Let's get the obvious out of the way: The answer to that exercise is yes. Hence it is completely unnecessary to do the second part. With the parameters given in the exercise I calculated a pressure of about 440 atmospheres using the ideal gas law.

But then since we're all here to learn, let's also have a look at the second part. The good news is that I can't see anything wrong with how you used the equation. The problem is already pointed out by orthocresol in the comments, which I will quote here for safekeeping (emphasis by me):

  Ok. Firstly, the constant $a$ has units of $\mathrm{L^2~atm~mol^{-2}}$ (not $\mathrm{L~atm~mol^{-2}}$), but I am guessing it is just a typo on your part. The issue seems to arise from the values given for $a$ and $b$. Wikipedia gives the constants as $1.352$ and $0.0387$ respectively (converted to the same units as yours). Using those values I get a pressure of 997.3 atm.

I would assume that nobody caught this mistake earlier since you are not actually supposed to do that part of the question.


"ionic bonds can be regarded as an extreme form of covalent bonds." They "can be regarded" as magic, too. What's your point? In what specific context is such regard useful? (I'm not questioning that there may be some specific context that you could regard them that way, rather I AM saying that it is not generally useful.)
 The octet rule is used to predict valence.  Obviously H doesn't obey it. It's quite useful for the chemistry of C,N, O, and F.  But consider ClF5. The Rule is not going to get you there. Nor is it useful, imho, for P4O10 --- but there are ways to bend the meme to "explain" the deviancy. see for example the discussion of hypervalent compounds in https://en.wikipedia.org/wiki/Octet_rule#Hypervalent_molecules. Once you get away from the 2nd & 3rd Period and the 13th - 17th Groups, it shouldn't be relied upon. So, its applicability is limited, but fortunately it's applicability is for a lot of elements which are interesting for us in our environment here on Earth. Since d-orbital (and f-) contributions to hybridizations is generally small, the octet rule is generally useful. What sort of confidence should you have in using it to predict valence? Well, there's two ways to answer this: 1. as applicable to a student learning basic chemistry or 2. as applicable to experts who have in addition to the rule a vast literature on the subject. It is useful for the first and nearly useless for the 2nd.


The surface of glass has many Si-O-H and Si-O-Si bonds; deeper inside would be Na and Ca and Al. An active metal might engage those surface atoms. Na might be too reactive (forming a dross of NaOH on the surface of your alloy, or reacting with water at the glass surface). Aluminum or zinc or perhaps magnesium might be active enough to contribute some sort of adhesion to glass.

Tin oxide is bound to glass surfaces for electrical conductivity. Perhaps it could be an intermediate layer, bonding glass and your alloy. And silver is deposited on glass to make mirrors. Your alloy might well stick to a silver deposit. So, while this doesn't actually make your alloy wetter, it could solve your problem of adhering to glass.


How is it Possible to Measure Electrode Potential?

It is the same as measurement of voltage difference at any electronic circuit. The fact the metallic measuring points are attached to electrochemical electrodes is not relevant here.
Yes, there is metal contact potential difference, but it very low and very probably mutually canceling.


In the heterogenous permanganate oxidation of styrene, that is when the reaction is performed in dichloromethane with potassium permanganate on a solid support (such as aluminium oxide), the reaction stops at the stage of the aldehyde.

The method was published in Synthesis 2001, 11, 1645-1648 and the authors report a yield of 90% for the oxidation of styrene to benzaldehyde.


How does the human body deal with free radicals it creates itself?

Some enzyme-catalyzed reactions produce free radicals as short-lived intermediates, but these are likely to be contained within the enzyme active site until they can react further to form product. Example: ribonucleotide reductase - which catalyzes the formation of deoxyribonucleotides and enables the production of DNA.

I can't think of any case where a biochemical process "intentionally" forms a free radical and then allows it to escape and cause damage elsewhere in the body. But biology is wonderfully varied and I would not want to say "never".


What is the structural difference between ketogenic amino acids and glucogenic amino acids?

“Standard” amino acids are degraded to one of seven metabolic
  intermediates: pyruvate, α-ketoglutarate, succinyl-CoA, fumarate,
  oxaloacetate, acetyl-CoA, or acetoacetate.

The amino acids may therefore be divided into two groups based on their catabolic pathways:

Glucogenic amino acids;

whose carbon skeletons are degraded to pyruvate, α-ketoglutarate, succinyl-CoA, fumarate, or oxaloacetate and are therefore glucose precursors.

Ketogenic amino acids;

whose carbon skeletons are broken down to acetyl-CoA or acetoacetate and can thus be converted to ketone bodies or fatty acids.

The amino acids degraded to  acetyl-CoA and acetoacetyl-CoA are used in the citric acid cycle, but mammals cannot synthesize glucose from acetyl-CoA. This fact is the source of the distinction between glucogenic and ketogenic amino acids. Glucogenic amino acids can be converted to glucose, with oxaloacetate as an intermediate, but ketogenic amino acids cannot be converted to glucose.

Strictly speaking, whether an amino acid is regarded as being glucogenic, ketogenic, or both depends partly on the observer. This classification is not universally accepted, because different quantitative criteria are applied will identify the degradation pathways by the entry point into metabolism.

I would also attribute to things like availability of feasible pathways, enzymes etc (for instance alanine is glucogenic because its transamination product, pyruvate can be converted to glucose via gluconeogenesis)

Some amino acids have more than one pathway for catabolism, which explains why four of the amino acids are listed as both glucogenic and ketogenic examples are Isoleucine, phenylalanine, threonine, tryptophan, and tyrosine):

Below I will include snippets of degradation products of named amino acids:

Five amino acids, alanine, cysteine, glycine, serine, and threonine, are broken down to yield pyruvate:

Asparagine and Aspartate Are Degraded to Oxaloacetate:

Arginine, Glutamate, Glutamine, Histidine, and Proline Are Degraded to α-Ketoglutarate:

The reason I included these pathways, is to allow you to visualise structure, feasible metabolic pathways and enzymes so you have a clear picture on the differences between these amino acids.

And so on..

Hope this helps..

References

Voet and Voet biochemistry
J.Berg Biochemistry
Campbell and Farrel Biochemistry


Determination of brominated aromatics by 13C and 1H NMR spectroscopy

I don't find any experimental spectra online, but I have been suggesting NMRDB.org to students and colleagues. I don't know how accurate the predictions are, but the site is easy to use and for the compounds you're suggesting, rule-based methods should be fairly reliable.

Predict $\ce{^1H}$ NMR
Predict $\ce{^13C}$ NMR

In the 4,4’-dibromo-N,N’-diphenylhydrazine, you should have two sets of "double doublets" (i.e., four peaks in each set of equal height).

In the 4-bromoazobenzene, you break symmetry, so there should be four signals which integrate to two protons, and one additional signal integrating to one proton (i.e., all ~7.3-7.8 ppm, so a big mess).

In the $\ce{^1H}$ spectra of 4-bromoazoxybenzene, the protons closest to the azoxy -N+-O- should be deshielded a bit (maybe ~7.8-7.9 ppm?) compared to the aza analogue. Certainly it's also going to be a fairly congested spectrum.

I'll leave the $\ce{^13C}$ NMR as an exercise (or to others to answer).


Comparison of energy release from different types of fuels

This is essentially a question concerning the trend of molecules in the same homologus series that have differing chain lengths. When we increase chain length, the number of C-C bonds increase in proportion to C-H bonds. Looking at the bond energies, the bond energy of a C-C bond is $347 kJ/mol$, while for C-H it is $414 kJ/mol$. This means to break a C-C bond compared to a C-H bond, less energy is required. Also when increasing the number of C-C bonds, while producing the same products ($CO_{2(g)} + H_2O_{(l)}$), more carbon dioxide is produced in proportion to the amount of water. The enthalpy of formation of carbon dioxide, results in it releasing more energy than that of water. Taking this into account with the lower energy requirement to break the reactants, the chain length is generally directly proportional to the energy produced. Thus, the COMPLETE combustion of Butanol would have a higher energy out put than Methanol and likewise Butane > Methane.
It is important to note that, as the compounds get more energy dense, i.e release more energy upon combustion, that it becomes harder for the compound to reach complete combustion; what I mean is that more oxygen is required for the reaction to release the full energy output. If not enough oxygen is supplied to the system soot ($C_{(s)}$) & Carbon Monoxide ($CO_{(g)}$) will be produced and the energy produced will be less than the energy produced in the complete combustion. (The degree to which this works, depends on how many moles of oxygen is supplied).


How does the size of a molecule affect acidity?

It is key to learn about the applicability of certain concepts — or why does a concept work in a certain case — along with the concepts themselves. Most notably, atom size (it should not be molecule size, but for $\ce{H2O/H2S}$ it does not matter due to hydrogen’s tinyness) is only of limited help in a few select cases to determine acidity.

When comparing the acidity of two compounds you should always actually look at the following equation and then compare the anionic products of two different acids:

$$\ce{HA + H2O <=> A- + H3O+}\tag{1}$$

Whichever generates the more stable anion is the stronger acid. Thence, a number of concepts have been developed to determine the more stable anion from a given list of alternatives — one of them is the atomic size. It is useful only when directly comparing two otherwise group-identical species — like oxygen and sulfur which are in the same group but different periods or chlorine and iodine. Basically, under otherwise identical conditions (same valency and same charge) the larger anion is able to stabilise the negative charge better. Thus, $\ce{H2S}$ is a stronger acid than $\ce{H2O}$ and $\ce{HI}$ is stronger than $\ce{HCl}$ but we cannot say anything about the difference between $\ce{H2S}$ and $\ce{HCl}$ or even $\ce{H2S}$ and $\ce{HF}$ from that.

All the other pairs you listed do not fit into that difference concept at all. The acidity of $\ce{Fe^n+}$ is not explained simply by looking at the metal ion. And the difference between acetic and formic acid is on a much more subtle scale since both are oxyanions. To reiterate: do not only learn concepts but learn what concepts are for.


Just browsing through this "Unanswered Question" now (in late 2014). I hope it's not too late?

In protein structure validation, Ramachandran outliers are a strong, but not absolute, indicator that the backbone confirmation of your protein structure deviates from all the high resolution structures known to date, and hence need your attention during the progress of refinement.

If all the phi-psi angles (backbone conformation) all fall into the "allowed/favoured" regions on the Ramachandran plot, your structure model is probably, but not necessarily, esp in the case of low resolution, correct. Hence there are other weaker criteria in protein structure validation, and Cbeta deviation is just one of them. In the light of high resolution structures out there, it turns out that Cbeta also has a favoured range of dihedral angles, suggesting that any misfit between your structure and the "probable" side chain-backbone conformation also deserves your attention during structure refinement.

In MolProbity, the justification and validation procedure are described here.


Finding the mM concentration of capric acid (decanoic acid) in coconut oil

Thank you very much for a well-specified question.  If 8% of coconut oil were free capric acid, then as noted in the comments the units of "%", i.e. w/w, w/v, or v/v, would affect the calculation somewhat, but since capric acid likely has a density that is very close to that of bulk coconut oil, the difference will be slight.  8% w/w, and assuming a density of 0.8 g/mL for coconut oil would yield $0.08~{\frac{\rm g~capric~acid}{\rm g~oil}}\times \frac{\rm mol~capric~acid}{172~\rm g~capric~acid}\times0.8\frac{\rm g~oil}{\rm mL}\times\frac{1000~\rm mL}{\rm L}=0.372\frac{\rm mol~capric~acid}{\rm L}=372~\rm{mM}$.

However, I suspect that the capric acid in coconut oil is not free acid.  It is part of triglycerides.  Instead of containing capric acid, the oil contains an ester of capric acid with glycerol.  Triglycerides are molecules where all of the available 3 hydroxyl groups of glycerol are esterified with a fatty acid.  Because triglycerides are large complex molecules, they were historically difficult for chemists to analyze directly.  The method of choice for analysis of plant oils was transesterification or ester hydrolysis, converting the triester triglyceride molecules into three smaller monoester molecules that were easier to detect by analysis.  The 8% figure you are using is probably the result of such an analysis.

The bottom line is that coconut oil that is "8% capric acid" actually contains very little capric acid, but contains a large amount of glycerol-esterified capric acid.  But only free capric acid is likely to be antimicrobial.  So the capric acid content of coconut oil is unlikely to make it antimicrobial.

The presence of free fatty acids in plant oils is usually a sign of degradation.  In olive oil, an acidity above 0.5% (i.e. of all the fatty acid chains in the olive oil, 1 in 200 are in the free form and 199 in 200 are in the esterified form) is a sign of lower quality.  I would imagine the for coconut oil the situation is similar, although the specific numbers used as quality thresholds may be different.


Products of the reaction between ethoxyethene and HI

Here is a mechanism for the cleavage. After proton transfer, the carbocation has a resonance contributor. Nucleophilic attack can occur on the ethyl group because there is a good leaving group.

The electrophilic addition can happen, with iodide attacking the carbocation. However, it is reversible. Iodide is a good leaving group, and the carbocation is resonance stabilized. The cleavage reaction is much less reversible.


Is there a list of known rate laws for reactions?

Integrated Rate Laws in Chemical Kinetics are very well known (http://www.csus.edu/indiv/m/mackj/chem142/kinetics.pdf).

Here is a good article with the list of the constants: Table of Recommended Rate Constants for Chemical Reactions Occurring in Combustion by Francis Westley


Gibbs energy and chemical equilibrium : false proof of spontaneous generation, but why?

Within the context of the equation in question, $\Delta G$ refers to the change in free energy between two specific thermodynamic equilibrium states:

State 1:  Stoichiometric quantities of pure reactants in separate containers at temperature T and individual specified pressures

State 2:  Corresponding stoichiometric quantities of pure products in separate containers at temperature T and individual specified pressures

In the case of $\Delta G^0$, the specified pressures are all 1 atmosphere.  This means that Q = 1, and $\Delta G =\Delta G^0$.

If the specified pressures in states 1 and 2 happen to correspond to the partial pressures of an equilibrium mixture of reactants and products, $\Delta G$ will be equal to zero.  In this case, if one had an equilibrium mixture of the reactants and products, the partial pressures in the equilibrium  mixture could also be held in equilibrium through individual semipermeable membranes with the gases in the separate containers of reactants and products (with no tendency for the reactants or products to enter or leave the reaction mixture through the semipermeable membranes).


Melting point of unsymmetrical dimethylhydrazine vs 1,2-dimethylhydrazine

According to the Carnelley’s Rule:

  That of two or more isomeric compounds, those whose
  atoms are the more symmetrically and the more compactly
  arranged melt higher than those in which the atomic
  arrangement is asymmetrical or in the form of long chains

If we attend to the structure of the 1,1-dimethylhydrazine, one can see that the steric hindrance due to the methyls groups inhibits the formation of H-bond in the central nitrogen atom for this isomer, in contrast to the 1,2-dimethylhydrazine which both nitrogen atoms are capable to form H-bonding.

For the thermodynamics point of view, the melting point of a solid is related to the enthalpy and entropy of melting by the following expression:

$$T_{\text{m}}=\frac{\Delta_{\text{m}} H}{\Delta_{\text{m}}S}$$

where the subscript "$\text{m}$" stands for "melting". Is difficult to estimate the impact of both enthalpy and entropy without further information than the molecular structure, but in most of the cases the process of melting is driving by enthalpic factors; therefore if the symmetric molecule can make more H-bonds than the unsymmetrical it's reasonable to assume that the symmetric molecule has a higher enthalpy of fusion and consequently a higher melting temperature provided the entropic effects are ignored.

References and further reading:

Effect of molecular symmetry on melting temperature and solubility

Melting Point and Molecular Symmetry

Carnelley’s Rule and the Prediction of Melting Point


How to prevent losing solvent into the vacuum pump in rotovap solvent collection?

Honestly, I never bothered to actually measure the volume of the condensed solvent in order to compare it with the original volume.

The concept described in the question, that is using a cooling trap with dry ice + acetone, is good practise and typically good enough to retain the evaporated solvent and protect the membrane pump. In rare cases, I have used a second, flat Dewar filled with dry ice + acetone on a "Laborboy" to cool the condensate-collecting flask.


13C NMR of bromobenzene ipso carbon shielding

The principle influence on chemical shift for 13C is the paramagnetic term, which essentially describes the electron density in the p orbitals, and this largely mirrors the influence of electronegativity that is seen in 1H chemical shifts.

However, there is something called the 'heavy atom effect' where the effect of substitution runs counter to electronegativity. This is the case for Bromine, Iodine, Tellurium etc. This comes from an increased level of diamagnetic shielding introduced by the electrons from the halogen atom.

To understand why this is the case, we can consider that the overall shielding on the ipso carbon relative to a methyl susbituent. Even though the bromine is more electronegative than the methyl group, it is so large that it still contributes significant electron density around the ipso carbon nucleus, and hence provides greater overall shielding than a methyl group.

While trends in electronegativity are useful tools to justify observations on chemical shift, it is always good to relate the effects of electronegativity, and other properties, on what is happening to the electron density around the nucleus on question. Ultimately, chemical shift is a measurement of electron shielding around the nucleus.


To clarify, I've reworded your questions. Please comment if I changed the meaning:

Calculate the volume of pure water that needs to be added to dilute a  0.2M HCl solution with a volume of 250cm3 to a 0.12M HCl solution

The first step would be to work out how many moles of HCl is in the original solution. As you have stated

c = n / v

c = 0.2M

v = 250cm3 or 0.250L.

Therefore n = 0.05m

Then work out how much water is needed to make a 0.12M solution.

c = 0.12M

n = 0.05m

Therefore v = 0.42L

To work out how much water needs to be added, subtract the original 0.25L from 0.42L, which equals 0.17L

Calculate the mass of Na needed to react with 400cm3 of water to form a 0.15M NaOH solution.

As you haven't progressed much on this question, I'll give a few hints.

Write a balanced equation

Work out how many moles of NaOH is needed to make 0.15M solution

Work out how many moles of Na is needed to form that many moles of NaOH

Convert from moles to mass.

Hope this helps.


What is the grey film/black sludge from mercury? How to remove it?

First, I would advise against using mercury in a home lab due to the ease which scattered droplets fall into cracks and evaporate.

That said, if distillation is out, then try bulb-pipetting from the center of the mass of mercury to avoid some of the surface contamination. Avoid dropping the mercury when lifting the pipette, since $\ce{Hg}$ is so dense it tends to run out as soon as lifted. Ideally, work in a hood with a large tray with raised sides underneath the apparatus.

If purity is important, electrolysis will yield a higher-grade product... at the risk of working with even more toxic compounds!

Better, get some gallium if you want to play with a liquid metal, and don't need $\ce{Hg}$ to make Grignard reagent or other esoteric substance.


The starch forms a loosely bonded network that traps water vapor and air into a foamy mass, which expands rapidly as it heats up.

Starch is made of glucose polymers (amylopectin is one of them, shown here):

Some of the chains are branched, some are linear, but they all have $\ce{-OH}$ groups which can form hydrogen bonds with each other.

Let's follow some starch molecules through the process and see what happens. In the beginning, the starch is dehydrated and tightly compacted - the chains are lined up in nice orderly structures with no water or air between them, maximizing the hydrogen bonds between starch polymers:

As the water heats up (or as you let the pasta soak), water molecules begin to "invade" the tightly packed polymer chains, forming their own hydrogen bonds with the starch:

Soon, the polymer chains are completely surrounded by water, and are free to move in solution (they have dissolved):

However, the water/starch solution is not completely uniform. In the middle of the pot of water, the concentration of starch is low compared to water. There are lots and lots of water molecules available to surround the starch chains and to keep them apart. Near the surface, when the water is boiling, the water molecules escape as vapor. This means that near the surface, the local concentration of starch increases. It increases so much as the water continues to boil, that the starch can collapse back in on itself and hydrogen bond to other starch molecules again. However, this time the orderly structure is broken and there is too much thermal motion to line up. Instead, they form a loosely packed network of molecules connected by hydrogen bonds and surrounding little pockets of water and air (bubbles):

This network is very weak, but it is strong enough to temporarily trap the air as it expands due to heating - thus, the bubbles puff up and a rapidly growing foam forms. Since they are very weak, it doesn't take much to disrupt them. Some oil in the water will inhibit the bubbles from breaking the surface as easily, and a wooden spoon across the top will break the network mechanically as soon as it touches it.

Many biomolecules will form these types of networks under different conditions. For example, gelatin is a protein (amino acid polymer) that will form elastic hydrogen-bonded networks in hot water. As the gelatin-water mixture cools, the gel solidifies, trapping the water inside to form what is called a sol-gel, or more specifically, a hydrogel.

Gluten in wheat is another example, although in this case the bonds are disulfide bonds. Gluten networks are stronger than hydrogen-bonded polysaccharide networks, and are responsible for the elasticity of bread (and of pasta).

DISCLAIMER:

pictures are not remotely to scale, starch is usually several hundred glucose monomers long, and the relative size of the molecules and atoms isn't shown.
there aren't nearly enough water molecules - in reality there would be too many to be able to see the polymer (1,000's).
the starch molecules aren't "twisty" enough or showing things like branching - the real network structure and conformations in solution would be much more complicated.

But, hopefully you get the idea!


Why does the derivation Gibbs-Helmholtz equation hold when the enthalpy is temperature-dependant

The derivation of the van't Hoff equation does not require that you specify the T-dependence of H. It arises from computing the derivative (limiting slope) of $G/T$ so you need to consider a polynomial expansion of $G/T$ to linear terms in $T$ around the temperature of interest (that is, an infinitely  small interval dT). At the risk of redundancy:

$$\begin{align}
 (d(G/T))_p &= ((1/T)dG-G(dT/T^2))_p \\
&= ((1/T)(VdP-SdT)-(H-TS)(dT/T^2))_p \\
&= (-(S/T)dT-(H/T^2-S/T)dT)_p \\
&= (-(H/T^2)dT)_p
\end{align}
$$

The answer is therefore found in differential Calculus.

Another comment explains that, if

$G(T)=H(T) -TS(T)=H_0  +C_p \Delta T -TS(T)$

where $H(T)$ is expressed as a linear polynomial in T ($C_p$ and $p$ assumed constant) then

$dG=C_p dT -SdT-TdS$

Since for a process at constant pressure,  $dq = C_p dT=TdS$ (since $C_p = (dH/dT)_p$ and $dq_p= dH$; also for a reversible process $dq= TdS$), it follows that

$\left(\frac{ \partial G}{\partial T} \right)_p=-S$

and the van't Hoff expression provided in the linked answer also follows.


Calculation of VOCs concentration (ppm) in the headspace of a bottle

Usage of ppm/ppb etc units is discouraged, as they are ambiguous without the explicit context. As generally, 1 ppm (w/w) <> 1 ppm (V/V) <> 1 ppm (n/n) <> 1 ppm (w/V).

It is recommended to use explicit units, like e.g mg/L.

Expecting ideal gas behaviour $pV = nRT$ is not reasonable.   Ideal gases are not in equilibrium with their liquid phase.

There should be rather used a real gas equation.The easiest to use is the van Der Waals equation, but even that has significant error near condensation state.

Additionally, the sample(!) temperature  has to be strictly controlled, otherwise $n$ would be variable.

Why not rather to use small volume of ethanol to be known to fully evaporate in given volume ? Or perhaps other solvent, unless ethanol is waterless.
Note that ethanol has significant deviation from the Raoult's law.

I prefer $R = \pu{8.31446 J K-1 mol-1}$

Note that by May 2019 redefinition,
 It's exact value is $\pu{8.31446261815324 J K-1 mol-1}$


When your book says that reducing character increases down the group, it PRIMARILY meant in gaseous state. What you have described above are all ionic reactions, i.e., in some solution.

NOW in gaseous state, as we go down the group, the bond energy of hydrides reduces (due to increasing atomic size as we go down the group). So they can be more easily broken to release nascent hydrogen (see my note at the bottom), which is the cause of reducing nature. So reducing ability increases down the group in gaseous state.

NOTE : Nascent hydrogen is referred to the hydrogen which has just evolved post some reaction, and has not yet formed its molecule. This is not the same as hydrogen atom so keep that in mind.

Reducing property in a solution

This is the catch. In aqueous solution, reduction takes place due to the group 15 ion, and not hydrogen ion. So it depends on how easily the group 15 ion can lose an electron (reduction is gain of electrons)
Down the group, electronegativity decreases and hence it becomes easier to remove electrons for reduction. So reduction ability decreases down the group in solutions too.

Now to answer your question, Sb(3+) will be oxidised to Sb(5+) as it is what is causing the reduction. You should be able to answer your following question now.

Btw, the E0 value for (H-/H2)is 0.00. Note that all oxidation reduction reactions of hydrogen gas have this value as 0.00


What would be the state of water in interstellar space or vaccum?

tl;dr: ice

Let's have a look at the phase diagram of water:

In space, both the temperature and pressure are very low. That would put beyond the bottom left corner of the diagram, well within the ice field. So simply speaking, water would be ice.

However, there is the concept of vapour pressure. Any condensed material (just a fancy word for both solids and liquids) has that same material in the vapour around it. That's why you can smell alcohol, even though it's liquid as one example. That's why a puddle of water will eventually dry out even though it's below the boiling point of water.

The way vapour pressure works is that if you look on the phase diagram, for a certain temperature, the liquid-vapour and the solid-vapour lines have certain pressures associated with them. For example, if you look on 50 C, then the pressure on the curve is 10 kPa. Roughly speaking, this means that in atmospheric pressure, one tenth of the air (which is at 100 kPa) has to be water vapour. If it's any less, water will keep evaporating until it reaches that amount. On Earth, wind and diffusion will disperse the water so it will basically evaporate until there are no more liquid water left.

Now let's think what would happen in space. The temperature is very low, so the vapour pressure is very low. Thus, it should evaporate also in space. However, because the vapour pressures at such low temperatures are extremely low, you might end up with a situation where the gravitation of that ice blob is enough to overcome evaporation, so that will be an agonizingly slow process.


Why do strong acids dissociate (almost) fully on a molecular level?

Deprotonation is always a reversible reaction. Hence why, even strong acids, have $\mathrm{p}K_\mathrm{a}$ values (which rely on the presence of equilibria).


Calculating the solubility of cuprous iodide

First of all, as MaxW pointed out in the above comment, your math seems to be wrong. From the the $K_\mathrm{sp}$ value given in your question:
$$s = \sqrt{K_\mathrm{sp}}= \sqrt{1.0 \times 10^{-12}}= \pu{1.0 \times 10^{-6} mol/L}= \pu{1.0 \times 10^{-6} mol/L} \times \frac{\pu{190.45 g}}{\pu{1 mol}}\times \frac{\pu{1000 mg}}{\pu{1 g}} \\ = \pu{0.190 mg/L}$$

Other than that, there seems to be a problem with online value of $K_\mathrm{sp}$ of $\ce{CuI}$. Some sites listed it as $1.1 \times 10^{-12}$ while other sites listed as $5.3 \times 10^{-12}$. However, I'd like the value $5.3 \times 10^{-12}$, because at least it's given in one of reliable sources (Ref.1).

If you use this value (instead of $1.1 \times 10^{-12}$), you'd get:

$$s = \sqrt{K_\mathrm{sp}}= \sqrt{5.3 \times 10^{-12}}= \pu{2.30 \times 10^{-6} mol/L}= \pu{2.30 \times 10^{-6} mol/L} \times \frac{\pu{190.45 g}}{\pu{1 mol}}\times \frac{\pu{1000 mg}}{\pu{1 g}} \\ = \pu{0.438 mg/L}$$

This value is in a good agreement with the literature value ($\pu{0.00042 g/L}$ at $\pu{25 ^\circ C}$; Ref.2), which listed the properties of $\ce{CuI}$ as:

  Copper (I) iodide is a dense, pure white solid, crystallizing with a zinc-blende structure below $300^\circ$. It is less sensitive to light than either the chloride or bromide, although passage of air over the solid at room temperature in daylight for $\pu{3 hr}$ results in the liberation of a small amount of iodine. It melts at $588^\circ$, boils at $1293^\circ$, and, unlike the other copper halides, is not associated in the vapor state. Being extremely insoluble ($\pu{0.00042 g/L}$ at $25^\circ$), it is not perceptibly decomposed by water. It is insoluble in dilute acids but dissolves in aqueous solutions of ammonia, potassium iodide, potassium cyanide, and sodium thiosulfate. It is decomposed by concentrated sulfuric and nitric acids.

References:

Steven S. Zumdahl, Donald J. DeCoste, In Chemical Principles; 8th Edition; Cengage Learning: Boston, MA, 2017, “Chapter 8: Applications of Aqueous Equilibria,” pp. 241–298 ($K_\mathrm{sp}$ of value $5.3 \times 10^{-12}$ is given for $\ce{CuI}$ in example 8.13, page 288).
George B. Kauffman, Lawrence Y. Fang (Checked by N. Viswanathan, G. Townsend), “Chapter 2: Transition metal Complexes and Compounds – 20. Purification of Copper (I) Iodide,” In Inorganic Syntheses: Volume 22; Smith L. Holt Jr., Ed.; John Wiley & Sons: New York, NY, 1983, pp. 103–107 (https://doi.org/10.1002/9780470132531.ch20).


Silver-promoted reaction of alkyl halide with diphenyl sulfide

In the presence of $\ce{AgBF4}$, diphenylsulfide is alkylated to 3-chloropropyl diphenylsulfonium tetrafluoroborate. The sulfonium salt is stable and can be isolated as a crystalline product.

This sufonium salt can be deprotonated (e.g. with $\ce{NaH}$ in $\ce{THF}$) to the corresponding sulfonium ylide.

In the latter, intramolecular displacement of $\ce{Cl-}$ yields cyclopropyl diphenylsulfonium tetrafluoroborate, which again is a stable solid.

This cyclopropyl derivative, once converted to its sulfonium ylide S may serve as a reagent in the Corey–Chaykovsky reaction.


Safety when handling cyclohexylamine in non-industrial amounts

I once did some work with some small chain alkyl amines and they were bad news.  They would actually burn you if some got on your skin.  I always used a respirator and worked in a hood when using them.

  as long as I don't feel irritation of my lungs or eyes, am I safe?

If this is a one time occurrence, given the small quantities, I'd say the odds are way in your favor, but they call it "odds" for a reason.  My advice would be not to do it again without a respirator and hood where the exhaust is treated.

  And if I do, is there enough time to clean up (i.e. the most essential
  things, like shutting off electricity and closing open jars of
  chemicals) and leave?

If you ask an industrial hygiene professional, I'm sure the answer would be "no".  If it were me and I just began to notice some very mild sensation, I think I'd hold my breath and try to shut things down.  I know it doesn't make sense, but that's what I'd do.  If I noticed anything beyond the onset of very minor discomfort, I'd clear out fast and call the fire department.

  Finally I accidentally touched a drop of cyclohexylamine (from the
  pipette) when I was cleaning up, with no gloves on.

I'd keep washing it every so often during the day and I'd put some burn cream on it in between the washings.  I'd also take a good shower.  No harm in overreacting.

  still haven't figured out if I can pour the diluted mixture of water
  and cyclohexylamine down the drain

I'm almost positive that the answer is no.  Besides you and the neighborhood, it might react with the pipes or what's stuck in the pipes and create longer term toxicity problems when you're in that room at some later date.   Where I live, we have hazardous chemical drop-off centers (mainly for paint, oil and household chemicals, but used for everything chemical).  I'd pay the relatively small amount they charge and drop it off there.

One final note, even when using the right protective gear, some people become "sensitized" to amines.  This means that future exposure at even lower levels to a wide array of amines can be more dangerous.  Also, I'd try to open up the exposed room to as much outside air as possible, use a room fan too if you have one.


Heavier isotopes make stronger bonds, therefore C-D and C-T bonds are stronger than C-H bonds. The difference between the bond strengths is however not very significant and can be ignored for most practical purposes.

Therefore, the extent of hyperconjugation can be given as: C-H > C-D > C-T (the difference however is not very significant).

The exact reason for this little difference involves a quantum mechanical explanation, which you should not bother about if you are studying high school chemsitry. If you want to read that, you can refer the links in the comment section.


In orbital hybridisation, what determines the number of orbitals that hybridise to be at the same energy level?

The difficulty in talking about hybridization lies in the fact that it tends to confuse atomic and molecular orbitals, or (equivalently?) one- and many-electron wave functions.

Mixing atomic orbitals generally only becomes possible when we do a mathematical approach which treats more than one electron at once. That, however, is mathematically rather complex, and the resulting wave functions are molecular orbitals, which are entirely different from atomic orbitals. Molecular orbitals are constructed as combinations of atomic orbitals, which means that it is possible to e.g. add s and p wave functions together to obtain something that looks like sp hybrids. But, and this is important, there are actually many different possibly ways for such combinations, and none of them should really be considered physical reality; and furthermore, this "mixing together" has to follow very strict mathematical rules which in the end affects the entire set of molecular orbitals at once. Describing molecular orbitals in terms of hybrid orbitals (or something similar) is possible if one does it thoroughly, in terms of a full and correct mathematical treatment. Still, even when applying the concept in this way, it is somewhat arbitrary, since it represents only one of many possible solutions (and depending on the problem, not even the best one).

In the simple and qualitative terms often introduced in chemistry (especially organic chemistry), it is a very crude concept which should be regarded with great caution and especially not as something that would be some sort of reality. Hybridization is a mathematical procedure for many-electron wave functions; nothing more, nothing less.

In more technical terms: The role of one- and many-electron wave functions in atoms is crucial here. Atomic orbitals are assumed to be hydrogen-like wave functions; in the one-electron atom, the orbitals are $\ell$-degenerate, so linear combinations of wave functions for different orbitals would mathematically still be solutions to the Schrödinger equation. In many-electron atoms, the degeneracy breaks down, so combining different orbital wave functions would be invalid, since they belong to different eigenvalues. We do, however, employ a many-electron Hamiltonian such as the Fock operator in this case, and unless I am mistaken, the electron-electron interactions lift this restriction. So, by some unitary transformation of the many-electron wave function of a single atom, I guess one might arrive at something like hybrid orbitals. Then again, this does not reflect physical reality.


What would be the effect of adding liquid beyond the calibration mark on a volumetric flask?

If you have plenty of solution and reagents, the solution should be prepared again, because the volume is not the nominal volume written on the flask. However, the proper trick to fill the volumetric flask is to use a dropper when you are about to reach the calibration line. Experienced hands can use a squeeze bottle until the end but the chances of this mistake are high.
However, if the solution was was just a single sample solution (provided by somebody else, or the compound is too expensive or rare), slightly overfilling the volumetric flask is not the end of the world. You can mark the liquid level after proceeding normally (dissolve and invert several times). With a help of a high quality volumetric syringe or gas tight syringe (e.g., 2.5 mL Hamilton used in gas chromatography injections) draw the excess liquid until the liquid level is back to the calibration line. Measure that volume in the syringe and add it to the nominal volume of the volumetric flask.
If the volumetric flask had a nominal volume = 100.0 mL and you had to withdraw 1.80 mL then the total volume would be 101.8 mL for molarity or other concentration calculation.


How does a lower pH change the alcohol concentration of wine?

This is a slightly broad question, but ill try to assist some. Google has a large assortment of articles describing pH, Le Chatelier's Principle, and EtOH.

Im not sure if you mean altering the pH during the fermentation process or altering the pH of the final product? It also makes a difference on the types of yeast you use. Some yeasts can handle different conditions and produce a different amount of alcohol.

Here is a great response from last year on the Stack Exchange, regarding fermentation without the addition of commercial yeast, using the change in pH as a catalyst for fermentation.

With the production of wine that uses commercial yeast for fermentation (which is most commercial wines), altering the pH too far outside of it's "designated" range will cause it to produce less alcohol. Some connoisseur winemakers will alter the pH slightly sometimes, varying on the types of grapes, to alter the terpene profile present. Different variables, equals different systems of equilibrium.


The most recent IUPAC recommendations[1] states that:

  P-61.9: Preferred IUPAC names are formed substitutively using the prefix ‘isocyano’ attached directly to a parent hydride.

This is a change from previous recommendations, and differs from some of your examples above.

To give a concrete example, $\ce{C6H5NC}$ should be named isocyanobenzene - this is known as a PIN by IUPAC (preferred IUPAC name). Phenylisocyanide is also acknowledged as an acceptable trivial name (a name that does not conform to the recommendations but is still commonly used).

[1]: IUPAC Recommendations for Preferred Names 2013, Chapter P-6. DOI:10.1039/9781849733069-00648


Do the compounds NaₓIrO₂ and NaOsO₂ exist?

For $\ce{Na_{x}IrO2}$ the current answer seems to be "no" it has not been made but it is discussed in the literature theoretically.

From Kitaev-Heisenberg models for iridates on the triangular, hyperkagome, kagome, fcc, and pyrochlore lattices:

  However, no triangular lattice iridate with the relevant edge sharing octahedra structure is currently available; a compound of the
  type NaxIrO2 may or may not turn out to be stabilized.

and in Chiral spin liquid and topological superconductivity in the undoped and doped
triangular Kitaev-Heisenberg model:

  In addition, a class of ABO2 (where A and B are alkali and transition metal ions, respectively) type transition metal compounds and the possible material NaxIrO2 may be potential candidates for realizing the triangular Kitaev-Heisenberg model.

(But that's your paper isn't it? I don't have any information beyond what you already know, so really this is a useless answer.)


Why is zinc oxide reduced at a temperature of 1400 degrees Celsius?

If the answer is $1400\ \mathrm{^\circ C}$ it must imply that the user does not want $\ce{CO}$ in their product. Based on the diagram, the intersection of $\ce{ZnO + C -> Zn + CO}$ occurs around $1000\ \mathrm{^\circ C}$. Full conversion of $\ce{CO}$ to $\ce{CO2}$ requires a temperature of around $1400\ \mathrm{^\circ C}$ as shown in the diagram.

As to why the reaction does not follow the $\ce{C -> CO2}$ curve (no $\ce{CO}$ involved), this is a kinetic reason. Subtract the $\Delta G$ values of each line at a specific temperature (where the lines intersect, the value is 0, meaning the reductions of the two species are in equilibrium). You will find that at the temperature of intersection between $\ce{2ZnO -> 2Zn + O2}$ and $\ce{C + O2 -> CO2}$ $(\Delta G = 0)$, there is sizeable energy driving force for $\ce{ZnO + C -> Zn + CO}$ $(G = -300\ \mathrm{kJ/mol}\ \ce{ZnO})$. The difference of the Gibbs energies is related to the ratios of the $\ce{CO}$ and $\ce{CO2}$ products by

$$\Delta G_\mathrm r(\ce{ZnO + C -> Zn + CO})-\Delta G_\mathrm r(\ce{2ZnO + C -> 2Zn + CO2}) =-RT\ln\frac{[\ce{CO}]}{[\ce{CO2}]}$$

Through plugging in the number it is a very high proportion that goes through the $\ce{CO}$ pathway. Again, the only way any of this is important is if it mentions that $\ce{CO}$ must be eliminated, else the reaction proceeds at low temperature as you say.


Extraction of a pyridine tetrazole from the aqueous phase

Concentrate under reduced pressure on the rotovap. Add a miscible co-solvent that does not solubilize the zwitterion at its isoelectric point but dissolves starting materials, like THF.

The pyridine is basic, the terazole is acidic. Catch it on an ion exchange resin column in water, then displace it with a polyvalent counterion.  Possibly down neutral alumina in water to slow it and concentrate it.  You can run a microcolumn in a disposable pipette with a tiny ball of glass wool inside at the taper.

Salt it out as a solid with saturated NaCl or potassium nitrate its isoelectric point. Precipitate it, perhaps as the tetraphenylborate or picrate (careful there).


Does migraine medicine Topiramate work by supplanting Pyridoxal phosphate in enzymes?

Background
Topiramate (TPM) is an anticonvulsive and antimigraine drug that helps prevent or reduce the frequency of epileptic seizures and that is used as a prophylaxis of migraine. (Shank and Maryanoff, 2008)
TPM not only inhibits kainate subtypes and α-amino-3-hydroxy-5-methylisoxazole-4-propionic acid (AMPA) subtypes of glutamate activated ionotrophic receptors, the modulatory effects on GABAA receptors and voltage-gated Ca2+ and Na+ channels (Gibbs et al., 2000), but also inhibits carbonic anhydrase enzymes (demonstrated in electrophysiologic and biochemical studies). see drugs.com
OP question
It is an intriguing possible correlation that “Topiramate supplants PLP in the enzyme glutamate decarboxylase [GAD]”, notably based on structural similarity of the alkyl-sulfamate to the alkyl-phosphate as well as the linked rings.
However, the literature does not seem to align with this premise:

Shank and Maryanoff et al.(2008) noted that “[Topiramate’s] sulfamate moiety is essential, but not sufficient, for its pharmacodynamic properties.”
Whereas PLP is a substitute pyridine (planar) derivative, TPM is a sulfamate, substituted derivative of β-D-fructopyranose (Maryanoff et al., 1987) where its substituted ring is in the boat conformation and its ring is puckered. See the crystal structure as entry SEQKAA02 at CCDC

TPM does not seem to inhibit glutamate decarboxylase. “Single-dose and repeated treatments with TPM were without effect on the activities of GABA-T and GAD (Fig. 2) in mouse brain at 4 h after administration.” (Sills et al., 2000)
GAD inhibition and Topiramate treatment have opposite effects. “Clinical evidence shows that lowered GAD activity is associated with several forms of epilepsy which are often treatment resistant.” (Zhang et al. 2017; Lloyd et al. 1986) Furthermore, Zhang et al. synthesized an ethyl ketopentenoate (EKP), a lipid-permeable GAD-inhibitor, and used it to induce refractory seizures in zebrafish larvae. (yes, that’s mean.. but informative here.)

Molecular basis of carbonic anhydrase inhibition
Topiramate is a carbonic anhydrase inhibitor (multiple classes), a Zn-dependent metalloenzyme. Below is how Topiramate (green) binds to human carbonic anhydrase II. The zinc ion is shown as a silver sphere. RCSB entry 3HKU.pdb DOI: 10.2210/pdb3HKU/pdb and Lopez et al. (2009):

Conclusion
Given the above, I wouldn’t think the apparent structural similarity of Topiramate to PLP is relevant to epilepsy and/or migraine treatment.
References
Lloyd, KG, Bossi, L, Morselli, PL, Munari, C, Rougier, M and Loiseau, H (1986) “Alterations of GABA-mediated synaptic transmission in human epilepsy” Adv. Neurol. 44:1033–1044. source
Lopez, M, Paul, B, Hofmann, A, Morizzi, J, Wu, QC, Charman, SA, Innocenti, A, Vullo, D, Supuran, CT and Poulsen, SA (2009) “S-Glycosyl Primary Sulfonamides−A New Structural Class for Selective Inhibition of Cancer-Associated Carbonic Anhydrases” J. Med. Chem. 52:6421-6432. DOI: 10.1021/jm900914e
Maryanoff BE, Nortey SO, Gardocki JF, Shank RP, Dodgson SP (1987) “Anticonvulsant O-alkyl sulfamates. 2,3:4,5-Bis-O-(1-methylethylidene)-β-D-fructopyranose sulfamate and related compounds” J. Med. Chem. 30:880–887. DOI: 10.1021/jm00388a023
Shank, RP and Maryanoff, BE (2008) “Molecular pharmacodynamics, clinical therapeutics, and pharmacokinetics of topiramate” CNS Neurosci. Ther. 14:120-142. DOI: 10.1111/j.1527-3458.2008.00041.x
Sills, GJ, Leach, JP, Kilpatrick, WS, Fraser, CM and Brodie, MJ (2000) “Concentration-Effect Studies with Topiramate on Selected Enzymes and Intermediates of the GABA Shunt” Epilepsia 41 (Suppl. 1):S30-S34. DOI: 10.1111/j.1528-1157.2000.tb02168.x


Ideal gas Law best describes the properties of which of the following gases at 0 degrees Celsius and 1 atm?

One of the defining characteristics of ideal gases is that ideal gas particles do not interact with each other, and there are no intermolecular forces between them. If a certain type of gas molecule has a molecular bipole, then each molecule will have ends with opposite partial charges. In a collection of gas particles, if the particles have a molecular dipole, these particles will attract and repel each other and thus will experience intermolecular forces. Because the two nitrogens in $\ce{N2}$ are equally electronegative, the bond and each $\ce{N2}$ gas particle will have minimal partial charges and thus minimal intermolecular forces. Thus, $\ce{N2}$ behaves very ideally.


In the determination of the $\mathrm{pK_a}$ values of really strong acids, water as a medium is irrelevant at a certain point.

Ivo Leito and coworkers have examined the acidity of various compounds in 1,2-dichloroethane (DCE).

In this medium the $\mathrm{pK_a}$ values for $\ce{HCl}$, $\ce{HNO3}$, $\ce{HBr}$, $\ce{HI}$, $\ce{HBF4}$, an $\ce{HClO4}$ have been determined as -0.4, -1.7, -4.9, -7.7, -10.3 and -13.0.

In this survey, 1,1,2,3,3-pentacyanopropene (1), was found to be even more acidic with a $\mathrm{pK_a}$ of -15.0!

EDIT
It is important to note that these values are likely to be correct in their relative order. But the values are based on picric acid, for which the $\mathrm{pK_a}$ was arbitrarily set to zero. This initially escaped my attention!

Thanks to Martin for pointing out this important detail in his comment below!

Being an oddball, the rather unstable perbromic acid ($\ce{HBrO4}$) was not in the list!

It was synthesized by Evan Appelman (even on a larger scale) by bubbling fluorine gas through an alkaline solution of 1M $\ce{NaBrO3}$ and 5M $\ce{NaOH}$ until the solution becomes acid.

Amazingly, the author further states in a footnote:

  The reaction is not smooth, and small explosions may take place in the vapor above the solution.


Both calcium hydroxide and ethanol have a hydroxide group, then why isn't ethanol ionic?

No, it is not ionic.
Your doubts are somehow legitimate if you are not aware of the concept of electronegativity. As a rule of thumb if two atoms have similar electronegativity is harder to have ionic bonds. You can use Pauling scale of electronegativity to have some insights on it.

A difference of electronegativity of over 1.7 is likely to be ionic,
and a difference of less than 1.7 is likely to be covalent.
Source: Atkins, Peter; Loretta Jones (1997). Chemistry: Molecules, Matter and Change. New York: W. H. Freeman & Co. pp. 294–295. ISBN 0-7167-3107-X.

This means you will often deal with ionic compounds made of one ion with a strong electronegativity and one with a weak electronegativity. In fact, ions are composed of anions and cations. With a lot of approximations, we can calculate the electronegativity of a group averaging the electronegativities of the components for $\ce{-OH}~\dfrac{3.44 + 2.82}2 = 3.13$.
In your example, $\ce{C2H5OH}$ ethanol the bond is with $\ce{C}$ ( electronegativity 2.55)  so the difference is 0.58 hence covalent. If the bond is with $\ce{Ca}$ (1.00) the difference would be 2.12 which is greater than 1.7 hence is categorized as ionic.
Of course is not always easy to determine if it's ionic or not so in complex cases we can say that it has an ionic character o covalent character.


What is the appropriate method to visualize electrostatic potentials?

One way to do it is to perform a quantum chemistry calculation of the electrostatic field generated by the molecule. Another is to determine (or assume) partial atomic charges for each of the atoms of the system, and use a Poisson solver (as I believe APBS is) to calculate the electrostatic field generated by this point charge distribution.


When we cannot use Friedel-Crafts acylation to insert alkyl group?

tBu-benzene is prepared exactly as you describe, see this preparation here. Follow this by nitration and reduction.
There is no obviously comparable strategy of acylation that can compete with this. Making a tBu group from acetophenone is not straightforward and is illogical when there is a perfectly good direct route.


Do volumes and pressures calculated from Boyle's Law depend on the temperature?

No,the volume would change this time  as by Gay Lussac law pressure is directly proportional to temperature  which has increased this time and can overcome the pressure you had applied earlier thus increasing the volume.
Also I think you have a misconception regarding the term constant temperature.It means that in Boyle s law pressure is inversely proportional to volume at a given temperature say 200K .But for different temperature the law will follow but with different values.For eg. at a particular temperature 200k it can be:- 200litres and 50 pascal or 100 litres and 100 pascal.For higher temperature it would be:- 200 litres and 100 pascal or 100 litres and 200 pascal.


Competing resonance and inductive effects in a substituted benzene

Often, but not always, mesomeric displacement leads to a shift in prevailing over the inductive effect.

Pi-electrons are at the peripheral orbitals. Association with the nucleus is less strong than at the sigma-electrons. The ionization potential of pi-electrons is smaller and chemical bond is more polarizable. Therefore, the dipole moment associated with the mesomeric effect can prevail over the dipole moment associated with the inductive effect.

However, there are groups in which the inductive effect prevails, such as halogens.

Also, you can read about hyperconjugation. Hyperconjugation can have influence to the mesomeric and inductive effects too.

Source: Reutov  O.A., Kurts A.L., Organic Chemistry, vol. 1 of 4, ch. 2.2.2, MSU


Determining Kc for the equilibrium involving iron thiocyanate using spectrophotometric data

The problem is a bit strange since the details of the absorbance measurement aren't detailed very well. I'll assume the book defines the relationship as:
$$A = \epsilon bc$$

where:

A=absorbance
$\epsilon$ = molar attenuation coefficient
$b$ = path length
$c$ = concentration

So the concentration, c, of the $\ce{[Fe(SCN)]^2+}$ is:

$$ c = \dfrac{A}{\epsilon b} = \dfrac{0.625}{3800} = 1.64\times10^{-4}$$

Considering the dilutions...

$\ce{[Fe^{+3}] + [Fe(SCN)]^2+} = 0.00110 \times \dfrac{12}{18} = 7.33\times10^{-4}$

$\ce{[SCN^-] + [Fe(SCN)]^2+} = 0.00140 \times \dfrac{6}{18} = 4.67\times10^{-4}$

But there is $1.64\times10^{-4}$ of $\ce{[Fe(SCN)]^2+}$ so:

$\ce{[Fe^{+3}]} = 7.33\times10^{-4} - 1.64\times10^{-4} = 5.69\times10^{-4}$

$\ce{[SCN^-]} = 4.67\times10^{-4} - 1.64\times10^{-4} = 3.03\times10^{-4}$

Therefore:

$K_c = \dfrac{1.64\times10^{-4}}{(5.69\times10^{-4})(3.03\times10^{-4})} = 951$

I'd assume the small difference is just due to a difference in rounding somewhere...


What is the balanced ionic redox reaction of nitrite ions with manganate(VII) ions in acidic solution?

Your suggestion is certainly part of the picture, but as is usually the case with nitric acids, a number of side reactions can and often do take place depending on concentration and temperature. In acid solutions, for example, nitrite spontaneously decompose giving nitric oxide together with dinitrogen tetroxide. My feeling is that under optimum conditions your suggested reaction would predominate


Please explain why dioxygen difluoride is so dangerous

XKCD's source material is an article from the blog of one of the funnier chemists I've read, Dr. Derek Lowe. The chemical in question made his "Things I Won't Work With" list, and the article is found here.

Dioxygen difluoride, $\ce{O_2F_2}$, sometimes evocatively labeled according to its atomic arrangement $\ce{FOOF}$, is first and foremost a vigorous oxidizer; in plain English, it starts roaring fires with almost anything it touches, even with things that simply will not burn in any other circumstances, like sand, concrete, brick, iron, copper, even asbestos and fiberglass. Both the oxygen and the fluorine are pretty highly electronegative (fluorine is the most electronegative element in the table, and oxygen's #2), and so they will readily react with anything even slightly more willing to donate an electron; carbon, silicon, metals, alkalis, most other gases, etc. When that happens, the high-energy bonds of this molecule (it can only be formed at 700*C, and that thermal energy remains "locked" in the structure of the molecule) are rearranged into structures that require less energy, and the excess energy is released. The resulting rapid, aggressive, highly exothermic reaction of dioxygen difluoride with just about any other material would not be described as anything less than an explosion.

Second, even in the absence of something to oxidize, the oxygen would much rather exist in its "native" diatomic gas state. Again, there's a lot of energy tied up in the molecular bonds of this chemical, and the lower the energy required to form a bond, the more stable the molecule. Therefore, even at extremely cold temperatures (-160*C) in complete isolation, dioxygen difluoride decomposes to oxygen difluoride ($\ce{OF_2}$) and $\ce{O_2}$ gas. At room temperature, this decomposition is uncontrollable (no-one's gotten it anywhere close to RT without it coming apart), and even at extremely cold temperatures, the heat produced by slow decomposition can cause what's called a "thermal runaway"; the heat produced by one molecule decomposing is absorbed by nearby molecules causing them to decompose, releasing more heat, and so on. When things rapidly decompose into more stable gaseous components, the event is normally accompanied by a loud noise and shrapnel, formed from the remains of whatever had been holding the substance and anything else nearby.

Of the products formed by the self-decomposition of FOOF, oxygen gas is a pretty strong oxidizer, but it's the least of our worries here. The other product, oxygen difluoride, is almost as strong an oxidizer as dioxygen difluoride, which is only stronger because it's oxygen difluoride plus more oxygen. The "shared" electron pairs forming the bonds between the oxygen and each fluorine atom are much more closely attracted to the fluorine, as opposed to a structure like water where the oxygen is the more attractive. This gives the oxygen a formally positive oxidation state of +2.

By its very nature, oxygen yearns to flip that sign and achieve its "natural" -2 ox state by bonding with something less electronegative (which, as previously mentioned, is almost everything else in the periodic table besides fluorine). However, unlike the other oxygen atom in the FOOF molecule which flew the coop, this second oxygen can't escape, because the two fluorine atoms won't let it go. The fluorines prefer to bond to it rather than to each other... until something else, anything else, is introduced that's even more attractive.

$\ce{OF_2}$, like its FOOF parent, reacts very strongly and exothermically with almost anything, especially water. It will literally burn ice, rapidly; that's one of the reactions guaranteed to produce a powerful explosion. The most stable ultimate products of that reaction are hydrogen fluoride gas and more oxygen, and when you call HF gas a "stable" product of any reaction you are speaking in very relative terms; a release of HF gas into the air is one of those "drop everything and run" types of industrial accidents. HF gas dissolves in water, including the mucus of your lungs, to produce hydrofluoric acid, a persistent, toxic, corrosive acid that will shred just about any organic chemical it comes across from hemoglobin to proteins to DNA, and you really want to keep those kinds of things intact.

Basically, there are a lot of things in an inorganic chemistry lab that have a Medusa-like ability to kill at a glance, but fluorine deserves special notice, in part because of the number of notable chemists who were poisoned or blown up in the early heady days of isolating it. When you combine it with other elements on the right hand side of the periodic table in "lightweight" combinations (between two and about ten atoms per molecule), your day is about to get reeeeealy interesting; see the work of A.G. Streng, who was tasked not only with producing large quantities of dioxygen difluoride, but also with reacting it with things that normal chemists would keep very far away from it (and in fact "normal chemists" stay very far away from FOOF itself for the obvious reasons).

Fluorine only starts becoming more stable and easier to work with as an ionic salt, such as sodium or calcium fluoride (in which form it naturally occurs as the mineral fluorite; still poisonous in sufficient doses but much less of an explosion/reactivity hazard), or when you start throwing in elements of the carbon group (the carbon-fluorine bond is among the strongest known to chemistry, and fluorocarbons have a variety of uses as propellants, refrigerants and polymers, including Teflon), or heavier chalcogens like sulfur (sulfur hexafluoride, $\ce{SF_6}$, has virtually no reaction chemistry and is favored as a dielectric and to blanket air-sensitive materials, though other sulfur-fluoride compounds like sulfur tetrafluoride and disulfur decafluoride are highly toxic).


dissociation of aluminum sulfate in water

Aluminum sulfate has special properites and uses. It is true that aluminum sulfate is soluble in water, and it has specific solubility values for certain temperatures:

$31.2\ \mathrm g$ per $100\ \mathrm{mL}$ at $0\ \mathrm{^\circ C}$
$36.4\ \mathrm g$ per $100\ \mathrm{mL}$ at $20\ \mathrm{^\circ C}$
$89.0\ \mathrm g$ per $100\ \mathrm{mL}$ at $100\ \mathrm{^\circ C}$

You stated you used $12\ \mathrm g$ in $100\ \mathrm{mL}$, so it would seem to completely dissolve. However, it this situation, since it isn't being dissolved in very large amounts of water, it doesn't dissociate and form aluminum hydroxide. Aluminum sulfate can be used in water purification in small amounts, because it causes impurities to group up into larger particles and then settle to the bottom of the container (or be filtered out) more easily. The solid you see sometimes are the impurities of the water that have been grouped up and look like precipitate. If you use more pure water, you will see less of these impure solids, or possibly nothing if the water is pure enough (such as if it's distilled water).


Why do so many biochemical reactions require enzymes?

all 10 steps in glycolysis require an enzyme

We count it as 10 steps because there are 10 enzymes. There might be other steps that are fast (like a product acting as an acid or base) that we don't count because they don't need an enzyme.

  I’d have thought that evolution would have selected for “simpler” pathways.

What is simple is that everything is happening in the same reaction vessel, the cytosol (well, there are more compartments, but just a handful). This is different from the organic chemist, who works with very reactive substances in pure form, carefully mixed in separate reaction vessels with distinct solvents refluxed at distinct temperatures for hours.

  Is the very fact that enzymes are required make life more probable since the cells ability to control their quantities make homeostasis and regulation possible?

Yes, that is true. It also allows for a web of reactions rather than a single pathway (you can funnel fructose or galactose into glycolysis, and you can turn pyruvate into lactate instead of decarboxylating it to carbon dioxide and acetic acid).

Stable molecules require catalysts

One requirement for life as we understand it is a way to store, replicate and access genetic information. In our living world, that requirement is met by having nucleic acids. Under physiological conditions, RNA is fairly stable (good enough for a virus) and DNA is very stable (to the degree that sometimes DNA sample from extinct organisms preserve genetic information to this day). If these molecules do not form or decay on their own, there must be enzymes to act on them.

Granted, there could be other pockets of metabolism that work without catalysis. The other requirement, though, is to capture free energy to drive reactions away from equilibrium (e.g. to make the carriers of genetic information mentioned above). This is done through coupled reactions, where e.g. the free energy from food being oxidized (with oxygen as terminal electron acceptor) is captured in proton gradients across membranes and in the phosphorylation of ADP to obtain ATP. If any of the reactions going toward equilibrium would occur without catalysis, the free energy would be lost instead of captured by a coupled enzyme reaction.

General thoughts on complexity

In any complex system (government, computer chip, industrial process etc.), as the complexity grows, the number of steps that seem inefficient or over-complicated will also grow. Many of these steps might not make sense from the present view, but were put in place (or evolved) earlier to meet a historic need. Once in place in a system that has to run continuously, there is often no path to swap it out for a more efficient way of doing things.

One example from biology is the human eye, where the nerve cells run through the optical path, giving rise to the blind spot. The octopus eye "lucked out" and evolved in a way that makes more sense, but human eye sight is just fine (there is a "software patch" in the brain to fill in the blind spot):

Source: https://external-preview.redd.it/xWn0rxCSyu9TrnXozPrsF3aq2_uILbUEJwzBv87TZo4.jpg?auto=webp&s=7a1bccf554ec19a813f9da4466c114ac84c426b7


Is there more than one equation for Delta T(reaction)?

If You are stating you have your initial water temperature and the final solution of your temperature, then your change in temperature is just your final temperature minus your initial temperature. No need for linear equations. The change in temperature for the reaction is the temperature change of the water due to the heat given off or absorbed by the system to the surroundings.


The ionization energy of H₂⁺ molecule ion

I would like to clarify the matter since from OP's own answer I have a feeling he/she does not quite understand it.

So, first of all, in general the ionisation energy has to be calculated as the difference in total energy and not just the electronic one.

Secondly, for molecules there are two types ionization energies: adiabatic and vertical:

  The adiabatic ionisation energy refers to the formation of the
  molecular ion in its ground vibrational state and the vertical
  ionization energy applies to the transition to the molecular ion
  without change in geometry. (From IUPAC Gold Book)

Note that in accordance with the Franck–Condon principle the most probable transition is the vertical one. Consequently, usually, when we say "ionisation energy" we actually mean the "vertical ionisation energy".

Finally, if the total energy is understood in the clamped nuclei approximation, i.e. as the electronic energy $E_{\mathrm{e}}$ together with the nuclear repulsion energy $V_{\mathrm{nn}}$, then for the vertical ionisation energy one could indeed use the difference in electronic energy rather than in total one simply because the nuclear repulsion energy stays the same. Mathematically, $V'_{\mathrm{nn}} = V_{\mathrm{nn}}$ since geometry is not changing, and thus,
$$
  E_{\mathrm{i}}
  =
  U' - U
  =
  (E'_{\mathrm{e}} + V'_{\mathrm{nn}})
  -
  (E_{\mathrm{e}} + V_{\mathrm{nn}})
  =
  E'_{\mathrm{e}} - E_{\mathrm{e}} \, ,
$$
where primed quantities refers to the resulting molecular specie and non-primed are that of the original molecular specie.

So, yes, for the vertical ionisation energy in the clamped nuclei approximation one can use the difference in electronic energy. And one can do so regardless of what are the molecular species in question. For the specific case of the vertical ionisation of $\ce{H2+}$ the electronic energy of the resulting $\ce{H2^2+}$ specie is obviously zero, so we specifically get what OP suggested
$$
  E_{\mathrm{i}}
  =
  E'_{\mathrm{e}} - E_{\mathrm{e}}
  =
  0 - E_{\mathrm{e}}
  =
  - E_{\mathrm{e}} \, .
$$
However, this is just a trivial simplification of a general formula for the vertical ionisation energy for the case of a zero-electron resulting specie.


Is (1R,3aR,4S,6aS)‐1,4‐dibromo‐octahydropentalene chiral or achiral?

It indeed is achiral, because the red dot is the center of symmetry, hence it is superimposible with its mirror image.
Having an internal plane of symmetry is not the only criterion, the presence of a centre of symmetry also leads to the compound being achiral.


What you see, when you open your checkpoint file, is all you can see. I assume you use TDDFT and with this method you can't obtain other orbitals than the ones from the ground state. Why is this so?

Because in LR-TDDFT you calculate your ground state and apply a perturbation to it. (You can look up on the behind theory in good books like Introduction to Computational Chemistry from Frank Jensen.) The result from this calculation is presented from Gaussian like this:

Excited State   1:      Singlet-A      1.0662 eV 1162.87 nm  f=0.0007  <S**2>=0.000
110 ->113         0.60155  (0.72)
110 ->115        -0.23139  (0.11)
110 ->116         0.22569  (0.10)
Excited State   2:      Singlet-A      1.2549 eV  988.02 nm  f=0.0002  <S**2>=0.000
109 ->113        -0.39288  (0.31)
110 ->114         0.51726  (0.54)
Excited State   3:      Singlet-A      1.2929 eV  958.99 nm  f=0.0027  <S**2>=0.000
109 ->113         0.50172  (0.50)
110 ->114         0.29522  (0.17)
(The weights in spaces are added by me.)

These are the configurations of three excited states. The first excited state, for example, consists mainly (72%) from an excitation from MO 110 (HOMO) to MO 113 (LUMO+2). There are some contributions with weights of about 10% into the MOs 115 and 116. (On how to get the weights, have a look at an answer by me on this site here.)

You ask, how you can see the excited state in GaussView. You can't ... directly. And this question is also a little bit tricky, because what is it that you want to see? Here are some options:

You could

calculate the electron density of the excited state TD(Root=1) Density(Current) and subtract the electron density from the ground state. Then this difference shows you the "hole" where now is less density and the "hole" where there is more density.
calculate Charge Density Differences (CDD) with Multiwfn or Electronic Difference Density Maps (EDDM) with GaussSum which show you the same but a little bit easier as you do not need an additional calculation because they are based on the excitations and the CI coefficients.
calculate the transition density for the excited state TD(Root=1) Transition=1
calculate Natural Transition Orbitals with, e.g., Gaussian or Multiwfn, for very diffuse excitated states as NTOs reduces/optimizes the number of contributing excitations for each root with Density=(Transition=1) Pop=(Minimal,NTO,SaveNTO)

Just have a look on the manual for TDDFT, Density, FAQ and Pop.


Does acid and base react directly, or do they react first with water?

With strong bases and acids, only $\ce{OH-}$ and $\ce{H3O+}$ react, right, because $\ce{AH}$ and $\ce{B}$ (practically) don't exist in aqueous solution.

For weak acids and bases, that's not true, but still there the water can "transport" protons much faster between AH and B than direct exchange after diffusion. As was noted below, that's called the Grotthuss mechanism.

So yes, you are right, but the formula (1) is a valid simplification which is, for most uses, as good.


Yes, the ‘hydrate’ label applies exclusively to solids. Consider the following experiment: Dissolve both anhydrous $\ce{CuSO4}$ and $\ce{CuSO4 . 5 H2O}$ in water. You will get the same, blue solution.

What happened on a microscopic level is that the crystal lattice which once included the water atoms is broken up and the ions dissolve separately. What used to be copper(II) sulphate pentahydrate is now turned into a hexaaquacopper(II) complex.

$$\ce{CuSO4.5 H2O + H2O -> [Cu(H2O)6]^2+ + SO4^{2-}_{~(aq)}}$$


First of all, for both products, you have not given the O atom in the carbonyl position. In organic chemistry, we assume that there is a C atom if no atom is mentioned specifically. Now for this reaction, the carbonyl group will be there as it is.
Now we can consider the reaction. The epoxide is away from the plane, i.e. below the plane. For the steric reason, it is more feasible for azide to attack from the upper side of the plane. Now again, azide can attack both C centre of the 3-membered epoxide ring. In that case, we'll be having two different products. But in both cases, we will have the azide group on the above of the plane and O atom in the below of the page.
Now if I assume that you have mistakenly skipped the O atom in your both products, then your first molecule matches with the first answer given below. But  the second answer given by you is not correct and doesn't match with the second answer. You can have the green and blue coloured bond in the same place and then you have to alternate the position of OH group and azide group. Then that will match the answer given below and the answer will be correct.


How does the screening effect work for orbital in the same shell?

Screening within the same shell is generally poor, which is why effective nuclear charge rises going across the Periodic Table: each added electron screens the others poorly from the added proton.  And, yes, as you suspect, the quality of screening varies with l.

In addition to the screening effect, in many-electron atom you have electron-electron repulsion, as well as quantum exchange-correlation effects between electrons.  Since these are interactions between electrons they depend on the distance between the electrons, which in turn depends not only on the distance between each electron and the nucleus (controlled by n) but also on the angular distribution of electrons around the nucleus (controlled by l).


In all cases we study the stability of the carbocation resulting from the hydride/methyl/phenyl shift. That shift will be more favored whose resulting carbocation is the most stable.


How can I measure a droplets contact angle without a setup?

In 2010, Borguet et al. published in Journal of Chemical Education, DOI: 10.1021/ed100468u an intentionally simplified setup which may be useful as a reference.  A Sony Cybershot camera was placed in front of a sample scrutinized under diffuse illumination:

The additional lens shown is focussed on the drop (both for water and $n$-hexadecane, the authors recommend a volume of $\pu{5 uL}$; focal length of the length used equals $\pu{50 mm}$):.

With photos recorded like this

the contact angle determination was performed with the contact angle plugin of imageJ.  As Fiji, it is open source, free ware, and java based/OS portable.  To quote from their publication:

  For each easurement, the user must choose two points to manually
  define the baseline and three points along the drop profile (see
  the supporting information that features a set of instructions
  supplying all the details on the procedure used to make the
  measurement). The program then fits the profile of the drop and
  calculates the contact angle using the sphere approximation
  or the ellipse approximation. In our study, the ellipse approximation gave consistent results for contact angles $> 40^\circ{}$. For
  drops with contact angles $< 40^\circ{}$, the sphere approximation was
  used.

In comparison with reference data, the authors judge contact angles recorded as well as the subsequently calculated surface tensions as good enough for instruction / lab classes and research lab.


What does the peak separation of carbon dioxide IR rotational-vibrational spectrum corresponds to and why?

(In this answer I am using the symbol $\require{begingroup} \begingroup \newcommand{\tb}[0]{\tilde{B}} \tb$ to represent a wavenumber instead of $B$, which I choose to define as an energy – not that we will be using it.)

  Why is the peak separation $4\tb$ instead of $2\tb$?

This has to do with nuclear spin statistics. Because oxygen-16 is a boson (spin 0), the wavefunction of the entire molecule (which comprises translational, rotational, vibrational, electronic, and nuclear spin components) must be symmetric upon exchange of the two oxygen-16 atoms.

The translational component is symmetric, as usual. The electronic ground state is ${}^1\Sigma_\mathrm{g}$, which is totally symmetric. The nuclear spin component must be symmetric by virtue of the fact that both oxygen atoms are spin 0; this means that there is only one possible spin state that they can be in (cf. hydrogen-1 has a spin of 1/2 and two possible spin states). Consequently, exchange of the oxygen atoms leads to no change in the nuclear spin wavefunction.

As a result, this means that the vibrational component times the rotational component must be symmetric; or in other words, the vibrational and rotational components must have the same parity.

Since the peak structure at $\pu{2349 cm-1}$ arises due to the antisymmetric stretch, the vibrational ground state $v = 0$ is symmetric with respect to exchange of oxygen, and must therefore be paired with rotational levels with even $J$, which are likewise symmetric. In a similar manner, the vibrational excited state $v = 1$ is antisymmetric, and must be paired with odd $J$ levels.

In $\ce{CO}$ where there are no nuclear spin statistics at work, the following transitions are all allowed in the $R$-branch ($\Delta J = +1$):
$$\begin{align}
(v,J) = (0,0) &\to (1,1) & \text{line 1}\\
(0,1) &\to (1,2) & \text{line 2}\\
(0,2) &\to (1,3) & \text{line 3}\\
(0,3) &\to (1,4) & \text{line 4}\\
\end{align}$$

and so on. However, in $\ce{CO2}$ the state $(v,J) = (0,1)$ is not permissible, and hence line 2 does not appear in the spectrum. Similarly, neither does line 4, or line 6, and so on; the separation between adjacent lines is therefore doubled.

  Why are the peak separations in the $P$-branch larger than in the $R$-branch?

This is because the rotational constant $\tb_0$ in the vibrational ground state is different from the rotational constant $\tb_1$ in the first vibrational excited state. The rotational energy levels (in wavenumber units) are therefore by:

$$\tilde{G}(v,J) = v\omega_0 + \tb_vJ(J+1)$$

where $\omega_0$ is the wavenumber of the pure vibrational transition (i.e. the band origin). For the transition $(v,J) = (0,J) \to (1,J-1)$ in the $P$-branch, the associated energy is therefore

$$\tilde{P}(J) = \omega_0 + \tb_1J(J-1) - \tb_0J(J+1)$$

The separation between adjacent lines in the $\ce{CO2}$ spectrum is then

$$\begin{align}
\tilde{P}(J) - \tilde{P}(J+2) &= \tb_1J(J-1) - \tb_0J(J+1) - \tb_1(J+1)(J+2) + \tb_0(J+2)(J+3) \\
&= 4J(\tb_0 - \tb_1) + 6\tb_0 - 2\tb_1
\end{align}$$

You can easily verify that if $\tb_0 = \tb_1 = \tb$, then this simplifies directly to $4\tb$, as expected.

However, because $\tb_0 \neq \tb_1$, the gap between adjacent lines acquires a $J$-dependency, because the term $(\tb_0 - \tb_1)$ is nonzero. In the case of the asymmetric stretch, it turns out that $\tb_0 > \tb_1$ (can you figure out why? It's best explained using the moment of inertia $I$, since $\tb = \hbar/8\pi cI$). Therefore, as $J$ increases (i.e. as we move farther left along the spectrum), the separation between lines increases.

An exactly analogous calculation will explain why the lines in the $R$-branch tend to "bunch":

$$\tilde{R}(J+2) - \tilde{R}(J) = 4J(\tb_1 - \tb_0) + 6\tb_1 - 2\tb_0$$

$\endgroup$


It is an example of the Leidenfrost effect. Ideally the water being expelled from the food would actually keep the oil from contacting the food itself by forming a barrier of steam (vaporizing water) around the food.


The setting of cement involves the reaction of clinker minerals with water, to form a system of interlocking crystals, which lock the materials together.
The setting of cement is primarily due to the reaction of tricalcium aluminate(C3A), which is the most of the four clinker minerals present in concrete. The set regulating action of gypsum is due primarily to its reaction with C3A.
In the absence of gypsum, the reaction of C3A with water is very rapid. The C3A is converted to calcium aluminate hydrate ($\ce{4CaO.Al2O3.19H2O}$) which forms interlynking bridges of solid material between the cement particles. These interlinking bridges reduce the mobility of the cement particles and produce a firm structure, which results in setting of cement mixture. The rapid setting is called a flash set and its irreversible.
The presence of gypsum in cement slows the reaction of the C3A. When mixed with water gypsum dissolves and reacts with C3A to form ettringite ($\ce{3CaO.Al2O3.3CaSO4.32H2O}$). This ettringite is initially formed as very fine-grained crystals, which form a coating on the surface of the C3A particles.

These crytals are too small to bridge the gaps between the particles of cement. The cement mix therefore remains plastic.
After the initial hydration reactions of cement, a dormant period occurs during which the rate of hydration of cement is greatly reduced. During this dormant period, ettrinite crystals continues to grow and eventually are large enough to impede the mobility of cement particles. At this stage setting occurs.


Apart from some Group 13 weirdness, hydrogen can only make one bond. Covalent bonds require pairs of electrons and hydrogen can only have two electrons bound in one covalent bond.


How to balance a redox equation in molecular form?

Judging from the picture you attached you have successfully identified the species that gets oxidised
$$\ce{Cu -> Cu^2+ + 2e-}\tag1$$
and reduced, i.e. $\ce{NO3-}$. While balancing this reaction you have made a few mistakes. You cannot leave it out as a spectator ion, since some of it is involved in the reaction.
Consider the oxidation state in nitrate and nitrogen monoxide. How many electrons have to be transferred?

   three, nitrogen goes from $+5$ to $+2$ (you have only written one)

How many water molecules will be produced? How many protons are needed?

   Two water, four protons

Write down the full half reactions

   $$\ce{\overset{+5}{N}O3- + 4H+ + 3e- -> \overset{+2}{N}O + 2H2O}\tag2$$

Now you need to find the factors to multiply $(1)$ and $(2)$ with to balance out the electrons.

   $3\times(1)$ and $2\times(2)$

This results in the following equation

   $$\require{cancel}\ce{Cu + 2NO3- + 8H+ + \cancel{6e- } -> 3Cu^2+ + \cancel{6e- } + 2NO + 4H2O}$$

How many more nitrate ions do you need to charge balance the protons? What is the final balanced equation?

   Six more nitrate ions, giving a total of eight
 $$\ce{Cu + 8HNO3 -> \underbrace{3Cu^2+ + 6NO3- }_{3Cu(NO3)2} + 2NO + 4H2O}$$


Is the behavior of a dilute strong acid the same as the behavior of a weak acid?

The pH is the same, yes, but the weak acid has undissociated molecules "in reserve", so to speak, and thus can neutralize a lot more of any base.

Indeed, this concept of weak acids and their conjugate bases having molecules "in reserve", reacting only when called upon, is the principle behind acid-base buffers.


Can we predict Dextrorotatory and levorotatory character of an organic molecule?

There is no simple set of rules that one can use to determine the absolute configuration from the specific rotation, or the other way around.

A quick web search did dig up this reference that deals with calculating the specific rotation for arbitrary molecules, but it requires some fairly sophisticated algorithms. In principle, you could reproduce their methods, but it would require a substantial time commitment unless you are familiar with the software they describe. (Which is Gaussian, if you're interested.)


HDEHP and complexes formed in extraction

We need some more details such as what diluent you used. Maybe you were using pure DEHPA as the organic phase which is something which I would suggest that you do not do.

I would like to know if you did triplicate experiments, sorry if I offend you but I have no idea how good you are at the measurement of D values by solvent extraction. I like to think I am good at it, but I always use triplicate experiments. I will shake three different vials for each set of conditions. This allows me to see how reproduceable my work is, it also guards against me doing something stupid with one of the vials.

If you only do single extractions then it is possible that a random error causes the experiment to give misleading results. I hold the view that a beginner needs triplicate shaking experiments more than anyone. When you are new to solvent extraction you are more likely to make errors, also when you have done it for a while you might get careless and then you would need the triplicate experiments again to guard against mistakes.

The triplicate experiments will not prevent you making mistakes, but they will help to alert you to the fact that something has gone wrong.

I also worry about the use of very low acid concentrations, the Eu will tend to form hydroxides.

I think that you need to pay attention to the activity coefficents. I would be much more happy if you had used sodium nitrate to keep the ionic strength of the aqueous phase constant. I would want to do the experiment with a constant nitrate concentration to allow you to separate the effects of changing the nitrate concentration from that of the solvated protons.

I would like to point out that at very high nitrate concentrations the extraction mechanism for lanthanides with DEHPA can change from the normal ion exchange mechanism to a solvation mechanism which is similar to the extraction of plutonium from nitric acid by tributyl phospahte (PUREX process),


I dunno if the author meant exactly this, but it is a well-known fact that pairing of electrons actually consumes energy.

For example, free nitrogen atom in ground state has three unpaired electrons. To form triple bond in dinitrogen molecule, three pairs of electrons must be, well, paired. The process of pairing does consume energy.  However, the energy freed in the bond formation is so big, that it overwhelms pairing energy by huge margin. On the other hand, in oxygen molecule two highest-energy electrons has two orbitals to occupy, and they stay unpaired, so dioxygen is paramagnetic.

So, by no means it is the fact of being unpaired that drives electrons to form bonds. In fact, it prevents formation of weak bonds. However, when two atoms come closer to each other and have two half-occupied orbitals that can form a bond, the fact that electrons may be paired allows this two orbitals to combine into occupied bonding orbital, forming a bond.


Why is this so?

Is this so?

Following is a tabulation of the ionization potentials (IP) and electron affinities (EA) for some alkali metals and halogens (kJ/mol):

\begin{array}{|c|c|c|c|c|c|c|}\hline
& IP & EA & && IP & EA          \\ \hline
F & 1681 & 331 && Li & 520 & 60  \\ \hline
Cl & 1251 & 352 && Na & 496 & 53 \\ \hline
Br & 1140 & 328 && K & 419 & 49  \\ \hline
I & 1008 & 298 && Rb & 403 & 47  \\ \hline
\end{array}

A good reducing agent readily donates electrons and would be expected to have a low IP.  Sodium would be expected to act as a good reducing agent and chlorine as a poor one, and, indeed the IP of sodium is much lower than that of chlorine.

A good oxidizing agent readily accepts electrons and would be expected to have a large EA.  Chlorine would be expected to act as a good oxidizing agent and sodium as a poor one, and, indeed, the EA of chlorine is much higher than that of chlorine.

This NIST webpage tabulates experimental IP's and EA's for the methyl radical.  The IP is given as (9.84 eV) 945 kJ/mol and the EA appears to be less than (0.6 eV) 58 kJ/mol.

Let's compare the IP and EA of the methyl radical to values in the above table for strong oxidizing and reducing agents.  The IP of the methyl radical is significantly larger than those reported for the alkali metals and a bit less, but relatively close to the IP's for all of the halogens except fluorine. This comparison suggests that methyl radical is, at best, a weak reducing agent. The EA of methyl radical is significantly less than those reported for the halogens and right around those reported for the alkali metals. This comparison suggests that methyl radical is, at best, a very weak oxidizing agent.


Why do surfactants lower the surface tension

We usually say that at an interface, molecules will orient themselves at a "preferred" position, but we usually ignore that such a "preferred position" can actually be their "least unfavorable" position.

Energetically speaking, the interaction of a lipid chain with a gas is less unfavorable than interaction of a polar group with gas, because interactions from a lipid chain is mainly based on van der Waals interactions which are quite weak compared to polar (i.e. electromagnetic) interactions with a polar group. Also, they cannot make stabilizing bonds with the medium, i.e. water usually.

It must also be noted that the surface tension is linked to the interactions of molecules at the interface.

So "why do surfactants lower the surface tension"?

If you add some "surfactant", aka "tensioactive" agents, in water, its "least unfavorable" position would be at the interface between water and the other medium (usually gas). It will bring down the cohesive energy at the surface, compared to water alone, because it relies on van der Walls interaction. And that is why it lowers surface tension.


In theory, yes it is possible. In practice, not really.

First of all, the anorthite on the moon is never pure anorthite ($\ce{CaAl2Si2O8}$) but rather an anorthite-albite ($\ce{NaAlSi3O8}$) solid solution. That is, the single crystal contains both components. The diagram you added there, where it says "mole %"? The other stuff is albite.

Now, back to glass making. Just about any crystallite solid silicate can be made into a glass. The process is simple: heat it up above the melting point, and then cool it down fast enough so it quenches to glass instead of crystallising. Let's look at these points in details.

Melting the stuff: Anorthite has a very high melting point: 1553 °C. Forsterite (the Mg-rich end member of the olivine group) has an even higher melting point: 1890 °C. To melt the stuff you're going to need a furnace that can go that high, and your energy bill is going to be extremely high. Here's a liquidus phase diagram of the $\ce{CaAl2Si2O8-Mg2SiO4-CaMgSiO6}$ system so you can get an idea of the temperatures required:

Luckily, your material is not pure anorthite. It also has albite! That's good because it takes the melting point down:

However, 1100 °C is still pretty high. The glasses that are commonly used in industry today have melting points of below 1000 °C. This is achieved by having high contents of sodium in the glass, which is perfect in taking the melting point of the material down. However, materials on the lunar surface are not that rich in sodium, and that's a problem.

Adding iron to forsterite (i.e. the fayalite component of the olivine group) will take the melting point of it down, but pure fayalite melts at around 1200 °C which is no good, and the iron will colour the glass green to dark brown (seen as black if thick enough), depending on how much oxygen is around. More about iron later.

Cooling the stuff down: Silicate glasses that we know and love contain abundant silica, commonly above 70%. This makes the glass very viscous because of Si-O polymerisation. A viscous glass can be easily left to cool in air (by the way - not readily available on the moon) and it will remain glass. Cooling a non-viscous glass that has below ~50% silica in air will cause crystals to form in it (anorthite in your case) and it will not be perfectly clear anymore. Therefore, you need to cool it very rapidly, by quenching in water. The problem now is that the glass is very brittle and will most likely crack during cooling or very soon after (this is from my own personal experience in trying to make glasses of these compositions).

So far we have two problems: melting it, and cooling it. There are other problems. You were worried about impurities and rightfully so. Iron is your biggest enemy. Even trace amounts of iron (< 1%) can make your glass dark green or brown. Not only that, if there is oxygen around you might even have magnetite crystals in your glass and it's going to be completely opaque. You need a way to get rid of the iron. Nickel is also a minor problem. As for other "impurities", such as Al, Na, K, Ca, Mg, that's not a problem. They don't give any colour to the glass. It's the transition elements that you should fear.

Yet another problem is that you are not going to get pure minerals. The lunar surface does not contain blocks of anorthite or forsterite or anything else. The regolith is a powder made by weathering down of the minerals, and it contains a horrible mixture of everything. If you want to separate the anorthite or transition-metal-free stuff, you will need to figure out a way to do it.


What all information can be gathered qualitatively from the radial distribution functions of a molecule?

You can get alot from the RDF... an amazing amount actually. It is true though that people like printing out the RDF and thinking their paper now has substance, but there is so much more to the pair correlation function.

Integral equations are based on g(r). Kirkwood-Buff solution theory is too. Kirkwood-Buff solution theory will make you dizzy with what it can calculate.

See wikipedia for more on Kirkwood-Buff solution theory. See Arieh Ben-Naim's Molecular Theory of Solutions for the real deal explanation. What a good read. I get excited everytime I get to link it. Ben-Naim, amazing book


what actually happens inside the battery once it is fully discharged?

Alkaline batteries use the exchange of electrons from zinc to manganese dioxide to produce electricity

$$\begin{array}{rcl}\\
\ce{Zn (s) +2 OH- (aq)}& \ce{->}& \ce{ZnO (s) + H2O (l) + 2 e-}\qquad \text{(anode)}\\
\text{(high potential anode)}\quad \ce{2e-}&\ce{->[load][\ ]}&\ce{2e-}\quad\text{(low potential cathode)}\\
\ce{2MnO2 (s)+ 2e- + H2O (l)} &\ce{->}&\ce{Mn2O3 (s) + OH- (aq)}\qquad \quad\text{(cathode)}\\
\end{array}$$

typically a battery voltage would decline according to:

$$\mathscr{E} = \mathscr{E}^{\circ} - \frac{RT}{nF}\ln Q$$

but in alkaline batteries all of the products and reactants are solids, which without going into heavy detail of the thermochemistry means that $Q = 1$ thus $\frac{RT}{nF}\ln Q = 0$ which means cell potential remains fairly constant ($\mathscr{E} = \mathscr{E}^{\circ}$). But this is only in theory as we can see alkaline batteries do lose voltage:

This is partly because the internal resistance of the cell will increase with time meaning that for a given external load the total load will increase with time reducing current and apparent voltage. But even with no load voltage drops, the the zinc oxide will dissolve in alkaline medium meaning the reactions depart from ideal equations:

$$\ce{ZnO (s)  ->[+ H2O (l)] Zn(OH)2 ->[+2OH- (aq)] [Zn(OH)4]^2-}$$

this is not to say that $\mathscr{E} = \mathscr{E}^{\circ} - \frac{RT}{nF}\ln Q$ is an unreliable equation, but it does not account well for the complexity of the system and does not apply well to alkaline batteries. Additionally it cannot explain the vertical slope of the discharge curve. What can explain the vertical slope is that once all of either zinc or manganese dioxide is consumed, there is no more matterial to react and the battery is considered fully discharged.


What is the meaning when it is said "Temperature of an Oxyacetylene Torch is 3500 degrees Celsius"?

Oxyacetylene is a mixture of acetylene (ethyne) and oxygen (not air). The mentioned temperature of about $T=3500\ \mathrm{^\circ C}$ is the adiabatic flame temperature for the combustion reaction of acetylene and oxygen.

An adiabatic process is one that occurs without transfer of heat. The adiabatic flame temperature is a theoretical temperature, which would be achieved in a combustion system in which there are no heat losses (e.g. no radiation losses from the flame). Note that this cannot be achieved in a real flame; therefore, adiabatic flame temperatures are calculated values.

The heat $Q$ released from the combustion reaction of acetylene and oxygen is the heat of combustion. If all of the energy released by this chemical reaction is used to increase the temperature $T$ of the products ($\ce{CO2}$ and $\ce{H2O}$) with no heat losses, the resulting adiabatic flame temperature can be calculated from the heat capacity $C$ of the products.

Hence, the adiabatic flame temperature represents the maximum possible theoretical temperature for a particular fuel/oxidant combustion.