The equilibrium constant does only contain the activities/concentrations of the species that are actually available to the reaction, e.g. in solution it would only contain species that are actually dissolved in the solvent or that are present at the interface between the solution and another phase. So, solids would not enter the equation with all there moles but only with the small percentage that is acutally available to the reaction: if some of the solid is dissolved and those dissolved species take part in the reaction then it is those (and only those) molecules that enter into the equation for the equilibrium constant; in the case of a heterogeneous reaction the solid can only enter the equation via its available surface sites. So, the bulk of a solid would not contribute to the equilibrium constant in a reaction taking part in a liquid phase that shares and interface with the solid.

Now, concerning the main question: If the equilibrium constant in a reaction is greater than 1 is it possible that there are more moles of reactants than products? This can certainly be the case if you already have some of the products present and keep their concentrations constant. Look at the example reaction

\begin{equation}
  \ce{A + B <=>> C + D}
\end{equation}

which shall have an equilibrium constant of $K_{c} = 10$. Let's assume you keep the products at a concentration of 1 mol/L each, so $[\ce{C}] = 1 \, \mathrm{mol}/\mathrm{L}$ and $[\ce{D}] = 1 \, \mathrm{mol}/\mathrm{L}$. The equation for the equilibrium constant is

\begin{equation}
  K_{c} = \frac{[\ce{C}] [\ce{D}]}{[\ce{A}] [\ce{B}]}
\end{equation}

Since you've already fixed the product $[\ce{C}] [\ce{D}]$ at $1 \, \mathrm{mol}^{2}/\mathrm{L}^{2}$ the product of your reactants must fulfill

\begin{equation}
  [\ce{A}] [\ce{B}] = \frac{[\ce{C}] [\ce{D}]}{K_{c}} = 0.1 \, \mathrm{mol}^{2}/\mathrm{L}^{2}
\end{equation}

So, if you use only very little $\ce{B}$, say $[\ce{B}] = 0.01 \, \mathrm{mol}/\mathrm{L}$, your concentration of $\ce{A}$ in the solution will be

\begin{equation}
  [\ce{A}] = \frac{[\ce{C}] [\ce{D}]}{K_{c} [\ce{B}]} = 10 \, \mathrm{mol}/\mathrm{L}
\end{equation}

Now, if you conduct this reaction in a vessel with a volume of 1 L you'll find a total of 2 mol of products and 10.01 mol of reactants in your reaction mixture, i.e. more reactants than products.


The solubility of calcium salts is highly dependent on pH. For example, let’s look at tricalcium phosphate, which presents the solubility equilibrium:
$$\ce{Ca3(PO4)2(s) + H2O <=> 3Ca^2+(aq) + 2PO4^3-(aq)}$$
The solubility product constant for this equilibrium ($K_\mathrm{sp}$) is vastly varied from source to source. Some listed it as $\pu{2.0 × 10^–29 mol^5L{–5}}$, and another listed it as $\pu{2.07 × 10^−33 mol^5L^{–5}}$ at $\pu{25 ^{\circ}C}$ and $\mathrm{pH}~7.00$. However, I found a reliable $K_\mathrm{sp}$ value in literature, which reported that $K_\mathrm{sp}$ of $\beta$-$\ce{Ca3(PO4)2}$ in the $\mathrm{pH}$ range $6.0-7.5$ is $\pu{1.2 × 10^–29 mol^5L^{–5}}$ at $\pu{25 ^{\circ}C}$ and $\pu{2.83 × 10^–30 mol^5L^{–5}}$ at $\pu{37 ^{\circ}C}$ (a negative thermal coefficient of solubility; Ref.1). Therefore, the quantitative analysis of $\ce{Ca3(PO4)2}$ is near impossible.

Still, we can qualitatively look at the problem: If you decrease the $\mathrm{pH}$ of the solution in above equilibrium by adding acid to the solution, some of the phosphate ions would protonate and transform into $\ce{HPO4^2–}$ ions, as what happens in phosphate buffer solutions:
$$\ce{PO4^3-(aq) + H3O+ (aq) <=> HPO4^2-(aq) + H2O}$$
As a result, the concentration of $\ce{PO4^3-}$ ion would be reduced. The system would respond to this reduction by producing more $\ce{PO4^3-}$ ions, according to the Le Chatelier principle. To do so, some solid $\ce{Ca3(PO4)2}$ would dissolve in first equilibrium, and the equilibrium will be shifted to the right.

Keep in mind that in first equilibrium, $\ce{Ca/P}$ ratio is 1.5. The solubility of calcium phosphates is also depend on $\ce{Ca/P}$ (Ref.2). For example, both monocalcium phosphate ($\ce{Ca(H2PO4)2}$) and monocalcium phosphate monohydrate ($\ce{Ca(H2PO4)2.H2O}$) are highly soluble in water at $\pu{25 ^{\circ}C}$ ($\ce{Ca/P}$ is 0.5). On the other hand, $K_\mathrm{sp}$s of dicalcium phosphate ($\ce{CaHPO4}$) and dicalcium phosphate dihydrate ($\ce{CaHPO4.2H2O}$) are $1.26 × 10^{–7}$ and $\pu{2.57 × 10^{–7} mol^2L^{–2}}$, respectively at $\pu{25 ^{\circ}C}$ (Ref.2). Based on these values, when $\mathrm{pH}$ is gradually reduced to a point such that all calcium phosphates would dissolve.

References:

T. M. Gregory, E. C. Moreno, J. M. Patel, W. E. Brown, “Solubility of $\beta$-$\ce{Ca3(PO4)2}$ in the system $\ce{Ca(OH)2-H3PO4-H2O}$ at $5, 15, 25,$ and $\pu{37 ^{\circ}C}$,” Journal of Research of the National Bureau of Standards–A. Physics and Chemistry 1974, 78A(6), 667–674 (http://dx.doi.org/10.6028/jres.078A.042).
L.C. Chow, “Solubility of Calcium Phosphate,” In Octacalcium Phosplrate; L. C. Chow, E. D. Eanes, Vol. Eds., Volume 18 of Monographs in Oral Science (G. M. Whitford, Ed.) Karger Publishers: Basel, Switzerland, 2001, pp. 94-111 (DOI:10.1159/000061650).


Difference between the absolute thermodynamic scale and ITS-90

The absolute thermodynamic temperature scale is a scale fixed by absolute zero and (at least for the kelvin, I'm not sure about the Rankine scale) the triple point of water, where absolute zero is defined as 0 K and the triple point of water is defined as 273.16 K. The rest of the scale is just a straight line from zero through 273.16 K off to infinity. However, this scale relies on only two defined points (only one of which can actually be measured, absolute zero being established from thermodynamic relations).

In practical use, for the purposes of calibrating thermometres and the like, this is not a great scale as the two calibration points may be far away from the temperature region one is interested in actually measuring. The usual way to deal with this is to use other reference points outside the actual definition. So that different people do this as consistently as possible, ITS-90 was created as a practical implementation of the absolute scale, setting additional reference points (all phase transitions) to defined values, interpolating between them to get other values. It's technically not a scale in itself, just a standard for calibrating temperatures that are as close to the absolute scale as possible. By making sure everyone uses the same reference points, measurements can be more consistent than if reference points are chosen arbitrarily and possibly assigned different values.

For example, if one wanted to make a thermometre for general lab use, one might pick the triple point of mercury, the triple point of water, and the melting point of gallium from the ITS-90 standard for reference points, knowing that other thermometres in this range conforming to the standard would also be calibrated this way, rather than choosing say the melting and boiling points of water, increasing the likelihood of consistent measurements with other thermometres.


Are there any solids that undergo diffusion?

Diffusion also occurs in solids. For example, (intentional or undesirable) movement of impurity atoms or other crystal defects can be very important for the fabrication of semiconductors.

The classical experiment to demonstrate diffusion (and to measure the diffusion coefficient) in solids is the diffusion of gold in lead. A thin layer of gold that is labelled with $\ce{^198Au}$ is applied to the surface of a lead sample using electrolysis. The sample is heated, and the penetration of gold atoms into the lead sample is observed using the β radiation from $\ce{^198Au}$.

You could use a similar experiment to measure self-diffusion (e.g. of gold atoms in gold).


Does the prefix Ortho have some significance?

The reference given by Andselisk does not bring an answer to the question about the origin of the prefix "ortho" in inorganic nomenclature.

The prefix "ortho" is added before the name of an acid $H_aX_bO_c$ if first $b = 1$ and if it corresponds to the highest possible number of $OH$ groups attached to the central X atom. For example $H_4SiO_4$ is the orthosilicic acid, $H_3BO_3$ is the orthoboric acid. $H_4CO_4$ is the orthocarbonic acid (which only exists in esters). $H_3PO_4$ is the orthophosphoric acid (where the prefix is often skipped). $H_3AsO_4$ is the orthoarsenic acid. The use of the prefix does not depend on the oxidation number of $X$. For example, if $P$ and $As$ are taken with a lower oxidation number, one obtains $H_3PO_3$ as the orthophosphorous acid, and $H_3AsO_3$ as the orthoarsenous acid.

The prefix "meta" is added to the name of an acid, if its formula is formed by removing one molecule $H_2O$ from the formula of the ortho-acid. $H_3BO_3$ - $H_2O$ = $HBO_2$ and this is the metaboric acid. $H_3PO_4$ - $H_2O$ = $HPO_3$ which  is the metaphosphoric acid. Repeating this process, one gets :$H_2CO_3$ is the metacarbonic acid (often written without the prefix meta). $H_2SiO_3$ is the metasilicic acid. As with ortho-acids, the use of "meta-"  does not depend of the oxidation number. $HPO_2$ is the metaphosphorous acid. $HAsO_2$ is the metaarsenous acid.

It should also be mentioned that removing one $H_2O$ out of two molecules of an ortho- acid gives a pyro- acid. Example :
$2 H_3PO_4 - H_2O = H_4P_2O_7$ and this is the pyrophosphoric acid. $H_4As_2O_7$ is the pyroarsenic acid. This prefix pyro- is even used for acids existing only under the ortho- definition. For example, a pyro derivate from the sulfuric and sulfurous acids $H_2SO_4$  and $H_2SO_3$ exist. They are the pyrosulfuric acid $H_2S_2O_7$. and the pyrosulfurous acid $H_2S_2O_5$

It should also be mentioned that the prefixes ortho-, meta-, and pyro- are never used with acids having already a prefix like hypo- and per-.

All these rules have been edited in the middle of the 19th century.


Can I use one alkaline battery to recharge another?

First, some batteries are rechargeable, but some are not. However, even if we talk about rechargeable one, then
1) some degradation during recharge usually occurs
2) deep equality will require very long time to achieve


Applying salt to roads to melt ice--is it special?

Yes, applying salt to melt ice it is special:

Salt is a great ice melter because it causes “freezing point depression.” This means that salt helps in lowering the freezing point and, consequently, the melting point of water (the main component of snow and ice). In its pure state, water freezes at 0°C or 32°F. By using salt, that temperature can be lowered, which forces the ice to melt and prevents the water from freezing or re-freezing. This does not happen with  sand.
After melting, no additional effort is needed to clean the roads, because salt mixes with water. If you use sand, additional work will be needed to remove it from the streets afterwards.
Salt is much cheaper than other chemicals, although arguably more expensive than sand, depending on location.
Salt is recyclable, so no wastage or decompositon of salt is there in such process. Since salt becomes soluted in the melted ice, it gets washed away and absorbed by the ground. This makes salt hard to retrieve after use, and harms the environment by damaging the plants in the area.


Electrolysis of water in 0.1 M hydrochloric acid

Possible reactions :

At Cathode:

$\ce{2H2O{(l)} + 2e- -> H2{(g)} + 2OH-{(aq)}}$

$\ce{2H+{(aq)} + 2e- -> H2{(g)}}$

At anode:

$\ce{2H2O_{(l)} -> O2_{(g)} + 4H+_{(aq)} + 4e-}$

$\ce{2Cl-{(aq)} -> Cl2{(g)} + 2e-}$

General rules of electrolysis of aqueous solution using inert electrodes :-

Reduction potential of $\ce{H2O}$ > reduction potential of cations of 1, 2 and 13 groups of periodic table. So reduction of $\ce{H2O}$ takes place at cathode.
Reduction potential of $\ce{H2O}$ < reduction potential of other cations (than cations of 1, 2 and 13 group). So reduction of $\ce{H2O}$ does not occur at cathode.
Oxidation potential of $\ce{H2O}$ > oxidation potential of fluoride, sulphite, sulphate, nitrate, etc. So oxidation of $\ce{H2O}$ occurs at anode.
Oxidation potential of $\ce{H2O}$ < oxidation potential of chloride, bromide and iodide, etc. So oxidation of these ions takes place.

Keeping these rules in mind, at cathode reduction of hydrogen ions will take place and hence $\ce{H2}$ gas will be evolved. At anode, oxidation of chloride ions will take place and chlorine gas will be evolved and $\ce{O2}$ won't be evolved.

Although the oxidation potential of $\ce{H2O}$ > oxidation potential of chloride ion, chloride ion gets oxidized because of over potential of $\ce{O2}$ i.e. extra potential is required to initiate a reaction at a required rate. And formation of $\ce{O2}$ from $\ce{H2O}$ is kinetically very slow.

Therefore oxidation of chloride ion takes place.


Why is ascorbic acid (vitamin c) readily oxidised in the presence of oxygen?

According to Juan Du, Joseph J. Cullen and Garry R. Buettnera[1]:

Ascorbic acid ($\ce{AscH2}$, vitamin C) is a water-soluble ketolactone with two ionizable hydroxyl groups. It has two $\mathrm{p}K_\mathrm{a}$’s, $\mathrm{p}K_1$ is 4.2 and $\mathrm{p}K_2$ is 11.6; thus, the ascorbate monoanion, $\ce{AscH−}$, is the dominant form at physiological pH. Ascorbate is an excellent reducing agent and readily undergoes two consecutive, one-electron oxidations to form ascorbate radical ($\ce{Asc•−}$) and dehydroascorbic acid (DHA). The ascorbate radical is relatively unreactive due to resonance stabilization of the unpaired electron; it readily dismutes to ascorbate and DHA ($K_\mathrm{obs} =\pu{2×10^5 M−1 s−1}$ at $\mathrm{pH} = 7.0$).

Also,

Ascorbate readily undergoes pH-dependent autoxidation producing hydrogen peroxide ($\ce{H2O2}$).

So, in presence of oxygen, this process is accelerated due to formation of hydrogen peroxide. It is also known that this reaction may happen even faster with a metal as a catalyst[2]

References:

Du, J.; Cullen, J. J.; Buettner, G. R. Ascorbic acid: Chemistry, biology and the treatment of cancer. Biochim Biophys Acta. 2012, 1826 (2), 443–457. DOI: 10.1016/j.bbcan.2012.06.003.

Jiang, D.; Li, X.; Liu, L.; Yagnik, G. B.; Zhou, F. Reaction Rates and Mechanism of the Ascorbic Acid Oxidation by Molecular Oxygen Facilitated by Cu(II)-Containing Amyloid-β Complexes and Aggregates. J. Phys. Chem. B 2010, 114 (14), 4896–4903. DOI: 10.1021/jp9095375.


In the ethyl cation, one can write the following resonance structures:

The structure on the right can be actually written 3 times since there are 3 hydrogens available. Note also that the "ethylenic-like" structure on the right involves a $\mathrm{sp^3-p}$ bond (the $\ce{C-H}$ bond is $\mathrm{sp^3}$ hybridized and the cationic carbon has a $\mathrm{p}$ orbital), so it is not a true pi bond ($\mathrm{p-p}$) like that found in ethylene.  The resonance structure on the right is meant to show the "hyperconjugation" effect of the $\ce{C-H}$ bond ($\mathrm{sp^3-s}$) with the $\mathrm{p}$ orbital located on the adjacent cationic carbon.  The drawing is a shorthand way of saying that when the $\ce{C-H}$ bond is rotated into the same plane as the $\mathrm{p}$ orbital on the cationic carbon, there can be a stabilizing overlap - and it is called "hyperconjugation." In the 2-propyl cation, there are 6 of these hyperconjugative resonance structures (there are 6 $\ce{C-H}$ bonds on the 2 methyl groups adjacent to the cationic carbon and in the t-butyl cation there are 9 such resonance structures. This can be used to explain why tertiary carbonium ions are more stable than secondary carbonium ions, which in turn are more stable than primary carbonium ions.


Reaction of Diols with HX hydrohalic acids, Competing SN2 Reactions

Bob:
Here is the likely process. The alcohol 1 is protonated to provide intermediate 2 which undergoes SN2 displacement. The closure of 3 to  tetrahydropyran 6 is unlikely under the reaction conditions. Rather a second displacement via 4 would lead to diiodide 5. If the tetrahydropyran did form, ethers are cleaved by HI. The conversion of 3 to 6 is normally accomplished in the presence of base (NaH/THF).


The purity of the iron is less important here than the conditions of the atmosphere and other surroundings.  Rusting of iron involves the formation of hydrated oxides and so requires the presence of moisture, either as water vapor or as liquid.  In a dry environment iron is indeed more resistant to corrosion because it forms anhydrous oxides which, as long as moisture is kept away, do provide some passivation.


Which materials/compounds expand in response to cold temperatures?

Perhaps one of the most studied materials to exhibit negative thermal expansion is cubic zirconium tungstate $(\ce{ZrW2O8}).$


Why do ethanethiol molecules have stronger London dispersion forces than those of ethanol?

Oooh, interesting question. Let me have a go at this and we'll see if there's someone else who can put it in better terms.

So, the way I tried to reason this out is to remind myself that every single bond, in each of these two molecules, is the same except the ones where C is bonded to O or S and the ones where H is bonded to O or S. We then pay attention to these bonds, particularly.

Now, at this point, we also have to remind ourselves that O is more electronegative than S and therefore, pulls electrons towards itself much more than S does and both of these atoms are a tad bit more electronegative than C or H. Therefore, the region of higher electron density in ethanol is O and the region of higher electron density in ethanethiol is S. We also have to remember that O has a more negative partial charge than S does. We also need to remind ourselves that S is larger than O which, by the way, is also a contributing factor as to why O is more electronegative than S.

I then uses these facts and try to make sure my reasoning is sound; If we look at the way london dispersion forces work, we realize that it is easier for electrons to fluctuate between molecules when they are within the bonds than when they are within the atoms constituting the molecules. In other words, when the electron density is concentrated in the bond, as is the case when there is 0 electronegativity difference, then london dispersion forces are stronger. (Electron fluctuation, by the way, is the basis for dispersion forces)

Using the reasoning above, it becomes natural to think that the dispersion forces in ethanthiol are stronger, since the electron density is more concentrated in the bonds than it is when you consider ethanol. Therefore, it is easier for the electrons to fluctuate between two ethanethiol molecules than between two ethanol molecules, which is why ethanethiol molecules have stronger dispersion forces between themselves as opposed to the ethanol molecules.

Note: However, ethanol molecules have a much more powerful intermolecular force available to them, which is hydrogen bonding. So, clearly, ethanol molecules have a higher boiling point or whatnot than ethanethiol. Just to make sure that this is clear enough.


What makes Kleenex Cool Touch tissues cool to the touch, and is it safe?

My first guess as to the nature of this behaviour is similar to water. Hypothermia is a realistic risk if you are drenched in water. This is because water can absorb large amounts of energy (it has a large specific heat capacity) to only rise in temperature slightly. Water itself evaporates into steam spontaneously, if some water is spilled then it will evaporate readily, regardless of the temperature. Likewise, the solids in the tissue are probably coming closer to melting temperature, which I believe is higher than your body temperature, but quickly absorbing a lot of energy from your body in the process.

Something similar to properties of water are probably characteristic to the tissue. Leave the tissue in the open in a cool room, along with a glass of water. After awhile, touching the glass of water and the tissue should invoke the same sense of coolness. This is because your body is at a higher temperature than the items in the room.

There is likely another property of matter besides specific heat capacity that is a measurement of how quickly heat is transferred, this is likely to play a role as well. I do not know what this property is called and it was never covered in my classes so far, but take metal for example. Metal has a very small specific heat capacity, it can not hold much energy. However, touching a metal object invokes a sense of coolness. I believe this is because heat is transferred very quickly through metal, although it warms very quickly due to its limited heat capacity. If you put a metal cup and a porcelain cup on the stove, the metal will burn you much quicker if you grab it than grabbing the porcelain/ceramic cup.

If you were very interested and inquistive you could test the specific heat of the tissue through experimentation. Also, try holding the napkin for a few seconds to try and find out how quickly the tissue warms to the touch. Chances are, it warms relatively fast and can absorb a lot of energy, giving you the impression that it is 'cooler' than most objects, to which marketers will be overwrought with the joy of hearing.


Why is Iron very rare in seawater compared to many other elements?

I'll elaborate this answer as I learn.

It's an effect owed to the low solubility of iron in the modern oxic ocean and the tendency for iron to be ‘scavenged’ from the water column by sinking particles.

Brand, L. E., Sunda, W. G. & Guillard, R. R. L. Limitation of marine
phytoplankton reproductive rates by zinc, manganese and iron. Limnol.
Oceanogr. 28, 1182–1198 (1983).

Cooper, L. H. N. Iron in the sea and in marine plankton. Proc. R. Soc. Lond. B
118, 419–438 (1935).

Goldberg, E. D. Marine geochemistry. 1. Chemical scavengers of the sea.
J. Geol. 62, 249–265 (1954).


Given proper conditions, any element could have multiple allotropes For example, under sufficient pressure, hydrogen has a metallic phase, which may have been observed at Harvard recently.

The extreme example is degenerate matter, formed under the pressure of a collapsed star, though it could be argued that the elements are no longer the same, since nucleons themselves are eventually pooled in a "sea".

If you mean by your question "having allotropes at standard temperature and pressure (STP)", then the list is shorter, of course. Since this depends on the electronic configuration of an element and on the minimal-energy configuration of atomic arrangement, calculation is difficult, but not impossible, and there is software available for crystal structure prediction.

Now if I could only find a more convenient way to change graphite to diamond, without Superman's help...


Why is iodine the heaviest bonded element in circumstellar/interstellar space?

I disagree with the assertion that Sodium Iodide has been found in circumstellar or interstellar space.

Sodium Iodide is not on the list Molecules in Space kept updated by University of Köln.

Instead it is likely that someone misunderstood an article such as Interstellar NaI absorption towards the stellar association ARA OB1, where NaI does not mean sodium iodide, but instead means neutral atomic sodium.

Also, LaO (lanthanum oxide) has been found in stellar atmospheres, and lanthanum is heavier than iodine. See table 5 here: http://www.ucl.ac.uk/phys/amopp/people/jonathan_tennyson/papers/311.pdf

as well as the 1948 article On the Occurrence of Lanthanum Oxide in S-Type Stars.


Can we compare the vapour pressure of solutions whose concentrations and temperature is known?

The point of the question is to test your understanding
i. of Raoult's law (that increasing the solute concentration decreases the vapor pressure and that the identity of the solute is not important in an ideal solution; the mole fraction of solute determines $\Delta p$: $\Delta p = p-p^\circ = -\chi_\textrm{solute}p^\circ$
ii. that vapor pressure increases with T
Other things you might want to remember are that molality and molarity are linearly proportional for dilute solutions (ie then $m \propto c_M$), and that ideality is approached for dilute solutions of small solutes (typically at low $\pu{mM}$ concentrations such as in this problem).
You can answer the question based entirely  on consideration of the van't Hoff factor to compute total concentrations, and by assuming that the vapor pressure increases with T.
First compute effective total solute concentrations $im$ (equal to either the solute concentration for a non-dissociating solute such as sucrose, or the total ion concentration for the electrolyte solutions):
$$\begin{array}{|c|c|c|} \hline \text{species} &T (\pu{^\circ C})& c_M (\pu{M})& i & ic_M(\pu{M}) \\\hline  1.~\ce{NaCl} & 50 & 0.02 & 2 & 0.04\\
2.~\text{sucrose} & 15 & 0.02 & 1 & 0.02\\
3.~\ce{CaCl2 } & 15 & 0.005 & 3 & 0.015\\
4.~\ce{CaCl2 } & 50 & 0.005 & 3 & 0.015\\\hline
\end{array}$$
Inspection should quickly eliminate all options but row 4 since they have a higher effective concentration at the same T, or since they are at a lower T at the same effective concentration.
Actual vapor pressures are not stated in the problem. In the absence of that information there is no point in attempting to compare things in more detail, and that's not the point of the problem. However, provided Raoult's law holds it can be used to determine the change in vapor pressure regardless of T. Raoult's law assumes the identity of the solute does not matter, that the effect of added solute on the activity of solvent is entirely entropic (accounted for by ideal mixing entropy) and that the gas is ideal. In general, to predict vapor pressure at any T and solute concentration you need a model of the activity of the solvent as a function of the solute concentration, and of the activity in the gas phase. You can determine the vapor pressure by finding the conditions that render the activities identical.


What explains the stark white colour of my iodometric titration analyte solution?

You’re correct that this is just a suspension of copper(I) iodide. If CuI were soluble in water, the solution would be clear; the white colour is a sign of (very small) particles of the solid suspended in solution.


Why is guanidine considered a strong base?

Wikipedia notes that if a base (in water solution) has a conjugate acid whose $\mathrm{p}K_\mathrm{a}$ exceeds $13,$ it's a strong base.  The criterion is (apparently) based on complete or at least predominant dissociation up to $0.1$ molar.  Guanidine makes it, ergo a strong base.


Replacing Salt Bridge with Platinum Sheet

In continuation of the discussion above, I have another opinion:

  What would happen, if in a galvanic cell, the salt bridge is replaced
  by an inert platinum sheet?

  The cell would stop working in most cases. Opposite charges would accumulate in the two > half cells, opposing the reduction and oxidation half reaction.

I am afraid the statement is not correct. The set up in which Cu and Zn are partitioned by a platinum sheet, is equivalent to two galvanic cells connected in series.

The two cells
(a) Cu-Pt cell
(b) Pt-Zn cell

which are connected via conductor. In this case, the conductor is also platinum.

I just was reading LeBlanc's historical textbook on electrochemistry (lovely book from German translation) for another purpose. He shows a nice picture of two cells connected in series.

Consider a similar situation in case of Pt. However we may not be able to calculate the potentials because Pt is not dipping in Pt ions in either beaker. Instead, one end is in copper ions and the other end in the zinc ions. It still may show a positive potential (Ecell) just like how a lemon battery does (=Zn and Copper rods immersed in a citric acid solution).

This has been discussed before in physics. https://www.physicsforums.com/threads/copper-wire-as-salt-bridge.146868/ The cell still works! Charges can never accumulate like static electricity in a galvanic cell in the presence of a conductor.


Why is the melting point of PCl3 less than that of PCl5?

Because $\ce{PCl5}$ does something which is not immediately obvious from its molecular formula: it autoionizes and becomes an ionic solid $\ce{PCl4+PCl6-}$. As such, it has much stronger interactions than $\ce{PCl3}$ with its mere dipole-dipole attractions, hence the higher melting point.

If not for that fact, you deduction should have worked just fine.


Doesn't that also have to do with chemical reactions but inside living organisms?

Exactly! That's why there is one scientific society for both phenomena: ISBC - International Society for Bioluminescence and Chemiluminescence ;-)

The thermal reactions, that lead to an excited species which eventually emits a photon are quite similar. From a reactive compound, typically called a luciferine, a four-membered ring, often a dioxetanone is formed. This decarboxylates and gives rise to a molecule in an electronically excited state. In reality, the decarboxylation is often not just a spontaneous process but requires an activator. W. J. Baader at IQUSP in Sao Paulo, Brazil has examined that in great detail.

The luciferines can be quite different compounds, such as

diaryloxalates (these are used in glow sticks, where the diferent colours come from diferent fluorphores that absorb the initial photon and emit at different wavelength)
benzothiazole derivatives (firefly)
triazine derivatives (marine organisms)


How does the cyclopropyl group influence conjugation and aromaticity?

About the molecule that you remember cited in your textbook, there's a study in ref.1. According to these authors:

  The delocalization of the C1-C2 and Cl-C3 electrons of spiro[2.4]hepta-4,6-diene (1a) and partial rehybridization at C1, C2, and C3
  result in a contributing structure which is perhaps best represented as a $\pi$ complex (lb) of cyclopentadienylidene and ethylene.

From NMR studies they concluded that:

  The comparison of the predicted downfield shift
  of approximately 6.6 ppm for the complex and the
  observed shift of approximately 1.0 ppm leads to the
  the conclusion that the $\pi$ complex may contribute as
  much as 15 % to the total electronic structure of 1.

The most recent reference I have found is from 1997(ref.2). They studied these structures:

  In the series of the investigated compounds 1-5, spi- ro[2.4]hepta-4,6-diene (1) shows the strongest  influence of cyclopropyl  conjugation on molecular structure in the group of doubly  unsaturated   cyclopropyl  hydrocarbons

The conjugation properties of cyclopropane are subject of debate and other references don't support this. You can find a blog about this on the royal society of chemistry website(http://my.rsc.org/blogs/84/1009).

I have found another paper in ref.3 and according to this:

  In one chemical attempt to test the transmission
  of conjugation by cyclopropane rings, Cannon, et al.were unable to alkylate or acylate the methyl group in diethyl 2-methylcyclopropane-l, 1-dicarboxylate and concluded that transmission of conjugative effects in the transition state for carbanion formation are small compared with analogous, unsaturated compounds.

Also:

  Trachtenberg and Odian concluded from their review
  of the various conflicting reports that a cyclopropane
  ring is incapable of transmitting conjugation in the
  ground state but may be able to do so in some excited
  states. It is likely that, during chemical reactions,
  such transmission occurs only in a transition state,
  i.e., after bond breaking of the ring has commenced.

From what I have read, even assuming a conjugate effect, this would be much less evident than a C-C double bond.

About your questions:
Is there always a conjugation between a cyclopropyl group and an adjacent double bond? In other words, do cyclopropane's "bent" bonds always have extensive π-character and can, therefore, act as an alkene?
 Partially, the effect is not comparable to a true double bond. This hypothesized conjugation effect was found to be stronger in piro[2.4]hepta‐4,6‐diene(1) compared to the other molecules studied in ref.2(molecules 2 to 6 in the second picture).

1a. Are electronic structures 1 and 1a; 2 and 2a; 3 and 3a similar, and to what extent?
Considering that the supposed conjugation contribution of the cyclopropane ring is much less than a C-C double bond, I think they differ substantially. Unfortunately, I can't find any spectra or other additional data supporting this.

Can we use Hückel's 4n+2 and 4n rules for cyclopropyl-containing monocyclic rings? If not, how to estimate aromaticity/anti-aromaticity in such system?The Huckel rule is not valid in these complex cases. The Hückel's rule is based on calculations using the Hückel method and therefore has a limited validity. Here you need advanced MO calculations and as you have seen from all the references mentioned in such systems there's a lot of debate.

References:
1) Aromaticity via cyclopropyl conjugation. Electronic structure of spiro[2.4]hepta-4,6-diene, J. Am. Chem. Soc.197092154736-4738 https://pubs.acs.org/doi/abs/10.1021/ja00718a047

2) Structural Aspects of Cyclopropyl Conjugation: Experimental Studies and Ab Initio Calculations.  https://doi.org/10.1002/jlac.199719971007

3)Transmission of conjugation by the cyclopropane ring, Stewart JM, Pagenkopf GK, The Journal of Organic Chemistry, 1969. https://pubs.acs.org/doi/pdf/10.1021/jo00838a003?rand=vimssm6d


Somebody messed up the scheme completely!

As you have already imagined, there is no magical $\ce{NO2}$ to $\ce{CH3}$ conversion in this classical reaction ;-)

In the times before modern spectroscopy, these reactions were valuable analytical tools: the melting points of the dinitrophenylhydrazones and/or other derivatives were used to identify the carbonyl compounds.


What is a prosthetic group of protein? What are prosthetic groups?

A prosthetic group is a smaller organic molecule strongly bound to a protein. Typical examples are

haemoglobin (bound to haeme)
biotin (bound to carboxylases)
flavines (bound to flavoproteines)


If you look at the (average) composition of seawater:

you will find that NaCl is the main solute of this complex solution. Its concentration is 10 times higher than that of the next constituents (magnesium and sulfate). So, it seems to me like a simple NaCl solution is a good approximation for your experiments.

Regarding testing salinity of water, I don't have a definite answer on that, I do have two suggestions to offer:

Measure salinity by electric conductance of the solution, e.g. by making a simple conductometer and testing it with solutions of known concentration of salt.
Measure salinity by the quantity of salt (solid residue) once all water is evaporated. You can speed the evaporation by heating up.


Reconsider if a more electronegative element bound to carbon and bond to carbon will -- formally -- take the electrons, or donate the electrons in shared the bond with carbon, and how this is reflected in the oxidation state.  In each instance, watch out the signs, please.

After getting familiar with the basics, look up references focussing on organic chemistry and redox chemistry.  Albeit organic chemists tend to prefer moving electrons by hooks and arrows, there are references like this (hint: it includes an answer you may compare with the one you should find yourself, too) or even video classes like this.

Eventually, "left" and "right" will advantageously become obsolete when you gain enough credit to upload structural schemes on SE, too.


A bond is formed between the oxygen of water and the phosphorus of the gamma-phosphate.  Here is a good link.

Bonds are both broken and made in chemical reactions but many biology teachers and textbooks state that "Breaking ATP bonds releases energy."  In reactions bonds are broken and made.  If the strength of the bonds formed exceeds the strength of the bonds broken the reaction is exothermic. Under physiological conditions (with respect to pH and magnesium concentration), the reaction is exergonic as well (i.e. negative Gibbs free energy of reaction).


Typically, ring strain energies (RSE) are calculated as the difference of the heat of formation of cyclic starting materials and less cyclic products in homodesmic reactions.

For cyclopropane, the RSE would be calculated using the following reaction:

$$\ce{(CH2)3 + 3 CH3-CH3 -> 3 CH3-CH2-CH3}$$


Unfortunately the reason solder works is that it partially dissolves copper as shown in Figure 1 to form $\ce{Cu3Sn}$ and $\ce{Cu6Sn5}$ phases (also known as bronze) shown in Figure 2. Because of this you cannot simply melt solder off the solder wick to recover it as the tin will continue to dissolve the copper forming bronze. There is also the issue of needing rosin in your recovered solder to be useable. This 1960's video from NASA is an excellent video for learning more about the soldering process.

Figure 1. Reactions of the Copper-Solder Interface

Figure 2. Copper-Tin Binary Phase Diagram

To address the acid method. Physically, it could work (I would not recomend doing so with leaded solder) as most ROHS compliant electronics are $95\%+$ tin solders with either silver or antimony added as alloying agents to prevent tin whiskers from forming. That said it economically dissolving in acid then precipitating then reducing tin in a way which does not also produce copper would be a tedious and time consuming process. You are far better off just buying solder which does not cost that much and has rosin needed to flux the solder.


Can Gas diffuse according to its partial pressure, not according to its concentration in gas mixture?

Strictly speaking, diffusion occurs due to gradients in chemical potential.
One implication of this is that sometimes diffusion can occur from lower to higher concentration.
See, for example, this nice class summary

If the total pressure is different, the transport will also occur by fast flow convection instead of diffusion until total (mechanical) pressures are equal (unless thermal transpiration effects are significant).


Are there any fluorophores with lifetimes in the micro to milli second range?

According to the article Indicators for optical oxygen sensors; Michela Quaranta, Sergey M. Borisov, Ingo Klimant

  Among all the indicators that have been used for optical oxygen sensors, Pt(II) and Pd(II) porphyrins are the mostpopular luminophores since they possess strong phosphorescence at room temperature, moderate to high molar absorption coefficients and large Stokes’shifts. Additionally, phosphorescence lifetimes are rather long (microsecond to millisecond) and can be tuned by varying the nature of the central atom.

They then go on to include a table (Table 3 on page 131) of 36 dyes based on that chemistry, 35 with lifetimes in the microseconds and 1 (PdTFPP) with a lifetime of 1.65 milliseconds.

In relation to my quest for a data source to compare fluorophors I managed to stumble upon the website fluorophores.org that currently includes 952 substances and can be filtered by excitation, emission, quantum yield, and lifetime parameters.


Hyperconjugation and the stability of alkenes

Ordinarily without the contribution of the π* orbital, we would expect the interaction of two filled orbitals to be destabilising, in the sense that $E_2 > E_1$ (as drawn in the diagram). This is the same rationale as why simple MO theory predicts the helium dimer to be unstable with respect to dissociation, for example.

It might be slightly unclear, but my interpretation of the text is that the empty π* orbital contributes a little bit to both MOs $\psi_1$ and $\psi_2$. This leads to a tiny stabilisation of both of these two MOs, such that the destabilisation $E_2$ is a bit smaller and the stabilisation $E_1$ a bit larger. Although it is not very obvious, you can see this from the figure.

It could probably have been made clearer by drawing extra dotted lines from the π* orbital to those MOs. My guess as to why Fleming hasn't done that is because he wanted to use dotted lines to illustrate the most significant orbital interactions (those between orbitals of similar energy). Note how he emphasises in the text that the σ and π* orbitals are not very close in energy (which implies that the interaction is small).


Why is absorbance spectrum continuous and not quantized?

The image shows the absorption and emission spektra of molecules. For isolated atoms, you would indeed observe sharp lines. Think in the Fraunhofer lines.

UV/VIS absorption of molecules

In the case of molecules, absorption occurs from the vibrationally relaxed electronic ground state ($S_0$) to various vibration levels of the electronically excited $S_1$ state. This results in a broad absorption band instead of a sharp line.

The manifold of $S_1$ with different vibrational levels does not live forever. Within the shortest time, relaxation to the vibrational ground state of $S_1$ occurs. The electron distribution of the excited state is typically significantly different from the $S_0$ state. This results in a completely changed reactivity.

Radiationless deactivation

If you don't observe emission from an $S_1$ or $T_1$ state, the excited state is deactivated by internal conversion, i.e. by vibrational exchange with the bulk (solvent).


Unlike silk or cotton, wool does not have a homogeneous structure. Its core is considered to be made of keratin, a generic type of protein found in human hair and epidermis. A detailed PDF on Merino wool can be downloaded from here.

Thanks for encouraging me, through your question, to read that PDF, which explains both the chemical and physical properties in a straight-forward fashion.


Mechanism of conversion of Cr from FeCr2O4 (chromite)

The first reaction is a double oxidation. Both Fe and Cr are oxidized by O2 from the air. The reaction is made of two half-reactions. The first half-reaction is not easy to establish, because two elements (Fe, Cr) are oxidized simultaneously, Fe from +II to +III, and Cr from +III to +VI. Sorry to say it : It is one of the most difficult half-equations I ever had to state. It is :    $$\ce{FeCr_2O_4 + 11 OH^- -> Fe(OH)_3 + 2 CrO_4^{2-} + 4 H_2O + 7 e-}$$ The second half-equation is easier to write.
$$\ce{O_2 + 2H_2O + 4 e-  -> 4 OH^-}$$ The overall equation is obtained by multiplying the first equation by $4$ and the second by $7$ and adding the whole. After simplification, it yields :
$$\ce{4 FeCr_2O_4 + 16 OH^- + 7 O_2 -> 4 Fe(OH)_3 + 8 CrO_4^{2-} + 2 H_2O}$$
or, without ions :
$$\ce{4 FeCr_2O_4 + 16 NaOH + 7 O_2 -> 4 Fe(OH)_3 + 8 Na_2CrO_4 + 2 H_2O}$$
This was the first step, defining how to pass from the mineral $\ce{FeCr_2O_4}$ to $\ce{Na_2CrO_4}$, at high temperature, and in the presence of air and $\ce{NaOH}$. The final mixture can be washed with water, which dissolves easily $\ce{Na_2CrO_4}$, as $\ce{Fe(OH)_3}$ is insoluble in water.
Then $\ce{Na_2CrO_4}$ is transformed in $\ce{CrO_3}$ by adding moderately concentrated sulfuric acid : $$\ce{Na_2CrO_4 + H_2SO_4 -> CrO_3 + Na_2SO_4 + H_2O}$$ $\ce{CrO_3}$ is not soluble and can be separated by filtration. Then it is mixed with aluminum powder to get a exothermic reaction when engaged by a match : $$\ce{2 Al + CrO_3 -> Al2O3 + Cr}$$


Why doesn't Boron form Diborane with water?

I assume the hypothetical reaction you are asking about is:

$\ce{3H2O + 2B -> B2H6 + 3/2O2}$

under standard conditions.

The standard enthalpy of formation of water is −285.8 kJ/mol, whereas that of diborane is 35.4 kJ/mol; this reaction would therefore be very endothermic and hence unfavourable. You're breaking up strong O-H bonds and B-B bonds to form B-H bonds (including two 3c2e bonds, which are very weak) and weak O-O bonds. In contrast the formation of boron trioxide by reaction of boron and water is very exothermic, and has been considered in potential applications of boron as a rocket fuel.


These are all mathematically identical. Here’s why:

$$\frac{\text{mass of one atom}}{\frac1{12}\text{ mass of one atom of }\ce{^12C}}\\=\frac{\text{mass of one atom}\times N_\mathrm{A}}{\frac1{12}\text{ mass of one atom of }\ce{^12C}\times N_\mathrm{A}}\\=\frac{\text{mass of one mole of substance}}{\frac1{12}\text{ mass of one mole of }\ce{^12C}}$$


Equilibrium expression for dissolution of carbon dioxide

In the case of heterogenous equilibria when molar concentration of pure solid or liquid involved, water in this case is constant or is independent on the amount present its concentration is omitted as per the general norm. The equilibrium constant is just modified to

$$K_c' = K_c[\ce{H2O}]$$

So the book and you are referring to different constants.


Predicting order of nucleophilic substitution reactivity

Acetone a is aprotic polar solvent which is often used as solvent for $\ce{S_{N}2}$ reactions (notably in conjunction with $\ce{NaI}$ as in the Finkelstein reaction). For $\ce{S_{N}2}$ reactions steric hindrance and leaving group capability is of great importance, so primary halides will be usually easier replaced by iodine then tertiary.
$\ce{AgNO3}$ will probably react as a base forming nitric acid. Even though nitric acid is a rather strong acid this is more likely than $\ce{NO_3-}$ acting as nucleophile. Besides that, $\ce{Ag+}$ is known to have a unique selectivity for halogens and will attack them readily, forming $\ce{AgX}$ and a carbocation, hence promoting $\ce{E 1}$/$\ce{S_{N}1}$.


It has been noted in the comments and I shall repeat that the premise of the question is already bogus. The question does not consider two different ions but two mesomeric depiction of the same ion. This is also evident by the mesomeric arrow between the two forms. Since mesomery strictly means that no geometrical change takes place, both depictions are identical in all aspects and have the same (resonance-influenced) energy. Thus, any book claiming that either of the two have a lower energy is wrong.

However, what is commonly done is to determine a more likely resonance structure. This is, for example, commonly used to explain the strongly stabilised nature of certain aromatic systems, e.g. when DMAP (4-N,N′-dimethylaminopyridine) is used as a carboxylic activating agent. We can attempt to do that for your structure, too. If we compare the two we see — of course — that a lot of features are identical. The principal differences are:

A has a trisubstituted double bond while B only as a disubstituted (and terminal) double bond
B has a secondary carbocation while A has a primary one (both, however, are allylic).

Since allylic stabilisation for cations is much stronger than hyperconjugation, we can discard the second bullet point as being essentially equal for both resonance structures. Thus, we should argue with the location of the double bond. Both double bonds can benefit from hyperconjugation twice (the primary cation of A cannot hyperconjugate due to geometry), but in A the two hyperconjugations stabilise different carbons. Along with the general double bond stability given as:

$$\text{monosubstituted} < \text{terminal disubstituted} < \textit{cis} < \textit{trans} < \text{trisubstituted} < \text{tetrasubstituted}$$

we can identify A as being slightly better than B and therefore contributing slightly more towards the overall structure.

You mentioned in the comments that your book states:

  structure B is not planar and hence conjugation is not possible in it

Your book probably wants you to assume that the following depiction in figure 1 is adequate for the two resonance forms:

Figure 1: structures B and A under the assumption of a chair-configured cyclohexyl structure for B.

Under the underlying assumption of a different geometry for both structures, this may be valid. However, there is no way that cation B will adopt a chair cyclohexane configuration. Instead, as I mentioned above, the following structures in figure 2 are much more likely and even show that we are dealing with resonance:

Figure 2: depiction of the mesomeric forms without geometry change.

You can easily see that the argument of non-conjugation is bogus, since there very well is a way to optimise geometry so that the orbitals may form an allyl system.


How to improve efficiency of the diamine synthesis from fluorescein?

What I think is happening is that you are not getting enough mixing between your reagents, the solid remaning is unreacted potassium carbonate. I think you can reduce the amount of potassium carbonate you are using - you only need 1 eq as it is dibasic, and use finely powdered if you have access to it. Run the reaction more dilute with a more powerful stirrer (have you an overhead stirrer?) so that material does not stick to the side and follow it by TLC. If this does not improve the outcome I would consider using a strong organic base such as DIPEA or DBU in place of the potassium carbonate so that everything is in solution.


The precipitation of lead phosphate would be due to the following reaction.

$$\ce{3Pb^{2+} + 2PO4^{3-} <=> Pb3(PO4)2}$$

In acidic solution the phosphate ions would be protonated, thus reducing the concentration of $\ce{PO4^{3-}}$ and increasing the solubility of $\ce{Pb^{2+}}$.

But the precipitation of the phosphate is not lead's only reaction in alkaline solution.  Lead can also form a hydroxide precipitate. In strongly alkaline solution lead can also form various soluble hydroxide complexes which would reduce the concentration of $\ce{Pb^{2+}}$ thus increasing the solubility of $\ce{Pb3(PO4)2}$.


Diamond has the shape of diamond for precisely the same reason why any other crystal has the shape it has. Some crystal faces have lower surface energy than others, so the crystal grows in these directions, as if it wanted to develop these faces and avoid developing others.

This, BTW, is not determined by crystal family alone. Think of pyrite: it also has cubic unit cell, but it mostly develops the face (001) (see Miller indices) and hence makes those nice cubic crystals. On the contrary, diamond prefers the face (111), which together with its symmetry equivalents produces an octahedron. And that's why natural diamonds are typically found in this shape (not counting scratches and fractures).

Because of its low energy, the (111) face is also the hardest face of a diamond, to the point that jewelers deliberately avoid it when cutting brilliants, for it can't be polished quite as good as any other arbitrary plane.

So it goes.


What is the trend in hydration entropy of alkali metal ions?

Yes, there is. The following is a plot of the hydration enthalpy versus row in the Periodic Table. Data was extracted from the Wikipedia, not the primary source (Ref 1).
Your thinking is right. The hydration entropy is evidently negative due to structuring of water (including loss of rotational degrees of freedom), becoming less negative for the larger ions (smallest loss of entropy for the cesium cation).

References

Burgess, John (1978). Metal Ions in Solution. Chichester: Ellis Horwood. ISBN 978-0-85312-027-8.


Can decarboxylation of carboxylic acids be effected by bases other than soda lime?

Decarboxylation can be effected by pretty much any base strong enough to produce the anion plus a high enough temperature. I've used DBU to effect an aromatic decarboxylation.
The role of the $\ce{NaOH}$ is this reaction is to capture the $\ce{CO2}$ as sodium carbonate.


Let $\Phi, \Psi$ be normalized wave functions.

Then $0 \le|\langle\Phi\vert\Psi\rangle| \le 1$, with $\langle\Phi\vert\Phi\rangle = 1$ and $\langle\Phi\vert\Psi\rangle = 0$ for orthogonal $\Phi$ and $\Psi$.

Since we can argue that any wave function completely contains itself and for two orthogonal wave functions none of them contains a part of the other, we can interpret the scalar product $\langle\Phi\vert\Psi\rangle$ as a measure for overlap.


Does concentration remain constant in an electrochemical cell's half cells?

But I was wondering that in a simple cell like Daniel cell if concentration of $\ce{Zn^{2+}}$ keeps increasing and that of $\ce{Cu^{2+}}$ keeps decreasing, won't the value of $\mathrm{Q_c}$ keep increasing?  In that case, won't that lead to a decreasing emf?

Have you ever wondered how battery meters on phones, laptops and other electronics can tell how charged your battery is? The question you just asked is the answer. As energy from the battery is consumed, the voltage will drop which can be correlated to a percent charge if known.

  If it does, then what's the use of such a cell that yields decreasing emf?

These cell are still useful.  Astute engineers know this and design around it using batteries that provide more voltage than needed when charged. If I have a $\pu{3.3 V}$ circuit, I can use a $\pu{3.7 V}$ Lithium battery (note $\pu{3.7V}$ is the charged voltage) to run the device without damaging it and will continue to work until the battery declines to below $\pu{3.3V}$ (actual/rated voltage). Similarly car batteries are typically rated for $\pu{12V}$ but if you measure it with a galvanometer (voltmeter) you will find the voltage is on the order of $\pu{13-14V}$.


Why does orthocresol behave as a skin irritant?

It is the phenol moiety of ortho-cresol that causes this. Phenol is classified as toxic because of the way it can penetrate the skin and cause damage.

In its neutral state, the phenol molecule is a slighlty polar but generally organic compound. Since our skin is designed to mainly keep water and water-soluble compounds out, it is lipophilic; this in turn helps lipophilic compounds to penetrate it and be absorbed by the body. As soon as phenol has been absorbed, it is again in an aquaeous environment. Here, phenol’s acidic properties ($\mathrm pK_\mathrm a = 9.95$) start playing a role: it can be deprotonated to the phenolate anion. This is hydrophilic and so will be able to dissolve in the water inside your body. If any membrane needs to be crossed, just add a proton to regenerate the lipophilic free phenol.

Another way in which phenol can often react is the single electron transfer mechanism to generate a phenol radical which can then cause havoc in cells. And finally, phenol should also be able to intercalate into DNA further causing damage.

So the stingy sensation is caused by the slight burns that you encounter once phenol has penetrated your skin. since neither ethanol nor vinegar can penetrate your skin as well as phenols can (they are too polar), these react differently and give you a different sensation.


Extracting iron oxide from ceramics waste

Assuming the iron oxide is just mixed and has not been fused with the other oxides you could use a magnet to pull it from the other oxides.

If it is fused with the other oxides, you could smelt the iron out with carbon. This will also smelt out the tin, which mixes with iron quite well. If you re-oxidized the iron in a furnace with air the iron will selectively oxidize first leaving tin metal when done. This is likely an impractical and cost-inefficient procedure as iron oxide is incredibly cheap relative to the cost of separation.


Will silicon dioxide react with hydrochloric acid?

According to the paper by Chung and Sturm [1], $\ce{HCl}$ will do a very small amount of etching at high enough temperatures (525 °C will give you about 1 nm/min), but the presence of hydrogen limits the reaction because the formation of $\ce{HCl}$ is preferred to the formation of $\ce{SiCl4}.$

References

Chung, K. H.; Sturm, J. C. Chlorine Etching for In-Situ Low-Temperature Silicon Surface Cleaning for Epitaxy Applications. In ECS Transactions; ECS: Chicago, Illinois, 2007; Vol. 6, pp 401–407. DOI: 10.1149/1.2727426.


Covalent Bonding in odd electron sharing molecules

I will answer to your question, but first let me show you that NO has a $2.5$ bonds, and I will use what I made for NO to answer your question. Here we go ! :)

Proof that NO has a $2.5$ bonds :

First the electron configurations of these two elements are :

$\ce{{}_7 N :1s^2 2s^2 2p^3}$ which have $5$ valence electrons in $4$ orbitals (one $\ce{s}$ and three $\ce{p}$)

$\ce{{}_8 O :1s^2 2s^2 2p^4}$ which have $6$ valence electrons in $4$ orbitals (one $\ce{s}$ and three $\ce{p}$)

Then we can build the orbitals-diagram of $\ce{NO}$ and I will suppose that its not a correlated diagram, and because of electronegativity of these elements verifies $\ce{\chi(N)<\chi(O)}$ then we have:

Then the electon configuration of $\ce{NO}$ is $\ce{(1\sigma)^2 (1 \sigma^{\star})^2 (2\sigma)^2 (1 \pi_{x})^2 (1 \pi_{y})^2 (1 \pi_{x}^{\star})}$

The bond order $i$ is : $$i=\frac{\text{number of electrons in bonding MOs}-\text{number of electrons in anti-bonding MOs}}{2}$$

Here we have $$i=\frac{2\times 4-3}{2}=\frac{5}{2}=2.5$$

Quod Erat Demonstrandum.

So now I did it for $\ce{NO}$ I have this configuration : $\ce{(1\sigma)^2 (1 \sigma^{\star})^2 (2\sigma)^2 (1 \pi_{x})^2 (1 \pi_{y})^2 (1 \pi_{x}^{\star})}$.

Then you said $3$ electrons bond considered has $0.5$ bond. As I understand your question, its the case if the configuration is $\ce{(1\sigma)^2 (1 \sigma^{\star})^2 (2\sigma)}$ (for example but it could be an other) then you have $3$ electrons in bonding bonds and $2$ in anti-bonding bond.

Then the bond order of your molecule is : $$i=\frac{3-2}{2}=0.5$$

Which conclude my answer.

NB:I'm seeking common examples of this kind.


Difference between internal energy of combustion and enthalpy of combustion?

Combustion releases energy from the system usually as heat and/or work to the surroundings, that's why the standard energy of combustion has a negative sign. The bomb is the system in this case (alternatively you can see the chemical as the system, but you will need to include the gaseous products in your system too, changing the volume of the system, which I find a less preferable viewpoint).

The value for the enthalpy will be lower since energy is trapped in the form of increased pressure introduced by the reactions products (mainly by $\ce{CO_2}$) in the bomb. This pressure can be used to do work on the surroundings (in an open system this will be usually on the atmosphere). If you determine the final pressure, you can calculate your $\Delta U$ back ($\Delta U = \Delta H - (p_{end}-p_{atm})V_{bomb}$).

What book do you use for the lectures?


How to differentiate between iron and sodium flames?

Sodium flame is yellow, but all the light is due to two lines in the yellow region. If you look a sodium flame through a purple glass (cobalt glass), the yellow color is absorbed, and you do not see the sodium flame any more. The flame looks dark and colorless through a cobalt glass.
On the contrary iron flame looks yellow, but it is made of a huge number of lines belonging to all regions of the spectrum. The sum of these colors look yellow, but it's a visual effect. The flame contains all colors. If you look at such an iron flame through a cobalt glass, the flame is visible. It looks purple or violet, but it is visible.


Can "Common Ion Effect" suppress the dissociation of water molecules in acidulated water?

You have a couple of concepts mixed together in the wrong way.

(1) The ionization of water is typically written as

$\text{K}_\text{w} = \ce{[H+] \cdot [OH^-]}$

The assumption is that the ionization is so low that the concentration of water remains a constant. For a solution which is half water and half acetic acid that wouldn't be true of course.

(2) The dissociation and recombination of water molecules is a very fast reaction and occurs continuously in water.

$\ce{2H2O <=> H3O+ + OH-}$

(3) In reality the autoprotolysis of water is dependent on the activities of the $\ce{H+}$ and $\ce{OH-}$. So the relation is more properly written as

$\text{K}_\text{w} = a_\ce{H+} \cdot a_\ce{OH^-}$

In dilute solutions the activity is essentially the same as the concentration, but in more concentrated solutions the activity drops as the concentration increases because of the formation of clusters of ion pairs in solution.

Also the activity drops as the overall ionic strength of the solution increases for the same reason. So if you add $\ce{NaCl}$ to the solution, then the ionic strength of the solution increases and the activity of the hydronium  ions and hydroxide ions decreases. So in concentrated $\ce{NaCl}$ the $\ce{Na+}$ and $\ce{Cl-}$ ions aren't really just "spectators."

(4) If you add $\ce{H2SO4}$ to distilled water then the acid will dissociate according to the following reactions:

$\ce{H2SO4 + H2O <=> H3O+ + HSO4-}\quad \quad \text{pK}_{\alpha1} = -3$

$\ce{HSO4- + H2O <=> H3O+ + SO4^{2-}}\quad \quad \text{pK}_{\alpha2} = 1.99$

The final pH will depend on the final concentration of the sulfate species which is commonly expressed as the molarity of the sulfuric acid. So if we diluting concentrated sulfuric acid we typically use

$\text{m}_1 \cdot \text{V}_1 = \text{m}_2 \cdot \text{V}_2$

where $\text{m}$ is the molarity of the solution and $\text{V}$ is the volume of the solution. This ignores the fact that in dilute solution the sulfuric acid isn't just $\ce{[H2SO4]}$ but that the overall sulfuric acid concentration is the sum of $\ce{[H2SO4]}$, $\ce{[HSO4^-]}$, and $\ce{[SO4^{2-}]}$.

(5) Acidifying an electrolysis solution isn't really about ion concentration since adding NaCl isn't as effective as adding HCl. The gist is that protons can move through solution very quickly since a water molecule can accept a proton from one direction to form hydronium and kick off a different proton in the other direction to form water again. So a net charge can move through solution quickly as a result of the chain reaction of such transfers. Thus any individual proton doesn't have to travel far.

CONCLUSION

The overall notion would be better expressed as that adding sulfuric acid to water increases the $[\ce{H+}]$ and reduces the $[\ce{OH-}]$ thus shifting the equilibrium for the autoprotolysis of water.


Does PCl3 react with ester as it does with carboxylic acid to produce an acid halide?

Yes it does. All carboxylic acid derivatives undergo nucleophilic substitution reactions. So esters, acid anyhyrides, amides all on reacting with $\ce{PCl3}$ will give acid chlorides.


The bond enthalpy is conventionally the enthalpy change of bond breaking, not of bond forming, in the contrary to formation enthalpy of compounds.
This way they are all positive.
The consequence is, the order of products and reactants in computation the enthalpy difference has to be switched.
So if the summary bond enthalpy of products is higher than of reactants, the difference of enthalpy as the reaction enthalpy is negative, not positive, and the reaction is exothermic.
Formation enthalpies:
$$\Delta H_\mathrm{r}=\Sigma  \left( \Delta H_\mathrm{f, products} \right) - \Sigma \left( \Delta H_\mathrm{f, reactants} \right)   $$
Bond enthalpies:
$$\Delta H_\mathrm{r}=  \Sigma   \left(  \Delta H_\mathrm{b, reactants} \right)   - \Sigma  \left(  \Delta H_\mathrm{b, products}  \right) $$


Is the photoelectric effect possible for semiconductors?

Given that Jaques Pankove's excellent book on 'Optical Processes in Semiconductors' devotes chapter 13 to 'Photoelectric Emission', you can rest assured that the photoelectric effect is alive and well in semiconductors.  While somewhat more complex than emission from simple metals, there is no need for two-photon processes. The 'tricky' part is that you have to look at where the electron states are, and then how much energy is needed to take it out of the semiconductor.

The work function is the energy difference from the Fermi energy to the vacuum level.  There may not be any states at the Fermi energy though.

The electron affinity is the energy difference from the bottom of the conduction band to the vacuum level.

One has to take the band structure into account, since direct and indirect transitions will likely have different energies required.
Also, surface band bending may make interpretation of experimental results more difficult. Again, perusing Pankove may prove useful.


Theoretical principles of molecular simulations

Computational chemistry is a synthesis of many different subjects, each of which is very complex on its own. So, if you really want to learn the theoretical basis, you should be prepared to put in a lot of time and effort. I don't want to discourage you, I just want you to have realistic expectations.

The fundamentals of computational chemistry are physical chemistry and computer science. The fundamentals of those are math, physics, and logic. So, depending on what your background is, you might have to start with math and physics.

Math:

Calculus
Vector Calculus
Differential Equations
Statistics

Physics:

Classical Dynamics
Classical Electrodynamics

Once you have taken those courses (if you haven't already) you can start with physical chemistry. You could conceivably teach yourself those subjects, but it would be very difficult - a free online course would be a good option if you can't take a university course. I believe Khan academy covers all of these subjects.

Physical Chemistry

Statistical Mechanics
Quantum Mechanics

I always liked Atkin's physical chemistry book (although I don't know if the newer editions are still good). McQuarrie's quantum chemistry and statistical mechanics are very good as well and are classics in the field.

You asked:

  Where do I start with the chemistry part?

  solving the Schrödinger equation
  Hartree–Fock method
  Density functional theory

Quantum mechanics in physical chemistry will cover solutions to Schrodinger's equation for all of the cases where a closed-form solution exists. This will give you an understanding of the basic principles, so you should start with that. To solve "real" systems, you need more advanced approximation techniques, and Hartree-Fock and DFT will get you started. That is a good order to study them in, as well.

For computer science, a lot depends on which software you will be using or what languages you want to use. Most of the core, highly optimized algorithms are written in fortran or C. Some go as far as assembly for the most important parts. Most of the software I have seen (non-core) is written in fortran or C++. My experience is a little out of date (I last worked in the field about 7 years ago) so that may no longer be true. At any rate, you will need at least a basic programming course in the language you want to use (C would be good to start with, since you will really learn the fundamentals). I would also take some object-oriented courses, so courses involving more advanced C++ and java would be good as well.

Key concepts (beyond programming and architecture) that you will need are:

Stochastic methods (Monte Carlo)
Linear Algebra
Numerical integration

With all of the above, you will have a good set of base skills and knowledge on which you can build.

You asked:

  Are there comprehensive literature for beginners and what would you recommend? A good book would certainly be the best start.

I am only familiar with one book that might be suitable for beginners in computational chemistry, so perhaps others can give better suggestions. The one I used was Molecular Modelling by Leach. It gives a good overview of common techniques, but doesn't go too in-depth. However, it was last updated in 2001 - so it may be out of date. The state of the art moves pretty quickly in terms of detailed application, although the fundamentals tend to stay pretty consistent over time. Another classic text which covers the basics in some depth is Allen & Tildesley's Computer Simulation of Liquids.

Both of these books focus more on the modeling aspects of molecules as opposed to chemical reactions, so if you are more interested in simulating reactions as opposed to statistical mechanics or protein-ligand interactions, you would need different books.


What would follow in the series sigma, pi and delta bonds?

tl;dr The next in the series is called φ bond. There is even a tiny Wikipedia article about it.

Nicolau pointed me to the Wikipedia article, that had at the time a tiny section about the φ symmetry of the bond. Ben also kindly agreed with my naming proposition. I'd like to back up just a little bit an quote one sentence of this article:

  The type of interaction between atomic orbitals can be further categorized by the molecular-orbital symmetry labels σ (sigma), π (pi), δ (delta), φ (phi), γ (gamma) etc. paralleling the symmetry of the atomic orbitals s, p, d, f and g. The number of nodal planes containing the internuclear axis between the atoms concerned is zero for σ MOs, one for π, two for δ, etc.

Basically this was what I was already thinking, but since the section is quite tiny overlooked it accordingly. But since this is a reference request, I'll go a little deeper.

The concept of the bond is actually quite old and I was able to track it down to at least 1984, where Bruce E. Bursten and Geoffrey A. Ozin "demonstrate[d] for the first time the existance of φ bonds between metal atoms."[1] A little further back, 1978, N. Rösch and A. Streitwieser already performed SCF-Xα calculations on thorocene and uranocene, where they came across f orbital contributions in bonding. Yet, they did not coin the phrase φ bond.[2] Following that, there have been more theoretical studies on the diuranium molecule, with increasing level of theory.[3] One of the more recent studies is by Laura Gagliardi and Björn O. Roos, which is available free of charge (see below).[4] Xin Wu and Xin Lu also studied the phenomenon in endohedral matallofullerenes, i.e. diuranium in a Buckminster fullerene.[5]

It is quite safe to assume, that there are more studies about this research topic or in a more general aspect the research on molecules with high bond orders. For this purpose I will include a couple of good reads.[6]

References and further reading

Bruce E. Bursten, Geoffrey A. Ozin, Inorg. Chem., 1984, 23 (18), 2910–2911.
(a) Rösch, N., Streitwieser, A., Jr., J. Organomet. Chem., 1978, 115, 195-200. (b) Rösch, N., Streitwieser, A., Jr., J. Am. Chem. Soc. 1983, 105, 7237-40.
Melanie Pepper, Bruce E. Bursten, J. Am. Chem. Soc., 1990, 112 (21),  7803–7804.
Laura Gagliardi, Björn O. Roos, Nature, 2005, 433, 848-851. Available free at Archive ouverte UNIGE.
Xin Wu, Xin Lu, J. Am. Chem. Soc., 2007, 129 (7), 2171–2177.
Without order: (a) Gernot Frenking, Nature, 2007, 446, 276-277.
(b) Gernot Frenking, Science, 2005, 310, 796-797.
(c) Gabriel Merino, Kelling J. Donald, Jason S. D'Acchioli, Roald Hoffmann, J. Am. Chem. Soc., 2007, 129 (49), 15295–15302.
(d) Fernando Ruipérez, Gabriel Merino, Jesus M. Ugalde, Ivan Infante, Inorg. Chem., 2013, 52 (6), 2838–2843.
(e) Jia Zhou, Jason L. Sonnenberg, H. Bernhard Schlegel, Inorg. Chem., 2010, 49 (14), 6545–6551.
(f) Awal Noor, Rhett Kempe, Inorg. Cimica Acta, 2015, 424, 75-82.
(g) Frank R. Wagner, Awal Noor, Rhett Kempe, Nature Chemistry, 2009, 1, 529-536.
(h) Björn O. Roos, Antonio C. Borin, Laura Gagliardi, Angew. Chem. Int. Ed., 2007, 46 (9). 1469-1472.


Ethylbenzene yields acetophenone when it undergoes Étard reaction.
The mechanism is as follows:


Why does Birch reduction of naphthalene with isoamyl alcohol form tetralin instead of 1,4-dihydrodialin (as formed with C2H5OH)?

As I noted in a Comment that there should be no significant difference between the use of ethanol or isoamyl alcohol in the reduction of naphthalene with sodium in that both alcohols are primary. The difference lies in the conditions of the reaction. The sequence is a classic case of kinetic vs. thermodynamic conditions.
The reaction conditions in the diagram are from this paper.
Naphthalene (1) reversibly adds an electron from sodium under ambient conditions to form the resonance stablized radical anion 2 that is shown in a canonical form. This species is readily protonated by an alcohol, be it primary or tertiary as in this instance, at the site of highest electron density to afford the resonance stabilized radical 3. The addition of a second electron to 3 in this 2-electron reduction gives rise to the resonance stabilized anion 4, which, in turn, is protonated to form the deconjugated 1,4-dihydronaphthalene 5, the kinetically formed product. Just like cyclohexene is unable to reduce under these reaction conditions, neither can alkene 5.  During the course of this reduction, something else is formed -- namely -- a base, sodium tert-butoxide.
To achieve reduction of the double bond in 5 it is not sufficient to conduct the reaction at reflux but rather to conjugate the double bond with the aromatic ring. This is the role of the base at elevated temperature. Deprotonation of 5 at the benzylic site provides resonance stabilized anion 4 again, which can reversibly protonate to form 1,2-dihydronaphthalene (dialin) 6. This styrene-like double bond has a low reduction potential. Addition of an electron produces the radical anion 7 that protonates to form stabilized benzyl radical 8. Subsequent addition of a second electron affords the benzylic anion 9. This anion is readily protonated by the alcohol to form 1,2,3,4-tetrahydronaphthalene 10 (tetralin).


What is the main difference between Quantum Clusters and Nanoclusters?

Short answer: "quantum clusters" is not widespread and likely confusing. I'd suggest "small nanoparticles" or perhaps "metal nano clusters."

I can't comment on optical shifts, but many of these properties are distinct to the particular material and the onset of quantized energy levels in that material. So I doubt you can simply combine properties of quantum dots and nanoparticles.

I asked a colleague who synthesizes gold nanoparticles. She hasn’t heard anyone use the term "quantum clusters" and thinks people would find it very confusing.

Typically the usage:

Quantum Dots refer to semiconductor nanoparticles, e.g., $\ce{CdS}$ or $\ce{PbSe}$ etc.
Nanoparticles refer to nanometer-sized particles of basically anything (proteins, polymers, metals, etc.)
Clusters refer to very small particles (i.e., <1-2 nm diameter) with well-defined numbers of atoms (i.e., a few hundred at most). There is an IUPAC Gold Book definition:

  A number of metal centres grouped close together which can have direct metal bonding interactions or interactions through a bridging ligand, but are not necessarily held together by these interactions.

For reference, Wikipedia suggests nanoclusters are aggregates of nanoparticles. I don't think that's in common use, but certainly it suggests your suggested term would be very confusing.


Why do some element symbols contain a single letter while others contain two letters?

There is a bit of history behind this.

In the website Development of the chemical symbols and the Periodic Table, it is stated that the Swedish chemist Berzelius updated Dalton's previous method.  He established a protocol for the symbols, still in use today:

  In the class which I call metalloids, I shall employ the initial letter only, even when this letter is common to the metalloid and some metal.
  In the class of metals, I shall distinguish those that have the same initials with another metal, or a metalloid, by writing the first two letters of the word.
  If the first two letters be common to two metals, I shall, in that case, add to the initial letter the first consonant which they have not in common:

  for example, S = sulphur, Si = silicium, Sb = stibium (antimony), Sn =
  stannum (tin), C = carbonicum, Co = cobaltum (cobalt), Cu = cuprum
  (copper), O = oxygen, Os = osmium, &c

The basis of the symbols are the Latin word for the elements.

A notes as what Berzellius was refering to when he used the term 'metalloid', from the Wikipedia page Origin and use of the term metalloid:

  In 1811, Berzelius referred to nonmetallic elements as metalloids, in reference to their ability to form oxyanions.

.


Why isn't organic chemistry taught in 3D?

Organic chemistry is often taught in 3D.  There have long been 3D molecular models used in classrooms, as well as Fischer projections.

Sometimes 2D is enough for a given situation.  Maybe curved 4-dimensional spacetime is needed for some situations, vicinity of neutron stars or black holes lets say.

Use the simplist description that includes all the relavent concepts.


Is the half-full rule and full rule followed in the 6th and 7th periods?

Short answer: no! The first known exception to occur in the periodic table is in period 5:  niobium’s ground state electronic configuration is

  $\ce{Nb: [Kr] 5s^1 4d^4}$

which is not warranted by the “usual” rules for determining electronic configuration. So, the “half-full rule” is not sufficient in periods 5 and higher.

Let’s look at Wikipedia’s list of exceptions to Madelung’s rule. In period 4, the only exceptions are $\ce{Cr}$ and $\ce{Cu}$, which are accounted for by the “half-full rule”. (Don’t bother too much about the dispute for Ni.)

In period 5, the exceptions to Madelung’s rule are classified in three groups:

$\ce{Mo}$ and $\ce{Ag}$: they are the analogues of Cu and Cr
$\ce{Pd}$ is $\ce{5s^0 4d^10}$: a different type of application of the explanation that “fully filled shells are particularly stable”, because in this case two electrons from the s shell were pulled into the d shell. Not what you usually learn.
$\ce{Nb}$ is $\ce{5s^1 4d^4}$,  and $\ce{Ru}$ is $\ce{5s^1 4d^7}$: these cannot be explained by simple rules, and one has to perform complex quantum chemistry calculations to understand these electronic configurations.

Period 6 contains further inconsistencies, because of the introduction of $\ce{f}$ orbitals into the mix.


Can a substance donate a proton without the proton getting accepted by another substance?

The closest thing I can find to this involves a most unlikely proton donor.  Given an appropriate base, sodium hydride donates a proton for encapsulation in the base (rather than forming a localised covalent bond), leaving the valence electrons on sodium (!) in inverse sodium hydride.  The proton is not free (a completely free proton is possible only in plasma), but analogous to complexed metal ion or solvated electron.


Carbonation of Grignard reagent with a carbonyl group present in it

You should look at two things:

1)  Will the reaction stay on course?  If you add water to the Grignard reagent first, without the carbon dioxide initially, what do you get?  Is it something that will get you your carboxylic acid when you add the carbon dioxide?

2)  Can you make the Grignard reagent to begin with?  Here again you are asking the reaction to stay on course; if the Grignard function can react with another functional group in the molecule, it will do so either within the molecule or as an intermolecular reaction between two of the molecules.
Then you no longer have your Grignard reagent.

Only one of the given possibilities works against both of these questions, so there is your answer.


To expand upon the comments, the y axis in this graph is simply the log of the concentration in pg/ml.  To convert you use the equation $$C = 10^{y}$$ where y is the y value and C is the concentration.  Note that in older literature, log was oftentimes assumed to be base e whereas nowadays we assume base 10.  The figure (thankfully) clarifies this potential discrepancy.

A final note: log values are unitless so the y axis has no units.  The units are reported here so that you know what the appropriate units of the converted value should be.


Why does increasing the temperature for a reaction at equillibrium cause the reaction to shift to the endorthermic side?

Think about it this way:

There are two people. One guy likes chocolates and one guy hates them. They decide to transfer the chocolates. If you start throwing chocolates randomly to both of them, in which side will the reaction go? The guy who hates chocolates would keep giving it to the guy who likes them.

In a similar way, if you provide heat to a reaction, it would move in the direction of the where the heat is absorbed, i.e the endothermic pathway.


What chemical could be used as carbon monoxide scrubber?

Air reacting with charcoal will produce $\ce{N2},$ $\ce{CO}$ and $\ce{CO2}.$ $\ce{CO2}$ will be absorbed by calcium hydroxide or sodium hydroxide.
However, $\ce{CO}$ cannot be absorbed at ordinary temperatures and pressures. Only at $\pu{200 °C}$ and $\pu{10 atm},$ it reacts with $\ce{NaOH}$ to produce sodium formate $\ce{HCOONa}.$
Finally, $\ce{CO}$ can be oxidized into $\ce{CO2}$ by passing over hot $\ce{MnO2}$ or $\ce{Fe2O3},$ and then be absorbed by a hydroxide.


Direction of external cell when voltage is supplied by it to Daniel cell to oppose cell potential

Suppose two identical Daniell cells are connected with the zinc electrodes electrically connected through an amperemeter, and the two copper plates also electrically connected. Of course no currant will be measured by the ammeter. But the tension between the plate is independent on the connections. Now if one of the cell has a higher nominal voltage, maybe because of aa lower zinc ion concentration, this "stronger cell" imposes its capacity of producing electrons to the other one. Electrons will be flying through the ammeter. So the weakest cell has to reverse the sense of its reactions on its plates. The cell will work as in an electrolysis. Zinc ions $\ce{Zn^{2+}}$ will be reduced into metallic zinc in this cell. Slowly the two zinc ion concentrations will tend to equalize themselves in both cells.


What is the mechanism for the oxidative cleavage of diketones via permanganate under basic/hot conditions?

In J. Org. Chem. 1979, 44 (15), 2726–2730, the authors Lee and Chang have investigated the permanganate oxidation of alkynes to α-diketones. In addition, they investigated the oxidative cleavage of several α-diketones to carboxylic acids. In the case of hexadecane-8,9-dione, permanganate oxidation under at least three sets of conditions provided n-octanoic acid 8 (major) and n-heptanoic acid 7 (minor). They suggested that the oxidation of the dione 1 leads only to octanoic acid 8 via intermediate 4 while oxidation of enol 2 affords 2-oxononanoic acid 6 and ultimately octanoic acid 8 via oxidative decarboxylation. In addition, oxidation of heptanal 5 affords heptanoic acid 7. The three solvent systems employed were acetone/aqueous HOAc, aqueous H2SO4 and water. Starting material was recovered in each experiment. Tetradecane-7,8-dione showed similar behavior.
For the mechanism of permanganate oxidation of alkynes to α-diketones, see the accompanying post by @Raul Luciano.


Why did they provide the value for the enthalpy change of fusion?

You calculated the energy required to change solid gallium at 20°C to solid gallium at 29.76°C.  Once solid gallium reaches 29.76°C, additional energy is required to melt it.  The energy required is given by the $\Delta H_{\text{fusion}}$.  So to complete the question, transform the $\Delta H$ units to a grams basis instead of a mole basis, multiply it by the weight of gallium in the problem, and then add that result to the one you already calculated.  That will be the energy required to change solid gallium at 20°C to liquid gallium at 29.76°C.


What does it mean for an equillibrium to move?

Let's keep this simple. Think about adding vinegar (acetic acid) to water.

$$\ce{CH3COOH + H2O <=> H3O^+ + CH3COO^-}$$

The equilibrium equation is

$$\ce{K_a} = \dfrac{\ce{[H^+][CH3COO^-]}}{\ce{[CH3COOH]}}$$

Le Chatelier's principle simply says that if I add more acetic acid to the water then the solution gets more acidic. However if I add more water, then the solution gets less acidic. Hence the equilibrium moves.(In this case it works to a point. Adding a drop of water to glacial acetic acid isn't an aqueous solution...)


Oxidation and reduction in organic chemistry

Addition of oxygen and hydrogen can be useful to predict oxidation and reduction in most of the cases but this criteria fails to classify a reaction as oxidation or reduction when there are no oxygen or hydrogen atoms involved. This method is very primitive.

A better perspective :

Oxidation : Increase in oxidation state (by loss of electrons) of central atom.

Reduction : Decrease in oxidation state (by gain of electrons) of central atom.

If you already know how to find the oxidation state of the central atom (in your case carbon), then I hope you will be able to get your answer now.

The answer to the question why oxidation of alcohol to carboxylic acid takes 2 oxidation reactions, lies in the reaction mechanism. These mechanisms cannot be challenged, they are what they are. We can only observe and provide a acceptable explanation to why they happen like the way they do.


You have done everything correctly so far. Your calculator might not be able to factor the expression as it is written. It is a cubic equation with three roots and two discontinuous points. You may be used to ICE-type equilibrium problems where you can simplify parts of the expression because $x$ is small compared to the initial concentrations. This is not the case in this problem. Factoring the cubic polynomial is not always easy.

Your mass action expression can be rearranged to the following cubic equation so long as $x \ne -0.02$ and $x \ne -0.0625$, both of which are nonsense anyway. If your calculator cannot factor this expression, try using Wolfram Alpha.

$$0.3 x^3-3.9565 x^2+0.201922 x-0.00247656 = 0$$

Trying solving this equation. It has three roots, two of which are nonsense since they yield negative final concentrations. The remaining root is the answer.


Does the stoichiometry effect the value of the equilibrium constant of an equilibrium reaction?

$$K_c = \mathrm{products \over reactants}$$ So this is done by multiplying the concentrations of the products and dividing by the concentrations of the reactants. The concentrations are put to the power of their coefficient. So in your problems at equilibrium $K_c$ would be:
\begin{align}
  K_c &= \frac{[\ce{C}]^1}{[\ce{A}]^{\frac{1}{2}}[\ce{B}]^{\frac{3}{2}}}\tag1\\
  K_c &= \frac{[\ce{C}]^2}{[\ce{A}]^1[\ce{B}]^3}\tag2\\
\end{align}

Note that $[\ce{A}]$ means concentration of $\ce{A}$.


Why is meta-nitrophenol less acidic than para-nitrophenol?

The acidity of nitrophenols (or any acid for that matter) is determined by the stability of the conjugate base. In the case of m-nitrophenol and p-nitrophenol, the relative stability can be determined by looking at the resonance structures.

You can see that p-nitrophenol has an additional resonance structure where the negative charge is delocalised onto the oxygens of the nitro group, which are highly electronegative and therefore stabilise the charge effectively. This stabilisation is not possible in the case of m-nitrophenol because there is no opportunity via resonance for delocalisation onto the nitro group.


What is the exact meaning of the term 'isostructural' molecules?

Two molecules are isostructural if they have a similar number of atoms arranged in a similar structure (spatial coordinates). In your example, both $\ce{NO3^-}$ and $\ce{SO3}$ have same number of atoms, and they have similar spatial arrangement as well. So they are isostructural. Sometimes using hybridization of the central atom works in finding out, but not always. For example $\ce{SCl3}$ and $\ce{BF3}$ are isostructural, even though the former is $\text{sp}^3\text{d}$, while the latter is $\text{sp}^2$.


Reuse 20 litre plastic drums for drinking water?

Regardless of what it takes to remove any traces of hydrochloric acid from the drum, the container is probably not certified as food-grade. So in addition to any risks from the acid itself (including things like heavy metals, which are common contaminants in mineral acids), the plastic is not certified to be free of toxic substances left over from manufacturing or that the plastic will not degrade to release such substances over its lifetime.


You can create distance constraints between two atoms (set them to the distance you expect for the hydrogen bonds), and then do a force field minimisation.

It's in the menu called Geometry, and then "Constrain distance". You pick the atoms, and click the little lock in the lower right-hand corner, and then enter the distance you want next to the lock.


How do you work out whether reduction would occur based on reduction potentials

First, using observations, since the E value for fluorine's reduction is higher, fluorine more preferentially undergoes reduction compared to chlorine. Furthermore, if you know your periodic table, fluorine being higher in the group more readily gains a valence electron (with fewer electron shells and hence a stronger force of attraction) and would be reduced preferentially and oxidise chloride ions.

Now using the numbers

We know the following formula: $$E^o_{cell}=E^o_{cathode}-E^o_{anode}$$
And also that oxidation is at the anode and reduction at the cathode.

Now, if and only if $E^o_{cell}$ is positive would we have a spontaneous reaction.
All that is left is to consider the two cases and see which would work.

Case 1: $\ce{2 F- +  Cl2 -> 2Cl- + F2 }$

Reduction Equation (Chlorine): $\ce{Cl2 + 2e- -> 2Cl-  }$,  $E^o_{cathode} = 1.36\,V$

Oxidation Equation (Fluorine): $\ce{2F- -> F2 +2e-  }$,  $E^o_{anode} = 2.87\,V$

Then $$E^o_{cell}=1.36\,V-2.87\,V=-1.51\,V\lt0\,V$$

Hence this reaction would  not be spontaneous and not take place in this direction.

Case 2: $\ce{2 Cl- +  F2 -> 2F- + Cl2 }$

Reduction Equation (Fluorine): $\ce{F2 + 2e- -> 2Cl-  }$,  $E^o_{cathode} = 2.87\,V$

Oxidation Equation (Chlorine): $\ce{2Cl- -> Cl2 +2e-  }$,  $E^o_{anode} = 1.36\,V$

Then $$E^o_{cell}=2.87\,V-1.36\,V=1.51\,V\gt0\,V$$

Hence this reaction would be spontaneous and take place in this direction.


Does salt on the snow and ice covered roads and sidewalks make the nearby air temperature cooler?

As water melts due to adding a salt (AKA melting point depression), it absorbs energy from the environment and the temperature of the salt/ice/water mixture does, indeed, drop. Whether that is sufficient to cool the air above the ice perceptibly, though, is a matter of opinion.

Another effect that might make the air feel colder is increased humidity, particularly as the salt slush is splashed into the air by vehicles.

You could perform an experiment yourself to determine the air temperature above two containers, one with fresh snow and the other with snow and salt. If you do, please post your results as an answer!


Nope. Carbonate is $\ce{CO}_3^\color{red}{2-}$ (an ion), not $\ce{CO3}$. Carbonite is $\ce{CO}_2^\color{red}{2-}$.

Carbonation involves dissolving $\ce{CO2}$ gas in water. It turns out that $\ce{CO2}$ reacts (maybe not the best term to use here*) with water via the following reaction $$\ce{CO2 + H2O -> CO3^{2-} + 2H+}$$

So you have effectively carbonated the soft drink.

When you depressurize the bottle by opening it, the reverse reaction occurs and you get carbon dioxide.

*Such dissociation is normal and integral to dissolution of polar solutes in water, so generally it is considered as part of the dissolution


Can we justify that "For sublimation of a solid at 1 atm  $\Delta U>0$ at low temperature and $\Delta U<0$ at high temperature?"

No.
$\Delta U>0$, always, for sublimation, because of the energy needed to separate the atoms or molecules in changing from the solid to the gas phase.
As for the enthalpy,
$$H = U +PV \Rightarrow \Delta H = \Delta U+\Delta (PV)=\Delta U+ P\Delta V+ V \Delta P$$
Thus at constant pressure:
$$\Delta H = \Delta U+ P\Delta V \approx  \Delta U+ nRT$$
[Here I've used the ideal gas law to approximate the volume of the gas, and ignored the volume of the solid, which (at 1 atm, room temperature) is about 3 orders of magnitude smaller.]
I.e., $\Delta H$ will be even more positive than $\Delta U$, because of the pV-work required to make space for the gas.
The reason why sublimation becomes more favorable as the temperature increases is because sublimation has a positive $\Delta S$, and $\Delta G = \Delta H - T \Delta S$.  But, irrespective of temperature, at any temperature and pressure at which the solid can exist,  $\Delta U_{solid->gas} >0 $.
$\Delta U >0$ for liquid->gas and solid->liquid phase transitions as well (with the possible exception of the solid->liquid phase transitions for helium-3 and helium-4 at extremely low temperatures which, at least according to https://en.wikipedia.org/wiki/Enthalpy_of_fusion, have $\Delta H < 0$; but solid helium doesn't exist at the 1 atm pressure specified by the OP).
*Yes, you might be able to come up some extreme hypothetical mechanical system where the solid is under such great pressure that the intermolecular forces are so far into the repulsive part of their potentials that $\Delta U_{solid->gas} <0 $.  But that's clearly not what the OP had in mind, since s/he specified 1 atm.  And, in addition, such an arrangement would require that only the solid, but not the gas, be at that extreme pressure.  So, strictly speaking, that $\Delta U$ would not be for the sublimation alone, it would be for the sublimation plus the pressure change.


Aetiology of the word "fullerene" in Chemistry

Fullerenes are distinct (meaning every fullerene type has a well defined structure) carbon-only molecules, where all atoms have three bonds ($sp^2$, or nearly $sp^2$, as they're not exactly planar).

Leaving out very freakish topologies (which are unlikely to be stable and not fullerenes), this leads to a closed shell-like structure where all surface elements are pentagons or hexagons. (Which are not perfect, bond lengths differ slightly if the edge is shared by two pentagons, two hexagons, or one-and-one.)

The simpler, smaller forms are spherical (C60) or prolate (C70, ...), larger ones can probably have a surface that is not completely convex. (anyone seen oblate fullerene structures?)

A carbon nanotube with a well defined, closed end structure can be a fullerene, but they usually haven't got one.


Does 1,3‐dichloroallene show geometrical isomerism?

This is a chiral compound; it has two enantiomers.  This source defines "geometrical 8somerism" as isomerism involving different geometrical arrangements, and a pair of nonsuperimposable mirror images would seem to qualify.  But in the body they talk about about cis/trans or E/Z isomers, not chiral compounds.  The latter, which applies here, should be described by more precise terms like "enantiomers".


Find the Ionisation Potential and Electron affinity of X

The Ionisation Energy of an atom is defined as the minimum amount of energy required to remove the most loosely bound electron of an isolated neutral gaseous atom or molecule. In this case, since $E_1$ amount of energy is used to ionise $N_0/2$ atoms, the ionisation energy of each atom is $\frac{E_1}{N_0/2} = \frac{2E_1}{N_0}$.
The Electron Affinity is defined as the amount of energy released when an electron is attached to a neutral atom or molecule in the gaseous state to form a negative ion. In this case, $E_2$ amount of energy is released when $N_0/2$ atoms accept electrons to form negative ions, hence electron affinity of each atom is $\frac{E_2}{N_0/2} = \frac{2E_2}{N_0}$.
Your answers are correct. The textbook must have made a typo in the second question. For the textbook's answer to be correct, the second question should read

$N_0/2$  atoms of $\ce{X+_{(g)}}$ are converted into $\ce{X−_{(g)}}$ by energy $E_2$

This gives the answer as $\frac{E_2 - E_1}{N_0/2} = \frac{2(E_2 - E_1)}{N_0}$ for electron affinity.


How to predict if hydrogen atoms could be placed in the interstials of a lattice

Your intuition is correct; $\ce{NbH_{x}}$ retains the bcc Nb lattice, with the hydrides in the tetrahedral interstital sites, up to x = 2, when the compound switches to a fcc Nb lattice (the Fluorite structure). This is pretty typical behaviour for a transition metal hydride.

I don't think radius ratio rules are going to be useful in predicting the behaviour of these systems; there is a lot of covalency going on (hydride-hydride interactions and Nb-Nb interactions), and the metal lattices will end up being distorted by the correlated interstital occupancies of the hydrides. Hydride has a radius of around 130 pm while Niobium has a radius of around 150 pm so you wouldn't predict the observed behaviour based on radius ratios.

More information on this particular system can be found in Hauck (1976), Ordering of Hydrogen in Niobium Hydride phases.


How do metal's density affects the corrosion rate?

The reason you need density is to determine the mass lost by the system, assuming that the equations you are talking about look something like:

$$
m_{loss} = k\rho S_{area} t
$$

Where $k$ is a constant defining how fast the corrosion is occurring in your system (in this case k would be in units of [length]/[time]), $\rho$ is the density of the material, $S_{area}$ is the surface area, and $t$ is time: the equation outputs the total mass of material lost.

In more comprehensive corrosion equations $k$ would be substituted with equations describing the system causing the corrosion and may be dependent on (not a complete list): pH, temperature, surface activity, liquid flow, concentration of certain chemicals, passive layer growth, ect.


Is this a phase change in a gas under Thermodynamic Equilibrium?

I don't think it necessarily suggests a phase change.  It just seems to me that, at the higher pressures, we are seeing non-ideal gas behavior setting in.


ligand pi system irreducible representation for trigonal bipyramidal transition metal complex

I believe you have the wrong character for the $\hat{C_3}$ rotation. None of the p orbitals in the equatorial positions remain in place so they all contribute zero. The axial p orbitals are rotated by $120^°$ which means they each contribute $\cos(120^°)=-\frac{1}{2}$. Since there are two of them at each of the two axial position, $\chi(\hat{C_3})=-2$. Using this new value for the character (assuming all the others are right, which they seemed to be), I obtained: $a_2',a_2'',2e',2e''$


When is phosphorus a polyatomic molecule?

Phosphorus is most common as white phosphorus (but more stable as red phosphorus) and in this form, it exists as a tetrahedron, or $\ce{P4}$. Chlorine is most stable as a diatomic molecule, as you may have learnt it with the other common gases that exists as diatomic molecules


Is this Newman projection energy profile for 2,3-dihydroxypropanal correct?

There seems to be only one problem. The 240° conformer* will be lower in energy as compared to the 120° conformer as there will be hydrogen bonding between the two hydroxyl groups
.
Rest all looks good.