How can I compute molar concentration if I am given parts per trillion?

I have the feeling, that this is a superbly stupid question. And I also have the feeling that it is quite possible, that my answer will be wrong.
First of all the parts-per-notation should be avoided as it is not compliant with SI and highly ambiguous.
Secondly, it generally may only refer to a unitless number - a molar concentration has a unit ($\mathrm{mol/L}$).

The here used unit may refer to a couple of things. In general its meaning should most likely refer to $\mathrm{PPTV = 10^{-12}L/L}$. Sometimes the use of $\mathrm{PPTV = 10^{-12}g/L}$

From here you can probably make your best guesses, determining how many moles of compound do you have:
\begin{align}
pV &=n\mathcal{R}T\\
n  &= \frac{pV}{\mathcal{R}T}\\
n  &= \frac{101325~\mathrm{Pa}\cdot 261\cdot10^{-12}~\mathrm{L}}{8.314~\mathrm{\frac{J}{mol\cdot K}}283.15~\mathrm{K}}\\
n  &= 1.12\cdot10^{-8}~\mathrm{mol} &\implies c&=  1.12\cdot10^{-8}~\mathrm{\frac{mol}{L}}
\end{align}

If you have the mass given, then you need to know which gas it is, as $n=\frac{m}{M}$.

Another possible use could be the particles by volume notation, so it could also refer to $\mathrm{PPTV = 10^{-12}/m^3}$. (Apparently this is somewhat common in ecological air measurements.) Here you simply have to use $n=\frac{N}{\mathcal{N}_\mathrm{A}}$ to figure out the concentration in $\mathrm{mol/L}$.


Are there any acid-base indicators whose protonated and deprotonated forms have the same number of resonance structures?

In the examples you drew, these resonance structures are not ‘opened up’; they are there in both the protonated and the deprotonated form. However, since the protonated form requires charge separation to access these resonance structures, they are less important.

It rarely makes sense to do resonance structure bookkeeping like you are trying to implement in your question.

Indicators need to display a colour change upon protonation/deprotonation. that can either happen by modifying the structure (e.g. bromocresol green) or by the electronic changes induced by the charge occurring. The latter case will mean some shifting in the orbital energies — most importantly in HOMO and LUMO — and can practically always be explained in a ‘resonance way’.


Reactions of tin (IV) iodide with potassium iodide and water

If iodide is oxidized to iodine (the cause of the yellow color), then something needs to be reduced in turn. Potassium is already in its most stable oxidation state (+I), so tin will be reduced, from $\ce{Sn}$(IV) to $\ce{Sn}$(II).

$$\ce{SnI4 + 2H2O -> SnO2 + 4HI}$$

$$\ce{SnO2 + 2HI -> I2 + SnO + H2O}$$

With $\ce{KI}$, a yellow-brown solution of $\ce{KI3}$ will be formed, which contains the triiodide anion $\ce{I3-}$.

$$\ce{I2 + KI -> KI3}$$

When acetone is added, the yellow color of $\ce{I2}$ will slowly disappear as iodoacetone and $\ce{HI}$ are formed in an acid-catalyzed reaction (source):

$$\ce{(CH3)2CO + I2 -> CH3C(O)CH2I + HI}$$


Good question!  As the following figure illustrates, there are 2 ways we can protonate an allene.

(image source)

The top line in the drawing starts by protonating at the central allene carbon.  This generates a very unstable primary carbocation.  Initially, this primary carbocation is not stabilized by the adjacent double bond.  It is not until after this carbocation rotates 90° that its p orbital is lined up with the p orbitals in the double bond and allylic stabilization can occur.  Since there is no immediate allylic stabilization when the carbocation is formed, and since it is a primary carbocation, the pathway leading to it is very high energy.

On the other hand, protonation at the terminal allene carbon (bottom line of figure) produces a relatively stabilized secondary carbocation.  Formation of a secondary carbocation is a much lower energy process.  Therefore, protonation at the terminal allene carbon is lower energy and is what will occur.

Edit: Response to OP's comment

  I always thought that a positive charge appearing on sp3 carbon (which
  then becomes sp2) is more stable than a positive charge appearing on
  sp2 carbon

Generally we think of vinylic carbocations as high energy.  That is true, but what is meant by "vinylic" carbocation is different from what we have here.  Here there is an empty p orbital.  In a vinylic carbocation the empty orbital is not a p orbital, but rather something like $\ce{sp^2}$ or $\ce{sp}$ hybridized.

  is the secondary carbocation more stable because of inductive effect,
  the rotational barrier in the primary carbocation, or both

The secondary carbocation is more stable for the same reasons that we have the  tertiary > secondary > primary carbocation stability order.

Inductive effects, $\ce{sp^3}$ and $\ce{sp^2}$ carbons are electron releasing towards an $\ce{sp}$ carbon due to electronegativity differences
Resonance effects from hyperconjugation involving hydrogens on the carbon next to the carbocation center.


How can a vacancy defect be a stoichiometry defect?

Yes, you are not completely wrong but try to answer these questions to check your understanding (mouse over for the answers):

First of all, what is a Non-stoichiometric compound?

   Non-stoichiometric compounds are chemical compounds, almost always
 solid inorganic compounds, having an elemental composition whose
 proportions cannot be represented by integers.

In which way a vacancy affects the ratio of the elemental composition?

   If the compound it's a "pure" metal the vacancy will not affect at all the ratio of the elemental composition. However, it could affect the ratio if, for example, we have $\ce{NaCl}$ crystal where a lot of $\ce{Na+}$ ions are missing.

How is likely to have an ionic compound where a lot of anion or cation are missing?

   Not at all, the ionic compound, usually, maintain an overall neutral charge see Schottky defect.


Equation to find specific gravity (or density) from a solution of sugar and water

Here an empirical approach:

1) A simple way to calculate density: mass/volume

mass = x g of sugar + (y L of water*1000 mL/L*1g/mL)

1 g/mL = water density

Then measure the total volume (total mL) of the mix:

density = (x g of sugar + (y L of water * 1000 mL/L * 1g/mL))/(total mL)

2) For the reverse calculation, you can determine density as a function of sugar content by preparing solutions with different sugar concentrations and use this to interpolate your specific density value to a sugar concentration.

You can also use preexisting tables or graphs as Felipe suggested


Are the C-C bond lengths different in 2-methyl propene

The cases of benzene and 2-methyl propene are different.

In benzene, all carbon atoms are equivalent. One can also imagine the electron resonance.

In case of $\ce{CH3-C(CH3)=CH2}$ , the electrons have no place to move to create a delocalization cloud.

Additionally,  it would have to switch a hydrogen atoms between 3 atoms,
or create highly unstable $\ce{C=CH3}$

Such molecule would be highly reactive.


$K_a$ is not unitless (in this case). Some equilibrium constants may have their units totally cancel, but the acidity constant has units of concentration:

$$K_a = \dfrac{[\ce{H+}][\ce{A-}]}{[\ce{HA}]}\implies\dfrac{\frac{\text{mol}}{\text{L}}\cdot\frac{\text{mol}}{\text{L}}}{\frac{\text{mol}}{\text{L}}}=\frac{\text{mol}}{\text{L}}$$

I have used moles per liter (molarity) as it is a common concentration unit.

In this case, $K_a$ and $[\ce{H+}]$ both have concentration units, as does $C_a$, so your initial equation $[\ce{H+}]^2 + K_a [\ce{H+}]+ K_a C_a =0$ is valid (at least for units.

If you recast $K_a$ and $C_a$ in terms of activities, then you need to replace $[\ce{H+}]$ with $\alpha_{\ce{H+}}$ (which is unitless).


We actually do mean $\ce{Mg^+}$ here.  The idea is to show what factors are involved when an alkaline earth metal forms stable compounds with $\ce{M^{2+}}$ ions whereas an alkali metal favors $\ce{M^+}$.  In this case, as pointed out in some of the comments, $\ce{Mg^+}$ is bulked up by the electron that remains in the relatively diffuse, loosely held $3s$ subshell.  You need to remove that electron, making $\ce{Mg^{2+}}$, to get a compact ion that gives good lattice energies in ionic crystals.


From The Nitrosyl Azide Potential Energy Hypersurface: A
High-Energy-Density Boom or Bust? J. Am. Chem. Soc. 1996, 118, 4860-4870 :

  [Abstract] ...While the ring isomer is predicted to be the most stable structure on the
  hypersurface, the barrier to dissociation is most likely between 1 and 2 kcal mol-1 (including zero-point vibrational
  energy [ZPVE], the existence of any barrier becomes questionable) making isolation theoretically possible but
  experimentally difficult. This small barrier also detracts from the attractiveness of the N4O ring structure as a high energy-density material. The trans-chain isomer, however, lies in an energy valley with higher sides, consistent
  with its previous experimental observation.

  [full text]...In an ideal five-membered ring
  with 6 $\pi$ electrons, the $\pi$ electrons would be distributed evenly
  among all the bonds. In the present case, the highly electronegative
  O atom prefers to keep electrons around itself, leading
  to a partial negative charge on the oxygen. Energetically, there
  is a certain degree of stability associated with the ring isomer,
  although not on the order of common aromatic systems. The
  ring isomer is predicted to be at most (DZP CISD) 20.9 kcal
  mol-1 more stable than the trans-chain isomer; however, this
  value decreases to 13.2 kcal mol-1 with TZ2P CCSD.
  The ring isomer TS to dissociation into N2 and N2O is shown
  in Figure 5. ... Energetically, the barrier to dissociation
  is at most 15.3 kcal mol-1 (DZP CISD) and drops lower
  with improvements in both basis set and correlation scheme.
  In going from a DZP to a TZ2P basis set, for example, this
  barrier drops by 5.2 kcal mol-1 for CISD and 4.3 kcal mol-1
  for CCSD. Assuming a similar trend in moving from DZP
  CCSD(T) to TZ2P CCSD(T), the ring dissociation barrier is
  expected to drop to 1-2 kcal mol-1 with the addition of f-type
  functions possibly making it even lower. A barrier of this size
  lies below the ZPVE, throwing doubt on the existence of the
  N4O ring isomer.

So thermodynamically, yes, the ring form is the lowest energy isomer.  However, all the isomers are unstable to decomposition and the linear isomer is in a deeper potential energy well than the ring isomer. Therefore, the linear isomer is easier to observe experimentally for kinetic reasons.

There may be additional information in Theoretical study on structures and stabilities of N4X (X = O, S, Se, Te) series International Journal of Quantum Chemistry Volume 109, pages 226–235.


How 'heavy' should an element be, to be a "Heavy Metal"?

There is no true, accepted definition of heavy metal. I was taught to apply the option a metal that has density equal to or over $5.0\ \mathrm{g/cm^3}$.

Other variants include a different density range, specific gravity over density, environmental impact, atomic number, toxicity, or atomic mass, even chemical properties. See here$^{[1]}$ for further information and references.

$[1]$ John H. Duffus. '"Heavy metals"—a meaningless term?'. IUPAC technical report. Pure and Applied Chemistry, $(2002)$, 74(5), pp 793$-$807.


The simple version:

Petroleum-based rubbers can be dissolved by petroleum jelly eating away at the mechanical integrity.

The slightly longer version:

Petroleum jelly incorporates its way into the rubber, changing the plasticity (ie how rubbery or stiff) the rubber is.  Changes in plasticity change the mechanical strength and toughness such that the rubber is then outside the material property specifications of the product it is with/on/in, and hence fails.

Even longer version:

Most rubbers have small molecules inside them that help the molecular chain is the rubber move.  These molecules are called plasticizers (and occasionally tackifiers).  They are small molecules usually 200-500 MW which is less than the several hundred thousand of the rubber itself.  Petrleum jelly is basically hydrocarbon of roughly the size of the tackifiers.  The petroleum jelly molecules can 'wiggle' their way into the rubber, moving and adjusting how the existing plasticizers interact with the rubber.  In some cases the plasticizers may be extracted out, also changing the interaction, fundamentally changing the energies involves in intermolecular motion which is then related to toughness and strength (many steps in that derivation omitted).  In extreme cases in poor quality rubbers, some of the rubber molecules themselves can be dissolved by the petroleum jelly.  Changes in toughness, plasticity and mechanical integrity are common when petroleum jelly interacts with rubbers, and hence it should be avoided as a lubricant.

As you have pointed out silicone oil is a costly replacement, but not unbearable more expensive for most jobs.


The rate is determined by the slowest step, as you correctly assumed. The transition state of C->P is assumed to occur significantly faster than the first step in Organic Chemistry at least.
Unless you have an energy diagram for the reaction or any Le Chat, I think you have it right as rate = k[A][B].


Is there a Material that is High Heat Resistent and soluble?

I cannot guarantee the materials in this answer will meet all your requirements, so it would still be very helpful if anyone else could pitch in (especially an inorganic chemist or someone with hands-on knowledge of refractory materials). I will describe my thought process, which may ultimately be more useful than the suggestions themselves.

Immediately, the requirement to resist temperatures up to 1000 °C implies organic materials are very unlikely to be suitable. Almost all reasonable organic compounds would boil, decompose or burn (in air) well before then.

So now we focus on inorganic materials. Simple examples of very heat-resistant compounds are quartz ($\ce{SiO2}$) and alumina ($\ce{Al2O3}$), both melting only past 1500 °C and with no decomposition. However, these materials also resist dissolution; quartz dissolves only in concentrated hot aqueous base, whereas alumina dissolves in both of concentrated hot aqueous base or aqueous acid.

That said, it is possible to add other compounds to these materials and change their properties. For example, mixing $\ce{SiO2}$ with a number of other inorganic substances causes its melting point to lower to more manageable temperatures. This is how we make most kinds of glass. Particularly interesting is that the addition of sufficient alkali metal ions can turn the glass water-soluble. For example, there are a variety of sodium silicates, whose exact composition and properties will vary with the sodium content, and at high sodium content they become very water-soluble. However, it appears these silicates melt close to 1000 °C, which may preclude their use.

Fortunately the same approach works with alumina; increasing its alkali metal ion content eventually renders the material highly water-soluble. Sodium aluminate with a 1:1 mol ratio of sodium to aluminium forms $\ce{NaAlO2}$, which appears to be very water soluble and still melts at >1600 °C. In principle, this material could be made into a concentrated, thick aqueous solution or slurry, poured into a mould, dried at low heat to solidify then calcined to obtain a sodium aluminate object of the desired shape. This could then be used for your applications, and redissolved in water when necessary.

In both of the previous examples, it's certainly possible to add further inorganic compounds (e.g. borax, lime, etc.) to tweak properties further. However, much of this kind of research has already been performed, so I would recommend digging into glass fabrication/cement research rather than rediscovering the wheel.

I should also mention there are ways in which my above suggestions may be inadequate. Alkali aluminates/silicates appear to be strongly hygroscopic, which may make their use more cumbersome. More worryingly, at high alkali metal content they become strongly basic, which could corrode some of the metals you plan to use, both during dissolution in water, or by direct reaction with the aluminate/silicate surface at high temperatures. You'll have to see whether this is a potential deal-breaker.


Sorbitol's solubility in ethanol seems to be very temperature dependent according to the Merck Index$^{[1]}$

  Freely sol in water (up to 83%); sol in methanol, isopropanol, butanol, cyclohexanol, phenol, acetone, acetic acid, dimethylformamide, pyridine, acetamide solutions; practically insoluble in most other org solvents; quite soluble in hot alcohol, sparingly soluble in cold alcohol

1) O'Neil, M.J. (ed.). The Merck Index - An Encyclopedia of Chemicals, Drugs, and Biologicals. Whitehouse Station, NJ: Merck and Co., Inc., 2006., p. 1498


What factors could lower the voltage of a galvanic cell

Although it seems simple to make a galvanic cell ( i.e. a battery) it is really fairly complex.

The theoretical voltage is given by the Nernst equation.
The actual measured voltage also depends on the current. The equation
$V=iR$ still applies. So various restance sources in the cell cause voltage drops.

    The current must be very low so a high impedance voltmeter must be    used. (Not really a problem these days with modern semiconductor
    voltmeters as opposed to galvanometers.)
    The salt bridge offers resistance, so it has a voltage drop.
    Impure metals for the anode or cathode can cause a voltage change.
    Surface coating can cause a voltage drop.
    The acid or base concentration can cause a voltage change.
    The ionic strength of the solution can change the voltage.


Do all reducing sugars show mutarotation?

The compounds having hemiacetal in their structure give mutarotation.
Generally, all monosaccharides (glucose, fructose, ribose, arabinose, …) and disaccharides (lactose, maltose, …) give mutarotation, except sucrose.
Sucrose, despite being a disaccharide, fails to give mutarotation because it doesn't contain hemiacetal.


The cyclopropenyl (1) cation is believed to be aromatic as it demonstrates abnormal stability and NMR-shifts typical for aromatic compounds. Moreover, there are a good number of derived systems, where aromatic conjugation is maintained despite extra methylene insertions into the ring, such as is typical for the nornbornadienyl (2) cation. Some of these cations can be isolated in salts or observed via NMR in solution. Signs of aromaticity are found also for the cyclooctatetraene dianion and the cyclononatetraene dianion, both produced by reduction of the corresponding alkene with metallic sodium. Remember, please, that even though all systems noted above are 'aromatic' in some sense and are abnormally stable for their family of species, they are still very reactive, because they all belong to a very reactive family of species. While the neutral boron analog of the cyclopropenyl cation has not been isolated to my knowledege, a similar structure was found in organometalic compounds. Aromaticity of some ring systems may result in some weird properties of some compounds, like squaric acid (abnormally strong for ketoenol and a stable enol), high polarity and basicity of cyclononatrienol and cyclopropenone and so on.

(1) - cyclopropenyl cation

(2) nornbornadienyl cation


Why is the pairing energy not considered when calculating the crystal field splitting energy?

It is actually the two single-paired electrons are more likely to be promoted than the two paired electrons.
Promoting the electrons always cost energy whether the electrons in question are paired or not. In the classical examples of hybridization (where they might actually promote a paired electron), this energy cost is compensated by the formation of stable covalent bonds, but in this case, a complex, no bonds formed here.
So why the single electrons are promoted before the paired ones?
The spin rule in the coordination complexes tell that the allowed d-d transitions must not involve the change in the total spin of the electrons.
In your case of a low-spin d4, promoting the the paired electrons will leave the d-orbitals with single electrons with opposite spin, which is highly unfavorable. The changing of the spin of the opposite-spin electron is forbidden per the spin rule since it will change the total spin of the system.


When is it better to use coupled-cluster (CC) over configuration interaction (CI) calculations?

CCSD is preferred in nearly all cases because size-extensivity/size-consistency is a critical property to have for practical chemistry, and on top of that, more of the correlation energy (the part of the energy missing from Hartree-Fock) is captured by CCSD. CISD only accounts correlation energy for up to double excitations and not at all for higher excitations. CCSD accounts for correlation up to double excitations and then approximates all higher excitations as combinations of the single and double excitations it has already calculated. This gives much more accurate numbers for pretty much all systems and provides for size-extensivity/size-consistency. The only time you would really be better off doing CISD than CCSD is if you wanted to do full CI on a 2-electron system, in which case CCSD is not as efficient.

So in short, CCSD is preferred because it almost always gives better, more usable numbers than CISD for similar or better costs (they both scale as ~$n^6$).


How does X-Ray Crystallography determine the components (as opposed to the structure) of a molecule?

That is because the solution to the structure isn't done just via the plate. The signal you see on the plate is what is called "structure factors" of the experiment. They are associated with the fourier transform of the electron density. There are computational ways of obtaining the electron densiy once you have the structure factors. When you do that, you arrive at something like this:

Then, once you have something like that, at least for proteins, you need to find the best possible geometry that fits the density field. So if you succeed you might end up with something like this:

The trick here is that the molecular biologists and physicists who work with proteins already know something of what they are looking for. They don't try to fit a haphazard collection of atoms to the electron density map, they start with a polypeptide chain, often one for which the sequence has already been experimentally determined so they know a lot about the composition of the molecule well before any structure resolution takes place.
This is the tl;dr version of the thing. In reality there are a number of things I have omitted. For instance, it isn't always necessary to obtain the electron density in order to resolve the structure. There are ways to theoretically compute the structure factors from a test geometry such that a structure is considered resolved if the theoretical and the actual form factors coincide up to a certain tolerance.
The structure is also resolved in steps. Often when you crystallize something, a lot of elluents and additives are added to the mix in order to make the crystallization possible, particularly for proteins, which have very fragile and sensitive crystallization behaviors. So researchers first try to resolve the geometry of the main protein chain, after that is done there will be residuals which can be adjusted by trying to find molecules of the additives and elluents in the unit cell. That is how we end up with sulfates, waters and other small molecules and analytes floating amidst the protein structure for the majority of the PDB structures resolved via X-Ray crystallography.
That said, for very small molecules and an electron density with a very high resolution, it is possible to distinguish one atom from the other by the shell structure around them. But, again, composition is often an information that the researcher has a priori, not something the diffraction is expected to provide.
P.S. If you have time, you can check the Yale Course of Professor McBride Freshman Organic Chemistry I. It is totally free an in the internet. In it, he does a very pedagogical and awe-inspiring introduction to the ins and outs of X-Ray diffraction (and other experimental techniques) that go well beyond the scope of this answer. He does not focus on proteins, but on the basics, small organic molecules, but it does wonders to clear the fog. The strictly relevant classes are classes 5 and 6, but I desperately urge you to go through everything that came before instead of just jumping in.


Understanding why it happens unfortunately will not help her to remember. This is the classic case where you have to find a dirty way to remember them. Maybe you can make some practical experiment I remember most of them with a practical application of the salts.

There are a lot of online resource regarding Mnemonic Tricks to remember solubility rules, these are two video on youtubes.

Memonic tricks

Song!

Quoting the song:

  This is solubility it's so crazy, don't be a fool and follow these rules!


Bomb calorimeter and heat capacity of calorimeter

The heat capacity of a calorimeter, $C_{calorimeter}$ is given in units of kJ/°C, with no units of mass included.  This is because the calorimeter's mass is a constant and each calorimeter's heat capacity (and mass) is different.

The general procedure that is used to determine the heat capacity of a calorimeter is as follows:

Place a carefully measured quantity of a compound whose combustion energy is well known and that is available in a pure form in the bomb, usually benzoic acid.
Seal the bomb and add oxygen so that the sample will burn completely.
Submerge the bomb and ignite the sample.
Measure the temperature change of the water bath in which the bomb is submerged.

Since you know the energy of the reaction and you know the heat energy in the water, you can find the heat energy of the calorimeter.

$\Delta E = q_{water} + q_{calorimeter}$

The heat of the calorimeter is the difference between the energy we know we put into the system through the combustion of our standard and the heat that was measured in the water bath.


Why do strong acids have weak conjugate bases?

All chemical processes are reversible at least to some extent.

Keeping this in mind, a strong acid must have a weak conjugate base (I think you mean strong acids have weak conjugate bases).

Why? Let's try to prove by contradiction. If a strong acid had a strong conjugate base, then the base would quickly re-associate itself with a hydrogen proton. This reverses what the acid just did - dissociate itself from a hydrogen proton!

To illustrate this concept, consider hydrochloric acid - the oft-cited strong acid. $\ce{HCl}$ is a strong acid because in water solution it dissociates (nearly) completely into chloride ion and hydrogen protons (which are then solvated by water).

$\ce{HCl + H_2O ->H_3O^+ + Cl^-}$

That is the reaction of hydrochloric acid with water. Note the one-way arrow; the reaction is a one way street. $\ce{HCl}$ is the acid; $\ce{Cl^-}$ is the conjugate base; water is the base and hydronium ion is the conjugate acid. In other words, this is the generalized reaction:

$\ce{(strong)~acid + base -> conjugate ~acid + conjugate ~base}$

In bite-size pieces:

$\ce{(strong)~acid -> conjugate ~base}$

and

$\ce{base -> conjugate ~acid }$

Add the above two "half-reactions" together and you get a complete Brønsted acid/base reaction.

So back to the dissociation of $\ce{HCl}$.

If this reaction (the reverse reaction) happened to a significant extent:

$\ce{HCl + H_2O <- H_3O^+ + Cl^-}$

Then that would be antagonistic to the dissociation of $\ce{HCl}$! $\ce{HCl}$ would no longer be nearly completely deprotonated in water solution since the reverse reaction is significant!

Therefore, we conclude that if one has a strong acid, then its conjugate base must be weak. The converse is also true; if an acid's conjugate base is weak, then the acid must be strong.


How many dipeptides are possible by the reaction of glycine and alanine?

The number of isomers are ${9}$ as shown below.

AlaAla has ${4}$ isomers.
AlaGly has ${2}$ isomers.
GlyAla has ${2}$ isomers.
GlyGly has ${1}$ isomer.


The technique of Circular dichroism (CD) is used to detect changes and the presence of chirality in peptides and can detect both helical chirality in polypeptide chains that form helices and in individual chiral centers.

Polarimetry. The optical activity of a polypeptide (oligopeptide) has a specific rotation. A change in the optical rotation of such a peptide. Chirality is a property of a molecule as a whole. Such a change will alter the optical rotation of the peptide.


Electrophilicities of carboxylic acid derivatives

Background

When a carbonyl compound reacts with a nucleophile we form a tetrahedral intermediate.  Since the carbonyl carbon becomes $\ce{sp^3}$ hybridized in the tetrahedral intermediate there is no significant resonance stabilization - the various tetrahedral intermediates have (approximately) the same energy.  However, strong resonance effects can occur in the starting carbonyl compound, potentially stabilizing the carbonyl compound.

In other words, while a substituent attached to the carbonyl carbon can stabilize the starting carbonyl compound, it should have little effect on the relative energetic position of the tetrahedral intermediate.

If we place the carbonyl compound and the tetrahedral intermediate on a reaction coordinate we can see that anything that stabilizes the carbonyl compound will increase the activation energy for reaction, slowing down the reaction rate.

Below are the resonance structures we can draw for our carbonyl compound.

If structures II and  III contribute significantly to the overall description of the compound, then we would expect that carbonyl compound to be more stable, lower in energy, than a carbonyl compound where structures II and III contribute less.

Answer

Three of your compounds are esters, one is an amide.  We know that resonance structure III is more important in an amide than in an ester (due to electronegativity differences, positive charge on nitrogen is better than positive charge on oxygen).  Therefore the amide (B) will react the slowest.

Now to separate the three esters.  You are right that ester C will be the most reactive.  This is because the inductively electron withdrawing chlorine substituent destabilizes resonance structure II.

In compound D we can draw stabilizing resonance structures like the following;  the para-hydroxy substituted aromatic ring is very effective at moving electron density towards the electron deficient carbonyl carbon thereby stabilizing it.

The methyl group in compound A also donates electron density to the carbonyl carbon, but not nearly as effectively as the aromatic ring.

Therefore our overall order of carbonyl reactivity is C > A > D > B


Atkins did a good job on explaining what parametric dependence of the electronic wave function (and electronic energy) on nuclear coordinates is, but failed to mention that in the derivation of the Born-Oppenheimer approximation this dependence is assumed to be continuous and differentiable and that both first and second derivatives of this quantities with respect to nuclear coordinates are in general non-zero.
In particular, for the system described in the text, we have
$$
  \frac{\partial \psi}{\partial Z_j} \neq 0 \, , \quad
  \frac{\partial^2 \psi}{\partial Z_j^2} \neq 0\, .
$$

Now, when the solution of the form (8.3) is substituted into the Schrödinger equation (8.2), the term involving $T_\mathrm{e}$ is trivial: since the nuclear wave function $\chi$ is not a function of electronic coordinates it is just a constant when differentiating with respect to them, so we get
$$
  T_\mathrm{e} (\psi \chi) = \chi T_\mathrm{e} \psi \, .
$$
But the term involving $T_\mathrm{N}$ does not trivially transform in a similar way,
$$
  T_\mathrm{N} (\psi \chi) \neq \psi T_\mathrm{N} \chi
$$
since both $\psi$ and $\chi$ depend on the nuclear coordinates. Rather, applying the product rule twice, we get
\begin{align}
  \frac{\partial^2}{\partial Z_j^2} (\psi \chi)
  &=
  \frac{\partial}{\partial Z_j} \left( \frac{\partial}{\partial Z_j} (\psi \chi) \right) \\
  &=
  \frac{\partial}{\partial Z_j} \left( \psi \frac{\partial \chi}{\partial Z_j} + \chi \frac{\partial \psi}{\partial Z_j} \right) \\
  &=
  \psi \frac{\partial^2 \chi}{\partial Z_j^2}
  +
  2 \frac{\partial \psi}{\partial Z_j} \frac{\partial \chi}{\partial Z_j}
  +
  \chi \frac{\partial^2 \psi}{\partial Z_j^2} \, ,
\end{align}
so that
$$
  T_\mathrm{N} (\psi \chi)
  =
  - \psi \sum\limits_{j=1,2} \frac{\hbar^2}{2 m_j} \frac{\partial^2 \chi}{\partial Z_j^2}
  -
  \sum\limits_{j=1,2} \frac{\hbar^2}{2 m_j} \left(
  2 \frac{\partial \psi}{\partial Z_j} \frac{\partial \chi}{\partial Z_j}
  +
  \chi \frac{\partial^2 \psi}{\partial Z_j^2} \right) \, ,
$$
where the first term is nothing but $\psi T_\mathrm{N} \chi$ and the remainings are designated as $W$.


Why is the solution of (NH4)2[Ce(NO3)6] used as an oxidizing agent?

Most lanthanides display stability in their +3 oxidation states. Cerium does the same, which is why it has a very high reduction potential. The reduction potential for the $\ce{Ce^{4+}|Ce^{3+}}$ couple is 1.61 V (for comparison, $\ce{Cl2}$ has a potential of 1.32 V)

Cerium in the form of ceric ammonium nitrate is also quite stable compared to other oxidizing agents and can be stored on the shelf for a longish period of time. It's also quite usable: the bright orange colour is decolourized to a pale yellow on reduction of $\ce{Ce^{4+}}$. This is why it's used as a common oxidizing agent.


Elimination reaction with bad leaving groups

Alkyl halides with bad-leaving groups (like $\ce{F-}$) having EWG at $\ce{\beta}$-position undergoes these reactions, also known as $\ce{E_{1CB}}$, viz. elimination unimolecular conjugate base.

Example:

aldol condensation

photo-induced decarboxylation

Reagent: bulky or strong bases or both (like $\ce{K+(^tBu)-, LDA, NaOH, KNH2}$, etc) along with some heating.

Mechanism:


Plasma, Fourth State of Matter, and Magnetic Fields and Electric Fields

There is not necessarily a preexisting magnetic or electric field in a plasma, since the charges can be randomly distributed. Think of plasma as similar to a metallic state conductor -- a sea of mobile electrons (and, of course, ions in plasma, so perhaps it better approximates a semiconductor, with electrons and holes). A plasma in thermal equilibrium, not exposed to an external electromagnetic field, should be uncharged.

That said, unbalances occur for a number of reasons. For example, electrons are thousands of times less massive and therefore more mobile than ions, so charge separation occurs, leading to a plasma potential.. Lasers can generate a charged plasma, accelerating particles with the wakefield effect.


This explanation is true for most organic indicators.  Litmus is a complex mixture of organic molecules.  The molecule responsible for the color change (the chromophore), is 7-hydroxyphenoxazone (image taken from Wikipedia, public domain):

See that $\ce{OH}$ group at the lower left?  That's where the so-called "acidic proton" is.  This molecule is actually a weak acid, meaning that in the presence of a strong base, the $\ce{OH}$ group can lose a proton, which reacts with an hydroxide ion ($\ce{OH^{-}}$) to form a water.  The 7-hydroxyphenoxazone is now a negatively charged ion: it looks the same except it is missing the $\ce{H}$ on the $\ce{OH}$ group, and it leaves behind an electron.

To think about the equation that describes what's going on, pretend that the entire 7-hydroxyphenoxazone molecule is represented by the formula $HLit$, where $\ce{H}$ is the hydrogen on the $\ce{OH}$ group and $Lit$ is everything else (this is common procedure: only the hydrogen matters in the acid-base equation).  Then the acid-base equation is:

$$\ce{OH^{-} + HLit <=>H2O + Lit^{-}}$$

As more strong base is added, more hydroxide ion is present, driving the equilibrium to the right and producing more $Lit^-$ ion.  And now, the color change: $HLit$ is red, but $Lit^-$ is bue.  So the more acidic your solution, the more the equilibrium is to the left and thus there is more $HLit$ than $Lit^-$.

Litmus is pretty good for checking rough acidity or basicity, because its midpoint pH is close to 7 (neutral).  Litmus is as red as it gets at a pH of 4.5, and is as blue as it gets at a pH of 8.3, so the midpoint is about 6.4, a bit more acidic than pure water.


Why do isotopes have different chemical reaction rates?

Isotopes differ in mass.
Otherwise the same molecules but with the different mass have the different mean speed.
The different speed means the different molecular collision rate.
The different collision rate means the different kinetic constants.
There are some indirect effects of different isotop mass on the kinetics, but these are much smaller than the direct kinematic effect.


Choosing a vacuum pump for fractional distillation

If you are keeping your hands on you can't beat a water aspirator for cost, simplicity, and chemical compatibility.

Here is an aspirator assembly with a gauge and valves I put together for about $50.  Connecting it to a standard hose bib it can pull up to 27mmHg.  The water valve controls the vacuum rate, and if you have a closed system once you reach the desired vacuum you can shut one of the other valves to hold it.

Note: This was inspired by Lysander's answer to my similar question last year, which has additional useful details.


What is the relation between ionization energy and electropositivity?

Electropositivity is rarely used by itself. It is much more commonly seen as just the opposite of electronegativity. While electronegative is more or less defined to the point where numeric values can be found for each element in a given system (sometimes excluding noble gases), electropositivity has no such scale I am aware of.

Ionisation energy or potential are clearly defined energies that can be experimentally directly determined and are tabulated across physical chemistry textbooks with little to no variance (the main variance being experimental error).

The two terms are hardly related; neither by definition/physical meaning nor by value trend.


Aldehydes can be oxidized by bromine under acidic conditions per the mechanism is given below.  The haloform reaction, however, occurs under basic conditions and requires a methyl ketone.  Your professor probably forgot that the haloform is for methyl ketones in basic conditions and called the reaction below a haloform even though it isn't. Given the similarity in the oxidation of the carbonyl group to a carboxylate with a halogen, (and the thousands of organic reactions to remember,) it seems like a simple easy mistake.


Obtaining values for electrical mobility from ion transport numbers in a Hittorf cell

You can't directly derive electrical mobilities from ions from their transport numbers; transport numbers are a relative measurement of charge transfer that, precisely, reduces absolute mobilities for different ions to contributing fractions. Absolute mobilities are lost when we use transport numbers, so if all we know about an electrochemical system is transport numbers, we cannot recover ionic mobilities.

However, we can recover drift velocities if we have additional information on the system - specifically, total current $I_\mathrm{total}$ and distance between plates in the cell $d$. Alternatively, if we're interested in ion mobilities, we will also need to know voltage $V$.

Let's look at a simple system with only one electrolyte involved in the electrochemical reaction. Total current will be the sum of the contributions of the anion and the cation:

$$ I_\mathrm{total} = I_+ + I_- =  \cfrac{n_+ z_+ \nu_+ e}{d} +  \cfrac{n_- z_- \nu_- e}{d} $$

where $n_i$ is the number of ions $i$, $z_i$ is the charge of ion $i$, $\nu_i$ is the drift velocity of ion $i$, $e$ is the electron charge and $d$ is the distance between the plates. Per the requirement of electroneutrality,

$$ n_+ z_+ = n_- z_- $$

so

$$ I_\mathrm{total} = \cfrac{ n_+ z_+ e \left( \nu_+ + \nu_- \right) }{d} = \cfrac{ n_- z_- e \left( \nu_+ + \nu_- \right) }{d} \tag{1}$$

We can calculate transport numbers as

$$ t_+ = \cfrac{I_+}{I_\mathrm{total}} = \cfrac{n_+ z_+ \nu_+ e /d}{n_+ z_+ \left( \nu_+ + \nu_- \right) e / d} = \cfrac{\nu_+}{\nu_+ + \nu_-} $$

$$ t_- = \cfrac{I_-}{I_\mathrm{total}} = \cfrac{n_+ z_+ \nu_- e /d}{n_+ z_+ \left( \nu_+ + \nu_- \right) e / d} = \cfrac{\nu_-}{\nu_+ + \nu_-} $$

Note that we could use the definition of transport numbers to obtain a specific drift velocity, $\nu_i$, from the total drift velocity, $\nu_+ + \nu_-$:

$$ \nu_+ = t_+ \left( \nu_+ + \nu_- \right) $$

$$ \nu_- = t_- \left( \nu_+ + \nu_- \right) $$

Let's go back to $\mathrm{(1)}$ to derive an expression for total drift velocity, $\nu_+ + \nu_-$:

$$ \nu_+ + \nu_- = \cfrac{d}{n_+ z_+ e} I_\mathrm{total} = \cfrac{d}{n_- z_- e} I_\mathrm{total} $$

We can therefore derive drift velocity for an ion, $\nu_i$, from its transport number, $t_i$:

$$ \nu_+ = \cfrac{ t_+ d }{ n_+ z_+ e } I_\mathrm{total} $$

$$ \nu_- = \cfrac{ t_- d }{ n_- z_- e } I_\mathrm{total} $$

Note that it is necessary to know current $I_\mathrm{total}$ and cell size $d$ since precisely these values introduce the time and distance dependence necessary to recover drift velocity $\nu_i$ from transport number $t_i$.

Assuming the system can be approximated by parallel plates, i.e. assuming electric field is constant, ion mobility, $\mu_i$ can be derived from drift velocity through voltage, $V$, and distance between plates, $d$:

$$ \mu_i = \cfrac{\nu_i}{E} = \cfrac{\nu_i d}{V} $$

So the expressions for cation and anion mobility would be

$$ \mu_+ = \cfrac{t_+ d^2}{n_+ z_+ e V} I_\mathrm{total} $$

$$ \mu_- = \cfrac{t_- d^2}{n_- z_- e V} I_\mathrm{total} $$

For a more complex system with $N$ ions contributing to current,

$$ I_\mathrm{total} = \cfrac{e \sum\limits^{N} n_i z_i \nu_i }{d} $$

$$ t_i = \cfrac{ n_i z_i \nu_i }{\sum^{N} n_j z_j \nu_j } $$

and drift velocity and mobility can be computed as

$$ \nu_i = \cfrac{ t_i d }{ n_i z_i e} I_\mathrm{total} $$

$$ \mu_i = \cfrac{ t_i d^2 }{ n_i z_i e V} I_\mathrm{total} $$


Since aqueous acids release Hydrogen cations into the solution, will an electric current create Hydrogen gas?

Yes.  In fact the first electrolysis was done with acidified water.  Pure water is only slightly ionized and is a poor conductor.  Adding ions to water increases its conductivity and speeds up the electrolysis process.  So by adding a source of protons to pure water the following reaction will occur at the cathode $$\ce{2H+ (aq) + 2e^{-}->H2(g)}$$  Of course a parallel reaction will occur at the anode $$\ce{4OH- ->2H2O + O2(g) + 4e-}$$ otherwise a current wouldn't flow and the electrolysis would not continue.


Single Displacement Reactions: A + AX =?

Nothing. All you would have is just a mixture of white powder and lead metal.

In general, lead (II) nitrate is synthesized by:

$$\ce{PbO (s) + 2 HNO3 (aq) -> Pb(NO3)2 (aq) + H2O (l)}$$

Lead (II) nitrate is soluble in water.

Alternatively, you could do:

$$\ce{PbCO3(s) + 2HNO3(aq) -> Pb(NO3)2(aq) + H2O(l) + CO2(g)} $$

Single displacement reactions are a kind of redox reaction. For single displacement reactions to occur:

1.) A and B must either be different metals OR

$$ \ce{2AgNO3(aq) + Zn(s) -> 2Ag(s) + Zn(NO3)2(aq)}$$

2.) A and B must be a halogen

$$\ce{Mg(s) + 2 HCl(aq) → MgCl2(aq) + H2(g)}$$

This is necessary as for the redox reaction to occur, you need two differing metals on the reactivity series in order for the single replacement to work, or alternatively, use compounds that contain halogens.

Even so, in combining two compounds with metals or halogens, if the element is less reactive than the element in the compound, the single replacement reaction will not proceed.

Some examples are:

$$\ce{Ag (s) + Cu(NO3)2 (aq) -> } \text{No reaction}$$

$$ \ce{I2 + 2KBr ->} \text{No reaction} $$


It is tricky wording. Usually when you compress something you want to do so as efficiently as possible. The minimum amount of work would be required if you compressed this reversibly.

If for some reason, as in the question, you want to waste your energy and overkill the work required (i.e., do things the hard way), then compress the gas IRREVERSIBLY. that will make it take the maximum amount of energy to get the desired compression.

The engineer in my cringes at this :)


Carbanions and Carbocations are perfectly opposite in order of stability because they are opposite in order of charge. The order is preferred of carbocations the same as acidity by Lewis' nucleophile/electrophile definition. Therefore, because a positive charge is stabilized better by a tertiary carbon, it is preferred to host a carbocation because the electron shells of its substituent carbons help stabilize the charge.
In the case of a carbanion, the negative charge is best stabilized when not in contact with neighboring nuclei that would resonate the negative charge away from the nucleophilic site. If you think about it like sterics, the further away the negative charge can be from other substituents, the better in the case of carbanions.


Is there a characterisation for conjugated dienes similar to cis-trans?

The current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) supports various stereodescriptors that are used to denote specific conformers, in particular concerning conjugated dienes:

  P-94.3.1.3 Stereodescriptors s-cis and s-trans

  The spatial arrangement of two conjugated double bonds about the intervening single bond is described as ‘s-cis’ if ‘synperiplanar’ and ‘s-trans’ if ‘antiperiplanar’. This descriptor should not be applied to other systems such as N-alkyl amides (use ‘E/Z’ or ‘sp/ap’).


Could coating silver inside of a cuvette be used for SERS?

from theory this should work what you have scatched here. When it comes to SERS a main point is, that you got a surface which is rough enough that you can create mini hot-spots for the electric field. This will give you the enhancement.
Maybe you must put a little bit of work in you set up. The scattering is needed to arrive at the detector. maybe you try a Raman probe or do a other angle for the wall with silver on it.
Best regards!


How do electron donor groups affect the polarity of an O-H bond in alcohol?

Oxygen tries to draw electron density from the two moieties it's bound to, thus making both bonds more polarized than the normal carbon-carbon bond. Some functional groups offer more resistance than others to this polarization depending on their electronegativity. For this reason, if one of the groups "puts up more of a fight" than the other, oxygen will draw more electron density from the weaker one.

In your case, hydrogen is always the same since we're always talking about alcohols. Alkyl substituents donate electron density to the oxygen atom, so it is less eager to draw electron denstity from the hydrogen bond and will make it less polar. Change the alkyl for a primary alkyl chloride for example and the effect will be reversed.


You measure an electric potential difference. It can be considered sum of three different effects. If you could isolate them, you will have the electromotive force of both reference electrodes. If they can be isolated, each of them will generate a potential difference at open circuit (when the stationary state is reached) equal in magnitude to the emf. You always will have this difference of potential plus the potential difference at both sides of the membrane. Just add potential differences due to the stationary condition reached because of the emf.


This method works because one of the reaction products, $\ce{CO}$, is gaseous (at the stated reaction temperature) and is continually removed. Following Le Chatelier's Principle, the equilibrium:

$$\ce{ZnO(s) + C(s) <=> Zn(l) + CO(g)}$$

is pushed to the right and the reaction proceeds.

There are numerous of these reactions where Ellingham data don't tell the full story, notably also the reduction of magnesia ($\ce{MgO}$) by coke, is in accordance with the same principle.

One of the most interesting cases is a lab preparation of caesium metal that takes place according to:

$$\ce{CsCl(s) + Li(l) <=> Cs(g) + LiCl(s)}$$

The reaction is carried out under vacuum and at about $970\ \mathrm{K}$, conditions under which $\ce{Cs}$ is much more volatile than $\ce{Li}$. The former thus distils off and is caught in a condenser, pushing the reaction equilibrium to the right.


How to compute 2-electron integral for Hartree-Fock code?

You can also write this down in a similar way as for the one-electron integral. You already had:

\begin{equation}
\mathbf{S} = \mathbf{D_2}' * \mathbf{S}_{\rm prim} * \mathbf{D_2}
\end{equation}

where $\mathbf{S}$ is $(2\times2)$, $\mathbf{D_2}$ is $(6\times2)$ and $\mathbf{S}_{\rm prim}$ is $(6\times6)$.

For the 2-electron integral we now need

\begin{equation}
\mathbf{V} = \mathbf{D_4}' * \mathbf{V}_{\rm prim} * \mathbf{D_4}
\end{equation}

where $\mathbf{V}$ is $(2\times2\times2\times2)$, $\mathbf{D_4}$ is $(6\times2\times6\times2)$ and $\mathbf{V}_{\rm prim}$ is $(6\times6\times6\times6)$.

You can get $\mathbf{D_4}$ from $\mathbf{D_2}$:
\begin{equation}
\mathbf{D_4} = \mathbf{D_2} * \mathbf{D_2}
\end{equation}
where you do not sum over any index in order to get the 4-index tensor.

The equation for $\mathbf{V}$ is a bit more tricky, since you need to pay attention over which indices you sum (which is actually not specified by the notation). I add an example on how you can do this with Python/Numpy with Einstein summations.

# 1-electron integral
S = numpy.einsum('ba,bc,cd', D2, Sprim, D2)

# 2-electron integral Option 1: form (6x2x6x2) tensor
D4 = numpy.tensordot(D2, D2, 0)
V = numpy.einsum('badc,bedf,egfh->agch', D4, Vprim, D4)

# 2-electron integral Option 2: transform indices one-by-one
V = Vprim.copy()
V = numpy.einsum('ba,bcde->acde', D2, V)    # i
V = numpy.einsum('abcd,be->aecd', V, D2)    # j
V = numpy.einsum('da,bcde->bcae', D2, V)    # k
V = numpy.einsum('abcd,de->abce', V, D2)    # l

You will need to give D2, Sprim and Vprim as input.

(If anyone knows how to write this down in a formal notation, please edit.)

You were asking how to get $\mathbf{V}$ as a 2-index $(4\times4)$ matrix. For this we can simply reshape the above 4-index tensors. So $\mathbf{V}_{\rm prim}$ becomes $(36\times36)$. Now this means $\mathbf{D_4} $ needs to be $(36\times4)$, and not $(12\times12)$. To these reshaped 2-index matrices we can apply again the same transformation equation from above. Here is the code for my example:

# 2-electron integral Option 3: reshape to 2-index matrices
Vprim = Vprim.reshape((Nprim**2, Nprim**2))
D4 = numpy.swapaxes(D4, 1, 2)    # making sure to combine the correct axes
D4 = D4.reshape((Nprim**2, Ncntr**2))
V = numpy.einsum('ba,bc,cd', D4, Vprim, D4)

Now $\mathbf V$ will be:
\begin{bmatrix} (\phi_1\phi_1|\phi_1\phi_1)  & (\phi_1\phi_1|\phi_1\phi_2) &  (\phi_1\phi_1|\phi_2\phi_1) & (\phi_1\phi_1|\phi_2\phi_2)\\      (\phi_1\phi_2|\phi_1\phi_1)  & (\phi_1\phi_2|\phi_1\phi_2) &  (\phi_1\phi_2|\phi_2\phi_1) & (\phi_1\phi_2|\phi_2\phi_2) \\      (\phi_2\phi_1|\phi_1\phi_1)  & (\phi_2\phi_1|\phi_1\phi_2) &  (\phi_2\phi_1|\phi_2\phi_1) & (\phi_2\phi_1|\phi_2\phi_2) \\      (\phi_2\phi_2|\phi_1\phi_1)  & (\phi_2\phi_2|\phi_1\phi_2) &  (\phi_2\phi_2|\phi_2\phi_1) & (\phi_2\phi_2|\phi_2\phi_2)   \end{bmatrix}

And here is the structure of $\mathbf{D_4}$:
\begin{equation}
\mathbf{D_4} = \mathbf{D_2} * \mathbf{D_2} = \begin{bmatrix}
d_{11}*\mathbf{D_2} & 0*\mathbf{D_2} \\
d_{12}*\mathbf{D_2} & 0*\mathbf{D_2} \\
d_{13}*\mathbf{D_2} & 0*\mathbf{D_2} \\
0*\mathbf{D_2} & d_{21}*\mathbf{D_2} \\
0*\mathbf{D_2} & d_{22}*\mathbf{D_2} \\
0*\mathbf{D_2} & d_{23}*\mathbf{D_2} \\
\end{bmatrix}
\end{equation}


Why does the boiling and freezing-point of non-polar molecules rise with increasing molecular mass?

In general, the larger the molecule, the larger the accessible surface area of the molecule and the more surface polarisable electron density. There are more atoms that can participate in dispersion interactions and the effect stacks up. To melt or boil a substance you need to overcome the intermolecular forces holding the molecules together and dispersion forces -  whilst weak even with respect to hydrogen bonding - can add up to large interaction energies. One of the more interesting examples of strong dispersion interactions is the gecko's foot, which permits the reptile to easily stick to sheer surfaces due to the exceptional surface area of the footpad, which is covered in many many tiny hairlike protrusions.


Which are more reactive lanthanides or aluminium?

Since the lanthanides are neither in the same period nor in the same group as aluminium, comparision is actually not feasible. Factors like atomic size and effective nuclear charge tend to predominate such comparisions, rather than ionization enthalpy or reduction potential.

Bear in mind that due to presence of electrons in f-orbital for lanthanides provide poor shielding of electrons farther away from the nucleus. This poor shielding is due to the fact that f-orbitals are quite diffused in nature. Yet, we cannot say anything about reactivity of lanthanides when compared to aluminium.
If you are still a high-school student, I suggest you do not ask such hypothetical questions.


Can the following compound having a centre of symmetry be called both optically inactive and meso?

You are correct that the molecule is not chiral. It is indeed a meso compound due to the internal symmetry plane. Also, don't worry about free-rotation of single bonds.


Can Aluminum Hydroxide (Al(OH)3) be heated to it's melting temperature before decomposing?

TL.DR: Aluminium oxide (sometimes known as alumina) is made by heating the aluminium hydroxide to a temperature of about 1100 - 1200°C. (Chemguide)

$$\ce{Al(OH)3 ->[\Delta] Al2O3 + 3H2O}$$

As @andselisk said, the temperature Wikipedia mentions is aluminium hydroxide mixture in a fire retardant which might have other compounds. Decomposition of pure aluminium hydroxide has been studied in various papers and its decomposition reaction mechanism, temperature, reaction kinetics has been exhaustively studied.

A paper examines$\ce{^{[1]}}$ the decomposition of aluminium hydroxide at a temperature of 973-1123 K(700 °C-850 °C).

Another paper$\ce{^{[2]}}$ shows aluminium hydroxide being heated from room temperature to 1200 K(926.85 °C) where the exact temperature of decomposition products i.e $\ce{Al2O3}$ is noted:

  [...] we find the first and the smallest endothermic peak at 519
  K, which is due to the partial dehydroxylation of gibbsite
  ($\ce{Al(OH)3}$) and formation of boehmite ($\ce{AlOOH}$). Endothermic
  peak at 585 K corresponds to two processes: (i) transformation of
  gibbsite to phase $\ce{χ-Al2O3}$ and (ii) additional conversion of
  gibbsite to boehmite. The first process is in accordance with the
  results obtained. Another endothermic peak at 815 K is due to
  decomposition of boehmite and formation of alumina $\ce{γ-Al2O3}$. The
  obtained TG curves show clearly three steps of weight loss. The weight
  loss in the first step (about 5 wt.%) is due to the partial
  transformation of gibbsite to boehmite; the second step (about 25
  wt.%) corresponds to decomposition of gibbsite to boehmite, otherwise
  to $\ce{χ-Al2O3}$. The last step of around 3 wt.% relates to the
  formation of $\ce{γ-Al2O3}$. Further observations show that the total weight
  loss is equal to 33 wt.%. All these transformations of gibbsite and
  the chemical composition are confirmed by the XRD analysis.

References

Chemical kinetics and reaction mechanism of thermal decomposition of aluminum hydroxide and magnesium hydroxide at high temperature (973-1123 K) by Ienwhei Chen, Shuh Kwei Hwang, and Shyan Chen, Ind. Eng. Chem. Res. 1989, 28, 6, 738–742 DOI: https://doi.org/10.1021/ie00090a015
http://przyrbwn.icm.edu.pl/APP/PDF/131/a131z3p62.pdf
Bhattacharya, Indra & Das, S. & Mukherjee, P. & Paul, Subir & Mitra, P.. (2004). Thermal Decomposition of Precipitated Fine Aluminium Trihydroxide. Scandinavian Journal of Metallurgy. 33. 211 - 219. 10.1111/j.1600-0692.2004.00686.x.(DOI) (Here aluminium hydroxide is subjected to dehydration upto a temperature of 1440 °C)


3 component (ternary) system in equilibrium

If there are 3 components, using the formula, F=M+2-P and solving for P under equilibrium conditions there would be a total of 5 phases.

I understand from that you set $F=0$. Equilibrium does not necessarily mean that you have no degrees of freedom. In this case, you actually have two degrees of freedom: the molar fractions of two components, for example $X(\rm{Cl}_2)$ and $X(\rm{I}_2)$.

$X(\rm{F}_2) = 1 - X(\rm{Cl}_2) -X(\rm{I}_2)$ and cannot vary independently, so it is not a degree of freedom. Once you consider $F=2$, you only need three phases.

What about liquids and solid? Let's say that you vary temperature, and you reach the melting/condensation/etc curve in $T$ space. On the curve itself you will have four phases, three gases and (for instance) liquid chlorine. This causes you to lose a degree of freedom, which in this case is temperature. These four phases can exist only in that specific temperature.

EDIT

Following the comments on the original question, I would like to note that I was addressing a simplified case where the only phases possible are the ones listed in the original question. Once you have intermediate compounds (ICl for example) it's a whole different story.


Liquification of gas to liquid at critical temperature

Good question.  I think you got the right answer but by using some pretty suspect logic.

First, there isn't enough information to solve the problem in general because pressure isn't specified.  Is the cooling of the gas isobaric?  Or adiabatic?  Or ... ?  Let's assume its isobaric, i.e., it happens at constant pressure.  If so, what's the pressure?  If the pressure is below 5.1 atm, then $\ce{CO2}$ will never condense into a liquid, instead it would transition directly to a solid phase (dry ice).  Ammonia's triple point is roughly 0.06 atm.  But your question says "cooled to the critical temperature", which implies not below the critical temperature, which means the only way any of them will liquefy is if the pressure is at or above the critical pressure.  So let's assume the pressure is isobaric and is at least 11.8 MPa, the critical pressure of ammonia, the gas with the highest critical pressure among the possibilities.  But if we do that, the question is easy.  The gas that liquefies first is the gas with the highest critical temperature, which would be ammonia.

Relaxing this assumption makes the question much harder.  Complex calculations would be required.  The theory of corresponding states would likely be useful.

Your logic is OK, but in your equation, $w$, the weight, and $V_c$, the critical volume, also vary.   So the statement that

  $T_c$  is directly proportional to $M$

is not true, because the other factors are not constant.$%edit$


What is the best approach to balance this equation? Why I cannot do it efficiently?

This may be a bit dumb method, but sometimes I do use it as it is pretty straightforward and doesn't require any guessing. I setup a few equations and try solving over integers.
 $$\ce{aCO2 + bH2O <--> cC2H2 + dO2}$$

Balancing $\ce{C}$ atoms both sides,
$$a = 2c\\$$
Balancing $\ce{H}$ atoms both sides,
$$2b=2c\\$$
Balancing $\ce{O}$ atoms both sides,

$$2a+b=2d$$

Solving gives this. Not hard to see that $a=4$ does the job.


Insight into covalent bonds from electrostatic perspective

I assume you mean from the classical electrodynamics side only, not from quantum electrodynamics side. By the former, even H atom cannot exist, as the electron would fall along a spiral curve on the nucleus, continually emitting radiation being radially accelerated.
For a hydrogen molecule, both electrons move around both protons, they are not dedicated to the one they came with.
Consider a similar case of charged macro objects, where applying the pre-relativistic classical electrostatics is justified. Imagine 4 respectively charged balls in gravityless vacuum.
The scenario in the bottom picture is one of the less preferred positions, having rather higher energy due short distance of negative charges. But there is plenty of different positions where the whole system would have lower energy than if they were 2 and 2 separately.
Note that even classical system have no general analytic solution for the motion equations for more than 2 objects.


Why do Fullerenes have even number of carbon atoms?

The structure of (closed) fullerenes is governed by Euler’s polyhedron formula: vertices (carbon atoms) + edges (bonds) - faces = 2. Geometric constraints determine that at least some of the faces must be pentagons or heptagons, and that the number of atoms must be even.


Why is the size of Al3+ less than that of Li+?

Now, it seems obvious that the size of Al3+ should be greater than that of Li+

No it does not. It is true that electrons repulse each other, however, they all are attracted towards the nucleus.  The real 'size' of an atom or ion is a result of compromise between electron-electron repulsion and electron-nuclear attraction.

  I think that it is a kind of exception

Obviously, it is not.

  Why is this?

Commonly, there are three consideration, used to rationalize basic trends in atomic/ionic radii.

electron-electron repulsion (negative charges repulse each other)
electron-nucleus attraction (negative charges are attracted towards positive ones)
electron shielding (electrons in shells closer to nucleus 'shield' outer electrons from some charge of the nucleus.

While electrons repulse each other, increasing nuclear charge more then compensates it; so if not for shell organisation of electrons, atoms would monotonically decrease in size toward the end of the periodic table, as it is observable in each period 1—3.

However, inner electronic shells effectively shield the outer electrons from the nucleus, so they 'see' a nucleus that is a) larger in size and b) has charge equal to nuclear charge minus number of electrons in inner shells. Given that, outer electrons 'see' a nucleus with low charge density, so outer shells progressively increase in size within the same column.

In addition to that, electrons still repulse each other, so positive ions are smaller than neutral atoms, and neutral atoms are smaller than negative ions.

Now, consider $\ce{Li+}$ and $\ce{Al^{3+}}$. For the former, 2 outer electrons feel the charge of three protons. For the latter, 8 outer electrons feel the charge of 11 protons in the nucleus. Such great charge easily beats all other factors, so the latter cation is much smaller than the former.


Is there a direct relationship between the adsorption energy and the catalytic activty

Adsorption energy is indeed related to catalytic activity, but not "greater energy means greater activity".

As you were guessing, there is an optimum window in adsorption energies for which catalytic activity is maximum, this is explained by the Sabatier principle:

  The Sabatier principle is a qualitative concept in chemical catalysis named after the French chemist Paul Sabatier. It states that the interactions between the catalyst and the substrate should be "just right"; that is, neither too strong nor too weak. If the interaction is too weak, the substrate will fail to bind to the catalyst and no reaction will take place. On the other hand, if the interaction is too strong, the product fails to dissociate.

There is a nice recent paper in which you can find much more mechanistic details.


TL;DR: Don't taste pure elements. They either taste of nothing, taste foul, or kill you. Or all of the above.

Edit: Also, for clarification, elements don't have a taste. Taste is a biological reaction to chemical interactions happening with an element, not a property of the element itself. So, like everything else in chemistry, the taste is decided by the electron configuration of the element.

Taste as we know it happens in chemical reactions between our taste buds and the chemical species in question. Most of the taste is actually governed by the nose, where very specific receptors interact with very specific chemicals to send very specific signals to the brain, the chemistry of which is, while not trivial, also not interesting. It's your basic, run of the mill presumptive test.

In the taste buds, however, the receptors look for certain specific characteristics of the species being tasted. As an example, acids are sour, indicating that our sour taste buds are actually pH meters (although there might be more to it than this).

Similarly, most sugars have similar characteristics, that can be detected by a relatively simple chemical detector.

If, however, you go around tasting elements, you'll find two very interesting things.

First, you'll find that you're dying from all kinds of poisonings. At a rough count, all of the periodic table is poisonous in its pure form, with only a few exceptions (carbon, oxygen, hydrogen, nitrogen, silicon, titanium, noble gases, gold and platinum, possibly a few others).

Second, you'll find that the pure elements don't taste much at all, if anything. Taste is a completely biological concept, and these elements simply aren't found in nature, meaning that the tongue and nose have no frame of reference.

Most metals will taste metallic, due to tarnishes forming on their surface, mostly oxides, which all have the familiar, metallic taste. The non-metals are mostly fairly reactive, and taste absolutely foul as they react to the air in your moist mouth.

Gases like chlorine and fluorine may react and form chloride and fluoride, which taste sour, and then dissolve your mouth and nose, and then kill you.

The more radioactive metals will seem to taste of blood, but that's actually just acute radiation poisoning setting in, actually filling your mouth with blood. If they didn't kill you, they'd probably taste metallic like the rest of the metals, but they do, so they don't.

Finally, the alkali metals like lithium and its friends will taste bitter, also kill you.

"But wait!", I hear you scream, "What of, say, lithium in pharmaceuticals?"

When elements like lithium are ingested, they are done so in a rather more controlled fashion. You don't get a brick of pure, elemental lithium and told to go crazy when you go crazy, but rather, delivered in pills, in conjunction with a delivery agent, in this case, lithium citrate.

Most elements have some biological role or other, but if they are in the kind of quantity you could taste, they are most likely poisonous, and in many cases, fatally so.


Why we don't we save water by boiling the contaminated water to get the pure steam?

Boiling a water mixture doesn't always produce pure water.

Boiling a solution of salt and water (or evaporating the water) will separate the salt from the water, but those are compounds with very different properties.

In the making of distilled alcohols (Ethanol), the vapor phase has more alcohol than water.  Repeated distillations can increase the concentration of the alcohol.  But, the leftover undistilled solution is not pure water.  Ethanol and water are very "sticky" to each other.

Ethanol is just an example.  Household 'grey' water (waste water that doesn't include human waste) will have a whole mess of compounds.  Some you can isolate through boiling, some, like ethanol, will 'come along for the ride'.  Wastewater treatment is complicated.  Look at the system that had to be engineered for the International Space Station.  Actually, a version of this system has been tested for use in the types of communities you are describing.

Purifying grey water isn't a simple process.  There are, however, plenty of things that can be done with grey water.  Using grey water to flush toilets is perfectly fine.  With some simple amount of treatment, grey water can be perfectly suitable for watering crops.

So, since it's cheaper and easier to repurpose grey water, and harder to purify it back to drinking-water, reuse is more efficient than trying to completely treat it at home.


Why does mutarotation occur only in specific media?

Essentially, it is just the formation of the open chain form, and then back to the other cyclic anomer, right?

Precisely, mutarotation involves interconversion of the α-and β-forms (cyclic anomers) of the monosaccharide with intermediate formation of the linear aldehyde (or ketone).

The mutarotation of glucose is a general acid- base-catalysed reaction which involves two proton transfers. Studies have proposed that these proton transfers are either carried out in a stepwise or concerted manner, with the latter alternative possibly incorporating several water molecules in a cyclic transition state.

Two possible mechanisms of mutatoration have been investigated:

The first involves an intramolecular proton transfer i.e acid-catalyzed reaction.
The second uses one solvent water molecule to assist proton transfer i.e water-catalyzed process.

There is some experimental  evidence1 for the solvent participation in the mutarotation reaction that appears from the studies of primary solvent effects on the rates of glucose and tetramethylglucose observed in $\ce{H2O + D2O}$ mixtures. In addition the absence of intramolecular proton transfer can be deduced from the small values found for the mutarotation rate of glucose in organic solvents.

  Eigen1 proposed that the two proton transfers could proceed
  in aqueous solution via a pathway involving two or more water
  molecules in a cyclic hydrogen-bonded transition state, where the
  intramolecular proton transfer does not take place in aqueous
  solution.

Previous studies2 have shown that the reaction order with respect to water is zero for an acid-catalyzed reaction involving a stepwise mechanism, however in aqueous solution without added catalyst, water molecules must act as an acid or base and it is possible that the mutarotation reaction does proceed via a cyclic-concerted mechanism involving about three water molecules.

  The results obtained in experimental work in the  newer
  studies1 have shown that mutarotation is a
  solvent-assisted process that obeys pseudo first-order kinetics. Besides that, the first water molecule from the solvent acts as a
  catalyst, and plays a decisive role, since it completely changes the
  mutarotation mechanism.

So a solvent is necessary, the evidence been seen  in proton transfers in  reactions that do occur via cyclic concerted mechanism as proposed in these studies.

References:

A theoretical study of glucose mutarotation in aqueous solution. A. M. Silva et al. / Carbohydrate Research 341 (2006) 1029-1040. Carbohydrate Research 341 (2006) 1029-1040.
The Mutarotation of Glucose in Dimethylsulfoxide and Water Mixtures. Ballash and Robertson CAN. J. CHEM VOL. 51. 1973.
CM Garrett, Grisham Biochemistry


Finding cubic ice CIF files in crystal database

One resource you can use is the Crystallography Open Database. Search for structures with two elements O and H, and named “ice”. You will then find several experimental structures for cubic ice:

with downloadable structure files in CIF format: 1, 2. These were done from X-ray diffraction studies and only feature heavy atoms (oxygen) positions.

Although I could not find in any open database a CIF file for cubic ice with hydrogen positions, it is easy to make one. From the paper “High-resolution neutron powder diffraction study of ice Ic”, Le Journal de Physique Colloques, 48.C1 (1987), C1-631:

you can create the CIF file yourself:

data_Ice_Ic
_audit_creation_method         'generated by CrystalMaker 9.2.7'
_publ_section_comment
;
Chemical formula (sum): H2 O

Systematic name: H2 O

Crystal structure data from: W.F. Kuhs, D.V. Bliss, J.L. Finney,  High-resolution neutron powder diffraction study of ice Ic, Journal de Physique Colloques, 1987, 48, C1-631

;
_cell_length_a                   6.358(3)
_cell_length_b                   6.358(3)
_cell_length_c                   6.358(3)
_cell_angle_alpha               90.000(0)
_cell_angle_beta                90.000(0)
_cell_angle_gamma               90.000(0)

_symmetry_space_group_name_H-M     'F 41/d -3 2/m'
_symmetry_Int_Tables_number         227
_symmetry_cell_setting             cubic
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loop_
_atom_site_label
_atom_site_type_symbol
_atom_site_occupancy
_atom_site_fract_x
_atom_site_fract_y
_atom_site_fract_z
    H1    H  0.5000      0.4670(19) 0.4670(19)  0.4670(19)
    O1    O  1.0000      0.3750     0.3750      0.3750


I would write down the half equations for each reaction:
A

$\ce{ HNO3 + H+ -> H2NO3+ }$ (Base)
$\ce{H2SO4 -> HSO4- + H+}$  (Acid)

B

$\ce{  HSiO3- -> SiO2 + OH- }$  (Base)
$\ce{HCN  -> CN- + H+ }$ (Acid)

C

$\ce{HNO2 -> NO2- + H+ }$ (Acid)
$\ce{HCO3- -> CO2 + OH- }$(Base)

D

$\ce{C6H5O- + H+ -> C6H5OH }$ (Base)
$\ce{CH2ClCO2H -> CH2ClCO2- + H+ }$(Acid)


Distinguishing Test for Organic Compounds

Since this sounds like an assignment, I am not going to give it all away.

Does you have to use qualitative chemical tests? I don't even teach those anymore. They were supplanted by spectroscopic techniques in most professional labs decades ago.

IR spectroscopy would answer both questions quickly.

Benzamide would have peaks characterizing an aromatic ring, for example:  $\ce{C}_{sp^2}\ce{-H}$ stretches around $3050-3100\ \mathrm{cm^{-1}}$. Ethanamide will not have these peaks, but will instead have $\ce{C}_{sp^3}\ce{-H}$ peaks $2800-3000\ \mathrm{cm^{-1}}$. There are other differences as well.
There are differences in the position (and sometimes number!) of carbonyl peaks between an anhydride and an acid halide. What are they?

So would mass spectrometry. Benzamide and ethanamide have different molar masses and different fragmentation patterns. Ethanoyl chloride... chlorine has two major isotopes. How can that help?

NMR spectroscopy would also be useful. However, IR is cheap, fast, and easy.


All of these elements can form compounds in other oxidation states. Aluminium forms some compounds in the +1 state (e.g. see the section in https://en.wikipedia.org/wiki/Aluminium_iodide), as does Zinc (see the section in https://en.wikipedia.org/wiki/Compounds_of_zinc) and Cadmium (e.g. https://en.wikipedia.org/wiki/Cadmium(I)_tetrachloroaluminate). But Silver is the element that shows this most often in the list given - for instance it forms fluorides in the +1, +2 and +3 oxidation states (https://en.wikipedia.org/wiki/Silver_fluoride). As such I wouldn't say including the oxidation states in the formula is wrong, but on the other hand if you omit it everybody who knows there is an ambiguity will understand what is implied.


Stability of alkenes tri>di>mono, but how to explain tetra

I agree with the OP and have seen numerous examples where tetrasubstituted olefins are indeed quite difficult to make, both in the literature and in my own experience, leading to the formation of Anti-Zaitzev (Hoffman) elimination products.
Many question writers are reluctant to make a solution where the tetra-substituted option is the correct answer. It is not as simple as the trend in carbenium ion stability where hyperconjugation and conjugation dominate around a trigonal planar (less sterically encumbered) center.
A tetrasubstituted olefin is too sterically encumbered due to 1) two A(1,2) interactions and 2) two exo substituents (gem-dimethyl) interactions.

Such proof for this failure in the stability trend is included in the original question as the kinetic stability of such an olefin is unusually and unexpectedly high. As far as the the thermodynamic stability, torsional strain makes effective hyperconjugation less possible in the transition state which varies only marginally upon the strength of base and the lability of the bond to the leaving group.

This trend: mono is less stable than exo-di is less stable than cis-di is less stable than trans-di is less stable than tri is....that is about all you can correctly say about the trend. Consider why cis-disubstituted is less stable than a trans-disubstituted olefin. Now consider a tetra-substituted alkene.

Zaitsev's "rule" like Markownikoff's "rule" is a just fast and dirty guideline for undergraduates and which both have numerous exceptions. It does not follow that tetrasubstituted must be more thermodynamically stable than tri-substituted and product ratios indicate this repeatedly in the literature. Stereoselectivity of E over Z is a non-problem compared to the difficulty of and abysmal yields when creating a tetrasubstituted olefin.

Simply put an ellipsis ("...") after the trisubstituted alkene as that is all that can be stated to an undergraduate class without presenting a falsehood.


Difference between the terms "autocatalysis" and "(branching) chain reaction"

An autocatalytic reaction is a reaction where the product catalyzes the reaction. An autocatalytic reaction takes the following form (or more complex variations of it).

$\ce{A + B -> 2B}$

An autocatalytic reaction must produce the exact species $\ce{B}$ that cataylzes the reaction. More about autocatalytic reactions, including examples, can be found at this question and this answer.

A chain reaction is one where every step of the mechanism produces another activated intermediate of some kind that can continue to react under the same conditions. Chain reactions continue until they terminate by means specific to the type of chain reaction. Chain reactions may appear to be autocatalytic, but there is a net consumption of both reagents, and the end product does not tend to cataylze the reaction:

$$\ce{A + B}* \ce{ -> C}$$
$$\ce{C + B -> D + B}*$$
Net:
$$\ce{ A + B -> D}$$

In the generic example above, the second step produces more $\ce{B}*$, which can reenter the first step. Overall, however, the reaction is not autocatalytic, since no more $\ce{B}$ is being produced and the concentration of $\ce{B}*$ is likely unchanging (steady state). All chain reactions have this feature. While the reactive species may appear on both the reactant side and the product side, there is no net change in the number of reactive species present in the reaction.

A classic introductory chain reaction is the radical halogenation of an alkane like methane. This reaction's mechanism is divided into three regimes: 1) Initiation (generation of radicals where there were none previously), 2) propagation (the chain part), and 3) termination (in which there is a net reduction in radicals, usually by recombination).

Initiation:
$$\ce{Cl2 ->[hv] 2Cl*}$$

Propagation:
$$1) \ \ \ \ \ \ce{Cl* + CH4 -> HCl + CH3*}$$
$$2) \ \ \ \ \ \ce{CH3* + Cl2 -> CH3Cl + Cl*}$$

Termination:
$$\ce{CH3* + Cl* -> CH3Cl}$$
$$\ce{2CH3*  -> CH3CH3}$$

The above example is a cyclic chain reaction. Chain reactions can be linear, for example chain-growth polymerization. In this kind of chain reaction, two species, where one is the activated end of a polymer chain $\ce{P}*$ and the other is a new monomer $\ce{M}$, react to form a new activated chain end $\ce{P}'*$ that is one monomer unit longer. $\ce{P}'*$ can then reenter the reaction and react with another $\ce{M}$. Again, this reaction is not autocatalytic. The number of activated chain ends is not increasing (or decreasing).

$$\ce{P}* +\ce{M -> P'}*$$
$$\ce{P'}* +\ce{M -> P''}*$$

The activated chain ends, $\ce{P}*$, can be radicals, anions, cations, the products of ring-opening, or organometallic complexes.


Why does the barium ion in barium titanate move in its tetragonal crystal structure?

This is known to be because the $\ce{Ba^2+}$ ion sits off center in the middle of the tetragonal crystal structure

First of all this is not correct. It is the $\ce{Ti^4+}$ ion that sits in a tetragonal off center configuration (black atom in perovskite structure below). Also this hold for any material with perovskite structure with the $\ce{ABC3}$ general formula where $\ce B$ is your center site, $\ce A$ is the corner sites and $\ce C$ is your face site counter ions $(\ce{BaTiO3, LaAlO3, NaMgI3, ...})$.

  Additionally, does the $\require{enclose}\enclose{horizontalstrike}{\ce{Ba^2+}} \ \ce{Ti^4+}$ ion naturally sit off-center in this configuration, or does it only behave that way when introduced to a charge?

It does sit off center all of the time in any of 6 directions on a $(001)$ plane but the many centers are randomly oriented which produces no net dipole moment. When charge is introduced, the $\ce{Ti^4+}$ are attracted toward the negative plate and will shift toward that plate creating a net dipole moment.

  How much freedom of movement does this ion have in a solid crystalline structure?

Very little, the $\ce{Ti^4+}$ is located between the plane of 4 oxygens (red atoms) and an oxygen atom perpendicular to the 4-atom plane. It can move between the 6 possibilities of this shift $(\pm x, \pm y, \pm z)$ but not between cells.

  How does this $\require{enclose}\enclose{horizontalstrike}{\ce{Ba^2+}} \ \ce{Ti^4+}$ ion, or even the crystal structure itself, move in order to accompany a larger amount of charge?

When the electric field is applied, the titanium ion receives enough force to squeeze between oxygen atoms to a new lower energy position (e.g. $-x$ to $+y$).


Calculating concentration of an ionic compound

Keep it as simple as possible :)

If both initial solutions are not saturated and all the $\ce{NaCl}$ is dissolved, this will not change upon mixing.

If in doubt with the result of a calculation, have a look at the units!

Concentrations are given as the quotient of moles and volume, $c = \frac{n}{V}$, the unit apparently is $\mathrm{mol \cdot l^{-1}}$. The absolute amount of $\ce{NaCl}$ in the first solution is therefore given as $n = c\cdot V$. (Note that this is not the n in your question ;).

The new $\ce{NaCl}$ concentration is the sum of the moles, divided by the new total volume:
$c = \frac{(k\cdot n)+(l\cdot m)}{m+n}$

Unless given further information, it is safe to assume that $\ce{NaCl}$ still is fully dissociated after mixing.


What are the CIP rules for cyclic substituents?

I presume you want to know the E/Z geometry of the double bond. You need to determine on each end of the double bond which ring has priority. Intuitively one may argue that cyclohexane is larger than cyclopentane and cyclobutane is larger than cyclopropane. In this simple case, this is true and the double bond has the Z-configuration. However, digraphs are used in more complex cases. Your structure below has four colored dots. Each attached ring bond is "cut" and stretched out until one returns to the original atom. The three red atoms from the cyclohexane ring are identical. The same is true of the other three rings. Now treat the digraph as you would any acyclic alkene. The digraph has the Z-configuration.


There is nothing magical about light; it just brings in some energy. What's important is that it brings a huge lump of energy to one point, at once. If you'll heat the system in the hope that the heat will provide the necessary energy for the reaction, then you might or might not succeed. In the worst case, before initiating your reaction, heat will initiate some other reactions you don't want.

I expect the radical bromination of some huge branched alkane to be a particularly bad case. Put it under sunlight, and it will go smoothly; heat it, and you'll have all sorts of side reactions, fragmentations, isomerizations going on at once. Of your examples, decomposition of silver bromide is relatively "well-behaved" (after all, $\ce{AgBr}$ must decompose since it doesn't have many other things to do), and $\ce{CH4 + Cl2}$ is somewhere in between.

If you try it the other way around, it is the same: some thermal reactions can be started by using sunlight instead of heat, some can't.


How to calculate the heat capacity of a calorimeter based on a neutralisation reaction?

First, you have to find out the concentrations so you can calculate the amount of heat released. Since the density is given for a 1 M solution, I will assume this concentration.

Next, the mass of our solution:
$$ m = \rho \times V = 103.7~\mathrm{g}$$

Now we have to calculate the amount of energy that gets released through the reaction, but viewed from the environment and not the reaction (that's the reason for the negative sign before $q$):
$$ -q = \Delta_\text{r}H \times n = -58.3 \times 0.1 ~\mathrm{kJ\, mol^{-1}\, mol} = -5.83~\mathrm{kJ}$$

The total heat capacity of your system is given as
$$ c_\text{tot} = c_\text{s, sol}\times m + c_\text{cal}$$
where $c_\text{s, sol}$ is the specific heat capacity of your solution and $c_\text{cal}$ is the heat capacity of the calorimeter.

Now we know all the quantities needed for the following equation:
$$ q = c_\text{tot}\Delta T$$

We expand:
$$ q =( c_\text{s, sol}\times m + c_\text{cal}) \times \Delta T$$

Solve for $c_\text{cal}$:
$$ c_\text{cal} = \frac{q-c_\text{s, sol}m\Delta T}{\Delta T} = \frac{5.83~\mathrm{kJ} - 3.91~\mathrm{J\, g^{-1}\, K^{-1}} \times 103.7~\mathrm{g} \times 11.5~\mathrm{K}}{11.5~\mathrm{K}} = 101~\mathrm{J\, K^{-1}}$$

And now you know the heat capacity of the calorimeter.


Is it possible to have only liquid in a closed container?

In a tight, solid vessel, you can of course have a liquid (or any condensed phase) completely fill it.

There are basically three possibilities

the inside is at (potentially very) high pressure. no problem, water is compressible, just not very much, and also the vessel has a finite $E$ modulus
the inside is just at the vapour pressure of its content, but (because of the different expansion coefficients of vessel and content) the temperature must be kept perfectly constant at exactly the level where this is the case
the inside is below vapour pressure, but no gas bubble has evolved yet, that's principally an unstable condition, but if the deviation isn't large, it can stay practically forever.

Because of surface tension, there is even a small pressure range where the last case is actually stable.

I say your statement is basically right: You can have a completely filled vessel, but you can't do anything useful in it before it either bursts or bubbles.


Do L-sugars (enantiomers of natural sugars) have a sweet taste?

Take a look at tagatose. It is a 6-carbon sugar (a ketohexose like fructose) but is inefficiently metabolized and has about 30% the caloric content of sucrose. It is on the FDA GRAS (generally recognized as safe) list. Unlike common artificial sweeteners, you can cook with it like sugar. It has been industrially produced as a sugar alternative, but it is not clear if it commercially available to consumers at this time.


How to avoid oxidation of Si using a sub nanometer layer

Fluorine termination might be worth a look, if you have the facility for it. Sunada et al., in 1990 Jpn. J. Appl. Phys. 29 L24082, write in the abstract:

  The principal role of silicon-fluorine bonds in the chemical nature of HF-etched Si surfaces has been investigated by angle-resolved X-ray photoelectron spectroscopy. The native oxide growth thickness and the fluorine coverage have been systematically measured as functions of HF concentration, pure water rinse time, air or N2+O2 gas exposure time and gas phase H2O concentration. It is shown that the native oxide growth is strongly suppressed by the existence of Si-F bonds of about 0.12 monolayers on the surface. This is explained by a model in which Si-F bonds chemically stabilize the surface reactive sites such as atomic steps as supported by the result of the layer-by-layer oxidation of Si.

(My bolding).
I haven't read the full text, and I suspect that "strongly suppressed" really means that there was a very small amount of oxidation, rather than none at all. And note that while silicon surfaces may be stabilised by fluorination, scientists tend to get violently destabilised by the same reaction. This is a potentially hazardous process and not to be undertaken lightly.

In any case, "stability" really depends on what the surface is going to be exposed to. F-termination may oxidise only very slowly in air, but if you get water in there, it's a different story. In 1993 Jpn. J. Appl. Phys. 32 3125, Nakamura et al. write that:

  The 1.0 monolayer of Si-F bonds was found to be replaced with the 0.7 monolayer of Si-H bonds and the 0.2-0.3 monolayer of Si-OH bonds after dipping into water for 1 min.

So there may not be a "permanent" solution, just one that resists further oxidation for a while. What are you actually trying to achieve?

One other thing you could investigate is growing a self-assembled monolayer (SAM), probably on an H-terminated Si surface. It's been a while, but IIRC some pretty stable coatings can be generated that way. A SAM forest may offer a decent steric barrier to further reaction.


@Voldemort you learn the valencies rather than calculate them.  Valency is a fact of experiment whilst electron configuration gives an underlying explanation for the observations from experiment.

Iron has two different valencies (and oxidation states) which are a consequence of the stability of the electron configuration that remains when electrons are lost upon reaction.  So, iron can loose 2 electrons to leave a stable $\ce{Fe++}$ species; in your reaction those 2 electrons are gained by oxygen to give an anion that has 8 electrons in its outer shell which is a stable configuration.

Iron can also loose 3 electrons to leave a stable system and again, in your system, the electrons are picked up by the appropriate number of oxygen atoms to give oxygen anions that have 8 electrons in their outer shell.

If you look at the either side of the periodic table, say sodium reacting with a halogen, say chlorine (to give sodium chloride); you can see the 8 electron rule working.  But you're looking at a system where the iron is found in the middle of the periodic table and is a lot more difficult to predict when compared to the 8 electron rule.


Help in drawing resonance structures in case of polybasic acids

It depends on what specifically the question is asking you to do.

In reality, a polyprotic acid such as $\ce{H3PO4}$ becomes deprotonated in a sequential way; the first proton is the easiest to remove (most acidic), then the second, then finally the third is hardest to remove (least acidic).

So there are four different species that the question might be asking you about: $\ce{H3PO4}$, $\ce{H2PO4-}$, $\ce{HPO4^2-}$, and $\ce{PO4^3-}$

Each of the three anionic species above has multiple resonance structures.


I warn you that I've virtually no knowledge of the composition of grape juice - I didn't even remember that it contained tartaric acid, for example. My admimission of my ignorance is to guard you from letting my technical vocabulary  fool you into thinking that just because I might sound knowledgeable, that what I say is correct concerning grape juice and sea shells.

Calcium tartrate, when pure, is colorless (its crystals are "white"). When made from grape juice, I would expect that the crystals are colored. This would be due to impurities which co-precipitate & are incorporated into the crystal structure. I was able to find some pictures of crystals of calcium tartrate using Google which were colored (purplish red in one case, orange-red in another) and which might be consistent with what you saw. See also https://en.wikipedia.org/wiki/Tartrate.

On the other hand, grape juice contains other organic acids which might also form colored crystals. Two families in particular I am thinking about: tannins and anthrocyanins. Both are polyphenols, which means they have two or three -OH (hydroxyl) groups attached to one aromatic ring (the simplest example of an aromatic ring is benzene).

Sheen is generally not a property of a chemical compound's pure crystal. The prototypical example is the gemstone opal, which is composed of microscopic silica spheres packed into a regular array. The point here is that "sheen" is more likely due to microscopic particles included in the crystal rather than an inherent property of a pure crystal. There are exceptions to this, however.

The reason I mention the polyphenols is that they typically form very highly colored complexes (salts) with iron. Often they are colored brown, purple or black (and the that their crystals might have a sheen.) It appears to me that the shell has turned purple-black due to the polyphenols reacting with trace quantities of iron. A way to check this would be to take an iron nail (ungalvanized, unplated, uncoated) and let it rust in some grape juice and see what you get - if its just the typical orange-red rust, then there probably aren't many polyphenols present. But, note that in anaerobic (low oxygen,eg. in unmixed water) conditions, iron will rust to a black color, which will turn slowly red-brown when allowed to air dry. Wine is well known to contain tannins, which come from grape skins.

Anthrocyanins are known to change color with changes in acidity (pH). So, you could check to see if the crystals change color in water with added acid (vinegar, say) and base (ammonia, say). With supervision, Draino in water is also an option; it contains a very strong base, sodium hydroxide, but it definitely shouldn't be used by a child - it can cause severe burns to the skin and worse to the eyes. Most powdered detergents also are basic.

Grape juice contains large amounts of tartaric, malic, and citric acids. Neither of the latter two are likely to be the source of your crystals, imho. Two other experiments you could try are: 1. Add some rubbing alcohol (40% or even better 70%) to a bit of the juice. You might get some crystals (sediment).
2. Put a shell into some Cream of tartar in water. (The Cr. of T. will be more soluble with some vinegar or lemon juice added, making it more acidic). If the crystals you saw were indeed calcium tartrate, then this should give you the same thing. When I saw the picture of the black seashell, I immediately assumed tannins (and iron) were present, fwiw.


How to make 0.3% ammonia water from 28-30% ammonia hydroxide

Hint : Ammonia water is actually diluted Ammonium Hydroxide ruled by the following equilibrium :

$$\ce{NH3 + H2O <=> NH4+ + HO-}$$