What is the group reagent for the zero group (ammonium cation) in inorganic salt analysis?

Ammonium salts are identified by the release of ammonia upon warming with base.
$$\ce{NH4+ + OH- ->[\Delta] NH3 ^ + H2O}$$
Vogel's Qualitative Inorganic Analysis 5th ed., p 400 writes

6. Test for ammonium ions Boil a little (c. 0.1 g) of the substance with sodium hydroxide solution. The evolution of ammonia, detected by its odour and its action upon red litmus paper and upon filter paper soaked in mercury(I) nitrate solution,* indicates the presence of an ammonium salt.
* Drop-reaction paper, treated with tannic acid and silver nitrate solution, may also be used.


Define basicity. For Brønsted-Lowry, it's all about $\ce{H+}$ and the conjugate bases. In Brønsted theory, acceptance of a hydrogen ion is what a base does. If you didn't know before, you should know that the hydrogen halides acid strength is $\ce{HF < HCl < HBr < HI}$. So for the conjugate bases, you just flip the sign $\ce{F- > Cl- > Br- > I-}$.

You way overthought the question. You should also think about where you got this idea about "electron donation". (Lewis acid-base theory deals with electron pair donation/acceptance but there's a huge difference between electron donation and electron pair donation.) You claim for some species $\ce{X-}$, that if $\ce{X- -> X^0 + e-}$ easily, then $\ce{X-}$ is a stronger base (than some $\ce{Y- -> Y^0 + e-}$ which is less easy). (By 'easily', assume I mean in the thermodynamic (energetic) sense)

Well, take a look at that and ask yourself whether it is relevant to Brønsted acid-base theory. I say no, it's not. Whether it is true or not, it is not relevant to the base strength as defined by Brønsted-Lowry (at least it has no clear, direct relevance).


Separation techniques for different organic compounds

You have correctly identified the carboxylic acid in pair 1. The second conpound is 9-chloroanthracene. Calling it a chlorodiene isn't correct. It's an aromatic compound. Anyway, your conclusion to use acid-base extraction is reasonable.

In pair 2, you seem to overestimate the polarity of the first compound. But there is another, more striking difference between the dipentyl ether and n-hexane.

   Dipentylether, $\ce{C5H11-O-C5H11}$, has a much higher molar mass than n-hexane, $\ce{C6H12}$. As a rule of thumb, you can assume that the compound with the higer molar mass will have a much higher boiling point. (There will be exceptions when strong interactions between molecules come into play, such as in the case of hydrogen bonds. But this isn't the case here!) If you look up the boiling points (bp) for both compounds, you'll see that our rule of thumb is supported here. The bp of n-hexane is 68 °C, that of the much heavier ether is > 180 °C. Consequently, distillation (under reduced pressure) is the way to go here.

In pair 3, you seem to misinterpret the nature of the second compound. It's not a diene, but an aromatic hydrocarbon named naphthaline. You might want to read about the properties of these compound and/or the comcept of aromaticity some other time. But for now, let's concentrate on another detail in the structure of the first compound in this pair.

   There is an $\ce{OH}$ group in the the first molecule. Here, the term polar is justified! Think about the typical setup of column chromatography. The stationary phase is usually silica (or aluminium oxide), while the mobile phase typically consists of mixtures of not-so-polar organic solvents. Compound 1 can strongly interact with the silica through hydrogen bonding, while compound 2 is an unsubstituted aromatic hydrocarbon. Separation by colunm chromatography is a very promising method for this pair.


What has stronger antioxidant potential: BHT or BHA?

That's a good question. In literature there are some contrasting opinions about the antioxidant power of BHA vs BHT. TBHQ (2-tert-butyl hydroquinone) is also used as antioxidant, although maybe is less common.
There is a paper from 1997 in which the antioxidant activity of BHA, BHT and TBHQ are examined with Miller's test (M. Karamac and R. Amarowicz, Grasas y Aceites, 1997, 48 (2), 83-86), and in this test BHA and BHT gave similar results, although in the initial phase of the experiment BHA seemed to have a higher antioxidant activity than BHT.
In the paper precedent studies are also reported. Quoting from the paper:

Higher antioxidative activity of BHA than of BHT and lower for TBHQ was
demonstrated in thin layer model system of methyl linoleate on cellulose (J. Am. Oil Chem. Soc. 52,174-178). Berner et al. (J. Am. Oil Chem. Soc. 51, 292-296) reported antioxidative index in lard emulsion 30-fold higher for BHA than for TBHQ. Yet, in active oxygen method, antioxidant index for TBHQ was 3-fold higher than for BHA and BHT. Higher antioxidative activity of TBHQ compared with BHA and BHT in model studies with soybean oil and chicken fat in emulsion versus dry oil tests was reported by Cort et al. (J. Am. Oil Chem. Soc. 52,174-178). Relative effectiveness of antioxidants-linoleic acid monolayer on silica gel for TBHQ was almost the same as for BHT, while for BHA almost 3-fold higher (J. Food Sci. 42, 1533-1535).


Hosoya Z Index and Correlation with Boiling Point

For a given hydrocarbon C$_n$H$_{n+2}$ more branched hydrocarbons have lower boiling point. This is because they have less surface area to have Van der Waals interactions to entangle with each other (ropes have more chances to entangle with each other than balls). For example C(CH3)4 (n-pentane) is a gas, but all other pentanes are liquids.

Their method is qualitative only. For instance, o-p-nitrophenol can be steam distilled and has much higher vapor pressure at a given temperature than p-nitrophenol, but the table would predict that they have the same properties. The difference is due to intramolecular h-bond present in o-, but ansent in p- isomer.


Octane is a specific molecule with 8 carbons normally found among the many different molecules in gasoline. Pure iso-octane (2,2,4-trimethyl pentane) was chosen as the reference fuel standard for resistance to preignition for gasoline fuels. So the octane number of a fuel is a rating of the preignition potential of that fuel compared to a fuel containing 100 %  iso-octane molecules. Since iso-octane is a mouthfull, the "iso" part was dropped and people just call it octane.  This is why some vendor calls his product petrol or gasoline or gas or octane or high test, etc...


Where can I find the total enthalpy and heat of dissolution in water of a given components?

It's very unlikely you actually need the enthalpy.  Like energy (of which it is a variation), absolute enthalpy is almost unmeasureable, normally one deals only in enthalpy changes.  In your case I would hazard a guess that you're  interested in things like dH_rxn, the heat of reaction for the decomposition of H2O2.  You can readily calculate this from the dH0 of the components.  The latter are called the "standard heats of formation" and they are combined with "Hess's Law."  Googling either of those terms should lead you to explanations of how the data you can find can be combined into the data you want. How exactly you do it will depend on what quantity you want, and the thermodynamic conditions (e.g. T and p) under which you want it, and the accuracy you need.

Yes, H is generally a function of pressure and temperature.  The starting search term there would probably be "derivatives of the enthalpy" because you would generally calculate H at one (T,p) from H0 at some other (T0,po). You'll find the derivatives can be related to material properties like the heat capacity, compressibility, and coefficient of thermal expansion.  For changes over significant temperature ranges these can be integrated, provided they are known as functions of temperature, or you are willing to use approximate models (like the ideal gas model) that have formulae.


Which of the following has highest solubility in water?

You can't predict solubility reasoning, you have to remember solubility rules in your case this one:

  Most hydroxide salts are only slightly soluble. Hydroxide salts of Group I elements are soluble. Hydroxide salts of Group II elements (Ca, Sr, and Ba) are slightly soluble. Hydroxide salts of transition metals and $\ce{Al^3+}$ are insoluble. Thus, $\ce{Fe(OH)3, Al(OH)3, Co(OH)2}$ are not soluble.

So in this case $\ce{Ca(OH)2}$ even if slightly soluble theoretically should be the right answer.


What properties a compound should have to be called an "ice"?

Ice is a term usually given to $\ce{H2O}$ in the solid state. Solid carbon dioxide, what is sometimes called "dry ice" is actually a misnomer — although the gas is frozen, it is not truly ice, in the chemical sense. It is called dry ice because instead of melting into liquid carbon dioxide (like actual ice, which is frozen water to liquid water), dry ice sublimates into it's gaseous state, skipping the liquid phase (almost — still goes through liquid phase, but this happens so fast it's basically negligible) entirely.

As for properties, you basically have all of them listed.


Using significant figures when measuring with an instrument with marks other than powers of ten

I think the rule of thumb for the "reading error" being $\pm\,0.1$ of the smallest division applies here, so for your case, it'd be $5\;\mathrm{mL}\pm\, 0.5\,\mathrm{mL}$. See this page, entitled "Math Skills - Scientific Notation", from Texas A&M. Specifically, it is noted that

  A rule of thumb: read the volume to 1/10 or 0.1 of the smallest division. (This rule applies to any measurement.) This means that the error in reading (called the reading error) is 1/10 or 0.1 of the smallest division on the glassware. If you are less sure of yourself, you can read to 1/5 or 0.2 of the smallest division.


Quick way to find magnetic nature of compounds or ions

Well, for some compounds it's easy, because there are an odd number of electrons - thus there's no way to pair them all.

Otherwise, there is no quick-fire way without doing the MO diagram.

Consider $O_2$ which has an even number of electrons, but is a ground-state triplet with two unpaired electrons. The only way to know is to look at the MO diagram.

If you practice, you'll realize certain patterns and develop more of an intuition.


Reactions that occur at the anode electrode of electrolysis dilute H2SO4

How to know that reaction that occur at the anode electrode ...

There are two kinds of anions: oxidizable (ones that contain central element in lower oxidation states), like $\ce{ClO3-}$, $\ce{Cl-}$, $\ce{NO2-}$, and unoxidaziable, where central element is in hightest oxidation state ($\ce{ClO4-}$, $\ce{NO3-}$. Fluoride obviously is unoxidizable as well.  As a rule of thumb, it is usually assumed that unoxidizable anions do not react in electrolysis. A rationalization for it is that in such cases an oxygen should be oxidized, and oxygen in water reacts faster. For the sake of basic chemistry course it is enough.

However, it is not entirely true: electrolysis of cold solution of ammonium sulfate in sulfuric acid results in formation of ammonium persulfate $\ce{(NH4)2S2O8}$, where anion has structure $\ce{ (O3SOOSO3)^{2-}}$ Similiarly, electrolysis of very cold solutions of carbonates produces various percarbonates. A synthetic method exists, that is based on electrolysis of carboxylic acid. The anions on oxidation form $\ce{RCOO\cdot}$, that further loose $\ce{CO2}$ and resulting radicals pairs, resulting in net reaction $\ce{2RCOO- \rightarrow R2 + 2CO2 + 2 e-}$ However, in general, formation of peroxocompounds requires low temperatures, usually below zero.

The exact mechanism of the process highly depends on the nature of the electrodes, but likely includes formation and transformation of surface particles on the electrode, making reasoning difficult and electrode-specific. Since commonly anodes used have oxide coating, it can be more-or-less safely assumed, that the process includes dissociative adsorption of water with subsequent oxidation of adsorbed (hydro)oxide atoms.


Weight Percent of Solutions Given Without Units

I have assumed this to mean: (mass of solute) / (mass of solution) * 100

Most likely you are correct in this assumption. Percentage compositions of ionic solutes in water are nearly always implicitly expressed in percent-by-weight ("% w/w").

$6\%$ bleach is relatively dilute and will have a density very near to $1\ \frac{\mathrm{g}}{\mathrm{mL}}$.  So, $1\ \mathrm{L} \approx 1\ \mathrm{kg\ solution}$, and thus one can expect to have approximately $60\ \mathrm{g}\ \ce{NaClO}$ per liter of such a solution.


To start off, all chalcogens are known to form binary hydrides of the form $\ce{H2X}$ which can then behave as binary acids, however weak or strong, by the equation $\ce{H2X <=> HX^- + H+}$.

Now to compare the strength of the acids. The most straightforward way of doing this is comparing gas phase acidities, which removes the complication of solvation effects. For this, we look at the proton affinities of the conjugate bases $\ce{HX^-}$. The definition of a substance's proton affinity is the $\Delta G$ when it is protonated in the gas phase (always an exergonic reaction for neutral species and anions), so gas phase acidity is simply the inverse reaction, with the opposite sign for the $\Delta G$. According to these two sources, the gas phase acidities of the hydrogen chalcogenides are:

$$\ce{H2O_{(g)} <=> HO^{-}_{(g)} + H^{+}_{(g)}}\ \ \ \ \ \ \ \ \ \ \ \ \Delta  G = +1635\  kJ\  mol^{-1},\ K^{25ºC} = 2.5\times 10^{-287}$$
$$\ce{H2S_{(g)} <=> HS^{-}_{(g)} + H^{+}_{(g)}}\ \ \ \ \ \ \ \ \ \ \ \ \Delta G = +1477\  kJ\  mol^{-1},\ K^{25ºC} = 1.2\times 10^{-259} $$
$$\ce{H2Se_{(g)} <=> HSe^{-}_{(g)} + H^{+}_{(g)}}\ \ \ \ \ \ \ \ \ \ \ \ \Delta G = +1417\  kJ\  mol^{-1},\ K^{25ºC} = 4.1\times 10^{-249} $$
$$\ce{H2Te_{(g)} <=> HTe^{-}_{(g)} + H^{+}_{(g)}}\ \ \ \ \ \ \ \ \ \ \ \ \Delta G \simeq +1386\  kJ\  mol^{-1},\  K^{25ºC} \simeq 1.1\times 10^{-243} $$

As you can see, when going down the group, deprotonation becomes less endergonic, so the binary hydrides definitely become stronger acids, even if against what one might expect from electronegativity considerations. Calculating the equilibrium constants allows a more direct comparison. Don't worry that the constants are all extremely small, what matters to us is the ratio of the equilibrium constants. Thus, in the gas phase, $\ce{H2S}$ is a whopping 28 orders of magnitude more acidic than $\ce{H2O}$, while $\ce{H2Se}$ and $\ce{H2Te}$ are 38 and 44 orders of magnitude more acidic than $\ce{H2O}$, respectively. It is certainly noteworthy that $\ce{H2O}$ is a much, much weaker acid than its heavier homologues.

With the data presented, we ask, why does acidity increase down the group? There are two main factors presented in different sources. One of them is that the chalcogen-hydrogen bond becomes weaker for heavier chalcogens, and this is ascribed to the poor overlap between the tiny $\ce{H}\ 1s$ orbital and the increasingly larger chalcogen orbitals which weakens the covalent bond and presumably makes ionization easier. This can be seen from the dissociation energies of the $\ce{X-H}$ bonds ($\ce{O-H} = 428\  kJ\  mol^{-1}$ | $\ce{S-H} = 344 \  kJ\  mol^{-1}$ | $\ce{Se-H} = 305\  kJ\  mol^{-1}$ | $\ce{Te-H} = 268\  kJ\  mol^{-1}$). The second factor is the size of the chalcogen atoms. The heavier atoms are larger, meaning they spread the same charge in the conjugate base $\ce{HX^-}$ over a larger volume, which makes reprotonation less favourable. This latter explanation seems to be the most common.

Up to now we've been talking about acidity in the gas phase, but now let's add the solvent, specifically water. The presence of a solvent stabilizes ions immensely relative to gaseous ions (especially protons), and water is a particularly good solvent for ions, so all the dissociation equilibria will be shifted massively to the right compared to the gas phase. Let us compare the acid dissociation constants in water:

$$\ce{H2O_{(l)} <=> HO^{-}_{(aq)} + H^{+}_{(aq)}}\ \ \ \ \ \ \ \ \ \ \ \ \ k_w^{25ºC} = 1.0\times 10^{-14}$$
$$\ce{H2S_{(aq)} <=> HS^{-}_{(aq)} + H^{+}_{(aq)}}\ \ \ \ \ \ \ \ \ \ \ \ \ K_a^{25ºC} = 1.0\times 10^{-7} $$
$$\ce{H2Se_{(aq)} <=> HSe^{-}_{(aq)} + H^{+}_{(aq)}}\ \ \ \ \ \ \ \ \ \ \ \ \ K_a^{25ºC} = 1.3\times 10^{-4} $$
$$\ce{H2Te_{(aq)} <=> HTe^{-}_{(aq)} + H^{+}_{(aq)}}\ \ \ \ \ \ \ \ \ \ \ \   K_a^{25ºC} = 2.3\times 10^{-3} $$

Some things to note:

All the binary chalcogen hydrides still increase in acidity after solvation when going down the group. This is not necessarily always the case. A classic example of relative acidities changing on solvation is the dimethylammonium ($\ce{Me_2NH2^+}$) and trimethylammonium ($\ce{Me_3NH^+}$) cations; in the gas phase the former is more acidic, but in aqueous solution the latter is more acidic as the $\ce{Me_3NH^+}$ ion is not as well solvated.
As I mention in the comments, aqueous hydrogen telluride ($pK_a=2.6$) is about as acidic as aqueous phosphoric acid ($pK_{a_1}=2.15$). Compare also with a common weak acid such as acetic acid ($pK_a=4.76$), a borderline strong acid such as nitric acid ($pK_a \simeq -1$), and a recognized strong acid such as hydrochloric acid ($pK_a \simeq -6$).
The $K_a$ of water is a little tricky, since it is both the reactant and the solvent. Its value depends on the definition used. I quoted the $k_w$, but many people use a $pK_a$ of 15.74 which implies $K_a = 1.8\times 10^{-16}$. The difference is not too important here.
$\ce{H2O}$ is a much stronger acid than expected from gas phase data, relative to its heavier homologues. This is ascribed to the exceptionally strong aqueous solvation possible only for $\ce{OH^{-}}$ among the listed conjugate bases, stemming from strong hydrogen bonding interactions, which helps weigh the equilibrium to the right side.


Effective diffusion coefficient of a species in a heterogeneous mixture

This greatly depends on macro structure. If the phases were say arranged in a sheet-like structure (like a diffusion front) then the low diffusivity phase would be the dominating determinate of the diffusion rate. For this macrostructure, if we assume steady state with no change in phase or length from diffusion or time, then mass transport of the bulk can be approximated by:

$$\frac{\mathrm dn}{\mathrm dt} = \frac{A \times \nabla C}{\sum\frac{L_i}{D_i}}$$

If the high diffusivity phase is continuous then that could be the dominating determinate of the diffusion rate. That does not mean it is easily calculable, just noting that dominating factors can change. If no phase is continuous or other phases are continuous then this also becomes convoluted. For non-sheet-like structures calculating diffusion requires a computer model of the grain structure and finite element analysis to resolve an approximate solution.


Why does a silanized glass surface become hydrophobic?

Water has many, rather strong intramolecular bonds called hydrogen bonds, which influences many of its properties, for example being fluid even at high temperatures.

Wheter some thing is hydrophilic/hydrophobic is influenced mostly by the availability of places where hydrogen bonds can happen, these are either
1. bonds between hydrogen and an element with high electronegativity (e.g. O; N)
2. an orbital already filled with 2 electrons.

Because water molecules are also polar, charged or polar particles are also more hydrophilic. (e.g. salts)
Hydrophobic substances simply lack those features.

A good example illuminating this is that even small alkanes are quite insoluble in water whereas starch, a large molecule with many hydroxy groups is still soluble.

Of course water "tries" to minimize  its  surface energy for example by forming spherical (least surface are) drops. If a surface has many possibilities for  bonds on the other hand, water wets it.

A silanized surface has many long alkane chains, and therefore few possibilities to form bond, ergo it is quite hydrophobic. (In contrary to the normal glass, which  has many atoms with (formal) charge.)


How to rationalise the difference in the melting points of acids and alcohols with inter molecular forces?

The ultimate goal of melting-point reasoning is to explain how intermolecular forces lead to better packing of molecules in the solid phase.
Both alcohols and organic acids have hydrogen bonds, but organic acids have stronger hydrogen bonds due to the withdrawing carbonyl group, $\ce{COO-}$. This causes the acids to interact stronger with each other. More importantly, organic acids form dimers (usually) with their hydrogen bonds (acetic acid dimer, for example). For this reason, the packing of the acids in the solid phase is significantly better than that of alcohol; hence the higher melting point. Acetic acid: $16~^\circ\mathrm{C}$ vs. $-114~^\circ\mathrm{C}$ for ethanol.


What is the principle of “scrubbing” in liquid extraction or solvent extraction?

"Scrubbing" in this context is the treatment of the organic phase to remove impurity metals. This is as opposed to "back extraction" which is to remove the metal of interest from the organic phase.


Maximum density of heavy water at 11.6 degree Celsius?

The maximum density of $\ce{H2O}$ occurs when cooled to about $\pu{4^oC}$. At this point, the thermal motion of the molecules has slowed down enough such that water molecules can start to orient themselves in a manner more resembling that found in the expanded crystalline ice structure, and density thus starts to decrease with decreasing temperature. This pre-freezing expansion happens at a somewhat higher temperature for $\ce{D2O}$ due to the greater hydrogen bonding strength of $\ce{D2O}$ at any given temperature.

Also note that $\ce{D2O}$ freezes at about $\pu{4^oC}$, so clearly the maximum density of $\ce{D2O}$ must occur at a greater temperature than that of $\ce{H2O}$. In the case of $\ce{H2O}$, the temperature of maximum density occurs at about $\pu{4^oC}$ above it's freezing point, where for $\ce{D2O}$ the temperature of maximum density occurs at about $\pu{7^oC}$ above it's freezing point.


Ok, after having poked around a bit I doubt that $\ce{Pb(CN)2}$ forms.

I'm guessing that you're collecting the froth to get a concentrate of the galena $\ce{PbS}$ out of some finely crushed ore.

An aspect of the operation seems to be chemically reducing the concentration of $\ce{ZnS}$ by dissolving via the following reaction:

$$\ce{ZnS(s) + 4NaCN(l) -> Zn(CN)4^{2-}(l) + 2Na2S }$$

Now there are two ways that $\ce{ZnS}$ could be mixed in with the $\ce{PbS}$.

(1) The individual $\ce{Zn^{2+}}$ cations replace individual $\ce{Pb^{2+}}$ cations in the $\ce{PbS}$. In this case the random distribution of the $\ce{Zn^{2+}}$ cations wouldn't allow much extraction of $\ce{ZnS}$ as $\ce{Zn(CN)4^{2-}}$. The $\ce{Zn^{2+}}$ cations just are not going to diffuse through the solid galena particles to any appreciable extent.

(2) There could be also be that the crushing creates a mixture of relatively pure $\ce{PbS}$ and $\ce{ZnS}$ particles which are mixed together. In this case an appreciable amount of the Zn impurity could be removed.

Now for what seems to be the question. $\ce{PbS}$ is very insoluble with a $K_\mathrm{sp} = 9.04\times10^{−29}$. I couldn't find a solubility product for $\ce{Pb(CN)2}$ but it is doubtful that the solubility would be enough to cause the lead sulfide to dissolve and then form lead cyanide.


What is the product of alpha decay of curium-226?

The question is flawed (possibly a misprint), your reasoning is sound (given the information provided to you).
Curium-226 does not exist, according to the Web Elements page on the isotopes of curium, the 'smallest' isotope is curium-240, which undergoes $\alpha$ decay to become plutonium-236.
$$\ce{^{240}_{96}Cm -> ^{236}_{94}Pu + ^{4}_{2}\alpha}$$
The most likely candidate of what the question should have been asking about is radium-226 decaying to radon-222,
$$\ce{^{226}_{88}Ra -> ^{222}_{86}Rn + ^{4}_{2}\alpha}$$
which is part of the uranium-series decay chain:

This is the most likely scenario, as the smallest isotope of thorium is thorium-227 (WebElements) and for radium, it is radium-223 (WebElements), both have too high a mass for anything with an atomic mass of 226 to directly decay into by alpha decay (as per the question).


How are the two formulas for enantiomeric excess/optical purity equal?

The equations are trivially equal since each is equal to the enantiomeric excess. To prove this, we need only show each equal to the enantiomeric excess.

Start with the second equation. Because both enantiomers have the same molar mass, we can divide both numerator and denominator by the molar mass of the compound and rewrite the equation as $$\text{EE} = \frac{|\text{difference in moles of enantiomers}|}{\text{total moles of enantiomers}} = |\text{difference in mole fractions of enantiomers}|,$$ which is the definition of enantiomeric excess. This equation is not very useful, since most of the time the masses of each of the enantiomers in solution are not known.

The first equation, as you have it, is incorrect. Following Ben Norris's suggestion, we write $$\text{EE} = \frac{|[\alpha]_\text{obs}|}{[\alpha]_\text{pure}} = \frac{|[\alpha]_\text{obs}|}{[\alpha]_\text{pure}} = \frac{|\chi[\alpha]_\text{pure}-(1-\chi)[\alpha]_\text{pure}|}{[\alpha]_\text{pure}} = |\chi-(1-\chi)|,$$
where $\chi$ represents the mole fraction of one enantiomer, and we once again obtain the same result.

Had your original equation been right, $L$ would have cancelled in your derivation and you would have obtained $(c_d-c_l)/(c_d+c_l)$, which is readily brought to equation $(2)$.


What is the difference between the reactivity of Lead subacetate and Lead II acetate undergoing a hydrothermal process?

Compared to lead(II) acetate, lead subacetate contains additionally $\ce{2 Pb(OH)2}$. The hydroxide can either be neutralized by protons from EDTA or take part in the hydrolysis of thiourea:

$$\ce{SC(NH2)2 + OH- \rightleftharpoons OC(NH2)2 + HS-}$$

$$\ce{HS- + OH- \rightleftharpoons S^{2-} + H2O}$$

The pH of the solution should be higher unless all hydroxide is neutralized or consumed by the hydrolysis reaction.


Why do boiling points of alkanes and alcohols become closer with increased chain length

It is simple. Longer the chains higher the similarity between them. In other words the OH group are getting "diluted".
It is more difficult for the donor to find the accepting counterpart, and the number density of the hydrogen bonds, both for volume and mass, gets lower.
The interaction between molecules becomes dominated by VdW forces.

To visualize this think of methanol and of a very very long primary alcohol. In the second case it is somehow "more difficult" to spot the difference versus the parent alkane.


Energy profile of cyclohexene conformations

I found the answer in the book: "Practical Aspects of Computational Chemistry"
. In chapter 19: Unusual Properties of Usual Molecules. Conformational Analysis of Cyclohexene, Its Derivatives and Heterocyclic Analogues.

The value for ring inversion of cyclohexene is smaller than the one of cyclohexane (which is near 10 kcal/mol):

  Value of ring inversion barrier is within 4.2–
  7.9 kcal/mol for molecular mechanics and 5.2–6.6 kcal/mol for quantum chemical
  methods.

The inversion proceeds as follows:

  The ring inversion process in cyclohexene includes three stage. The first
  stage involves the transition from an equilibrium half-chair conformation to a twist-boat conformation with a significant increase in energy. The second stage
  may be described as a very easy transformation of one twist-boat conformation to
  another via boat almost without any changes in energy. The third stage includes the
  transition from a twist-boat conformation to a half-chair conformation accompanied
  by a significant decrease in energy. The difference in energy between the twist-boat and the boat
  conformation is less than 0.01 kcal/mol. The boat conformation
  is just central point of this plateau and it is only the most convenient point for a description of the saddle point for the ring inversion
  .

The interconversion diagram should be something like this:


How to properly work with potassium HMDS

If you would look up the properties of KHMDS from a chemical supplier, you will find it will react violently with with water. So do keep it under inert atmosphere.

As for solubility, it is a non-nucleophilic base, so probable no substitution and addition like reactions will occur. I wont suspect it will be a problem there.

A comment on the use though. If you plan to use this, explore the dangers and possibilities of a substance like this a bit further than asking it here. For safety considerations. Look up the MSDS sheets from your supplier and be sure to only use this in a properly equipped lab.

Hope this helped.


Biopolymers as alternative for PVC and PU?

Ethylene vinyl acetate (also known as EVA) is the copolymer of ethylene and vinyl acetate and is a replacement for PVC.

Biomass > (fermentation) > Bioethanol > Ethene

Ethylene + acetic acid + 1/2 O2 → Vinyl acetate + H2O

Tung oil can be used to create a plastic like water proofing film as a replacement for PU


Is it possible to make a fuel cell that runs on triglyceride?

It is certainly possible to oxidize fats electrochemically, but there are a few problems with the idea that I can think of:

Triglycerides do not flow freely through the blood as they are too hydrophobic and are instead carried by lipoproteins that would have to be removed. Similarly, free fatty acids are transported bound to serum albumin.
The blood concentration of triglycerides and fatty acids is relatively low most of the time in healthy individuals.
Existing fuel cells that work on fatty acids seem to be mostly microbial fuel cells, which are quite slow.

I think it would be more practical to use glucose from blood instead, using an enzymatic fuel cell, as blood glucose levels are relatively high and the oxidation is easier to accomplish (though there are still a lot of technical challenges to be resolved—biofouling, efficiency, device lifetime, etc.).


Will a container made of ice hold an alcoholic beverage that is below freezing in an environment that is also below freezing?

Depends on the temperature and alcohol content. At 0°C, the container will definitively melt.

Below, your container will melt from the inside until the liquor is diluted so much that its freezing point is raised to the current temperature. (Or the other way round, water ice precipitates from your booze, concentrating it until its freezing point has dropped to the temperature of the surrounding.)

You can easily calculate the equillibrium alcohol content for a given temperature, https://en.wikipedia.org/wiki/Freezing-point_depression#Calculation
(The result will be slightly off, because this calculation is derived for infinite dilution.)


Why are hydrogen bonds in an antiparallel beta sheet stronger than those in parallel beta sheets?

They are probably not. The difference is small, and depends on sequence context. Also, the diagrams do not reflect the typical conformation of the backbone in beta sheets. To complicate matters, most sheets are mixed rather than purely parallel or anti-parallel, and many are irregular (different length of strands, bulges, one strand part of two sheets etc).

"hydrogen bonds are aligned directly opposite each other, making for stronger and more stable bonds."

No, the geometry around the hydrogen bond is fairly similar. Here is a depiction of structures from an ab initio study showing the geometry:

[OP ..] shouldn't the ideal, strongest, H-bond is orientated at an angle (specifically 60∘) relative to the C=O bond?

Based on analyzing known structures with a single hydrogen bond for a given acceptor, the hydrogen lies in the plane of the carbonyl group but has a fairly smooth distribution with many examples having a 180 degree angle (carbonyl carbon, carbonyl oxygen, hydrogen).

A possible explanation would be that the bond length in parallel strands would be longer, thereby weakening it. But that's not how most textbooks explain it, so what am I missing out here?

The ab initio study talks about the primary effect (having a hydrogen bond), secondary effects (repulsion between partial charges of oxygen atoms, repulsion between partial charge of hydrogen atoms, C-H...O hydrogen bond in the parallel sheet) and tertiary effects (induced dipoles due to formation of hydrogen bonds). You have to consider all of these; as the differences in hydrogen bonding geometry are subtle, the secondary and tertiary effects will become significant.


What if solute and solvent exist in equal quantities?

As far as I can recollect, I think in such a situation you can arbitrarily select one of them as solute and one as solvent. However, according to convention, it is better to assign the solvent as a substance which is usually used as a solvent in other situations.


Why does formation of bond releases energy and where does that energy come from?

It's a matter of perspective. You say you understand why bond breaking requires energy. In other words:
$$\ce{A2 + energy -> 2A}$$
But consider the reverse reaction:
$$\ce{2A -> A2 + energy}$$
This is also valid and can be understood in the following way: Bonds stabilize the electrons. In the unbound or atomic state, the electrons have a bit more energy. You can consider that atoms have some potential energy, like a ball on a hill. By forming a bond, the atoms "go down the hill" into a more stable state and thus lose this potential energy. This energy must go somewhere, and it is thus released in one form or another (heat, light, etc.) To break the bond is to push the atoms "up the hill" again, which requires energy.


Yes, the Ksp for $\ce{PbCl2}$ is ~$10^{-5}$ while for $\ce{PbS}$ is ~$10^{-28}$

So if the concentration of $\ce{Pb^2+}$ is too low to detect by dilute $\ce{HCl}$ precipitation, it is still possible it could be detected as the sulfide precipitate.


In an isolated molecule /atom the intrinsic width of the transition is limited by the lifetime of the upper level and this can be huge ( seconds) and so the line-width can be minuscule by time-energy 'uncertainty'. If the decay is exponential the spectral line shape is Lorentzian by Fourier Transform theory.

In a medium, the surrounding solvent or gas molecules repeatedly interact with our victim molecule by collisions and by electric and magnetic dipoles etc, and do so  in a random fashion and thus cause fluctuation in its energy levels. These changes are then observed in the transition energy and so broaden the observed transition in the ensemble of molecules. So far this is quite general. This paper examines one model of this, linear + quadratic coupling and the quadratic part is presumably why the peak is split - but I only skimmed the paper so you will have to check this.  The detailed maths is just the technical part, ignore this and try to get to the basic physics first.

(The term 'thermal bath' is just jargon for all the random interactions from solvent/buffer gas surrounding the victim molecule, 'modulation' just means changes, so the first sentence could read 'We describe how random interactions with a solvent or buffer gas changes the observed spectroscopic line shape ....')


Carbamates are usually reduced to N-methyl groups. There are numerous examples:

J. Am. Chem. Soc. 2012, 134 (16), 6936–6939

Org. Lett. 2012, 14 (18), 4834–4837

But it is not always a given. In this next example, the nitrogen is part of a three-membered ring (aziridine). These nitrogens are better leaving groups than usual, cf. Ketone/aldehyde synthesis from N-acylazetidines or aziridines where the same kind of reactivity is observed:

Angew. Chem. Int. Ed. 2002, 41 (24), 4683–4685


Does Cl- Have more -I effect or +M effect ( Resonance effect)?

Chloro group is considered a deactivating group according to Hammett plots of each possible reaction. The $\sigma_\mathrm{para}$ of $\ce{Cl}$-substituent is listed as $+0.227$ while $\sigma_\mathrm{meta}$ of $\ce{Cl}$ is listed as $+0.373$. It is known fact that $\sigma_\mathrm{meta}$ is an indicative of how much inductive effect contribute to the reaction, while $\sigma_\mathrm{para}$ is a combination of the strength of both inductive and mesomeric effects (more resonance effect than inductive because the substitution is one carbon away compared to meta-position). For comparison, when the substitution is $\ce{H}$, both effects are zero while when the substitution is $\ce{NO2}$ group, $\sigma_\mathrm{meta}$ and $\sigma_\mathrm{para}$ are $+0.710$ and $+0.778$, respectively (resonance effect predominates). Based on Hammett's original definition, when $\sigma$ of any group has positive sign,it is electron withdrawing group (EWG) and the numerical value is an indication of hoe big electron withdrawing ability compared to that of $\ce{H}$ (value of which is always zero). Note that Hammett has assigned these sign and value according to his research on the acidity of substituted aromatic carboxylic acids (EWG substitution makes acid stronger).
Contrary to carboxylic acids, EWG substitution at aromatic nucleus makes amino bases weaker because EWG dilutes the electron density on active $\ce{N}$ atom by inductive and/or resonance effects. Therefore, largere the positive number of $\sigma_\mathrm{meta}$ and $\sigma_\mathrm{para}$ of EWG, weaker the base. Also keep in mind that $\sigma_\mathrm{meta}$ and $\sigma_\mathrm{para}$ of $\ce{NH2}$ group are $-0.161$ and $-0.66$, respectively. Hence, basic strength of given benzamidine are decreased in order of their 4-substitution of $\ce{NH2} \gt \ce{H} \gt \ce{Cl} \gt \ce{NO2}$ meaning your correct answer is $\bf{(D)}$. It is noteworthy that this order remains the same even if these substitutions are at 3-position of the aromatic nucleus (based on $\sigma_\mathrm{meta}$ values).
I can't really find the $\mathrm{p}K_\mathrm{a}$ of relevant compounds. However, I found $\mathrm{p}K_\mathrm{a}$ values of some guanidines, $\ce{X-C6H4-N=N(N(CH3)2)N(N(CH3)2}$ (parent compound: 2-phenyl-1,1,3,3-tetramethylguanidine, $\ce{C6H5-N=N(N(CH3)2)N(N(CH3)2}$) which would prove my point (Ref.1):
$$
\begin{array}{c|ccc}
\hline
\ce{X} & \sigma_{(m,p)}^a & \mathrm{p}K_\mathrm{a} \ \text{of 2-Ar-guanidine} & \mathrm{p}K_\mathrm{a} \ \text{of aniline}^b  \\
\hline
\ce{H} & 0 & 12.18 & 4.62 \\
\text{3-}\ce{CH3} & -0.06 & 12.25 & 4.70 \\
\text{4-}\ce{CH3} & -0.14 & 12.37 & 5.11 \\
\text{3-}\ce{OCH3} & 0.11 & 11.96 & 4.23 \\
\text{4-}\ce{OCH3} & -0.28 & 12.57 & 5.34 \\
\text{3-}\ce{Cl} & 0.37 & 11.47 & 3.52 \\
\text{4-}\ce{Cl} & 0.24 & 11.70 & 3.98 \\
\text{4-}\ce{NO2} & 1.26 & 9.78 & 0.99 \\
\hline
\end{array}\\
^a: \text{Ref.2 and } ^b: \text{Ref.3.}
$$

References:

Przemyslaw Pruszynski, "Synthesis and properties of phenyl substituted derivatives of 2-phenyl-1,1,3,3-tetramethylguanidine," Canadian Journal of Chemistry 1987, 65(3), 626-629 (https://doi.org/10.1139/v87-107).
D. D. Perrin, Boyd Dempsey, E. P. Serjeant, In $\mathrm{p}K_\mathrm{a}$ Prediction for organic acids and bases; Chapman and Hall: London, England, 1981 (ISBN 978-94-009-5885-2).
D. D. Perrin, In Dissociation constants of organic bases in aqueous solution, Volume 1; International Union of Pure and Applied Chemistry. Commission on Electroanalytical Chemistry, Butterworths: London, England, 1965 (Suppliment 1972)(ISBN-13: 978-0080208275).


Which alcohol will most easily react with HCl to form an alkyl halide?

Tertiary alcohols have greater reactivity with hydrogen halides than secondary alcohols, which in turn have comparatively more reactivity than primary alcohols. For tertiary alcohols, the reaction proceeds by an SN1 mechanism which prefers a more substituted alcohol.
Therefore answer is C


How can diesel fuel produce less carbon monoxide but more carbon particulates?

It seems that the mechanism of particulate formation is not yet well known. This article (preview) states some basics about the composition and formation of Diesel Particulate Matter (DPM):

  Diesel particulates form a very complex aerosol system. Despite considerable amount of basic research, neither the formation of PM in the engine cylinder, nor its physical and chemical properties or human health effects are fully understood.
  [...]
  The basic fractions of DPM are elemental carbon, heavy hydrocarbons derived from the fuel and lubricating oil, and hydrated sulfuric acid derived from the fuel sulfur. DPM contains a large portion of the polynuclear aromatic hydrocarbons (PAH) found in diesel exhaust

Also, it seems that the particulates are formed both during combustion and from gas during dilution. During combustion, carbonaceous agglomorates (soot), metallic ash from lubricating oil, and precursors to particle formation during dilution (sulfur oxides and partially burned hydrocarbons from fuel and lubricating oil) are formed. During dilution, materials such as sulfuric acid and other sulfates, as well as heavy hydrocarbons and hydrocarbon derivatives, are for the most part adsorbed onto the carbonaceous agglomorates formed during combustion. Some material may also nucleate by itself, but most nucleation occurs on the agglomorates.

PDF with explanation of ultrafine particle formation mechanisms

Unfortunately, i found it difficult to find any reason why this is such a big problem with diesel and not with gasoline. Perhaps the composition of the diesel is to blame, not temperature or oxygen supply.


Trifluoroacetic acid, Formic acid and dichloromethane

How to correctly wash the glassware after use?

The glassware must be rinsed with a non-toxic (or less-toxic) solvent such as acetone or isopropanol before washing to carry over the bulk of hazardous materials.  You should the wash the glassware with soap and water, rinse with water, followed by a rinse with DI water, isopropanol, or ethanol to eliminate water spots.

  How to correctly dispose of wasted solvents?

Your solvents should go into an appropriate waste container and be marked as hazardous waste with the contents listed.  You also need to list the companies EPA number to dispose of the waste.  Have a reputable waste disposal service receive your waste and take it to a disposal facility.

  Should I leave the chemical fume hood on after using it and, if so, for how long?

If anything hazardous is still in the hood you must leave it on. If there is nothing in the hood there is no reason you must leave it running.


Freeware that converts structure in a conformational model?

Short answer: yes, the dioxane has a minimum when it's in a twist-boat form.

I use Avogadro which offers a "auto-optimize" tool. You can grab atoms and throw them around to force rings into different conformations while the force field is minimizing.

So here's what I did.

File -> Import -> Fetch by chemical name: (2R,4R)-2-isobutyl-4-methyl-1,5-dioxaspiro[5.5]undecane. This grabs the 3D geometry from the NIH Resolver (i.e., PubChem)
Auto-Optimize Tool: Force Field MMFF94

Then I started minimizing and looking at the energies. I pulled various atoms around to try to minimize steric clashes and flip rings from chair to boat, etc.

For the figures that follow, I've removed the hydrogens for clarity, but they exist as far as the energies and minimization. I just removed them to save the graphic.

Chair/Chair: 24 kJ/mol

Twist-boat/Chair: 14 kJ/mol

So yes, there's a decent stabilization in the twisted form of the twist-boat. You have to get the hydrogens right, or you'll end up with a bunch of eclipsed interactions between the dioxane oxygens and hydrogens on the cyclohexane.


What are B and n in this expression for charge repulsion?

There isn't an exact form for the Pauli Repulsion term, but it is often approximated as a $\frac{1}{r^n}$ term for simplicity.  For example, in the Lennard-Jones expression for Van Der Waals interactions, $$E(r) = \epsilon \times[(\frac{\sigma}{r})^6-(\frac{\sigma}{r})^{12}]$$, the $\frac{1}{r^6}$ term is the London force term (which in reality, does go as $\frac{1}{r^6}$), and the $\frac{1}{r^{12}}$ is the Pauli repulsion term (just an approximation, but in this approximation, $n=12$).
In the OPLS force field (which uses a Lennard-Jones expression for Van Der Waals interactions), the parameters for $\ce{Na+}$ and $\ce{Cl-}$ are:

$\ce{Na+}$: $\sigma = 3.500 A$, $\epsilon = 0.0660\ \ce{kcal/mol}$

$\ce{Cl-}$: $\sigma = 4.180 A$, $\epsilon = 0.11779\ \ce{kcal/mol}$.

OPLS is parameterized for a geometric mixing rule, meaning that for an interaction between the two, the $\epsilon$ and $\sigma$ parameters are equal to the geometric mean of the parameters for each atom.  Thus, for an $\ce{Na+ - Cl-}$ interaction:

$\sigma = 3.8249 A, \epsilon = 0.08817\ \ce{kcal/mol}$

Now, the Pauli part of the Lennard Jones expression is $\epsilon \times (\frac {\sigma}{r})^{12}$, so $B$ in this case is $\epsilon \times \sigma^{12} = 864554\ \ce{kcal A^12 /mol}$

However, there are other force fields parameterized differently, and some of these approximate the Pauli repulsion using a different value of $n$ (and some like $MM3$ don't even assume a $\frac{1}{r^n}$ dependency).  So since $n$ is assumed (and since different force fields are parameterized differently), there are other possible values for $B$ and $n$ to use in the approximation.


Structure diagrams indicate plausible structures?

Structural diagrammes illustrate connectivity in molecules but they are often very limited in their ability to represent the real 3D shape of complex structures.

A drawing on a 2D surface can never properly represent a 3D structure. This is as true for maps as it is for chemical structures (which is why all map projections contain distortion, e.g. the Mercator projection seriously distorts areas).

Chemical structure drawings also omit lots of detail from molecules like the relative size of atoms and the relative lengths of bonds. What they focus on is the connectivity of the molecule, showing us what connects to what. For many structures this gives a poor representation of the 3D shape, especially (as in the example in the question) where there are multiple ring systems. The flat diagram itself doesn't tell you much about the stability of the molecule. That is one reason why chemists need modelling kits that get the 3D geometry right. Even then, the kits are imperfect. Structural kits sometimes just do the connectivity and bonding (in effect a 3D version of the 2D connectivity picture). But, sometimes, you need to understand the spatial interactions that take into account the size of atoms. There are 3D space filling kits for that sort of modelling.

Having said all this, most modelling is done via computers not by creating hand-built models which are relatively expensive and limited. Computers can allow for extra calculations to represent the interactions of bonds and spatial factors to give plausible 3D geometries and can sometimes do detailed calculations of energy levels. Many of the software packages (e.g. Chemdoodle) that allow you to draw 2D structures will automatically attempt to generate plausible 3D structures based on those to give you better intuition about the geometry of the molecule.


The IUPAC Compendium of Chemical Terminology defines the mass fraction $w$ as the

  mass of a constituent divided by the total mass of all constituents in the mixture

$$w_i = \frac{m_i}{\sum\limits_{i=1}^{N}m_i} \tag{1}$$

This means that $w_i$ can adopt values between 0 ( = 0 %) when the compound in question is not contained in the sample), and 1 ( = 100 %) when the sample only consists of the compound.

If your sample isn't totally dry, the mass of water adds to the total mass of the sample.

As a result, the denominator in $(1)$ will become larger and the mass fraction $w_i$ for any component (except the water) will decrease.


Why does the reference electrode in potentiometry have a diaphragm?

This device is basically a galvanic cell (and an amplifier and voltmeter). Galvanic cells are quite similar to electrolytic systems, the difference being wheter one has an external voltage source.  If there is no “salt bridge” (i.e. some way for ions to wander between the two half-cells), the cell immediately stops working. This is certainly true, since one can directly observe it.

This makes sense if one considers that in (about all?) systems only one kind of metal ions is involved at each electrode. If you disconnect the two half-cells, there is a potential difference, but only between the electrode and the electrolyte. (How do you measure such a thing without intruducing another electrode??) The ions would simply stick around near the electrode until the charge makes the reaction unlikely to occur; only if the ions in the electrolyte can mix somewhat can the reaction continue. Keep in mind that the voltmeter does not have infinite resistance but "only" 10MOhms or so. This animation might be quite helpful.
In different terms: One needs a small current (at least for a short time) to measure voltage (keep in mind that the amplifier will have some capacitance) and for this to occur the circuit has to be closed, on one hand via the voltmeter, on the other via the salt bridge.

But, since you do not want to influence the measured system too much, you only provide this almost tight diaphragm — not a large(ish) diameter hole — and almost no current flows. I see no theoretical reason why a very small hole would not work equally well, but I assume this is more annoying for technical reasons (drilling it, clogging, etc.). One could argue that very small currents can flow even without a salt bridge because separate containers are still not infinitely far apart and their fields can influence the other one, but this seems to be negligible; normal glass is already thick enough to limit capacity to a few pF.


The problem is that you are starting the derivation from conditions that ensure reversibility or spontaneity, but you are not considering conditions that lead to the third possibility.
The following expression from your post is generally true:
$dq-dq_{rev}=dw_{rev}-dw $
I've dropped the "irreversible" labels and assume that $dq$ and $dw$ refer to some process that you have dreamed up.
Now the condition that leads to non-spontaneous processes must be
$dw_{rev}-dw>0$
since you have shown that the other conditions are for spontaneous processes, which leads to
$dq−dq_{rev}>0$
and therefore
$dG>0$
which is what you are looking for.
If you forget about $dG$ for a second and focus on  how you can get to the equation $dq−dq_{rev}>0$ (since you already showed the next step to an inequality in terms of $dG$), you see that to arrive at this equation you have to say that the system can do more work in a process you dreamed up than in a reversible process  ($dw_{rev}-dw>0$), which violates the 2nd law. To carry out a process in which $dG>0$, you have to correct this violation and make up the work difference by having the surroundings do additional (more) work on the system than was assumed in the original analysis.
By the way, you should be careful using PV-work in your examples, as for an ideal gas in the absence of reactions, Boyle's law says that the pressure will change during an isothermal expansion and Charles' law says that the temperature changes in an isobaric process. In neither case would the Gibbs free energy provide the criterion for spontaneity.


Is it possible to improve electrolysis rate without increasing heat?

Consider the electrolysis apparatus as a resistor in series with an EMF. Ideally, the cell voltage should be just enough to overcome the EMF (potential needed to split water), but some over-voltage is inevitable to have a decent current flow. Much of the heating is due to this wasted over-voltage. Then, to increase the efficiency and reduce heating, it's necessary to reduce the series resistance.

$\ce{NaOH}$ is already highly ionized, so adding more, beyond a certain point, is not the solution (no pun intended;). Close spacing helps, but you must leave room to physically separate $\ce{H2}$ from $\ce{O2}$. The remaining variable is electrode area: make the electrodes as large as feasible. You can further increase area by using porous sintered, or at least coarsely-sanded electrodes.

Quantum Sphere Inc. claims to produce a more active electrode from $\ce{Ni-Fe}$ particles, though they're probably a bit pricey for filling hydrogen balloons. Perhaps they offer samples?

BTW, I assume you're using a moisture trap for the gas, otherwise bubbling hot caustic electrolyte would inadvertently wind up in the balloon, reducing lift, damaging the balloon and possibly causing severe chemical burns. $\ce{NaOH}$ is called "caustic soda" for good reason!


What prevents everyday objects from reacting?

As I see it, two factors: thermodynamics and kinetics.

Thermodynamics: Many of the reactions that "don't happen" in everyday life are not favored thermodynamically, we call them non-spontaneous. This means that the products would be higher in energy than the reactants. Such reactions can certainly occur if you supply external energy (charging your cell phone battery, for instance, is a non-spontaneous process). But, without applied energy, non-spontaneous processes generally will not occur. Reactions that we observe in nature are generally spontaneous, favored by thermodynamics (with lower-energy products). Some thermodynamically spontaneous  processes are not observed to occur, even when we might naively expected them to. Example: Sugars decomposing to $CO_2$ and water is a very energetically favorable reaction, in fact the energy released from this process fuels nearly all known life. However, we do not observe our table sugar decomposing in its container because without outside help like enzymes and applied heat, the process is terribly slow kinetically.

Kinetics: For us to observe a reaction happening, it must proceed at rate that is fast relative to our human scale of time. Perhaps your fingers fusing to your keyboard is thermodynamically favored/spontaneous, but if it takes hundreds of years to occur then for our human purposes we may regard it as "no reaction". Indeed, the decomposition of most types of the biomolecules constituting our bodies is favored thermodynamically, but the kinetics are so slow (440 years for table sugar)°. There are a few ways to increase the rate at which a reaction occurs. To occur, a reaction requires a certain amount of activation energy. Think of this as hill you have to climb before you can sled down to the products (or up, potentially). The reaction does not occur if the starting material molecules do not possess enough energy to climb the activation energy barrier. You may help them overcome the barrier by supplying heat; thus, heating your table sugar will allow it to decompose into $CO_2$ and water before your eyes (the water will be released as steam of course). The other main way to increase the kinetics of the reaction is to lower the hill, instead of adding the energy to climb it. That is, use a catalyst like an enzyme to lower the activation energy barrier of the reaction. Without such external factors like applied heat and use of a catalyst, many reactions that are thermodynamically favored will be disfavored by kinetics, and not proceed at an observables rate.

So the reason the universe isn't a useless, constantly rearranging chemical soup boils down to thermodynamics and kinetics. Even if a process is thermodynamically spontaneous, it may be kinetically slow. I should point out that in a lot of places, such as the centers of stars, there is indeed a copious abundance of uncontrollable chemical and nuclear reactions. And planets like Jupiter pretty much are a useless chemical soup (no offense to Jupiterians). But in our favorite places on Earth, at room temperature and atmospheric pressure, the carbon in our bodies isn't going to undergo nuclear fission and become lead (thankfully).

By the way, this is in not a silly question. In fact, if a chemistry student doesn't wonder this, one may question if they are actually learning the subject. I applaud you for your curiosity.


Overlap between computer science and chemistry?

Have a look at list of computational chemistry software. It gives you overview, how the computers are used in chemistry. Many of them fall more into the chemical physics field, some into the chemoinformatics, to name just few. So definitely there are lot of places to contribute to use of computers in chemistry. If you are able to mix the software engineering and the science itself in a good way, possibilities are immense.

As for the last question - almost whatever field of chemistry you choose, you can contribute to the programs used within it. There is no such thing like completed scientific code. And university level is a necessary prerequisite to do reasonable science. Just concentrate on the models and mathematics behind the scenes and everything will follow.


The notion of chemical element in modern sense arised at the end of 19 century.

Before this, the term "element" was obviously too fuzzy to discuss here, as well as methods and proofs involved.

After this, the optical spectroscopy was method of choice to find new elements and prove their existence,
because of distinct and reproducible spectra.
See the history of helium, for example.

Of course, besides of spectroscopy, chemists proved the finding of new elements by separation of individual substances.
For example, metals are expected to form oxides and salts with (at least roughly) expected properties and mutual transformations.
In particular, salts, even in microamounts, may be recrystallized to form visually distinct crystals of individual substance.
If such a substance possessed properties not seen before, this was indication/proof of the presence of new element.


Possible Water peak in PMMA nmr spectrum

Your peak at δ1.6 in $\ce{CDCl3}$ is indeed water. This is typical of the appearance of water in $\ce{CDCl3}$ at the sample concentration you present. Water can sometimes be broad; this depends on the overall sample concentration, acidity of the solution and relative concentration of other labile protons. This looks like a fairly new bottle of $\ce{CDCl3}$.

You will notice that the residual chloroform peak at δ7.26 has $\ce{^{13}C}$ satellites; intensity approximately 0.55% of the parent peak, with a splitting of ~210Hz (so about 105Hz either side of the parent $\ce{^{12}C}$ peak). You will then also see that the water peak, although of similar intensity, does not have any carbon satellites. You could also confirm this by adding a small amount of $\ce{D2O}$, and observing a change in chemical shift for this peak.


Is it possible to estimate a substance's water solubility from data like the substance's partition coefficient, dissociation constants, etc.?

It could certainly be estimated from those properties, the structure, etc.  A lot of companies have developed proprietary algorithms for things like that:  solubility, diffusion coefficient, polarity.... (For example of a company, see "ChemAxon")

The water solubility for many chemicals is already known, though.  Search for a chemical ChemSpider, and look under the "Properties" tag.

They've even got an API, in case you're interested in using water solubility in an app.


Why is calcium fluoride insoluble in water?

This answer to a question about the solubility of silver nitrate and silver chloride will give you a good starting point to answer your question.

In order to dissolve a salt, you have to break apart the ions and hydrate them in solution. You can use the enthalpies of hydration of the ions, and the crystal lattice energy of the solid, to predict which compounds will dissolve. There is quite a nice discussion about how to think about solubilities of salts, and what factors affect solubility using a Born-Haber cycle, in this reference. The values for enthalpies of hydration and crystal lattice energies of sodium chloride and calcium fluoride can be found in standard chemistry texts or with an on-line search.

If you have more questions after looking at this material, come back and ask for clarification.


Balancing complex redox reaction of gold treated with sodium cyanide

In principle, the oxidation of gold is the half-reaction $$\ce{Au ->  Au^+ + e^-}$$But as this oxidation is done in a cyanide medium, $\ce{Au^+}$ is included in a complex ion : $$\ce{Au  +  2 CN^- -> Au(CN)_2^- + e^-}$$ The second half-equation is $$\ce{2H2O + O2 + 4 e^- -> 4 OH^-}$$ So the final equation describing the dissolution of gold in an aerated cyanide solution is the sum of the two previous equations, after multiplying the first by 4 : $$\ce{4 Au + 8 CN^- + 2 H2O + O2 -> 4 Au(CN)2^- + 4 OH^-}$$ And if you prefer equations with neutral substances, you may write it this way $$\ce{4 Au + 8 NaCN + 2 H2O + O2 -> 4 NaAu(CN)2 + 4 NaOH}$$


This is a bad question.  But the answer is probably supposed to be (d) because:

Water is more dense than ice.
Thus when ice melts, the total volume of "substance" goes down.
This means that atmospheric pressure did PV work on the melting ice.

Of course, heat transfer would also be required to melt the ice but the question isn't asking about that.


Writing the chemical equation given molarity

Have you considered the possibility that your reaction is wrong.  where, for example, is the barium on the product side.  I'll give you one hint here:  barium chloride dissociates in water to barium and chloride ions.  Ammonia reacts with water to produce some ammonium ion and the same amount of hydroxide ion.  However, barium hydroxide is totally ionized.

Can you take it from there?


How to Run GAMESS and Avogadro on Command Line?

Contact the HPC helpdesk at your university or search for some instructions compiled by them.

We cannot know whether your local cluster uses PBS (TORQUE/MAUI), SLURM or anything else to reserve cores, nodes, memory and cpu time to run jobs.


Because you are manipulating the pressure using a vacuum pump, the altitude change should not matter. If you look at the phase diagram for water in the attached link, you can see that if you keep your system below 0C, say at -10C, than you will cross the solid-vapor equilibrium line and your ice will start to vaporize when the vacuum pump decreases the pressure to ~200 pa.

https://en.wikipedia.org/wiki/Phase_diagram#/media/File:Phase_diagram_of_water.svg

To convert from pa to in Hg, I suggest searching for the conversion factor. From google: 1 in Hg = 3386.39 pa. So if you have a reading of say, 2 in Hg on your gauge then: $ 2 inHg*(3386.9 pa)/(1 in Hg) = 6772.78 pa $


Mechanism for Chromyl Chloride Formation

The answer may (and does) lie in the oxidation potential of the halogens. As we move down the group 17 the tendency for oxidation increases.

Refer:
$$\ce{H2SO4 + HBr -> Br2 + SO2} \tag{unbalanced}$$
but
$$\ce{H2SO4 + HCl -> X} \tag{does not occur}$$

$\ce{K2Cr2O7}$ is not able to oxidise $\ce{Cl-}$ to $\ce{Cl2}$ but it does so with $\ce{Br2}$, thus we observe reddish brown fumes of $\ce{Br2}$ as if $\ce{Br-}$ had been present, which are then passed through $\ce{NaOH}$ to confirm the presence of chloride or bromide.

Refer:

\begin{align*}
\ce{CrO2Cl2 + NaOH ->& Na2CrO4} \tag{yellow soln,unbalanced} \\
\ce{above + H+ ->& (Cr2O7)2-} \tag{orange solution} \\
\ce{Br2 + NaOH ->& NaBr + NaBrO3} \tag{light yellow soln}
\end{align*}


Can we use liquid hydrogen to power a conventional engine?

The efficiency that matters is the efficiency of the vehicle not the efficiency of the engine

The problem of designing more efficient transport is bigger than the problem of designing better engines. Conventional combustion engines can be adapted to burn a wide variety of fuels (alcohols, natural gas and hydrogen are not too big a challenge). And there are alternative ways to move vehicles than combustion engines (fuel cells, for example, which run on hydrogen). So it looks like we can solve the engine problem.

But that isn't enough. To have a viable vehicle we have to solve other problems as well. How to store and deliver the fuel to the engine, for example. And how to distribute the fuel across the country so vehicles can be refilled. The second problem is probably easier to solve if the first problem–a viable vehicle–has a solution. Unfortunately, it isn't that easy.

To have a hydrogen powered car we would need a way to store enough hydrogen in the vehicle to allow it to drive several hundred miles.

This is easy to do with liquid hydrocarbon fuels in conventional vehicles (petrol/gasoline and diesel have very high energy densities and can be stored in a simple unpressurised tank). The weight of the fuel alone is larger then the weight of the tank that contains it and the volume is small as the energy density is very high.

It isn't so easy with hydrogen. While there are several existing ways to store hydrogen and some proposed solutions that are not yet operational, all of them involve big issues with volume and weight. Compressed hydrogen will involve very heavy thick-walled tanks and even then the energy density of the hydrogen gas alone will be lower than a tank full of petrol. The weight and volume of the tanks required to store as much energy as a tank of petrol will very large. Liquid hydrogen has a different set of compromises but isn't much better as it requires cryogenic system to keep the liquid hydrogen cool. There is considerable interest in clever chemical ways to store hydrogen (clathrates, metal organic frameworks and various molecular systems than can reversibly store hydrogen are all being researched).

To get an idea of the scale of the challenge it is worth looking at the mass and volumetric characteristics required to compete with conventional fuel. Diesel (slightly more efficient than petrol) carries about 12kWh/kg and about 10kWh/L. The current targets for research on hydrogen storage for the Fuel cell Technology Office are 1.5kWh/kg and 1kWh/L. This is about 10 times worse than diesel even ignoring the fact that fuel-cell power vehicles might have other advantages partially compensating the overall vehicle when compared to combustion engines. And the best technology solutions are much better than liquid hydrogen storage.

So you can use liquid hydrogen (or other ways of storing hydrogen) to power an conventional engine or vehicle (and such vehicles exist). But the cost and design compromises to make this work for a whole vehicle given that you need 10 time the volume and weight to store the equivalent energy, make such solutions not competitive for mass market vehicles.


Distilled water to clean greasy stains: Why does that work?

There may be two different factors at play here.

First, it isn't necessarily the water that is doing most of the cleaning. I presume the water is used to wet a cloth and the cloth then wipes the surface. So the cloth may then physically remove surface contaminants. The role of the water may be just to improve the cloth's surface contact hence improving its ability to physically remove the contaminants.

Secondly, distilled water might do a better job than tap water as it has fewer dissolved solids which would leave other stains on the surface when the water evaporates.

However, the only way to be sure is to do a comparative experiment using different cleaning fluids. You might try tap water, distilled water and distilled water with a surfactant (e.g. a liquid soap). You might find that the best result uses the surfactant but that pure distilled water does a less good, but good enough job. Science isn't about single isolated experiments: we usually need to compare alternatives to know what is best.


Electrolysis - What is the basis for the discharge rules?

The redox potential is a measure of tendency of  redox system to exist rather in oxidized form ( negative potential values) or reduced form ( positive values ).

Therefore the higher the standard redox potential is, the stronger oxidation effect and weaker reduction effect have the respective oxidized/ reduced forms of the redox system.

And vice versa.

Particularly, on the opposite sites of the standard redox potential table are
$$\begin{align}\ce{
Li &<=>> Li+ + e- \\
F2 + 2e- &<=>> 2 F-  \\}
\end{align}$$

The former is very strong reducting agent, while latter is very strong oxidating agent.

I will illustrate it on anode case, with the cathode case it is the same, but with the reversed potential order.

When the anode potential is externally forced to raise up, if first start to oxidate the reducted form of the redox system with lowest redox potential.

If the current is already saturated by diffusion rate of redox reagents, futher increase of potential does not increase the current until it reaches potential region of next easiest to oxidize redox system.

If the redox systems have similar values, the next system start to react on the electrode before the system

The result is, the curve current versus anode potential has a typical look of the  rounded stepwise  voltammetry curve.

See also Voltammetry,  Polarography and example of the cathodic polarographic curve.


where is sulphate going here? electrolysis of magnesium sulphate

In electrolysis of aqueous solutions sulfate ions just sit there.  Protic hydrogen (or the metal in the cation if this metal is not highly electropositive like magnesium) is reduced more easily at the cathode, and water is more easily oxidized at the anode.

Sulfate ion (actually, bisulfate ion) can be oxidized electrolytically at the anode if you use sulfuric acid instead of water as the solvent.  See the preparation methods for potassium peroxydisulfate.


Why are there only 7 periods in the periodic table?

There is no theoretical limit to the number of rows, but...

On one hand, as Tom Lehrer states, there may be many others but they haven't been discovard.
On the other hand, after all s, p, d, f orbitals are filled in the seventh period, there might be a new row, or there might be an extension of the seventh row, as a new type of orbital (g? Anyone know it there's a name?). Each orbital has double an odd number (double because the Pauli Exclusion Principle allows pairs of electrons of opposite spins), so the next orbital would have 18 electrons. Is it possible? Inner electrons would be moving at very high speeds, and they'd also shield outer electrons from attraction by the positive nucleus.
On the gripping hand, transuranics may last a few seconds... or a few attoseconds, so getting spectra might be problematic.


How do I interpret characters that are not 1 or -1 in a point group table?

When you observe that in the characer table of a symmetry group there are irreducible representations where the character of the identity operation (E) is 2 or higher, you may find that the character of some symmetry operations is 0, +2, -2, etc. This is due to the fact that this IR are degenerate.

To illustrate this idea, consider that you trying to determine the symmetry of the MO of bezene (D6h). You will find that some of these orbitals have the same energy (are degenerate) for example

When you apply the symmetry operations of the D6h point group you can check that what you obtain is a linear combination of these orbitals. If you express this linear combinations in matrix form, the character of the matrix will give you the value that appears in the character table for the symmetry of that orbital. In this case these orbitals have a E1g symmetry.


Each electron shell can hold up to 2n² electrons. Is there a geometric reason for this?

A good insight!

The number of electrons in each orbital are double the odd numbers. The s orbital can hold 2, p holds 6, d holds 10 and f has 14: double 1, 3, 5 and 7. If you add the odd numbers, you get a square number:

1 = 1

1 + 3 = 4

1 + 3 + 5 = 9

1 + 3 + 5 + 7 = 16

The doubling is due to the Pauli exclusion principle; in each orbital, an electron of a specific energy can be in one of two spin states.

It is interesting to see spectral lines split in a magnetic field (Zeeman effect) or an electrostatic field (Stark effect): one line may become 3, 5 or 7 lines. It clearly shows how orbital theory was derived.


Effects of potassium hydroxide & carbonate on metals?

Your conjecture is at least partially correct, in that there was a mix of $\ce{KOH}$ and $\ce{K2CO3}$. However, there is only ~400 parts per million (ppm) $\ce{CO2}$ in air, so at first, the $\ce{KOH}$ was free to react with metal (and, possibly, the green lacquer on the PC board), until enough $\ce{CO2}$ entered the cabinet to form the carbonate. It is possible some raw $\ce{KOH}$ is still present, under a film of carbonate, so be careful. Note that there is ventilation in a PC cabinet, so $\ce{CO2}$ is never really depleted in it.

Far more damage, though, may be due to electrolytic corrosion. Wikipedia has a photo similar to yours, showing "Severe PCB corrosion from a leaky PCB mounted Ni-Cd battery". The liquid electrolyte creates bridges across the bare IC pins and across the PC board traces, and the continual current from the power supply and the not-quite-dead battery cause metal to be removed from one place and deposited in another.

After removing the battery, rinsing many times with distilled water, and tooth-brushing the board, you may find that surface traces need to be rebuilt (such as by soldering a bit of braid across a gap). This is likely a multilayer board, and if an internal trace is gone, it would be difficult to find and bridge the gap. Note that any such patches will affect timing, due to the impedance mismatch, so you may find that after all repairs, crashes occur with some frequency.

Best wishes for your restoration project. It's nice to know that at least one venerable computer is not going to landfill!


Solubility of Lead in water from leaded pipes in Roman aqueducts

Let me give an extreme example of chemical analysis. It is an anecdote. Some analytical chemists were using a very sensitive technique to analyze metals in blood. It is called neutron activation analysis. What they saw was that a very small quantity of molybdenum (Mo) was present in small quantities persistently. Nobody expected that. It turned out after a lot of hard work that Mo was coming from the stainless steel syringe needle used for drawing blood. Lesson of the story: There is nothing which is perfectly insoluble in water. It all depends on the analytical technique. If you try to weigh a hair on a "weighing machine" it will tell you that the mass is zero kg, but if you use a fine analytical balance it will tell you that hair has a mass.

The same goes for lead pipes. Natural water has plenty of minerals, dissolved oxygen anions, cations and don't forget the microorganisms. The bacteria do a lot of chemical transformations in water. As a result, Pb would slowly leach into the water at parts per million or parts per billion levels. Don't assume that Romans were drinking lead salt solutions in high concentration ranges. Not only water lead was present in water, it was also present in alcoholic beverages.

Read more here Lead in ancient Rome’s city waters

Lead in water is still a serious problem in several areas of the modern world. It is not an ancient problem.


how are the frequencies at a local maximum of PES like?

The potential energy surface (PES) is a very complex thing and it is hard to be understood (if we can understand it at all). With quantum chemical approaches we can map out this surface. We differentiate between positive, real and imaginary (usually shown as negative) frequencies. There are no negative frequencies. What you may encounter analyzing the PSE, are one of the following extreme points.

Local/Global Minimum
All derivatives have to be positive, the number of imaginary frequencies (NImag) is zero $i=0$. This corresponds to an observable state of the system/ molecule, i.e. it can be trapped and analysed with spectroscopy. Exciting any of these frequencies, the energy of the molecule will increase. But you already knew that.
Transition State
A transition state is a first order saddle point on the PES. The Nimag is exactly one, $i=1$. It connects two local minima with each other. Try to imagine a molecule, like ammonia. This can undergo inversion, i.e. nitrogen inversion. The two pyramidal forms are local minima, which are connected via a planar form (nitrogen is $\ce{sp^2}$ hybridised), which is the transition state. At that point one of the frequencies is imaginary, which means, that it corresponds to lowering the energy while exciting this frequency. All other vibrations would still increase the energy of a molecule.
You cannot observe a state like this via common spectroscopic methods, but it still has a physical meaning. There are various theories in use to describe reactions of molecules. The most popular is the Transition State Theory, which was developed by H. Eyring and M. Polanyi.
Second Order Saddle point
It formally corresponds to a transition between transition states. Its NImag is $i=2$, so there are two vibrations that lower the energy. It usually has no physical meaning, as the molecule would probably decompose before reaching this state.
If one of these vibrations correspond to a rotation of a moiety, e.g. a methyl group, and is therefore close to zero, it corresponds to the so called rotational space and might be within range of the transition state.
$n$-th Order Saddle point
It is like the above a transition between two $(n-1)$-th order saddle points and has even less physical meaning. You sometimes encounter these saddle points as artefacts of the used calculation model. It is strongly advised to not use these states for any explanation.
Maximum
This is one cannot be found. The PES is usually like a well. The borders of this would be total fragmentation of the molecule. Imagine the homolytic dissociation of dihydrogen (a 2D PES). The maximum value of the PES corresponds to superimposed atoms and its energy value is at infinity.

An additional note. A transition state always connects two local minima. And vice versa, two local minima will always be connected via a series of transition states and local minima. It does not make much sense, looking at any other points than these, as they can most likely not interpreted in a physical meaningful way.

Comment by Ath

  What I mean is not global maximum, but local maximum, which doesn't necessarily represent bond breaking...The local maximum I'm interested in is actually the rotation of part of a molecule. I wonder whether the frequencies are all positive or all negative at that point.

There are no maxima on the PES, since the maximum energy corresponds to superimposed atoms. A rotation of moieties is a transition state, or a first order saddle point, and has exactly one imaginary mode.


Can hydrides react with hydrocarbons? And why?

The pKa of sodium or lithium hydride is (solvent dependent) about 35, the pKa of Butyl lithium is around 50 so clearly H- is not strong enough to tear a proton off an unactivated alkane pKa tables here


What is the solid crud that appears inside my kettle?

This is limescale and you have it because your tap water is rich in minerals (mainly, calcium, as you note). You could remove these minerals with an ion exchanger, but you wouldn't to it with a tap water since these minerals give it a better taste. Limescale is not dangerous to ingest, but certainly not pleasant either (and I see on the photo that there is a slot for a mesh that should filter scales). Thick layer of limescale is also bad for the heating element – it hinders heat transfer to water and leads to overheating.

It is a good idea to remove limescale once in a while. You can dissolve limescale with any weak acid, like vinegar, but vinegar smells, so I suggest buying a descaling agent.


Is it possible for the remaining solute of a saturated solution to form a single even phase from top to bottom (similar to that of a colloid)?

The density of the solute relative to that of the solvent will primarily determine whether the excess solute will float, sink or tend to disperse throughout the solution.  Additionally, the particle size and porosity will play a role in the rate of settling or floating, as will the viscosity of the solvent, but it is the density that determines whether the particles will float or sink.

But, the answer to your question is no. Even if the density of the undissolved solute is essentially the same as that of the solvent, and the undissolved solute is very fine, then you could have a uniform looking solution, but it would still be a two-phase solution, with a solid phase suspended in a liquid phase.


Why is mass proportional to the strength of a dipole-dipole attraction? meaning more mass more strength

Increasing mass increases the availability for molecules to attract through London dispersion forces (induced dipole-dipole). A longer chain has more atoms and hence more electrons and therefore experiences stronger bonding.

When you say 'polar' I assume you mean permenant dipole-dipole and hydrogen bonding which is actually not affected by the 'mass' or length of a carbon chain in a molecule. However if you take ethanol and pentanol as an example they both have hydrogen intermolecular bonding on account of the -OH group but pentanol also experiences greater London dispersion forces than ethanol which results in its significantly higher melting and boiling points.

So to recap increase mass is not a direct causal link to charge. It is in fact more mass meaning more atoms meaning more electrons meaning induced dipoles leading to greater intermolecular forces. I hope that's clear (:


Experiment to demonstrate Dalton's Law of Multiple Proportions

Because of toxicity, I'd avoid middle or high school student experimentation with $\ce{CO}$.
You might consider $\ce{H2O}$ and $\ce{H2O2}$, if you stick to 6% (20 vol) or weaker. There is the complication of working with a dilute solution, though, in getting precise measurements on proportions.
Another possibility is demonstrating the various oxides of iron. However, there is another issue: you can produce fairly pure pure red iron(III) oxide, $\ce{Fe2O3}$, by the electrolysis of sodium bicarbonate, $\ce{NaHCO3}$ with an iron anode, but rather than pure black iron(II) oxide, $\ce{FeO}$, thermal decomposition of iron(II) oxalate in an inert atmosphere, makes a mix, wustite (a nonstoichiometric compound) that is primarily $\ce{FeO}$.
Chromium compounds, e.g. oxides, of varying proportions might make an interesting demonstration because of obvious color changes, but the toxicity of chromium compounds precludes use by students, and even safe disposal of the demo material may be problematic.
BTW, an interesting effect is observed in podzol soils, as found in pine forests. If there is iron in the soil, the reducing nature of freshly-exposed soil makes it look greenish; but after exposure to air, the iron is oxidized to a reddish color (I've observed the change in less than 15 minutes.).


When is a dipole-dipole interaction strongest?

An electric dipole is formed in a molecule as the vectorial sum of all the partial charges on the atoms multiplied by their separation. Thus if, these charges are large and well separated and not symmetrically situated about the molecule the dipole will be large. For example chloro-benzene has a far larger dipole than toluene. However, 1,4-dichlorbenzene has zero dipole as the charges cancel out. (Molecular point groups can be used to decide if a molecule has a dipole, but not its magnitude. Only molecules in groups $C_1, C_s,C_n,C_{nv}, n>1$ have dipoles). Hydrogen bonds have no special significance with respect to dipoles.

Dipoles interact through space, and the strength of their interaction depends on the distance between the dipoles,  relative orientation and relative permittivity (dielectric constant) of the intervening medium. Each of these have a dramatic effect for any given pair of dipoles.

The interaction energy E for dipole vector $\mu_1, \mu_2$ at separation R and where $\bar R$ is the vector joining the centres of the dipoles is

$$E = \frac{1}{4\pi\epsilon_0\epsilon} \left( \frac{\mu_1\cdot\mu_2}{R^3} - 3\frac{(\mu_1\cdot \bar R)(\mu_2\cdot \bar R)}{R^5}    \right)  $$

The dot product between two vectors e.g. $\mu_1\cdot \bar R = |\mu_1||\bar R|\cos(\theta_1)$ where $\theta_1$ is the angle between the vectors, see figure.

If the vectors are now represented as unit vectors e.g. $\bar \mu_1=\bar u_1 d_1$ where d is the magnitude of the dipole, and $\bar R = \bar u_R R$ then the energy becomes
$$E = \frac{K}{4\pi\epsilon_0\epsilon}\frac{d_1d_2}{R^3}   $$
where
$$K= \bar u_1\cdot \bar u_2 -3(\bar u_1 \cdot \bar u_R)(\bar u_2 \cdot \bar u_R) $$

The figure shows the geometry.

If the dot product $\bar \mu_1 \cdot \bar \mu_2 =0 $ and one  of the two dot products with $\bar u_R $ are $0$ then the energy is zero, and this can occur when dipoles are at 90 degrees and one points to the centre of the other. Physically this makes sense as each end of a dipole sees equal positive and negative charges on the other dipole.


Which acid to choose when designing a world with acid seas?

Coca Cola or a mixture of Coca Cola and Pepsi Cola would probably be a good choice.

It's not immediately fatal for carbon life forms that fall into it. Only prolonged exposure will cause digestion and damage to bones and teeth.

Due to the phosphoric acid ships with iron hulls would withstand corrosion due to passivation through the formation of iron phosphate (also a molluscicide to protect from shipworms).

In agriculture phosphate is also a potent fertilizer.

For propulsion purposes, Cola together with Mentos is a well-known hybrid rocket propellant.


Electrolysis of concentrated sulphuric acid

This is because in concentrated sulfuric acid solution equilibrium condition will favour more water molecules than oxidation of water into protons. So there will be an over potential required (to go against the equilibrium) , that is extra potential beyond the theoretical reduction potential derived from thermodynamics to complete the reaction. So the total potential required will be theoretical Plus overpotential. In case of oxidation of sulphate reduction potential will be much less that for water ,thus oxidation of sulphate happens.


What are physical properties and chemical properties?

This is a tricky question.

I have to concur with Paul - it's difficult to give a truly hard and fast definition of the distinction between a chemical change and a physical change as there is a fairly smooth continuum between 'physical' and 'chemical' interactions. However, a good rule of thumb is that chemical changes involve changes in electronic structure, usually in the form of chemical bonding or redox. The problem is that traditionally 'physical' properties may have big implications for chemistry - I'll give some examples.

The Van der Waals and dipole-dipole interactions are mediated respectively by 'instantaneous' and permanent polarisations of electron density with respect to positive charges. Dipole-dipole interactions such as those found in water are quite significant and give rise, for instance, to water's frankly freaky liquid range as well as water's ability to mediate a wide range of chemical processes. These interactions don't involve significant changes in electronic structure however they have chemical implications in, for example, making a choice between liquid water or liquid nitrogen as a solvent.

The atomic mass (and hence the isotope) of an element is typically considered a physical property, however it has implications for reaction kinetics. This can be illustrated (expensively) by giving living organisms nothing but heavy water to drink. They will eventually experience toxicity due to the slowing of key biochemical reactions.

The magnetic properties of an atomic nucleus are typically considered to be physical, however they can be exploited in the observation that e.g. different isotopes of uranium have slightly differing hyperfine splittings and hence different ionisation energies. This allows isotopes to be enriched by selective ionisation.

Regarding the egg:

The dramatic changes that occur when you cook an egg are the result of denaturation of the proteins in the egg. Proteins are long chains of amino acids that take on their characteristic active conformations through a process called 'folding', so called because that's what it looks like (if you've heard of Folding@Home, that simulates the folding of proteins). If you raise the temperature much higher than physiological temperature, the protein degrades through unfolding into a different conformation. This involves the Van der Waals and dipole-dipole interactions holding together the secondary/tertiary/quaternary structure of the protein being broken en mass as well as the formation and/or scission of covalent bonds.


Why the term 'unified' in "unified mass unit"?

Your question is historically important. Nobody has the energy to write an answer which would require several pages of an article. In short, it is a can of worms. For almost 5-6 decades chemists and physicists used a different scale for the atomic masses, so the atomic masses in the publications/books used by physicists did not exactly match what the chemists used for their own publications. Note that it is just not a matter of academic egos, the entire chemical industry is dependent on atomic weights (for calculations and trade). Money matters. This is why both the governments and the scientific community wanted to sort this out.

Both parties were not willing to budge because that would have affected their millions of pages of research. However a compromise was reached to use a common scale by two three sensible men i.e., they proposed to choose C-12 in the 1960s instead of oxygen. This process sort of "unified" the community of physicists and chemists and it did not affect the values of either party by much.

More history is given here, The AME2003 atomic mass evaluation (I). Evaluation of input data, adjustment procedures, Nuclear Physics A 729 (2003) 129–336

There are scores of other articles from the 1960-70s on the choice of C-12. Use Google Scholar. The original work is in French I guess, I did not check

T.P. Kohman, J.H.E. Mattauch and A.H. Wapstra, J. de Chimie Physique 55 (1958)
393.


Why do heavier elements have smaller specific heat values.

The more particles in a system, the more modes of vibration (and rotation), so one gram of $\ce{H2}$ has about $235 \times$ more particles than one gram of U. The heat is "stored" in the motions of the atoms and electrons, until reaching millions of kelvin, at which point dissociation of the nucleus would be significant in specific heat. See these links and  for more information.


Why do two atomic orbitals form two molecular orbitals?

Well, one important thing to keep in mind is that atomic and molecular orbitals are not "real" things, they are mathematical constructions that are useful in helping understand and predict how electrons behave in atoms and molecules.  They are like a set of rules that tell us all of the ways in which it is possible for the electrons with a particular set of quantum numbers to behave.  For example, a H-atom in the ground state has only one single electron in a $1s$ orbital (quantum numbers: $n=1$, $l=0$, $m_l=0$), but we also know that if the H-atom absorbs some energy, the electron can end up in a higher electronic state, for example the $3\mathrm p_z$ orbital (quantum numbers: $n=3$, $l=1$, $m_l=0$).

Now, when you are dealing with molecular orbital theory, you are talking about the possible ways in which electrons in atomic orbitals can combine in a covalent interaction.  So it's not that the two atomic orbitals combine "constructively and destructively at the same time", it's just that those are the two ways in which its possible for the two orbitals to combine: in the region where they overlap, they can either add their amplitudes, or subtract them.  The "constructive" case where they add produces an molecular orbital that is lower in energy (that is to say, an MO that describes a state where the electrons will have lower energies) than the "destructive" case, so by the aufbau principle, electrons will occupy the lower energy "bonding" orbital first.  In a case like $\ce{H2}$, where there are only two electrons, only the bonding orbital is occupied, but because the Pauli exclusion principle tells us we can only have two electrons in a given orbital, if you add another electron (as in $\ce{He2+}$), it will occupy the higher-lying "anti bonding" orbital.


"Spontaneous" means different things in different contexts
Your penultimate paragraph captures a key idea. The explanation for why this is right requires a recognition of the context of the term "spontaneous".
The context of the statement at the start of the question $\Delta G=\Delta H-T\Delta S$ is negative is thermodynamic stability. But this is somewhat at variance with the more natural use of the term which implies "things happen without being pushed". This idea is closer to the idea of kinetic stability in chemistry.
You correctly identify the need to add energy to get the reaction past the transition state. Even if the reaction overall releases energy (thermodynamically spontaneous) the reaction won't just happen if there is a huge barrier to getting over the transition state. There are big kinetic barriers that stop the reaction "just happening". Oxygen and gasoline will react to release energy but this doesn't happen without the push given by the spark plug in the engine of a car.
That barrier is so low in some cases a compound will react with nearly everything with little excuse (chlorine trifluoride will set fire to asbestos). That is a spontaneous reaction in any context. Luckily, few thermodynamically spontaneous reactions are also kinetically spontaneous or humans would catch fire in air.
So when you see the term "spontaneous" ask what is the context: thermodynamic or kinetic? And don't confuse them.


Conception and manufacture of the Avogadro Project's (nearly) perfect silicon sphere (and, what is “atomic feeling”?)

Short answer:  The success of the project benefitted from many contributors.

According to BIPM and its Avogadro project, there are two spheres (reference).  The page equally refers to a publiction about the 2011 $\ce{^{28}Si}$ atom count (paywall) and the report in 2017 (here, open access).  From the later paper, you may identify the origin of the two spheres in the Czochralski crystal obtained initially:

as well as where samples were taken to determine the amount of carbon, oxygen, nitrogen, and boron (loc. cit., table 1), the molar mass of silicon (table 2), concentrations of point defects in the lattice (table 3).  Equally, the analysis took into consideration the $\ce{SiO2}$ layer naturally formed if the sample is exposed to air, as well as adsorbed water; for both, the upper limit of layer width as well as the mass are provided.  As an example, the deposit of oxygen on the spheres was mapped by X-ray photon spectroscopy (XPS):

The article describes that the diameter of the spheres was assessed by a sphere interferometer

in vacuum at $20\,^\circ{}\mathrm{C}$ within a tolerance of $\pm 3\,\pu{mK}$ to allow the statement

  The measured diameter was the apparent diameter, which is not corrected for the phase shift due to the surface layers. The mean apparent diameter was 93 723.723 61(61) μm.

As for imperfections, this property equally was mapped:

With papers titled like The self-weight deformation of an x-ray interferometer the dedicated issue of Metrologia 48(2), April 2011 shed some light on monitoring the quality of the samples and the used tools for the analyses.

Additions:
A video of channel veritassium is dedicated to the reference spheres, too.  The section starting at 7:00 min focusses on carving and polishing them (8:11 min, at 8:19 min likely showing Leistner himself).

A now open-access publication by Fujii et al. equally briefly describes the stages to carve-out the spheres:

Starting from the ingot, a section is cut.  A lathe is used to obtain a sphere already with an shape accuracy of $1\,\pu{mm}$.  To smoothen the surface, these spheres are lapped with alumina particles in water with finer and finer grains; eventually (fine lapping) within a margin of $100\,\pu{nm}$.  To polish the surface, collodial alumina and eventually titanium dioxide yielded a shape deviation of less than $50\,\pu{nm}$ and a surface roughness below $0.2\,\pu{nm}$.  This publication explicitly refers to Leistner's work including publications authored by him (example, example) and then established protocols easing to manufacture such samples reliably.

PTB continues both further development of the interferometers (here) and the deployment of these spheres as substitute of the old standards (example).  (As of writing, the mass standard's group seeks collaborators for Si spheres (reference, perma).)  Comments by Maurice and GM suggest the advancement of mechanical engineering and this new definition of the kilogram might offer smaller mass references which do not need to be as large and heavy as $1\,\pu{kg}$.