfate (Na2SO4) / sodium sulfate decahydrate (Na2SO4 * 10 H2O) / Water might be an example.

Dissolving Na2SO4 in water is exothermic due to hydration.
Dissolving Na2SO4 * 10 H2O is endothermic (entropy effect).

Below 32.384°C Na2SO4 * 10 H2O crystallizes from aqueous solution, above 32.384°C water free Na2SO4.

The solubility of sodium sulphate shows the following unusual shape with a maximum at 32.384°C.

If you would now prepare a saturated solution of sodium sulphate at 32.384°C, and would then be able increase the temperature without precipitation of Na2SO4, you would get a supersaturated solution of sodium sulphate.

Putting a seed crystal of Na2SO4 into it should result in crystallization with the solution getting cooler during this process.

Note that the crystallization frees water molecules that were previously bonded through hydration. This will lead to an increase of entropy.


Does washing soda (sodium carbonate) turn into baking soda (sodium bicarbonate) over time?

Industrially, sodium bicarbonate is produced by reacting sodium carbonate with water and carbon dioxide.  However, under normal household conditions sodium carbonate simply reacts with water or moisture in the air (it's hygroscopic) to form a hydrate - it doesn not form sodium bicarbonate. The water of hydration can be removed by mild heating (~100 C).  At high temperatures (>850 C) sodium carbonate can decompose according to the following equation.$$\ce{Na2CO3(s) -> CO2(g) + Na2O(s)}$$


Which poison was used in the failed attempt to assassinate Khaled Mashal of Hamas?

According to several sources online the chemical weapon used in the failed assissination of Mashal was levofentanyl.

This Wikipedia article states:

  On September 25, 1997, Mashal was injected in the ear with a toxin
  (thought to have been a derivative of the synthetic opiate Fentanyl
  called Levofentanyl).

Another source reads:

  In September 1997, an Israeli hit squad was sent to Amman to
  assassinate Khalid Mishal, the Hamas leader. The chosen instrument was
  levofentanyl, a deadly poison that leaves no traces and produces
  effects like a heart attack. It was administered by a slight physical
  touch.

And another:

  ...The agents entered the city using Canadian passports and waited for
  Mashal at his office. As he was entering, one agent injected a
  chemically modified toxin called levofentanyl into his left ear.

Concerning the structure I don't think it was ever publicised considering the case was high profile (perhaps all traces were taken offline or can be found in deep when-no one knows)

I've been trying to figure out how fentanyl could have been modified, but have no clue either.


Finally found what I was looking for, especially paired with Mithoron's comments.

"Currently, purine nucleotides are obtained at industrial level mainly by microbial fermentation".

https://microbialcellfactories.biomedcentral.com/articles/10.1186/s12934-015-0234-4

Ledesma-Amaro, Rodrigo & Jiménez, Alberto & Santos, María de los Ángeles & Revuelta, José. (2013). Biotechnological production of feed nucleotides by microbial strain improvement. Process Biochemistry. 48. 1263-1270. 10.1016/j.procbio.2013.06.025.


Why is tin(II) fluoride more effective in turning apatite into fluorapatite?

Found an answer to this question in a book called 'Pediatric Dentistry'.

Apparently, when Stannous Fluoride reacts with dental enamel (hydroxyapatite), both the fluoride and the tin react to form Stannous Trifluorophosphate $(\ce{Sn3F3PO4})$ -- This is highly resistant to acid attack.

At high concentrations of stannous fluoride (I don't know what counts as "high", the book doesn't say), apparently the following compound is formed as well: Trifluorostannate -- This is also highly acid resistant.

In addition, Calcium Fluoride is also produced, which then further reacts with hydroxyapatite forming Fluorhydroxyapatite (which is also rather acid resistant, although it seems that the stannous trifluorophosphate is more resistant.)

Sodium fluoride only produces Fluorhydroxyapatite as its reaction.

It appears that Stannous Fluoride is a superior agent because it produces 1 or 2 additional compounds (that become part of the surface enamel) that are more acid resistant than just the one that Sodium Fluoride produces. These other compounds (which become a part of the surface of the tooth) also have an anti-microbial effect, thus inhibiting the growth of the very bacteria which are directly responsible for producing the majority of the acid found in dental decay.


Bond order of H2O and NH3 using group theory to construct MOs

It depends on your definition of bond order. If you are using the simple formula

$$\text{Bond order} = \frac{\text{Number of bonding electrons} - \text{Number of antibonding electrons}}{2}$$

then, yes, it won't be able to capture the subtleties that you describe. After all, when you use that, you're categorising electrons into bonding vs antibonding in a black and white manner. So it's really no surprise that it doesn't reflect "weakly bonding" MOs.

You gave the example of water. Well, for that weakly bonding MO, you need to make a choice: is that considered a bonding MO, leading to a bond order of 1.5 per O–H bond in water (doesn't make much sense)? Or is it considered a nonbonding MO, leading to a bond order of 1 (makes more sense but then you have this problem you brought up)? Or is it somewhere in between, leading to a bond order between 1 and 1.5 – and crucially, if this is the route you want to go down, how do you quantify the bonding character of the MO?

I don't have any experience using different definitions of bond order, but if you used one of these, I would not be surprised to see an O–H bond order that deviates ever so slightly from 1.


Orbitals SPDF. Why they named like that?

From the abstract of Structure of the Line Spectra of the Elements as published in the 1890 Journal of the Chemical Society.

  There are three kinds of series — principal, sharp (well-defined), and diffuse (ill- defined). The principal series form the most vivid lines in the spectra, and only occur in the first, periodic group ; next come the diffuse (really double) lines ...

See also On the Structure of the Line Spectra of the Chemical Elements, also from 1890 by Rydberg:

  Liveing and Dewar were the first to remark the existence of the series, as well as their different appearances. They are of three kinds : diffuse, sharp, and principal series. The first two are formed by the above-mentioned doublets or triplets ...

Later, "fundamental" was added.

For more information see The Origin of the s, p, d, f Orbital Labels J. Chem. Educ., 2007, 84, 757-758


My verdict on $0.064$ is three decimal places, two significant figures. The underlying philosophy is: how many figures would you give if you converted this to "scientific" notation - which would be $6.4 \cdot 10^{-2}$. Since the results must be independent of representation and nobody would argue about the number of significant figures or decimal places here, this is the way to go for me.

I received some chemistry education both in the US and Germany. Unfortunately, my experience is that in Germany these two concepts are often confused, even on university level and by lecturers. All high school level books on Chemistry I looked at in the US contained a section on this and were clear and adamant.


Why do we need sulfuric acid in creating alkyl bromide and not in creating alkyl iodide from alcohol and hydrogen halide?

I suspect $\ce{HBr}$ is produced in situ from $\ce{KBr}$ and $\ce{H2SO4}$.

Same process isn't particularly suitable for $\ce{HI}$ generation as it reacts with $\ce{H2SO4}$:

$$\ce{2HI + H2SO4 -> H2SO3 + H2O + I2}$$

I actually seriously doubt second reaction will take place at all. Hydroiodic acid can also act as a reducing agent, converting the resulting alkyl iodides to alkanes. Instead of $\ce{HI}$ is often being used $\ce{I2}$ with red phosphorous, or halide salts (Finkelstein reaction).

As @orthocresol noticed in the comment, the following two reactions are more preferable for alkyl iodides synthesis:

$$\ce{R - OH ->[(C6H5O)3P+ CH3I-] R - I}$$

$$\ce{R - OH ->[ (CH3 - S - CH = N+ (CH3)2) I- , THF] R - I}$$

Also, chapter 2.01.5.4 Alkyl Iodides from Alcohols, p. 16 in Comprehensive Organic Functional Group Transformations II (1) accumulates numerous ways of iodoalkanes synthesis:

  The adduct of triphenylphosphine with elemental iodine ($\ce{Ph3PI2}$)
  is a classic reagent used to convert alcohols cleanly and with
  inversion to iodides. Addition of imidazole is well known to promote
  this conversion with high yields. Furthermore, treatment of alcohols
  with triphenylphosphine and cyanogen iodide affords the corresponding
  iodides in good yields. The reagent $\ce{KI/BF3 * Et2O}$ in dioxane is
  highly selective and effective for the transformation of allylic and
  benzylic alcohols to iodides.

  A mild and effective procedure for directly converting secondary,
  tertiary, and benzylic alcohols into the corresponding iodides
  involves treatment with iodine in refluxing petroleum ether. The
  reaction proceeds with inversion of configuration:

  Alkyl iodides can also be prepared in a single step from the
  corresponding alcohols upon treatment with the standard reagent
  $\ce{P-I2}$. A less well used reagent,
  1,2-bis-(diphenylphoshino)ethane, can be used to prepare iodides from
  primary alcohols in the presence of iodide.

  Cerium(III) chloride, a Lewis acid imparting high regio- and
  chemoselectivity in various chemical transformations, can be used in
  combination with sodium iodide in refluxing acetonitrile to replace a
  hydroxy by an iodo group:

  This method cannot be applied to tertiary alcohols. In this case an
  alkene is derived by dehydration of the alcohol.

  Finally, a classical method for the transformation of alcohols to
  halides is the well-known two- step procedure via a sulfonate ester,
  commonly tosylate (p-toluenesulfonate) or mesylate (methanesulfonate).

  Treatment of the sulfonate esters with $\ce{NaI}$ or $\ce{LiI}$ gives
  iodides in good yields.

(1) Katritzky, A. R.; Ramsden, C.; Taylor, R. J. K. Carbon with One Heteroatom Attached by a Single Bond; Comprehensive Organic Functional Group Transformations II; Elsevier Science Ltd, 2004; Vol. 2. ISBN: 978-0-08-044253-2


Opposite effects of Le Chatelier's Principle

It is important to know the externally caused change the system is dealing with.
If the primary change is increasing the pressure, then the equilibrium shifts to the lower volume, in case of gases the lower molar amount.
If the primary change is increasing the temperature, than the equilibrium shifts away from the direction that evolves thermal energy.
If external causes are combined, like pressurizing the system AND externally initiated heat transfer, the resulting direction of equilibrium shift would depend on what effect is stronger. As the Le Chatelier's rule is qualitative, it cannot provide answer here, but it must be quantitatively calculated via the equilibrium constant at particular temperature.


I believe the chemical you are talking about that can break down the red blood cells/hemoglobin into heme and globins is carboxymethyl cellulose (CMC; see Ref). There is a US patent claiming a method for separating blood hemoglobin into heme and globin from blood cells. The abstract states that:

  The invention relates to a method of dividing blood hemoglobin into heme and globin by suspending blood cells, separated from blood, in an acid aqueous solution, by precipitating the heme by means of carboxymethyl cellulose, and by separating the precipitated heme-CMC complex out from the aqueous solution of globin.

Title of the patent: Method for dividing blood hemoglobin into heme and globin

Patent: Karin Autio, Martti Kiesvaara, and Yrjö Malkki, US Patent 4,518,525, 1985 (11 pages)


How are orbitals displayed in computational Chemistry software?

Only the program that you used to calculate the electronic structure knows all the idiosyncrasies of the computation to correctly evaluate the orbitals. Therefore, what most people do is to export the evaluation of the orbitals in a grid. Then, a plotting program performs an extraction of a polygonal mesh with a marching cubes (or similar) algorithm.


Why does delocalization of π electrons bring stability?

In simple terms electrons repel each other, so distributing them over a geater number of bonds will reduce the repulsive force between them so resulting in a lower energy state. In the case of phenol the oxygen p orbital lone pair at right angles to the ring can overlap with the delocalised p orbital above and below the ring enabling a degree of delocalisation into the ring system so leaving a slight + charge on the oxygen. This in turn will reduce the electron density in the OH bond so weakening it and accounting for the weak acidity of phenol.


How to differentiate diastereotopic protons by NMR in flexible groups?

I don’t think you can distinguish which is which in a freely rotatable group like the one you drew. They will be different, both signals will split up as dd (potentially ddd, if coupling to the hydroxy proton is observed). There will be significant roofing. Careful analysis of the spectrum should reveal two different shifts coupling to each other but very close.

Rotation around the $\ce{C-C}$ axis should be rapid on the NMR timescale resulting in the couplings $^3J_{\ce{HaHc}}$ and $^3J_{\ce{HbHc}}$ averaging out to one common value and the signal of $\ce{Hc}$ having a dddt-type pattern (three doublets to the neighbouring ring protons, a triplet to the side-chain protons).

All NMR methods I know to determine relative stereochemistry (e.g. NOESY, Murata, …) rely on rotation being slow on the NMR timescale.

Potentially cooling your sample down to very low temperatures ($-50~\mathrm{^\circ C}$ or less) could inhibit the rotation and allow for intermittent resolution of the protons. However, that should also produce a more distinct chemical shift difference, so it may actually not help.


Why does the local density approximation (LDA) overestimate atomization energy?

What is the cause?

The LDA exchange hole is spherically symmetric and centered on a reference electron, which is a fairly good approximation for the case of molecules, but not for the case of atoms. For instance, in an atom the exact exchange hole is displaced toward the nucleus, while the LDA exchange hole is always centered on its reference electron. This inability of LDA exchange hole to appropriately describe the situation in atoms leads to significant deviations in energies upon bond formation or breaking.

  Also, how can inclusion of gradient can help solve this problem?

In general, the electron density in atoms is more inhomogeneous than in molecules, and GGA is ntroduced exactly in order to account for the non-homogeneity. In particular, the gradients in the separated atoms are usually bigger than the gradients in the corresponding molecule, and therefore gradient corrections lower the energy of the atoms more than the energy of the molecule, which leads to a reduction of the overbinding.

See, e.g. Ernzerhof, M., Perdew, J. P. and Burke, K. (1997), Coupling-constant dependence of atomization energies. Int. J. Quantum Chem., 64: 285–295. DOI: 10.1002/(SICI)1097-461X(1997)64:3<285::AID-QUA2>3.0.CO;2-S


Will an aqueous sodium hydroxide solution degrade over time?

Usually, the biggest problems with preventing a sodium hydroxide solution from changing in $\ce{NaOH}$ concentration, or pH, over time are adsorption of $\ce{CO2}$ from the air, and evaporation of water via permeation through the container holding the solution.

If we can really say that it is inside a closed system as you've specified, then any degradation over reasonable time-scales (i.e. months to years) should be negligible.

However, you need to be sure about your "closed system". Is it glass, which could slowly react with the $\ce{NaOH}$? Is it a plastic that could be slightly permeable to water vapor, air (and thus $\ce{CO2}$), etc.? Is it, heaven forbid, a metal that would react and degrade your $\ce{NaOH}$? You need to think about these things at every contact point (vapor or liquid) throughout your system.


Why do electron-rich aryl groups have greater migratory aptitudes?

The answer varies slightly depending on whether the migration is one-step or two-step via an intermediate.
In either case, the respective transition state or intermediate looks like this:

You can see that the middle structure would be stabilized most by a group that is able to stabilize the ring via resonance, for example, an alkoxy substituent.
If the mechanism is single-step, then that substitution would stabilize the transition state and make this migration the fastest.
If the mechanism is two-step, then we can invoke the Hammond postulate to say that the transition state leading to the higher energy intermediate will resemble it in structure and thus also be stabilized by the same kind of substitution, leading to the same conclusion.
We therefore suggest that the propensity of migration is X=OMe > X=H > X=NO₂.


It really depends on the initial temperature of the ice packs and their composition. If they start at a temperature far below the freezing point of water, they could conceivably freeze the water as their container comes to thermal equilibrium. That said, they’re going to have to be quite cold.

To cool by $1°\mathrm{C}$, liquid water requires the removal of roughly double the amount of energy that ice would give up in moving up $1°\mathrm{C}$ (per gram.) If your bags of medicine started at $5°\mathrm{C}$, (stored in a refrigerator) then the ice will warm by about $3°\mathrm{C}$ as it brings the medicine down to its temperature because it has three times the mass and gives up half as much heat per degree gram.

Once the water has reached $0°\mathrm{C}$, it will begin freezing. For every gram of frozen water, about $160~\mathrm{g}$ of ice will warm by $1°\mathrm{C}$. The entire mass of ice will be enough to freeze ~$2.3~\mathrm{g}$ of water for each degree it is below freezing.

Assuming the ice packs came from a normal freezer, their starting temperature is about $−20°\mathrm{C}$. They would warm to about $−17°\mathrm{C}$ in bringing the water to its freezing point. From there the ice could then freeze about $40~\mathrm{g}$ of the medicine before it all came to equilibrium.

Note: There are significant assumptions (notably, that the heat capacity of the ice pack is equal to that of ice and that its melting point is equal to $0°\mathrm{C}$) and some healthy rounding here. If the ice packs melted at a temperature significantly lower than $0°\mathrm{C}$, their heat of fusion could be sufficient to freeze the water, given time.


Calculating temperature change of helium due to Joule-Thomson effect

To do this analysis, one needs data up to 20000 psi (1400 bars) on either enthalpy vs temperature and pressure (a graph) or compressibility factor Z vs temperature and pressure.  The only graphs I have found of the former type go up to only 100 bars, which is a factor of 14 too low.  However, I have found data on the compressibility factor Z at temperatures up to 300 K and pressures up to 1000 bars:  https://cds.cern.ch/record/1444601/files/978-1-4419-9979-5_BookBackMatter.pdf
Although this is still a factor of 1.4 too low, it might provide some idea of the temperature rise that might be expected in the valve.

So here is how the data would be used.

The effect of pressure on enthalpy (per mole) of gas is given by $$\left(\frac{\partial H}{\partial P}\right)_T=V-T\left(\frac{\partial V}{\partial T}\right)_P\tag{1}$$For a real gas, the equation of state in terms of the compressibility factor Z=Z(P,T) is given by$$PV=ZRT\tag{2}$$If we substitute Eqn. 2 into Eqn. 1, we obtain:  $$\left(\frac{\partial H}{\partial P}\right)_T=-\frac{RT^2}{P}\left(\frac{\partial Z}{\partial T}\right)_P\tag{3}$$Integrating Eqn. 3 between P=0 and arbitrary P at constant temperature yields the so-called Residual Enthalpy $H^R$:$$H^R(P,T)=-RT^2\int_0^P{\left(\frac{\partial Z}{\partial T}\right)_{P'}\frac{dP'}{P'}}=-RT^2\frac{\partial}{\partial T}\left(\int_0^P{(Z(T,P')-1)\frac{dP'}{P'}}\right)\tag{4}$$where P' is a dummy variable of integration.

If the final pressure coming out of the valve is low (so that the gas exiting the valve is in the ideal gas region), we can write:  $$\Delta H=-H^R+C_p\Delta T=0$$where, for a monoatomic gas like Helium, $C_p=\frac{5}{2}R$.  Therefore, $$\Delta T=-\frac{2}{5}T^2\frac{\partial}{\partial T}\left(\int_0^P{(Z(T,P')-1)\frac{dP'}{P'}}\right)\tag{5}$$
This expression would be evaluated using the data presented in the reference above.

If a reference can be found with data going out to 1400 bar, that would be even better.


The answer will be (a) Acyl chloride.
The acyl carbon in an acyl chloride is connected to the chlorine atom. While you have acknowledged this fact, you have missed the point that the +M (resonance) effect of Chlorine is overcome by the -I (inductive) effect. Thus, the electrophilicity of the acyl carbon is actually greater than the electrophilicity of the carbonyl carbon in an aldehyde. This jump in electrophilicity is so significant that the difference in steric hindrance is neglected.
In fact, this the major reason why acyl chlorides are considered to be one of the best electrophiles in all of organic chemistry.
If Acyl bromides were used, then they would have an even greater electrophilicity and it would be the answer to this question.
Hope this answers your question!


Some years ago I did a (relaxed) rotational scan of n-butane for the purpose of showing that fatty acids are all-trans. That is also why I found the graph in Awesome's answer some kind of disturbing - it should form a sinus like wave. I include this graph here:

Edited out the error noticed by ron. Thanks again.


Preparation of sulfuric acid through Contact process

You can call it whatever you like. Personally I would say you are preparing sulphuric acid because you end up with more than you started with - 1 mole is used to form 2 moles.


How to obtain the radial probability distribution function of a given orbital from a quantum chemical calculation?

This is actually a lot simpler than I initially thought. I'll be using the same example as previously explained in How to obtain the radial probability distribution function from a quantum chemical calculation? My test setup has slightly changed since then, i.e. I've tried it with Gaussian 16 B.01 and Multiwfn 3.6.0 and it worked the same.

We start the same way as in the linked question, until you have your ar.fchk and load that into Multiwfn, you should see something like this:

 Multiwfn -- A Multifunctional Wavefunction Analyzer
 Version 3.6, release date: 2019-May-21
 Project leader: Tian Lu (Beijing Kein Research Center for Natural Sciences)
 Below paper *MUST BE CITED* if Multiwfn is utilized in your work:
          Tian Lu, Feiwu Chen, J. Comput. Chem., 33, 580-592 (2012)
 Multiwfn official website: http://sobereva.com/multiwfn
 Multiwfn English forum: http://sobereva.com/wfnbbs
 Multiwfn Chinese forum: http://bbs.keinsci.com/wfn

>>

 Loaded ar.fchk successfully!
 Formula: Ar1
 Molecule weight:        39.94800

We need to modify the wave function now:

                    ************ Main function menu ************
 0 Show molecular structure and view orbitals
 1 Output all properties at a point
 2 Topology analysis
 3 Output and plot specific property in a line
 4 Output and plot specific property in a plane
 5 Output and plot specific property within a spatial region (calc. grid data)
 6 Check & modify wavefunction
 7 Population analysis and atomic charges
 8 Orbital composition analysis
 9 Bond order analysis
 10 Plot total DOS, partial DOS, OPDOS, local DOS and photoelectron spectrum
 11 Plot IR/Raman/UV-Vis/ECD/VCD/ROA spectrum
 12 Quantitative analysis of molecular surface
 13 Process grid data (No grid data is presented currently)
 14 Adaptive natural density partitioning (AdNDP) analysis
 15 Fuzzy atomic space analysis
 16 Charge decomposition analysis (CDA) and extended CDA (ECDA)
 17 Basin analysis
 18 Electron excitation analysis
 19 Orbital localization analysis
 20 Visual study of weak interaction
 21 Energy decomposition analysis
 100 Other functions (Part1)        200 Other functions (Part2)

Chose 6:

           ============ Modify & Check wavefunction ============
 Number of GTFs:    27, Orb:     9, Atoms:    1, A/B elec:   9.000   9.000
 -4 Exclude contribution of some atoms to real space functions
 -3 Only retain contribution of some atoms to real space functions
 -1 Return
 0 Save the modified wavefunction to a new .wfn file
 1 List all primitive function      2 List all basis function
 3 List all orbitals                4 Print detail information of an orbital
 5 Print coefficient matrix in basis function
 6 Print density matrix in basis function
 7 Print various kinds of integral matrix between basis functions
 11 Swap some information of two primitive functions
 21 Set center of a primitive       22 Set type of a primitive
 23 Set exponent of a primitive     24 Set coefficient of a primitive
 25 Set coefficients of GTFs/basis functions that satisfied certain conditions
 26 Set occupation number of some orbitals
 27 Set type of some orbitals       28 Set energy of some orbitals
 31 Translate the system            32 Translate and duplicate the system
 33 Rotate wavefunction, namely X->Y, Y->Z, Z->X
 34 Set occupation number of inner orbitals to zero
 36 Invert phase of some orbitals

As you can see there is plenty you can do. For this example I will reset the occupation of some orbitals (26). More precisely, I'll transform the wave function so that only the 2s orbital is doubly occupied. Here is how I did it:

           ============ Modify & Check wavefunction ============
 Number of GTFs:    27, Orb:     9, Atoms:    1, A/B elec:   9.000   9.000
 -4 Exclude contribution of some atoms to real space functions
 -3 Only retain contribution of some atoms to real space functions
 -1 Return
 0 Save the modified wavefunction to a new .wfn file
 1 List all primitive function      2 List all basis function
 3 List all orbitals                4 Print detail information of an orbital
 5 Print coefficient matrix in basis function
 6 Print density matrix in basis function
 7 Print various kinds of integral matrix between basis functions
 11 Swap some information of two primitive functions
 21 Set center of a primitive       22 Set type of a primitive
 23 Set exponent of a primitive     24 Set coefficient of a primitive
 25 Set coefficients of GTFs/basis functions that satisfied certain conditions
 26 Set occupation number of some orbitals
 27 Set type of some orbitals       28 Set energy of some orbitals
 31 Translate the system            32 Translate and duplicate the system
 33 Rotate wavefunction, namely X->Y, Y->Z, Z->X
 34 Set occupation number of inner orbitals to zero
 36 Invert phase of some orbitals
26

 Select the orbitals for which the occupation numbers are needed to be changed
 e.g. 2,4,13-16,20 means selecting orbitals 2,4,13,14,15,16,20
 Input 0 can select all orbitals, input q or 00 can return
0
 Set to which value? e.g. 1.2
 Note:
 You can also input for example "+1.1" "-1.1" "*1.1" "/1.1" to add, minus, multiply and divide the occupation numbers by 1.1
 To recover the initial occupation numbers, input "i"
 To generate occupation state for calculating odd electron density, input "odd"
0
 Done!

 Select the orbitals for which the occupation numbers are needed to be changed
 e.g. 2,4,13-16,20 means selecting orbitals 2,4,13,14,15,16,20
 Input 0 can select all orbitals, input q or 00 can return
2
 Set to which value? e.g. 1.2
 Note:
 You can also input for example "+1.1" "-1.1" "*1.1" "/1.1" to add, minus, multiply and divide the occupation numbers by 1.1
 To recover the initial occupation numbers, input "i"
 To generate occupation state for calculating odd electron density, input "odd"
2
 Done!

 Select the orbitals for which the occupation numbers are needed to be changed
 e.g. 2,4,13-16,20 means selecting orbitals 2,4,13,14,15,16,20
 Input 0 can select all orbitals, input q or 00 can return
q

           ============ Modify & Check wavefunction ============
 Number of GTFs:    27, Orb:     9, Atoms:    1, A/B elec:   1.000   1.000
 -4 Exclude contribution of some atoms to real space functions
 -3 Only retain contribution of some atoms to real space functions
 -1 Return
 0 Save the modified wavefunction to a new .wfn file
 1 List all primitive function      2 List all basis function
 3 List all orbitals                4 Print detail information of an orbital
 5 Print coefficient matrix in basis function
 6 Print density matrix in basis function
 7 Print various kinds of integral matrix between basis functions
 11 Swap some information of two primitive functions
 21 Set center of a primitive       22 Set type of a primitive
 23 Set exponent of a primitive     24 Set coefficient of a primitive
 25 Set coefficients of GTFs/basis functions that satisfied certain conditions
 26 Set occupation number of some orbitals
 27 Set type of some orbitals       28 Set energy of some orbitals
 31 Translate the system            32 Translate and duplicate the system
 33 Rotate wavefunction, namely X->Y, Y->Z, Z->X
 36 Invert phase of some orbitals
3

 Basic information of all orbitals:
 Orb:     1 Ene(au/eV):  -117.263218   -3190.8944 Occ: 0.000000 Type:A+B
 Orb:     2 Ene(au/eV):   -12.054866    -328.0296 Occ: 2.000000 Type:A+B
 Orb:     3 Ene(au/eV):    -9.308431    -253.2953 Occ: 0.000000 Type:A+B
 Orb:     4 Ene(au/eV):    -9.308431    -253.2953 Occ: 0.000000 Type:A+B
 Orb:     5 Ene(au/eV):    -9.308431    -253.2953 Occ: 0.000000 Type:A+B
 Orb:     6 Ene(au/eV):    -1.154432     -31.4137 Occ: 0.000000 Type:A+B
 Orb:     7 Ene(au/eV):    -0.495941     -13.4953 Occ: 0.000000 Type:A+B
 Orb:     8 Ene(au/eV):    -0.495941     -13.4953 Occ: 0.000000 Type:A+B
 Orb:     9 Ene(au/eV):    -0.495941     -13.4953 Occ: 0.000000 Type:A+B

           ============ Modify & Check wavefunction ============
 Number of GTFs:    27, Orb:     9, Atoms:    1, A/B elec:   1.000   1.000
 -4 Exclude contribution of some atoms to real space functions
 -3 Only retain contribution of some atoms to real space functions
 -1 Return
 0 Save the modified wavefunction to a new .wfn file
 1 List all primitive function      2 List all basis function
 3 List all orbitals                4 Print detail information of an orbital
 5 Print coefficient matrix in basis function
 6 Print density matrix in basis function
 7 Print various kinds of integral matrix between basis functions
 11 Swap some information of two primitive functions
 21 Set center of a primitive       22 Set type of a primitive
 23 Set exponent of a primitive     24 Set coefficient of a primitive
 25 Set coefficients of GTFs/basis functions that satisfied certain conditions
 26 Set occupation number of some orbitals
 27 Set type of some orbitals       28 Set energy of some orbitals
 31 Translate the system            32 Translate and duplicate the system
 33 Rotate wavefunction, namely X->Y, Y->Z, Z->X
 36 Invert phase of some orbitals

(The values I have typed in all start in the first column.)

Now leave this screen. You may want to safe with 0 the resulting wave function as new.wfn; but it is not necessary. To get back to the main menu simply type -1.

Now that you have modified the wave function, you can again follow the steps from the linked post.

TL;DR:

200 // other functions
5   // RDF
0   // calc
1   // plot
5   // Export to RDF.txt

I'm sure there  is a lot of fancy stuff you can do with that, I used the same gnuplot file to plot the result:

Shameless self-plug:

I have written some startup scripts for Multiwfn, which make some things easier or extend the functionality to run calculations within a queueing system (e.g. QTAIM charges). Find it on GitHub: https://github.com/polyluxus/runMultiwfn.bash


The additional groups are referred as methyl and methoxy groups. There is no category for these groups. It appears that the methoxy groups stabilizes the intermediates of the oxidation or reduction reactions.


Why copper sulphate is a catalyst in this reaction

The reaction rate increases because the copper is deposited on particles of zinc, creating a galvanic couple causing galvanic (aka electrolytic) corrosion. [It could be argued that this is not catalysis, because the copper sulfate is entering into a reaction.]


Can I get an accurate temperature reading from a flask with a built-in thermometer well?

The thermal conductivity of air isn't very high, when compared to organic solvents or silicon oils used for heating baths. Consequently, one doesn't place thermometers in the vapour phase above the liquid when monitoring or controlling temperatures. (That is, of course, except for distillations, where you want to measure exactly that - the temperature of the vapour phase in the head.)

In your case, it might help to fill the well with the same silicon oil that you use for the heating bath. In order to keep the rod thermometer in place (rather than letting it float), you might want to use a Quickfit adapter (similar to this one) that fits the joint of the well.


Substances that pass visible light but absorb microwaves

Sure - ethanol and methanol (among others) are listed here as being "high absorbers" of microwave energy.


When a candle burns, why does the top of wick glow if bottom of flame is hottest?

The wick temperature does not have to be the same as the flame temperature.The flame is hottest at the bottom, but the wick is hottest at the top.

For a candle, the wick burning isn't the intended purpose of the wick; light comes from burning wax (more generally: fuel), you want to burn the wax not the wick. Rather the purpose of a wick is to help fuel evaporate by soaking up wax and allowing the radiant energy from the flame to heat the wax causing it to evaporate and burn also.

As wax travels up the wick, it evaporates and less wax is in the wick the further up you go. Eventually the wax dries up and the radiant energy is heating a wick without any wax. Eventually the wick gets so hot at the tip, that it will glow due to black-body radiation.

In summary: though the blue is the hottest part of the flame, the wick can evaporate wax to cool towards the bottom. There is no wax at the top and thus as the radiant energy of the flame causes it to get hotter until it starts to glow.

Extra: I will note that the top of the wick does not burn when lit as the gas around the wick has too little oxygen to burn (unless the wick slumps out side the flame). When you blow out the candle the low oxygen environment of a flame is gone and thus the hot wick will smoulder.


The bond dissociation energy is the energy needed for the reaction leading to the rupture the $\ce{C-H}$ bond. I.e. it is the enthalpy of the reaction:

$$\ce{CH_3-H \rightarrow CH_3^{\bullet} + H^{\bullet}}$$

In the case of the reaction in the question, formation of the $\ce{H-Cl}$ bond will counterbalance the energy cost for breaking the $\ce{C-H}$ bond.


Can the green spots in these coins be copper chloride?

This phenomenon is relatively common among jewellers and silversmiths and they call it verdigris. This verdigris can be copper oxide, copper acetate, copper carbonate and even copper chloride. The former options are more likely, but copper chloride can actually be a possibility if you live close to the sea.

Perhaps relevant to your father: verdigris can be cleaned by applying a mildly acidic solution to it, such as lemon juice or vinegar. The formation of verdigris can be prevented by keeping the metal dry as this kind of corrosion is typically formed by contact of the metal with moisture.


Oxidation of an alcohol (with potassium dichromate) gives a ketone when the alcohol is secondary, like isopropanol.  So, which ester out of the given choices has a secondary alcohol component?

You see that A is correct as claimed.


How dangerous are these soil organic carbon digestion methods?

$\pu{0.1 M}~\ce{NaOH}$ is potentially corrosive when in contact with metals or human skin (see here), so I advise precautions such as wearing protective standard gloves and keep metals away from feasible damage.

$\ce{HCl}$ and $\ce{H2SO4}$, especially when they're hot and are in higher concentrations, are very corrosive acids that are dangerous in contact with many materials. They can also release noxious gases in the reactions, but their weakness is glass. Due to reasons irrelevant, these acids can't corrode glass. I recommend extreme precaution. You can work under a "glassy" protection and as Yomen mentions, under a fume hood. Gloves are very important here. Here's what Caltech thinks about precautions using $\ce{HCl}$ and here's what Berekley does about sulfuric acid.

$\ce{KMNO4}$ is a very strong oxidizer. It may cause fire in contact with different materials (we had a volcano experiment in elementary school where we mixed it with glycerine and the result was very exothermic), and it has biohazards such as kidney damage. Here it's explained more comprehensively.


What is the difference between Cyclic Process and Reversible Process?

The key concept in a reversible process is that the system is in equilibrium with it's surroundings at every instant. A cyclic process is different: in a cyclic process, the initial and final state are the same.
The graph that you have drawn consists of a reversible isothermal expansion(AB) and a reversible isothermal contraction(BA). Overall, the process A->B->A is a cyclic process made up of two reversible processes.


Do dipole dipole interactions need to be added explicitly to vdW forces while doing DFT simulations for organic molecules?

(Permanent dipole)-(permanent dipole) interactions are covered by the electrostatics of DFT (and Hartree-Fock, for that matter, because no [Coulomb] electron correlation is involved). In my opinion, interactions between a permanent dipole (such as the one of HCl) and an instantaneous dipole (methane) are covered by the parametrization of dispersion-accounting methods in an averaging fashion.


Interaction of the oscillating magnetic field of electromagnetic radiation with a molecule or atom

The magnetic dipole transition intensity is typically about $10^5$ times smaller than the allowed electric dipole but as this can vary by ten orders of magnitude, second order transitions (with terms $x^2,y^2$ etc ) as well as magnetic dipole transitions can occasionally be important. The electric dipole matrix element  is $\langle a|\hat r|b\rangle$, where $\hat \mu =e\hat r$, the  second order dipole $\langle a|\hat r \hat r|b\rangle$ and magnetic dipole $\langle a|\hat J|b\rangle$ where $\hat J$ is the angular momentum operator
$\displaystyle \sim \langle a|z\frac{\partial}{\partial x} - x\frac{\partial}{\partial z} |b\rangle $.
The transition intensity is the square of these terms and the selection rules differ for each type of transition. A derivation is given over several pages by Steinfeld 'Molecules and Radiation' chapter 1, 2nd ed (publ MIT Press).


Why does d/l-2,3-dibromobutane on reaction with NaI/acetone give cis-2-butene?

The process of elimination of two bromides which you are thinking is not the correct way how the reaction occurs in this case. First there will be a nucleophilic substitution (preferably SN2) by $\ce{I-}$ on the two carbon atoms containing bromine. As iodide is a better nucleophile than bromide, this substitution is majorly driven forward, and after the formation of vicinal di-iodide, the compound now undergoes bond rotations if required and make the two iodine atoms orient themselves anti-periplanar for a E2 elimination.

So, essentially not only elimination, but substitution also occurs prior to elimination, and thus meso-2,3-dibromobutane never actually gets converted into (d/l)-2,3-dibromobutane during the reaction. Also, going by Sawhorse Projection formula is particularly useful in this case rather than Fisher projection. The former gives you a better visualisation of the syn or anti-periplanarity.

Here is the mechanism of the reaction with the meso-compound as starting material:

Similarly, you can draw the mechanism with the optically active 2,3-dibromobutane as your initial compound:

Now you can easily see how the corresponding alkenes are formed depending on the substrates.


How to determine the equivalent weight of carbon dioxide correctly?

Metallic Carbonate means $\ce{M^_x(CO3)_y}$ .

For example there are $\ce{Li^_2(CO3)}$, $\ce{Ca(CO3)}$, and $\ce{Al^_2(CO3)_3}$

and $\ce{M^_x(CO3)_y -> M^_xO_y + yCO2}$

$PV=nRT$

n= 0.0500 moles carbon dioxide evolved

mass of the evolved carbon dioxide is 0.0500 moles $\times$ 44.0 g/mol = 2.20 grams

mass of the remaining metal oxide is 4.215 g - 2.20 g = 2.015  grams.

Of the 2.015 grams of the remaining metal oxide, 0.0500 moles $\times$ 16.00 g/mol = 0.800 grams are oxide, while the mass of metal is 1.215 grams.

There are 24.3 grams of metal per mole of carbon dioxide evolved.

Because the ratio x:y is unknown, it is unknown whether 1.215 grams corresponds to 0.05 moles, 0.1 moles or some other number of moles.

However, since there is one mole of carbon dioxide evolved per carbonate ion, and carbonate has a charge of -2, the molar mass of metal per charge of the metal ion will be 24.3/2 = 12.15 grams.  This is what is meant by equivalent weight of the metal in the question.


Why cotton and wool clothes shrink after washing them?

Wool, in particular, has very "scaly" fibers that mat together forming a dense felt. Natural oils prevent this (no shrunken sheep), but cleaned wool, agitated or shaken, shrinks due to the fibers gripping each other.

"Cotton shrinks because of the tension that is applied to its yarn and fabrics during the construction of the clothing. The tension is released by heat from the washer, dryer, steam and even sunlight, which causes the fabric to be reduced to its natural size." However, mercerization, formaldehyde and other reagents can "pre-shrink" cotton and prevent subsequent shrinkage in washing.


At which temperature does a concentrated aqueous solution of sodium chloride begin to boil?

The boiling temperature always increases when adding salt into water. It looks as if salt was attracting water molecules and preventing them from quitting the liquid. So that it is necessary to overheat the solution to let water molecules get into the vapor phase. I have often made this measurements in class and always found about $\pu{104°C}$.


Are atom coordinates in molfile necessary to produce a valid InChI identifier

Quoting from "InChI, the IUPAC International Chemical Identifier" article [Heller et al. Journal of Cheminformatics (2015) 7:23]:

Each atom is described by a number of properties: its
chemical element name; x,y,z- coordinates (all or any of
them may be zero);

When I edit all coordinates to zero, inchi-1 produces a valid InChI identifier:
InChI=1S/C7H6O2/c8-7(9)6-4-2-1-3-5-6/h1-5H,(H,8,9)

and auxiliary info, which reflect no valid atom coordinates available:
AuxInfo=1/1/N:6,3,7,1,4,2,5,8,9/E:(2,3)(4,5)(8,9)/rA:9nCCCCCCCOO/rB:d1;s1;s2;s2;d3;d4s6;s5;d5;/rC:;;;;;;;;;

I did a similar test earlier, but I didn't retain the strict 10 character column width for each coordinate, which produced an error:
Error 34 (no InChI; Cannot interpret atom block line:     0.0000  0.0000    0.0000


Supporting electrolytes and increased conductivity?

Remember that in electrochemistry whenever a direct current is passing through the solution there must be electrolysis occurring at the electrode. Now, all you want to see in electrochemical experiment, is current due to the oxidation or the reduction of the desired solute which is diffusing to the electrode in an unstirred cell.
The gist of your excellent and thought provoking question is:

My question is how electroinactive ions of the SE "increase the conductivity" of the solution, as is commonly stated, (when if they are electroinactive they do not produce any (Faradaic) current themselves).

After giving it a little thought and some literature search, it is clear that supporting electrolytes increase conductivity is an erroneous conclusion which is rampant everywhere. In the electrochemical experiment, the flux of the supporting electrolyte must be zero, it should not be carrying any current, because by definition, it is non-reactive in that potential window. The mass transport of the electro-inactive species must be zero.
*Yes, separately, if we were to measure the conductivity of a solution to which an supporting electrolyte, the conductivity would be higher than the pure solvent..but that is a separate measurement. Remember that conductivity is measured by very high frequency AC current.
The book, Electrochemical Engineering by Thomas F. Fuller, John N. Harb, explains it better, pg 69:


NMR: Magnetic and chemical equivalence in benzene

All of the protons in $\ce{^{12}C_6{}^{1}H_6}$ benzene are equivalent by symmetry, therefore they are chemical shift equivalent.

You wrote, "I understand that for two nuclei to be magnetically equivalent, they need to have equivalent coupling to all other nuclei in the molecule."

I think the correct statement would be, for two nuclei to be magnetically equivalent, they need to have equivalent coupling to all other non-chemical shift equivalent nuclei in the molecule.  If there are no other non-chemical shift equivalent nuclei in the molecule, then the chemical shift equivalent nuclei are also magnetically equivalent.

In the case of $\ce{^{12}C_6{}^{1}H_6}$ benzene, all of the protons are equivalent by symmetry, therefore they are chemically equivalent and since there are no "different" nuclei (non-chemical shift equivalent nuclei) for them to couple with, they are also magnetically equivalent.

In the case of $\ce{^{13}C_6{}^{1}H_6}$ benzene all of the protons are equivalent by symmetry so they are all chemical shift equivalent.  Since all of the carbons are $\ce{^{13}C}$ (spin=1/2), there are now "different" (the $\ce{^{13}C}$ nuclei are non-chemical shift equivalent) nuclei for the chemical shift equivalent protons to couple with; therefore the protons will be magnetically non-equivalent.  The spectrum is now an AA'A'A"A"A"'XX'X'X"X"X"' (A=proton, X=carbon-13) spectrum.  Since the protons are not magnetically equivalent (at least some of) the couplings will be observable.


From Ashcroft and Mermin's Solid State Physics:

  A fundamental concept in the description of any crystalline solid is that of the Bravais lattice, which specifies the periodic array in which the repeated units of the crystal are arranged. The units themselves may be single atoms, groups of atoms, molecules, ions, etc., but the Bravais lattice summarizes only the geometry of the underlying periodic structure, regardless of what the actual units may be.

So, a real crystal structure is defined by the Bravais lattice, and the unit placed at each point on the Bravais lattice.  Now, it turns out that the hexagonal close packed structure (hcp) is not, in fact, a Bravais lattice.  Hcp is based on the simple hexagonal Bravais lattice, but with two atoms as the basis (one could also call it two interpenetrating simple hexagonal Bravais lattices).  The bottom line is the hcp is not itself a Bravais lattice (see further details in Ashcroft and Mermin). I'll skip any discussion of point groups and space groups, since those are just further refinements of what it means to combine the symmetry of a Bravais lattice with the symmetry of the repeating unit.

There is a hierarchy of symmetry - 7 crystal systems, 14 Bravais lattices, 32 crystallographic point groups, and 230 space groups. For hcp, the point is that it can be represented as a simple hexagonal Bravais lattice with a two-atom unit, so having hcp as another "Bravais lattice" would totally change the definition of what a Bravais lattice is - the fundamental symmetry is simple hexagonal.

The general idea is to start with something you can repeat in a specified way to fill space.  For example, starting with a cube, it is pretty easy to see how to stack them up to fill space.  If, instead of being cubes they are stretched along one axis to get a rectangular solid, you can still fill space, but it will look different depending on what direction you are looking.

So, we get the seven crystal systems: Cubic (with simple cubic, body centered, and face centered cubic Bravais lattices); Tetragonal, which is a cube stretched along one side (with simple tetragonal and centered tetragonal Bravais lattices); Orthorhombic, a unit with perpendicular faces but unequal lengths on each side (with 4 Bravais lattices); Monoclinic, which tilts an orthorhombic unit to make one angle non-90 degrees (2 Bravais lattices); Triclinic, which has no two faces perpendicular to each other (1 Bravais lattice); Trigonal, which is obtained by taking a cube and stretching it along a body diagonal (1 Bravais lattice), and Hexagonal, which has one Bravais lattice.

You get those 7 crystal systems and 14 Bravais lattices by being able to stack the appropriate units to fill space.  Now, while this concept is pretty fundamental, there are further symmetries imposed on the crystal by what you stack up.  Lets consider something fairly simple, simple cubic.  If you start with a big pile of white cubes, the resulting object you put together will look the same regardless of which direction you look at it, or how you rotate it around.  BUT - if you start with a pile of white cubes with one face painted red, you just changed things. To conform with the Bravais lattice, you would stack them so that all the red faces pointed in the same direction.  You would still get a big cube made of the smaller cubes.  But now it would look different depending on what direction you looked at it - you have reduced the symmetry of the crystal.

For simple hexagonal, one starts with a hexagonal right prism (hexagon base, some height perpendicular to the base).  Stacking those one on top of each other results in a simple hexagonal Bravais lattice.  Lets make the hexagonal prisms out of clear acrylic, but place a marble at the center of the prism.  Now when you have stacked them up you would see a simple hexagonal lattice of marbles, or heck, lets call them atoms.  Now, exactly where you put the marble in the clear prism doesn't matter - you would still see the exact same arrangement of the marbles in space.  OK, lets place the first model in the center of the base of the prism, and place a second marble halfway up one of the edges.  When you stack them up, you will now get an arrangement of marbles in space that looks like the hexagonal close packed structure.  But you stacked the prisms following the rules of the simple hexagonal Bravais lattice.  Placing that second marble in the Bravais prism generated a two atom basis.  It clearly is different from simple hexagonal, but the reason is not how you stacked the prisms, it is what the prism contains.  The underlying Bravais lattice does not change, but the symmetry of the resulting crystal is changed by the basis unit used.

Now, use different colored marbles in the two positions, and it changes once again.  Throw three marbles in in different places in the prism and it changes yet again.

So, the Bravais lattice occupies a fundamental level in understanding the crystal structure - it is the rules for stacking things up (and only the rules for stacking things).  The objects you stack up may have their own properties, meaning that two things stacked using the same Bravais lattice may in the end look very different.  This combination of the lattice and the unit are what create the 230 space groups.


Why bond energy of acetone is more though it is more polar than formaldehyde?

the more polar the bond is, the less is its bond energy

I'm not sure where you are getting this from.

Take Acid halides as an example series. As the bond get more polar ( HI (0.4 D) < HBr (0.7 D) < HCl (0.9 D) < HF (1.9 D) ) the bond energies follow accordingly ( HI (295 kJ/mol) < HBr (362 kJ/mol) < HCl (428 kJ/mol) < HF (565 kJ/mol) )

As the charge separation gets larger (and therfore more polar) there is a stronger will to not have that large charge separation, as nature wants to naturally come to an equilibrium - and thus the bonds are more energetic.

Does this clear up your confusion?


Does a complex that has no overall charge still have color?

Tetrakis(triphenylphosphano)palladium(0) or $\ce{[Pd(PPh3)4]}$ is a common catalyst in organic chemistry that features a palladium(0) core with four triphenylphosphane ligands. It is a bright yellow, sometimes described as canary yellow. See the image below (taken from Wikipedia, where a full list of authors is available); however, I feel the picture doesn’t quite do the actual compound justice.

Colour is a feature of electronic transitions. Different orbitals have different energies. If the energy difference between an occupied and an unoccupied orbital happens to correspond to the energy of a photon ($E=h\nu=\frac{hc}{\lambda}$), the photon can be absorbed and the electron excited (subject to a couple of forbidden exchanges as usually defined in the Laporte rule and the spin rule). This absorption of photons is what we perceive as colour if $\pu{400 nm}<\lambda<\pu{700nm}$.

While energy levels of orbitals are affected by the overall charge of a complex, a correct energy difference can happen independent of partial or entire charge.


Pressure-dependence of electrolysis voltage

Of course, we're starting by acknowledging a perpetual motion machine is impossible.  The question, then, is how do we understand, through chemical thermodynamics, why your specific set of steps can't constitute a perpetual motion machine.

The answer is that what you're not accounting for is that the electrical energy required for your step 2 is greater than you get back from your step 4, even if we don't have losses.  That's because, everything else being equal, the work needed to create gases against a high pressure is greater than that needed to create gases against a low pressure. I.e., $|w_2| > |w_4|$.

To understand why, let's simplify things by considering the pressure-dependence of the free energy of a pure substance at constant T: dG = VdP.  In step 2, because the pressure is so, high, you have increased the free energies of both the reactants and the products relative to what they'd be at atmospheric pressure.  To get the actual changes, you would need to calculate $\int^{P_f}_{P_i}V(P) dP$ for the reactants and products.  Since the products are gases (oxygen and hydrogen), and the volume of gases is much larger than that of liquids, the free energies of the products would be raised far more than that of the reactant (a liquid).  Thus the separation between the free energies of the reactants and products, (i.e.,  the magnitude of $\Delta G_r$) for step 2 (which is at high pressure), will be substantially greater than it is for step 4 (which is at low pressure).

You could also understand it using Le Chatelier's principle:  The volume of the products is higher than that of the reactants, and thus higher pressure will favor the reactants over the products.


Order of reactivity of nucleophilic addition at carbonyl group with Grignard reagent

All the compounds are planar, steric hinderance might have played a more significant role if the attack of nucleophile was in the plane. Nucleophilic addition at the carbonyl group takes place at an angle of 107° with the plane.!

Addition of grignard reagents at the carbonyl group are primarily governed by electronic effects. More the positive charge at the carbonyl carbon more will be the rate of reaction.

The two phenyl groups reduce the $\delta +$ most significantly by resonance, next the dimethyl ketone by two methyl groups which reduce the positive charge by hyper conjugation and inductive effects.

The order can be finalized as:

$$1>2>3$$

References

Clayden, J.; Greeves, N.; Warren, S. Organic Chemistry, 2nd ed.; Oxford UP: Oxford, U.K., 2012; pp. 130


Are there any other elements that get less dense in their solid state?

Your terminology is a little off: water isn't an element.

There isn't really a simple single explanation of why some solids are less dense than their associated liquids. The general explanation (which doesn't really explain much) is that the solid has a structure and, sometimes, that structure takes us more space than the average structure that occurs in the liquid.

In water, ice has a definite structure where each oxygen sits at the approximate centre of a tetrahedron with two bonds to hydrogen and two hydrogen bonds to the hydrogens of other water molecules. In liquid water the structure is far more fluid and averages out to be slightly more dense that the (more ordered) solid. Exactly why this is true in this case requires a lot more than could fit into an answer here. If water was frozen extremely quickly you might be able to freeze the structure in the liquid; normal cooling allows enough time for the bonding to become more organised.

But the phenomenon isn't unique to water, though the specifics will be very different for other substances. Among elements Gallium, for example, also expands on freezing (don't store it in glass bottles: see this other answer here).

Other examples of elements are germanium and silicon. Acetic acid also forms a less dense solid. In all cases the substances form a relatively ordered crystal structure which just happens to be less dense than the (less ordered) liquid phases.


Why did roasting of copper(II) sulfate pentahydrate hurt my nose?

Roasting of copper(II) sulfate at elevated temperatures generates sulfur trioxide, which would hurt your nose.

$$\begin{align}
\ce{$\underset{\text{(blue)}}{\ce{CuSO4.5H2O}}$ &->[\hphantom{\gt\ 200\ \mathrm{^\circ C}}][100\ \mathrm{^\circ C}] CuSO4.H2O + 4H2O}\\[6pt]
\ce{CuSO4.H2O &->[\hphantom{\gt\ 200\ \mathrm{^\circ C}}][\gt\ 200\ \mathrm{^\circ C}] $\underset{\text{(white)}}{\ce{CuSO4}}$ + H2O}\\[6pt]
\ce{CuSO4 &->[\hphantom{\gt\ 200\ \mathrm{^\circ C}}][] CuO + SO3}
\end{align}$$


How to write a wave function for infinite potential well with different width than from 0 to a?

Well, yes; the original length $a$ is just a symbol and you can replace it with whatever other quantity you wish. The relevant wavefunctions are thus just

$$\psi_n = \sqrt{\frac{1}{a}}\sin\left(\frac{n\pi x}{2a}\right)$$

You can verify that these wavefunctions are still normalised correctly by explicit integration.


Calculating equilibrium constant given Ka value?

Never mind, I figured it out. All I had to do is reciprocate the value since Ka is the reciprocal of the equilibrium constant for that reaction.


It is true that rate equations typically use molar concentrations for each species, and that for theoretical rate equations you can make sure the units work out by choosing the appropriate units for the rate constant.

There are two things that are important to understand about rate equations, though.

The first is that the "real" rate equations use activity, not concentration. Concentration is usually a "good enough" approximation that this doesn't matter in practice, but when you start thinking about the units of the reaction constant for things like fractional exponents in empirical rate laws, understanding that the concentrations should really be dimensionless activities makes the problem a lot easier.

The second thing to understand is that real reactions don't always follow theoretical rate equations. As a result, what you are often looking at in real life is an empirical rate law. With empirical rate laws, we don't usually worry about the physical significance of fractional exponents, and in turn, how they affect the units of the rate constant. The units of the rate constant become whatever they need to be to make the equation right. In other words, if you have

$\frac{-\mathrm{d}[A]}{\mathrm{d}t} = k [B]^{0.6}[A]^{1.3}$

as an empirically derived rate law that gives $\rm{mol_A}\rm{L}^{-1}\rm{min}^{-1}$, then $k$ would need to have units of $\rm{mol_A}^{-0.3}\rm{mol_B}^{-0.6}\rm{L}^{-1}\rm{min}^{-1}$. These physically nonsensical units are ok because the equation is entirely empirical - we are just describing what we observe on a macroscopic level, without attempting to give it physical significance beyond that.


You basically said it.

  They can approach from anywhere.

Well, I can define the axes to be anywhere.

An isolated atom in vacuum with degenerate d-orbitals does not know where $x,y$ or $z$ are. Hence, the $\mathrm{d}_{z^2}$ orbital does not have the ‘orientation’ you know from pictures — it only gets that once in some way axes are defined. You could now go ahead and define that one of your octahedral ligand approaches the central metal (placed at $(0,0,0)$) from the coordinates $(1,1,1)$. Get out a large sheet of paper and a new pencil and do the maths, calculating what the final orbital distribution will be. The maths will be a lot more complicated, but the result will be the same. Hence, chemists decide to not do complicated maths and just have octahedral ligands approach along the coordinate axes.

Note that in tetrahedral complexes, the ligands would indeed approach from $(1,1,1)$ and the symmetry-equivalent positions of a tetrahedron.


How can the formation of an alkene be explained during the hydrolysis of an ester when the alkoxy moiety is tertiary?

By the normal mechanism given in books and in various sites I am not able to explain its formation

That's right, and it means that there must be a different mechanism predominating.

This is the "standard" mechanism for ester hydrolysis in acid:

If you look closely at the colour-coding, you will notice that it is the blue bond in the ester which connects the $\ce{C=O}$ to the $\ce{OR'}$ that is broken. This is called the acyl bond. The rate-determining step involves two molecules - it is a bimolecular reaction. Therefore, this ester hydrolysis mechanism is called $\mathrm{A_{AC}2}$: $\mathrm{A}$ for acid, $_\mathrm{AC}$ for acyl bond, and $2$ for bimolecular.

Let's use $\ce{R'} = \ce{C(CH3)3}$, the prototypical example of a tertiary alkyl group. The alternative mechanism goes as follows:

Notice here it is the green bond, between the $\ce{O}$ and the $\ce{R'}$, that is broken. This is called the alkyl bond. The slow step is unimolecular. So, following the same logic as before, this hydrolysis is called an $\mathrm{A_{AL}1}$ mechanism. (It doesn't matter which of the red and blue bonds ends up as the carbonyl group and which one ends up as the $\ce{OH}$ group - they are interchangeable via a simple proton transfer.)

As you correctly suspected, the formation of the $\ce{C(CH3)3}$ carbocation occurs. Depending on the reaction conditions, it could be attacked by a nucleophile (e.g. if you use ethanol as the solvent, you might get some ether byproduct), or it could just simply lose a proton to give isobutene.

What factors may work to favour the $\mathrm{A_{AL}1}$ pathway? Well, the most obvious thing is the formation of the relatively stable tertiary carbocation. Another possibility is steric hindrance arising from the bulky $\ce{R'}$ group, which disfavours nucleophilic attack of water in the "normal" pathway. In fact, t-butyl esters like the one shown above are so resistant to nucleophilic attack (under basic conditions) that they are sometimes used as protecting groups for carboxylic acids!


If you want 10 ppm copper in 2 liters of water, the best way to make it would be to start with some soluble copper salt, like $\ce{CuSO4.5H2O}$, make up a convenient concentration (like 1%), then dilute it to 10 ppm.

As was mentioned in the comments, copper metal is not noticeably soluble in water. When rainwater washing thru copper downspouts leaves a blue stain on concrete, the copper metal is dissolving because of tiny solubility of copper oxides or carbonates. It might be 10 ppm, or 1 ppm in a lot more water.


Your answer for question #3 is correct. However, there is a fundamental misconception you used in #1 and #2 that led to an incorrect answer.

In your notes, it seems that electrons from the metal ($\ce{Zn}$ and $\ce{Mg}$) somehow "attack" the alkyl halide and generate the methine ion ($\ce{CH3-}$). This does not happen for two reasons: Firstly, electrons are not free species in solution (like ions and molecules), they are only transferred from one molecule (reductant) to another (oxidant), so they cannot attack anything by themselves. Secondly, the methine ion is an extremely strong base (the pKa for methane is around 50, so the methine ion, its conjugate base, has to be VERY basic), which could only be produced by using an even stronger base to abstract a proton from it.

What actually happens in both cases is what is called metallic insertion. The metal ($\ce{Zn}$ or $\ce{Mg}$) is inserted between the carbon-halide bond, forcing the carbon atom to hold a partial negative charge (the author of this picture forgot to include the $\ce X$ atom bound to $\ce M$ in the organometal compound, but it is not that relevant here).

https://www2.chemistry.msu.edu/faculty/reusch/virttxtjml/alhalrx4.htm

When it happens, you end up having a nucleophilic carbon, and it could indeed react with a $\ce{CH3Cl}$ molecule to yield $\ce{CH3-CH3}$ under a unimolecular substitution mechanism ($\ce{S_N1}$). However, such substitution reactions are very slow compared to acid-base reactions, and since you have acidic hydrogens available in both cases ($\ce{HCl}$ in question #1 and $\ce{H2O}$ in #2) the acid-base reaction (abstraction of a H atom from $\ce{HCl}$ or water) would happen fast enough to prevent the formation of $\ce{CH3-CH3}$ adduct.