Note that $\ce{W3O8}$ is a non-stoichiometric compound and it is not that simple compound you may think. It has a very complex structure as such its oxidation state is very difficult to calculate. I am not going in details about the structure of the compound and I will only mention it a non-stoichiometric compound actually a oxygen-deficient cluster.

Please give a good read to Xin Huang et.al., J. Phys. Chem. A 2006, 110, 85-92, the link that @JM97 has posted. It says that the $\ce{O-O}$ is much longer than normal oxide bond and so it is assumed to be a peroxo bond $\ce{O2^2-}$. Now, peroxide has oxidation state of -1. So, assuming $\ce{W3O8}$ to be a simple compound (no stoichiometric defect or any kind of defect), the oxidation state is:

$$3x + 8(-1) = 0$$
$$x = +{8/3}$$


In a sample of radioactive substance all the atoms don't disintegrate simultaneously. Why?

The basic concept of radioactive decay kinetics is that every single nucleus has the same probability of decaying at any single given time, regardless of the physical-chemical surroundings (it does not know anything about the other atoms). This of course means that they do not disintegrate all at once, because the probability of all of them "deciding to decay" at the exact same time is infinitesimally small.


How to calculate a rough estimate for the volume of a nitric oxide molecule?

Take the volume per mole value (b) from a van der waals constants table (0.0289 L/mol) and divide by $6.022 \times 10^{23}$ molecules per mole.

This will give you a volume that you can compare to the box.


Demonstrating decomposition of hydrogen peroxide using iron(III) nitrate catalyst

Detection of oxygen: Detection of $\ce{O2}$ by a glowing splint is a good way to detect the oxygen. Also you could capture the gas by a simple fixture e.g.  and demonstrate the volume change in the receiver. This way you can actually measure moles $\ce{O2}$ produced (by $PV=nRT$) then moles $\ce{H2O2}$ decomposed stoichiometrically by the formula you've written.

Choice of catalyst: Using $\ce{MnO2}$ would be better if you want to make sure you have a catalyst. $\ce{MnO2}$ will not be consumed during the decomposition; I'm not sure about iron(III) nitrate.

Or, detecting change in $\ce{Fe(NO3)3}$ concentration: Addition of a very small concentration of potassium thiocyanate, $\ce{KSCN}$ (say 1/100 of your $\ce{Fe^{+3}}$ concentration) will yield a deep red product, iron thiocyanate ($\ce{Fe(SCN)^{+2}}$). If you have access to a spectrometer you can measure absorbance of the initial solution's product and the product after decomposing $\ce{H2O2}$.

Prove water was produced: The best way I can think of is to measure the (subtle) change of density of your solution before and after decomposition. Using $\ce{MnO2}$ would make this easy because you can remove/filter it as a solid after decomposition and thus measure mass/volume of your solution before and after. $\ce{H2O2}$ and $\ce{H2O}$ have small albeit detectably different densities at RT. $\ce{H2O2}$ is more dense than water so your density should decrease.


(i) that's correct

(ii) is aromatic, it fit's the 4n+2 rule with n=1.  Here is a drawing of furan showing 2 lone pair electrons in a p-orbital that completes the cyclic aromatic ring. Furan's other lone pair of electrons is perpendicular to the pi system and does not play a role in furan's aromaticity.

(image source)

(iii) Edit to incorporate @jerepierre's comment:

Just like phenol can exist in equilibrium with its non-aromatic tautomeric keto form

so can derivatives of pyrimidinedione (or uracil).  However, unlike phenol were the equilibrium lies far to the aromatic phenol side, these compounds usually (it depends on the substituents) exist in the keto form.  The strength of the keto and amide bonds outweigh the weaker aromaticity of the pyrimidine skeleton and shift the equilibrium to the keto side.

(image source)

The question remains, is the keto form aromatic?  As jerepierre points out, you can draw the following resonance structure for the keto form.  In this double-dipolar resonance structure there are 6 pi electrons 2 in the double bond and 2 [a lone pair] on each nitrogen) suggesting that this resonance structure would be aromatic.  Since this resonance structure contributes to the overall description of the molecule, it would be fair to say that this molecule has some aromatic character; certainly not as much as benzene or furan - but some.


Calculating vacancies in crystal structure

That's it, you have the right answer. Conservation of charge (electroneutrality) is what you use to determine the number of vacancies.


How to separate HCl and NaOH from an aqueous solution of NaCl?

The solution can be electrolyzed to produce chlorine and sodium hydroxide solution.  The chlorine can be added to water to produce HCl and HOCl.


Alright, I will keep this extremely brief. You know the following things I presume:

Ionic bonding - formed when electrons are transferred from one atom to another, forming a cation and anion.

Covalent bonding - formed when two atoms share an electron pair

Dative covalent bonding - formed when two atoms share an electron pair, but this electron pair comes entirely from one of the atoms

For example, in a Hydrogen to Hydrogen covalent bond:

H-H

They share a pair of electrons, one electron coming from each.

However,in the carbon monoxide molecule, there are two normal covalent bonds and one dative bond. In the dative bond, both electrons are initially from the oxygen, and are then shared between the two.

Hope this helps


Reducible representation of planar molecule N2H2 with bond lengths as a basis

Instead of using the $\Delta r$ as in your figure redraw it with a vector  shown as an arrow parallel to each NH bond and label them 1 and 2. You know from the point group that the NH bonds are the same length and at the same angle so these vectors are similar but point in different directions.

Operate on the whole diagram including arrows (with labels) according to the operations in the point group. If, after an operation the figure is indistinguishable from your starting point, then count $1$ for each vector otherwise count $0$.

In your figure the point in the point group is at the centre of the NN bond. The $C_2$ axis points out of the page centred at this point. The $\sigma_h$ plane is the plane of the figure.

The  identity, which is an 'I exist' or 'do nothing' operation count 2. The $C_2$ operation is rotation by 180 degrees and so interchanges the vectors 1 and 2 so is not indistinguishable  and so each vector counts 0. Similarly for the inversion $i$ through the centre. Reflection in the mirror plane (the plane of the image) leaves the molecule indistinguishable so counts 2. Thus you have the reducible representation you give in your question.


Do IUPAC names for haloalkanes require numerals?

You must use the numbers unless your substance in a mixture of isomers; otherwise your molecules structure is ambiguous.  For chloropentane this name alone does not specify weather the chlorine is on the 1,2,or 3 position which is important as the position of a chlorine can dictated physical and chemical properties as well as reactivity in various organic reactions.  Same is true for trichloropentane.  there are there chlorine but without numbers I dont know if the chlorines are all on one carbon, if they are in the 2,3,4; 1,3,5; 1,2,5; 1,2,3; 2,3,5; 1,1,2; 1,3,3 1,1,4 positions or any of the many more possibilities.  all of these molecules give different products in various organic reactions.

Again though if is a mixture of isomers (as is typical of hexane solvents) it is not important to give number


How to determine which ions will pair to form a compound in an aqueous solution?

Your question arises because you believe that ions in an aqueous solution remain bonded with each other. That isn't true though! In fact, ions in an aqueous solution freely roam around throughout the solution.

Note that most ionic compounds have ions bonded to each other in solid state only. Once you dissolve that ionic compound in water, the ions quickly dissociate (the ionic solid dissolves) and become solvated (surrounded by water molecules).

So, to answer your current question: no, there won't be any compound formed, and all the ions would be separate from each other.

Though, another possibility is of precipitation. In that case, the ions involved ($\ce{AD}$ for example) would separate out of the solution and settle down as a solid. Indeed, there may be more than one possible precipitates. So, to decide between any given number of compounds, as to which precipitates first, we'll need to check if their ionic product exceeds their $K_\mathrm{sp}$. The compound for which this happens earlier, due to lower $K_\mathrm{sp}$, would precipitate earlier.

Bonus content: Your question would make more sense if the ions were given in gas phase instead. Now, when the oppositely charged ions would come together, they would stabilize each other via the large amount of electrostatic energy released, and thus forming a stable compound.

Whether $\ce{A+}$ would combine with $\ce{C-}$ or $\ce{D-}$, would depend on their energy of interaction $\left(=\displaystyle -\frac{Kq_1q_2}{r}\right)$. More the energy released for a particular bond, more would be the probability of formation of that bond.


What happens to the t-butyl cation in the TFA deprotection of a t-butyl ester?

The reaction is, in principle, catalytic in TFA. If t-butyltrifluoroacetate is the by-product, then the catalyst is consumed. If only isobutylene is formed, catalytic is OK. Using excess TFA covers all contingencies. In the initial step, protonation is more likely on the C=O oxygen. Higher electron density.


Can a double bond extend beyond the plane and if so, how would it be drawn?

Yes, chemical bonds of higher order can be extended from the plane in some rare cases.
There are IUPAC recommendations existing at least for double bonds suggesting to depict them simply as doubled wedge bonds if it's really necessary.
From [1, p. 1930] (emphasis mine):

  ST-1.5 Tetrahedral stereocenters including higher-order bonds

  Tetrahedral configuration requires a tetrahedral arrangement of ligands around a central point. It does not require that all of those ligands be connected to a central atom using single bonds. It is perfectly reasonable to have one or more ligands connected to the central atom using higher-order bonds. This is most commonly seen with chiral sulfoximides, phosphates, and related compounds, each of which is
  commonly drawn with one or more double bonds. Stereocenters of this type should be indicated by the appropriate placement of bold and/or hashed wedged bonds connecting the central atom to the singly bonded ligands. Although wedged and hashed double bonds have appeared in the literature, they are rarely required for chemical clarity, and their use is infrequent enough to be confusing to the casual
  viewer (see ST-2.5).

Further from [1, pp. 1940-1942] (emphasis mine):

  ST-2.5 Trigonal pyramidal Polyhedral symbol: TPY-3 Number of ligands: 3

  […]

  In rare cases, a trigonal pyramidal atom with at least one double bond may be flanked by other stereocenters at the other ends of all of its single bonds. In such cases, one of the plain bonds should be changed to a wedged bond even though it joins two stereocenters. It is also acceptable to use a solid wedged or hashed wedged double bond, but that would not be preferred because wedged double bonds are so rarely seen in the literature.

References

Brecher, J. Graphical Representation of Stereochemical Configuration (IUPAC Recommendations 2006). Pure and Applied Chemistry 2009, 78 (10), 1897–1970. https://doi.org/10.1351/pac200678101897. (Free Access)


This is a very interesting question. Provided that the materials used in making papers aren't the same around the globe, this is a very broad case of study. However, a study has been conducted in which the main goal was to identify the compounds that are the cause of the smell; VOCs:[1]

  Volatile organic compounds (VOCs) are organic chemicals that have a high vapor pressure at ordinary room temperature. Their high vapor pressure results from a low boiling point, which causes large numbers of molecules to evaporate or sublimate from the liquid or solid form of the compound and enter the surrounding air.

So, quoting the article:

  Using supervised and unsuper-vised methods of multivariate data analysis, we were able to quantitatively correlate volatile degradation products with properties important for the preservation of historic paper: rosin, lignin and carbonyl group content, degree of polymerization of cellulose, and paper acidity.

  It is a result of the several hundred identified volatile and semi-volatile organic compounds (VOCs) off-gassing from paper and the object in general. The particular blend of compounds is a result of a network of degradation pathways and is dependent on the original composition of the object including paper substrate, applied media, and binding.

  The 15 most abundant VOCs present in all chromatograms were selected for further analyses: acetic acid, benzaldehyde, 2,3-butanedione, butanol, decanal, 2,3-dihydrofuran, 2-ethylhexanol, furfural, hexadecane, hexanal, methoxyphenyloxime, nonanal, octanal, pentanal, and undecane.

Strlič, M., Thomas, J., Trafela, T., Cséfalvayová, L., Kralj Cigić, I., Kolar, J., Cassar, M. (2009). Material degradomics: on the smell of old books. Analytical Chemistry 2009, 81 (20), 8617-8622.
The main link to the article might be behind a paywall, so for the interested, the relevant ResearchGate article is accessible.


NaHCO3 required to raise pH of weak HCl solution from 3.5 to 7.2 (doubting my calculations)

The approach I would suggest is a bit different.

First I would calculate the nominal concentration of HCl (i.e. the number of moles of HCl added, divided by the total volume of the solution after the addition, regardless of what happens to the resulting chloride ions and protons) in your initial solution, using all the necessary equations (with the approximation that activity equals concentation, although the figures are a bit large here).

1) dissociation constant of lidocaine:

$K_{a,lido}=\frac{[L] \cdot [H^+]}{[LH^+]}$

2) the total nominal concentration of lidocaine must be the sum of all its forms:

$C_{lido}= [L]+[LH^+]$

3) charge balance:

$[Cl^-] + [OH^-] = [H^+]+[LH^+]$

4) the total concentration of chloride is the same as the nominal concentration of HCl, as the dissociation is complete and $Cl^-$ does not react with anything else:

$C_{HCl}= [Cl^-]$

5) the ion product of water:

$K_w=[H^+] \cdot [OH^-]$

The system of these 5 equations can be solved for $C_{HCl}$, after eliminating all chemical species (except $[H^+]$, so in total 4 unknowns):

$C_{HCl}= [H^+] - \frac {K_w}{[H^+]} + \frac {C_{lido}}{1+ \frac {K_{lido}}{[H^+]}}$

Plugging in the numbers:

$C_{lido}=\frac {20 g/L}{234.34 g/mol} = 0.085346 M, K_{a,lido}=10^{-7.9} M, [H^+]=10^{-3.5} M, K_w = 10^{-14} M^2$

results in:

$C_{HCl}= 0.08566 M$

i.e. to reach a pH of 3.5 starting from freebase, we used just slighly more than the stoichiometric amount of $HCl$. Here we are assuming as well that the added volume of HCl has not changed the nominal concentration of lidocaine very much. If it has, then you need to account for that.

Now, when you add sodium bicarbonate, all the equations above except 3 are still applicable; you need to add the dissociation constants for carbonic acid (I'll consider it like hydrated $CO_2$, and that none of it is escaping as gas, to keep things simple) and hydrogenocarbonate, and the mass balances for sodium and the various carbonate species.
Here they are:

$K_{a,carbonic acid}=\frac{[HCO_3^-] \cdot [H^+]}{[H_2CO_3]}$

$K_{a,hydrogcarbonate}=\frac{[CO_3^{2-}] \cdot [H^+]}{[HCO_3^-]}$

$C_{bicarb}= [Na^+]$

$C_{bicarb}= [H_2CO_3] + [HCO_3^-] + [CO_3^{2-}]$

The charge balance now includes more ions:

$[Cl^-] + [OH^-] + [HCO_3^-] + 2 \cdot [CO_3^{2-}] = [Na^+]+[H^+]+[LH^+]$

We have now 9 equations (5-1+5). Like before, eliminating all chemical species except $[H^+]$  (i.e. 8 unknowns), gives a single expression. It is a bit complicated, but essentially it contains only $[H^+]$, $C_{bicarb}$ and $C_{HCl}$, and known constants.

In our case, as we know what the nominal concentration of $HCl$ is from the previous calculation, and the final pH we want to reach (7.2), we just need to solve for $C_{bicarb}$ and plug in the numbers:

$C_{lido}=\frac {20 g/L}{234.34 g/mol} = 0.085346 M, K_{a,lido}=10^{-7.9} M, [H^+]=10^{-7.2} M, C_{HCl}=0.08566 M, K_{a,carbonic acid}=4.3 \cdot 10^{-7}, K_{a,hydrogcarbonate}=5.6 \cdot 10^{-11} $

resulting in:

$C_{bicarb}=0.1142 M$

So for each litre of your initial solution you should add 0.1142 mol of $NaHCO_3$, which is 9.59 g.

This is closer to the 'official' quantity you mentioned (even quite a bit more in fact), although I have trouble following all the figures. Maybe they did this experimentally and found that some $CO_2$ escapes, making the pH rise without the need for more base. Just a hypothesis.

In a way, even without all these involved calculations, one could say: if you have a 0.085 M solution of lidocaine hydrochloride and want to react all the $HCl$ with sodium bicarbonate, use $0.085 mol/L \cdot 84 g/mol = 7.14 g/L$ of $NaHCO_3$. The excess used in your official method compared to this figure is probably due to the slight excess of $HCl$ in the initial solution, and the need to take the pH closer to the desired 'physiological' value you mentioned.

EDIT: addendum to answer further questions by the OP

The OP asked what would happen by adding more sodium bicarbonate.

The answer is: as long as there is no solid precipitate of sodium bicarbonate (and always assuming that lidocaine stays in solution), the above equations remain valid; as soon as the nominal concentration of sodium bicarbonate is sufficiently high for it to precipitate, new equations apply, where all the above equations still apply, except for a correction of the mass balances related to $C_{bicarb}$, in particular by adding the contribution of the solid:

$C_{bicarb}= [Na^+] + [NaHCO_3]_s$

$C_{bicarb}= [H_2CO_3] + [HCO_3^-] + [CO_3^{2-}] + [NaHCO_3]_s$

(where $[NaHCO_3]_s$ represents the solid salt expressed in the units of concentration, for homogeneity with the other terms).

And the equation for the solubility product of $NaHCO_3$ must be added:

$K_{s,bicarb}=[Na^+] \cdot [HCO_3^-]$

From the data table in this site it seems that $K_{s,bicarb}=10^{-0.55} M^2 = 0.2818 M^2$ (although from my own calculations based on a solubility of 96 g/L at 20°C it should be $1.277 M^2$, but OK, I will use the 'literature' value).

First, the limiting value of $C_{bicarb}$ after which residual solid bicarbonate is observed at equilibrium must be calculated.

This can be done by solving (numerically) the system of all 10 equations described above for all the chemical species plus $C_{bicarb}$, after setting $[NaHCO_3]_s=0$. The result is:

$[C_{bicarb}=0.5467 M, [Na^+]=0.5467 M, [HCO_3^-]=0.5155 M, [H^+]=2.504 \cdot 10^{-8} M, [H_2CO_3]=0.03002 M, [CO_3^{2-}]=0.00113 M, [Cl^-]=0.08566 M, [OH^-]=3.9936 \cdot 10^{-7} M, [LH^+]=0.0568 M, [L]=0.02855 M]$

meaning that this system is only valid for $C_{bicarb} > 0.5467 M$.

Then the system can be solved using any value of $C_{bicarb}$ greater than the above threshold, and, interestingly, it gives always the same values, except for $[NaHCO_3]_s$, that increases accounting for the undissolved salt.

$C_{bicarb}=0.7 M$

$[[NaHCO_3]_s=0.1533 M, [Na^+]=0.5467 M, [HCO_3^-]=0.5155 M, [H^+]=2.504 \cdot 10^{-8} M, [H_2CO_3]=0.03002 M, [CO_3^{2-}]=0.00113 M, [Cl^-]=0.08566 M, [OH^-]=3.9936 \cdot 10^{-7} M, [LH^+]=0.0568 M, [L]=0.02855 M]$

$C_{bicarb}=7 M$

$[[NaHCO_3]_s=6.453 M, [Na^+]=0.5467 M, [HCO_3^-]=0.5155 M, [H^+]=2.504 \cdot 10^{-8} M, [H_2CO_3]=0.03002 M, [CO_3^{2-}]=0.00113 M, [Cl^-]=0.08566 M, [OH^-]=3.9936 \cdot 10^{-7} M, [LH^+]=0.0568 M, [L]=0.02855 M]]$

$C_{bicarb}=70 M$

$[[NaHCO_3]_s=69.453 M, [Na^+]=0.5467 M, [HCO_3^-]=0.5155 M, [H^+]=2.504 \cdot 10^{-8} M, [H_2CO_3]=0.03002 M, [CO_3^{2-}]=0.00113 M, [Cl^-]=0.08566 M, [OH^-]=3.9936 \cdot 10^{-7} M, [LH^+]=0.0568 M, [L]=0.02855 M]]$

So it appears that, once solid sodium bicarbonate is present, the system is determined in a specific state, where all chemical species have a fixed concentration, regardless of how much solid there is.

In particular, the pH seems to be fixed at $7.60$. This is significantly lower than the pH of a saturated solution of $NaHCO_3$ ($8.31$) in water, probably because of the $H_2CO_3$ liberated in the reaction with $HCl$ (which, we are assuming, is remaining in solution).


Syringe pipette pump.

Tri-valve is not as easy to use as syringe one. And the little metal sphere in it can be thrown while using. The syringe is more comfortable, it is just required to have 3 types for different volumes.

Additionally, when the pipette is broken at the top or has similar problems, you can somehow adapt the syringe, like inserting pipette deeper or tightening the syringe. It is also not expensive and easy to find from others when you need more.


I suggest to pick one gaseous component, say, $\ce{N2O4}$, and, applying ideal gas law, express $K_\mathrm{c}$ as a function of the partial pressure of the chosen gas. Since at the equilibrium the ratio $n(\ce{N2O4}):n(\ce{NO2})=1:2$ is exact (e.g. molar fractions are $1/3$ and $2/3$, respectively), and the volume doesn't change, you can immediately deduce that

$$[\ce{N2O4}] = 2[\ce{NO2}];$$
$$p(\ce{N2H4}) = \frac{p_\mathrm{tot}}{3}$$

where $p_\mathrm{tot}$ is the known total pressure of the system. Using ideal gas law

$$[\ce{N2O4}] = \frac{p(\ce{N2H4})}{RT} = \frac{p_\mathrm{tot}}{3RT}$$

and the law of mass action

$$K_\mathrm{c} = \frac{[\ce{NO2}]^2}{[\ce{N2O4}]} = \frac{(2[\ce{N2O4}])^2}{[\ce{N2O4}]} = 4[\ce{N2O4}] = \frac{4p_\mathrm{tot}}{3RT},$$

one can plug in the parameters and calculate $K_\mathrm{c}$:

$$K_\mathrm{c} = \frac{4\cdot\pu{101325 Pa}}{3\cdot\pu{8.314 J mol-1 K-1}\cdot\pu{333 K}} = \pu{48.8 mol m-3} = \pu{4.88e-2 mol L-1}$$


While thermochromicity already was known for me, for example as thermometer for fish tanks like here, a search based on "thermochromic fluorescent fluid" as keyword yielded this video.

Beside safety, yet indeed, it is easier to fill the sculptures with a liquid / a solution of a fluorescent dye excited by black light from the inner of the (glass) sculpture and to wrap the container with thermochromic dyes. These are even commercialized as here and here.  "Black light" is an other name for ultraviolett light, and there are even hand-size battery powered lamps found in the chemistry lab, or in civil life to check banknotes.

Disclaimer: I have no relationships to the said corporation.

Addendum:  Examples of fluorescence triggered solely by the gentle warmth of a hand are not known to me (yet).


Miller indices. What determines the appropriate axis?

There is no "appropriate" translation. You must remember that everything includes translational symmetry, including the Miller planes.

The Wikipedia page on crystal structures and Miller indices go into much more detail, but some key points:

The unit cells possess translational symmetry, so there are a set of Miller planes that also differ by translational symmetry.
The Miller indices (hkl) (which yes should be expressed in smallest possible integers) define the normal of the Miller plane.

  That is, (hkℓ) simply indicates a normal to the planes in the basis of the primitive reciprocal lattice vectors. Because the coordinates are integers, this normal is itself always a reciprocal lattice vector. The requirement of lowest terms means that it is the shortest reciprocal lattice vector in the given direction.

So in your case, you have to remember if you "move the origin", you have to also consider the symmetrically identical unit cells around the one in your picture. As I said above, there are an infinite set of identical Miller planes that simply differ by the translational symmetry of the lattice.


Kröger-Vink notation for defect reactions

Yes, the Kröger-Vink notation you are asking about, implies that the metal has lost or gained one electron, relative to the case of the perfect crystal. This situation can be explained by non-stoichiometry.

A non-stoichiometric compound is one of which the proportions of the atoms can not be represented by integers.
Consider the case of $\ce{Fe}_{0.95}\ce{O}$. Indicated by the subscript 0.95, there are some metal atoms missing, which is expressed as the defect $\ce{$V$^{''}_{Fe}}$. This has a double negative charge, which can be compensated by 'replacing' $3~\ce{Fe^2+}$ by $2~\ce{Fe^3+}$ ions. That is, 2 iron ions need to be oxidized and 1 needs to disappear.

This can be accomplished by dissolving $\ce{Fe_2O_3(s)}$ into $\ce{FeO(s)}$, as in:

$$\ce{Fe_2O_3(s) ->[FeO] 2Fe^{.}_{Fe} + 3O_{O} + $V$^{''}_{Fe}}$$

Alternatively:

$$\ce{2Fe_{Fe} + 2O_{O} + 1/2 O_2(g) ->[FeO] 2Fe^{.}_{Fe} +3O_{O} + $V$^{''}_{Fe}}$$

More examples are described in my source, chapter 11 of the book 'Ceramic Materials' by Carter & Norton.


Finding the volume needed to make a new pH

$$\ce{NaOH + HCl -> NaCl + H2O}$$

Let $x$ Litres be the amount of $\ce{NaOH}$ needed.

$$-\log[\ce{H+}] = 2$$

$$-\log\left(\frac{(0.5 \times 0.02 - x \times 0.5)}{(0.02 + x)}\right) = 2$$

$$x = \frac{49}{2550}$$

$$x \approx 0.0192 \; \text{(3 sig. fig.)}$$

Thus, $\pu{0.0192 L}$ of $\ce{NaOH}$ is needed.


If Oxygen has a lower electron affinity than Sulfur, then why doesn't sulfur has a lower electron affinity than selenium?

This is likely due to second period elements' being quite small, so electron-electron repulsion is much more significant than in a third period element. The general trend is that EA is more positive as you move down the periodic table since effective nuclear charge is the same, but the electrons are farther away from the nucleus. Maybe it's easier to consider the second period the anomaly...


POI is a bent molecule similar to water. How are the two molecules different it terms of bond angles and what causes this?

Does this mean that:

    Due to the extra lone pair $\color{red}{\text{it}}$ $\color{limegreen}{\text{creates less resistance}}$ resulting in $\color{blue}{\text{a greater angle}}$?

More like:

Due to the extra lone pair $\color{red}{\text{in } \ce{H2O} \text{ the pendant atoms}}$ $\color{limegreen}{\text{experience more repulsion}}$ resulting in $\color{blue}{\text{a smaller angle}}$ $\color{magenta}{\text{compared to }\ce{POI}}$?

  Also, does it having three different elements in the equation do anything?

No, I think the three different elements in the question were there to evaluate if you were:

estute to the fact that oxygen forms a double bond thus
phosphorous has 1 unbonded pair of electrons (not 2) and
if the double bond would hang you up in obtaining an answer.


Is it possible to quantify the effects of a chelating agent?

Generally, a good starting point is the Formation Constant (β) between the metal and the ligand. The greater the constant, the better the ligand is be at chelating the metal center (as evidenced by the free energy change).


How does ammonium nitrate explode on its own?

It is known that ammonium nitrate decompose exothermically when heated to form nitrous oxide and water. This paper1 notes that the irreversible decomposition of ammonium nitrate occurs at the temperature range of $\pu{230-260 ^\circ C}$.
$$\ce{NH4NO3 ->[t >230 ^\circ C] N2O + 2H2O}$$
They also further noted that beyond $\pu{280 ^\circ C}$, $\ce{NH4NO3}$ is  capable of rapid, self-accelerating decomposition (to the point of detonation).
But at the detonation temperature, $\mathrm{t_d}$ (the temperature at which compounds detonate), ammonium nitrate fully decomposes to nitrogen, oxygen and water releasing a tremendous amount of energy.
$$\ce{2NH4NO3 ->[t_d] 2N2 + O2 + 4H2O}$$
In the context of Beirut explosion, the question that raised was "when did ammonium nitrate reached detonation temperature, and why did it suddenly explode?". According to a news report from cnet.com:

When heated to above 170 degrees Fahrenheit, ammonium nitrate begins
to undergo decomposition. But with rapid heating or detonation, a
chemical reaction can occur that converts ammonium nitrate to nitrogen
and oxygen gas and water vapor. The products of the reaction are
harmless -- they're found in our atmosphere -- but the process
releases huge amounts of energy.[...]
Additionally, in the explosion, not all of the ammonium nitrate is
used up and exploded. Some of it decomposes slowly creating toxic
gases like nitrogen oxides. It's these gases that are responsible for
the red-brown plume of smoke seen in the aftermath of the Beirut
explosion, Rae said.

So, my theory is that ammonium nitrate started heating (from the fire) releasing all sorts of nitrogen oxides (the red fumes). This fire further accelerated the reaction, further heating the remaining ammonium nitrate to the point of detonation and that's when ammonium nitrate exploded instantaneously releasing tremendous amount of energy which send shockwaves around the site along with a white mushroom shaped cloud (from @DDuck's comment) which could probably be nitrogen and/or water vapours where the humid air (water vapor laden air) condensed due to the explosion(@StianYttervik) with release of nitrogen. It is a sad and quite devastating incident.
References

On the Thermal Decomposition of Ammonium Nitrate. Steady-state Reaction
Temperatures and Reaction Rate By George Feick and R. M. Hainer, 1954 (PDF)
Reaction Rates of Ammonium Nitrate in Detonation
Melvin A. Cook, Earle B. Mayfield, and William S. Partridge
The Journal of Physical Chemistry 1955 59 (8), 675-680
DOI: 10.1021/j150530a002 (PDF)
https://en.wikipedia.org/wiki/2020_Beirut_explosions


How do I make a simple home distillation device?

This depends on what you need this setup to accomplish, as the amount of solutions that this setup can manipulate vary greatly.

The basic consensus is that if you would want an optimal apparatus you would need:

Round Bottom Flask
Thermometer
A Column
Condenser (with Water In/Water Out nozzles)
Collection Flask


Are fossil fuels and petroleum the same thing?

Fossil fuels are those that originate from ‘fossilised’, anaerobically partially degraded prehistoric living species (mostly plants and animals) and that take a few million years to form hence are not considered renewable.

Depending on how old the deposit is and which conditions it encountered across Earth’s history, the deposit will either turn into something gaseous, liquid or solid — and these categories are typically labelled ‘natural gas’, ‘crude oil’/‘petroleum’ and ‘coal’. Thus, petroleum is only one type of fossil fuel: the one that presently is in the liquid aggregate state.

However, equating petroleum and fossil fuels isn’t entirely unwarranted, since petroleum is both the most important fuel and the scarcest today.


Mechanism for Friedel-Crafts Acylation in Synthesis of Sertraline

The alleged original RSC source for this reaction was not accessible online.1. However, Olah et al. have reported2 that α-naphthol 1 in superacid medium forms a dication species 4. The use of excess AlCl3 provides a mechanism for the formation of superacid 3. Benzylic dication 4 undergoes substitution of o-dichlorobenzene 5 which ultimately forms the sertraline precursor, tetralone 8. For reaction conditions, see ref. 3. No reducing agent is required.

1) Kavitha, C. V.; Mantelingu, K.; Sarala, G.; Naveen, S.;
Anandalwar, S. M.; Prasad, J. S.; Rangappa, K. S. J. Chem. Res. 2006,
11, 730.
2) Koltunov, K. Y.; Chernov, A. N.; Prakash, G. K. S.; Olah, G. A. Chem. Pharm. Bull. 2012, 60, 722.
3) Lee, S. H.; Kim, I. S.; Qing, R. L.; Dong, G. R.; Jeong, L. K.; Jung, Y. H. J. Org. Chem., 2011, 76, 10011.


eˈlute, v.
trans. To wash out, cleanse. Now spec.  [after German eluieren, in same sense] to wash (adsorbed matter) away from the substance that has adsorbed it.

It's transitive, so you elute something, rather than something eluting. Your original "were eluted" is correct.


Why is no bromine liberated in the reaction of potassium bromide and concentrated phosphoric acid?

Firstly, $\ce{H3PO4}$ is not a strong enough oxidizing agent to remove the electrons from the $\ce{Br-}$ ions in order for them to then form $\ce{Br2}$, which exists as a liquid at room temperature. Assuming we are working at room temperature, using a strong enough oxidizing agent such as $\ce{H2SO4}$ or a more reactive halogen like $\ce{Cl2_{(g)}}$ or $\ce{F2_{(g)}}$ would be enough to oxidize the $\ce{Br-}$ to $\ce{Br2_{(l)}}$.

If we desired to evolve $\ce{Br2_{(g)}}$, we would have to heat $\ce{Br2_{(l)}}$ to its boiling point of $58.8\mathrm{^oC}$ and then continue to supply it with enough heat to vaporize it. As docscience stated, however, some oxidizing reactions might be exothermic enough to vaporize $\ce{Br2_{(l)}}$ itself. In solution this would depend on $[\ce{HBr}]$ as to whether or not the heat evolved is enough to vaporize the $\ce{Br2_{(l)}}$, but in the solid form $\ce{Br2_{(l)}}$ is vaporized instantly in the extremely exothermic reaction of $\ce{KBr}$ with either $\ce{F2_{(g)}}$ or $\ce{Cl2_{(g)}}$.


Why can't helium nuclei (alpha particles) react with fluorine?

Why, many atoms would readily create a bond with helium, when it comes in a form of alpha particle (just cooled down enough to chemically interact with). There is comparatively stable $\ce{HeH}^+$ and other similar particles, too. But a positive particle is not a compound yet. And when you try to form a compound out of it, that is, to balance it with some negative ions - well, that's where the problem arises. It would violently react with absolutely anything else, form some compound, and happily release the neutral helium atom.


How do we assign alpha/beta status in polysaccharides?

Here is an explanation for alpha- and beta- glucose.
What it doesn't say is that the alpha-glucose can convert to beta-glucose and back again. Once the linkage is formed, it cannot change from alpha to beta.

Alpha linkage has the oxygen (on the aldehyde or ketone) below the ring and the beta has it above the ring.

Down the page, it shows beta-Maltose. The label beta (for the right most oxygen) is unimportant since the link has to be alpha for it to be maltose. A beta link would result in a molecule of cellobiose.


Overall enthalpy balance on a mixing process

Let’s use a trivial case to test the two possible results. If we set $T_\ce{A}$, $T_\ce{B}$, and $T_p$ all to room temperature, the first and third term would be zero (sign does not matter), and you should simply get the enthalpy of mixing as a result. The result you arrived at does that, but the answer you were given changes the sign.

Unless there are some unexpected conventions about the signs of the terms, your result makes more sense.


For bases, you can derive the $\mathrm{p}K_\text{a}$ from the $\mathrm{p}K_\text{b}$.

Knowing that

$$\mathrm{p}K_\text{a} + \mathrm{p}K_\text{b} = \mathrm{p}K_\text{w} = 14$$

Let’s examine ammonia.

Ammonia has a $K_\text{b}$ of $1.8 \times 10^{-5}$.

Taking the negative logarithm, we can derive $\mathrm{p}K_\text{b}$:

$$\mathrm{p}K_\text{b} = -\log( 1.8 \times 10^{-5}) = 4.74$$.

Knowing that $\mathrm{p}K_\text{a} + \mathrm{p}K_\text{b} = \mathrm{p}K_\text{w} = 14$:

We can find $\mathrm{p}K_\text{a}$ by manipulating that expression, thus:

$$\mathrm{p}K_\text{a} = 14 - \mathrm{p}K_\text{b} $$

$$\mathrm{p}K_\text{a} = 14 - 4.74 $$

$$\mathrm{p}K_\text{a} = 9.26 $$

Even in very basic solutions, there will be minute quantities of hydronium ions. By taking the anti-logarithm of the $\mathrm{p}K_\text{a}$, you can derive the $K_\text{a}$.


What are the effects of heating paraffin wax above it's melting point

In short: A brief heating to 100 C (instead to 60 C) probably did not destroy your paraffin.

It is not too surprising the paraffin accidentally was warmed to 100 C, instead to 60 C. Especially in the solid state it does conduct heat not very well, and has a considerable heat capacity, too.  If it melts around 60 C,  heating to 100 C won't lead to vigorous boiling (like boiling water at this temperature).  In addition, I assume once you noticed the temperature higher than necessary / suitable for the embedding, you rapidly down-regulated the heater, too.

Maybe the manufacturer of the paraffin provided some properties of the batch delivered to you, beside melting point an indication of the decomposition temperature, too; which however should be much higher than 100 C (say 150...175 C).  I think it is unlikely that you observed fumes similar to the ones by an overheated frying pan, which were indicative for decomposition of the paraffin.

The reasoning: Many of the contemporary candles are equally made of paraffin, and when the children make new ones, chunks of old candles and new paraffin are equally thrown into the same melting pot, surrounded by (gently) boiling water, too. And while they do so for hours, I still have to see one new candle that became brown or black because the vessel was heated for so long time.


Why is the excitation energy different from the energy difference between the orbitals involved in transition?

Just look at the Hartree-Fock equations that define orbitals
$$
  \hat{F} \psi_{i}(1)
  =
  \varepsilon_{i} \psi_{i}(1) \, ,
  \quad i = 1, 2, \dotsc \, ,
$$
where $\hat{F}$, the Fock operator, is given by
$$
  \hat{F}
  =
  \hat{H}_{\mathrm{core}}
  +
  \sum\limits_{j=1}^{n}
  \big(
    \hat{J}_{j} - \hat{K}_{j}
  \big) \, .
$$
First, notice that summation in the series goes over the number of electrons, or, in other words, over all occupied orbitals. Secondly, note that $\hat{J}_{i} \psi_{i}(1) = \hat{K}_{i} \psi_{i}(1)$, and thus, the Fock operator for the $i$-th occupied orbital does not include the Coulomb and exchange contribution from the $i$-th occupied orbital itself. The so-called self-interaction is perfectly canceled in the HF model and this is a rather nice feature of this model.

It also means that an electron in each and every occupied orbital "feels" only $n - 1$ electrons in other occupied orbitals, while an electron in a virtual orbital "feels" all $n$ electrons in all occupied orbitals without any exceptions. As a result a virtual orbital is useful only for describing the system in which an electron was added to it but none of the occupied orbitals were depopulated, so that electron in the previously unoccupied orbital still "feels" all $n$ electrons.

Now think about exciting an electron from an occupied orbital $\psi_o$ to a virtual one $\psi_u$. Before excitation an electron in $\psi_u$ would "feel" all $n$ electrons, but after excitation it "feels" only $n - 1$ electrons since $\psi_u$ is not unoccupied anymore. That is clearly a contradiction. So, indeed, $\varepsilon_u - \varepsilon_o$ should not be used to calculate the excitation energy, because exciting an electron from $\psi_o$ to $\psi_u$ leaves behind a hole in $\psi_o$ and $\varepsilon_u$ describes the case when one puts an electron into $\psi_u$ but does not remove one from $\psi_o$.


Is 2,3-dimethylprop-1-ene a possible organic structure?

2-methylbut-1-ene.

For future reference: an alkyl substituent at the 3 position of a propyl group probably indicates you did not select the longest carbon chain to name the parent molecule.

So your answer was not 'formally' correct.  That being said there was only possible structure to be drawn from the name you used, so correct or otherwise, it is effective communication.


Table salt (NaCl) is diamagnetic. For paramagnetism, one needs free unpaired spins.
The NaCl lattice is built up from closed shell cations and anion, there is no such unpaired spin pure NaCl.

Also: experimentally it is difficult to measure paramagnetism, and you generally cannot just try it with a magnet and observe it without some tricks.


Is there a general / simple formula for Slater's rules?

Sadly, no. It wouldn't be too hard to program a "get Slater shielding" script. But it wouldn't be terribly useful.

First off, let's back up briefly. Slater's rules allow you to estimate the effective nuclear charge $Z_{eff}$ from the real number of protons in the nucleus and the effective shielding of electrons in each orbital "shell" (e.g., to compare the effective nuclear charge and shielding for say $3d$ vs. $4s$ in transition metals).

Slater wanted to use these for quantum calculations, such that the radial wavefunction would be something like:

$$\psi_{ns}(r) = r^{n-1}e^{-\frac{(Z-\sigma)r}{n}}$$

So the rules he devised were fairly simple and produced fairly accurate predictions of things like the electron configurations of transition metals and ionization potentials.

Later, others performed better optimizations of $\sigma$ and $Z_{eff}$ using variational Hartree-Fock methods. For example, Clementi and Raimondi published "Atomic Screening Constants from SCF Functions." J Chem Phys (1963) 38, 2686–2689. (That article covers elements up to Kr.)

It's pretty obvious that no simple formula will cover these curves.

These days, basis set optimization is done by computer, and I doubt many people even glance at the exponents.

In my mind, Slater's rules serve two main purposes at this point:

They're relatively simple to teach and learn and allow quick qualitative predictions, particularly for electron configurations in the transition metals.
They emphasize that exponents for multi-electron atoms must be empirically derived.

I still teach them, particularly, for the first reason. To a lesser degree, they also reinforce the idea of effective nuclear charge and provide some semi-quantitive basis for that concept.


What are possible hazards of fog machines?

Given that these machines have been used extensively in the entertainment industry for decades, and one (Rosco Fogger) even has an Academy Award, common sense says that they aren't dangerous.

Breathing in copious quantities of the fog causes irritation, but that applies to a lot of things. Dry ice fog[1] is basically water vapour - fill the room with it and you die from asphyxiation from the CO2

looks better in many cases as it's "fog" and not "smoke"


Standard electrode potential for electrolytic vs galvanic cells

Potential refers to the amount of energy produced or consumed by the reaction. If the sign for potential is (+) than the reaction is spontaneous (moves on its own). If the sign is (-) then the reaction requires energy to proceed.

In the case of electrolytic cells, the cell is used to decompose a compound. this can require energy depending on the reaction. Think of water and how much energy it takes to produce hydrogen gas and oxygen, that's an example of electrolytic cell doing work.

The definition of an anode is an electrode in which a positive charge flows into an electrical device, this is the direction of the electrons.

For both types of cells direction of electron flow would remain the same, but with electrolytic cells, there would be an additional object, the electrolyte, in the diagram.

Here is a decent picture i found that also does some explanation.


Calculating the change in pH upon diluting an acid

As Ivan said in the comments, $\ce{HCl}$ is a strong acid. It dissociates as follows,
$$\ce{HCl -> H+ + Cl-}$$
Now, let us assume the intial pH to be x. Since, $$\mathrm{pH} = \ce{-log[H+]}$$ $\ce{[H+]_i}$ = $\mathrm{10^{-x}}$, where subscript i means initial. Likewise, since final pH= x+1, $\ce{[H+]_f}$ = $\mathrm{10^{-(x+1)}}$ where f means final. Since $\mathrm{V_i = 400 mL}$, since total moles remain the same,
$$\mathrm{M_iV_i = M_fV_f}$$
we get $\mathrm{V_f = 4000 mL}$.
Hence amount of water to be added is $\mathrm{4000-400 = 3600 mL}$.


IUPAC nomenclature of molecules containing alkoxide and carboxylate groups

The principal characteristic group is expressed as a suffix. The name of the carboxylate anion alone (without the alcoholate function) would be ‘butanoate’. The name of the anion derived from the hydroxy compound (without the carboxylate function) would be ‘butan-1-olate’.

For the compound that is given in the question, one anion is chosen as the parent anion and the other one expressed as an anionic substitutent group. Since the carboxylate anion is the senior characteristic group, the parent structure is butanoate.

According to Subsection P-72.6.2 of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), the prefix for the anionic group $\ce{-O^-}$ is ‘oxido’. (The prefixes is derived from the name oxide by changing the final letter ‘e’ to ‘o’.) Therefore, the name of the dianion that is shown in the question is 4-oxidobutanoate. The complete name of the salt is disodium 4-oxidobutanoate.


Copper is a metallic element, so it is likely to be?

Yes, It is C.

Metals are:

-Generally good conductors of electricity
-Substances with high melting points since they are mostly in liquid/solid states. Only liquid metal I can recall offhand is probably mercury, at room temperature and pressure of course.
-They form positive ions because they usually lose electrons to gain octet configuration.

well, Wikipedia says "Metals are described as malleable (can be beaten into sheets) and ductile (can be pulled out into wires). This is because of the ability of the atoms to roll over each other into new positions without breaking the metallic bond."

Other examples of malleable metals are Gold, Silver and Iron.


What are the limits of analytical chemistry?

The trouble with your question is the use of the phrase in general. Most analytical chemistry is directed towards specific goals and isn't about trying to find what is in a sample in general.

This means that the range of techniques applied are highly dependent on the problem you want to solve. If you want to detect, for example, pesticide residues in foodstuffs you probably already know the list of things you want to detect and the range of techniques appropriate to identifying them. If you seek the active ingredient of a herbal remedy, a different range of techniques apply. Sometimes you want to verify that a pure substance you have made is what you think it is, this involves a different range of techniques again.

One common thread in much analytical chemistry is separation: if you seek to identify something in a mixture you want to separate the mixture into its different components. So a variety of types of chromatography may be used in the bulk or on a micro scale. Identifying very small amounts of contaminants in food (e.g. pesticides) probably involves some form of extraction into a solvent followed by separation using analytical chromatography (gas or high pressure liquid chromatography (HPLC)). Where the range of compounds is not known in advance this can be supplemented by mass spectroscopy (the chromatography separates the components, the mass spectrograph helps identify the fragments of the molecules in each component).

If you don't know what you are looking for in advance, it isn't easy to elect the right separation technique and a great deal of experimentation is likely to be required.

Another consideration is that some things are much easier to detect than others. Chromatographic and detection tools like the electron capture detector can identify organohalides to below parts per trillion levels. Atomic emission spectroscopy can identify some metals easily and at low levels. Other techniques can help identify structure; NMR is (sometimes) good at identifying structures, but usually needs pure compounds in relatively large quantities. X-ray techniques can be useful for minerals. Some compounds have very distinct features in their IR spectra or their UV absorption. And so on.

The bottom line is there are no general techniques. There are a wide range of tools to apply depending on the circumstances and the problem you are trying to solve. Almost everything starts with separation. Then, for simple problems, you may be able to apply a single known technique to find out what you want. For harder problems, a rage of tools may be required but the tools will be fairly specific to the problem.


Does salt hydrolysis takes place in buffer solution?

Water autohydrolyzes, or autodissociates, or autoionizes, to produce a very small amount of H+ and OH- ions. The product of their concentrations is 10e-14.
Most salts that dissolve in water dissociate into ions. Often, one of these ions will have a strong attraction for H+ or OH- and tends to reduce the concentration of that ion in the solution. But since the autodissociation of water is an equilibrium process, if an ion (like acetate ion) combines with the H+, more water will dissociate to produce H+ and also OH-. When sodium acetate is added to water, it dissociates into Na+ and CH3CO2- ions; the Na+ ion attracts several water molecules around it without a preference for OH-. The CH3CO2- ion exhibits a strong preference for H+ over plain H2O, and so would reduce the H+ concentration in the solution. Water compensates by producing more H+ and OH- to keep their product equal to 10e-14.

Acetate ions continue to absorb H+ (it also has an equilibrium to maintain), and finally the acetate ion may be largely present as acetic acid (HCH3CO2). The extra OH- ions that came from the water hang around increasing the pH.

So, the answer to your question is yes, the salt hydrolyzes; in fact, the hydrolysis of the salt is what allows a buffer to be made. You establish a ratio of concentrations of the salt and its acid to give a desired pH. Instead of adding tiny amounts of acid to affect a tiny equilibrium product like 10e-14, you use the ratio of larger amounts of buffer ingredients, governed by the acid dissociation constant.

If the salt doesn't hydrolyze, like NaCl, which ionizes without changing the pH of water, it isn't useful as a buffer material - you can't use it to affect the concentration of H+ or OH- in water.


The Horwitz function 1, also known as the Horwitz “Trumpet”, is an empirical relationship that has, thus far, not been derivable from fundamental principles. As noted by Thompson 2, and as shown below in Fig. 1 (from 2), the relative standard deviation of reproducibility (RSDR) increases as analyte concentration decreases:

Horwitz formulated his expression after noticing a pattern in the relative standard deviations reported in many (thousands at this point) collaborative trial results. The following figure (a screenshot from Hanley [3, p. 12037]) provides the essential equations and background information:

As Hanley notes, the Horwitz function appears to be in the Tweedie family of exponential dispersion fluctuation scaling relationships (https://en.wikipedia.org/wiki/Tweedie_distribution ). Specifically, the Horwitz function may be a compound Poisson-Gamma process, since the exponent, $\alpha $ in equation 2, is approximately 0.85. (More accurately, Thompson 1 notes that it is $1 - (log2)/2$).

Answer to OP’s first question

Assume measurements are Gaussian distributed, with $\mu $ as the population mean and $\sigma $ as the population standard deviation, and assume both parameters are unknown. Let $X_{lab} $ be a single measurement, i.e., a random sample from the Gaussian distribution. Then $z = (X_{lab} - \mu )/\sigma $. So z, which is the so-called ‘z score’, is simply the signed difference parsed in units of $\sigma $. A small z score means the measured result is close to the true value, i.e., $\mu $, assuming no systematic error. A large z score means the measurement has relatively poor accuracy.

But both $\mu $ and $\sigma $ are unknown, so $X_{cons}$ is used as an estimate of $\mu $ and $S_{target} $ is used as an estimate of $\sigma $. The result is a rough, and possibly biased, estimate of z. Frankly, this is too crude for me.

Answer to OP’s second question

Also as noted by Hanley, attempts have been made to derive the Horwitz function from “the binomial distribution, Zipf’s law, and a heuristic approach”: see 4 and 5 below. However, these have not been generally been accepted and it is fair to say that the final word has not been written on the origin of the Horwitz function. As well, the Horwitz function potentially has a problem at zero analyte concentration, as noted by Holland et al. [6]. This is problematic because analytical blanks do not have infinite standard deviations.

References:

W. Horwitz, “Evaluation of analytical methods used for regulation of foods and drugs”, Anal. Chem. 54 (1982) 67-76.
M. Thompson, “The amazing Horwitz function”, AMC Technical Brief No. 17, July, 2004 Royal Society of Chemistry. (2 pages).
Q.S. Hanley, “Chemical Measurement and Fluctuation Scaling”, Anal. Chem., 88 (2016) 12036-12042 (Open Access).
R. Albert, W. Horwitz, “A Heuristic Derivation of the Horwitz Curve”, Anal. Chem., 69 (1997) 789−790.
P. Hall, B. Selinger, “A statistical justification to relating interlaboratory coefficients of variation with concentration levels”, Anal. Chem., 61 (1989) 1465−1466.
R. Holland, R. Rebmann, C.D. Williams, Q.S. Hanley, “Fluctuation Scaling, the Calibration of Dispersion, and the Detection of Differences”, Anal. Chem., 89 (2017) 11568-11575.


How to numerically solve a reaction network with fast equilibrium steps?

For A adsorption:
$$r_1=0$$
because the reaction is in equilibrium.So,
$$K_1=\frac{[A]}{P_A \times [*]}$$
From this equation you can relate $[A]$ to $[*]$.Similarly, using the last reaction you can relate $[B]$ to $[*]$. Your [A] and [B] are connected via reaction 2.
$$\frac{d[A]}{dt}=-\frac{d[B]}{dt}=-k_1[A]+k_{-1}[B]$$
This is a first order differential equation, so you need only one boundary condition. You have three unknown and three boundary conditions $([A]_0=[B]_0=0$ and $[*]=1.00)$.


Does bicyclo[3.2.0]heptane have a plane of symmetry?

It turns out that bicyclo[3.2.0]heptane does have a plane of symmetry (courtesy PubChem):

According to The Molecular Structures of Bicyclo[3.2.0]heptane and $\Delta^6$-Bicyclo[3.2.0]heptene. An Electron-diffraction Study of Gaseous $\ce{C7H12}$ and Molecular Mechanics Calculations on $\ce{C7H12}$ and $\ce{C7H10}$. Robert Glen, Grete Gundersen, Peter Murray-Rust, and David W. H. Rankin, here are the bond angles in degrees:

The link explains the experimental procedure in deduction of the actual structure.There are two possible stable structures/forms for the molecule namely endo and exo.The endo form is supposed to be more favourable than the exo form.

From here, we can conclude Cs symmetry.


According to latest IUPAC recommendations, lowest locant rule is the valid rule followed for organic nomenclature and presently, the lowest sum rule is no longer followed.

So, as you have discussed in the question, the name that follows the lowest locant rule is the proper name of the compound.

So, it is 1,6-dimethylcyclohexene.


You asked to keep it simple, which is why I’m skipping explaining some details. Check the bottom of the answer to see the parts with superscript numbers explained.

The signal at $7\,\mathrm{ppm}$ is not a quartet. It would only be a (typical; there are exceptions, however not in typical organic chemistry) quartet if both of the following two conditions are met:

The signal ratio is $1\,:\,3\,:\,3\,:\,1$ and
The distance between all four signals is equal (they have the same coupling constant[1])

Both reasons stem from the way quartets are created due to coupling.

In your case, the signal is approximately $3.5\,:\,6.5\,:\,6.5\,:\,3.5$ (violating the first condition) and the distances are (enlarging the image on my screen and measuring with a ruler, so only a very rough approximation) $\mathrm{7\,mm, 12\,mm, 7\,mm}$ (violating the second condition).

What you are in fact observing is a pair of doublets. (It could be a doublet of doublets[2] as suggested by bon, but it is not.) You can think of it as two signals, each being a doublet, that are very close together. And that fits perfectly with the theory: Each signal would have an integral of $2$ and would correspond to one set of aromatic protons:

The two on the aldehyde’s side at lower field (slightly higher shift)
The two on the ethyl group’s side at higher field (slightly lower shift).

They are doublets because each one couples with one hydrogen of the other group. All that is left to explain is why one of the two signals of the doublet is higher than the other, creating a quartet-like effect. This is due to an effect called roofing or tenting. The bottom line about roofing is that it occurs if two coupling protons have close chemical shifts, and that once you have identified its existence, you can basically ignore it (it doesn’t give you any information that you couldn’t have derived otherwise).[3] You can observe the same effect in the triplet at $\mathrm{1\,ppm}$ and the quartet at $\mathrm{2.5\,ppm}$: The right-hand side signals of the quartet are slightly stronger than those on the left-hand side and vice-versa for the triplet.

[1] As you know, coupling stems from two protons being close, and connected via a chain of bonds — usually three for hydrogens. The reason that nuclei couple is because they have a spin. For hydrogen (and carbon), the spin can be ‘up’ or ‘down’. Consider two protons next to each other: Their spins can point in the same or in the opposite direction. Both cases have a slight difference in energy which gives rise to slightly different chemical shifts — the coupling you observe.

For reasons that I won’t go into detail into here, this difference does not depend on the strength of your NMR magnet (while the difference between the signals of different protons does). The ppm-scale has been chosen to remove the differences between different proton’s signals (so $2\,\mathrm{ppm}$ will always be $2\,\mathrm{ppm}$) and therefore the coupling splits will vary with the strength of the NMR magnet, which is why they aren’t given in ppm, but rather in Hz.

If you have multiple protons of the same type, the coupling value (one says: coupling constant) will always be the same, so $7\,\mathrm{Hz}$ three times in the event of a true quartet. The reason why the signals then need to be $1\,:\,3\,:\,3\,:\,1$ lies within Pascal’s triangle: Starting from a central signal split it into two halfes with a certain distance from the centre: You will get two $1\,:\,1$ signals. Do the same thing with each signal and the same distance: You will get a $1\,:\,2\,:\,1$ signal, because the two central ones will overlap. Do the same again: Voilà, $1\,:\,3\,:\,3\,:\,1$. So all true quartets that stem from this will obey the $1\,:\,3\,:\,3\,:\,1$ rule.

[2] If you have two different protons on a neighbouring atom instead of two identical ones, you will receive a doublet of doublets instead of a triplet. Think of it this way: One proton has a coupling constant of (say) $\mathrm{5\,Hz}$ with the proton of interest, and would give a doublet. The second one has a coupling constant of (say) $\mathrm{7\,Hz}$ with the same proton, and by itself would give a doublet again. Together, these two doublets would not meet in the middle as in the above $1\,:\,2\,:\,1$ case, because they have different distances. Instead, you will see a $1\,:\,1\,:\,1\,:\,1$ set of four signals, where the distance between the first and the second corresponds to $\mathrm{5\,Hz}$ while the distance between the first and the third corresponds to $\mathrm{7\,Hz}$.

But this doesn’t work here, because we are only coupling to one proton (and I would estimate a coupling constant of $\mathrm{19\,Hz}$ for the second coupling which is impossible in the structure given, and isn’t seen anywhere else in the spectra. Estimate was based on the aromatic coupling being $\mathrm{7\,Hz}$.)

[3] Two spins pointing in the same direction are different in energy from two pointing in different directions I just pointed out. However, if the chemical shifts of both protons that couple are similar, the difference is also noticeable without an external magnet, so that even without the sample being in the NMR magnet the two energy levels would differ — hence the protons are slightly more aligned per se than even distribution would suggest. This in turn means that one of the sides of the signal (corresponding to an energy difference) shows stronger absorption than the other, and because it is always the inner one, one can assume to see a roof — hence roofing.

The only information that roofing can give you — that two protons are close by in the molecule and their shifts are similar — can be derived from other information in the spectrum as well (information that corresponds to numbers! Real, mathematically physical hard numbers! Hence preferred by scientists): The proximity by the coupling constants and 2D-COSY and the similarity of the shifts … well, science doesn’t care too much.


Change in Gibbs free energy of vaporization reaction

Yes. You are correct. $\Delta G^0_{rxn}$ is a function of temperature, and is equal to zero at 100 C for this reaction.


Intuitive explanation for depression of freezing point

Imagine a pure liquid. If part of it freezes, what happens to the rest of the solution? Well, it looks exactly the same since it's also pure, so nothing.

However, this is not true for the system you've described.

What happens if there's a solute mixed in liquid? When, now you're concentrating the solute in the remainder of the liquid, so you're creating an entropic barrier to freezing the liquid.

Freezing is a process with $\Delta H < 0$ and $\Delta S < 0$. This means that with $\Delta G = \Delta H - T\Delta S$, the process is spontaneous when $T$ balances the favorable enthalpy change against an unfavorable entropy change. This $T$ is the freezing point. However, notice that due to the above argument, $\Delta S$ is now more negative. This means that the same balance will be achieved with a smaller value of $T$, hence, a freezing point depression.


Is it true that calcium carbonate is more soluble in NaCl than deionized water?

You claim to have a solid solution of Calcium carbonate in solid (or molten? you didn't specify temperature!) sodium chloride. I doubt it.
Because $\ce{CaCl2}$ is more soluble in water than $\ce{CaCO3}$, addition of $\ce{Cl^-}$ ions would be expected to increase the solubility of $\ce{CaCO3}$ in water. In other words, "yes, it is possible." One crude way to think about it is that the "average" environment for the $\ce{Ca^{2+}}$ ions is more "friendly" in a mix of $\ce{CO3^{2-}}$ and $\ce{Cl^-}$ ions than in a mixture of $\ce{CO3^{2-}}$ions alone. The effect is called the "Common Ion" effect. Look it up on Wikipedia.


Does an increase in pressure necessarily lead to an increase in temperature?

I think what you are asking is "why can the temperature of the gas increase when you compress it, even if the cylinder is adiabatic so that no heat can enter the gas?"  When you move the piston to compress the gas, you are doing work on the gas at the interface with the piston.  The piston is moving toward the gas, and the molecules of gas that collide with the piston leave with a greater average velocity than when they arrived.  So their average kinetic energy is increasing.  If expansion were occurring, such that the piston were moving away from the gas, the colliding molecules would leave with lower average kinetic energy.


Changing the concentration of B has no effect on the results.

This statement could only mean one thing: B does not participate in the reaction. Obviously, if B does participate in the reaction, and you change the concentration of B to 0 the reaction won't happen.

The equation for the half life of a zero order reactant is
[Binital]2k.

Zero order reaction have no half-life. Only 1st order reaction has half-life.


Finding the nearest neighbour, from an atom within a crystal

I don't use VESTA and there are probably a few tools that can do this.

My suggestion would be to use the open source Python package ASE or "Atomic Simulation Environment." developed in part by the group of Karsten W. Jacobsen.

There are several algorithms available for generating neighbor lists and iterating through them:
https://wiki.fysik.dtu.dk/ase/ase/neighborlist.html

Here's an example from the documentation:

# get neighbors of atom 42
indices, offsets = nl.get_neighbors(42)
for i, offset in zip(indices, offsets):
    print(atoms.positions[i] + dot(offset, atoms.get_cell()))


How does an anode metal come into solution if it cannot react with an electrolyte?

I was just wondering then that when you have a galvanic cell which has a $\ce{Mg}$ metal anode submerged in sodium chloride solution, how the magnesium metal gives two of its electrons and come into solution (as magnesium ion) even though it does not react with (and cannot react with) sodium chloride?

Your misunderstanding stems from assuming that the electrode reaction has to be the same sort of single-displacement redox reaction that would occur with no electrodes present at all.

At a dissolving magnesium anode$^\dagger$, the only reaction that needs to occur is the one you wrote involving magnesium:

$$
\ce{Mg\!(s) -> Mg^2+\!(aq) + 2e-}
$$

The potential applied to the electrode pair is what enables the above reaction to occur without a paired reduction reaction at the anode surface.  Instead of being taken up in another chemical reaction, the electrons are drawn into the anode and on into the connected electrical circuit. Those electrons are then "pushed" into the cathodic reaction, whatever it might be.

  From what I understand, anode material is chosen so that it does not react with the anode electrolyte solution.

This is true, but does not have to strictly hold. MaxW is right: $\ce{Mg}$ does react directly with water. But, the reaction is slow enough that it's perfectly safe, and that one can conduct useful electrochemical operations even while it's going on (e.g., see here and here).

  Is it that the metal is more stable in its ion state and thus readily gives up its electrons?

Despite what one reads from a table of standard reduction potentials or a metal reactivity series, in practical electrochemistry, the above is true for most metals, except for the noblest of the activity series. In my experience, it's usually a lot easier to dissolve metals than to electrodeposit them.  You may have to use additional chemistry to make it happen (e.g., including $\ce{HF}$$^\ddagger$ to strip the passivating oxide from $\ce{Ti}$, $\ce{Nb}$, $\ce{Ta}$, etc.), but once the metals are oxidized/dissolved they tend to stay that way.

 $^\dagger$ Operating at $100\%$ faradaic efficiency. Other reactions that might occur in this system, depending on the applied potential, include the chloride oxidation reaction you noted and water electrolysis to oxygen gas.
 $^\ddagger$ $\ce{HF}$ is very dangerous, do not use at home!


According to the Canadian Centre for Occupational Health and Safety page Welding - Fumes And Gases, the welding fumes are

  a complex mixture of metallic oxides, silicates and fluorides. Fumes are formed when a metal is heated above its boiling point and its vapours condense into very fine, particles (solid particulates).

This composition varies dependent on the composition of metals being welded and fluxes.  The website lists several main types and their associated health risks.  Often associated with metallic oxides are some pungent/noticeable gases:

Gases produced from the thermal breakdown of coatings, from welding and cutting processes, stated in the article linked as being examples of "welding gases" (examples from the linked website include ozone)
And depending on the coating that may be on the metal, organic vapours (including aldehydes, diisocyanates, phosgene and phosphine)


IUPAC nomenclature of coordination complexes

I think IUPAC's Red Book in the 5th edition by 2005, is the most recent reference of this topic, IR-9 (pages 142 onward) about coordination compounds.  Corrections of errors in the print are published here.


Rate and mechanism of the reaction of Mg and HCl

Your rate law should be
$$ r = k_\text{obs} \times [\ce{H3O+}]^2 = A_\text{obs} \exp\left(\frac{E_\text{a, obs}}{RT}\right) \times [\ce{H3O+}]^2\; ,$$
which nicely fits your experimental data of a quadratic dependence on the concentration for the rate.

The exponent is necessary because the reaction is seemingly second-order in the concentration of $\ce{H+}$.


Why does the specific conductivity against concentration curve for NaCl curve more than KCl?

For equally charged ions a smaller radius implies a higher charge density, and therefore stronger Coulombic interactions with other charges in solution, including stronger ion-dipole interactions with water molecules, resulting in a more stable hydration sphere, and stronger interactions with counterions (here chloride). Such interactions reduce the mobility of the ion by opposing the effect of the applied external electric field, by either increasing solvent drag or generating an opposing electric field (see for instance this chem libretext for an explanation). This slowing down can be interpreted as an increase in the hydrodynamic or Stokes's radius, and also results in a reduced diffusion coefficient, as shown in the following table (source CRC, $\pu{ 25^\circ C}$):

Ion
$\Lambda_+ \pu{(10^-4 m2Smol^-1)}$
$D \pu{(10^-5 cm^2s^-1)}$

$\ce{Cs+}$
77 .2
2 .056

$\ce{Rb+}$
77 .8
2 .072

$\ce{K+    }$
73 .48
1 .957

$\ce{Na+}$
50 .08
1 .334

$\ce{Li+}$
38 .66
1 .029

$\frac12$ $\ce{Sr^{2+}}$
59 .4
0 .791

$\frac12$ $\ce{Ca^{2+}}$
59 .47
0 .792

$\frac12$ $\ce{Mg^{2+}}$
53 .0
0 .706

$\frac12$  $\ce{Be^{2+}}$
45
0 .599

Therefore the smaller monovalent cation, $\ce{Na^+}$, and divalent cation, $\ce{Mg^{2+}}$, have lower molar conductivity than the species lower in the periodic table ($\ce{K^+}$ and $\ce{Ca^{2+}}$).

Additional comments:
(1) Instead of discussing NaCl or KCl, start discussing individual cations and anions, recognizing that these salts completely dissociate into ions in solution. Notice that the different salts share a common anion (chloride, Cl-). Therefore any differences are due to the choice of cation. Sounds obvious, but then you can look at everything from the perspective of the cation's interactions (even if those of the anion will also change between salts, and affect conductivity).
(2) Distinguish between the size of an ion before and after hydration. Bare (unhydrated) $\ce{Na^+}$ is smaller than $\ce{K^+}$, but hydrated $\ce{Na^+}$ is bigger than $\ce{K^+}$.
(3) Bigger bare (unhydrated) cation means weaker interactions and therefore faster mobility.
(4) What drives the formation of a larger hydrated ionic radius is deep down the same thing that drives interactions with other ions. The hydration shell is not that tight. Interaction with other ions will readily override them.
(5) Stronger interactions with other ions means more deviation from ideality. Ideality here means that individual ions behave the same (their mobility is the same) at all concentrations.


How do I write a procedure for creating a buffer?

Initial note

The specific problem with your scenario is, that you need a buffer of buffer capacity progressively decreasing with increasing $\mathrm{pH}$, but capacity of simple buffers is for $\mathrm{pH} \lt \mathrm{p}K_\mathrm{a}$ increasing.

As consequence, its "breaking" capacity is too big, compared to "withstanding capacity".

General analysis

Let there is given the initial $\mathrm{pH_{ini}}$ and the acid dissociation constant
$\mathrm{p}K_\mathrm{a}$

Let there there is given "breaking volume" of $\ce{NaOH}$ $V_\mathrm{br}$, where the buffer losses it's buffering ability.

Let there is given "withstanding volume" $V_\mathrm{ws}$ when $\mathrm{pH}$ changes by value $x$.

Then:

$$\mathrm{pH_{ini}}=\mathrm{p}K_\mathrm{a} + \log {R}$$

where

$$R=\frac{ [\ce{A-}]_\mathrm{init}}
{[\ce{HA}]_\mathrm{init}}=10^{\mathrm{pH_{ini}}-\mathrm{p}K_\mathrm{a}}$$

$$\mathrm{pH_{ini}}+x=\mathrm{p}K_\mathrm{a} +\log  \frac {R+V_\mathrm{ws}/V_\mathrm{br}\cdot (1-R)}{1-V_\mathrm{ws}/V_\mathrm{br}}$$

If we consider
$$R_1=10^{\mathrm{pH_{ini}}+x-\mathrm{p}K_\mathrm{a}} $$
$$V_\mathrm{r}=V_\mathrm{ws}/V_\mathrm{br}$$

then
$$R_1=\frac {R+V_\mathrm{r}\cdot (1-R)}{1-V_\mathrm{r}}$$

$$V_\mathrm{r}=\frac {R_1-R}{R_1-R+1}$$

Goal

The desired $$V_\mathrm{r}\ge \frac{25}{35}=0.714$$

Acetic acid case

If we consider acetic acid with $\mathrm{p}K_\mathrm{a}=4.75$ then $$R=10^{3.75-4.75}=0.1$$

If the change should be $\Delta \mathrm{pH}=0.5$:

$$R_1=10^{3.75+0.5-4.75}=0.316$$

$$V_\mathrm{r} =\frac{0.316-0.1}{0.316-0.1+1}=0.177\lt 0.714$$

If the change should be $\Delta \mathrm{pH}=1.0$:

$$R_1=10^{3.75+1.0-4.75}=1$$

$$V_\mathrm{r} =\frac{1-0.1}{1-0.1+1}=0.475 \lt 0.714 $$

Formic acid case

If we consider as a replacement e.g. formic acid with $\mathrm{p}K_\mathrm{a}=3.75$ then $$R=10^{3.75-3.75}=1$$

If the change should be $\Delta \mathrm{pH}=0.5$:

$$R_1=10^{3.75+0.5-3.75}=3.16$$

$$V_\mathrm{r} =\frac{3.16-1}{3.16-1+1}=0.684\lt 0.714$$

If the change should be $\Delta \mathrm{pH}=1.0$:

$$R_1=10^{3.75+1.0-3.75}=10$$

$$V_\mathrm{r} =\frac{10-1}{10-1+1}=0.9 \gt 0.714$$

Summary

Acetic acid is unuseable for making a buffer to fit your requirements.
Formic acid can be used to make a buffer with "withstanding" parametr $\Delta~ \text{pH} \lt +1.0$, seen in the other question.
Formic acid very tighly missed the "withstanding" parametr $\Delta~ \text{pH} \lt +0.5$, so seems about usable.
Even better results could be achieved with an acid with $\mathrm{p}K_\mathrm{a} \lt 3.75$ .


Unfortunately this is not feasible
Current catalytic converters are designed to remove noxious pollutants from exhausts. Things like carbon monoxide (poisonous) and nitrogen oxides (noxious and irritating).
The trouble with doing this for carbon dioxide is fundamental. Carbon dioxide and water are the ultimate products of combustion of hydrocarbons because they are the thermodynamically deepest products of combustion (ie forming them releases the most energy from combustion). We are not worried about water. But the trouble with being at the bottom of the thermodynamic well is that, to turn the compound into something else, you need to add energy.
There are known chemical reactions that convert carbon dioxide into other, more useful and less greenhouse-contributing products. We can go the whole hog back to methane or, more commonly, methanol. But these reactions are very energy intensive and require fairly special catalysts and conditions. To do them as a way of reducing carbon dioxide emissions would end up using most of the energy from hydrocarbon burning to run the reaction not to move the vehicle. And there are always energy losses in such reactions so we might end up greatly increasing fuel consumption overall to get a small reduction in emission. It is probably far better to make engines and vehicles more efficient.
On an industrial scale the preferred alternative to carbon dioxide conversion is carbon capture and storage (CCS). Separating carbon dioxide is much cheaper than converting it to something else. Even so, CCS plants are in their infancy and none are remotely economical even when subsidised (or at any feasible carbon price). and these reactions require large chemical plants and could never be viable in a car.


I like bananas, but I almost went bananas doing some research for this question!

Let me post some quotes - too much info to post the sources:

1) Is it safe to store my bananas in the refrigerator?
Dr. Brecht: “Never, never, never put bananas in the refrigerator!

2) The answer, far from dissuading me from keeping bananas in the fridge, only confirms my habit. Banana-eaters are divided into two fairly equal camps: those who like the delicate flavour of the green (unripe?) banana and those who prefer the quite different strong flavour of the yellow (over?)ripe banana. Using cold to inhibit the enzymes which aid the ripening process is just what is needed by the 'greens' but should be avoided by the other camp.

3) If they're kept at a cold temperature, the enzymes that enable them to ripen are inhibited. And as those enzymes become inactive, other enzymes operate more efficiently. Some cause cell damage, while others (browning enzymes) cause the skin to blacken. Bananas do best if stored at around 10°C.

4) Bananas require gas, that is ethylene, to trigger the ripening process.

5) Cooler temperatures also slow down the chemical ripening process. Once bananas are ripe to your likening, store them in the fridge.

6) De gustibus non est disputandum

So it seems that bananas do change because of the cold, but the process is not exactly ripening. It does involve a chemical change, which is so complex that it can be predicted and analyzed, but is not useful as a guide to developing a tasty banana for a specific person. Too much variation in the people who eat bananas! I did an experiment some time ago, wondering if the economy of buying green-yellow bananas (with noticeably thick, firm skins) was significantly less than buying bananas with exactly the color for immediate eating (deep yellow for me), where the skins were noticeably thinner, softer. How much weight loss do I incur and pay for, but don't get the benefit of? Not much, it seems, about 10% of the weight of a banana, maybe 15% if you got a green one and let it go almost brown.

So banana color may not be the whole story: enzyme activity occurring at different temperatures may affect taste without being "ripening" - a banana may go from underripe to overripe without ever being exactly ripe! The process of marketing, from green to yellow, is described in some detail, and interestingly, ripening bananas develop heat (and I would guess, ethylene also), so the recommendation was to unpack the banana boxes quickly when they get to the grocery store to reduce the temperature (and C2H4) so the bananas keep longer.


What barriers prevent plastic from becoming a fiber optic, data transmission medium?

In fact, I've never even heard of such research making headlines on
  the more common tech news sites.

Tasks like that are called deveopment, not research, normally.
And no industrial company will talk much about such work
if its not sucessful.
In fact there are such organic fibers for minor purposes (eg the Laserjets
II and III used such a fiber to send a synchronising signal from the
laser scanning unit down to the control board, for about 10 inches.
30 years ago one could buy such fibres for experiments at radio shack)
If You read some literature on optical fibres, You will learn that
low scattering is the grail. And plastics scatter a lot, for
very basic reasons of polymer structure. Try to look throug a plate of
acrylic glass "edge on", and You will understand.


Have P-acyl phosphines ever been synthesized?

When I saw this question, I realized I couldn't remember ever seeing an acylphosphine, but I was curious and went looking. I found that they're actually quite easy to make. One common thing I saw was the use of trimethylsilylphosphines with acyl chlorides to form acylphosphines.

The easier method, however, was simply to react a substituted phosphine in the presence of triethylamine with an acyl chloride. The analogous reaction can be done with amines to form amides. In this paper, they describe the synthesis of P-acetyldiphenylphosphine using diphenylphosphine, triethylamine and acetyl chloride. They have some other examples in there, but this suggests that these compounds are typically fairly trivial to synthesize.

One more note: In my searches, I saw "acylphosphine," "acyl phosphide," and "phosphomide" as named. The paper I linked uses the term "phosphomide."


Uranium enrichment via effusion through a porous membrane

The second approach looks more correct to me. The first approach is using raw percentages without converting to a ratio between isotopes, and the ratio rates is going to be directly related to the ratio of isotopes, not raw percentages.

From the comments, ed:
Well, $1.0043$ is the ratio between rates, so each stage will cause the ratio between isotopes to change by this factor.
I simple set up:
$$\frac{0.72}{99.27} \times 1.0043^{n} = \frac{3}{97}$$ which is the same equation as method 2.

$\frac{0.72}{99.27}$ is the initial ratio (before any enrichment). You need to increase it successively by $1.0043$ until you get to $\frac{3}{97}$, the desired mol%.


Why are the boiling points of aldehydes less than those of ketones?

It isn't that good a generalization: always look at the data first.

Here is a table of most of the aldehydes and ketones with 6 or fewer carbons (the labels are used in the chart later):

Now plot this on a chart:

Branches is the number of branches in the carbon chain.

Note that while for 3 and 4 carbons the ketones do have higher boiling points, it is not clear this is true for 5 carbon compounds and certainly not true for 6-carbon ones.

So, I'd argue the pattern of boiling points is complicated and there is no simple pattern that needs to be explained.


Spin coat of Graphene oxide on glass slides

The procedure will vary with the size and shape of the glass, the concentration and viscosity of your graphene oxide dispersion, the solvent you're using, the wettability of your glass slide, and so on. You'll need to figure out the parameters yourself by varying the spinning conditions, measuring the resulting film thicknesses via profilometry/AFM/cross-sectional SEM, and then constructing spin curves (examples).


Can Grignards react with diatomic halogens as to perform an
  substitution reaction?

Yes, but it looks like a mixture of products results, including something explosive.  Here is a link to an abstract of some early work that reports,

"Chlorination of Grignard and Iotsich reagents, RMgX, (where X = Br, I) gives considerable quantities of RI and RBr together with RCl. Bromination of RMgI results in RI and RBr. Bromination of RMgCl leads to RBr, and iodination of RMgX (X =Cl, Br), to RI."

An "Iotsich" reagent is an acetylenic Grignard ($\ce{RC#CMgBr}$)

Judging by the fact that 1) the abstract suggests that product mixtures of the various possible alkyl\aryl halides are often formed and 2) no further reports claiming synthetic utlity have emerged, I suspect that the reaction is of limited synthetic value.  Nonetheless, your initial suspicion that reaction should occur seems supported.

One further note, many MSDS's and "Hazardous Laboratory Chemicals Disposal Guide, Third Edition" report that when phenylmagnesium bromide is reacted with chlorine, a solid is formed that explodes when shaken.  The reference provided is the same J. Organomet. Chem. paper cited above.  Given the lack of detail provided on the solid, my guess is that it wasn't characterized.


Explanation for the reactions in a saltwater battery with zinc and copper electrodes

This answer deals with the situation, when but electrodes are in the same salt water solution.

Assuming your salt solution contains sodium chloride
$$\ce{NaCl (s) ->[\ce{H2O}] Na+ (aq) + Cl- (aq)},$$
then your reaction at the zinc terminal will certainly be
$$\ce{Zn (s) + 2Cl- -> ZnCl2 (aq) + 2e-}.$$
On the copper terminal water will be electrolysed
$$\ce{H2O + 2e- ->[\ce{Cu}] 1/2 H2 ^ + {}^{-}OH},$$
or using autoprotolysis
\begin{aligned}\ce{
2H2O &<=> H3+O + {}^{-}OH\\
H3+O + e- &->[\ce{Cu}] 1/2 H2 ^ + H2O
}.\end{aligned}

This will happen since sodium has a much lower electron affinity than water/ hydronium
\begin{aligned}
\ce{Na+ + e- &-> Na} &E^\circ&=−2.71\:\mathrm{V}\\
\ce{2H2O + 2e- &-> H2 + 2 {}^{-}OH}&E^\circ&=−0.83\:\mathrm{V}\\
\ce{2H+ + 2e- &-> H2}&E^\circ&=0\:\mathrm{V}.
\end{aligned}

Depending on what salt solution you are using, different cations might be reduced. You can find a collection of values at wikipedia.

In a battery (Galvanic cell) there are usually coupled half cells, and in this case this would probably be $\ce{Zn|Zn^2+||Cu^2+|Cu}$.

This means, that the Zinc terminal is in a zinc containing solution and the copper terminal is in a copper containing solution. In this case your assumption is correct
\begin{aligned}
 \text{Zinc terminal:}&&\ce{Zn(s) + ZnCl2 (aq)&<=> 2Zn^{2+}(aq) + 2Cl- + 2e^-} \\
 \text{Copper Terminal:}&&\ce{CuCl2(aq) + Cu (s) + 2e^- &<=> 2Cu(s) + 2Cl-}
\end{aligned}
There has to be a salt bridge to neutralise charge differences (ion flow).
Note that in that case zinc ions will go into solution, while copper ions will be removed from the solution. If the concentration of copper drops close to zero, then no current can be measured. The galvanic cell can be charged by "reversing" the process, apply opposite current. Then zinc ions will be removed from solution and copper will go into solution.
In reality, this is a equilibrium process.


How to calculate mole ratio of two volatile compounds in distillation?

If you have the vapor-liquid phase diagram, then it is easy to determine the vapor composition above a boiling liquid mixture (which should approximate the composition of your first drop in a simple distillation).

I'll assume you diagram looks like the one below, which is from the Wikipedia article on vapor-liquid equilibrium(and is mostly at the physical chemistry level):

Let's say you have a mixture of components 1 and 2 that is 40 mole percent (mole fraction = 0.4) component one (approximately the blue line labeled liquid on the diagram). Draw a line vertically from that composition to the bubble point (or liquid) line. Then draw a horizontal isotherm at that point (red line on the diagram above). Where that isotherm crosses the dew point (or vapor) line, you can read down to get the composition of the vapor above that boiling liquid (blue line labeled vapor), about 15% Component 1 (and 85% Component 2). This represents a single vaporization-condensation cycle, which is what happens in a simple distillation.

In a fractional distillation, you can have multiple vaporization-condensation cycles (called theoretical plates). If you know how many theoretical plates your distillation column will have, then you repeat the process for the simple distillation several times. For example, if there are two theoretical plates, you have two vaporization-condensation cycles. After the first one, start over with a new liquid composition equal to the vapor composition after the first cycle. Draw a line up to the liquid line, draw the isotherm to the vapor line, and draw a line back down to composition. If you use the example I have been working with on the diagram I have shown, you should get about 5% Component 1. Repeat as necessary.

If you do not know how many theoretical plates you have in your fractionating column, then you can only make an educated guess.


Temperature relation for isomerism reactions

The van't Hoff equation would predict that the equilibrium constant is not dependent on temperature, since in this case:
$$\frac{\mathrm{d}}{\mathrm{d}T}\ln K_{eq} = \frac{\Delta H^\circ}{RT^2}=0$$
Strictly speaking the equilibrium constant is a function of fugacities, but it is probably safe to assume that the pressures are also independent of T, so the answer would be yes.


what's the name of this org. compound. RR'-C=NNH2

The (now deleted) answer by NotCorey has already explained that the compound that is shown in the picture is a hydrazone that is derived from acetone (propan-2-one).

According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), the preferred IUPAC name (PIN) for a hydrazone is generated substitutively as ‘ylidene’ derivatives of hydrazine rather than by functional class nomenclature as in previous recommendations.

  P-68.3.1.2.2 Hydrazones

  Compounds having the general structure $\ce{RCH=N-NH2}$ or $\ce{RR'C=N-NH2}$ are called ‘hydrazones’ and are named in two ways:

  (1) substitutively as derivatives of the parent hydride ‘hydrazine’, $\ce{H2N-NH2}$;

  (2) by functional class nomenclature using the class name ‘hydrazone’.

  Method (1) generates preferred IUPAC names.

Therefore, the PIN for the compound that is given in the question is (propan-2-ylidene)hydrazine.


These ionic charges actually do behave as you would expect from electric charges: cations will be drawn towards the negative pole in an electrolytic cell while anions will be drawn towards the positive pole. The same feature can be observed with α and β radiation (which are $\ce{^4_2He^2+}$ and electrons, respectively); these are deflected in an electric field as expected.

On a more philosophical level, the entire concept of one charge being ‘positive’ while the other is ‘negative’ is of course meaningless. We might as well have called them ‘East’ and ‘West’ charges akin to the North and South poles of a compass. The only physically observable is that the two act as opposites; two particles of like charge will repel, two of opposite will attract. Having positive and negative charges makes calculations much easier though.


Why isn't my electrochemical cell producing its potential voltage

Aluminum redox potential has nothing to do with the voltage of the cell you have build with it. Let me explain why, because it is not obvious.
Electrolytic cells made with aluminum anodes always yield rather low voltages. The reason is that usual pieces of aluminum are always covered by a thin, continuous and colorless layer of aluminum oxide $\ce{Al2O3}$. This oxide is an electrical insulator which makes a barrier between the metal and the solution. Aluminum ions cannot cross this barrier. There is no way of preventing this aluminum oxide formation. If you rub the aluminum surface to remove it, the pure metal so exposed will immediately react with air to remake this layer.
The only way to suppress this layer is to dip aluminum in a mercury chloride solution. Of course today this operation is forbidden by law. But this experiment was done before this legal decision. So let's see what was happening. Some mercury ions may cross the barrier and react with aluminum to produce an amalgam $\ce{Al(Hg)}$ plus some $\ce{Al^{3+}}$ ions getting in solution$$\ce{5 Al + 3 HgCl2 -> 3 Al(Hg) + 2 Al^{3+} + 6 Cl^-}$$ The formula Al(Hg) is not sure, as it is an alloy. Maybe the ratio Al:Hg is different from $1:1$. It does not matter for the present study. The only important thing now is that the aluminum oxide layer does not perfectly cover the amalgamated metal any more, and is partly destroyed. Therefore aluminum atoms are suddenly in contact with water, and they quickly react with it according to :  $$\ce{2 Al(Hg) + 6 H2O -> 2 Al(OH)3 + 3 H2 + 2Hg}$$ or$$\ce{2Al + 6 H2O ->  2 Al(OH)3 + 3 H2}$$This shows that unprotected aluminum metal quickly reacts with water, producing $\ce{H2}$ bubbles, and these $\ce{H2}$ bubbles are responsible of the real anodic half cell according to :
$$\ce{H2  -> 2 H^+ + 2 e-}$$ As a consequence, this "aluminum" electrode would have, at maybe $p$H$ 5$, a redox potential $\ce{E(H2/H+)}$ given by Nernst law : $$\ce{E(H_2/H^+) = E°(H_2/H^+) + 0.0296 V·log[H+]= 0 V + 0.0296 V·log 10^{-5} = - 0.148 V}$$ In your setup, this anode is coupled with an ordinary copper half-cell as cathode, whose standard redox potential is + $0.34$ V. The voltage of this cell is $$\ce{E(Cu/Cu^{2+}) - E(H2/H+) = 0.34 V -(-0.148 V) = 0.488 V}$$ This value is nearly the voltage that you have obtained in your cell ($0.50$ V).


What are these "certain chemical compounds" denser than osmium?

This text was added as part of a refactoring/cleanup of the page in this edit in March 2013, where it replaced the text "Osmium and iridium are the densest known elements at standard conditions for temperature and pressure but not the densest materials." This older text was tagged with the note "Which are denser, at standard conditions?" It in turn was added in this edit in May 2011 by an anonymous editor (the IP address used is apparently registered to the Wyoming Community College Commission). The original wording, "Osmium is [...] not the densest material", is ambiguous, but could be read as referring to exotic forms of matter such as neutronium, in spite of the earlier "standard temperature and pressure" caveat.

In short, this likely derives from a well-meaning but misguided editor attempting to note the existence of much higher-density, non-chemical substances in an inappropriate place.


Can carbon monoxide from a floor-heater in a poorly ventilated room eventually be ignited by the fire and explode?

Another possibility:  the "carbon monoxide" formed by combustion is not pure enough CO to get to flammability (12% total CO) in air.  The gas could be mostly carbon dioxide but still enough CO to poison you.


I don't think the point of freeze drying is to avoid liquid water -- I mean, there's liquid water already in the substance to be dried -- but to avoid heat.  Ordinarily, to dry something out to preserve it, you need to heat it to speed up the evaporation of water, certainly if you want to get rid of the water quickly (which you would want to do in any commercial setting, where time = money).  But heating causes all kinds of changes to foods, and while some foods are fine with it, others  suffer degradations of perceived quality, e.g. undesirable changes in color and texture.   The point of freeze-drying is that you can get rid of the water at a low temperature.


How to measure band gap of powder sample with using diffuse reflectance method

It's not too hard to determine band gap.  When the energy of a wavelength of light is higher than the energy of the band gap the light will be absorbed, any wavelengths that are lower in energy will reflect or transmit. Simply make the spectrum, there should be a cliff-looking cut-off in transmittance, determine the wavelength and corresponding energy and that is your band gap.


What molecules are produced in carbon arc lamp and how do they produce light?

Answers:

Consider the mass and kinetic energy of an electron or of an ion of nitrogen or oxygen, and the reactivity with carbon.
Good question! It would be interesting to see the relative abundance of allotropes of carbon deposited from a filament in vacuo. There is a metallic glisten to the deposit, so there might be graphite, carbon nanotubes or buckminsterfullerene. $\ce{C2}$ would not be there for long, though.
There are a great many chemical species produced by an arc lamp operating in air. Depending on the gases in which it operates, an arc can produce complex organic molecules. Even something as ornate as a buckyball can be made!
I've not seen the relative abundance of products of an arc lamp, but I assume carbon oxidizes rapidly to the monoxide and dioxide, depending on abundance of oxygen. Charcoal briquets oxidize at a far lower temperature.
Most of the light is black-body radiation, though there would also be some spectral lines produced by recombination, such as from nitrogen. With a spectroscope, you'd see a continuum with a few bright lines added (or dark lines from absorption, as the gases cool).


Temperature as an independent variable in an endothermic reaction

For this experiment to work, you need good control of your temperature profile.  In other words the temperature must be somewhat uniform.

If this is the case measure two temperatures.

$T_e$ is the isothermal temperature of the experiment.
$T_r$  is the temperature at the point of reaction.

If $T_e$ is 10°C, 20°C, 30°C and  $T_r$ is 9°C, 18°C, and 29°C  then the temperature change can be assumed to have a negligible effect. (You would want to use $T_r$ in the kinetics calculations.)

If $T_e$ is 10°C, 20°C, 30°C and  $T_r$ is 4°C, 6°C, and 8°C  then the effect is not negligible.  Accounting for the endothermic temperature change in the calculations is a poor option because it would add needless complication. A better approach is to keep things simple by revisiting the controls in the experimental design to see what can be done to minimize the endothermic effect.  One possibly would be using less reactants.


When an electron in any element absorbs energy it will be promoted to an excited state.  How much energy we supply determines how many \ which excited states will be created.  As the electron returns from these various excited states to the ground state light can be emitted, this is the basis of atomic emission spectroscopy.

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In the case of sodium (or any other atom with just one valence electron) many excited states are possible, so many emission lines are observed.  However, a very likely transition will be the one from the s orbital to the next highest energy orbital, the p orbital.

The magnetic energy of an electron spin is split into energy states with different total angular momentum due to the electron spin's interaction with the internal magnetic field set up by the orbital motion of the electron, this is termed spin-orbit coupling. In the case of a p orbital, it is split into two states (the total angular momentum j is given as j = l +/- s, where l is the angular momentum quantum number [l=1 for a p orbital] and s is the electron spin (1/2), therefore j has two values or states for an electron in a p orbital, 1/2 and 3/2). It is the light emission that occurs when the excited electron returns from these two p orbital energy states to the s orbital ground state that produces the doublet seen for all atoms with a ground state involving one electron in an s orbital.

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