"Sacrificial agent" usually refers to something variable, but with little value, that can be used to produce something specific or carry out a specific function. In this case, an alcohol (ethanol or glycerol) is sacrificed to act as a source of hydrogen gas, while the titanium species is a catalyst that is activated by light to allow generation of hydrogen.


Why will a strong acid neutralize as much base as a weak acid?

Strong acid is dissociated all but completely: you have a certain amount of $\ce H^+$ and start neutralizing them, one drop at a time. As you do so, their numbers steadily decrease. It feels like sawing a huge log with a hand saw: you see the amount of job, and you do it until it is all done.

Weak acid is dissociated only partially. You see a relatively small amount of $\ce H^+$, but as you start neutralizing them, more molecules of acid get dissociated and their $\ce H^+$ ions stand in place of those you've neutralized. It feels like sawing a thin log, but a peculiar one: it seems to regenerate under the saw.

So at the end of the day, it is the sheer amount of acid that matters, rather than immediate concentration of $\ce H^+$.


Yes, for example lysine.

Lysine has two amino groups and one carboxyl group. When dissolved in aprotic solvents, all of these will be uncharged so the carboxyl group will be $\ce{-COOH}$ and the two amino groups $\ce{-NH2}$. When dissolved in water, these three groups have distinct $\mathrm pK_\mathrm a$ values of $2.15, 9.16$ and $10.67$. Depending on the acidity of the solution, the following (acid-sided) equilibriums occur:

When dissolved in water, the carboxyl group’s proton will immediately be removed and used to immediately protonate the ε-amino group (the far end). However, the α-amino group is still more basic than water and thus will be partially protonated; meaning that the right-hand equilibrium will be happening predominantly if lysine is dissolved in water. Dissolved lysine will be a mixture of uncharged lysine and lysine cation.

When an acid such as hydrochloric acid is added, at first the α-amino group is completely protonated and then the left-hand equilibrium occurs predominantly as more and more lysine cation is converted to lysine dication.

If a base is added (such as sodium hydroxide), the ε-amino group’s proton can also be removed yielding lysine anion; I have not shown that equilibrium in the scheme above.

Lysine cation thus fits your question as it is an amphoteric cation, the conjugate acid of a dibasic compound.


Why do molecules have to have a change in dipole moment in order for them to be IR active?

There is an intuitive reason why a vibration is IR-active only if it involves a change in the dipole moment.

Recall that the typical wavelength of IR radiation ($\sim 10\mu\mathrm{m}$) is much larger than the typical size of a molecule ($\sim 1\mathrm{nm}$). Hence, to a very good approximation, the (time-varying) electric field of the IR radiation is spatially uniform within a molecule.

Now observe that under a spatially uniform electric field, all positive charges, regardless of their positions, are pushed to a common direction, and all negative charges are pushed to exactly the opposite direction. In this case, can the change in the total dipole moment, defined as $\Delta \vec{\mu} = \sum_{i} q_{i} \Delta \vec{r}_{i}$, be equal to zero?

Clearly, the answer is "No." The displacements $\{\Delta \vec{r}_{i}\}$ are parallel between like charges and antiparallel between opposite charges. Therefore, $\{q_{i}\Delta\vec{r}_{i}\}$ are all parallel to one another, and $\vec{\Delta\mu}$ cannot be equal to zero.

So far, I have argued that a motion induced by a spatially uniform electric field always involves a change in the dipole moment. A corollary is that any motion that does not involve a change in the dipole moment cannot be induced by a spatially uniform electric field (of IR radiation).

As a simple example, imagine applying a spatially uniform electric field to a $\ce{CO_{2}}$ molecule such that the direction of the field is parallel to the length of the molecule. Can this field ever induce a symmetric stretch, in which the two oxygen atoms move in opposite directions? Certainly no. Both oxygen atoms are negatively charged by exactly the same amount. Under a spatially uniform electric field, they are pushed to the same direction by the same magnitude, and there is no room for a symmetric stretch.


Change in gas phase reaction equilibrium when not all product and reactant are increased

Given the reaction $$\ce{2NOCl <=> 2NO + Cl2}$$  we can formulate the equilibrium as $$\mathrm{K} = \dfrac{\ce{[NO]^2[Cl2]}}{\ce{[NOCl]^2}}$$

The problem then asks "What will be the effect on equilibrium concentration of NOCl when equal moles of NOCl and NO are introduced in the mixture at constant temperature?"

Now if we add equal amounts of $\ce{NOCl}$ and $\ce{NO}$ we don't have to do it at the same time to end up with the final equilibrium.

So let's add the $\ce{NOCl}$ first. We now have two possibilities (assuming that it is very unlikely that K=1 exactly).

If K>1 then more $\ce{NO}$ and $\ce{Cl2}$ form.
If K<1 then less $\ce{NO}$ and $\ce{Cl2}$ form.

Now let's add the $\ce{NO}$. This has to push the reaction to the left.

So the net result is it depends. You have too have enough data to plug into the equilibrium expression and other equations to work the net result out.


What do s, m, w, m-w, and w-s mean in the context of infrared spectroscopy?

Given the general agreement surrounding my comment on the question and the similar abbreviations and descriptions in the chart on the page linked by Tyberius in his comment, the meanings appear to be:

strong
moderate (or medium)
weak

Notations such as m-w and w-s presumably indicate a peak intensity that is variable in the indicated range, depending on the properties of the particular molecule of interest.


Why is the melting point of tert-butyl alcohol 140 °C higher than that of sec-butyl alcohol?

I am not aware of a solid scientific publication on this matter, but the following points should elucidate what is happening:

The high melting point sort of proves that there is strong hydrogen bonding in tBuOH. I also don't see how the tBu-group could be so bulky that each OH group cannot form two hydrogen bonds.
n-, sec- and iso-butanol all have pairs of methyl- and/or OH-groups that can be in gauche or anti position with respect to each other. The  energy difference is in the range of 4 kJ/mol, which is not much higher than the thermal energy $k_BT$ at 300K. Those alcohols exists more or less as a mixture of conformational isomers, which precludes crystallisation. In tBuOH, there is only one conformation.
Rotation of the methyl groups has a low excitation barrier, but it can also rotate in the solid phase (there is some tunneling involved), so this does not preclude crystallisation.

(The energy of ~4kJ/mol is from n-butane, but I expect no large difference. The gauche conformation of the OH group is probably even lower, energetically.)


first-order binding kinetics with multiple identical binding sites

The scheme you need seems to be of the form
$$\ce{S + M_i  <=>[k_2][k_{-1}] M_{i+1}    }$$
where S is the molecule in solution and M the bead with i sites filled.
If there is no restriction on how many S may be associated with any bead and as the  S are indistinguishable molecules the number of ways of arranging the S among a total of M beads is
$$ W=\frac{M!}{\prod M_i!}$$
and the constrains are $\Sigma_0^N M_i = M$ and $\Sigma iM_i = N$ where N is the total number of S molecules.
Solving with these constraints is done using Lagrange multipliers, in the same manner as the Statistical Mechanics calculation of Boltzmann distribution, and leads to the distribution of finding i molecules of type S in a bead as,
$$p_i = \frac{\bar n ^i}{(1+\bar n)^{i+1}}$$
where $\bar n$ is the average occupancy of a bead. This distribution, which is clearly not Poissonian, has the form that it decreases as i increases even for moderate $\bar n$.

If the kinetic scheme is changed slightly to
$$\ce{S + M_i  <=>[k_2][(i + 1)k_{-1}] M_{i+1}    }$$
where the ‘off’ rate constant now depends on how many S are on a bead then the probability becomes Poissonian. This change/trick effectively makes the molecules distinguishable and now
$$p_i = \frac{\bar n^i}{i!}e^{-\bar n}$$
with $\bar n = k_2\ce{[S]}/k_{-1}$.

Which of these or other models applies will have to be determined by experiment.

If you want to solve the equations numerically then it is best to try a Master equation approach.

The set of equations for the population of M is written as
$$d\vec M/dt = \vec k \vec M$$
and has the formal solution $\vec M=\vec M_0e^{\vec kt}$ where $\vec M$ is a vector of concentrations of species i and $\vec k$ a matrix of rate constants such as

$$ \bar k=\left [\begin {matrix}
-Sk_2 & 2k_{-1} & 0 & 0 & 0 & \cdots\\
Sk_2 & -Sk_2 -2k_{-1} & 3k_{-1} & 0 & 0 & \cdots \\
\vdots& \vdots&\vdots & \vdots& \vdots& \\
\end {matrix}\, \right ]$$

where S is the concentration of S and is assumed to be in excess so that its value is constant. (If you want S to vary then you will need to do a numerical integration on all N equations instead of this method).

To solve, the secular determinant is set up and the eigenvalues $\lambda$ obtained. The populations at time t are given by
$$M(t)= x[e^{\lambda t}]x^{-1}M_0$$
where x is the matrix of eigenvectors corresponding to eigenvalues $\lambda$ and $[e^{\lambda t}] $ is a diagonal matrix of $e^{\lambda_1t}, e^{\lambda_2 t} \dots $ etc.


I guess I don't completely agree with the premise of your question. Gasoline is highly flammable, even potentially explosive, yet "amateurs" deal with it constantly. The key is that it is stored and used properly, from the pump to your gas tank. It becomes less safe if you pump it into a gas can, but gas cans also have certain requirements for storage safety.

So I think the issue may be what sorts of flammables would an amateur have a high chance of storing or using inappropriately. For example, any pyrophoric compound would be a good candidate to avoid, as most require a way to exclude air and moisture, which often requires specialist equipment.

Solvents with low flash points and high vapor pressure should also be treated with respect, but rubbing alcohol (isopropanol) or acetone might qualify for that designation. Storage in airtight fireproof containers in cool and dry places is often sufficient. Better yet, use a flammables cabinet. This simple step is something an amateur might skip for cost, but it may be the most sensible precaution. I have flammables cabinets, and I use them for ordinary hardware solvents along with experimental chemicals.

Finally, work safely! No flames, as they say in organic chemistry. Use hotplates or heating mantles rather than burners. If you must use flames for some reason, put away all flammables and keep them a safe distance. Pay careful attention to any spills and attend to them immediately before resuming work. And perhaps key for amateurs, small quantities! No need for an amateur to have gallons of flammables on hand. Even professional labs are trying to reduce quantities in the name of green chemistry.

EDIT: My go-to source for using and disposing of hazardous chemicals is Hazardous Laboratory Chemicals Disposal Guide by Margaret-Ann Armour. In addition to disposal techniques, each chemical has an extensive incompatibility section about what combinations can make a chemical even more hazardous, and all these entries have literature references. I use this book literally every time I do an experiment. It deals with much more than flammables, but flammables are definitely present.


Why do impurities lower the melting point of an isolated substance?

It's a very general statement, but it's not always true. I'll explain why it's often true, and give a counter-example at the end.

Your majority component B and the impurity (let's call it A) form a binary system. In most cases, such binary mixtures exhibit a solid–liquid phase diagram as follows:

(image taken from these lecture notes).

This binary phase diagram has pure A on the left, pure B on the right. A and B form, somewhere, a eutectic. It is the point here at concentration e and temperature y. Because the existence of a eutectic point is guaranteed for any A/B binary system, and because the eutectic corresponds to a lower temperature, your liquidus curve decreases with increasing impurity concentration, and the impurity thus lowers the melting point.

However, not all binary mixtures form a eutectic. In the words of Wikipedia:

  Not all binary alloys have a eutectic point; for example, in the silver-gold system the melt temperature (liquidus) and freeze temperature (solidus) both increase monotonically as the mix changes from pure silver to pure gold.

The corresponding phase diagram is as follows:


Does this 'Belousov-Zhabotinsky Reaction' kit contain the correct reagents?

No, it doesn't, at least not strictly.
According to Wikipedia (emphasis added):

One of the most common variations on this reaction uses malonic acid $(\ce{CH2(CO2H)2})$ as the acid and potassium bromate $(\ce{KBrO3})$ as the source of bromine. The overall equation is:
$$ \ce{3CH2(CO2H)2 + 4BrO3− → 4Br− + 9CO2 + 6H2O} $$
[ $\ldots$ ]
Many variants of the reaction exist. The only key chemical is the bromate oxidizer.

The kit doesn't contain any bromide or bromate, so by definition it will not allow demonstration of a Belousov–Zhabotinsky reaction.

However: It does appear that the kit may allow demonstration of an iodine clock reaction:

Hydrogen peroxide variation
This reaction starts from a solution of hydrogen peroxide with sulfuric acid. To this is added a solution containing potassium iodide, sodium thiosulfate, and starch.

The ingredients are not a perfect match, and actually appear to be something of a mishmash between the iodine clock reaction and the Belousov–Zhabotinsky reaction, but they're in the right ballpark. Metal ions and their oxides are well known as catalysts for hydrogen peroxide decomposition, and "Sulfamic acid" could be a typo that was supposed to be "sulfuric acid".


Exceptions to motif: nucleophile attacks electrophile

I believe that you are asking if there are any reactions that are essentially "non-polar". There are two classes that immediately come to mind: (1) radical reactions and (2) pericyclic reactions.

A common radical reaction studied in introductory organic chemistry is the radical halogenation of alkanes. Instead of cationic and anionic reactants/intermediates/products, there are radical intermediates that do not fall into the nucleophile/electrophile classification.

Pericyclic reactions involve the flow of pi-electrons and do not require any charged reactants/intermediates/products. Reaction 1 below is a Diels-Alder reaction and reaction 2 is an electrocyclization. Depending on your definition of nucleophile and electrophile, a Diels-Alder reaction might fit the motif of "nucleophile attacks electrophile." That is, the HOMO of the diene (1,3-butadiene in the example) interacts with the LUMO of the "dienophile" (ethylene). If you define a nucleophile as the reaction component providing the HOMO, then a diene is essentially a nucleophile in the Diels-Alder reaction (and vice-versa for the "dienophile"). However, since these reactions don't use the same types of nucleophiles as say SN2 reaction, they are often considered to fall outside of the usual "nucleophile attacks electrophile" motif. That's the way they were first introduced to me, at least.

I should add that many organometallic mediated reactions fall outside of the "nucleophile attacks electrophile" motif as well. My answer was confined to organic reactions.


What is the correct way to verify a structure's geometry, for example for benzene?

A deviation within the low picometres is nothing to worry about, there are many reasons for this.

Primary literature, like peer-reviewed journals, will always publish an analysis of the obtained values along with any error bars, i.e. how accurate the measurement is. When you have disagreeing values it is worthwhile to have a look at these, too.
Secondary literature, like encyclopaedias, will usually only quote values and might not include all details due to their general reference nature.

Quite often if you have different values, the conditions of the measurement simply were not the same. Also keep in mind, that within the recent years a lot of progress has been made for spectroscopic measurements. Hence you can find values from older publications that are significantly different from more modern values.

For example Tamagawa et al. reported $r_g(\ce{C-C}) = 139.9 \pm 0.1~\mathrm{pm}$.[1] They used a method, which combines diffraction and spectroscopic data. This article was published in 1976 and it was introduced with the following paragraph. I think that pretty much summarises the predicament of modern structural analysis.

  It is often found that the structural data of the most basic molecules are not very up-to-date because such molecules were studied in the early stages of the development of experimental methods for structure determination. Benzene is one such molecule.

And since then many more studies on benzene have probably been published. Chemspider.com lists (currently) seven data sets[2] and I am confident this is just the tip of the iceberg.

In a quite recent publication by Budzianowski and Katrusiak the crystal structure of benzene has been studied at various pressures. They pressure froze the molecule(s) in situ in phase I.[3]
We learn from this, that there is more than one phase in which benzene crystallises. And a little more digging leads us to a publication by Piermarini et al. dealing with the crystal structure of benzene II at 25 kbar.[4] But I will not go into that and rather focus on the former publication.
This paper illustrates nicely that there cannot be only one true value for the compound. I will reproduce a part of table 2 from the publication here.
$$
\text{Bond lengths (}\mathring{\mathrm{A}}\text{) and angles (}^\circ\text{) for benzene in phase I.}\\
\begin{array}{llll}
             & 0.3~\mathrm{GPa} & 0.7~\mathrm{GPa} & 1.1~\mathrm{GPa} \\\hline
\mathrm{C1—C2}     & 1.40 (4)   & 1.397 (8)        & 1.420 (16)       \\
\mathrm{C2—C3}     & 1.38 (8)   & 1.384 (10)       & 1.41 (3)         \\
\mathrm{C3—C1^i}   & 1.38 (5)   & 1.354 (10)       & 1.32 (2)         \\\hline
\end{array}\\
\text{Symmetry code: (i)} 1-x; -y; -z.
$$

As you can see even in a single crystal structures, the bond lengths are not equal. This is obviously due to packing effects and intermolecular interactions. When we talk about a single value for the carbon carbon bond length in benzene, then we assume an idealised, even rigid, molecule.

And there are more ways to obtain the geometry of a molecule: IR, microwave, Raman to name a few.[5] Nowadays we can employ computational chemistry and try to predict bond lengths, but there is a catch to that, too. These calculations take place at absolute zero and no pressure and we have a couple of other approximations in the mix. You can have a look at the Computational Chemistry Comparison and Benchmark DataBase, which lists values from 138.3 to 144.3 pm for the carbon carbon bond.[6] For example, my own calculation on the MP2/def2-SVP level of theory gives me a carbon carbon bond length of 140.0 pm. But that is a value that corresponds to a stationary point on the potential energy surface. We all know that this molecule has thirty vibrational modes, which are not yet accounted for. Below is just one, one that I find particularly funny.

TL;DR Don't worry about deviations of about 1 pm. When there are larger differences, look at the original publications about the conditions and trustworthiness of their measurements. Always consider the error margins on your own experiments when you compare values. It also depends on how accurate you need your value to be before you go and dig deep into publications from 50 years ago. (For what it is worth, you could round 139 pm to 140 pm.)

Notes and References

Tamagawa, K.; Iijima, T.; Kimura, M. Molecular structure of benzene. J. Mol. Struct. 1976, 30 (2), 243–253. DOI: 10.1016/0022-2860(76)87003-2.
CSID:236, http://www.chemspider.com/Chemical-Structure.236.html (accessed November 18, 2018)
Budzianowski, A.; Katrusiak, A. Pressure-frozen benzene I revisited. Acta Crystallogr Sect B 2006, 62 (1), 94–101. DOI: 10.1107/S010876810503747X.
Piermarini, G. J.; Mighell, A. D.; Weir, C. E.; Block, S. Crystal Structure of Benzene II at 25 Kilobars. Science 1969, 165 (3899), 1250–1255. DOI: 10.1126/science.165.3899.1250.
Wikipedia has a longer list: Molecular Geometry.
Unfortunately I am unable to generate a permalink from the database: http://cccbdb.nist.gov/ I welcome you to play around a little by yourself.
However, I'll include the summary from the geometry comparison between the C1–C2 bond length:

  distance is atom 1 (C) to atom 2 (C)
  Experimental bond length is 1.397 Å     Calculated bond lengths range from 1.383 to 1.443 Å
  The bar charts show differences (theory - experiment) and the range is −0.046 to 0.046 Å


Gauche correction in Bensen group increment

Here is how I do it.

Sight down each C-C bond along the backbone of the molecule and draw a Newman projection
Look at all 3 possible conformations in the Newman projection and pick the conformation with the fewest gauche interactions
Count the number of gauche interactions
Repeat with the next C-C bond along the backbone

Here is a drawing of the 3 Newman projections in 3,4-dimethylhexane that contain gauche interactions (there are no gauche interactions along the $\ce{C1-C2}$ or $\ce{C5-C6}$ bonds.  You can see that there are 4 gauche interactions in the molecule.


Claisen rearrangement in substituted ring

Another great example of how it's important to look at actual evidence. There have been various studies on Claisen rearrangement selectivity, here's a combined experimental/theoretical one on meta-substituted ethers specifically: Gozzo, F. C.; Fernandes, S. A.; Rodrigues, D. C.; Eberlin, M. N.; Marsaioli, A. J. Regioselectivity in Aromatic Claisen Rearrangements. J. Org. Chem. 2003, 68 (14), 5493–5499. DOI: 10.1021/jo026385g.

An important screenshot is shown here:

Based on this article at least, it turns out that for your specific reaction, it's likely that the regioisomers would be formed in equal amounts. (See allyl aryl ether 1). We further see that sterics are not as significant of a consideration as electronics; allyl aryl ether 6 actually prefers the "more sterically hindered" product 14 to 1.

The article concludes that "the preferred conformation of the reactants ... with no exception, was found to lead to the major isomeric product.... The regioselectivity for the Claisen rearrangement of meta-substituted allyl aryl ethers is mainly influenced, as expected, by the electronic effects of the meta substituent, which governs the relative stability of conformers, transition states, and intermediate to final products.


Mass percentage of potassium chlorate in decomposition reaction

Generally, it is preferable to write down the complete quantity equation before plugging in the numbers. If necessary, perform all mathematical operations on quantities symbolically. Do not plug in the numbers until you have only one equation for the desired result. In the following, we use this method to calculate the amount of oxygen $n_{\ce{O2}}$, the mass of potassium chlorate $m_{\ce{KClO3}}$, and the mass fraction of potassium chlorate $w_{\ce{KClO3}}$ directly from the parameter values given in the question (although only the last calculation is actually required to answer the question).

Amount of oxygen

For simplicity’s sake, you may use the ideal gas law to estimate the amount of oxygen $n_{\ce{O2}}$ as follows. The volume $V=685\ \mathrm{cm^3}=0.000\,685\ \mathrm{m^3}$, the temperature $T=30\ \mathrm{^\circ C}=303.15\ \mathrm K$, and the pressure $p=101\ \mathrm{kPa}=101\,000\ \mathrm{Pa}$ are given in the question. The molar gas constant is $R=8.314\,459\,8(48)\ \mathrm{J\ mol^{-1}\ K^{-1}}$.

$$\begin{align}
p\cdot V&=n_{\ce{O2}}\cdot R\cdot T\tag1\\[6pt]
n_{\ce{O2}}&=\frac{p\cdot V}{R\cdot T}\tag2\\[6pt]
&=\frac{101\,000\ \mathrm{Pa}\times0.000\,685\ \mathrm{m^3}}{8.314\,459\,8\ \mathrm{J\ mol^{-1}\ K^{-1}}\times303.15\ \mathrm K}\\[6pt]
&=0.027\,448\,609\ \mathrm{mol}\\[6pt]
&\approx0.027\,4\ \mathrm{mol}
\end{align}$$

By way of comparison, the amount of oxygen calculated using standard reference data for oxygen instead of using the ideal gas law yields $n_{\ce{O2}}=0.027\,47\ \mathrm{mol}$. Considering the estimated uncertainty of the result obtained by using the ideal gas law, the difference is not significant. Hence, the approximation using the ideal gas law is acceptable in this case.

Mass of potassium chlorate

The chemical equation
$$\ce{2KClO3 -> 3O2 + 2KCl}\tag7$$
indicates that $2\ \mathrm{mol}$ of potassium chlorate yield $3\ \mathrm{mol}$ of oxygen. Therefore,
$$n_{\ce{KClO3}}=\frac23\cdot n_{\ce{O2}}\tag8$$

The molar mass of potassium chlorate is $M_{\ce{KClO3}}=122.55\ \mathrm{g\ mol^{-1}}$. Since the molar mass $M_{\ce{KClO3}}$ is defined as
$$M_{\ce{KClO3}}=\frac {m_{\ce{KClO3}}}{n_{\ce{KClO3}}}\tag9$$
the mass of potassium chlorate $m_{\ce{KClO3}}$ is
$$\begin{align}
m_{\ce{KClO3}}&=M_{\ce{KClO3}}\cdot n_{\ce{KClO3}}\tag{10}\\[6pt]
&=M_{\ce{KClO3}}\cdot \frac23\cdot n_{\ce{O2}}\tag{11}\\[6pt]
&=M_{\ce{KClO3}}\cdot \frac23\cdot\frac{p\cdot V}{R\cdot T}\tag{12}\\[6pt]
&=122.55\ \mathrm{g\ mol^{-1}}\times\frac23\times\frac{101\,000\ \mathrm{Pa}\times0.000\,685\ \mathrm{m^3}}{8.314\,459\,8\ \mathrm{J\ mol^{-1}\ K^{-1}}\times303.15\ \mathrm K}\\[6pt]
&=2.242\,551\,324\ \mathrm g\\[6pt]
&\approx2.24\ \mathrm g
\end{align}$$

Mass fraction of potassium chlorate

The mass fraction of potassium chlorate $w_{\ce{KClO3}}$ is defined as
$$w_{\ce{KClO3}}=\frac{m_{\ce{KClO3}}}{m}\tag{13a}$$
where $m=3.40\ \mathrm g$ is the total mass of the sample. Therefore,
$$\begin{align}
w_{\ce{KClO3}}&=m_{\ce{KClO3}}\cdot\frac1m\tag{13b}\\[6pt]
&=M_{\ce{KClO3}}\cdot \frac23\cdot\frac{p\cdot V}{R\cdot T}\cdot\frac1m\tag{14}\\[6pt]
&=122.55\ \mathrm{g\ mol^{-1}}\times\frac23\times\frac{101\,000\ \mathrm{Pa}\times0.000\,685\ \mathrm{m^3}}{8.314\,459\,8\ \mathrm{J\ mol^{-1}\ K^{-1}}\times303.15\ \mathrm K}\times\frac1{3.40\ \mathrm g}\\[6pt]
&=0.659\,573\,919\\[6pt]
&\approx0.660=66.0\ \%
\end{align}$$


You can find this using elementary geometry. Also, since the unit cell may be positioned in different ways with respect to the coordinate system, there can be many different answers, all equally valid.

One possible answer implies alignment of the cell vectors $\vec a$ and $\vec c$ with the coordinate axes X and Z, which leads to the following:
$$\left\{\begin{align}
\vec{\bf a} =& \Big(|a|,0,0\Big)\\
\vec{\bf b} =& \left(-{1\over2}|a|,{\sqrt3\over2}|a|,0\right)\\
\vec{\bf c} =& \Big(0,0,|c|\Big)
\end{align}\right.
$$


Molecular dynamics (Mac OSX user): what software can I use to plot RDFs for NPT simulations? VMD only works for NVT (constant volume)

Thank you, Buck, for all your suggestions.  You're right that a good way to check the influence of the volume is to try varying it.  I did that by using the maximum and minimum volumes of the fluctuations in one simulation.  There was some variation, so the volume does matter!

I was lucky enough to hear about "travis," an amazingly simple-to-compile code (even for Macs) available online for free.  I thought I'd share that information here because other people are bound to run into this same problem.  This code, travis, takes my trajectory file and my volume file and then asks a number of questions interactively.  It outputs a nice CSV file that can be plotted.

Thanks again for your help, Buck.  Really helped me think this through.  I'd like to ask the moderators or whoever is in charge of tagging to please add some molecular dynamics-related tags if possible!  Thank you.


Can someone recommend me a book ( preferably written long ago) that includes all the details of high school chemistry?

"College Chemistry - An Introductory Textbook of General Chemistry" by Linus Pauling (2nd Ed) is available as a free download (pdf & other formats) from https://archive.org/details/CollegeChemistry.


Rationale for titant standardization in alkalinity measurements [ISO 9963-1:1994]

According to this listing, $\ce{Na2CO3}$ is on of the primary standards used in volumetric analysis, i.e. is easily available in a reproducible form and concentration, has a high molecular mass, is -- once the container is opened -- not (so much) hygroscopic; and is stable.
I speculate, a dilute solution of HCl (0.1 mol/L), if stored over weeks in a bottle recurrently opened, maybe stored on a window bench (seen so frequently in the undergrad's labs) subject to ventilation and underneath heating by the radiators does not have this resillience.
Using a cartrige to prepare a standard solution of HCl that you rinse into a volumetric flask and fill up with deionised water may be seen valid as alternative for a secondary standard by convention. (ISO standards are settled agreements; similar to ASTM standards, if you see things differently, you may contact them.)  Obviously, alteration of a single step within a whole chain of procedures may invalidate the protocol's fit into ISO; a less severe issue in a teaching lab than for a GMP-accredited lab.
Addendum
"How is the diluted solution of HCl obtained?" brought an other element into play that I did not consider earlier: the use of concentrated hydrochloric acid. If purchased and used freshly, it may be a batch of concentrated (37%) or even "fuming HCl" (about 40%), and both are hygroscopic.  Small errors about the concentration of the stem solution of this corrosive reagent may indeed yield larger systematic errors, down the road of repeated dilutions.  (Obviously not so important if the sole intend is to fill up again the bottle of 1 M or 2 M HCl for the extractive workups ...)


The simplest approach would be to degas a sample and measure its specific gravity with a vintner's hydrometer. It will have scales of specific gravity and Brix (sugar content) on it.The Brix (Bx) reading is the percent of sucrose, w/w, in a solution with the same density/specific gravity as the one you are testing. There are, of course, other things in coke besides sucrose (fructose, coca extract, phosphoric acid) but they should be in small enough concentration that the sugars will swamp everything else and all sugars behave amazingly the same with respect to density of their solutions).

Based on the reported sugar content of Coke it contains 108 grams of sugar per liter. Thus its specific gravity should be about 1.041 and the corresponding Bx reading about 10.4. The density of water at 20 °C is 0.998203 so the density corresponding to 1.041 SG is 1.039 and 10.4% of 1039 grams (weight of a liter of 10.4 Bx sucrose solution) is 108 grams.

If you don't have a hydrometer with the Bx (or Plato) scale or if you determine density by weighing a known volume (convert to specific gravity) you will need to be able to determine the sucrose content of a solution of given specific gravity. There is lots of opportunity here to go into the history of wine and beer making, which you male students will love (though I understand that in modern schools discussion of things related to alcohol may not be politically correct) but in any case determining the w/w sucrose content of solutions of given SG is a well studied problem. There are various tables but the easiest way to get w/w sucrose from specific gravity is a polynomial promulgated by the ASBC (American Society of Brewing Chemists). It is

°P = ((135.997*S -630.272)*S + 1111.14)*S -616.868

I believe that separating out the pure sugars at anything close to 100% recovery would be difficult (barring HPLC or something of that sort). Sugars seem to be assayed by indirect methods such as the gravimetric ones I proposed or measurement of refractive index.


Drinking water out of a 24 year old Ether Bottle

Mallinckrodt is the chemical manufacturer. The ethanol is added to the ether to prevent peroxides from forming in the ether (like an antioxidant). If you wash and rinse the container thoroughly a few times, then it should be fine.


What do the half-reactions need to be for balancing a redox reaction in an acidic medium?

Below, the number mentioned in the parenthesis is the oxidation number of sulfur.

Writing down half reactions:
\begin{align}
\ce{CuS^{(-2)}     &-> ^{(4)}SO2}&& \text{Oxidation}\\
\ce{^{(6)}SO4^{2-} &-> ^{(4)}SO2}&& \text{Reduction}
\end{align}
Adding electrons in both reactions:
\begin{align}
\ce{CuS^{(-2)}           &-> ^{(4)}SO2 + 6e-}&& \text{Oxidation}\\
\ce{^{(6)}SO4^{2-} + 2e- &-> ^{(4)}SO2}&& \text{Reduction}
\end{align}
Balancing $\ce{Cu}$:
\begin{align}
\ce{CuS^{(-2)}&-> ^{(4)}SO2 + 6e- + Cu^{+2}}&& \text{Oxidation}
\end{align}
Balancing charges of both half reactions:
\begin{align}
\ce{CuS^{(-2)}&-> ^{(4)}SO2 + 6e- +CuO + 6H+}&& \text{Oxidation}\\
\ce{^{(6)}SO4^{2-} + 2e- +4H+ &-> ^{(4)}SO2}&& \text{Reduction}
\end{align}
Balancing oxygen and hydrogen:
\begin{align}
\ce{CuS^{(-2)} + 3H2O&-> ^{(4)}SO2 + 6e- +CuO + 6H+}&& \text{Oxidation}\\
\ce{^{(6)}SO4^{2-} + 2e- +4H+ &-> ^{(4)}SO2 + 2H2O}&& \text{Reduction}
\end{align}
Adding 3 copies of reduction half reaction with oxidation half reaction, gives finally balanced redox reaction:
$$\ce{CuS + 3SO4^{2-}  + 6H+ -> 4SO2 +CuO + 3H2O}$$

(Thanks to Silvio Levy for helping me to solve this problem.)


I predicted the answer to 10 which indeed is $5\choose 3$ because maximum number of points in a plane will be 3 since the structure is tetrahedral.

Yes, 10 is the number of sets with three atoms, and each set defines a unique plane.

  But the book writes an answer 7 to the above problem

3 of those (HHSi, HHF, HHCl) don’t count, so 7.


How did early chemists determine the major product of a solution?

In absence of currently available (or, in principle available) spectroscopies like NMR, IR, Raman, UV-Vis; and mass spectrometry our ancestors tried first to separate the products obtained by sequential recrystallisation.  Or by extraction under acidic, neutral, basic conditions (including back extraction).

They then recorded other physical properties of the purified products, if accessible, which then were compared with reference data.  For example, melting point and crystal shape if the products were solid, otherwise boiling point, refractive indices of the products obtained.  Then, they knew a whole array of pretests to check for the presence / absence of elements in the products formed, like Beilstein test for halogens and Lasaigne test for nitrogen.  They would determine the molecular weight by means of cryoscopy or ebullioscopy.

From a chemical point of view, it was common to subject the products to combustion analysis to determine the molecular formula.  (One of the instruments for this, the  Kaliapparat is still present in the logo of The American Chemical Society.)  Till today, there are journals preferring this over a mass spectroscopic determination of the molecular weight.

They knew about reactions that would yield derivatives, for example from the reaction with picric acid (affording pircrates), or with toluenesulfonic acid (yielding amides); often crystalline with a sharp melting point, easing to compare with reference data.  If not, other reactions like the haloform reaction would sequentially degrade the products into fragments, and then these products obtained would be characterised by physical or chemical means.

And obviously, the balance was much more "sacred" than today, where often we "just" weight-in reactants, and weight-out products.


Why is silver chloride less soluble than silver nitrate?

In the comment to my previous answer, you asked for a theoretical reason for the solubilities, not considering energy data. Since I know from energy considerations that the issue is not the solvation of the anions, I can present a reason based on the strength of the ionic bond in the two compounds. This reference (as well as others) states the bonding in $\ce{AgCl}$ has an unusually high covalent character which makes it a tighter bond. The $\ce{Ag+}$ ion and the $\ce{Cl-}$ ion are close to the same size (with the silver ion being smaller), so they can approach each other quite closely. In silver nitrate, the $\ce{NO3-}$ ion is larger and does not allow as close an approach as the chloride ion, so the bond is weaker, easier to break up, and the salt is more soluble.


Mechanism involved in bromine water test

There are no carbanion in this reaction.
In bromine water, $\ce{Br2}$ is partially made of $\ce{Br+}$ and $\ce{Br-}$ ions. Then $\ce{Br+}$ attacks the double bond, producing a carbocation, with $\ce{Br}$ at the outer $\ce{C}$ atom, and the + charge on the central $\ce{C}$ atom. This is the most stable carbocation. Then it will react with the $\ce{Br-}$ ion which will be attached on the central $\ce{C}$ atom.

It also should be mentioned that a secondary reaction may happen where $\ce{H2O}$ reacts with the transition state carbocation. As a result, the molecule $\ce{CH3-CHOH-CH2Br}$ is produced.


Why is the second acid dissociation constant for sulphuric acid much lower than the first in water?

For the second deprotonation you are trying to remove a proton from something that is already negatively charged, a process far more difficult than removing a proton from something that is neutral.
Alternatively you can think that the $-2$ charged $\ce{SO4^2-}$ anion attracts really strongly the $\ce{H+}$ resulting in the equilibrium being towards $\ce{HSO4^-}$, making it a weak acid compared to $\ce{H2SO4}$.
It is similar thinking to why the second ionization energies are always much higher than the first.

There can always be more than one ways to explain an observed phenomenon. With this in mind, one might be tempted to believe that there is suppression of dissociation of the $\ce{HSO4^-}$ ion due to the common ion effect.  The reasoning goes along the lines of:

Note that even before $\ce{HSO4^-}$ dissociates, there is an equal concentration of $\ce{H+}$ ions already present in the solution. We know, by Le Chatelier's principle, that the increased concentration of the products favors the reactants. Hence for the second dissociation, since $\ce{H+}$ is one of the products, and is in large concentration, the second dissociation is actually favored in the reverse direction. Thus, deprotonation of $\ce{HSO4-}$ is not favorable.
Hence, the $K_\mathrm {a_2}$ of $\ce{H2SO4}$ is significantly lower than its $K_\mathrm {a_1}$.

However, this is a wrong reasoning. Note that $K_\mathrm{a}$ values are always a constant. The $K_\mathrm{a}$ value for the $\ce{HSO4-}$ ion would be the same whether we took $\ce{H2SO4}$ or $\ce{NaHSO4}$. However, in the case of $\ce{NaHSO4}$, there are no pre-existing $\ce{H+}$ ions in solution when $\ce{HSO4-}$ deprotonates. Hence, it would be wrong to invoke the common ion effect here.
Therefore, the common ion effect should not be invoked when explaining the low $K_\mathrm a$ of $\ce{HSO4-}$ ion as compared to $\ce{H2SO4}$.


There is actually an erroneous explanation in your textbook. There is how the nitrogen monoxide molecular structure is explained in school: as there are no extended methods known by pupils or moreover the teacher, it's ok children to think that third unpaired electron on nitrogen atom is resonating between two atomic kernels as it's presented on picture upstaris. But it's wrong. Of course, that third electron cannot entirely belong to an oxygen atom (look to the right structure of first picure), because it (ox. atom) simply doesn't have enough free "quantum cells" to take it up to.

Things became much easier ever since people could describe the situation using molecular orbitals. Consider the molecular orbital diagram below:

It's to say, that

As the modern theory of molecular orbitals considers, there is some inter-atomic space between two atoms, where is the bonding electron allocation is the most likely situated in by the moment. There is exacly where our "third" elecron situated towards to that molecule. It doesn't ever belong to oxygen atom, but, technically, we can consider that structer as a resonant. It would me more correct, if to represent it's structure like that:

As that molecule has one antibonding electron π*2px it is a molecular diradical and it has an unstable structure. That also follows because of it's real physical-chemical properties (semi-stable only under low temperatures, and is very chemically active molecule).


How can weaker nucleophiles kick out stronger ones in the reaction of chloroalkanes with sodium iodide in acetone?

Acetone is the key here. The solubility of NaI in acetone is about 50 g in 100 mL while NaCl is virtually insoluble in acetone, thus the reaction is driven towards the alkyliodide due to Le Châtelier’s principle, since NaCl will be precipitating from the reaction. This process is called a Finkestein Reaction. Also see this.


How do acids help 'solubilise' compounds?

There are multiple ways in which plant root and root hair cells increase the availability of nutrients and positively affect the plant’s ability to absorb nutrients from the soil.

One of these ways is by excretion of $\ce{H+}$ ions from malaic acid. This promotes a cation exchange process that results in cationic nutrients being brought into the cell in order to maintain charge balance.  This is the process to which you are referring in your question.

Additionally, enzymes, for example phosphotases, can convert chemically unavailable nutrients, like organic phosphorus, into bioavailable inorganic phosphates.  Phosphate uptake is a case in which acidic soils actually inhibit the availability of this nutrient to plants. (Annals of Botany, 112: 317 – 330, 2013)

There are also proteins excreted by root hairs that can complex metal ion nutrients like iron, facilitating the uptake of such nutrients.


The inert pair effect describes the preference of late p-block elements (elements of the 3rd to 6th main group, starting from the 4th period but getting really important for elements from the 6th period onward) to form ions whose oxidation state is 2 less than the group valency.

So much for the phenomenological part. But what's the reason for this preference? The 1s electrons of heavier elements have such high momenta that they move at speeds close to the speed of light which means relativistic corrections become important. This leads to an increase of the electron mass. Since it's known from the quantum mechanical calculations of the hydrogen atom that the electron mass is inversely proportional to the orbital radius, this results in a contraction of the 1s orbital. Now, this contraction of the 1s orbital leads to a decreased degree of shielding for the outer s electrons (the reason for this is a decreased "core repulsion" whose origin is explained in this answer of mine, see the part titled "Why do states with the same $n$ but lower $\ell$ values have lower energy eigenvalues?") which in turn leads to a cascade of contractions of those outer s orbitals. The result of this relativistic contraction of the s orbitals is that the valence s electrons behave less like valence electrons and more like core electrons, i.e. they are less likely to take part in chemical reactions and they are harder to remove via ionization, because the s orbitals' decreased size lessens the orbital overlap with potential reaction partners' orbitals and leads to a lower energy. So, while lighter p-block elements (like $\ce{Al}$) usually "give away" their s  and p electrons when they form chemical compounds, heavier p-block elements (like $\ce{Tl}$) tend to "give away" their p electrons but keep their s electrons. That's the reason why for example $\ce{Al(III)}$ is preferred over $\ce{Al(I)}$ but $\ce{Tl(I)}$ is preferred over $\ce{Tl(III)}$.


Let me start by providing a reference to a nice collection of van der Waals' data (actually the site provides a lot of other interesting data about the elements, you might want to poke around). As to your questions,

Yes, that is a general trend, but as you can see there are many exceptions.  The trend exists for the same reason that covalent radii generally shrink as we move from left to right across the periodic table, every time we add another proton and electron to an atom, the electrons do not effectively screen the outermost electrons from the positive nucleus so they become drawn in closer to the nucleus.
& 3. There are a lot of different ways to measure the van der
Waals radius of an atom, but in recent times the preferred method is
to examine the crystal structure of a molecule.  Using $\ce{F2}$ as
an example, the crystal would be composed of fluorine molecules.  if
we measure the crystallographic distance between two adjacent, but
non-bonded fluorine atoms and divide by two, we would have a single
value that could be used to estimate the van der Waals radius for
fluorine (we would probably want to measure a number of similar
values for fluorine from other molecules, maybe using $\ce{CH3F,
NF3, ClF,}$ etc., for example, and average them together). Our
crystal might look something like the drawing below.

Because diatomic and polyatomic molecules are polarizable, the bonded fluorine atoms will likely be slightly polarized when packed in the crystal in order to stabilize (lower the energy) the overall system.  This polarization will cause non-bonded fluorines to draw a bit closer to each other then they would if there were no polarization.  A crystal of a noble gas (let's say neon) will just be a collection of monatomic neon atoms, no bonds, no polarization.  Therefor there is no "attractive" force to draw adjacent neon atoms closer together like there was in the case of fluorine.  So even before we make our measurements we might expect the van der Waals radius for neon to be larger than that for fluorine, which turns out to be the case.


What are the key differences between Raoult’s Law and Henry’s Law

Raoult's law can be seen to be valid for ideal liquids. The assumption behind Raoult's law is that both the liquid phase and the vapour phase behave ideally. This means that the interaction between any two molecules in the liquid and the vapour have no interaction.
The Raolt's law just states the mole fraction of a component in the liquid phase is proportional to the mole fraction of the same component in the vapour phase. This is valid only for mixtures of ideal liquids.
When you add a very small amount of impurity in a solvent (equivalent to infinite dilution) the interaction between the impurity and the solvent molecule is minuscule or technically infinitesimal. Therefore, the liquid still behaves ideally. Hence Raoult's law is applicable for the solvent.

Whereas Henry's law is also defined for a ideal mixture only this time the assumption is that the mole fraction of the component is tending to zero. this means that when very few molecules a present in a vast space, the interaction between them is almost zero. This makes the system ideal. therefore, an infinitesimally small quantity of impurity does not feel any interaction with another molecule of the same impurity in the solvent. Therefore, Henry's law is used for the solute. That is why Henry's law constant have huge values. Huge values makes sure that even small amount of the solute is accounted for when calculating the mixture properties.

For further clarity, please refer to chapter 6 of 'Molecular Thermodynamics of Fluid phase equilibria' by John M.Prausnitz et. al.


Use the table mentioned in the linked article, §NT-1.1:

1   mono- or hen-*  10  deca-       100 hecta-      1000    kilia-
2   di- or do-*     20  icosa-**    200 dicta-      2000    dilia-
3   tri-            30  triaconta-  300 tricta-     3000    trilia-
4   tetra-          40  tetraconta- 400 tetracta-   4000    tetralia-
5   penta-          50  pentaconta- 500 pentacta-   5000    pentalia-
6   hexa-           60  hexaconta-  600 hexacta-    6000    hexalia-
7   hepta-          70  heptaconta- 700 heptacta-   7000    heptalia-
8   octa-           80  octaconta-  800 octacta-    8000    octalia-
9   nona-           90  nonaconta-  900 nonacta-    9000    nonalia-

just enumerate the digits of the carbon and add "-ne".

So $\ce{C21H44}$ becomes "hen-" "icosa-" "ne" = henicosane / heneicosane (the latter is just a spelling variation).

There are exceptions like propane ($\ce{C3H8}$) but these only occur for small molecules, up to $\ce{C4H10}$.


How can they "wash away" the caffeine from Nescafe Gold to produce that "de-caf" version?

In general they use different types of solvents and methods to wash away the caffeine (or at least a great percentage of it) from the coffee beans , you can find more information here : https://www.todayifoundout.com/index.php/2011/04/how-decaffeinated-coffee-is-made/


When the pH is lower than the pKa, the group will be protonated, and when the pH is higher than the pKa, the group will be deprotonated. Your explanation is correct that basic groups will be positively charged when protonated and neutral when deprotonated and that acidic groups will be neutral when protonated and negatively charged when deprotonated. Shown below is the compound in it's fully protonated state with the given pKa's.*

The given pH is 3.3 which means that:

N-terminus: pH < pKa, protonated basic group, +1
lys-sidechain: pH < pKa, protonated basic group, +1
asp-sidechain: pH < pKa, protonated acidic group, 0
his-sidechain: pH < pKa, protonated basic group, +1
arg-sidechain: pH < pKa, protonated basic group, +1
C-terminus: pH > pKa, deprotonated acidic group, -1

The overall charge is +3.

*An aside, the pKa's for the side chains are a gross approximation here. It turns out they are sufficient to answer the question, but given a higher pH, the answer wouldn't be correct.


Do dyed-alcohol thermometers tolerate cold/hot stems?

The very shortest answer is no. Since asking this question, I've learned that thermometers have an insertion depth they are calibrated to, presumably assuming the rest of the thermometer is somewhere between room temperature and the bath temperature. Some thermometers are "total immersion", meaning the conductivity of the fluid doesn't matter if used as intended, because it's expected to conduct through the sides as well.

Given what I've learned, the first half of the answer to this question is that if a thermometer is calibrate to be immersed 2 cm, then immersing it deeper will give a wrong reading.

This article explains further: https://www.coleparmer.com/tech-article/glass-thermometer-selection-guide. In particular, correction factors are given for when a thermometer is not inserted at its calibrated depth:

  Emergent Stem Corrections. Avoid inaccurate readings when a total
  immersion thermometer cannot be properly immersed. Determine the
  approximate stem correction with these formulas:

For mercury Fahrenheit thermometers:
°Correction = 0.00009°F x n x (T-t)
For mercury Celsius thermometers:
°Correction = 0.00016°C x n x (T-t)
For spirit-filled Fahrenheit thermometers:
°Correction = 0.0006°F x n x (T-t)
For spirit-filled Celsius thermometers:
°Correction = 0.001°C x n x (T-t)

  where T is the bath temperature (the temperature indicated on the
  thermometer), t is the average temperature of the emergent stem, and n
  is the number of emergent thermometer degrees. Hold an auxiliary
  thermometer next to emergent stem to determine t.

While these corrections aren't meant to be applied backwards (for a partial immersion thermometer that's immersed too deep), the numbers speak for themselves: a colored organic thermometer is not nearly as accurate when it's either immersed too far or too shallowly. I plugged in some numbers for a very nearly worst case scenario (cold room with a strong draft where the thermometer is calibrated to be fully inserted but is actually only inserted up to the 0⁰C mark):

T=100⁰C
t=20⁰C
n=100
mercury °C correction = 0.00016°C x n x (T-t) = 1.28⁰C
spirit °C correction = 0.001°C x n x (T-t) = 8⁰C

The final answer is that no, dyed-spirit thermometers do not tolerate being at the wrong insertion depth, at least not if your experiment's parameters are too far off from what the thermometer manufacturer intended.


Error with explaination of shape of weak acid strong base titration?

This buffer doesn't seem like it would help to reduce pH changes

You have to be clear what you want to compare it with

Adding a strong base to pure water:
If you start with pure water, the concentration of hydroxide will be pretty much equal to the concentration of strong base you add. The pH will drop quickly.
Adding a strong base to a strong acid:
If you start with strong acid, the first few drops of strong base will be neutralized. The pH does not change much because the strong acid is still in large stoichiometric excess, and the pH is a logarithmic scale.

  When a strong base is added to a weak acid, as more OH− is added this will react with the acid molecules in solutions and form more of the conjugate base, forming a buffer solution which I do agree with.

In the first approximation, the added hydroxide will be neutralized by the acetic acid, and the pH will not change. However, as you do that, the weak acid/weak base equilibrium will be disturbed, and the reaction will shift to the left, consuming some hydronium and increasing the pH. The more you change the ratio of acetic acid to acetate for a - say - milliliter of added strong base, the more the pH will change. The ratio changes fast at the beginning (where there is no acetate) and near the equivalence point (where there is almost no more acetic acid), so that is when the pH changes are largest because those are the steps where we disturb the equilibrium the most.

  They typically explain the little dip at the start and the following plateau as a result of creating a buffer solution.

If you compare the slope of the dip to that of scenarios 1. and 2., you will find that it is less than in 1. (strong base added to pure water) and more than in 2. (strong base added to strong acid).

  This buffer doesn't seem like it would help to reduce pH changes

The way I explained it was to say the strong base reacts with the weak acid. We could have said it reacts with the hydronium ion instead. Then the shift of the weak acid/base reaction would have been the other way. In the end, we are just rationalizing the result of two equilibria, that of water autoionization (taking a back seat here because hydronium is the dominant species and hydroxide is present at very low concentration) and that of the weak acid dissociation. The shape of the titration curve is what we get experimentally, and the two equilibria explain it fully. Any explanation on top of that is just to convince ourselves.


Synthesis of synthetic detergent with Friedel Crafts Acylation

Your synthesis looks reasonable, and it is correct to perform sulfonation as the very last step, because the alkanoyl group would be meta-directing for the reason you have provided.

You can simplify your synthesis by reducing the carbonyl group away in one step with the Clemmensen reduction, which works well for aryl-alkyl ketones produced by Friedel-Crafts acylation.

Alternatively, the Wolff-Kishner or Mozingo reaction could be used. Reducing the number of steps in your synthesis would in practice have the benefit of increased overall yield for the final product, as the yield of each single step is usually < 100 %.


Why hydrofluoric acid (HF) has two Raman bands?

I've been doing some digging and I think I have a possible answer so I'm gonna post it and if I'm misinterpreting information, I'm open to being told so!

As I said in my comment, this vibration has to be coming from some dimer which is formed because if one of these modes were the $\ce{HF}$ fundamental it would be up around $4000\ cm^{-1}$.

So, I found this paper which uses FTIR to analyze $\ce{H2O/HF}$ complexes. They find two librational modes (which I believe is just a solvated rotational/rocking mode basically) which they identify as coming from the $\ce{H2O--HF}$ dimer. They assign these two frequencies at $635.5\ cm^{-1}$ and $753.1\ cm^{-1}$. Those two values are extremely similar to the frequencies we're seeing in the spectra you posted which is why I think we might be on to something here. Unfortunately I don't have access to the paper so I can't go into any more detail really.

In answer to your question of how we could even be seeing multiple peaks in an $\ce{HF}$ spectrum, the answer is that we're not looking at just $\ce{HF}$ here!

We know that both $\ce{H2O}$ and $\ce{HF}$ are very strong hydrogen bonding molecules, so it's no surprise that we find peaks which are associated with the various vibrational modes of the dimers present. And, once we have a dimer, we expect to find all sorts of rotational and vibrational modes.

So, to conclude, I'm fairly certain that the modes we're seeing in this spectra are two (and I think the only two because my digging around says this dimer has $C_s$ symmetry) librational modes of the $\ce{H2O--HF}$ dimer, each rotating about a different axis of the water.

Source:

Andrews, L., & Johnson, G. L. (1983). FTIR spectra of water–hydrogen fluoride complexes in solid argon. Evidence for inversion doubling in the HF librational modes of H2O––HF. The Journal of chemical physics, 79(8), 3670-3677.


Basis sets that can be used for macrocyclic complexes

In order to check on the availability of basis sets you may want to look at the Basis Set Exchange. After determining which basis set suits your problem (this will likely involve some literature search, Nickel, DFT, basis set, in Web of Science should do the trick), you can enter basis sets into Gaussian using the Gen keyword. BSE will export any selected basis sets in Gaussian format, so that shouldn't be a problem.


Can I use a digital multimeter as substitute for a zero resistance ammeter?

Yes, using Ohm's law: I=V/R.

The only true "zero" resistance ammeter are superconductive. In practical use, the voltage drop across a small resistance is measured. If your DVM can measure down to a millivolt (mV), then a 1 mV drop across a 0.001 ohm resistor shows a flow of 1 ampere current. You'll need to consult the guide (~$45 online) for the maximum permissible voltage drop, and use an appropriate resistance.

Recently, I used the voltage drop across a piece of #16 copper wire to measure currents in the 1 to 5 A range... a switch in the circuit had greater resistance!


How can we write the wave function in quantum mechanics?

The wavefunction contains all the information about the system of interest.  This is a basic premise of quantum mechanics.

Within the Born-Oppenheimer approximation, we 'index' all the values required to describe where an electron is.  This includes the spatial coordinates, ($\textbf{r}$), and the spin coordinate, $\omega$.  Electrons are characterized by their spin ($\uparrow$ vs. $\downarrow$).  Another way to think about it is this.  The quantum numbers are used to describe everything we need to know about 'where' an electron is ($n$, $l$, $m_l$, $m_s$).  The spatial coordinates (e.g. Cartesian coordinates) take care of the first 3 quantum numbers.  We need the fourth coordinate to characterize $m_s$.


Imagine you have one googolplex ($10^{10^{100}}$) liters of water. Now let's dissolve one mole of sugar in your water. Setting aside where would you find so much water, the concentration would be way lower than the inverse of Avogadro's constant.

Long story short, there's no lower limit to the concentration of a substance. You can always add more solvent to the mixture thus making the concentration smaller.


Is the atom the smallest particle, which takes part in chemical reactions?

Protons definitely participate in chemical reactions.  Free protons are generally not present in liquid water because a free proton is extremely reactive, but in the upper atmosphere or in other situations where the density of matter is low there can be free protons which participate in chemical reactions.
However, I completely agree with Jori that an electron, which is much smaller than a proton, takes part in chemical reactions.  Gain or loss of one tiny electron can completely change the properties of a large metalloenzyme.
How small is an electron vs. a proton? In terms of mass, the electron is 1836 times lighter, but what about in terms of radius. According to Hans Dehelt's Nobel Lecture:

Proton radius [is] $0.86 \times 10^{-15}$m...
Today everybody “knows” the electron is an
indivisible atomon, a Dirac point particle with
radius R = 0....
But is it?

And goes on to explain that it is still possible the electron could have a radius on the order of $10^{-20}$ m.
See also Limit on Electron Substructure
Also, photons, which have no mass, participate in chemical reactions.


activating and deactivating groups and directing effect

I believe that if we ignore the inductive effects of the substituents for the time being, then the electron-donating group(which donates electrons via resonance effect) would manage to activate a few positions in one or more resonating structures, and the electron-withdrawing group can only take up this delocalized pair of $\pi$ electrons when it is present at exactly an ortho or para position where the $\ce{+M}$ effect operates, and the negative charge lies on the carbon alpha to the electron-withdrawing group.
It can be seen here that the negative charge in the resonance structures of $2,4$-dinitroaniline can be taken up by the nitro group only when the negative charge becomes alpha to it:

So the only way an electron-withdrawing group can "win" is if it placed against a very poor electron-donating group.
First, in the case of $\ce{-CH3}$ or halogens (who neither use $\pi$ conjugation for electron donation nor have an electron-donating nature), the electron density donated to the ring is unable to emerge in the intermediates without being delocalised by the EWG, or second when the EWG is present exactly at positions where the negative charge reaches by $\pi$ conjugation(in case of "strong" electron donators).
If the group donating via the $\pi$ bonds is even a relatively better donating group, then unless all the ortho and para positions are having electrons being withdrawn by the presence of electron withdrawing groups, there will always be resonance structures donated having some partial activation at ortho or para positions.
Hence, the electron-donating group "wins" in such situations unless there is no steric hindrance during the substitution taking place.


What is the liquid coming out of these Duracell batteries?

Well, it depends on which type of battery you have. Alkaline batteries such as Duracell Coppertop usually leaks KOH (potassium hydroxide) under conditions of misuse as a white foam. However, KOH is hygroscopic and will easily form a oily like liquid which is what you might be seeing in your batteries. Handle it with proper care since KOH is caustic and can cause tissue damage.


How does the Wurtz reaction produce alkynes?

@tox123: You have generated two radicals on the same carbon by a two-electron reductive α-elimination (1 --> 2). So far, so good. In other words, you have created a carbene (2). Carbenes can rearrange to alkenes, NOT alkynes. Think of the two orbitals on the carbenoid carbon as one bearing two-electrons (negative) and the other one vacant (positive) (3). When a vicinal hydride migrates to the vacant orbital, one has a negative charge on one carbon and a positive charge on the vicinal carbon (4). Structures 4 and 5 are identical. If dichloride 1 is treated with base, one gets a double elimination to an alkyne. There is no change in oxidation state. Therefore, dichloride 1 cannot give an alkyne via a 2-electron reduction.


Which theta should be taken from x-ray diffraction pattern for calculating Scherrer's equation?

Scherrer’s formula is a simple equation for the simple case of a single peak broadened only due to crystalline particle size. Application of the formula to other, more complex cases, gives only an estimate. Quoting Wikipedia:

  It is important to realize that the Scherrer formula provides a lower bound on the particle size. The reason for this is that a variety of factors can contribute to the width of a diffraction peak; besides crystallite size, the most important of these are usually inhomogeneous strain and instrumental effects. If all of these other contributions to the peak width were zero, then the peak width would be determined solely by the crystallite size and the Scherrer formula would apply. If the other contributions to the width are non-zero, then the crystallite size can be larger than that predicted by the Scherrer formula, with the "extra" peak width coming from the other factors.

Scherrer formula gives an estimate, an order of magnitude. Here, calculate the particle sizes for the two peaks, and they should be close. The differences are covered by the large uncertainty of your resulting value.


Is a gas leak considered incomplete combustion?

In climate science we call these leaks fugitive emissions. The chemical composition of the emissions are different. If you just have a leak, the emissions will consist of what ever your fuel is made out of, probably methane. If you burn your fuel and the combustion is not perfect you might sill have a littel bit of the original fuel left but most will have been used up and formed CO (incomplete combustion), $\ce{CO_2}$ (complete combustion) and $\ce{H_2O}$ .


How is chemical and dynamic equilibrium related?

Background: It is generally the case that there is an energy barrier between reactants and products, sometimes called an activation energy, and this is smaller for reactants than products so that a reaction proceeds towards products.

Suppose now that there is a reaction such that $\ce{ A \to B}$ and initially only A is present so that the rate of forward reaction is large, i.e. lots of A and almost no B make the forwards reaction faster than the reverse reaction. As soon as some B is present it can react back to make A, but $\ce{ B \to A}$ will be slow compared to $\ce{ A \to B}$ because there is less B (smaller concentration) and the activation barrier is larger so the 'rate constant' is smaller.

Eventually the amount of A is reduced and B increased until both rates are the same and equilibrium is achieved. Note that the concentrations need not be the same, in fact they usually are not. Also note that molecules of A continuously convert to B; $\ce{ A \to B}$ and similarly B convert to A; $\ce{ B \to A}$ which is to say that the normal state of affairs is the 'dynamic equilibrium' and simply using the word 'equilibrium' implicitly assumes that this is the case.


How can I find the smaller symmetric structure from big crystal unit cell?

The structure contains 4 independent atoms, Mg, Al, Si and O. The space group is a cubic one containing 96 asymmetric units. If you place an atom on a general position, it will appear 96 times in the unit cell. Mg, Al, and Si are located in various special positions, i.e. lie on symmetry elements. Placing the Mg and the Si atom will result in 24 atoms in unit cell, while placing the Al atom in its position (on the origin) will result in 16 atoms in the unit cell (on a three-fold and on the center of inversion).

So you could say that the asymmetric unit contains one oxygen atom, one quarter Mg and Si atom respectively (24/96), and one sixth Al atom (16/96). One of the sites you cited in a previous question gives the fractional coordinates as:

Si: 0.3750 0.0000 0.2500

Al: Origin

Mg: 0.1250 0.0000 0.2500

O: 0.0341 0.0490 0.6530

If you look at the space group, http://img.chem.ucl.ac.uk/sgp/large/230az1.gif,

you can see that Al, Mg, and Si are on special positions (you can either find the locations on the diagram or apply the 96 operations (48 combined with the I-centering) listed on the right to the coordinates and see how many unique locations you get), while oxygen is not.

  can I manually cut this unit cell in 1/4 size, 40 atom system with 1/2 size for x, y, and z direction, and run the minimization?

If you software is aware of crystallographic symmetry, it is a very simple system with 4 degrees of freedom (the oxygen position and the lattice parameter a). Everything else is fixed if you impose the symmetry of garnet.

  I have Pyrope (Mg3Al2(SiO4)3) crystal structure, downloaded from materialsproject: https://materialsproject.org/materials/mp-6073/#. It appeared as triclinic cell...

There are some inconsistencies in the way the site shows the data. The space group is (correctly) given as Ia3d. The 3D image does not show a body-centered cubic cell, but instead the primitive rhombohedral cell (the smallest unit that recreates the entire crystal through three translations). In one iteration ("primitive"), the coordinate file lists the symmetry as 1, applied the Ia3d operators to the atomic positions (maybe to "not to confuse" users unaware of crystal symmetry), and switched to the rhombohedral cell.


All hydrogenation reactions, not just for benzene, are exothermic. You can refer  Why are hydrogenation reactions exothermic .


Does the catalyst alter the Potential energy Surface?

As you said, catalysts do not alter the PES, but extend it to other dimensions, which usually provide new reaction pathways with lower reaction barriers. If you consider a surface catalyst and an impinged molecule. Considering the full systems PES, the pure molecule PES is a part of it (for large distances of the surface to the molecule). If we bring the molecule closer to the surface, we get new partial PES which have lower reaction barriers.


What kind of structure could give you a HNMR spectrum consisting of a singlet integrating to 5 protons?

The 5H singlet is likely due to five protons on a monosubstituted benzene ring.

The peak comes in at 7.5 ppm which is the aromatic region, immediately telling you that they are hydrogens on an aromatic ring. Benzene rings are the most common of these.

But why are they all equivalent? They actually aren't, but their shifts are so close that they appear to be equivalent. Sometimes they will appear as different peaks (usually one peak for the 2 hydrogens close to the substituent and one peak for the 3 hydrogens away from it) if there are strong enough electron withdrawing groups on the substituent that are close to the ring. It is likely that the singlet you're looking at isn't clean cut and looks a bit like a multiplet (see this question).


Why aryl halides are very less reactive towards nucleophilic substitution reactions?

No, that is a very incorrect explanation for why (unsubstituted) aryl halides do not participate in nucleophilic substitution reaction. Conjugating effect between the halide substituent and the benzene ring is not responsible for this. The problem is that there is simply no available mechanism to perform the substitution:

$\ce{S_{N}2}$ is impossible since  inversion would give an incredibly weird strained ring (even more weird than a bridgehead carbon that is already forbidden by Bredt's rule).
$\ce{S_{N}1}$ would leave a very unstable empty $\ce{sp^2}$ orbital. This can happen, but only for the very best leaving groups, like $\ce{N2}$.
There is no electron stabilizing group that is conjugated with the halide group. This renders an addition-elimination type mechanism impossible.


How do non-sublimable solids exist in equilibrium with vapour?

Thermodynamically, even cotton exists in equilibrium with its vapour, even though this is completely silly in any practical sense. The reason is that thermodynamics does not involve time; it may take $10^9$ years, but so what! The equilibrium may only consists of one molecule / km$^3$ but $technically$ its still equilibrium! This is because you can't arbitrarily pick and choose the cases you want thermodynamics to apply to, but you can apply reasonable physical insight based on observation rather than applying theory blindly.


Predicting the solution with the highest boiling point

The first aspect to consider is the van't Hoff factor. Thus #1 is first and #4 is last. Since both #2 and #3 have the same factor of 2, we must consider other properties.

A major part of BP is how many molecules of solvent are exposed on the surface layer. Therefore, the ions with the greater diameter should take up more surface area, decreasing the number of solvent molecules available to 'escape'.. decreasing the vapor pressure and increasing the BP elevation.
My guess is that the polyatomic ion is quite large in comparison and the deciding factor.

from lowest BP to highest
 #1,#2, #3, #4


Liquid that does not spoil for a mineral aquarium

There are three things to take into consideration here: volatility of the liquid you will be using (yes, the jar will be tightly closed, but some candidates might still evaporate slowly and/or have toxic or foul-smelling vapors), stability, and refraction index. Mineral oil is a great candidate, as Todd Minehardt commented. It's really stable, releases no vapors whatsoever, but its refractive index is a little higher than that of water, which might make your aquarium look funny. There are substances with similar properties that might be useful and also readily available, but with a refractive index closer to water's. Amyl alcohol (fusel oil) or ethylene glycol (car antifreeze) are good candidates and shouldn't be too hard to find. Glycerol and propylene glycol are also pretty good options, but their refractive indexes are closer to mineral oil than the other two.


Can benzene ring be the part of a acyl group

Yes, acyl groups can be derived from all kinds of oxoacids.

According to the Glossary of class names of organic compounds and reactivity intermediates based on structure (IUPAC Recommendations 1995), acyl groups are defined as follows:

  Groups formed by removing one or more hydroxy groups from oxoacids that have the general structure $\ce{R_$k$E(=O)_l(OH)_m}\ (l\neq0)$, and replacement analogues of such acyl groups. (…) In organic chemistry an unspecified acyl group is commonly a carboxylic acyl group.

Especially concerning acyl groups derived from carboxylic acids in organic chemistry, the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) defines:

  Carboacyl groups are $\ce{R-CO-{}}$, $\ce{-OC-R-CO-{}}$, or $\ce{-OC-R-[R'-CO-]_$x$-R''-CO-{}}$ groups and their functional replacement analogues, where $\ce{R}$, $\ce{R'}$, and $\ce{R''}$ are chains, rings, or ring systems, derived from carboxylic acids by the removal of the hydroxy group from each carboxylic acid group that is expressed by the suffix, and $x = 1 ,2,3$, etc.
  (…)

In particular, benzoyl is explicitly mentioned as an example for an acyl group.


When does ratiometric measurement account for signal variations?

National Instruments$^{TM}$ gives a good explanation here, in a paper titled "Measuring Bridge-Based Sensors with the Ratiometric Approach, Publish Date: Feb 11, 2016".  Here are a couple excerpts relevant to your question:

  1) "The fact that the mV/V ratio stays constant for changes in the excitation voltage increases measurement stability. Small changes in the environment due to temperature differences typically affect excitation voltages. Because the excitation voltage is sensed and fed back to the ADC, any changes play a smaller role in the accuracy and stability of the measurement."

  2) "By using the excitation as the source for the ADC reference, the output of the ADC provides measurements that are in units of V/V. This reduces the amount of scaling required to return the appropriate units."

So, the short of it is that noise, particularly that arising from power fluctuations, in either the source or detector in the measurement systems as you described is largely cancelled out when ratioed to a reference system.

This approach is most effective when you are rapidly switching, i.e. via a chopper mirror, between the sample measurement source and detector and that of your reference or blank measurement.  When the signal from the source is then ratioed to that of the reference system, then any noise common to the two is effectively cancelled out, increasing your signal to noise ratio.  This method addresses your examples of "sequential" measurements. In the system describe above, you are effectively taking your reference measurement simultaneously or in parallel rather than sequentially.


About alternative magnetic effect on porcelain (for induction furnace purposes)

On reason you don't see much information is that your plan will not work. The copper wire will get hot by resistance and conduction and at some temperature it will react with the refractory = fail. Copper has essentially no use as a heating element because the oxide spalls off after each heat cycle. Use Nichrome wire ( this alloy family has a few different names ). However , as I understand induction heating it needs an conductive target. I am guessing your dental porcelain has poor electric conductivity , so will not heat by induction.


Old sample of sodium tetraphenylborate—why it is red?

According to Decomposition of Sodium Tetraphenylborate, which relates to the Savanna River site for nuclear weapons material production,

  Stable aqueous NaTPB solutions are a transparent,
  flesh-toned color. During the initial stages of decomposition (or
  instability),the solution turns a golden color. As the quantity
  of TPB- that has decomposed increases, the solution turns dark
  red, then brown, and eventually black.

In the presence of oxygen "phenol and phenylboronic acid
are produced as well as benzene, biphenyl, and other benzene ring
derivatives".


When an expansion of a system does work on its surroundings, is the sign for work negative?

It means the change in internal energy is negative, so you are on the right track. Because the system is doing work on the surroundings, that $288\ \mathrm{J}$ is a loss for the system and a gain for the surroundings.

One way to intuitively work your way through this is to think about what the signs on the terms must be in order for the problem to make sense.

You know the final volume is greater than the initial (it's an expansion of the system into the surroundings), so $\Delta\mathrm{V}$ has to  be positive. You know that $w = -\mathrm{p_{ext}}\Delta\mathrm{V}$ is negative, because $\mathrm{p_{ext}}$ is positive. Thus, the sign of $\Delta\mathrm{E}$ is negative.

You want to solve

$$\Delta\mathrm{E} = -\mathrm{p_{ext}}\Delta\mathrm{V}$$

for $\Delta\mathrm{V}$ and add that to your initial volume of $0.25\ \mathrm{L}$.

Solving for $\Delta\mathrm{V}$:

$$-288\,\mathrm{J} = -2\ \mathrm{atm}\times\Delta\mathrm{V}$$

$$\Delta\mathrm{V} = 144\ {\mathrm{J\over atm}}$$

and then converting units

$$\Delta\mathrm{V} =
\left(144\ {\mathrm{J\over atm}}\right)\cdot
\left({\mathrm{L\cdot atm}\over 101.325\ \mathrm{J}}\right) =
1.42\ \mathrm{L}$$

we now solve for the final volume as (spoiler alert):

  $$\mathrm{V_{final}} = 0.25\ \mathrm{L} + 1.42\ \mathrm{L} = 1.67\ \mathrm{L}$$


How to accurately determine mass transfer using balances?

Accurate weighing requires you to be aware of enormous amount of factors. It is not possible to cover them all in one answer. For starters, where do you put your balances? Are they levelled? Are there any heat or vibration sources nearby? Are there any sudden air flows? Is your mains power reliable? Were the balances calibrated properly? Is the ambient temperature or humidity too high or too low? Do they change much? Is there much static electricity (depends on the materials you are working with and on the ambient humidity)? Do you use gloves? How much do you weigh? I mean, not you weigh, but isn't it something less than the minimum weight requirement for your balances? Et cetera, et cetera.

Returning to your questions: in general, sensitive balances can handle large masses which are within the specification. However, the measurement uncertainty depends on the mass. Then, if the receiving flask is not perfectly flat, there should not be any problem unless the flask is wobbly or unsteady. The location where you put the flask should be as close to the center of the weighing pan as possible. It matters.

As for the other things you could do, I'd like to recommend you to check out some weighing practice manuals. You can start with the balances manufacturers - Mettler has brief Good Weighing Practice whitepapers, maybe others like Sartorius have something similar. Analytical chemistry handbooks should have something to offer. Then, check out the manual for your balances. Aren't you trying to weigh something more accurately than the accuracy of the balances (remember, accuracy $\ne$ resolution)? Then, if you don't, just ensure that you are following the recommended weighing practices as closely as humanly possible.


Which substances have the highest/lowest enthalpy of formation per gram?

Beryllium oxide −23.96 kJ/g
Lithium fluoride −23.75 kJ/g
Beryllium fluoride −21.84 kJ/g
Beryllium hydroxide −21.01 kJ/g
Lithium hydroxide −20.24 kJ/g
Lithium oxide −19.93 kJ/g
Boron trioxide −18.29 kJ/g
Magnesium fluoride −18.04 kJ/g
Aluminum fluoride −17.98 kJ/g
Boron trifluoride −16.76 kJ/g
Aluminum oxide −16.43 kJ/g
Water −15.86 kJ/g
Magnesium hydroxide −15.85 kJ/g
Magnesium oxide −14.92 kJ/g
Hydrogen fluoride −13.66 kJ/g
Lithium hydride −11.4 kJ/g
Boron nitride −10.24 kJ/g
Carbon dioxide −8.941 kJ/g
Lithium nitride −4.73 kJ/g
Methane −4.6 kJ/g
Ammonia −2.71 kJ/g
Diborane 1.31 kJ/g
Cyclopropane 1.33 kJ/g
Cubane 5.73 kJ/g
Dicyanoacetylene 6.579 kJ/g
Various unstable nitrogen allotropes >10 kJ/g
Free hydrogen (not bound in diatomic molecules) 216.34 kJ/g


Cyclopropane instability to addition reactions

You're absolutely right: cyclopropane's maximum ring strain makes it highly prone to addition reactions. Keep the following definition in mind:

  An addition reaction, in organic chemistry, is in its simplest terms an organic reaction where two or more molecules combine to form a larger one (the adduct).

In your case:

Thus, it breaks down the ring into a straight chain, which makes the low stability of the ring a clear contributing factor to the reaction.

References: Cyclopropane, wikipedia||
Addition reactions, wikipedia


Gibbs free energy and electrical potential

Your $\Delta G$ is really $\Delta G_{\mathrm{reaction}}$, and the reaction does not happen alone: it is accompanied by charging of electrodes. The Gibbs free energy of charging is just the associated change in the electroststic energy. That is, it equals $zF$ (the amount of charge moving against the electric potential for 1 mole of reaction) times $E$. Therefore, at equilibrium, the total change of the Gibbs free energy satisfies
\begin{equation}
\begin{split}
\Delta G_{\mathrm{total}} &= \Delta G_{\mathrm{reaction}} + \Delta G_{\mathrm{charging}}\\
&= 0
\end{split}
\end{equation}
just as it should.

In fact, when deriving the expression for the cell potential $E$, the logic goes backwards. We know that the electrodes are charged until equilibrium is reached, in which case the sum of $\Delta G_{\mathrm{reaction}}$ and $\Delta G_{\mathrm{charging}} = zFE$ should vanish. Thus, the cell potential is given by
\begin{equation}
E = -\frac{\Delta G_{\mathrm{reaction}}}{zF}.
\end{equation}


Why does hydrogen phosphate act as a base?

When hydrogen phosphate salts are dissolved in water there are two main equilibria formed. This is based on the fact, that hydrogen phosphate can act as a Brønsted–Lowry base, i.e. accept protons, or as an acid, i.e. donate protons. For water the same is true. In addition to this it can react with itself, which is known as the autoprotolysis of water:
$$\ce{H2O + H2O <=> H3+O + {}^{-}OH}\tag1$$

With this knowledge you can write
\begin{align}
\ce{HPO4^2- + H2O &~<=> PO4^3- + H3+O}\tag2\\
\ce{HPO4^2- + H2O &~<=> H2PO4- + {}^{-}OH}\tag3\\
\end{align}

To a lesser extent there is also the following equilibrium happening:
$$\ce{HPO4^2- + 2H2O <=> H3PO4 + 2 {}^{-}OH}\tag4$$

From the acidity constant you know that $\ce{HPO4^2-}$ will react mostly as a base, i.e. $\mathrm{p}K_\mathrm{a}(\ce{Na2HPO4})=12.35$, so the equilibrium $(3)$ will be predominant.


What is the numbering of substituted benzene derivatives according to IUPAC nomenclature?

There are several criteria for the numbering of locants. The most important simplified criteria for the numbering for substituted benzene compounds are:

lower locants for the principal characteristic groups, which are expressed as a suffix or implied by the name of the functional parent compound (If there is a choice, the principal characteristic groups are assigned in accordance with the order of seniority of classes of compounds.)
lower locants for prefixes
lower locants for substituents cited first as a prefix in the name

The corresponding actual wording of the nomenclature rules (taken from Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book)) reads as follows:

  P-14.4 NUMBERING

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

  (…)

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

  (…)

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

  (g) lowest locants for the substituent cited first as a prefix in the name;

  (…)

The first compound given in the question has the functional parent compound phenol. The name ‘phenol’ implies the presence of the characteristic group $\ce{-OH}$. Since there are no other characteristic groups and low locants are assigned first to the principal characteristic group, the locant ‘1’ is assigned to the  substituent group $\ce{-OH}$. Therefore, the systematic name for this compound is 3,4-dimethylphenol.

The second compound given in the question does not have any principal characteristic group – neither expressed as a suffix nor implied by the name of a functional parent compound. Therefore, low locants are assigned to the prefixes ‘chloro’, ‘methyl’, and ‘nitro’, all considered together as a set in a series of increasing numerical order. Thus, the systematic name for this compound is 2-chloro-1-methyl-4-nitrobenzene and neither ‘1-chloro-2-methyl-5-nitrobenzene’ nor ‘3-chloro-4-methyl-1-nitrobenzene’ since the locant set ‘1,2,4’ is lower than ‘1,2,5’ or ‘1,3,4’.


Does "soluble in alcohol" imply ethanol?

The term "alcohol", without modification (i.e. "propyl" alcohol, etc.), in the context of something like a solubility table, almost always refers to ethanol (as it does in your sources for copper (II) acetate solubility).

Note, for example, that both pubchem and Wikipedia list alcohol as a synonym for ethanol. This is not the case for other alcohols like isopropanol.


What products do 4-nitrobenzaldehyde and 3-nitrobenzaldehyde give in a crossed Cannizzaro reaction and why?

The first thing to point out is that the products of the Cannizzaro reaction do not necessarily correspond to the most stable products, i.e. the reaction is not necessarily thermodynamically controlled. Instead, one should consider the transition state—a six-membered ring as taught generally in organic chemistry—and note that after the reaction the compound which formed the hydrate will end up as the carboxylic acid and the one which did not form a hydrate is reduced to the alcohol. The transition state is shown in figure 1 below.

Figure 1: transition state of the Cannizzaro reaction

Thus, the underlying question is which compound is more likely to form an aldehyde hydrate. You argued with the nitro group of 4-nitrobenzaldehyde exhibiting a strong $-M$ (mesomeric) effect. However, this mesomeric effect cannot, in fact, cross the $\ce{C-C(=O)}$ bond and thus cannot influence the carbonyl group as much as the simple $-M$ label implies.

Figure 2: some mesomeric structures of 4-nitrobenzaldehyde, demonstrating that the mesomeric effect cannot ‘cross the $\ce{C-C}$ bond’.

Now it is known that aldehydes with a more electron deficient carbon are more likely to form hydrates, hence why we are doing a mesomeric/inductive discussion in the first place. But since we actually do not have any $-M$ arguments that can help our case, we need to turn to $-I$ arguments. When considering inductive effects, distance matters much more so 3-nitrobenzaldehyde experiences a stronger $-I$ effect than 4-nitrobenzaldehyde. Therefore, this compound is hydrated more favourably and form 3-nitrobenzoic acid while the reduced product should be 4-nitrobenzyl alcohol.


Wow, equivalent weights keep popping up again and again. It is not a nebulous term, the point is you are checking recent webpages. You have to look up chemistry books from 1930 to 60. If corona were not an issue I would have suggested a library trip to read Vogel's Quantitative Chemical Analysis (oldest edition).
Equivalent weight has nothing to do with mol. It predates moles story because the unit originated centuries ago. Mixing mole with equivalents is perhaps the main source of confusion.
Equivalent is the weight of X in grams, that will furnish or combine with 1 gram H or 16 gram O atoms.
Let us take sulfuric acid, ask yourself, what weight (=mass) of sulfuric acid will provide 1 gram H. Note I am not involving any mole concept here. The answer is that if we take 49 g of sulfuric acid we will be able to "generate" 1 g hydrogen atoms.


According to the bronsted lowry definition of an acid, the sodium ion is not an acid. This is because the bronsted lowry definition of an acid states that an acid is a proton donor. The sodium ion is not a proton donor and so according to this definition it is not classed as an acid.

I hope this answers your question :)


Where can I find the positions of the infrared band for phosphorus compounds in general? Specifically sodium hexametaphosphate

Specific for your compound you find at least two public IR spectra, either in KBr:

(source)

or from one of the larger chemical vendors as .pdf

(source, Sigma-Aldrich, product number 305553, CAS-RN 68915-31-1)

More generally speaking for phosphates, you equally find some related entries in NIST webbook chemistry here, too.

For IR-spectra in general, beside the SDBS with focus on organic compounds and sources already mentioned, there are three big multi-volume collections: Aldrich (equally one large about NMR) which is why frequently chemicals in their online catalogue have a link to either IR or/and NMR spectrum, too. The Sadtler collection is equally broad and diverse, organizes the entries by field of application (e.g., pharmaceuticals, forensics).  At least in part Sadtler is electronically available via BioRad / KnowItAll, too.  Finally Hummel, with emphazis on polymers / plastics.  Because many products of chemical synthesis are characterized by IR spectroscopy, too, you may find them indexed in databases like Reaxys (Elsevier) and Scifinder (American Chemical Society), too.


Mass of monomer needed to obtain a polymer of a given molecular weight

You have a $2:1$ ratio. You need a $1:1$.

The $\bar{X}_m$ resulting (the number average degree of polymerisation) is derived from the Carothers equation:

$$\bar{X}_m = \frac{1+p}{1-p}$$

where $p$ is the mole ratio. For a $2:1$ this is $0.5$, and the effective $\bar{X}_m = 3$. If you mix at $2:1$, you'll have the $\ce{BAB}$ trimer as the major product with a $M_m$ of $\pu{297.1 g/mol}$.

If you have one component in $1\%$ excess ($1.01:1$) the $\bar{X}_m$ will be $199$, at $.1\%$ excess the $\bar{X}_m$ will be $1,999$ etc.


Compounds do not class themselves as either one thing or another. Looking at a specific compound, there will always be classes that it is not, but save very few very simple compounds all will fit at least two classes.

$\ce{Fe(OH)3}$ is both a base and a salt. And a transition metal compound and a hydroxide and …

Let’s take an analogy. To some, I am a friend. To others I am a student. To others again, I am a teacher. I can be all of those, depending on who looks at me.


Do we need to check for all kinds of symmetry in a compound to declare it achiral?

We need to check all 4 symmetries to determine whether a compound is optically active or not.
Consider the following compound. It has only Alternating Axis of Symmetry

Note that the rings are in perpendicular planes. The following diagrams illustrate the top view of the molecule.The stereochemistry of the methyl groups can be:

$S_4$ axis of symmetry is seen only in the following diastereomer where the methyl groups are anti to each other in each ring and the view looks like a Nazi cross from the top.


What are the conditions for using skeletal replacement ('a') nomenclature?

Rule P-51.4, not P-15.4, specifies the conditions under which skeletal replacement nomenclature generates preferred IUPAC names.

  P-51.4.1.1 Skeletal replacement (‘a’) nomenclature rather than substitutive or multiplicative names must be used to generate preferred IUPAC names for acyclic structures when four or more heterounits are present in an unbranched chain containing at least one carbon atom and when none of the heteroatoms constitute all or part of the principal characteristic group of the compound.

The former compound has only three oxygen heteroatoms, not enough to require the use of replacement nomenclature.


How many unique molecules exist? And how many exist only by synthesis?

Polymers make this question more or less unanswerable.

Consider human chromosome 1, which contains about 249,000,000 base pairs.  There is nothing that says we couldn't order those pairs in any way we like, so for DNA alone, in a quantity that exists in every person on Earth, there exists the possibility for 4249000000 different molecules.  Include things like alternative nucleotides (there are almost surely at least thousands of these that are possible) and longer chains and the number gets even more enormous, and that's only including DNA (and close relatives.)

There's no real way to include all of them, since there have to be an enormous number of ways to put these molecules together that we haven't found yet.

How many exist?  An awful lot.
How many only exist in our minds?  Almost all of them (the universe only contains about 1080 particles, after all.


Why is the IUPAC name of 1H-benzo[d]naphthalene not 1H-benzo[i]naphthalene?

Apparently, you are mixing two different numbering schemes.

For the construction of a fusion name, the two components are numbered separately (if necessary), using letters for the parent component and numbers for the attached component. The corresponding rule P-25.3.1.3 in the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) reads as follows.

  (…)
  Isomers are distinguished by lettering, continuously, each peripheral side of the parent component (including sides whose locants are distinguished by letters, for example, 2a,3a) using the italic letters a, b, c, etc., beginning with a for the side numbered ‘1 ,2’, b for ‘2,3’ etc. To the letter as early in the alphabet as possible that denotes the side where the fusion occurs are prefixed, if necessary, the numbers of the positions of attachment of the other component. These numbers are chosen to be as low as is consistent with the numbering of the compound and their order conforms to the direction of lettering of the parent component. (…)

The parent component for the example that is given in the question is naphthalene (not the complete benzonaphthalene). The letter as early in the alphabet as possible that denotes the side where the fusion occurs is d. Therefore, the correct name of the complete compound is
1H-benzo[d]naphthalene.

The other numbering scheme described in P-25.3.3.1.1 applies to the complete fused ring system.

  P-25.3.3.1.1 The numbering of peripheral atoms in the preferred orientation starts from the uppermost ring. If there is more than one uppermost ring, the ring furthest to the right is chosen. Numbering starts from the nonfused atom most counterclockwise in the ring selected and proceeds in a clockwise direction around the system, including fusion heteroatoms but not fusion carbon atoms. (…)

Thus, for the example that is given in the question, this numbering applies to the complete
1H-benzo[d]naphthalene. In particular, this numbering is used for the locant of the indicated hydrogen. It would also be used for any substituents of the compound.


Why is fluorine one of the most electronegative atoms even though it doesn't have the largest effective nuclear charge?

Fluorine is the most electronegative element because the definition of electronegativity makes it so. The electronengativity scales are defined based on experimentally determined properties of the elements. Fluorine has appropriate values for all of the common scales to ensure it has the highest electronegativity.

The Pauling scale, which is the first electronegativity scale proposed, relates the difference in electronegativity $\Delta\chi_{A,B}$ between two atoms $(A\ \& \ B)$ is related to the bond dissociation energies $(E_d)$ of the diatomic species $AA,\ BB,\ \& \ AB$:

$$\Delta \chi_{A,B}=|\chi_A - \chi_B| = (\text{eV})^{-1/2}\sqrt{E_d(AB)-[E_d(AA)+E_d(BB)]/2}$$

Since only differences can be determined, one element (hydrogen) was chosen to have an arbitrarily set value. On the Pauling Scale Hydrogen has a value of $2.20$. Since the $\ce{H-F}$ bond is much stronger $(E_d=586\ \text{kJ/mol})$ than the average of the  $\ce{H-H}$ bond $(E_d=436\ \text{kJ/mol})$ and the $\ce{F-F}$ bond $(E_d=157\ \text{kJ/mol})$. Data from here. In general, since fluorine forms stronger bonds to all other elements than it does to itself, fluorine will have the highest electronegativity.

The Mulliken scale relates electronegativity to two other properties: electron affinity $E_{ea}$ and ionization energy $E_i$:

$$\chi_A = \frac{E_{ea}(A)+E_i(A)}{2}$$

Fluorine has a high electron affinity (easy to gain an electron) and a high ionization energy (hard to lose an electron).