Why is gadolinium specifically used in MRI contrast agents?

The reasons may possibly be:

T1shortening
Relatively less adverse reactions
Increased contrast enhancement

  The most commonly used clinically approved contrast agents for MR
  imaging are gadolinium-based compounds that produce T1 shortening.

Tissue relaxation results from interactions between the unpaired electron of gadolinium and tissue hydrogen protons, which significantly decrease the T1 of the blood relative to the surrounding tissues.

Additionally, adverse reactions to this agent are far less frequent than those seen with iodinated compounds, with common reactions including nausea, vomiting, headache, paresthesias, or dizziness.

  T1weighted sequences are part of almost all MRI protocols and are
  best thought of as the most 'anatomical' of images, resulting in
  images that most closely approximate the appearances of tissues
  macroscopically, although even this is a gross simplification.

Figure 1-10. MR enterography on a patient with Crohn disease. A. Coronal T2-weighted image shows wall thickening
and stenotic ileum (arrow). B. Coronal gadolinium-enhanced fat suppressed T1-weighted image shows increased
contrast enhancement of one thickened segment of ileum (arrow).

Credits Basic Radiology Chapter 1 page 11

Update

Most of the answer above was based on Pharmacological/Medical aspects but some Chemistry aspects have been included below:

Also the article mentioned by@Linear Christmas contains valuable information on the subject

Contrast agents

  Contrast agents can be divided into two groups depending on whether
  they cause changes in either T1 (longitudinal relaxation – in simple
  terms, the time taken for the protons to realign with the external
  magnetic field) or T2 (transverse relaxation – in simple terms, the
  time taken for the protons to exchange energy with other nuclei)
  relaxation rates of the water protons, these being known as positive
  or negative agents respectively.

  The ability of an agent to affect T1 and T2is characterised by the
  concentration-normalised relaxivities r1 and r2respectively. These
  parameters refer to the amount of increase in 1/T1 and 1/T2
  respectively, per millimole of agent, and are normally quoted as a
  rate (mM21 s21).4 The values are used to determine the efficiency of a
  contrast agent, and they consist of contributions from both inner
  sphere and outer sphere relaxation mechanisms Negative contrast agents
  influence the signal intensity by shortening transverse relaxation
  (T2), thereby producing darker images as high T2 results in increased
  brightness of the images.

Gadolinium (III) vs other Lathanide (III) ions

  Gadolinium(III) reagents are commonly focused on, due to the coupling
  of a large magnetic moment with a long electron spin relaxation time
  of $\ce10^{-9}$s  at the magnetic field strengths used in MRI
  techniques.

Gadolinium (III) ion is unique in a number of respects:

It has the ability to shorten both the longitudinal and transverse relaxation times of water protons approximately to the same extent by relaxing all nearby protons. Other Ln(III) ions with large magnetic moments are less efficient in shortening T1.

  The Gadolinium ion is also unique among other Ln(III) because its
  symmetric seven electron ground state results in an electronic
  relaxation rate that is six orders of magnitude slower than the
  other Ln(III) ions. This in turn results in a strong magnetic
  anisotropy and fast electronic relaxation states with very short T1,
  on the order of $\ce10^{-13}$s.

Other Ln (III) ions tend to undergo Curie relaxation enhancement arising from the interaction of the nuclear spin with the thermal average of the electron spin. (The Curie-spin relxation effect is significant at lower temperatures and higher magnetic fields and for ions with a large magnetic moment).

It has been observed that the Curie spin relaxation effect affects the transverse relaxation more than the longitudinal relaxation. Consequently paramagnetic Ln(III) other than Gd (III) are less efficient T1 relaxation agents.

Use of other Lanthanide (III) ions as MRI contrast agents

Dysprosium(III) is another lanthanide ion that has been used in
MRI, being classed as a negative contrast agent. In the use of high magnetic fields,

  Gd(III) based contrast agents exhibit poor water relaxivity. As a
  result, the interest in dysprosium-based complexes is increasing as
  they display slow water exchange, due to the need to lengthen the
  residence time in order to optimise the r2 relaxivity. Dy(III) has a
  large magnetic susceptibility which induces local field gradients
  resulting in a lowering of T2.

Generally, dysprosium complexes where the water molecules have a long residence
time, possess potential application as negative contrast agents
at high magnetic fields due to their efficient transverse
relaxivity. However it has been demonstrated that lengthening the residence time of water
can actually be detrimental, because the transverse relaxivity
can then be limiting.

The overall conclusion from the use of dysprosium(III)
complexes as contrast agents is that due to the balancing of
factors required to optimise r2, design of a suitable molecular
structure is crucial. Fine tuning these residence times of the
water protons, and hence the relaxivity, may lead to promising
contrast agents for high field magnetic resonance imaging.

  Europium(II) analogues have been proposed as alternatives to Gd(III)
  because they are isoelectronic—each having seven unpaired electrons.

The only problem with this
complex was that the stability of the Eu(II) chelate was
107 times higher than that of the Eu(III) chelate, which
meant that the Eu(III) complex could be susceptible to
dissociation, possibly releasing toxic Eu(III) into the body.

Other strongly paramagnetic ions e.g Tb(III), Ho(III), Er(III) can have significant effect on line widths and such line broadening usually makes it difficult or even impossible to detect nuclei that are located within a certain distance of the lanthanide ion.

There is still promising use of other Lanthanide (III) compounds as contrast agents as alternatives to using T1 shortening contrast agents as contrast can originate from altering proton density or the total water signal that is detected (using another technique called Magnetisation Transfer (MT).In this case Gadolinium(III)cannot be used in this technique because the T1 of water would be too short, but other lanthanides, such as Eu(III), can be used because they have smaller magnetic moments, relaxing bulk water much less efficiently.

Other potential candidates are still being tested under laboratory conditions, Contrast agents used in the
body clearly should be biocompatible, but there are also other requisites need to be addressed as well. These include the requirements of rapid renal excretion, water solubility, stability in aqueous conditions, and a low osmotic potential when in solution for clinical work.

References

Basic Radiology Michael Y. M. Chen, MD
'Alternatives to Gadolinium-Based Metal Chelates for Magnetic Resonance Imaging' (Chem. Rev., 20102010, 110 (5), pp 2960–3018,
Lanthanides in magnetic resonance imaging. Melanie Bottrill, Lilian Kwok and Nicholas J. Long
Chemistry Society Reviews, DOI: 10.1039/b516376p


UV-Vis emission from charge-separated states?

There are at least cases in which charge recombination results in an excited species that deactivates via fluorescence.

Imagine the ionization of a hydrocarbon ($\ce{RH}$) to yield a radical cation and a solvated electron.
\[\ce{RH ->[\mathrm{e-beam}] RH^{+\cdot} + e-}\]

Trap the electron by reducing an aromatic hydrocarbon ($\ce{ArH}$) to the coresponding radical anion.

\[ \ce{ArH +e- -> ArH^{-\cdot}}\]

Charge recombination between the radical cation and the radical anion can yield $\ce{ArH}$ in an excited state.

\[\ce{RH^{+\cdot} + ArH^{-\cdot} -> RH + ArH^{\ast}} \]

Back in the days, Alexander D. Trifunac did a lot of work on that at ANL.


Why is dipositive dilithium more stable than neutral dilithium?

The comment asks "Why is Li2+ more stable than Li2". Li2 has a relatively low bond energy (gas phase) of 27 kcal/mol (Cotton and Wilkinson, Inorganic Chemistry). The energy cost to make one mole of Li gas is 37 kcal (CRC Handbook). The ionization energy of Li is 5.39 eV = 124 kcal/mol.

So, breaking a Li2 gas molecule into 2 Li (gas) costs 27 kcal/mol.

Breaking a Li2+ gas molecule into Li + Li+ involves not the separation of two uncharged atom, but the separation of a very small Li+ ion from a Li atom which provides some charge accomodation. The heat of formation of Li+ in water is 66 kcal/mol (presumably from aquation with 4 waters); the hydration bonding is worth about (124 + 37 + 66)/4 = 54 kcal/mole each. So I would estimate the bond between Li+ and Li to be perhaps 40 kcal/mol, at least a significant fraction of 54 kcal/mol, and probably more than 27 kcal/mol.

The important point is that the unusually small size of the Li+ ion makes it able to polarize a neutral atom to form a strong bond (and come in closer!), whereas the neutral atoms have only uncharged molecular orbitals to spread their electrons over. It is highly unlikely that larger atoms (Na, K) would show similar stability for an ion-molecule compared to a neutral molecule. Magnesium might be a similar exception, however, since both Li and Mg have small radii; it might be interesting to compare stabilities of Mg2 and Mg2+.


Characterisation of gold-silver alloy nanoparticle

I Think you are in right track using UV-vis spectroscopy as a tool to identify your alloys. Other than UV-vis spectroscopy, researchers around the world have used many other analytical techniques to evaluate the synthesized nanomaterials, including X-ray diffractometry (XRD), Fourier transform infrared spectroscopy (FTIR), X-ray photoelectron spectroscopy (XPS), dynamic light scattering (DLS), scanning electron microscopy (SEM), transmission electron microscopy (TEM), atomic force microscopy (AFM), and etc. (For review of these techniques: Read Ref.1).

According to your notes, it seems like your main concern is how to confirm that your samples are alloys. UV-vis spectroscopy is a very useful and powerful tool in that regards, and reliable technique for the primary characterization of synthesized nanoparticles as well. Silver absorbs around $\pu{400 nm}$ and gold around $\pu{550 nm}$, yet these values vary depends on particle sizes of the dispersion (See Figure 1A & 1B, respectively). For example, absorption spectra calculated using the quasi-static approximation for silver and gold nanospheres with diameters $\mathrm{D} = 26$ and $\pu{20 nm}$, respectively, have shown absorption maxima (surface plasmon resonance wavelength $\lambda_R$) at ~$\pu{375 nm}$  for $\ce{Ag}$ and ~$\pu{525 nm}$  for $\ce{Au}$ in water (Ref.2). This reference also discussed the behavior of $\lambda_R$ as a function of the dielectric constant of the surrounding matrix ($\epsilon_m$), which is 1.77 for water.

This phenomenon was used to the best in identifying $\ce{Ag-Au}$ alloys in various compositions in an article (Ref.3), abstract of which states:

  Gold−silver alloy nanoparticles with varying mole fractions are prepared in aqueous solution by the co-reduction of chlorauric acid $\ce{HAuCl4}$ and silver nitrate $\ce{AgNO3}$ with sodium citrate. As the optical absorption spectra of their solutions show only one plasmon absorption it is concluded that mixing of gold and silver leads to a homogeneous formation of alloy nanoparticles. The maximum of the plasmon band blue-shifts linearly with increasing silver content. This fact cannot be explained by a simple linear combination of the dielectric constants of gold and silver within the Mie theory. On the other hand, the extinction coefficient is found to decrease exponentially rather than linearly with increasing gold mole fraction $x_{\ce{Au}}$. Furthermore, the size distribution of the alloy nanoparticles is examined using transmission electron microscopy (TEM). High-resolution TEM (HRTEM) also confirms the formation of homogeneous gold−silver alloy nanocrystals.

According to this reference, only one plasmon band for particular alloy (Two bands would be expected for the case of a mixture of gold and silver nanoparticles). The band shows a blue shift with increasing $\ce{Ag}$ amount in the alloy (See Figure 2) and a linear relationship with increasing gold mole fraction, $x_{\ce{Au}}$ (See the insert in Figure 2):

References:

X.-F. Zhang, Z.-G. Liu, W. Shen, S. Gurunathan, “Silver Nanoparticles: Synthesis, Characterization, Properties, Applications, and Therapeutic Approaches,” Intl. J. Molecular Sci. 2016, 17, 1534 (34 pages) (doi:10.3390/ijms17091534).
F. Vallée, “Chapter 7: Optical Properties of Metallic Nanoparticles,” In Nanomaterials and Nanochemistry; C. Bréchignac, P. Houdy, M. Lahmani, Eds.; Springer-Verlag: Berlin, Germany, pp. 197-228, 2007.
S. Link, Z. L. Wang, M. A. El-Sayed, “Alloy Formation of Gold−Silver Nanoparticles and the Dependence of the Plasmon Absorption on Their Composition,” J. Phys. Chem. B 1999, 103(18), 3529–3533 (DOI: 10.1021/jp990387w).


Buffers can be made two ways.

Buffers

Weak Acid/Base + Salt of the Conjugate Base/Acid

Let’s take acetic acid ($\ce{HC2H3O2}$) and sodium acetate ($\ce{NaC2H3O2}$).

$$\ce{HC2H3O2 + H2O <=> H3O+ + C2H3O2-~~~~~~~~~~K_a=1.8 x 10^{-5}}$$

Acetic acid has a small $\ce{K_a}$ value, meaning that it does not dissociate much. Conversely, sodium acetate readily dissociates as sodium is an alkali metal. By having equimolar concentrations of both weak acid and conjugate base, the addition of an acid adds hydronium, which reacts with the acetate ion to shift the equilibrium to the left, and the addition of a base absorbs hydronium to shift the equilibrium to the right. Either way, the hydronium concentration, which determines the pH, remains close to the same.

Weak Acid/Base + Strong Base/Acid

This relies on the same principle as above but supplies the conjugate base in a different manner. Continuing with our acetic acid example, here, this would be added, alone. Then you would add half of the moles of acid in strong base, such as $\ce{NaOH}$. This would neutralize any available hydronium, shifting the equilbrium to the right until there was the same half-and-half of acid and conjugate base.

Your Solution

Strong Acid?

Unfortunately, strong acids do not make for good buffers as they fully dissociate -- the equilibrium is so far to the right that it is very unstable.

Salt of the Conjugate Base?

$\ce{Na2SO4}$ shares no common ion with $\ce{HCl}$, so it provides no conjugate base ($\ce{{SO4}^2-}$ is the conjugate base of the second dissociation of $\ce{H2SO4}$). Even if you used $\ce{NaCl}$, it still would not work because of the reason outlined above.

pH?

If you're still interested, the pH of the solution above is easily calculated. As $\ce{HCl}$ fully dissociates, its concentration (0.1 M, as you say above) is equal to the hydronium concentration. However, the addition of another liter of neutral solution halves the hydronium concentration to 0.05 M. Therefore,

$$\ce{-log(0.05)=pH=1.3}$$

Hope this helps!


Is the equilibrium constant in the expression based on pressure or concentration?

Neither. It is exactly what is given there $K$, the standard equilibrium constant, and the equation given is its definition,
$$\Delta G^\circ=-RT\ln K.\tag{1}\label{def}$$

A general equilibrium constant for any reaction
$$\ce{\nu_A A + \nu_B B + \nu_C C + \cdots
   <=>\nu_{A'} A' + \nu_{B'} B' + \nu_{C'} C' + \cdots  }$$
is given by
$$K_x = \frac{ x_{\ce{A'}}^{\ce{\nu_{A'}}}\cdot x_{\ce{B'}}^{\ce{\nu_{B'}}}\cdot
               x_{\ce{C'}}^{\ce{\nu_{C'}}}\cdots}{
               x_{\ce{A}}^{\ce{\nu_A}}\cdot x_{\ce{B}}^{\ce{\nu_B}}\cdot
               x_{\ce{C}}^{\ce{\nu_C}}\cdots},$$
or more general, gathering all components in $\mathbb{B}=\{\ce{A, B, C, \dots, A', B', C', \dots }\}$
$$K_x = \prod_{B~\in~\mathbb{B}} x_B^{\nu_B}.\tag2$$

If you insert that into \eqref{def}, then $x$ becomes the chemical potential of the component. You can derive the link to the activity of the component (see here), and you can link the activity to concentrations, pressures, etc.

Given the right boundary conditions, e.g. ideal dilution, $K$ essentially becomes $K_c$, etc.

$$\mu(i) \propto a(i) \propto c(i) \implies K \propto K_c\tag3$$

Obviously, this is a very sloppy generalisation, and you should study thermodynamics in more detail to fully understand the differences between the quantities used, and when to use which.


It depends, there are different kinds of densities: mass density, particle density, charge density, probability density ...

Depending on how your measure your matter, mass, number of particles, charge (e.g. for electrons), you could use the corresponding density.

"Matter per unit volume" would only work in a broader sense, since it does not specifically state what kind of measure you want to use. Most of the time one simply defaults on mass, but this depends on the specific field.


Is it possible to have an airborn fluorescent particle that reacts to lightning?

Maybe something that fluoresces in the presence of ozone? A quick search gave this article. Lightning is known to convert oxygen into ozone, so I think this should work or at least give a good starting point for future development.


Why is lithium plating irreversible in Li-ion cells?

It's true that when the battery is re-discharged, it's thermodynamically 'easier' to oxidize lithium than lithiated graphite. However, keep in mind that the lithium will react with electrolyte to form an SEI (solid electrolyte interphase) layer. This process consumes lithium ions, so on recharge you don't get all the lithium ions back. This is why the process is considered "irreversible."


Analysis of methylphenidate (Ritalin) HCl extended release tablets

If you are insistent on a picrate salt, I'd suggest first taking your methylphenidate and forming a known concentration of the picrate by adding a known amount of picric acid to a solution in which you know the base would be in excess. This ensures that all of the picric acids forms the picrate salt, and you would therefore know the picrate complex concentration (assuming there are no other compounds in solution that might react with the picrate). You could then measure the absorbances at a wavelength at which picric acid has only a negligible absorbance, provided one exists. Note that if no such wavelength exists, matters become rather more complex, and you will have to look for changes in absorbance on either side of an isosbestic point.

An alternative reaction to the use of picric acid is to instead use 1-fluoro-2,4-dinitrobenzene (FDNB), which reacts with primary and secondary amines at room temperature to form a highly-coloured 1-amino-2,4-dinitrobenzene (ADNB). The spectral overlap between FDNB and ADNB is typically pretty small, so using an excess of FDNB usually doesn't greatly change the absorbance at the maximal wavelength of ADNB. Again, to at least know the absorbance of the methylphenidate ADNB, you would have to first perform the reaction with an excess of methylphenidate, and assume a complete reaction to the ADNB.


This notation is mentioned in the IUPAC Nomenclature of Inorganic chemistry, Recommendations 2005 (‘Red Book’):

  IR-10.2.5.1 The eta (η) convention

  The special nature of the bonding of unsaturated hydrocarbons to metals via their π-electrons
  has led to the development of the ‘hapto’ nomenclature to designate unambiguously the
  unique bonding modes of the compounds so formed. (…) The Greek
  symbol η (eta) provides a topological description by indicating the connectivity between the
  ligand and the central atom. The number of contiguous atoms in the ligand coordinated to the
  metal is indicated by a right superscript numeral, e.g. η3 (‘eta three’ or ‘trihapto’), η4 (‘eta
  four’ or ‘tetrahapto’), η5 (‘eta five’ or ‘pentahapto’), etc.

  (…)

  If an unsaturated hydrocarbon serves as a bridging ligand, the prefix μ (see Sections
  IR-10.2.3.1 and IR-10.2.3.4) is combined with both η and κ, where necessary. The colon is
  used to separate the locants of the bridging ligand which indicate binding to different metal atoms.

  (…)

  Examples:

  21.

  (μ-η2:η2-but-2-yne)bis[(η5-cyclopentadienyl)nickel](Ni—Ni)

(This convention is not used for unsaturated hydrocarbons only, there is also a more complicated example (25) with carbonyl (CO) ligand.)

Use of this notation for formulae does not seem to be explicitly mentioned or “allowed”. If it is not an omission in the rules, I think that such construct could be called a name-formula hybrid, which might useful for detailed description the discussed structure detail in a shortest way possible.

The $\ce{[^{n}Bu4N]2[(UO2)2(\mu-\eta^2:\eta^2-O2)(NO3)2(\mu-Au(CN)2)2]}$ polymeric complex is described in [1], which talks about $\ce{U-O2-U}$ linkage with bidentate peroxo ($\ce{O2^{2-}}$) unit.

References:

Brown, M. L.; Ovens, J. S.; Leznoff, D. B. Dicyanoaurate-Based Heterobimetallic Uranyl Coordination Polymers. Dalton Transactions 2017, 46 (22), 7169–7180.


Does temperature affect the %oxygen in air by volume? (even though oxygen density varies inversely with temperature according to P/RT=n/V)

No, the concentration does not depend on density in a well-mixed ideal gas.

The key principle here is that the atmosphere is close to being an ideal gas. In an ideal gas the components are well mixed and the composition will not change unless there is a chemical change that acts selectively on one of the components.

The ideal gas equation applies equally to each component of the mixture of gases that is air. If you warm a vessel of fixed size, the pressure goes up. If you warm the mixture in a variable volume, the volume increases. But nothing in either process causes the composition of the gas to change, so the concentrations of the components will remain the same.

You seem to be applying the gas equation to just the oxygen and ignoring the other components (Nitrogen, argon etc.). Yes the warmer air will have a lower density, but the composition will be unaltered. So the relative amount of oxygen in the air remains the same. Warm air has an overall lower density (which is why hot air balloons work) and the volume density of all the components will decrease. So the percentage of oxygen per litre will be lower but so will the percentage of Nitrogen. Mostly when measuring gas composition chemists talk about the concentration relative to all the components in the mixture not to the volume (and in an ideal gas, the volume/pressure/temperature relationship is the same for all components). So, trivially, the volume density of all components is lower when the gas is warmer at constant pressure. There is no selective change for just one component.


Why should the product of the normality of the acid or base and its volume be constant?

$$\ce{N1V1 = N2V2}$$ The equation is used basically for dilution of solution and determining one of the parameters of acid or base.

In the case of dilution of solution, the equation holds true because the no of gm equivalents of the solute remains constant in the solution even the solution is diluted.

Initial no of gram equivalents of solute = Final no of gram equivalents of solute ....................eq (1)

And we know, $$\text{Normality} = \frac{\text{No of gram equivalents}}{\text{Volume in Litre}}$$
$$\text{No of gram equivalents}=\text{Normality}\times\text{Volume in Litre}$$
Now lets consider,

Initial Normality =$\ce{N1}$

Initial Volume=$\ce{V1}$

Final Normality=$\ce{N2}$

Final Volume = $\ce{V2}$

Then , eq (1) becomes

$$\text{Initial Normality}\times\text{Initial Volume}=\text{Final Normality}\times\text{Final Volume}$$
$$\ce{N1V1 = N2V2}$$

In case of acid and base reaction,
No of gram equivalents of acid will always react or consume equal no of gram equivalents of base and vice versa

No of gram equivalents of acid = No of gram equivalent of base .......eq (2)

Normality of acid = $\ce{N1}$

Volume of acid =$\ce{V1}$

Normality of base =$\ce{N2}$

Volume of base =$\ce{V2}$

Then eq (2) becomes,
$$\text{Normality of acid}\times\text{Volume of acid}=\text{Normality of base}\times\text{Volume of base}$$

$$\ce{N1V1 = N2V2}$$

The number of grams of a chemical element or compound equal to the weight of its chemical equivalent. The gram equivalent of a chemical element is numerically equal to the quotient resulting from dividing the element’s atomic weight by its valence. For example, the gram equivalent of iron (atomic weight 55.847) with a valence of 2 is 27.92 g and with a valence of 3, 18.62 g. The gram equivalent of acid is that amount of the acid in grams containing one gram equivalent of hydrogen capable of being displaced by metal with the formation of salt. The gram equivalent of a base is that amount of the base in grams that is necessary for its complete reaction with one gram equivalent of acid. The gram equivalent of salt is that amount of the salt in grams containing one gram equivalent of metal. The concept of gram equivalent is widely used in volumetric analysis.

http://encyclopedia2.thefreedictionary.com/Gram+Equivalent


How to calculate weight of CO₂ in one kg of fuel

You have 1 mol $\ce{C}$ reacting with 1 mol $\ce{O_2}$ to give 1 mol of $\ce{CO_2}$

What you need now is something to relate weight and mol. This is called molar mass, notated with the letter M, and has the unit g/mol. You'd just look that up on the internet, but it's calculated from the weight of the atoms in each molecule. If a molecule weighs 32.2 units, then it'd have a molar mass of 32.2 g/mol.

The calculations are pretty straight forward. How many moles of fuel per kilo? $1000g / M(\ce{C})$, where M, as noted, indicates molar mass.

You know that 1 mol carbon turned into 1 mol carbondioxide. So the number you just calculated is thus also the moles of carbondioxide. You need to find out how many grams those x mol carbondioxide corresponds to: $n(\ce{CO_2}) * M(\ce{CO_2}) = m(\ce{CO_2})$.

If you're not confident in this, pay attention to the units. n = mol; M = g/mol; m = g. Thus multiplying M by n yields a figure in grams. Is that clear?


Gibbs Free Energy of Solutes and Solutions

For the reaction
$$\ce{CaSO4 (s) <=> Ca^2+ (aq) + SO4^{2-} (aq)},$$
the appropriate equilibrium constant using activities can be set up as
$$K_\mathrm{eq}=\frac{a(\ce{Ca^2+})\,a(\ce{SO4^2-})}{a(\ce{CaSO4})}.$$

The activity $a$ of a pure substances in condensed phases is approximately one, since their behaviour can be often treated ideally, i.e. the chemical potential $\mu$ at the considered conditions is approximately the same as the chemical potential at standard conditions $\mu^\circ$. This derives directly from the definition of the relative activity: $$a = \exp\left\{\frac{\mu-\mu^\circ}{\mathcal{R}T}\right\}$$

For substances where the solubility is very low, concentrations of the ions are very low, so in a further approximation we can assume that in these solutions the activity coefficient is approximately one, therefore, we can write concentrations instead of activities and obtain for the equilibrium constant the expression, that is also known a solubility product.
$$K_\mathrm{eq} \approx K_\mathrm{sp} = c(\ce{Ca^2+})\,c(\ce{SO4^2-})$$

However, this is completely irrelevant to the question itself. Since the solution is saturated, it also means, that the solid is in equilibrium with the solution, which means that the kinetics of dissolution and precipitation are equal of magnitude. The Gibbs energy has a very distinct value for these equations.

   chemical equilibrium
 Reversible processes [processes which may be made to proceed in the forward or reverse direction by the (infinitesimal) change of one variable], ultimately reach a point where the rates in both directions are identical, so that the system gives the appearance of having a static composition at which the Gibbs energy,  $G$, is a minimum. At equilibrium the sum of the chemical potentials of the reactants equals that of the products, so that:
 $$\Delta G_\mathrm{r} = \Delta G_\mathrm{r}^\circ + \mathcal{R}T\,\ln\,K = 0\\\Delta G_\mathrm{r}^\circ = - \mathcal{R}T\,\ln\,K\\$$
 The equilibrium constant, $K$, is given by the mass-law effect.
 Source: IUPAC goldbook


How to make this citric acid substance I made not sticky?

A vacuum oven with a solvent trap is a really good way to get rid of persistent water in your product. I'm not sure what your overall objective is so I'd read up on how aggressively (or gently) you can heat your product first though.


How to perform a relaxed scan in GAMESS?

Unfortunately a feature like ModRedundant in Gaussian, which one can use to perform relaxed scans, is not available in GAMESS. Even the possibility to perform a rigid scan seems a bit complicated. You can request a calculation with RUNTYP=surface and you have to specify a $SURF group. See example 29 of the tests, which you should find in path/to/gamess/tests/standard/exam29.inp.

A relaxed scan is a bit harder to perform, and only straight forward for varying distances. You have to perform a series of single point calculations, with partially frozen coordinates. I describe the general procedure of doing that in How do I perform a partial optimisation in GAMESS?

Here I will outline it for your problem, based on what you can read there.

Before we do that however, you should be aware, that there is a python script by Andrew Hoos on github. I did not have the time to test it. It is quite old, and I cannot predict how the development of GAMESS will influence the functioning of rscan.

I assume you have read and understood the linked post, so in an effort to safe some space, I use a more automated approach here.

First, set up the general file, which you want to use for a check run. I have used your geometry.

 $CONTRL
   EXETYP=CHECK
   SCFTYP=RHF RUNTYP=ENERGY
   COORD=UNIQUE   ICHARG=1   MULT=1   NZVAR=15   $END
 $BASIS
   GBASIS=STO    NGAUSS=3   $END
 $ZMAT
   AUTO=.TRUE.   DLC=.TRUE.
   NONVDW(1)=1,5, 2,5
   $END
 $GUESS
   GUESS=HUCKEL    $END
 $DATA
   H5O2
   C1
   OXYGEN      8.0      0.086655000         0.133386000        -0.150088000
   OXYGEN      8.0     -1.636963000         0.213966000        -1.809353000
   HYDROGEN    1.0     -0.149749000         0.260762000         0.763793000
   HYDROGEN    1.0      0.802607000        -0.484925000        -0.261350000
   HYDROGEN    1.0     -0.423645000         0.140377000        -0.629197000
   HYDROGEN    1.0     -2.313930000        -0.430164000        -1.970372000
   HYDROGEN    1.0     -1.720592000         0.938467000        -2.416119000
   $END

It is again essential that you include NONVDW descriptions. Since the proton will move in between the two water molecules, I have included both pairs. Note that the 1-2 distance will not be included in the list of internal coordinates (even if you specify it directly). You can use NONVDW(1)=1,2 instead though.

Now that you have checked everything you can run the first optimisation:

 $CONTRL
   SCFTYP=RHF   RUNTYP=OPTIMIZE    COORD=UNIQUE
   ICHARG=1   MULT=1   NZVAR=15     $END
 $BASIS
   GBASIS=STO    NGAUSS=3   $END
 $ZMAT
   AUTO=.TRUE.   DLC=.TRUE.
   NONVDW(1)=1,5, 2,5
   IFZMAT(1)=1,1,2,     1,1,5
   FValue(1)=2.3938489, 0.700
   $END
 $GUESS
   GUESS=HUCKEL   $END
 $DATA
   H5O2
   C1
   OXYGEN      8.0      0.086655000         0.133386000        -0.150088000
   OXYGEN      8.0     -1.636963000         0.213966000        -1.809353000
   HYDROGEN    1.0     -0.149749000         0.260762000         0.763793000
   HYDROGEN    1.0      0.802607000        -0.484925000        -0.261350000
   HYDROGEN    1.0     -0.423645000         0.140377000        -0.629197000
   HYDROGEN    1.0     -2.313930000        -0.430164000        -1.970372000
   HYDROGEN    1.0     -1.720592000         0.938467000        -2.416119000
   $END

Note that I am keeping the automagic creation of the coordinates this time. To be on the safer side, you should probably follow more closely the other post. In any case, we are now ready to continue. Well use the optimised geometry of the last run, as well as the orbitals, which can be found in the auxiliary output.

 $CONTRL
   SCFTYP=RHF    RUNTYP=OPTIMIZE    COORD=UNIQUE
   ICHARG=1   MULT=1   NZVAR=15     $END
 $BASIS
   GBASIS=STO    NGAUSS=3   $END
 $ZMAT
   AUTO=.TRUE.   DLC=.TRUE.
   NONVDW(1)=1,5, 2,5
   IFZMAT(1)=1,1,2,     1,1,5
   FValue(1)=2.3938489, 0.800
   $END
 $GUESS
   GUESS=MOREAD
   NORB=15
   $END

! ----- RESULTS FROM SUCCESSFUL RHF      GEOMETRY SEARCH -----
! ----- COORDS, ORBS, GRADIENT, AND APPROX. HESSIAN -----
!  COORDINATES OF SYMMETRY UNIQUE ATOMS (ANGS)
!      ATOM  CHARGE       X              Y              Z
! ------------------------------------------------------------
 $DATA
   O2H5
   C1
   OXYGEN      8.0   0.0758752612   0.1029703596  -0.1380388056
   OXYGEN      8.0  -1.6329486914   0.2211188064  -1.8103049937
   HYDROGEN    1.0  -0.0725810430   0.4747088980   0.7504261439
   HYDROGEN    1.0   0.9330501410  -0.3174832331  -0.3330916306
   HYDROGEN    1.0  -0.4237766906   0.1376065810  -0.6270666776
   HYDROGEN    1.0  -2.3424944360  -0.4268887357  -2.0060894609
   HYDROGEN    1.0  -1.7921347121   0.9257714517  -2.4735439582
   $END

! --- OPTIMIZED RHF      MO-S --- GENERATED AT Fri Dec 15 00:48:25 2017
! E=     -150.1620919294, E(NUC)=   47.9559336673
 $VEC
 1  1 9.93613099E-01 3.20702947E-02-2.24634863E-03 1.56904217E-04-2.19957003E-03
 1  2 3.13313084E-04 9.04941985E-04 6.59436960E-04-4.52587984E-05 6.45901704E-04
 1  3-4.70970670E-03-4.70987324E-03-9.86284464E-03 1.88174472E-04 1.88109591E-04
 2  1-5.80880573E-04 1.71794788E-03-8.47977388E-04 5.87852394E-05-8.29926100E-04
 2  2 9.94111746E-01 2.64448946E-02-2.47981701E-03 1.59721737E-04-2.45660777E-03
 2  3 1.64987868E-04 1.65066081E-04-3.05915593E-03-5.57763329E-03-5.57725393E-03
 3  1 2.41238581E-01-8.54557950E-01 8.35552605E-02-5.83631489E-03 8.18285646E-02
 3  2 7.39936297E-03-2.26953665E-02-1.91551065E-03 1.44103019E-04-1.84324895E-03
 3  3-7.30179599E-02-7.30300125E-02-1.26560023E-01-2.80500756E-05-2.84225764E-05
 4  1-1.47611593E-02 6.57491207E-02 2.15954260E-02-1.48570875E-03 2.11365513E-02
 4  2 2.35829657E-01-8.53953087E-01 6.86900515E-02-4.36681579E-03 6.81986435E-02
 4  3 1.56173985E-02 1.56167812E-02-3.51168589E-02-1.45207104E-01-1.45193326E-01
 5  1 2.65621341E-02-1.38559271E-01-5.11829955E-01 3.54136523E-02-5.00838599E-01
 5  2 1.64839489E-02-6.35695788E-02 4.44897012E-02-3.02940337E-03 4.36454714E-02
 5  3-2.12387862E-01-2.12451288E-01 3.07944883E-01-3.22077160E-02-3.22120362E-02
 6  1-6.42384605E-06 3.19756766E-05-4.30678730E-01 3.39316521E-01 4.64142719E-01
 6  2-8.23546629E-06 3.50155053E-05-4.64034770E-03 3.65246668E-03 4.98286518E-03
 6  3 3.26546905E-01-3.26522265E-01-2.41992633E-05-2.15894395E-05 1.43171984E-05
 7  1 3.03726648E-05-1.46010812E-04 3.61316021E-01 8.78725612E-01-3.07129752E-01
 7  2-9.63734293E-07 4.12738444E-06 1.19803044E-02 2.92946887E-02-1.01831260E-02
 7  3 3.50564742E-05 4.37481046E-05 1.83831688E-05-2.02321039E-02 2.02282616E-02
 8  1 3.91710932E-06-1.95703405E-05 1.97311245E-02 4.80455996E-02-1.67388085E-02
 8  2 5.95251643E-06-3.08145320E-05-2.27533068E-01-5.59223458E-01 1.93232191E-01
 8  3 3.64914685E-05-2.06253221E-05 8.95712681E-08 4.24180458E-01-4.24177941E-01
 9  1-2.23062153E-02 1.54500963E-01-1.24786038E-01 8.66527729E-03-1.22090919E-01
 9  2 8.76656449E-02-4.57576664E-01-5.50893665E-01 3.63552632E-02-5.43466463E-01
 9  3-3.91196174E-02-3.91748912E-02-1.19065907E-01 2.64674392E-01 2.64677791E-01
10  1 1.09047481E-04-7.57403559E-04-4.23084857E-03 3.61227555E-03 5.66486586E-03
10  2 1.69891575E-04-8.14724604E-04 6.04758690E-01-4.68585751E-01-6.43938341E-01
10  3 5.24314248E-03-4.87589514E-03 5.88487922E-04 2.55282495E-04 2.68854270E-04
11  1 1.54833774E-01-1.06847947E+00-2.73466865E-01 1.87755601E-02-2.67446840E-01
11  2 2.46273283E-02-1.43176172E-01-9.48992300E-03 8.45045780E-04-8.70879842E-03
11  3 7.99564096E-01 7.99664119E-01 2.53819180E-01 7.71090350E-02 7.71448689E-02
12  1 5.72763567E-06-2.78066490E-05 5.58361863E-01-4.39877926E-01-6.01681240E-01
12  2-9.45283650E-06 6.04268124E-05-9.09590832E-05 2.17244079E-05 3.25012306E-05
12  3 8.69427097E-01-8.69318789E-01-3.23352070E-05-6.93390818E-05-3.27231202E-05
13  1 4.13030403E-03-7.72647532E-02 1.79424923E-01-1.24368684E-02 1.75491985E-01
13  2 1.34789464E-01-8.74151452E-01 4.71611018E-01-3.07542286E-02 4.66703957E-01
13  3-1.13602815E-01-1.13734341E-01 2.65565976E-01 7.69534468E-01 7.69759188E-01
14  1-1.26314109E-06 1.52361151E-05-7.19364333E-04-1.76212662E-03 5.75477468E-04
14  2 2.07150162E-05-1.49811355E-04 3.43071888E-01 8.42940764E-01-2.91210891E-01
14  3 2.75400846E-05-1.38879856E-05-3.40876355E-05 8.31340062E-01-8.31078482E-01
15  1-8.63113810E-02 9.44284502E-01-7.40065924E-01 5.12577143E-02-7.24284153E-01
15  2-1.65007257E-02 6.66544153E-02 2.89063586E-01-1.96228027E-02 2.83830925E-01
15  3 2.15524124E-01 2.15552519E-01-1.69321183E+00 1.62283414E-01 1.62340378E-01
 $END

And you'll have to do that for every step of you scan. If you are a bit talented with scripting, then you can surely automate the process.

I did not run the whole scan, but here are the energies of the ones I completed:

OH = 0.7   #1       NSERCH= 13     ENERGY=    -150.1620919
OH = 0.8   #2       NSERCH=  4     ENERGY=    -150.2905386
OH = 0.9   #3       NSERCH=  4     ENERGY=    -150.3504096
OH = 1.0   #4       NSERCH=  5     ENERGY=    -150.3755229
OH = 1.1   #5       NSERCH= 60     ENERGY=    -150.3875894
OH = 1.2   #6       NSERCH=  9     ENERGY=    -150.3898837
OH = 1.3   #7       NSERCH= 20     ENERGY=    -150.3872606
OH = 1.4   #8       NSERCH= 36     ENERGY=    -150.3774125


The mathematics of quantum mechanics predicts an unevenness of the observation, similar to wave equations in that there are points where high values of electron density/reaction and low (~zero) values occur. If you have places (~50% of the places) where "interference" has yielded zero electron density/reaction, then the other 50% of the places will have twice the expected (average) electron density/reaction.

The question should include the entire observation, the whole experiment. For example, it is easy to imagine a photon experiment where diffraction causes interference effects upon a photographic film: photons abandon their even spread by being diffracted, and don't go to certain places on the film, but do go (in twice the expected density) to other places. Two photons don't just cancel out in some places - they "cancel out" and also "add up" in other places. Oh, heck, just think of a sine wave (square its amplitude) plopped down on the film and where the high peaks are, the film turns dark and where the wave bottom is, nothing happens. Two photons, two sine waves; there has to be a slight difference in path so that when you superimpose them, they add up more in one place and subtract more in another.

That thought experiment was reasonably easy because we think of photons as sort of fuzzy, wavy, intangible things. We usually consider electrons to be touchable, therefore firm on some sort of basis, so how can they interfere and cancel out of existence? Well, the wave nature of electrons means that they have a spread in space which has some of the characteristics of a wave, and they can, under the right conditions, exhibit this effect strongly. Getting back to the photographic film experiment, electrons can be made to exhibit the same effects as diffraction does to photons. Where it becomes more difficult to understand is that while the photon (say, ~500 millimicrons wavelength) is much fuzzier than a silver ion (~0.25 millimicron diameter in AgCl), you would expect an electron to be, if anything, a bit smaller than an atom, since, after all, electrons (many of them!) will fit into an atom or ion.

So you can see that a photon with a spread of 500 millimicrons will smash down over close to a million Ag+ ions, but can only activate one, but you might expect a little electron to be like a bullet and just hit one silver ion. How could you miss? Consider the bullet to be wobbly (wavy) enough to hit either one Ag+ or its neighbor. The faster it goes, the wobblier (wavy) it gets. And these waves interfere so that some paths just don't happen, and others are travelled twice as frequently - this doesn't mean that two electrons cancel out of existence or form a bi-electron.

At least that's the way I look at it without the comfort and brevity of mathematical equations. But  words only poorly describe the situation and it takes forever to try to explain.


What is the magnetic moment of tris(oxalato)nickelate(IV)?

If the compound is really nickel(IV), then there are only two choices: a high-spin $\mathrm d^6$ or a low-spin $\mathrm d^6$ configuration, corresponding either to 4 or 0 unpaired electrons. As higher oxidation states make low-spin configurations more likely (due to the lower energy of metal orbitals which means the energy is more similar to the corresponding ligand orbitals leading to greater stabilisation and destabilisation of bonding/antibonding orbitals; $\mathrm{t_{2g}}$ is nonbonding while $\mathrm{e_g^*}$ is antibonding if π effects are ignored, so the energy difference increases), I would expect a nickel(IV) compound to be diamagnetic rather than paramagnetic.
However, oxalate is easily oxidised to carbon dioxide and likewise transition metals in unusually high oxidation states (which is true for nickel(IV)) are typically very good oxidation agents as their reduction is preferred. Therefore, I highly doubt this complex would be stable and suggest it immediately decomposes to carbon dioxide and a nickel(II) or nickel(III) complex. To quote Wikipedia:

Ni(IV) remains a rare oxidation state of nickel and very few compounds are known to date.

Thus, I am inclined to suspect some kind of typo. Two unpaired electrons would correspond to the well-known nickel(II) $\mathrm d^8$ system and tris(oxalato)nickelate(II) would seem much more likely.


As annav states, the labeling H1, H2 etc is labeling of hydrogen atoms within a graphene-like hexagonal ring structure. Another way of looking at it, is in the image below:

Image source

In an article published after the web article you link to, Aromaticity, closed-shell effects, and metallization of hydrogen. (Abstract only), the authors confirm that

  Recent theoretical and experimental studies reveal that compressed molecular hydrogen at 200-350 GPa transforms to layered structures consisting of distorted graphene sheets.

This structure, according to the article, forms a stable structure at those pressures.  As per the comments, this is form graphene-like hydrogen, the abstract goes further, discussing the process that would result in hydrogen metallization:

  Metallization occurs upon further compression by destroying the closed shell electronic structure, which is mainly determined by the 1s electrons via lowering of the bonding bands stemming from the unoccupied atomic 2s and 2p orbitals.

This metallization of hydrogen is, according to the article High temperature superconductivity in sulfur and selenium hydrides at high pressure theorised to occur at pressures of 400 GPa or greater.


Is sublimation caused by an environmental difference or an element's composition?

To a first approximation, any pure substance can sublimate. However, complex substances may decompose or change their structure before any appropriate sublimation conditions are reached.

The only (and very remarkable) exception is helium (in fact, helium is so special that helium-3 and helium-4 have quite different phase diagrams even though they're isotopes!). Note that due to the presence of superfluid phases, there is no direct transition between the solid and gas phases in either isotope.

As an aside, there is nothing prohibiting a substance to exhibit a solid-gas transition curve crossing a region of mild temperature and pressure conditions. Though unusual, an interesting example is iodine heptafluoride. It has a very narrow liquid range at ambient pressures (meaning a small decrease in pressure is enough to go below the triple point pressure), and the liquid range happens to fall near very reasonable temperatures.


Burning Of Coal and Black Body Radiation

Don't confuse the ignition temperature with the temperature of a burning coal. One is the temperature required to start the reaction, the other is the equilibrium temperature achieved when the substance is burning.

As for the colour of things at a given temperature, the colour will be broadly the same at the same temperature: this is black body radiation. For an interesting take on this in the context of movies, see this amusing video about why movies get it wrong on casting iron swords. A key point is that the glowing colour shown in most movies is far too cool (too orange red) for molten iron (which should be much whiter). Movies cheat and use aluminium which melts at less than 700 °C and glows orange. Iron needs ~1,600 °C and glows far whiter.


Although I've never heard this terminology before, I'm pretty sure I can see what it means.

As you can see from this diagram, each increase in the principal quantum number (n), adds a new type of orbital to that shell.

The electron configurations of the group 15 elements are:

\begin{array}{|c|c|} \hline
Element & Electron~configuration \\ \hline
\ce{N} & 1s^2~2s^2~2p^3 \\ \hline
\ce{P} & 1s^2~2s^2~2p^6~3s^2~3p^3 \\ \hline
\ce{As} & 1s^2~2s^2~2p^6~3s^2~3p^6~3d^{10}~4s^2~4p^3 \\ \hline
\ce{Sb} & 1s^2~2s^2~2p^6~3s^2~3p^6~3d^{10}~4s^2~4p^6~4d^{10}~5s^2~5p^3 \\ \hline
\ce{Bi} & 1s^2~2s^2~2p^6~3s^2~3p^6~3d^{10}~4s^2~4p^6~4d^{10}~4f^{14}~5s^2~5p^6~5d^{10}~6s^2~6p^3 \\ \hline
\end{array}

$\ce{N, P}$ and $\ce{As}$ all have full penultimate shells, whereas $\ce{Sb}$ and $\ce{Bi}$ do not -  the $4 f$ and $5 f$ orbitals are unnocupied respectively. Therefore 'saturated' seems to refer to elements where the penultimate shell is fully occupied, and 'unsaturated' to elements where it is not.


Surface energy for different crystallographic planes

You can think about this problem from the definitions of surface energy.

The surface energy quantifies the disruption of inter-molecular bonds that occurs when a surface is created

The composition or structure of a surface depends on how the bulk crystal has been cut.
The electronic structure (or number of dangling bonds) depends on surface composition
The reactivity of a surface depends on the electronic structure.

Imagine a battenberg cake:

The cake has been cut horizontally and the surface is comprised of tasty red and yellow squares.
If however we had cut the cake vertically straight down the middle, you would only see a red line next to a yellow line (Cut 1).
If you cut vertically and diagonally across the cake, you might only see either a purely red or purely yellow surface. In the case of the example image, you would only see a red surface (Cut 2).

I hope this example shows that (by relating the coloured squares to atoms), different cuts result in different surfaces with different properties.

When calculating the surface energy of a system:

You need to calculate the bulk crystal energy.
You need to calculate the energy of the surface slab. Make sure your model is converged with respect to slab thickness. The central layers should be representative of the bulk crystal. Look at magnetic change, charge difference, degree of ionic relaxation.
You then calculate the difference between the slab energy and the bulk crystal energy.

Note: In periodic calculations, a surface slab is constructed by cutting the bulk in a specified plane and introducing a vacuum gap. This vacuum gaps should be sufficient that the surface does not interact with its periodic mirror images. Around 15 Angstroms.

There are a number of codes that can help in the construction of surface slabs (METADISE, Materials Studio, Vesta)

If you wish to calculate the energy of a surface (and assuming you are using a periodic DFT code), I would highly recommend the paper by Wenhao Sun and Gerbrand Ceder (https://www.sciencedirect.com/science/article/pii/S003960281300160X)

Note: be careful of magnetic influences.


What is buffer capacity and how can a buffer "run out"?

If strong acid is added to a buffer system, conjugate base removes it.  Eventually, the concentration of conjugate base falls to such a low level that its effect on pH can be neglected.  At this point, the solution behaves more like a weak acid solution.  If even more acid is added, the system behaves like a strong acid system in terms of calculating pH.

If strong base is added to a buffer system, conjugate acid removes it.  Eventually the concentration of conjugate acid falls to such a low level that its effect on pH can be neglected.  At this point. the solution behaves more like a weak base solution.  If even more base is added, the system behaves like a strong base system in terms of calculating pH.

In the equation you wrote, $$\ce{A- + H2O <=> HA + OH-},$$ think about the effect of adding acid.  Acid reacts with the $\ce{OH-}$, producing water and removing $\ce{A-}$ from solution.  The $\ce{OH-}$ and $\ce{A-}$ cannot be removed from solution but their concentrations can be driven to negligible levels by the addition of sufficient quantities of acid.


This pretty much what happens (among other processes) in a three-way catalyst of a car:

$$\ce{2CO + 2 NO ->[\mathrm{cat}] 2CO2 +N2}$$


A molecule absorbs light having a specific wavelength, why doesn't it absorb shorter wavelengths?

Classical approach
Absorption occurs when there are "oscillators" that can resonate with the frequency of the electromagnetic wave. In classical weave physics, these frequencies are called natural frequencies.
These oscillators can involve electrons (UV, VIS, XR), bonds (MIR, FIR), nuclei (gamma rays). They are able to absorb electromagnetic waves changing their spins(NMR, ESR) orientation (microwave), configuration (infrared) nuclear configuration  ($\gamma$ rays) and finally electronic distribution (VIS, UV, X-Ray).
Of course, not all the materials have oscillators that absorb in all the spectrum. For wavelength between 10 nm and 100 pm the oscillators, that mainly involve electrons in these regions, have to absorb energy between $10^{7}$ and $10^{9}$ J/mol without being ionized, in this case only the inner electrons of some atoms (almost always metals) absorbs these wavelengths see XAS, so in our organic world seems that most compounds are X-Ray transparent but in fact there are a lot of inorganic compounds that absorb X-Ray.
When radiation is absorbed there are many ways to release his energy(see figure 1). We can divide them into two great categories radiative relaxation and non-radiative relaxation. In fact radiative relaxation is most common when we are dealing with shorter wavelength because with more energy is possible to excite an electron to a higher level and this can sometimes re-emit radiation when it goes from a higher level to the ground level, this as you know, is called fluorescence and is very common when you deal with UV, and X-Ray (see X-Ray Fluorescence spectroscopy). Eventually, there is also non-radiative relaxation process that are the main processes that occur from VIS (note that however is possible a VIS-NIR luminescence for some compounds) to the other spectral region with lower wavelength.

Quantum approach
Compound energy levels are quantized, so compounds can absorb only some discrete energy values. These values are determinate by transitions that respect strictly selection rules. If you know the energy of the transition, from Plank relation:
$$E=\frac{ch}{\lambda}$$
you can see that for every energy value there is a specific wavelength, so this means that compounds can absorb only at specific discrete wavelength. However if you see a UV-VIS spectrum you will see a continuous spectrum because roto-vibrational changes affect the energy levels of the compounds involved.


In general the answer would be no.

While there are many distillable compounds you could extract this way (not sure how many medicinal ones, though) there are many more that would decompose or simply not be easily extracted.

This is why there are many other ways chemists commonly extract things from mixture or from plants. For example, if you wanted to extract caffeine from coffee or from coffee beans, you could use solvent-based extraction (methylene chloride works well, for example, and has been used in industrial processes to create decaffeinated coffee). There is no single easy solution, though, and often much trial and error will be required to find the right solvent to differentially extract what you want.


How dependent is the molal depression constant on the nature of solute?

Some solutes form nearly ideal solutions up to moderate molalities, examples being glucose and sucrose in water. Such solutions allow fitting of freezing point temperature data to the following equation:

$$\mathrm{log}(1-x_s)=\frac{\Delta_{fus} H_m}{R}\left(\frac{1}{T_{m}}-\frac{1}{T}\right)$$

Taking various approximations which includes assuming that the heat of fusion is constant from $T$ to $T_m$ (the melting point of the pure solvent), that $x_s<<1$, and that $T_m\approx T$ you obtain the expression for the freezing point depression, $\Delta T = T_m - T$, in terms of the cryoscopic constant $K_f$ and solute molality $m_s$:

$$\Delta T = K_f m_s$$

where

$$K_f = \frac{M_wRT_m^2}{\Delta_{fus}H_m}$$

Since $K_f$ contains parameters that depend only  on the solvent (not on the particular solute) the equation can (to within the limitations imposed by the above approximations) be applied to any solutes with which the solvent forms ideal solutions. Which is why you can determine $K_f$ with one solute only to use that same constant to later determine the concentration of another solute.


Your convention is generally correct; as more complicated adducts and ions are formed (clusters, adducts, neutral losses, etc) things start to become a free for all where the clearest terminology wins.

As of now, just go with (where A means adduct):

$$\ce{[M + nA]^{n+}}\text{ or }\ce{[M - nA]^{n-}}$$

Protonated molecules can be denoted as $\ce{[MH]+}$ (common) or $\ce{[M + H]+}$ (technical). The $\ce{[M + H]+}$ convention is more precise because if you write $\ce{[MH]-}$ for a deprotonated ion the interpretation becomes muddled (i.e is this a hydride adduct or is this a deprotonated ion?).

One important think to keep in mind that is Molecular Ions ($\ce{M^{+.}, M^{-.}}$) are not the same as $\ce{[M + H]+}$ or $\ce{[M - H]-}$ and that the terminology is not interchangable.

$\ce{M^{+.}}$ or $\ce{M^{-.}}$ retains the elemental composition of the ion and is ionized through the removal or addition of an electron.
$\ce{[M + H]+}$ and $\ce{[M - H]-}$ are protonated and deprotonated molecules, respectively. They are not 'pseudomolecular' ions as there are no such thing as pseudo ionized specties.


Anomalous electronic configuration of platinum

The Madelung energy ordering rule gives the energy of the orbitals approximately:
$ \mathrm{1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d <
5p < 6s < 4f < 5d < 6p < 7s < 5f < 6d < 7p}$
That would speak for $\mathrm{[Xe] 4f^{14} 5d^{8} 6s^{2}}$ if you follow the Aufbau principle (from German Aufbau = setup). But $\ce{Pt}$ is an exception (as there are some). A rule of thumb is that half-filled shells are stabilized. So that means in the case of $\ce{Pt}$, $\mathrm{[Xe] 4f^{14} 5d^9 6s^1}$.
The real answer is much more complicated. It comes from relativistic effects, electron correlation and shielding effects. There is an interplay between the attraction of nucleus and the electrons, and the electron repulsion between all electrons. The heavier atoms become, the more important relativistic effects become, since the inner electrons are moving much faster as they are in a stronger electric field from the higher charge of the nucleus. Often it is said that the outer orbitals are less compact than the inner ones, which is true if one calculates and analyses them, but a stronger effect is that outer electrons are shielded and therefore the nucleus attraction is weaker. So there is no simple way, learn the exceptions or solve the Schrödinger/Dirac equations.
Also be careful when speaking about orbitals. An orbital is a one-electron wavefunction and they are more a chemical concept than reality in multi-electron atoms.


I would assume that the cathode region is a source of $\ce{O2(g)}$, which is probably the air in contact with the iron. Therefore the reacting should occur at the interface iron/air.


$$ \Delta G = \Delta G^\circ + RT \ln{\frac{a_{\ce{Sn^{4+}}}a_{\ce{H2O}}^{\;\;\;\;2}}{a_{\ce{SnO2}}a_{\ce{H+}}^{\;\;4}}} $$

At equilibrium, $\Delta G = 0$.

For $\Delta G^\circ = 2\times(-56.69) + 0.65 - (-123.20 - 4\times0) = +10.47\mathrm{\frac{kcal}{mol}}$

The question says "activity", but in the absence of any other information, let's assume that "activity" is equal to "concentration".  That is, let's assume activity coefficients are 1.

Thus, we get $$ \Delta G = 0 = +10.47  + RT \ln{\frac{a_{\ce{Sn^{4+}}}\cdot1^2}{1\cdot a_{\ce{H+}}^{\;\;4}}} $$

Thus, we get $$ {\left( \frac{-10.47}{RT}\right)}  = \ln{{10^{-10}} a_{\ce{H+}}^{\;\;-4}} = -4\ln{a_{\ce{H+}}} -10 \ln{10} $$

$$\frac{\frac{-10.47}{RT} + 10 \ln{10}}{-4} = \ln{a_{\ce{H+}}}$$


On the detection of molecules in electrospray ionisation (ESI)

I feel you have a misconception. ESI is an ionization technique not a detection technique. The sole purpose of the electrospray ionization is to impart a charge on a molecule in the gas phase. The molecule could be neutral, anionic or cationic in the solution phase. The inventor of ESI used to say that ESI makes the elephants fly, i.e., very large molecules could also be ionized in the gas phase.
The important point that you must learn is that if you have an ionized molecule in the solution, it is not necessary that when you bring the molecule in the gas phase, it will have a electrostatic charge. It may become a neutral molecule.
In order to "sort" molecules according to their mass to charge ratios in the gas phase, it is best to ionize them, i.e., remove an electron or add a proton so that the molecule of interest acquires an electrostatic charge in the gas phase. Note that I am emphasizing gas phase again and again. Forget what is the state in the solution. Almost 100 years ago, physicists had worked out how "charges" will move in an electric or magnetic field predictably. If the molecule is electrically neutral it will not interact effectively either with electric or magnetic field. Hence you would not be able to sort them according to their mass/charge rations. This is the job of the mass spectrometer.
Once your molecule has a charge, it is very easy to direct them here and there in the mass spectrometer and even focus all the ions at place  (one of the major advantage of ionized molecules), before exposing them electric fields or magnetic fields.


What does the 129Xe NMR spectrum of XeOF4 look like?

The $\ce{^{129}Xe}$ nmr spectrum of $\ce{XeOF4}$ is a sharp, quintuplet of peaks (reference, p. 445).  As you note, $\ce{XeOF4}$ adopts a square pyramidal geometry (reference), but the lone pair occupies an axial position.

Therefore, the 4 equatorial fluorines are equivalent and since they are spin 1/2 nuclei they will couple with the $\ce{^{129}Xe}$ nucleus (also spin 1/2) producing the observed quintuplet. $\ce{XeOF4}$ is an IUPAC suggested standard for $\ce{^{129}Xe}$ nmr spectroscopy (but xenon gas is more commonly used because $\ce{XeOF4}$ can decompose in the presence of moisture to produce $\ce{HF}$).


Why do alkali metals have an exothermic electron affinity?

I think you mean to ask about the electron affinity of the alkali metals. That is, the enthalpy gain or lost when adding an electron to an atom.

While the electron affinity for alkali metals are much smaller than the halogens, they are slightly positive. That is, quoting from Wikipedia:

  Electron capture for almost all non-noble gas atoms involves the release of energy and thus are exothermic.

The catch, is that a positive electron affinity (Eea) refers to an exo-thermic process:

  Confusion arises in mistaking Eea for a change in energy, ΔE, in which case the positive values listed in tables would be for an endo- not exo-thermic process.

You refer to the alkali metals as "electropositive" -- in this case, you're referring to their electronegativity relative to, say, halogens. Yes, a sodium atom is likely to become a sodium cation and give up an electron. But the electron configuration is $3s^1$ and consequently, can also gain an electron and become the more stable $3s^2$ state.

It's more likely that a sodium atom will lose an electron, and become a cation, but the electron affinity reflects the modest gain in stability as $3s^2$ after attaching an extra electron.


Effective core potential and Gaussian basis set

Sure, it can be valid physically. You can use the full-electron basis set, like 6-31G*, and the ECP with associated basis set, like Lanl2dz, for instance, in calculations of transition metal compounds or compounds with heavier non-metals (Sb, Te, etc.) as an example.
Important here would be to compare results obtained with available experimental data (such as bond distances/angles, reaction energies, UV-vis spectra, etc.) or results of similar calculations in order to see if the simultaneous use of ECP and full-electron basis sets is valid and would not give you some weird results.


What is the relation between partial pressure and concentration?

MATHEMATICALLY

At point A

The pressure according to ideal gas equation is $P_A = C_ART_A$. Here $C_A (= n/V)$ is concentration at A.

At point B

Similarly, the pressure according to ideal gas equation is $P_B = C_BRT_B$. Here again $C_B$ is concentration at B.

Now, $P_A$ > $P_B$

So, $C_ART_A$ > $C_BRT_B$

i.e. $C_A/C_B$  > $T_B/T_A$  .....equation 1

Now as the condition is that concentration at A is lower than the the concentration at B. Therefore we can say that

$C_A/C_B$ < $1$ .....equation 2

Now from equation 1 and 2 we have

$T_B/T_A$ < $1$

which gives us $T_A$ > $T_B$

So in the end, yes, your book was right. It is possible if the temperature at A is more than that at B.

"HOW TO FIND IT IN SECONDS" or "I HATE TO DERIVE BY USING EQUATIONS"

Think of two containers A and B which are connected by a pipe. Both are connected with the help of a pipe which is closed (i.e. it doesn't let the gases exchange). Now we know that concentration of gas in A is lower than the the concentration of gas in B. If we want more pressure of gas in A then the velocity of its particles should be more. Now it is possible if it has more thermal energy. So the temperature should be more in A.


The assumptions made in the Flory-Huggins theory are

Quasi-solid lattice in the liquid
Inter-changeability of segments (not necessarily the same as the polymer structure
units) of polymer and solvent molecules in the lattice
Independence of lattice constants on composition (artificial)
Polymer molecules are of same size
Average concentration of polymer segments in cells adjacent to cells unoccupied
by the polymeric solute is taken to be equal to the overall average concentration
The expected number of available positions for each successive segment is
overestimated in the formulation as the formula includes double counting of
segments, separated by 2 or more segments in the same chain, that will fall on the
same position twice

In particular the assumption that polymer molecules are of the same size is a very unrealistic one, which was later removed in the Flory-Krigbaum theory


Rule of thumb of how much vacuum does lab glass stand?

The basic rule is that the stress on an evacuated glass vessel is pretty much the same at 0.01 mbar (oil pump), 1 mbar (membrane) or 25 mbar (water jet) absolute internal pressure. 0.99999, 0.999 or 0.975 bar $\Delta p$ to the outside make very little difference if the thing goes pop.

And basically everything made out of borosilicate and with no flat surface is safe. Condensors definitely, they're not delicate at all. It's basic procedure to make them oxygen-free by evacuating.

Be careful with chromatography columns and flasks in the > 1 liter range. They're principally safe to be evacuated, too, but then are also a bomb, armed and ready to drop. If it goes off, you better not stand next to it. The danger from evacuated vessels is more or less proportional to their volume.


Why is potassium monoxide (K2O) coloured?

Probably as good an explanation as any can be found here.

https://uk.answers.yahoo.com/question/index?qid=20090729062525AA0gZlS

Basically, the electronic structure of an ionic compound like potassium oxide has "valence bands", which are based primarily on the anions (oxide ions), and empty "conduction bands" based on the cations  (potassium ions).  Typically, to excite electrons from the valence band to the conduction band you need to absorb ultraviolet photons and thus you get no color.  But potassium oxide has just a small enough energy gap between valence and conduction bands to allow some electrons to be excited with violet light.  That absorption gives the yellow color.

If we go further down the alkali metal oxides the band gap gets smaller, and more visible light with longer blue/green wavelengths may be absorbed.  Then the color becomes stronger and more reddish.


We can draw the 3 Lewis structures (or the corresponding resonance structures) pictured below for $\ce{O_2}$

Since an oxygen atom has 6 electrons,

A would correspond to a structure with a single bond between the oxygen atoms, 2 lone pairs on each oxygen and an unpaired electron on each oxygen; however A does not have an octet around each oxygen, in fact, each oxygen would only have 7 electrons
B would correspond to a structure with a double bond between the oxygen atoms, 2 lone pairs on each oxygen and no unpaired electrons on each oxygen; B does have an octet around each oxygen, but it is not a biradical
C would correspond to a structure with a triple bond between the oxygen atoms, 1 lone pair on each oxygen and an unpaired electron on each oxygen; however C does not have an octet around each oxygen, in fact, each oxygen would have 9 electrons and this would be impossible for oxygen

So while structure A would indicate a biradical, we wouldn't "expect" it to count for much since the oxygens do not have octets.  This inability to clearly predict the biradical nature of $\ce{O_2}$ illustrates one of the failings of both Lewis structures and resonance theory.
In order to correctly predict the biradical nature of $\ce{O_2}$ we must move up to molecular orbital theory.  Below is the molecular orbital diagram for $\ce{O_2}$.  As you can see it does predict that $\ce{O_2}$ should be a biradical with an unpaired electron in each of its degenerate, highest occupied molecular orbitals.

Edit: response to OP's comment

When I think of triple bond I don't think of 2 3 electron bonds(which
is what you drew). Rather I think of 3 2 electron bonds(1 sigma bond
and 2 pi bonds)

Structure C does represent 3 two-electron bonds (not 2 three-electron bonds), that's just how you draw the Lewis structure.

This type of triple bond would make the oxygen positive with 5
electrons around it.

No, the formal charge on the oxygen in structure C is
Z = 6 - 3 unshared - (1/2 * 6 shared)= 0,
there is no formal charge on oxygen in the "triple bond" structure and as I noted above, there are 9 electrons around it (not 5), which is impossible for oxygen.

I am assuming electrons are shared equally with half around 1 atom and
half around the other(which is the basis for formal charge

Yes, that's correct.


Skeletal formula for carbon with two double bonds

In chemical structure diagrams, carbon atoms that are bonded to at least two other atoms are usually left unlabeled. The carbon atom is indicated by the angle in the bonds.

However, according to the Graphical Representation Standards for Chemical Structure Diagrams (IUPAC Recommendations 2008), a carbon atom with two double bonds should be explicitly labeled:

  (…), any carbon atom with two identical collinear bonds should always be explicitly labeled, to remove the possibility of the two bonds being misinterpreted as one long bond.

  The use of a dot in place of an explicit carbon atom label is acceptable in allenes and related molecules with three consecutive carbon atoms. It is not acceptable to use a dot to represent a carbon atom when either of its adjacent atoms is other than a carbon atom.

  When a dot is used to represent a carbon atom, it must be depicted clearly and unambiguously. The dot should be large enough to be clearly seen by the reader, and must be visibly separated from the adjacent bonds.


I'm going to try and answer a bit of confusion in the original question "Does aluminium oxide ionize after melting point?". Aluminium Oxide is already ionised in the solid, or more pedantically the ionic model fits Aluminium Oxide sufficiently well that we can think of it comprised of $\ce{Al^{3+}}$ and $\ce{O^{2-}}$ ions and make reasonable predictions. The point is that in most solids, ionic or not, the constituent atoms and molecules are energetically constrained to vibrate about their "normal", so called equilibrium, position, and so never move very far, whereas in liquids they can move about reasonably freely. Now as conductivity involves the nett movement of charges most ionic solids are not conducting while the liquids are, as in the liquid the charged ions are free to move toward the appropriate electrode while in the solid they are not.

However it's not that quite that simple, but the above is probably good enough to University level. But for completeness I'll add that there are things called fast ionic conductors. These are "ionic" solids that conduct electricity well above a certain temperature. A well known example is Silver Iodide, which above 419.15 K start to conduct electricity pretty well. A simple model here is that at that temperature the Iodide sub-lattice stay rigid, so providing a solid like framework, but the Silver sub-lattice melts, and so the silver ions can move, and thus nett charge migration is possible, and so the solid conducts. For a bit more of an introduction, see Wikipedia article of fast ion conductor. Interestingly one form of Aluminium Oxide can act as a fast ion conductor, but in this case the charges that are moving are "impurities".


Which mechanism do ethers follow when cleaved by hydrogen halides?

The first step of this reaction is always a proton transfer:

Then, the substitution follows either the SN2 or SN1 pathway depending on whether R is methyl, primary, secondary, or tertiary.

Methyl and primary ethers undergo acidic cleavage by the SN2 mechanism:

Tertiary ethers undergo cleavage by the SN1 mechanism:

Secondary ethers probably undergo cleavage by both mechanisms competing with each other.

Note that unless R' is an aryl group, the alcohol R'OH that forms will also be converted into a halide by a similar mechanism.

Read more at any of the sites in this Google search.


If metals want to "lose" electrons, then why will copper ions take electrons from a zinc electrode?

Typically, when one uses terms such as "likes to lose" or "likes to gain", we should think of that phrase as "likes to lose relative to...".  In your case, when we think about electron configurations, it might be more helpful to think "metals like to lose rather than gain electrons to achieve a noble-gas-like configuration".  When comparing two metals, both of which tend to lose electrons, we need to look at different concepts to answer the question.

We can explain the transfer of electrons by looking at the the metal activity series such as the one shown here.  The way one reads this table is that a Galvanic cell created with two metals will have the metal higher on the list as its anode (where oxidation occurs) and the other metal will be the cathode.  Thus, we would expect Zn to be oxidized to Zn(II) and Cu(II) to be reduced to Cu.  These tables are based on standard reduction potentials:

$$\ce{Cu^{2+} + 2 e^- -> Cu(s)}\ \tag{E = 0.34 V}$$
$$\ce{Zn^{2+} + 2 e^- -> Zn(s)}\ \tag{E = -0.76  V}$$

By flipping the 2nd reaction (thus changing the sign of E) one obtains a cell potential of about 1.1 V; a positive number meaning that this reaction will happen spontaneously:

$$\ce{Cu^{2+} + Zn(s) -> Cu(s) + Zn^{2+}}\tag{E = 1.1V}$$

The standard reduction potentials tell us that the electrons would rather be on copper than on zinc.


Iron-II to iron-III change in complex formation

First iron(II) hydroxide $\ce{Fe(OH)2}$ is a precipitate made in aqueous solution, and it does not react with ammonia. Second, the iron(II) hydroxide $\ce{Fe(OH)2}$ is a green substance which is extremely sensitive to the oxygen of the air. In a couple of minutes, it gets brown, due to the formation of iron(III) hydroxide $\ce{Fe(OH)3}$ according to $$\ce{4 Fe(OH)2 + O2 + 2 H2O -> 4 Fe(OH)3}$$ If water is absent, the reaction may produce another compound of iron (III) $$\ce{4 Fe(OH)2 + O2 -> 4 FeOOH + 2 H2O}$$


As you are already aware, cathode rays are electron beams. You take evacuated glass tube, and  add two electrodes on either end. Once you apply a sufficiently high voltage, a stream of electrons flows towards the "cathode"; this is what you call a cathode ray. Inside the tube there exists near perfect vacuum, so there are barely any other gas molecules, ions etc.

Lightning is similar in the sense that you have a large potential difference that leads to electric discharge. The key difference is that this takes place in air, and not in vacuum. Since air is "insulating", it has to be ionised to for the electric discharge to take place. Indeed this is what happens, when the potential difference is sufficiently high. However, what this creates is a plasma, not a stream of electrons.

A plasma is the "fourth" of matter, and is created by ripping apart some electrons from gas molecules resulting in a collection of electrons and ions.


Enthalpy definitions. What are their main differences?

Explanation of notation:
$H$ is the enthalpy of the system.
$\Delta$ means change of, so $\Delta H$ means change of the enthalpy

This symbol means standard condition, standard condition is defined as a pressure of $100\text{kPa}$ and reactants and products are in their standard state, or concentration of solutions are $1\text{M}$. However, since LaTeX doesn't have this symbol, I'll substitute it for $^\ominus$.
Enthalpy changes:
Enthalpy of dilution – The enthalpy change when a solution containing one mole of a solute is diluted from one concentration to another.
Enthalpy of ($n\text{th}$) electron affinity – The enthalpy change when $n$ electrons are added to one mole of gaseous atoms.
$$\ce{Li(g) + e-(g) -> Li-(g) +60\ \text{kJ}}$$
$$\ce{F(g) + e-(g) ->  F-(g) +328\ \text{kJ}}$$
Enthalpy of ($n\text{th}$) ionization – The enthalpy change when $n$ electrons are removed from one mole of gaseous atoms. It is always positive.
$$\ce{Li(g) +520\ \text{kJ} -> Li+(g) +e-(g)}$$
$$\ce{He(g) +2372\ \text{kJ} -> He+(g) +e-(g)}$$
Enthalpy of lattice dissociation – The enthalpy change when one mole of an ionic lattice dissociates into isolated gaseous ions.
An example for sodium chloride which have an enthalpy of lattice dissociation, also known as lattice energy, of $787\ \mathrm{kJ/mol}$
$$\ce{NaCl(s) +787\ \text{kJ} -> Na+(g) + Cl-(g)}\qquad\Delta H=787\ \text{kJ/mol}$$
Enthalpy of lattice formation – The enthalpy change when one mole of solid crystal is formed from its scattered gaseous ions.
$$\ce{Na+(g) + Cl-(g) -> NaCl(s)} +787\ \text{kJ}\qquad\Delta H=-787\ \text{kJ/mol}$$
Which means Enthalpy of lattice dissociation$=-$Enthalpy of lattice formation
Enthalpy of hydration($\Delta_\text{hyd}H^\ominus$) – Enthalpy change when when one mole of ions undergo hydration.
Enthalpy of mixing – The enthalpy change from a substance when mixed.
Enthalpy of neutralisation – The enthalpy change when an acid is completely neutralised by a base.
For a strong acid, like $\ce{HCl}$ and strong base, like $\ce{NaOH}$, they disassociate almost completely $\ce{Cl-}$ and $\ce{Na+}$ are spectator ions so what is actually happening is $$\ce{H+(aq) + OH-(aq) -> H2O(l) + 58\ \text{kJ/mol}}$$
However, using a weak acid/base will have a lower enthalpy of neutralisation as normally most of the acid/base does not disassociate.
For example, mixing ethanoic acid and potassium hydroxide only has a enthalpy of neutralisation of $-11.7\ \text{kJ/mol}$
Enthalpy of precipitation – The enthalpy change when one mole of a sparingly soluble substance precipitates by mixing dilute solutions of suitable electrolytes.
Enthalpy of solution – Enthalpy change when 1 mole of an ionic substance dissolves in water to give a solution of infinite dilution.
Enthalpy of solution can be positive or negative as when a ionic substance dissolves, the dissolution can be broken into three steps

Breaking of solute-solute attraction (endothermic)
Breaking solvent-solvent attraction (endothermic), eg. hydrogen bonds, LDF
Forming solvent-solute attraction (exothermic)

An example of a positive enthalpy of solution is potassium chlorate which has an enthalpy of solution of $41.38\ \text{kJ/mol}$
Enthalpy of (Solid$\rightarrow$Liquid: $\Delta_\text{fus}H^\ominus$, Liquid$\rightarrow$Solid:$\Delta_\text{freezing}H^\ominus$, Liquid$\rightarrow$Gas: $\Delta_\text{vap}H^\ominus$, Gas$\rightarrow$Liquid: $\Delta_\text{cond}H^\ominus$, Solid$\rightarrow$Gas: $\Delta_\text{sub}H^\ominus$, Gas$\rightarrow$Solid: $\Delta_\text{deposition}H^\ominus$) – The enthalpy change from providing energy, to a specific quantity of the substance to change its state.
$$\Delta_\text{fus}H^\ominus=-\Delta_\text{freezing}H^\ominus,\Delta_\text{vap}H^\ominus=-\Delta_\text{cond}H^\ominus,\Delta_\text{sub}H^\ominus=-\Delta_\text{deposition}H^\ominus$$
Standard enthalpy of atomization($\Delta_\text{at}H_T^\ominus$) – Change when a compound's bonds are broken and the component atoms are reduced to individual atoms at at $T^\circ K$.
$$\ce{S_8 -> 8S}\qquad\Delta_{at}H^\ominus=278.7\ \text{kJ/mol}$$
Standard enthalpy of combustion($\Delta_\text{c}H_T^\ominus$) – The enthalpy change which occurs when one mole of the compound is burned completely in oxygen at $T^\circ K$ and $10\ \mathrm{kPa}$.
$$\ce{H2(g) +\frac{1}{2}O2(g) -> H2O(g)}+572\ \text{kJ}\qquad\Delta_cH^\ominus = -286\ \text{kJ/mol}$$
Standard enthalpy of formation($\Delta_\text{f}H_T^\ominus$) – Change in enthalpy during the formation of one mole of the compound from its constituent elements, with all substances in their standard states, and at a pressure of $100\ \mathrm{kPa}$ at $T^\circ K$.
It can be calculated using Hess's law if the reaction is hypothetical. An example is methane, $\ce{C}$ and $\ce{H2}$ will not normally react but the standard enthalpy of formation of methane is determined by Hess's law to be $-74.8\ \text{kJ/mol}$
$$\ce{\frac{1}{2}N2(g) +\frac{1}{2}O2(g) -> NO(g)}\qquad\Delta_\text{f}H^\ominus=90.25\ \text{kJ/mol}$$
Standard enthalpy of reaction($\Delta_\text{r}H^\ominus_T$) – Enthalpy change that when matter is transformed by a chemical reaction at $T^\circ K$ and $10\ \mathrm{kPa}$.
$$\ce{H2(g) +\frac{1}{2}O2(g) -> H2O(g)} +572\ \text{kJ}\qquad\Delta_\text{r}H^\ominus = −572\ \text{kJ/mol}$$
$$\Delta_\text{r}H^\ominus=\sum H^\ominus_\text{products}-\sum H^\ominus_\text{reactants}$$
$$\Delta_\text{r}H_\text{forward}=-\Delta_\text{r}H_\text{backwards}$$
Laws:
Le Chatelier's Principle – When an external change is made to a system in dynamic equilibrium, the system responds to minimise the effect of the change.
Yellow $\ce{Fe^3+}$ reacting with colorless thiocyanate ions $\ce{SCN-}$ to form deep red $\ce{[Fe(SCN)]^2+}$ ions:
$$\ce{Fe^3+(aq) + SCN-(aq) -> [Fe(SCN)]^2+(aq)}$$
When $\ce{NH4Cl}$ is added, $\ce{Cl-}$ reacts with $\ce{Fe^3+}$ ions to form $\ce{[FeCl4]-}$ ions. By the Le Chatelier's Principle, when $\ce{Cl-}$ is added into a solution of deep red $\ce{[Fe(SCN)]^2+(aq)}$ ions, the equilibrium will shift to $\ce{Fe^3+(aq) + SCN-(aq)}$, turning the solution pale red.
Kirchhoff's Law – Enthalpy of any substance increases with temperature.
Hess's Law – Total enthalpy change of a chemical reaction is independent of the number of steps the reaction takes.
Henry's Law – Amount of dissolved gas is proportional to its partial pressure in the gas phase.
Processes:
Constant Entropy – Isoentropic
Constant Pressure – Isobaric
Constant Volume – Isovolumetric
Side note: This is probably not a complete list as I may have missed some


Does silver oxide react with hydrogen sulfide?

Silver sulfide is formed by Ag and hydrogen sulfide not sulfur dioxide. You need reducing conditions where the latter can be reduced to

$$\ce{SO2 +reduction -> H2S}$$

Actually, sulfur dioxide chemisorbs on ultraclean silver surface, however heating can remove it. So this is reversible sorption, as suggested by Lassiter [1].

Just note that Auger electron spectroscopy is done under extremely clean environment. There is no trace of water, oxygen or any other component! Real atmosphere is far more complicated and tons of photochemical reactions occur in the atmosphere. A typical indoor air has plenty of undesirable components. How does $\ce{SO2}$ react with Ag must be another story because we cannot avoid or control other factors.

Coming to the second part of the query: If we have surface layer of silver oxide, will it prevent sulfide formation. In principle, possibly yes, because $\ce{Ag2O}$ is decent oxidizing agent. The moment traces of $\ce{H2S}$ come in contact with the oxide, it will reduce the oxide to elemental silver. This is my personal speculation.

As per Franey et al. [2]:

  Polycrystalline silver has been exposed to the atmospheric gases $\ce{H2S}$, $\ce{OCS}$, $\ce{CS2}$ and $\ce{SO2}$ in humidified air under carefully controlled laboratory conditions. $\ce{OCS}$ is shown to be an active corrodant while $\ce{CS2}$ is quite inactive. At room temperature, the rates of sulfidation by $\ce{H2S}$ and $\ce{OCS}$ are comparable, and are more than an order of magnitude greater than those of $\ce{CS2}$ and $\ce{SO2}$. It appears that $\ce{OCS}$ is the principal cause of atmospheric sulfidation of silver except near sources of $\ce{H2S}$ where high concentrations may render the latter gas important. At constant absolute humidity, the sulfidation rate of
  silver by both H2S and OCS decreases from 20 to 40 °C and then increases to 40 to 80 °C.

So may hydrogen sulfide may not be a major culprit!

References

Lassiter, W. S. Interaction of Sulfur Dioxide and Carbon Dioxide with Clean Silver in Ultrahigh Vacuum. J. Phys. Chem. 1972, 76 (9), 1289–1292. https://doi.org/10.1021/j100653a011.
Franey, J. P.; Kammlott, G. W.; Graedel, T. E. The Corrosion of Silver by Atmospheric Sulfurous Gases. Corrosion Science 1985, 25 (2), 133–143. https://doi.org/10.1016/0010-938X(85)90104-0.


Can I pour a solution into another vessel with no air bubbles?

There are two aspects to consider for the intended "air bubble free" (or at least, "air bubble less") transfer from your container where the liquid is prepared into the container where the liquid is supposed to solidify:

Your first container shown is a round bottom flask.  Initially, an increase in inclination of the flask will increase the flow of liquid out of the flask.  If however the inclination is too high, the air that aims to refill the inner volume of the container, initially occupied by the liquid, will no longer continously enter the flask.  Instead, there will be an alternation of sorting liquid and entering air; which may in turn yield air bubbles in the sorting liquid.  Depending on the viscosity of the liquid, the air bubbles may remain in the liquid for quite some time.  From the perspective to lower the potential air intake by the liquid during the transfer, a wide mouth bottle, or even a beaker were better suited than a round bottom flask.
The distance between the first and the receiving container.  For your purpose, it is not good to pour the liquid directly into the receiver. To lower splash and splatter in the transfer by decantation, an easy aide is using a glass rod, like shown here:

(picture source)

In your instance, it may be more convenient to use a slab of glass / a pane of glass that is just beneath the mouth of the first container, and bridges all the distance to the receiving container; as schematically drawn below:

Here, intentionally the slab of glass is partially immersed in the liquid already present in the receiving container.  If done in a coordinated team of two, one holding and moving gently the slab of glass, the second gently decantating the liquid onto this "chute", you might cast larger volumes into trough-shaped containers / fish tanks, too.  Once solidified, you would cut-out the block / shape you need.  Overall, it reminds me a little to ballistic gelatin...


Difference between Epimers and Diastereomers

This scheme I just drew up specifically for you should answer your question.

Diastereomers are stereoisomers that are not enantiomers of each other. That includes conformers (geometric isomers that derive from single bond rotation; usually interconverting rapidly) and atropisomers (under which I would subsume E/Z isomers; they derive from hindered rotation around a typically single bond and are separable) and anything with differeces in asymmetric carbons.

If you have exactly one asymmetric carbon with inverted stereochemistry, then the diastereomer is a special one: an epimer. Anomers (e.g. α- and β-D-glucose) should be considered special cases of epimers.

Let me use the menthol isomers (image taken from Wikipedia, where a full list of authors is available) to clarify the concept:

Two vertically aligned structures are always enantiomers. Two structures that are not vertically aligned are diastereomers.

(+)-Isomenthol is (+)-menthol’s C4-epimer, while ($-$)-neoisomenthol is (+)-menthol’s C1-epimer. And ($-$)-neomenthol is (+)-menthol’s C2-epimer. All of these are also diastereomers.

(+)-Neomenthol is not an epimer of (+)-menthol. It is still a diastereomer, however.


What does 0.12 N mean on a bottle of sulfuric acid?

0.12N is indeed 0.12 normal. In short, there are 0.12 moles of hydrogen ions per liter. Since this is sulfuric acid, it is 0.06 moles of sulfuric acid per liter.

0.12 moles of hydrogen ions per liter would give a pH just less than 1. To obtain a pH near 3, you would need to dilute 1 mL of this solution to 100 mL. For a pH of 4, dilute 1 mL to 1000 mL.


The spectral lines represent transitions between pairs of discrete energy levels in the atom; an electron is excited by absorbing the energy of a photon and a transition from one of these levels to another occurs as the photon is destroyed. Emission lines occur when the reverse process happens.

If enough energy is absorbed the atom can be ionised and an electron with un-quantised energy can be liberated and moves to a position of complete separation where attractive forces between electron and nucleus is insignificant. As the liberated electron's energy is un-quantised the absorption giving rise to ionisation is continuous. The absorption is truly continuous, no matter how high a spectral resolution is used discrete absorption lines will not be observed.

Black body radiation is continuous emission from a hot body. The analysis of this emission lead Plank in 1901 to propose, for the first time, the idea that the photons were quantised, thus avoiding the 'ultra-violet catastrophe'. He did this by allowing ultra-violet photons to have more energy than visible ones which in turn had more energy that infra-red ones. Nowadays we know that he energy is directly proportional to frequency via  the Planck constant, $E=hv$.

In the black body, all the numerous energy levels in the body are highly excited and transitions between them continuously occur as the body is being continuously heated. These transition energies overlap one another in energy so that the spectrum is continuous.

(In molecules there are also discrete energy levels, in the ground state these are due to the vibrational and rotational motion of the molecule.  A molecule may also have electronically excited states and these can also have discrete vibrational and rotational energy levels. Furthermore there are dissociative excited states that have continuous absorption spectra.)


Its not diffusion, it's effusion. The difference is that with diffusion the particles/molecules are migrating through a permeable medium where as for effusion the particles/molecules are migrating through a pore/hole.

For example, a scent travelling across a room is diffuision, a scent from the room traveling outside is effusion. Also fun fact: the helium escaping through a balloon skin is actually effusion not diffusion.

helpful source

Side note: I am starting to flashback to all the times my thermo professors said diffusion when they meant effusion.


Thermodynamics in Solution (solvation) from quantum chemistry

The best way is to take all the structures into account and use Maxwell-Boltzmann distribution to weigh them accordingly. Here is what I would do
For each chemical structures/isomers (here I used F to denote Gibbs free energy),
$ F_{liq} =  F_{gas} +  F_{solvation}$
Now, $ F_{gas} =  E_{scf} - k_BTlnQ$
Where, $Q=q_{translational} \times q_{vibrational} \times q_{rotational} $
Using appropriate method (implicit solvation or QM/MM) you can find $F_{solvation}$. Now you have $F_{liq}$ for each isomers. Then use the following method to weigh them according to Maxwell-Boltzmann distribution.

In the above image, I use $\epsilon_i$ to denote $F_i$.


The following disilyne has been prepared and found to be stable to ~100 C.  An X-ray crystal structure found that the two silicon atoms in the triple bond, Si(1) and Si(1'), along with the two attached silicon atoms, Si(2) and Si(2') are coplanar and the Si(1)-Si(1')-Si(2') angle is 137.44 degrees ("trans-bent").

The full text of the article describing the preparation, characterization and reactivity of the compound can be found here.  See figure 4 in the paper for a MO explanation as to why the trans-bent geometry occurs. Basically, the authors suggest that, "the bending is thought to be the result of the mixing of an in-plane $\ce{\pi}$-orbital with a $\ce{\sigma^{\ast}}$ orbital whose energies are close enough to cause the interaction."


How to calculate molarity and the number of molecules for a mixture in a simulation box?

I would use a slightly different approach than andselisk, based on the volume on your box, and that I don't think you can neglect the volume of your ammonia molecules. Given $V(\text{sol}) = \pu{17.576E-27 m3}$ and $c(\ce{NH3}) = \pu{15.0E3 mol  m-3}$ we can calculate the number of ammonia molecules we need:
\begin{align}
  && c(\ce{NH3}) &= \frac{n(\ce{NH3})}{V(\text{sol})};\\
  && n(\ce{NH3}) &= \frac{N(\ce{NH3})}{N_\mathrm{A}};\\
&\to&
     c(\ce{NH3}) &= \frac{N(\ce{NH3})}{N_\mathrm{A}\cdot V(\text{sol})}\\
&\Longleftrightarrow&
     N(\ce{NH3}) &= c(\ce{NH3})\cdot N_\mathrm{A}\cdot V(\text{sol}) \approx 159
\end{align}

Since you want to limit the number of molecules to $512$, we know that all other molecules have to be water:
$$N(\ce{H2O}) = 512 - N(\ce{Gly}) - N(\ce{NH3}) = 352$$

Lastly, let's check whether the density is anywhere close to what we would expect, we take $M_\mathrm{m}(\ce{Gly}) = \pu{75 u}$, $M_\mathrm{m}(\ce{H2O}) = \pu{18 u}$, and $M_\mathrm{m}(\ce{NH3})= \pu{17 u}$.
\begin{align}
  \rho(\text{sol}) &= \frac{m(\text{sol})}{V(\text{sol})}\\
  &= \frac{\left(
       N(\ce{Gly})M_\mathrm{m}(\ce{Gly})
     + N(\ce{H2O})M_\mathrm{m}(\ce{H2O})
     + N(\ce{NH3})M_\mathrm{m}(\ce{NH3})\right)\frac{m_\mathrm{u}}{\pu{u}}
     }{V(\text{sol})}\\
  \rho(\text{sol}) &\approx \pu{861 kg m-3} = \pu{0.861 g cm-3}
\end{align}

Citing Wikipedia's entry of ammonia I would say you are quite close:

  Solvent properties
  Ammonia is miscible with water. [...] The maximum concentration of ammonia in water (a saturated solution) has a density of $\pu{0.880 g cm-3}$ and is often known as '.880 ammonia'. [...]

The Handbook of Chemistry and Physics has a concentration table (91st Edition p. 8-53; currently 98th Edition, also online), and I'll reproduce a couple of values for comparison, at $\pu{20 ^\circ C}$.

\begin{array}{rrr}
  \text{Mass\%} & c / \pu{mol L-1} & \rho / \pu{g cm-3}\\\hline
    5.0   &   2.868    &   0.9770\\
    9.0   &   5.080    &   0.9613\\
   12.0   &   6.695    &   0.9502\\
   18.0   &   9.823    &   0.9294\\
   24.0   &  12.826    &   0.9102\\
   28.0   &  14.764    &   0.8920\\\hline
\end{array}

In conclusion, you are a few water molecules short, or your box is slightly too large, or your solution is not dense enough. I would expect some trade-offs when doing such a study and keeping the named constraints, so I guess you should have a close look at everything and settle for the optimum you can afford.


Entropy of universe for non spontaneous processes

Non-spontaneous processes do not occur. You will only see a decrease in the entropy of the universe in your theoretical calculations, never experimentally.


Why is 1-Ethyl-1-methylcyclohexane achiral?

The functional groups are on the same carbon, so either configuration can be rotated into the other. If the functional groups were on different carbons, then depending on their relative locations the molecule could be chiral.


First, from Boundless.com (Chemical Composition of Urine), "Urine is an aqueous solution of greater than 95% water, with the remaining constituents, in order of decreasing concentration urea 9.3 g/L, chloride 1.87 g/L, sodium 1.17 g/L, potassium 0.750 g/L, creatinine 0.670 g/L and other dissolved ions, inorganic and organic compounds (proteins, hormones, metabolites)."

Urine color is generally a result of the degradation of the heme molecule from blood into the urinary products, called bilirubin, urobilinogen, urochrome, urobilin and uroerythrin. In essence, they are all tetrapyrroles amounting to something on the order of 1mg per 100ml (or roughly 0.001% of urine)... so this is a very minor part of urine. Also, it should be noted that over 3,000 chemicals have been found in urine, so there's something to consider.

Activated carbon filtration might be able to be used to preferentially remove the tetrapyrroles and some other organic molecules while allowing urea to pass (for the most part).

There are various forms of activated carbon (of course), and so the amount of urea absorption (by carbon) will be variable; the adsorption capacity of the activated carbon for urea will (likely) be different for activated carbon from different sources. Absorbance of urea from one report was about 1.69 mg urea/g carbon, where the urea concentration was 100 mg/L (see picture below). At greater concentrations (like 930mg urea per liter), lesser amounts of acivated carbon will become saturated; in other words the activated carbon will not be able to absorb more urea, but it might be able to effectively remove the tetrapyrroles or some other minor stuff.

But I think that the first thing to do is deal with the concentration. Urea is only about 0.93% of urine. So you only have about 0.93g urea per 100ml of urine.

To concentrate urea: boil off the water (which will also kill bacteria). Urea doesn't boil. The melting point for pure urea is 132.7 °C and urea will begin to decompose at around 150 °C. The residue (which should look creamy) will have a relatively high urea concentration which can then be extracted with (warm/hot 60 °C) ethanol or diacetone alcohol. Salt and protein are significantly less soluble in alcohol(s), so they should hopefully be (mostly) removed at this step.

The solubility of urea in 95% alcohol (100 g/L), absolute ethanol (50 g/L), and methanol (166 g/L) at 20 °C (room temperature) sounds pretty fair. And from TECHNICAL MEMORANDUM No. 13/M/67 (Investigations of Crystal Modification of Urea) "in the case of diacetone alcohol the solubility [of urea] ranged from 2 per cent at 20 °C to about 10 per cent at 100 °C, offering possible development of a process yielding free-flowing crystalline urea."

The solubility of salt (at 25 °C) in ethanol (0.65 g/L) and methanol (14.9 g/L) is significantly lower.

Now, the solubility of a hodgepodge of stuff including tetrapyrroles is something that you will have to experiment with. But, many similar compounds are readily soluble in acetone and tetrachloroethylene (often had as brake cleaner). Urea should not be highly soluble in these two solvents, but I'm a afraid I can't find a reliable solubility listed online, and I happen to be out of acetone right now. Cleaning an alcoholic extract of boiled urine with tetrachloroethylene might require using a separatory funnel.

On a historical note, (around 1727) Herman Boerhaave was able to extract urea from urine by working with the concentated-by-boiling residue (having the consistency of "fresh cream"). After filter pressing the residue, he allowed it to dry for a year... until he obtained a solid mass with an unknown oily substance (on top).  He discarded the oil and rinsed the crystals (filtride) with cold water to remove salt, leaving urea. He then recrystallized the urea from hot water. Previously (in 1669), Hennig Brand discovered the philosophers stone (phosphorus) by cooking the bejeezus out of (lots of) urine.

Personally, instead of waiting a year, I would take a chance on extracting urea from the cream. Also, I read today that boiling urine causes irreparable stink-damage to a home; so get a hot plate and an exension cord and boil it outside. Another recipe (as if I had nitric acid to spare), calls for reacting concentrated urea (say 100g in 100ml water) to make urea nitrate. The addition of nitric acid is exothermic and urea nitrate is explosive, so make sure the solution doesn't boil off (ice bath)... and urea nitrate is precipitated in the freezer. Then after you have clean urea nitrate, you can neutralize the acid under alcohol to recover urea. I would reserve this (nitric acid) method for the final work up, using relatively clean urea.

Also, here's a couple of pay-walled articles that might be useful to you if you really want to do this: Solubility of urea in water-alcohol mixtures and Urea in the history of organic chemistry: Isolation from natural sources