Using baking powder to make baked food crispy

Not exactly an academic source, but a few articles on Serious Eats suggest that baking powder's gas releasing reaction forms small bubbles of protein on the surface that add surface area that can quickly dehydrate (a prerequisite for crispiness) and solidify.
It's also a pretty common ingredient in things like batter for fried chicken, where I imagine the expanding bubbles do a similar thing—allow easier removal of water from the batter by increasing the surface area so it can crisp up.

The pH change may also help break down the skin, if applied in advance.


Is the following paragraph about steam distillation accurate?

You have it right on all counts. But, to clarify...

i) Steam distillation is used to separate steam-volatile organic components of a mixture from non-volatile ones (not from each other, however). It is most useful for higher boiling, relatively non-polar organic compounds that are immiscible with water.


In antacids like Eno that fizz up when mixed in water, why is the citric acid important?

It's the citric acid that reacts with the sodium bicarbonate. When the tablet hits the water, the citric acid and bicarbonate react, forming $\ce{CO2}$ and sodium citrate. This causes the tablet to break up, thereby much speeding up its dissolution.

The table will contain a stoichiometric excess of bicarbonate, with respect to the citric acid, so it's anti-acid properties remain present, even after bubbling/dissolution. Dissolving a tablet that doesn't disintegrate due to the acid-base reaction and its fizz would take much longer.

The fizzing effect probably also has some aesthetic/marketing value, separate from anti-acid properties of the tablet.


Does SiO₂ have polar covalent bonds or coordinate covalent bonds?

The bonds are polar covalent.

'Given' is a strange term in this context and generally only of relevance for ionic compounds.

Coordinate covalent bonds imply the donation of two electrons to a bond with another atom.  In the case of $\ce{ SiO2 }$, The structure is actually tetrahedral, with an oxygen between and two silicon atoms, and where each oxygen forms two bonds and each silicon forms four bonds.  In these bonds, each of the four valence electrons for silicon is shared singly with a single electron from one of the oxygens.  Thus, in each bond, one of the electrons is from silicon and one from oxygen, which is why we don't normally consider this a coordinate covalent or dative bond.

This is one of those examples where the empirical formula, $\ce{ SiO2 }$, is highly misleading.


Software for compound database creation\enumeration

This is typically called library (or scaffold) enumeration. Doing it in SMILES is usually pretty easy by script, but there are a few other options:

KNIME
SMILIB

But it's very easy to write a script like this, e.g. (in python)

r1 = [ "", "C", "N" ]
r2 = [ "H", "C", "c3ccccc3", "C(=O)O" ]
r3 = [ "", "C", "C#N", "O" ]
scaffold = "Xc(cc1c2)ccc1c(Y)cc2-C(C=C1)C=C1Z"

for x in r1:
    for y in r2:
        for z in r3:
            print scaffold.replace('X', x).replace('Y', y).replace('Z', z)

You can then process the resulting SMILES strings with your program of choice (e.g., Open Babel).


What is the reaction mechanism of a selective catalytic reaction with urea?

After researching this topic, I've found the mechanism to be:


why do we obtain a sigmoid curve in vapour pressure versus temperature graph

the answer already present over here is absolutely excellent but, as an extention i would add some points to it

  the equation of $PV=nRT$ is made for gases and for a solution of gas and liquid in which the volatile solvent shows some vapour pressure on its solution

the curve sigmoid curve you get is through the Clausius Clapeyron equation which suits this condition [i.e. the situation of solutions]

after the whole solution is evaporated then you get pure solvent in the vapour phase after which your equation for an ideal gas $PV=nRT$ is applicable and you will get a striaght line there after


Which process is used to separate Hydrogen and Oxygen from Water?

Problem in your question: should be be O and H, because $\ce{H2O}$ (water) is made of H (Hydrogen) and O (Oxygen). $\ce{CO2}$ is carbon dioxide, for reference.

And yes, electrolysis can be performed on many compounds, not just water.

An electrolytic reaction is where an electric current is used to drive an otherwise non-spontaneous chemical reaction. Splitting water is just one of such reactions.

You can read more here.


Why is phenyl shift preferred over methyl shift in cumene hydroperoxide reaction?

Despite the mechanism shown in your question, which is commonly used to depict a phenyl shift, phenyl shifts occur by a slightly different mechanism.

Phenyl shifts do not usually happen by the normal 1,2-sigmatropic "hop". Instead, the phenyl group shifts by more of a "walk". It's not really a sigmatropic rearrangement. First, one of the $\pi$-bonds in the phenyl group can attack the electrophilic center. This leads to formation of a spirobicyclic intermediate where the cation is delocalized throughout the rest of the ring. Then, the three-member ring breaks open at the expense of a carbon-carbon bond to rearomatize the phenyl ring and form the new resonance stabilized cation.


Charge, Size, and Movement in Aqeuous Ions as Electrolytes

Would a pair of aqueous $+2$ and $−2$ ions necessarily conduct electricity better than the same concentration of $+1$ and $−1$ ions?

Not necessarily. Taking the concentration of 0.001 aqueous solution of $\ce{KOH}$ and $\ce{MgSO4}$ at $18$ degree Celsius, we find that the current conducted by $\ce{KOH}$ is $2.34 \space mA$ while the current conducted by $\ce{MgSO4}$ is $2.00 \space mA$. Why did this happen?

It happened because of the mass. And therefore here comes the answer to your second question.

  Would more massive ions not conduct as well as less massive ions because they cannot travel as rapidly?

Yup, more massive ion means less current. It's because of simple equation (and I hope you are aware of these physics equations)$$qV=\dfrac{mv^2}{2}$$

Here $q$ is the charge on the ion. $V$ is potential difference between both the solutions. $m$ is the mass of the ion. And $v$ is the drift velocity. We know that $qV$ is constant. So we get $$v \propto \dfrac{1}{\sqrt{m}}$$ We can easily analyse that more the mass we have, less the speed we get.

  If ions are moving, why don't they ever achieve the most stable position and stop moving?

They do achieve the most stable position and that's why batteries die.


Crystallization of zinc sulfate monohydrate

TL; DR: obtaining a reasonably big single crystal of $\ce{ZnSO4 * H2O}$ seems to be a tricky task, whereas one can easily obtain $\ce{ZnSO4 * H2O}$ as a result of thermal dehydration of heptahydtare above $\pu{70 ^\circ C}$ in a form of melt, and above $\pu{120 ^\circ C}$ – as a polycrystalline solid material.

I finally got across to suitable references to back up my comments. In nature zinc sulfate monohydrate occurs as mineral gunningite, always in a form of efflorescence deposition on the surface of sphalerite and other minerals and doesn't form single crystals [1]:

  Surface and ground waters carrying dissolved oxygen attack
  the sphalerite, yielding soluble sulphates of zinc [...]:

  $$\ce{ZnS + 2 O2 -> ZnSO4}$$

  [...] In the presence of abundant water all of the soluble salts are removed. If, however, the conditions are such that the supply of water is restricted or evaporation takes place, the metal-bearing solutions become supersaturated with the consequent precipitation of a number of hydrates of zinc, manganese, and iron of which gunningite, $\ce{ZnSO4 * H2O}$, is one.

Thermal dehydration among zinc sulfate hydrates has been studied with DTA in several publications. The most relevant and thorough investigation [2, p. 471] suggests the following scheme:

$$\ce{ZnSO4 * 7 H2O (s) ->[\pu{37 ^\circ C}] ZnSO4 * 6 H2O (s) ->[\pu{70 ^\circ C}] ZnSO4 * H2O (l) ->[\pu{120 ^\circ C}] ZnSO4 * H2O (s) ->[\pu{285 ^\circ C}] ZnSO4 (s)}$$

Original text in German:

  Festgestellt werden vier endotherme Effekte: bei etwa $\pu{37 ^\circ C}$ verwandelt sich das Heptahydrat in Hexahydrat; bei etwa $\pu{70 ^\circ C}$ geht das Hexahydrat in Monohydrat über, wobei teilweise Schmelzen im Kristallwasser auftritt. Bei etwa $\pu{120 ^\circ C}$ siedet die Schmelze, Wasser scheidet
  sich ab, bis das feste Monohydrat zurückbleibt; bei etwa $\pu{285 ^\circ C}$ erfolgt der Übergang des Monohydrates in wasserfreies Salz.

The tricky part is that monohydrate can be isolated in crystal form, but it initially isolates as a melt, and only above $\pu{120 ^\circ C}$ makes a polycrystalline solid phase (occurs as white powder or granules).

Aqueous solubility $\omega_2$ for zinc sulfate $\ce{ZnSO4}$ in mass% of a solute ($\omega_2 = m_2/(m_1 + m_2)$, where $m_2$ is the mass of solute and $m_1$ the mass of water) as a function of temperature ($T$) from CRC Handbook [3, p. 5-172]:

\begin{array}{r|rrrrrrrr}
T,\,\pu{^\circ C} & 0 & 10 & 20 & 25 & 30 & 40 & 50 & 60 & 70 & 80 & 90 & 100 \\
\hline
\omega_2,~\% & 29.1 & 32.0 & 35.0 & 36.6 & 38.2 & 41.3 & 43.0 & 42.1 & 41.0 & 39.9 & 38.8 & 37.6
\end{array}

With this in mind, what can be suggested is to prepare a saturated at $\pu{70 ^\circ C}$ solution, and then slowly increase the temperature (e.g. using a thermostat), hoping that zinc sulfate monohydrate starts to crystallize (using the decrease in solubility starting from $\pu{60 ^\circ C}$).

According to [4, entry 3575, p. 470], solubility of $\ce{ZnSO4 * H2O}$ also decreases in this interval: $\pu{101 g}/\pu{100 g}~\ce{H2O}$ at $\pu{70 ^\circ C}$, and $\pu{87 g}/\pu{100 g}~\ce{H2O}$ at $\pu{105 ^\circ C}$. The difference is rather small, so I wouldn't expect rapidly-grown large crystals.

References

Jambor, J. L.; Boyle, R. W. The Canadian Mineralogist 1962, 7 (2), 209–218.
Balarew, C.; Trendafelov, D.; Gerganova, M. Monatshefte für Chemie 1971, 102 (2), 465–473. DOI 10.1007/BF00909340 (in German).
Haynes, W. M.; Lide, D. R.; Bruno, T. J. CRC handbook of chemistry and physics: a ready-reference book of chemical and physical data.; 2017; Vol. 97.
Perry, D. L. Handbook of inorganic compounds, 2nd ed.; Taylor & Francis: Boca Raton, 2011. ISBN 978-1-4398-1461-1.


For example ,when a person drops a book. When he or she lifts a book, he or she gives potential energy to the book (energy absorbed). However, once the he or she drops the book, the potential energy converts itself to kinetic energy and comes in the form of sound once it hits the ground (energy released).

Electron affinity is defined as the change in energy (in kJ/mol) of a neutral atom (in the gaseous phase) when an electron is added to the atom to form a negative ion. In other words, the neutral atom's likelihood of gaining an electron.

When an electron is generally added to a neutral atom, i.e first electron affinity, energy is released; thus, the first electron affinities are negative. However, when an electron is added to a negative ion, i.e. second electron affinity, more energy is required. Thus, more energy is released to add the electron into an ion because the negative ion has to force the electron to go into its electron orbital; thus, the second electron affinities are positive.

First Electron Affinity (negative energy because energy released):

$\ce{X(g) + e- -> X- (g)}$

Second Electron Affinity (positive energy because energy needed is more than gained):

$\ce{X- (g) + e- -> X^2- (g)}$

This is because you are practically trying to forcefully add an electron to it.

Coming to your question: Remember that greater the distance, the less of an attraction; thus, less energy is released when an electron is added to the outside orbital. In addition, the more valence electrons an element has, the more likely it is to gain electrons to form a stable octet. The less valence electrons an atom has, the least likely it will gain electrons.

One might think that since the number of valence electrons increase going down the group, the element should be more stable and have higher electron affinity. One fails to account for the shielding affect. As one goes down the period, the shielding effect increases, thus repulsion occurs between the electrons. This is why the attraction between the electron and the nucleus decreases as one goes down the group in the periodic table.

As you go down the group, first electron affinities become less (in the sense that less energy is evolved when the negative ions are formed). Fluorine breaks that pattern, and will have to be accounted for separately.)

The electron affinity is a measure of the attraction between the incoming electron and the nucleus — the stronger the attraction, the more energy is released. The factors which affect this attraction are exactly the same as those relating to ionization energies — nuclear charge, distance and screening.

The increased nuclear charge as you go down the group is offset by extra screening electrons. Each outer electron in effect feels a pull of 7+ from the center of the atom, irrespective of which element you are talking about.

So,

as atomic radius increases, the nuclear attraction on outtermost electron decreases — this leads to decrease in IE — but the shielding also increases — less energy is released when an electron is added to the outside orbital — decreasing the EA.


How does hyperconjugation explain the stability of alkenes?

As always in chemistry one phenomenon (experimental observation) can be explained with different models, that live in different "physical model worlds", and more often than not these models are not consistent among themselves and might be even bare of a physical reality, especially if they are analyzed in rigorous way, nevertheless might be very useful tools, that take the justification of existence out of the fact that they work in many cases.

That said,  especially the concept of hyperconjugation itself can be employed at different levels. A very primitive model, but one that works astoundingly well, is to assign a positive inductive (electron releasing effect) to alkyl chains that is increasing with the degree of substitution of the ipso atom. That leads to the idea that this (however small) charge that is pushed to the sp$^2$ carbon atom stabilizes its slight positive polarization (since the majority of electrons in the double bond is above and below the plane of the atoms engaged attached to the doubly bound atoms. The physical reality of this model if often drawn into doubt, which is good to know, but still its simple and powerful and thus there is no reason to abandon it.

At a higher level of model, using a MO picture the idea is that the (empty, anti-bonding) $\sigma^*$ orbitals that come inevitably with the C-C and C-H $\sigma$ bonds can act as kind of "acceptors" for the $\pi$-electrons of the double bonds if in the right orientation. It is important to note that in this idea to actual overlap that mediates the interaction is thought to be established with the part of $\sigma^*$ orbital that is opposing the bound atom. That donor-acceptor interaction ultimately leads to a stabilization of the system. Again the physical reality is also here at least disputed, but however less than in the former case. But still both have in common that the Substituent acts as the donor and that this causes stabilization.

So yes in this higher level of description of the hyperconjugation the reason for the stability is the increasing number of possibilities to form stabilizing overlap.


Calculation of concentration at equilibrium for extraction

I'd like first to answer your question:

My question is, when writing the formula for $K_\mathrm{D}$, is the organic phase always in the numerator and the aqueous phase in the denominator?

The answer is yes. The IUPAC Recommendations 1993 (Ref.1) defines Partition Ratio ($K_\mathrm{D}$) as follows (also see the Goldbook):

Partition Ratio ($K_\mathrm{D}$): The ratio of the concentration of a substance in a single definite form, $\mathrm{A}$, in the extract to its concentration in the same form in the other phase at equilibrium, e.g. for an aqueous/organic system.
$$(K_\mathrm{D})_\mathrm{A} = \frac{[\mathrm{A}]_\mathrm{org}}{[\mathrm{A}]_\mathrm{aq}}$$

Thus, if you have used equal volumes of organic and aqueous phases, the equation you can set up to solve is:
$$(K_\mathrm{D})_\mathrm{A} = \frac{[\mathrm{A}]_\mathrm{org}}{[\mathrm{A}]_\mathrm{aq}}= \frac{[\mathrm{PhOH}]_\mathrm{org,eq}}{[\mathrm{PhOH}]_\mathrm{aq, initial}-\mathrm{PhOH}]_\mathrm{org,eq}} \\ =\frac{[\mathrm{PhOH}]_\mathrm{org,eq}}{\pu{1.5 \times 10^{-2} M}-[\mathrm{PhOH}]_\mathrm{org,eq}}= 14 \tag{1}$$
When you solve the equation $(1)$ for $[\mathrm{PhOH}]_\mathrm{org,eq}$, you get your answer:
$$[\mathrm{PhOH}]_\mathrm{org,eq} =\frac{\pu{1.5 \times 10^{-2} M} \times 14}{15}= \pu{1.4 \times 10^{-2} M} $$
Hence, $[\mathrm{PhOH}]_\mathrm{aq,eq} = \pu{1.5 \times 10^{-2} M} - \pu{1.4 \times 10^{-2} M} =  \pu{1.0 \times 10^{-3} M}$
Note: If you perform extraction in different volumes of aqueous and organic phases, you should read this Wikipedia article to get help for solution.
References:

N. M. Rice, H. M. N. H. Irving, M. A. Leonard, "Nomenclature for liquid-liquid distribution (solvent extraction) (IUPAC Recommendations 1993)," Pure & App. Chem. 1993, 65(11), 2373-2396 (DOI: https://doi.org/10.1351/pac199365112373).


I am trying to find the solubility of benzoic acid in different solvents to prepare a liquid-liquid extraction

Funny you're looking at solubility of benzoic acid. A few years ago, there was an Open Notebook Science challenge to measure solubility:

  The first round of this challenge calls upon people with access to materials and equipment to measure the solubility of compounds (aldehydes, amines and carboxylic acids are a priority) in organic solvents and report their findings using Open Notebook Science

So the raw data is available through Google Sheets here

There is a general web service for predicting solubility.

For example, you can predict the solubility of benzoic acid using Abraham solvation parameters

There used to be a search mechanism, but I'll have to ask the authors - it's been a few years.


Zinc is used effectively in the dehalogention of vicinal dihalides because a carbon zinc complex is formed in the mechanism which is probably more stable. This reaction is also known as Simmon' Smith reaction.

The stability and usefulness of the Simmons-Smith reagent (zinc metal)
may be attributed in part to the higher covalency of the carbon-zinc
bond together with solvation and internal coordination of the zinc (which makes the zinc complex stable). (Source)
Reduction of vicinal dihalides by zinc metal proceeds by way of
oxidative insertion of Zn(0) into a C-X bond, producing an organozinc
species (stable) which undergoes dehalogenation:

(source)


Dissolvant for the black spots around the rear mirror of the front window of my car

The part of the windshield that you are talking about is called the frit. This is typically made up of some form of high-heat enamel, and as far as I was able to find, it is intended to increase the surface roughness of the inside of the glass to increase the ability for the adhesive to stick to it.

I was not able to find any specific mention for what type of enamel frits are actually made of - that is whether it is an enamel glass or an enamel paint -  and as such I cannot give any specific recommendations for what solvent to use. If it is enamel glass, as I would hazard to guess it is - given that some sources indicate that it is applied prior to the windshield being bent - then you will likely not be able to remove the frit without also damaging the glass. If it is an enamel paint then you may be able to remove it with an organic solvent. In general, I like to start with acetone as a general organic solvent.

All of this said, given both the purpose of the frit, and its proximity to the adhesive that holds your windshield on, I strongly recommend that you consider moving your dashcam rather than attempting to dissolve the frit.

Best of luck.


Effect of increasing product concentration on time of completion of reaction

In equilibrium reactions the equilibrium is approached exponentially in time. The lifetime of the exponential is the reciprocal of sum of the forwards and back rate constants multiplied by the equilibrium concentration of the species, thus for reaction $\ce{A + B \leftrightharpoons C + D }$ the lifetime of the reaction is $\tau = 1/[k_1 (A_e+B_e) + k_{-1} (C_e+D_e) ] $ where $k_1$ is the forward rate constant and $k_{-1}$ the backwards one, and subscript $e$ means equilibrium amount of a species. Thus if you add more material you change the equilibrium concentrations  and so the lifetime gets shorter as its a reciprocal in the concentrations

The lifetime is the time taken to change to $1/e$ of the initial amount, approx to 38%, whereas the half-life is the time taken to halve the initial amount. The lifetime is the reciprocal of the total rate constant for reaction which is why it is used instead of the half-life.

(The exception is the simple $\ce{A\leftrightharpoons B}$ reaction when the lifetime is independent of the equilibrium amounts ). In equilibrium reactions the lifetime is also called the relaxation time.


Estimation of pressure and kinetic energy density of stellar interior using kinetic theory of gases

I would like to first point out that the pressure you've determined is too low, most likely because you used only molar mass of protons $\ce{H+}$ instead of average mass for star matter in order to fulfill the condition of plasma quasineutrality:

$$\bar{M} = \frac{M(\ce{H+}) + M(\ce{e-})}{2} = \frac{\pu{(1 + 0) g mol-1}}{2} = \pu{0.5 g mol-1} = \pu{5e-4 kg mol-1} \tag{1.1}$$

Referring to the question from Atkins' Elements of Physical Chemistry [1, p. 40], where this question has been taken from, we obtain necessary data to complete the solution:

  In the Sun [...] the density is 150 times that of liquid water at its centre and comparable to that of water about half-way to its surface

and

  (a) Calculate the pressure half-way to the centre of the Sun, assuming that the interior consists of ionized hydrogen atoms, the temperature is $\pu{3.6 MK}$, and the mass density is $\pu{1.20 g cm−3}$ (slightly higher than the density of water)

the pressure of star plasma (that can be treated as an ideal gas) can be found as following:

$$p = \frac{\rho RT}{\bar{M}} = \frac{\pu{1.2e3 kg m-3} \cdot \pu{8.314 J K-1 mol-1} \cdot \pu{3.6e6 K}}{\pu{5e-4 kg mol-1}} = \pu{7.18e13 Pa} \label{eqn:1.2}\tag{1.2}$$

Now for the second part, kinetic energy density is defined as kinetic energy per volume:

$$\rho_\text{k} = \frac{E_\text{k}}{V} \label{eqn:2.1}\tag{2.1}$$

Kinetic energy for a set of entities (ion, atom, molecule) is

$$E_\text{k} = \frac{Nm\bar{v^2}}{2} \label{eqn:2.2}\tag{2.2}$$

where $N$ is the number of particles, $m$ is the particle's mass, $\bar{v}$ is an average (root mean square) particle velocity. From the kinetic model of an ideal gas, the pressure obtained from \eqref{eqn:1.2} is also

$$p = \frac{Nm\bar{v^2}}{3V} \label{eqn:2.3}\tag{2.3}$$

Equation \eqref{eqn:2.3} implies $Nm\bar{v^2} = 3pV$, so kinetic energy from \eqref{eqn:2.2} can be rewritten as $E_\text{k} = \displaystyle\frac{3pV}{2}$. Finally, kinetic energy density, taking \eqref{eqn:2.1} into account, is

$$\rho_\text{k} = \frac{3p}{2} = \frac{3 \cdot \pu{7.18e13 Pa}}{2} = \pu{1.08e14 J m-3} \quad (\pu{1 J} = \pu{1 Pa}\cdot \pu{1 m3})$$

Comparison between kinetic energy densities $\rho_\text{k}(1/2r_\odot)$ (halfway to the Sun's center) and $\rho_\text{k}(\text{atm})$ (Earth's atmosphere) is really a matter of comparing the pressures:

$$\frac{\rho_\text{k}(1/2r_\odot)}{\rho_\text{k}(\text{atm})} = \frac{p(1/2r_\odot)}{p(\text{atm})} = \frac{\pu{7.18e13 Pa}}{\pu{101325 Pa}} = 7.09\cdot10^8$$

References

Atkins, P. W.; De Paula, J. Elements of physical chemistry, 5th ed.; Oxford University Press: Oxford; New York, 2009. ISBN 978-1-4292-1813-9.


If you spill nitric acid onto the table, you yourself are unharmed and you can use a cool head to decide what to do next. If the spill is small, pour water on it to both dilute it and dilute the heat — remember water has a high specific heat capacity. If the spill is larger, I would probably try adding sodium hydrogencarbonate to react away the acid before wiping it up. Adding larger amounts of water will work, too, but you need to get them from somewhere and then need to get rid of the resulting puddle.

If the spill is on you you want to get the acid off your clothes and body immediately. The quickest and most effective solution is to pour water over your head and lots of it. This is what the emergency showers are for. Lots, here, has two functions: it not only quickly dilutes the acid to a less harmful concentration but also (because it is washing down) serves well to actually remove the heat — remember again that water has a high specific heat capacity. Emergency showers have a high throughput for exactly this reason.


Kohlrausch in 1874 he demonstrated that an electrolyte has a definite and constant amount of electrical resistance. From 1875 to 1879, he examined numerous salt solutions, acids and solutions of other materials. This resulted in the law of the independent migration of ions, that is, each type of migrating ion has a specific electrical resistance no matter what its original molecular combination may have been.So this means that the electrical resistance of the ions in the solution that are migrating, and conductivity is just the inverse of this.

CH3COOH which is a weak electrolyte, so does not dissociate completely. Since it's only the migrating ions that affect it, only the dissociated ions affect conductivity and specific electrical resistance no matter what its original molecular combination may have been.

At lower concentrations, since there are more water molecules for every CH3COOH molecule, more of these dissociate so there are more migrating ions meaning a larger conductivity. The molar conductivity is defined as the solution conductivity normalized by the total ionic concentration. So concentration increases without any significant increase in conductivity meaning the molar conductivity decreases rapidly.


What are similarities and differences among shells, orbitals, subshells, and energy levels?

I am using the chemistry textbook by Masterton/Hurley (Cengage) as an outline for this answer. It has a side note that reads "Our model for electronic structures is a pragmatic blend of theory and experiment". In the spirit of that quote, there is some experimental data mixed in with the concepts.

  Each row on the periodic table has a new shell of electrons

Yes, and you see that in the plot of atomic properties vs. atomic number. For example, atomic radii (image source)increase significantly as you go from noble gas (last element in a period) to group 1 metals (first element with the next shell occupied). The quantum number that defines the shell is the principal quantum number n.

  A subshell is e.g. 1s, 2s, 2p, 3s, 3p, etc.

Yes. The periodic table is organized into s-block and p-block (main group elements), d-block (transition metals) and f-block (lanthanides and actinides) according to which subshell is partially filled (or just filled). The quantum number that defines the subshell is the second quantum number, $\ell$.

  Are energy levels the same as shells?

This is true only for the hydrogen atom and other one-electron systems ($\ce{He+}, \ce{Li^{2+}}$) in the absence of a magnetic field:

$$ E \propto -\frac{Z^2}{n^2} $$

with $Z$ the charge of the nucleus and $n$ the principal quantum number.

For systems with more than one electron, distinct subshells in the same shell correspond to slightly different energy levels. One method to measure these energy levels is photoelectron spectroscopy, with some data on aluminium 2s and 2p shown below (image source).

In general, the energy levels depend on the atomic number (number of protons in the nucleus) and the electron configuration (how many electrons are present, and whether they are at ground state or in an excited state). Going from a hydrogen atom to a larger multi-electron atom, the energy levels of distinct shells can be roughly described as an electron bound to a nucleus surrounded by inner electrons with a combined effective charge $Z_\mathrm{eff}$ (smaller than the actual charge of the nucleus because the attraction by the nucleus is offset to a degree by the repulsion by the other electrons):

$$ E \propto -\frac{Z_\mathrm{eff}^{2}}{n^2} $$

For the innermost shells (n = 1), $Z_\mathrm{eff}$ is almost the same as the nuclear charge, whereas for the outermost shells, $Z_\mathrm{eff}$ can be much lower.

For example, comparing 1s electrons across the periodic table, the binding is tighter and tighter as the atomic number increases. On the other hand, the outermost shell (valence shell) in elements of the same group (e.g. Be, Mg, Ca, Sr, Ba, Ra) corresponds to similar energy levels, even though they differ in the principal quantum number. This explains why elements in a group have similar chemical behavior.

  Are orbitals the same as subshells?

No, there are seven different orbitals in the f-subshell, five different orbitals in the d-subshell and three different orbitals in the p-subshell (compare size of blocks in the periodic table). The three orbitals in the p-subshell, for example, are either designated (+1), (0), and (-1), or alternatively $p_x$, $p_y$, and $p_z$. For the s-subshell, there is a single orbital, so only in this special case is the subshell the same as an orbital. The quantum number defining the orbital in a subshell is the third quantum number $m_\ell$.

While the different orbitals of the p, d, and f subshells have distinct, sometimes intricate shapes, a filled subshell corresponds to a spherical distribution of electrons.

  How do these concepts help to understand chemical bonds?

We study electronic structure because it explains properties and reactivity of atoms and molecules. It might seem that atomic orbitals of hydrogen are of little use in understanding the electronic structure of larger atoms and of molecules. It turns out, however, that they can be used as building blocks for the more complicated systems, both qualitatively (talking about sigma and pi bonds and lone pairs) and quantitatively (computational methods use atomic orbitals as basis functions).

Surprisingly, the changes in electronic structure going from atoms to molecules are moderate, with spherical electron distribution around the atoms explaining most of the electron density observed, for example, in crystallographic studies. The small deviations from this are called electron deformation densities, and they give an experimental view of the subtle shifts in electron density upon bond formation. The picture below shows a contour diagram of the total electron density on the left (near the atoms, there would be many more contour lines, but they are not shown), and the deformation density on the right (lines: higher density than atomic density, dashed: lower density than atomic density, dot-dash: same density as atomic density).

(Image source)


Does the term 'Cation' always refer to a positively charged particle?

Yes, cations always have a positive charge and anions always have a negative one.

The difficulty is that the term cathode and anode do not always correspond to the same pole. The cathode is that pole of an electrolytic/electrochemical cell where reduction takes place (cathodic reduction) while the anode is where oxidation takes place (anodic oxidation). Since it depends on the type of cell which will be oxidised and which will be reduced, the terms cathode and anode mean different things in different cells.


What are contrast agents in MRI and how do they work?

1) Contrast agent is used to create a sharp contrast on an otherwise fuzzy image. Consequently, its distribution must have sharp gradient, i.e it must have strong affinity to some type of tissue or physiological liquid. The target may be widely represented in the body, but it still should be a clear target. Or, alternatively, it should be injected into a physiological liquid and have no time to distribute evenly.

2) Wikipedia provides simplistic description. The idea is that to relax efficiently, a magnetic nucleus (including a proton from a hydrogen atom) needs to interact efficiently with chaotic microscopically inhomogeneous magnetic field. Such field may be produces by another microscopic magnetic objects. $\ce{Gd^{+3}}$ has seven unpaired electrons, giving the atom the strongest magnetic field of all single atoms available in reasonable chemical conditions.


What are the height and width of the large and small nodes of the sp3 hybridized orbitals of carbon and silicon?

So, after I get past the shock and awe that you are asking us to describe a fictional object,

this page has some pictures which will be suitable for you to construct visualization: ].

I eyeball it as about 4 times larger in all three dimensions for the 2s/2p level.  The radial distribution function for 3s/3p has its highest density about 2.2 times further away from nucleus as the 2s/2p level
]


Just calculating from the head by the simplifying formula $\mathrm{pH} = \mathrm{p}K_\mathrm{a} + \log{(\frac {[\ce{A-}]}{[\ce{HA}]})}$, $\mathrm{pH}$ of $\pu{1 M } \ce{CaCl2}$ is approximately 1/2( 12.6 - log 1)=6.3.
So concentration $\ce{[CaOH+]}$ is about million times smaller then $\ce{[Ca^2+]}$. Hydrolysis happens, but is fully negligible. Note that the real $\mathrm{pH}$ depends on impurities and dissolved $\ce{CO2}$.
About the calculation of respective fractions, we can use the formula for gradual dissociation of multiprotic acid:
If we formally suppose $$\ce{H2A <=>  HA- + H+ <=> A^2- + 2 H+}$$ then respective molar fractions for particular acid forms are
$$x_{\ce{H2A}} = x_{\ce{Ca^2+}} =\frac {[\ce{H+}]^2}{[\ce{H+}]^2 + K_\mathrm{a1} \cdot [\ce{H+}] + K_\mathrm{a1} \cdot K_\mathrm{a2}}$$
$$x_{\ce{HA-}} = x_{\ce{Ca(OH)+}} =\frac {K_\mathrm{a1} \cdot [\ce{H+}]}{[\ce{H+}]^2 + K_\mathrm{a1} \cdot [\ce{H+}] + K_\mathrm{a1} \cdot K_\mathrm{a2}}$$
$$x_{\ce{A^2-}} = x_{\ce{Ca(OH)2}} =\frac {K_\mathrm{a1} \cdot K_\mathrm{a2}}{[\ce{H+}]^2 + K_\mathrm{a1} \cdot [\ce{H+}] + K_\mathrm{a1} \cdot K_\mathrm{a2}}$$
where $K_\mathrm{a1} = K_\mathrm{w} / K_{\mathrm{b1},\ce{Ca(OH)2}}$ and $K_\mathrm{a2} = K_\mathrm{w} / K_{\mathrm{b2},\ce{Ca(OH)2}}$


Lemon juice, the ultimate anti-foaming agent?

A lemon has an oil that is present mostly in the skin. For instance if you squeeze lemon into hot tea in a styrofoam cup and let the tea sit, then you'll see a surface roughness at the liquid level. The oil actually dissolves the styrofoam. The oil acts a defoamer. So with a styrofoam cup you're suppose to use "commercial" lemon juice which has no oil (lest the cup dissolve and spill the liquid).


Where exactly do the orbital angular momentum get their numbers e.g. +1/2

$$
\newcommand{\linop}[1]{\hat{#1}}
\newcommand{\ramuno}{\mathrm{i}}
\newcommand{\ket}[1]{\,|{#1}\rangle}
$$

Classical Mechanics

In classical mechanics, the angular momentum, $\vec{L}$, of a particle with respect to a chosen origin is given by the vector product of its position vector relative to the origin, $\vec{r}$, and its linear momentum, $\vec{p}$,
\begin{equation*}
    \vec{L} := \vec{r} \times \vec{p} \, .
\end{equation*}
By expanding $\vec{r}$ and $\vec{p}$ over a standard basis $\{ \vec{i}, \vec{j}, \vec{k} \}$ of a Cartesian coordinate system as $\vec{r} = x \vec{i} + y \vec{j} + z \vec{k}$ and $\vec{p} = p_x \vec{i} + p_y \vec{j} + p_z \vec{k}$ and then writing the cross product as the formal determinant [1] and expanding it one could obtain
\begin{equation}
    \vec{r} \times \vec{p}
    =
    \begin{vmatrix}
     \vec{i} & \vec{j} & \vec{k} \\
           x &       y &       z \\
         p_x &     p_y &     p_z \\
    \end{vmatrix}
    =
    (y p_z - z p_y) \vec{i} - (x p_z - z p_x) \vec{j} + (x p_y - y p_x) \vec{k} \, ,
\end{equation}
which implies that components $(L_x, L_y, L_z)$ of the angular momentum in terms of which it can be expressed over a standard basis as $\vec{L} = L_x \vec{i} + L_y \vec{j} + L_z \vec{k}$ are given as follows
\begin{equation}
 L_x = (y p_z - z p_y) \, ,
 \quad
 L_y = (z p_x - x p_z) \, ,
 \quad
 L_z = (x p_y - y p_x) \, .
\end{equation}
The magnitude, $L$, of the angular momentum is given as usual as the square root of the scalar product of the vector with itself, $L := \sqrt{\vec{L}\vec{L}}$, or, in terms of components,
\begin{equation*}
    L := \sqrt{L_x^2 + L_y^2 + L_z^2} \, .
\end{equation*}

Quantum Mechanics

Operators

The quantum mechanical analogue of classical angular momentum is known as the orbital angular momentum and the operators for the components of orbital angular momentum can be obtained by simply replacing the components of position and momentum in classical mechanical expressions by the corresponding operators
\begin{equation}
    \linop{L}_{x} = (\linop{y} \linop{p}_z - \linop{z} \linop{p}_y) \, ,
    \quad
    \linop{L}_y = (\linop{z} \linop{p}_x - \linop{x} \linop{p}_z) \, ,
    \quad
    \linop{L}_{z} = (\linop{x} \linop{p}_y - \linop{y} \linop{p}_x) \, .
\end{equation}
and the square of the magnitude of the angular momentum operator can be expressed in terms of these operators as follows
\begin{equation}
    \linop{L}^2 = \linop{L}_{x}^2 + \linop{L}_y^2 + \linop{L}_{z}^2 \, .
\end{equation}
Then the following commutation relations of the angular momentum operators can be easily established:
\begin{equation}
    [ \linop{L}_{x}, \linop{L}_{y} ] = \ramuno \hbar \linop{L}_{z} \, ,
    \quad
    [ \linop{L}_{y}, \linop{L}_{z} ] = \ramuno \hbar \linop{L}_{x} \, ,
    \quad
    [ \linop{L}_{z}, \linop{L}_{x} ] = \ramuno \hbar \linop{L}_{y} \, ,
\end{equation}
\begin{equation}
    [\linop{L}^2, \linop{L}_{x}] = [\linop{L}^2, \linop{L}_y] = [\linop{L}^2, \linop{L}_{z}] = 0 \, .
\end{equation}

Interestingly, in quantum mechanics there exist a different kind of angular momentum, the intrinsic angular momentum, or spin, which does not have a counterpart in classical mechanics.
While the intrinsic angular momentum components operators does have a different form from that of orbital angular momentum, the commutation relations between them and the square of the intrinsic angular momentum magnitude are exactly the same.

More interestingly, the angular momentum can be treated in such a way that the treatment relies entirely on the commutation relations just introduced, and as such, valid for any kind of angular momentum.
Thus, for generality we can use the letter $J$ for any kind of angular momentum reserving the letter $L$ specifically for orbital one.
\begin{equation}
    [ \linop{J}_{x}, \linop{J}_{y} ] = \ramuno \hbar \linop{J}_{z} \, ,
    \quad
    [ \linop{J}_{y}, \linop{J}_{z} ] = \ramuno \hbar \linop{J}_{x} \, ,
    \quad
    [ \linop{J}_{z}, \linop{J}_{x} ] = \ramuno \hbar \linop{J}_{y} \, ,
\end{equation}
\begin{equation}
    [\linop{J}^2, \linop{J}_{x}] = [\linop{J}^2, \linop{J}_y] = [\linop{J}^2, \linop{J}_{z}] = 0 \, .
\end{equation}

Eigenvectors and eigenvalues

According to the commutation relations above, $\linop{J}^2$ commutes with all three components $\linop{J}_{x}, \linop{J}_y, \linop{J}_{z}$ while the components do not commute with each other.
Thus, $\linop{J}^2$ and one (and only one) of the components operators, conventionally chosen to be $\linop{J}_{z}$, admit the set of simultaneous eigenvectors, and using the so-called ladder operators technique it can be shown that,
\begin{align}
    \linop{J}^2 \ket{j(j+1),m} &= j(j+1) \hbar^{2} \ket{j(j+1),m} \, , \\
    \linop{J}_{z} \ket{j(j+1),m} &= m \hbar \ket{j(j+1),m} \, ,
\end{align}
where $\ket{j(j+1),m}$ is a simultaneous eigenvector of $\linop{J}^2$ and $\linop{J}_{z}$, $j$ could take non-negative integer and half-integer values,
\begin{equation*}
    j = 0, 1/2, 1, 3/2, \cdots \, ,
\end{equation*}
and $m$ ranges from $-j$ to $j$ in integer steps
\begin{equation}
    m = -j, -j+1, \cdots, j-1, j \cdots \, .
\end{equation}

Then, as I already said in my comment with respect to half-integer values of $j$, for orbital angular momentum, where the Born interpretation requires cyclic boundary conditions to be satisfied, only integer values are admissible, but where cyclic boundary conditions are not relevant, as for the intrinsic angular momentum known as spin, the half-integer values might be appropriate.

Epilogue

I'm still not sure what this question is about, but if it specifically about the spin of an electron, i.e. "How do we now that its component in any direction is $\pm 1/2$ in the units of $\hbar$?", then we know it both from the experiment (the Stern–Gerlach one) and the theory (the quantum field theory). And yes, for electron only these values are possible, but for other particles there might be different possibilities. Thus, for instance, a photon have component of the spin in any direction  $\pm 1$ in the units of $\hbar$.

[1] Here, "formal" means that this notation has the form of a determinant, but does not strictly adhere to the definition; it is a mnemonic used to remember the expansion of the cross product.


Why does freezing point decrease when vapour pressure decreases in dilute solutions?

Consider graphs of the vapor pressure of the liquid and of the solid, both as a function of temperature. Intuitively, both graphs increase with temperature, and intersect at the freezing point. By lowering the vapor pressure of the liquid, we lower the corresponding vapor pressure curve and hence the temperature at the point of intersection.


Chemical equilibria and changes in concentration

Assuming the following reaction:

$$\ce{A + B <=> C + D}$$

Where:

$$K_\text{c}=1.23\times10^{-2}$$

We say that:

$$K_\text{c}=\frac{\ce{[C]*[D]}}{\ce{[A]*[B]}}=1.23\times10^{-2}$$

And the reaction will be at equilibrium for any concentrations (actually, it should be activities, which are in essence effective concentrations) of A, B, C and D that satisfy the above.

I can understand your confusion. What they mean is... If you have a reaction at equilibrium, and remove half the reactants, then the reaction will proceed towards reactants, until equilibrium is once again reached.

  Their way of saying it:

You remove reactants, and the position of equilibrium is shifted towards the left, because the reaction will now proceed towards the left hand side to restore equilibrium.

It's confusing, because equilibrium occurs at the same conditions as before, so you haven't really shifted the equilibrium position in the way it sounds like.


What do 1T and 2H mean in front of compounds?

These are notations that are used in identifying polytypes in layered compounds.
2H indicates 2 layers per Hexagonal unit cell (H for hexagonal).
1T indicates 1 layer per Trigonal unit cell (T for trigonal).
At least, that's my understanding.


Why does pH increase as a weak acid becomes more dissociated?

A weak acid (HA) in water: $\ce{HA + H2O <=> H3O+ + A-}$, and thus,

$$K_\mathrm{a} = \frac{\ce{[H3O+][A-]}}{\ce{[HA]}}$$

By Le Chatelier's principle, if you add some $\ce{H3O+}$ from an external source (see Mithoron's comment above) to increase $\mathrm{pH}$, the equilibrium favors the backward reaction in order to reduce some $\ce{A-}$. Consequently, the amount of the dissociation of acid decreases and $K_\mathrm{a}$ will remain constant.

In similar manor, if you decrease $\mathrm{pH}$ of the solution by adding some $\ce{OH-}$, that will react with equal amount of $\ce{H3O+}$. Therefore, the equilibrium favors the forward reaction in order to produce some $\ce{H3O+}$ to keep $K_\mathrm{a}$ constant. Accordingly, $\ce{[A-]}$ in the solution increases while $\ce{[HA]}$ decreases, meaning the amount of the dissociation of acid increases. At one point, $\ce{[A-]} = \ce{[HA]}$, and hence, $K_\mathrm{a} = \ce{[H3O+]}$. At that point, therefore, $\mathrm{pH} = \mathrm{p}K_\mathrm{a}$.

Applying this to your example ($\ce{HA} = \ce{HF}$), $\mathrm{p}K_\mathrm{a}$ of $\ce{HF}$ is $3.14$ ($K_\mathrm{a} = 7.2 \times 10^{-4}$).

When $\pu{1.0 mol}$ of $\ce{HF}$ is dissolve in $\pu{1.0 L}$ of deionized water (no added external acid to adjust $\mathrm{pH}$), $\ce{[HF]} = \pu{1.0 mol\cdot L^{-1}}$ and $\ce{[F-]} = \ce{[H3O+]} = \alpha \gg 1.0$. Thus,
$$\ce{[H3O+][A-]} = K_\mathrm{a} \cdot \ce{[HA]}$$
$$\alpha^2 = K_\mathrm{a} \cdot 1.0~~~~\text{or}~~~~\alpha = \sqrt{K_\mathrm{a}}= \sqrt{7.2 \times 10^{-4}}= 2.7 \times 10^{-2}$$

Therefore $\mathrm{pH}$ of the solution is as follows:

$$\mathrm{pH} = -\log \ce{[H3O+]} = -\log \alpha = -\log 2.7 \times 10^{-2}= 1.57$$

Thus, at $\mathrm{pH} = 1.57$, the ionization of the solution is: $\ce{[F-]} = \ce{[H3O+]}= \pu{2.7 \times 10^{-2} mol \cdot L^{-1}}$.

Now, if you added an external base to the solution to increase its $\mathrm{pH}$ until it becomes $\mathrm{pH} = 3.14$, then, at $\mathrm{pH} = 3.14$, $\ce{[F-]} = \ce{[HA]}= \pu{0.5 mol \cdot L^{-1}}$.


Are orbitals always filled in from closest to nucleus to farthest away?

As you have studied for the quiz, you should know the expected answer.

In full-scope reality, the question has nothing to do with the two explanations you offered, but with the shape of atomic orbitals:

Radial distribution of selected orbitals in $\ce{Sm^{3+}}$, http://pubs.rsc.org/en/content/articlehtml/2004/dt/b407173e


How can I make a black precipitate without a sulfide?

Hint:  Manganese dioxide is pretty black.

Look up potassium permanganate and note its reactions especially the reduction in neutral solutions.


How to check whether intramolecular hydrogen bonding takes place in cyclic compounds?

From a structure diagram alone it is hard to determine these interactions. The reason for this is that important structural parameters are stretched or shortened in order to flatten the molecule for a 2D drawing.

In a first order approximation a molecular modelling kit based on balls and sticks could already be very helpful. It'll let you approximate the 3D structure and get a 'feeling' (also known as chemical intuition) about where to expect close contacts, or what bonds are strained, etc.; just to play around with it will give you a better visualisation of the shapes of molecules in general.

More advanced is using computers. While this might seem intimidating at first, it is actually something you can do at home. I've previously written extensively about how to do these identifications in this Q&A: How to identify hydrogen bonds and other non-covalent interactions from structure considerations? There are quite a few programs out there that let you build ball and stick models of compounds using tabulated bond lengths. That is probably already enough to identify likely positions for these interactions. Many of these molecular editor programs can run optimisations on a quick but low level, too. My personal favourite is Molden (very lightweight), but Avogadro does the job equally well, and probably has a nicer user interface. (Geoff, if you read this, we are all still hoping for a major update.)

Obviously you can increase the level at which you are running the calculations; for more complex structures you may need to. A lot of this can still be done with a home computer. You can start out with semi-empirical models, of which I recommend the fantastic open-source xtb, go on to ab-initio wave function methods (esp. HF), climb Jacob's ladder (AIP Conference Proceedings 2001, or see these slides by Perdew via helper.ipam.ucla.edu), and finally make your way to the various post-HF methods. (the last one you probably don't want to do on a laptop computer.) In this context I would like to point you towards Orca.

For the two molecules you mentioned, I've done the analysis with xtb and Multiwfn.

With these two molecules it is still quite simple, because they are flat (mostly at least; they are of course not static). Here is a model of the 3D structures with the hydrogen bond measured. For ortho-fluorophenol it is 2.25 Å and for ortho-nitrophenol it is 1.73 Å. You can see a clear difference already.

And here are the non-covalent interactions from Multiwfn analysis (Option 20, 2, visualised with ChemCraft at contour value 0.2 and in range ±0.05):

The red coloured blobs correspond to repulsive interactions, while the green and blue coloured values correspond to attractive interactions. As you can see, there is no hydrogen bonding in ortho-fluorophenol, but there is some in o-nitrophenol.


You don't need the Ideal Gas Law. You can use it (and I'll show you how). The question asks for partial pressure. You want Dalton's Law of Partial Pressures.

  Dalton's Law

Dalton's Law states that the total pressure in a system is equal to the sum of the partial pressures of the components:

$$P_T=P_A + P_B + P_C + ...$$

At constant volume and temperature (which you have, even if you don't know the volume), the ideal gas law can be rearranged so that:

$$\frac{P}{n}=\frac{RT}{V}$$
$$\frac{P_T}{n_T}=\frac{P_A}{n_A}$$
$$P_A=\frac{n_A}{n_T}P_T$$

  Ideal Gas Law

You can find the volume of the system because you have the total moles $n_T$ and the total pressure $P_T$:

$$V=\frac{n_T RT}{P_T}$$

You can then replace the number of moles with $n_A$ and solve for $P_A$;

$$P_A=\frac{n_A RT}{V}$$

This approach is basically equivalent to the previous except:

Using the ideal gas law you have to do two calculation (volume and partial pressure), while you only need to do one with Dalton's Law
Fewer math steps means less rounding, which means less error and higher precision.
If you use Dalton's Law, you don't have to worry about the units of temperature or pressure. Temperature is not in the equation, and the pressures are a ratio. You can use whatever pressure unit you want without converting.
You don't need to worry about $R$ and its units in the Dalton's Law approach.

  Solve problems as ratios whenever you have the chance.


Notation for specific/volumetric entropy

Specific entropy

According to IUPAC "Green Book" Quantities, units, and symbols in physical chemistry, specific entropy is denoted as lowercase latin "s": $s$ [1, p. 56], whereas $\bar{S}$ would refer to molar entropy:

  \begin{array}{lll}
\text{Name} & \text{Symbol} & \text{Definition} & \text{SI unit} & \text{Notes} \\
\hline
[...]\\
\text{molar quantity}~X & X_\mathrm{m}, (\bar{X}) & X_\mathrm{m} = X/n & [X]/\pu{mol} & 5,6 \\
\text{specific quantity}~X & x & x = X/m & [X]/\pu{kg} & 5,6 \\
[...]\\
\end{array}
  [...]

  $(5)$ The definition applies to pure substance. However, the concept of molar and specific quantities (see Section 1.4. p. 6) may also be applied to mixtures, n is the amount of substance (see Section 2.10, notes 1 and 2, p. 47).

  $(6)$ $X$ is an extensive quantity, whose SI unit is $[X]$. In the case of molar quantities the entities should be specified.

  Example $V_\mathrm{m,\ce{B}} = V_\mathrm{m}(\ce{B}) = V/n$ denotes the molar volume of $\ce{В}$.

Just as specific heat capacity $c$, specific entropy $s$ is measured in $\pu{J K-1 kg-1}$ [1, p. 90].

General note [1, p. 6]:

  The adjective specific before the name of an extensive quantity is used to mean divided by mass. When the symbol for the extensive quantity is a capital letter, the symbol used for the specific quantity is often the corresponding lower case letter.

Volumetric entropy

Quite often there is no special notation used, and volumetric entropy is denoted with $S$. Specifically, volumetric entropy generation rate for the convective heat transfer in flowing viscous fluid is often denoted with S-prime notation, e.g. from classical paper "A Study of Entropy Generation in Fundamental Convective Heat Transfer" [2]:

  \begin{align}
& S' & [\pu{W K-1 m-1}] \\
& S'' & [\pu{W K-1 m-2}] \\
& S''' & [\pu{W K-1 m-3}]
\end{align}

References

IUPAC “Green Book”. Quantities, units, and symbols in physical chemistry, 3rd ed.; Cohen, R. E., Mills, I., Eds.; IUPAC Recommendations; RSC Pub: Cambridge, UK, 2007. ISBN 978-0-85404-433-7.
Bejan, A. J. Heat Transfer 1979, 101 (4), 718–725. DOI: 10.1115/1.3451063.


Is this the right conclusion for a thought experiment?

This conclusion assumes that the enthalpy of the water remains entirely constant, which is not the case. Given that the water is being uniformly heated, but no end to the heating was specified, the enthalpy of the water will continue to increase, and thus force more and more of the liquid water into vapor form. Additionally, as you specified a very large surrounding, any water that enters the gaseous form would not largely impact the air pressure of the environment, so there would be few things acting to maintain some kind of equilibrium here.


Yes, it is possible, but only to a very small amount.

The $\mathrm pK_\mathrm a$ of the hydrazinium (2+) ion in water is $-1$ (CRC Handbook of Chemistry and Physics, Ed. D.R. Lide, CRC Press, Boca Raton, FL, 2005), so the equilibrium

$$\ce{N2H5+ + H2O <<=> N2H6^2+ + OH-}$$

is almost completely on the left side.


Why does raising the temperature affect the mass of dissolved solute?

For many (though not all) solids dissolved in water, solubility increases with temperature. The increase in kinetic energy that comes with higher temperatures allows the solvent molecules to more effectively break apart the intermolecular forces of the solute molecules.

For many salts and some gasses there are visual representations of solubility variation with temperature called a solubility curve, for which the units are $\frac{\mathrm{g~solute}}{\mathrm{100~g~solvent}}$. I could not find one for $\ce{KOH}$, but the Wikipedia page for $\ce{KOH}$ lists the solubility of $\ce{KOH}$ at various temperatures, and it is clear the trend is that it is more soluble at higher temperatures.

As a result of Le Châtelier's principle, substances that release heat upon being dissolved tend to dissolve better in colder water. This accounts for gasses and some salts.

You should note, however, that although $\ce{KOH}$ does evolve a large amount of heat upon being dissolved, as the concentration of $\ce{KOH}$ increases, the process becomes less exothermic, and eventually near the saturation point it becomes endothermic (it becomes less favorable to dissolve more $\ce{KOH}$), which explains why it is more soluble at higher temperatures.


Measuring a high Michaelis constant using fluorescence

Based on your description, it sounds like you are measuring the production of hydrogen peroxide by an oxidase enzyme. In that case, it is not necessary for the Amplex Red to be at the same concentration as the substrate. You only need enough reagent to determine the initial rate of reaction at each substrate concentration, ie when only a small portion of the substrate has been consumed.
If $\pu{5 μM}$ is not sufficient for that purpose, you can determine the resorufin (produced by reaction of Amplex Red with $\ce{H2O2})$ by absorbance rather than fluorescence for lower sensitivity and higher maximum. According to the Thermo Fisher assay kit info, resorufin has an extinction coefficient of $\pu{58000 M^-1 cm^-1},$ so with a standard $\pu{1 cm}$ path length cuvette, your linear range would probably go up to around $\pu{50 μM}$ depending on your spectrophotometer. With a short path cuvette (these are available with $\pu{0.2 cm}$ and $\pu{0.1 cm}$ path lengths), you could go quite a bit higher, certainly more than sufficient for your needs.
As for why you cannot just scale the concentrations down, consider the limiting case where the Michaelis constant is equal to the $K_\mathrm{d}$ of the substrate. We have that
$$K_\mathrm{d} = \frac{[\ce{E}][\ce{S}]}{[\ce{ES}]},$$
so $[\ce{S}] \approx K_\mathrm{d}$ when $[\ce{E}] = [\ce{ES}]$ (assuming $[\ce{S}]_\mathrm{tot} \gg [\ce{ES}]).$ Thus, the absolute concentration of $\ce{E}$ is not a factor, as long as it is small enough that the amount of $\ce{S}$ bound does not substantially change the concentration of free $\ce{S}.$ You should get the same result for $K_\mathrm{M}$ regardless of what concentration of enzyme you use.


Is there a way to oxidize salicyl alcohol to salicylic acid without using chromium trioxide?

@Waylander: A general Procedure given in J. Org. Chem. 1999, 64(7), 2564–2566 as follows:

  Typical Procedure. A mixture of alcohol 1 ($\pu{40 mmol}$), TEMPO ($\pu{436 mg}$, $\pu{2.8 mmol}$), $\ce{MeCN}$ ($\pu{200 mL}$), and sodium phosphate buffer ($\pu{150 mL}$, $\pu{0.67 M}$, $\pu{pH = 6.7}$) is heated to $\pu{35 °C}$. Then mixtures of $\pu{9.14 g}$ of 80% sodium chlorite ($\ce{NaClO2}$, $\pu{80.0 mmol}$) in $\pu{40 mL}$ of water and 2.0 mol% dilute bleach ($\pu{1.06 mL}$ of 5.25% $\ce{NaOCl}$ diluted into $\pu{20 mL}$ with water) are added simultaneously over $\pu{2 h}$ (Caution! Do not mix bleach and $\ce{NaClO2}$ before being added to the reaction mixture). The mixture is stirred at $\pu{35 °C}$ until the reaction is complete (<2% starting material, $\pu{2−5 h}$ reaction time), then cooled to room temperature. Water ($\pu{300 mL}$) is added, and the $\mathrm{pH}$ is adjusted to $8.0$ with $\pu{2.0 N}$ $\ce{NaOH}$ (about $\pu{48 mL}$). The reaction is quenched by pouring into cold ($\pu{0 °C}$) $\ce{Na2SO3}$ solution ($\pu{12.2 g}$ in $\pu{200 mL}$ of water) maintained at $\pu{<20 °C}$. The $\mathrm{pH}$ of the aqueous layer should be $8.5$−$9.0$. After stirring for $\pu{0.5 h}$ at room temperature, methyl tert-butyl ether (MTBE, $\pu{200 mL}$) is added. The organic layer is separated and discarded. More MTBE ($\pu{300 mL}$) is added, and the aqueous layer is acidified with $\pu{2.0 N}$ $\ce{HCl}$ (~$\pu{100 mL}$) to $\pu{pH = 3−4}$. The organic layer is separated, washed with water ($2 \times \pu{100 mL}$) and brine ($\pu{150 mL}$), and then concentrated to give the crude carboxylic acid 2 in 85−100% yield. The products may be purified by silica gel column chromatography or crystallization.


Clemmensen reduction vs. Wolff-Kishner reduction

A key consideration for the Wolff-Kishner reduction is thermal stability of the substrate. High temperature is required for the classical reaction using hydrazine and KOH, sometimes up to 200 °C. This can rule out plenty of compounds.
The Clemmensen reduction may be fine for the laboratory,  but fundamentally many industries wouldn't use it due to scaling up the use of mercury.
Scalable alternatives are molecular hydrogen with a catalyst under acidic conditions or, the trifluoroacetic acid-silane combination, such as triethylsilane or tetramethyldisiloxane.


Analysis of salts using color change when heated

No when cold they can be of different colours as can be seen from the following table:


Can I use a 1:1 mixture of isopentyl alcohol and water as mobile phase to carry out paper chromatography of a mixture of red and blue ink?

The principle of chromatography is that you can always try (as per the question in the title). This can be beneficial: in my PhD thesis I had two diastereomers that were hard to separate on TLC using hexanes and ethyl acetate, but a hexanes/toluene mixture worked wonders.

However, in your case you should not expect better results. For one, as the commentators have pointed out the two liquids are not sufficiently soluble in each other. For two, your separation is already pretty good and it seems unlikely that it can be improved even more.


NMR Splitting for Adjacent, Equivalent Hydrogens?

There are two ways to look at this.

a) All transitions you can see are between a state where the spins are parallel and a state where they are antiparallel.

$aa <-->  ab$  or  $ab <--> bb$

Their distances are identical, as are the states $ab$ and $ba$, so, as porphyrin said in his comment, you only see one line.

b) If the two protons were chemically different, you would see two doublets.

If the difference in CS were small, you'd see the "roof effect": An AB spin system. The two inner lines come closer and grow at the cost of the outer lines.

But there is no difference in CS, and so the outer lines vanish, and the two inner lines coalesce into one singlet.


How to deterime the number of equivalents of Grignard reagent used for this reaction?

No you are right it should be five. there are three dissociative protons thus one from the secondary amine, and two from the hydroxy groups.  These must be reacted first then you need two to react with the ester. The question was made by a human and the book answer of 2 is probably just an error from such.


Amylase can’t digest glycogen because of its inability to attack the branching (1→6) linkages.

Perhaps, another very important reason is controlling the rate of glycogen metabolism through glycogen phosphorylase. Just like any other biological system, regulation of metabolic substrates and/products is crucial to maintaining the balance (homeostasis) so to prevent excess glucose production from glycogen metabolism or to little according the needs of the organism.

Tl;dr

There are  actually three forms of amylases:

α-amylase (an endoglycosidase, which can hydrolyze a glycosidic linkage anywhere along the chain to produce glucose and maltose).
β-amylase (an exoglycosidase that cleaves from the nonreducing end of the polymer).
gamma amylase

Now starch consists of two main components amylose and amylopectin;

Amylose is a linear polymer of several thousand glucose residues linked by (1→4)
bonds:

Amylopectin consists mainly of (1→4)-linked glucose residues but is a branched molecule with (1→6) branch points every 24 to 30 glucose residues on average.
The primary structure of glycogen resembles that of amylopectin(not amylose), but glycogen is more highly branched, with branch points occurring every 8 to 14 glucose residues:

Digestion

The digestion of starch, the main carbohydrate source in the human diet, begins in the mouth. Saliva contains α-amylase, which randomly hydrolyzes all the (1→4) glycosidic bonds of starch except its outermost bonds and those next to branches.
On the other hand glycogen being a highly branched molecule, it is evident that amylase won’t be a good enzyme to digest it due to physical structure constraints.
 Substrate specificity of the enzymes ensure that the appropriate substrate which is highly optimized for the active site will be broken down.

Although glycogen posseses parts similar to starch's amlyopectin its structure is not optimized to be broken down by amylases. Note: even these amylases only partially break down starch to different oligosaccharides


What fluorescent materials absorb visible light and emit UVC light?

It is not possible in a single molecule as energy conservation will forbid it. There is not enough thermal energy, $k_BT \approx 210 $ wavenumbers, visible to uv thousands of wavenumbers.  It is sort of possible using triplet-triplet annihilation ($\ce{T + T \to S^* + S}$ ) but this needs significant initial excitation and long lived triplets, which means de-oxygenated solution, and produces the same singlets as were excited initially. There are also inorganic up-conversion materials, for example some lanthanides, such as ytterbium based ones will convert near ir light to visible light.


In the Henderson-Hasselbalch equation, $K_\mathrm{a}$ is a product of concentrations and considered a constant.

In reality, $K_\mathrm{a}$, when defined as a product of concentrations, is not a constant:

Upon dilution (decrease in ionic strength) the $\mathrm{p}K_\mathrm{a}$ will change, and therefore the pH of the solution will change.

In addition to the above reason, pH will always approach 7 at extreme dilution as it approaches being pure water.


Concentration is an intensive property. The value of the property does not change with scale. Let me give you an example:

  Let us say you had a homogenous mixture (solution) of sodium carbonate in water prepared from 112 g of sodium carbonate dissolved in 1031 g of water.

The concentration (in mass percent, or mass of solute per mass of solution) is:

$$c=\frac{112\text{ g solute}}{(112+1031)\text{g solution}}=0.09799 =9.799\%\text{ sodium carbonate by mass}$$

The concentration is the ratio of sodium carbonate to the total mass of the solution, which does not change if you are dealing with the entire 1143 g of the solution or if you dispense some of that solution into another vessel.

  If you dispense 11.7 g of that solution into a flask for a reaction, what is the concentration of sodium carbonate in that flask?

It is still 9.799% by mass. The ratio of the mass of sodium carbonate present to the total mass present has not changed. The actual mass of sodium carbonate has changed:

$$0.09799\dfrac{\text{g solute}}{\text{g solution}}\times11.7\text{ g solution}=1.15\text{ g solute}$$

The concentration is a property dependent only on the concentration of the solution, not the amount of solution you have. The concentration of a solution with defined composition is independent of the size of the system.

In general, any property that is a ratio of two extensive properties becomes an intensive property, since both extensive properties will scale similarly with increasing or decreasing size of the system.

Some examples include:

Concentration (including molarity) - ratio of amount of solute (mass, volume, or moles) to amount of solution (mass or volume usually)
Density - ratio of mass of a sample to the volume of the sample
Specific heat - ratio of heat transferred to a sample to the amount of the sample (mass or moles usually, but volume also)

Each of these intensive properties is a ratio of an extensive property we care about (amount of solute, mass of sample, heat transferred) divided by the scale of the system (amount of stuff usually). This is like finding the slope of a graph showing the relationship between two extensive properties. The graph is linear and the value of slope does not change based on how much stuff you have - thus the slope (the ratio) is an intensive property.


Does the electrolysis of water produce chlorine gas?

It's the tablespoon of salt he added, giving the chlorine when electrolysis is done. I have done electrolysis with salt-free water (though a bit acidic to get the water to electrolyze faster) and chlorine was nowhere to be found.

Electrolysis of saltwater produces hydrogen and chlorine; and leaves sodium hydroxide behind.


What is the correct Lewis structure of diazomethane?

The image above depicts three mesomeric structures of diazomethane ($\ce{CH2N2}$). When you count the valence electrons for every atom, you will find that only the left and middle structure satisfy the octet / duet (for hydrogen) rule, i.e. that every $\ce{C}$ and $\ce{N}$ atom has 8 valence electrons, and every $\ce{H}$ has two valence electrons. The structure on the right side does not obey the octet / duet rule because the positively-charged nitrogen only has an electron sextet. The same is true for the two structures you have drawn with an $\ce{N-N}$ single bond. In both cases, at least one nitrogen has only 6 valence electrons.


This is because the Gibbs free energy also depends on the concentration of the reactants and products. Hence, for a reaction, if the energy change is negative it moves forwards, forming more product, which also changes the concentrations of the species involved. The Gibbs energy change at this new concentration will be less negative than that in the initial case, but the reaction still proceeds since the reaction is accompanied by a negative change of free energy.

A point comes, where the concentrations of the reactions and the products is such, that the Gibbs energy change now becomes zero for the particular reaction. This is the equilibrium condition (minimum free energy, maximum entropy). If the energy change is positive, the reverse reaction becomes favorable, again inching towards the zero energy change condition, i.e. equilibrium. So, the reaction is already at the minimum Gibbs energy when it is at equilibrium. Further proceeding of the reaction will be accompanied by a positive change, and hence a rise in energy.

For how it depends on the concentrations, see these questions

Why does the reaction quotient use the products (multiplications) of reactants and products, rather than their respective sums?
Derivation of relationship between equilibrium constant and Gibbs free energy change


Why are alkali hydrides strong reducing agents?

Just like I said in the answer to you broader question, it's all about elements and their individual preferences. Hydrides (of any metals, not necessarily alkaline) contain hydride ion $\ce{H-}$, which is basically a hydrogen having one extra electron (or, in other words, a hydrogen in oxidation state -1). Now, hydrogen's electronegativity is relatively low for a non-metal, and so is its electron affinity. It can be said that hydrogen takes that extra electron almost reluctantly, and is ready to give it away to any oxidizer, even a weak one. This makes the hydride ion (and any compound which it is a part of) a powerful reducing agent.

In due time you will learn about the redox potentials which are a quantitative measure of the compound's oxidizing or reducing power. (Yes, they may depend on temperature and other things.) I'd rather not touch on that right now, for it's a long story.


Analytical Chemistry Textbook Recommendation

Fundamentals of Analytical Chemistry is a fine book and it does have an student solution.


On an industrial scale, the preferred method for the production of potassium hydroxide from potassium chloride is electrolysis (diaphragm, mercury, or membrane process).

$$\begin{align}
\ce{2Cl- &-> Cl2 + 2e-}\\
\ce{2H2O + 2e- &-> H2 + 2OH-}\\
\hline
\ce{2Cl- + 2H2O &-> Cl2 + H2 + 2OH-}
\end{align}$$

or

$$\ce{2KCl + 2H2O -> 2KOH + Cl2 + H2}$$


Why are complex basis functions not used in electronic structure calculations?

The basic answer is rather simple: In the non-relativistic limit it is just not necessary. There is no complex operator and hence nothing which could change the complex component of your basis functions. Using complex basis functions would only double the required work (i.e., computer time). There are very efficient schemes in order to calculate the standard integrals (McMurchie-Davidson, Obara-Saika, ...). I don't think that the representability of MOs is a reason for using real basis functions, since, strictly speaking, you can interprete only the square of the MOs and not their amplitude.

However, there are situations, where your MOs can be complex. This can happen when you have a complex operator in your equation. An example are the Pauli spin matrices of which $\sigma_y$ has only an imaginary compound. They occur, e.g., the Dirac equation (see here one formulation where they explicitly state the incorporation of the Pauli matrices). Complex MOs can be obtained either by using complex basis functions and real MO coefficients or by using real basis functions with complex coefficients. Starting from a non-relativistic implementation it might be simpler just to keep your basis functions real and to choose the MO coefficients complex.


Order of citation of alkyl and halide prefixes in IUPAC names

You correctly figured out that the prefixes that represent the substituent groups are arranged alphabetically. However, note that the order of citation and the numbering are laid down in two different rules.
With regard to the alphabetical order of citation of substituent prefixes, the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) reads as follows:

P-14.5 ALPHANUMERICAL ORDER
Alphanumerical order has been commonly called ‘alphabetical order’. As these ordering principles do involve ordering both letters and numbers, in a strict sense, it is best called ‘alphanumerical order’ in order to convey the message that both letters and numbers are involved
Alphanumerical order is used to establish the order of citation of detachable substituent prefixes (not the detachable saturation prefixes, hydro and dehydro), and the numbering of a chain, ring, or ring system when a choice is possible.
(…)
P-14.5.1 Simple prefixes (i.e., those describing atoms and unsubstituted substituents) are arranged alphabetically; multiplicative prefixes, if necessary, are then inserted and do not alter the alphabetical order already established.

Therefore, the alphanumerical order in your examples is correct:
x-ethyl-y-iodobenzene
x-iodo-y-propylbenzene
x-butyl-y-iodobenzene
x-iodo-y-pentylbenzene
The relevant rules concerning the numbering of locants for substituent prefixes are:

P-14.3.5 Lowest set of locants
The lowest set of locants is defined as the set that, when compared term by term with
other locant sets, each cited in order of increasing value, has the lowest term at the first
point of difference; (…)

and

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:
(…)
(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;
(…)

Note that Rule (f) takes precedence over Rule (g).
In accordance with Rule (f), your first example could be named as 1-ethyl-2-iodobenzene as well as 2-ethyl-1-iodobenzene since both names correspond to the locant set ‘1,2’. However, according to Rule (g), this example is named as 1-ethyl-2-iodobenzene rather than 2-ethyl-1-iodobenzene since ethyl is cited first as a prefix in the name.
Hence, the numbering in your examples is correct:
1-ethyl-2-iodobenzene
1-iodo-2-propylbenzene
1-butyl-2-iodobenzene
1-iodo-2-pentylbenzene


What's the standard deviation in bond angle for water molecules at a particular temperature?

I tried to answer by proposing a simple model. The model didn't work as well as I hoped, but the general procedure might serve as an answer.

A harmonic model

As a first approximation, you might consider a simple model for the energy of a single water molecule as a function of its angle. One possibility might be a harmonic function

$$E(\theta) = \frac{k \theta^2}{2}$$

where $k$ is an unknown parameter and $\theta$ is the angle. This is very simple and leaves out many complexities, of course.

If you had $E(\theta)$, the energy of a single water molecule at a particular angle, and if you assume an independent particle model (which may break down for a liquid), you could apply Boltzmann weighting

$$p(\theta) = \frac{1}{U}\exp\left(-\frac{E(\theta)}{R T}\right)$$

where $p(\theta)$ is the probability of finding a water molecule at a particular angle and $U = \int_0^\pi d\theta \exp\left(-\frac{E(\theta)}{R T}\right)$ is a normalization factor.

If you had a probability distribution $p(\theta)$ describing the spread of angles at a certain temperature, getting the mean and standard deviation is easy:

$$\langle\theta\rangle = \int_0^\pi d\theta\ \theta\ p(\theta)\\
\langle\theta^2\rangle = \int_0^\pi d\theta\ \theta^2\ p(\theta)\\
\sigma_\theta^2 = \int_0^\pi d\theta\ \left(\theta - \langle\theta\rangle\right)^2\ p(\theta) = \langle\theta^2\rangle - \langle\theta\rangle^2$$

So this would be how to get $\sigma_\theta^2$.
Below is my numerical attempt.
Beware: this model is too simple.

Numerical results

I did some calculations (mostly with SymPy) with the above and I came up with the following:

$$p(\theta) = \frac{1}{U} \exp\left(-\frac{k \theta^2}{2 R T}\right)\\
U = \sqrt{\frac{\pi R T}{2 k}} erf \left(\frac{\pi}{2} \sqrt{\frac{2 k}{R T}} \right)$$

$$\langle\theta\rangle = \frac{k}{R T U}\left(\exp\left( \frac{\pi^2 k}{2 R T} \right) -
 1 \right) \exp\left( -\frac{\pi^2 k}{2 R T} \right)\\
\sigma_\theta^2 = \frac{R T}{k} - \left[ \frac{\pi}{U} \exp\left( -\frac{\pi^2 k}{2 R T} \right) +
\left(\frac{RT}{kU}\right)^2 \left(\exp\left( \frac{\pi^2 k}{2 R T} \right) -
 1 \right)^2 \exp\left( -\frac{\pi^2 k}{R T} \right)
\right]$$

By substituting constants and temperature ($\pi = 3.14159265$, $R = 8.314459848$ J K$^{-1}$mol$^{-1}$ and $T = 293.15$ K) and setting the average above to as close as possible to 104.45° and solving numerically the complete least squares problem, I found an approximate solution for $k$: $384.05105591$ J/rad².

Sadly, this value is associated with an average angle of around 96°, which means the model is very poor.

Using this best $k$ gives us a $\sqrt{\sigma_\theta^2}$ of

$$\sqrt{0.508770928037803} = 0.713281801280393 \text{ rad}$$

which is around 41°. Again, a too simplistic model: the best prediction gives angles as 96°$\pm$41°!

A bad model

By expanding the average in series, I found the following

$$\langle\theta\rangle \approx 1.570796325 + 0.000530047067361739 k + O(k^2)$$

The second order term above was $- 7.86976333117302\times 10^{-7} k^2$, so I dropped it.
Observe the first term is pretty much 90° in radians. All other terms decay too quickly.

By doing the same expansion for the standard deviation the following comes out:

$$\sigma_\theta^2 \approx - \frac{4.54747350886464 \times 10^{-13}}{k} + 0.411233515772252 + 0.000388544793228128 k + O(k^2)$$

The next term in the series above was $- 4.34803406790969 \times 10^{-7} k^2$ so I dropped it. Again, a (wrong) zeroth order term dominates.


Firstly you should consider this simple model of atoms: There is a compact nucleus and smeared out electron clouds. In H this cloud is simply spherical. In many other atoms (e.g. O, C, N, S, etc.) the clouds are usually arranged like this (I did not find english site for some reason) (imagine tying 4 ballons together, forming a tetrahedron). This also explains the general geometry of water (and ice).

As you know in water the hydrogen is covalently bonded to oxygen. But due to the different properties of O and H the electrons of these bonds are not distributed symmetrically but much nearer the oxygen, leaving the hydrogen slightly positively charged. (Note that different elements cause different levels of asymmetry, a higher difference in electronegativity corresponds to a stronger asymmetry. This is why hydrocarbons ( C-H-bonds are rather apolar) are not very soluble in water.)

Now these positive H nuclei can bond to "full" electron clouds (e.g. the two of oxygen in water that are not bonded to hydrogen)

For something to be suluble in water you need to have many of those polar bonds and "free electron clouds", so e.g. sugars are easily soluble but e.g. alkanes are not.

These illustrations hopefully make now more sense. Sadly I could not find an english text about the tetrahedron model, which I consider didactically very useful.

Edit: Note that although hydrogen bonds are due to the electrostatic attraction, they are quite distinct from ionic bonds in the aspect that they have a definitive orientation, i.e. in ice the bonds still form tetrahedrons and not a cubic lattice or something like that.
I do not know a very good explanation but I think this is simply because the scale is much smaller; if the angle changes just a bit, the negatively charged part of the hydrogen atom interferes already.


Why does HgCl2 react with HCO3^- but not with CO3^2-?

Anirudh, which book is teaching this stuff? Sorry, I can only find this discussion in 1890s and the latest in my favorite Mellor's Treatise from 1930s.

Coming to 2020:

After doing a simple acid test for carbonates, a nice safe & sound method to distinguish carbonate vs. bicarbonate salt is

(a) to react it with a neutral barium chloride solution. If the solution is turbid, carbonate is present, otherwise bicarbonate. Of course the methods fail with a mixture of the two.

(b) Sodium Carbonate-Phenolphthalein Test: phenolphthalein is turned pink by soluble carbonates and but remains colourless by soluble bicarbonates

Your textbook is not completely correct when it states that bicarbonate does not react with $\ce{HgCl2}$. For your reference from A treatise on chemistry, Volume 2
by Henry Enfield Roscoe, Carl Schorlemmer, 1913, pg 699


Lye is very corrosive, just like acid, except opposite (hope that makes sense). Although lye is a very powerful chemical, and it is sufficient for many blockages, oxidizing agents (like bleach) may be helpful for breaking down paper/cellulose blockages.

Mixing bleach with acid will cause chlorine gas evolution... bad, bad, bad.

EDIT- The oxidative powers of hypochlorous acid and hypochlorite ion(s) compared:

Source: Descriptive Inorganic Chemistry, Third Edition (By Geoff Rayner-Canham, Tina Overton)

So in terms of power, yes hypochlorous acid is a stronger oxidizer, but it's not wildly different.

On a related note (see comments below) hypochlorous acid is a much better microbicide than the hypochlorite ion, because it penetrates cell walls more easily. However, regarding a drain cleaner, the oxidative difference is much less pronounced. And it's worth mentioning that at above pH 12, a cell will lyse (burst open) and become saponified, so that obviates whatever microbial effect hypochlorite or hypochlorous acid might have on bacteria... like being thrown into a poisonous wood chipper.


Part of the assumptions set of an ideal gas is that there are no attractive or repulsive forces between them.

Also, part of the assumption is that the volume of molecules obtain a negligible amount when compared to the overall volume of the gas.


Why does a reaction stop when turning into a solid?

The reason is due to the surface area of calcium carbonate available to react.

When reacting a block of solid calcium carbonate with dilute sulphuric acid, solid calcium sulphate is produced which is deposited on the surface of the calcium carbonate block (since that is where the reaction occurs). The calcium sulphate is insoluble in dilute sulphuric acid and so you end up with calcium carbonate coated in a layer of calcium sulphate, which stops the acid from getting at the carbonate and so stops the reaction.

In contrast, calcium nitrate, produced from the reaction with nitric acid, is highly soluble in dilute nitric acid and so no solid layer forms around the carbonate block and the reaction can continue until one of the reagents is used up.


Are there any early industrial age methods capable of determining diamond or graphite's crystalline structure?

Would 1850's level spectroscopy or other technology of the time be able to demonstrate the crystal structures of carbon (specifically graphite and diamond) without a 1900's understanding of atomic structure?

Yes, using visible light and petrographic microscopes. The field of optical mineralogy was mature enough by the mid-1800's for one to determine the crystal structure of minerals. In particular, the work of Sir William Nicol is worth investigating (thin but useful Wikipedia article).

I will add that there's a pretty simple way to determine whether or not the crystal structure of a mineral is the same or different than that of another mineral - the so-called crystal habit.

Finally, diamond exhibits conchoidal fracture, and graphite does not. This might be the easiest way to tell that the two minerals have different crystal structures.


Change in standard Gibbs energy is always zero?

No. $K$ is not a ratio of instantaneous activities! Let's say you hypothetically prepare a system where all components have unit activity. In that case you would have the reaction quotient

$$Q = \prod_i a_i^{\nu_i} = 1$$

but $K$ is the ratio of activities at equilibrium. If that hypothetical system is not at equilibrium, then it will go towards equilibrium. When it reaches equilibrium, the activities are no longer going to be equal to unity, and $K$ will not be equal to 1.


It is definitely synthetic, a trimethyl-ammonium derivative of polystyrene.

The first to use it as a human medication seems to be Stanley S. Bergen, Jr. et al.

See Effect of an Ion Exchange Resin on Serum Cholesterol in Man Proceedings of the Society for Experimental Biology and Medicine, Volume 102, 1959, Pages 676–679:

  It is well known that bile acids are a major
  end product of cholesterol metabolism and
  that cholesterol is probably the only precursor
  of bile acids([references]1-3 ) . Rate of oxidation of cholesterol
  to bile acids is believed to be regulated
  by need for bile acids for digestive purposes,
  cholesterol normally being oxidized at
  a rate sufficient to replace the bile acid that
  escapes reabsorption and is lost in the feces
  ([references] 4 - 5). If an agent could be found that would
  sequester bile acids and thereby promote their
  excretion in the feces, 2 consequences of clinical
  interest might be anticipated: (1) increase
  in rate of oxidative degradation of cholesterol.
  and (2) decrease in serum cholesterol level.
  Tennent ([reference] 6) described lowering of serum cholesterol
  levels in animals fed the chloride salt
  of a basic anion exchange resin. This material
  (MK-135) has marked affinity for bile
  acids in vitro and is believed to exert its serum
  cholesterol-lowering effect by virtue of this
  property( 6 ) . The present report summarizes
  results obtained when MK 135 was administered
  to 26 patients, many with elevated serum
  cholesterol levels.

Reference 6 is Tennent, D. M., Siegel, H., Zanetti, M. E., Kuron, C. W., Ott, W. H., Wolf, F. J., Circulation, 1959, v20, 996.

Even though Bergen's paper only uses the term "MK 135", according to the editor's note on the current online version, this paper is the first use of cholestyramine as a medication in humans.


Is it really appropriate to measure entropy change using enthalpy?

Yes.  So to get the change in entropy, you need to identify a reversible path between thermodynamic equilibrium state A and thermodynamic equilibrium state B.  If it is a phase change, you start out with liquid  at temperature T and equilibrium vapor pressure P(T) (state A), and end up with vapor at the same temperature and pressure (state B).  This can be done reversibly by putting one mole of a liquid in a cylinder with a massless frictionless piston and holding the external pressure on the piston constant at P while adding heat very gradually (reversibly) until all the liquid changes to vapor.  During this process, the volume of material in the cylinder increases from that of one mole of liquid to that of one mole of vapor.  Since the pressure is held constant throughout, the work done is $P(V_v-V_l)$.  From the first law, the change in internal energy is equal to the heat added minus the work done:  $(U_v-U_l)=Q-P(V_v-V_l)$.  Rearranging this equation gives:$$(U_v+PV_v)-(U_l+PV_v)=H_v-H_l=\Delta H_{lv}=Q$$So the heat that you add reversibly is equal to the change in enthalpy in going from the liquid at T and equilibrium vapor pressure P(T) to the vapor at T and equilibrium vapor pressure P(T).  Since this is a reversible process at constant temperature, $\Delta S=\frac{Q}{T}=\frac{\Delta H_{lv}}{T}$.  This completes what I wanted to say about the phase change situation.  I will move on to the chemical reaction situation after you have had a chance to look this over and ask questions.


Not really. Like lots of things with polar bonds, sulfuric acid can undergo ion transfer reactions in the liquid and make ionic species in solution.  And it is indeed protons that get transferred just like water.  But as with water, at least under ordinary conditions, these ionic species are only a minor portion of the liquid which is mostly $\ce{H2SO4}$ molecules.

Strong acids such as sulfuric acid are indeed acidic, capable of acting as acids.  But they still need a base with which they can react, to actually act as acids.  In the case of diluting sulfuric acid, water serves as such a base.


Calicheamicins are natural products, which were isolated from the fermentation broth of Micromonospora echinospora subsp. calichensis. These antibiotics exhibited significant activity against Gram-positive and Gram-negative bacteria in vitro. Calicheamicin $\gamma_1$ (Ref.1) demonstrated antitumor activity against P388 leukemia and B16 melanoma in vitro (Ref.2).

The structure of Calicheamicin $\gamma_1$ contains biologically active part of bicycloring system, which is described in the following diagram:

The rest of the structure contains four modified sugar units and one aromatic nucleus. The sugar 1 unit is a 6-deoxy-4-aminohexose unit (that's why it is called a modified sugar). Its anomeric oxygen (hemiacetyl $\ce{OH}$) bound to the $\ce{C}$9 position of the endiyne unit in $\beta$-linkage. The sugar 2 unit is a 2,6-dideoxy-4-thiohexose unit, hence it is also a modified sugar. This linkage is unusual among other sugar linkages in natural origin. The inkage is a hydroxyl amine bond ($\ce{-NH-O-}$) without eliminating $\beta$-$\ce{OH}$ of 2nd sugar unit. According to proposed biosynthetic pathway, these two sugars have been added to endiyne unit in first and second steps of the pathway:

The third step of the biosynthesis is addition of the aromatic nucleus (5-iodo-4-hydroxy-2,3-methoxy-6-methylbenzoate unit) to 4-thial group of sugar 2 as a thioester bond. The forth and fifth additions of other sugar units can be alternative. Nonetheless, one sugar (a 6-deoxy-3-O-methylhexose unit) attached to the free 4-phenolic $\ce{OH}$ group of aromatic nucleus in $\alpha$-linkage. The last sugar unit is attached with its $\alpha$-$\ce{OH}$ of the anomeric center to the free 2-$\ce{OH}$ group of the first sugar in $2 \rightarrow 1$ linkage. The unusual about this sugar is it is a pentose (a 2-deoxy-3-O-methyl-4-N-ethylaninopentose unit) instead of hexose among other three.

You can find a good description of mechanism of action of Calicheamicin $\gamma_1$ in Ref.3.

I hope this is what you want to know about this molecule.

References:

May D. Lee, Theresa S. Dunne, Marshall M. Siegel, Conway C. Chang, George O. Morton, Donald B. Borders, “Calichemicins, a Novel Family of Antitumor Antibiotics. 1. Chemistry and Partial Structure of Calichemicin  $\gamma_1$,” J. Am. Chem. Soc. 1987, 109(11), 3464-3466 (https://doi.org/10.1021/ja00245a050).
W. M. Maiese, M. P. Lechevalier, H. A. Lechevalier, J. Korshalla, N. Kuck, A.Fantini, M.J. Wildey, J. Thomas, M. Greenstein,  “Calicheamicins, a novel family of antitumor antibiotics: taxonomy, fermentation and biological properties,” The Journal of Antibiotics 1989, 42(4), 558-563 (DOI: 10.7164/antibiotics.42.558).
Miklós Simonyi, ”The Concept of Chiral Conformers and its Significance in Molecular Pharmacology,” In Advances in Drug Research: Volume 30; Bernard Testa, Urs A. Meyer, Eds.; Academic Press: San Diego, CA, 1997, pp. 73-111 (ISBN: 0-12-013330-X).


What is the mechanism of trialkylborane oxidation with hydroperoxide?

One aspect is the release of the stable anion $\ce{OH-}$.

Another aspect is that $\ce{B-O}$ bonds are much stronger than $\ce{B-C}$ bonds because the empty p-orbital of boron overlaps with one p-orbital with a lone electron pair of oxygen.

$\ce{B-C}$ $D_{298}^ \circ  = 448 \pm 29\ \mathrm{kJ\ mol^{-1}}$

$\ce{B-O}$ $D_{298}^ \circ  = 809\ \mathrm{kJ\ mol^{-1}}$

values taken from “Bond Dissociation Energies”, in CRC Handbook of Chemistry and Physics, 90th Edition (CD-ROM Version 2010), David R. Lide, ed., CRC Press/Taylor and Francis, Boca Raton, FL.


Disclaimer: I will be talking about the shear viscosity. I think that most of what I write is also applicable to extensional viscosity, although with different numerical constants, but I am not fully sure so I will limit myself to shear viscosity.

Most theoretical models predict that the shear viscosity of both solutions will be equal. The reason is that these models describe the viscosity of the suspension $\mu$ as the viscosity of the 'empty' liquid $\mu_0$ $+$ a correction term that depends on the volume concentration of solids, $\phi$.

The most famous one is the Einstein viscosity equation that simply reads:
$$\frac{\mu}{\mu_0}=1+2.5\phi $$

This model works for low solid concentrations when inter-particle interactions can be neglected. Later Batchelor showed that including long-range hydrodynamic interactions an additional correction of $+6.2\phi^2$ should be included, but still only a dependence on the volume concentration is there.

A nice summary of more models can be found in this paper: K.D Danov, J. Colloid Interface Sci., 235, 144–149 (2001). That paper also includes a discussion that shows that the factor 2.5 should be dependent on the mobility of the particle. This would mean that it will be higher for smaller particles, so in your case that the solution with part B will be somewhat more viscosity.

Another overview of models, including models which have an increasing viscosity with increasing particle size can be found here. So that would point to part A giving the more viscous solution.

So to summarize: to first order (i.e. low concentrations) the two viscosities will be the same. At higher orders you have competing effects of the particle radius that should increase the viscosity and the particle number that should also increase it. Which effect is dominant is probably hard to tell theoretically.


What is the analogue of "molecule" for ionic compounds?

The term you are looking for is formula unit, I think. Wikipedia doesn't really describe it super well, but just to give an example, you could write the sentence

  $\pu{58.44 g}$ of $\ce{NaCl}$ contains $6.022 \times 10^{23}$ formula units of sodium chloride

and it would be pretty well understood.

See also: What is the difference between a Chemical Formula and Formula Unit?


Determine and draw planes and axes of symmetry of a molecule using software (VESTA or Mercury)

I do not know VESTA well enough to provide an answer for this program.

However because your data are about a crystal structure I suggest to export your model in the .cif format.  CCDC's freely available version of CCDC's Mercury is able to read this file type, and add to its visualizations the symmetry elements of the unit cell (Display -> Symmetry Elements).  There will be a new menu where you may choose among the symmetry elements the program is able to recognize, and how they are displayed:

For example only the centres of inversion:

The export options in Mercury include bitmap .png, ray-tracing PovRay .pov, and .vrml / .stl.

Given lattice constants and space group $P6_3/mmc$, do not forget to compare your model with the literature known models for example in the freely accessible COD about experimental data (entries #1010993, #1011286, #1531960, #9007660, #9009144); and consider contributing to its sibling TCOD about theoretical data and MPOD.


How is the oxidation number related to the group number?

For main group elements (groups 1, 2, 13-18), the highest possible oxidation state $O_\text{max}$ is given by the group number $G$ (for groups 1 and 2) or by the group number - 10 (for the rest):
$$ O_\text{max} = \begin{cases} G &\text{if } G \in \{1,2\} \\
G - 10 &\text{if } 13 \leq G \leq 18 \end{cases} $$

This means that for your compound $\ce{\stackrel{VI}{X}O3}$, so with the oxidation state VI, the corresponding group would be group 16.


It is possible to obtain $\ce{Cu2S}$ by heating a mixture of finely-powdered elemental copper and sulfur in a crucible up to temperatures of above 450 °C. The powdered form of the educts ensures that the reaction starts as soon as the sulfur has melted. Alternatively, the mixture can be milled in a planetary ball mill for 2 hours, which yields $\alpha$-$\ce{Cu2S}$. However, $\ce{Cu2S}$ produced by these methods has limited long-term stability, as it slowly converts into $\ce{CuS}$ even when left in an inert atmosphere (reference).


There was a story in my days about a physical chemist who was asked to explain some effect, illustrated by a poster on the wall. He did that, after which someone noticed that the poster was hanging upside down, so the effect appeared reversed in sign. Undaunted, the guy immediately explained it the other way around, just as convincingly as he did the first time.

Cooking up explanations on the spot is a respectable sport, but your teacher went a bit too far. What's with that Charles' law? See, it is a gas law; it is about gases. And even then it is but an approximation. To make it exact, you have to make your gas ideal, which can't be done. As you lower the temperature, all gases become less and less ideal. And then they condense, and we're left to deal with liquids and solids, to which the said law never applied, not even as a very poor approximation. Appealing to this law when we are near the absolute zero is about as sensible as ruling out certain reaction mechanism on the grounds that it requires atoms to move faster than allowed by the road speed limit in the state of Hawaii.

The energy argument is even more ridiculous. We don't have to remove all energy, but only the kinetic energy. The $E=mc^2$ part remains there, so the mass is never going anywhere.

All that being said, there is no physical law forbidding the existence of matter at absolute zero. It's not like its existence will cause the world to go down with error 500. It's just that the closer you get to it, the more effort it takes, like with other ideal things (ideal vacuum, ideally pure compound, crystal without defects, etc). If anything, we're doing a pretty decent job at it. Using sophisticated techniques like laser cooling or magnetic evaporative cooling, we've long surpassed the nature's record in coldness.