I have an idea through utilizing soda lime or other basic substances to help solve climate change

My worry is that the energy requirements for making soda lime, caustic soda or some other carbon dioxide adsorber will be so high that overall it will result in the emission of more green house gases.

Years ago I was touring a waste disposal site where they react waste sulfuric acid with limestone. The boss of the site told me that some years ago someone had asked if he could scrub out the carbon dioxide from the reaction tank with sodium hydroxide or a similar reagent. He gave the same answer as I am giving to us when he recalled the tale.


Rate of exothermic/endothermic reactions

A reaction being endothermic or exothermic depends on thermochemistry while the rate of a reaction depends on its chemical kinetics. Hence, there is no relationship between the activation energy of a reaction and its enthalpy change. Also there is no relationship between the change enthalpy and the rate of a reaction. A rate of reaction is dependent on the concentration of the reactants which are to a certain power which is multiplied by a certain constant, k, which is called the reaction rate constant. For example, for the general reaction: $$\ce{A + B -> C}$$ the rate of the reaction is given by the following equation: $$ rate = k[A]^a[B]^b$$
The reaction rate constant is not dependent on whether a reaction is exothermic or endothermic. However is does depend on the temperature and and also the activation energy of the reaction, given by the equation: $$k = Ae^\frac{-E_a}{RT}$$ So you are right to say that a reaction will be slower if it has a higher activation energy, but you are wrong to say that endothermic reactions have a higher activation energy than exothermic reactions.

There are actually several endothermic reactions that are fast, for example, dissolving common table salt (NaCl) occurs pretty quickly, but it is actually a endothermic reaction. Meanwhile, the spontaneously decomposition of diamond to graphite is actually a exothermic reaction, however it is extremely slow. It is so slow that it not even observed after several decades.


Since the chemicals will not be in contact with the plastic for very long, very little will diffuse through the plastic.

If you are concerned about fire, you could wrap the outside (not the inside) with aluminum foil; this should keep the fire contained. Have a container of sand handy in case the fire is extremely hot and the aluminum starts to burn. In fact, sand would be handy for most fires since it will decrease the amount of oxygen getting to the flames.

As long as you have a high air flow and the volatiles are coming from a small area, there should be no problem. Added: The concentration of the volatiles will not be high enough for a spark to ignite.

I can't really comment about the environmental effect unless you provide more detail. If the MSDS doesn't mention a hazardous exposure from the fumes, then you probably don't need a fume hood. You can get some activated charcoal (from a pet store) and pass the fumes through the charcoal, but I don't know how effective that would be.


What is special about the molecule luciferin that it can emit light?

[...] the protein luciferin is oxidized by the enzyme luciferase

Luciferin is not a protein, but a benzothiazole with a thiazole attached to the carbon atom between nitrogen an sulfur.

Upon oxidation at the thiazole ring, an oxetanone is formed. This four-membered ring breaks, releases carbon dioxide and turns the thiazole into a thiazolone. The resulting molecule is called oxy-luciferine.

Amazingly, the oxy-luciferine generated through this reaction is in an excited state! While typically excited states are the result of the absorption of a photon, here excited state is generated chemically!

Anyway, the fate of this excited state is as to be expected: it relaxes to the ground state by emitting a photon.

In general, these reactions are known as chemoluminescence.

Another known example is the peroxyoxalate reaction of glow sticks.

Phosphorescence at ~450 nm has been observed from the comproportionation of cyclohexanoxy radicals in polymer matrices, when hydrogen atom atom transfer between two radicals produce cyclohexanol and cyclohexanone (in an excited triplet state).


How to calculate the temperature at which a reaction becomes spontaneous?

You have the right equation.  But the problem already gives you the temperature at which $\Delta G$ is 8.9 kJ.

Assume that $\Delta H$ and $\Delta S$ are constant and do not vary with temperature.  $\Delta G$ still will though because of the $T$ in the $\Delta G = \Delta H - T\Delta S$ equation.

So plug in the values for $\Delta H$ and $\Delta S$, and also the $\Delta G$ value that you think represents the transition from spontaneous to non-spontaneous.  You should be able to take it from there.


What is the process of turning ionic silver solution into colloidal silver solution?

If you provide a scheme of this mysterious  "colloidal silver generator" would be very interesting, from what I see from googling I see only things like this.

Up to my knowledge this "colloidal silver generator" has no sense. From instructions notes it seems there are two silver bar in a solution of pure water, as you say, this in fact should create a very slow (because there are no electrolytes) electrolysis of water into $O_2$ and $H_2$ but I don't see any reasons for the producing of colloidal silver.
In chemistry laboratory to produce colloidal silver usually you start from a solution of ionic silver $Ag^{+}$ and reduce it to metallic silver $Ag$.
$$\ce{Ag^{+} + e^{-} -> Ag} $$
Caution the following is a reaction not a recipe do not ingest the ultimate product of this reaction!!!
Once time I've made a true colloidal silver with Lee-Meisel method (J.Phys. CHem 1982,86,3391-3395) if you look at the original paper there are many methods however the following (method c) is used most.

(c) $AgNO_3$ ($90 mg$) was dissolved in $500 mL$ of $H_2O$ and brought to boiling. A solution of $1 %$ sodium citrate ($10 mL$) was added. The solution was kept on boiling for cs. 1h.

The solution should be greenish yellow. If you examine the reaction the $AgNO_3$ is the source of $Ag^{+}$, sodium citrate is the reducing agent that reduce $Ag^{+}$ to $Ag$ and stabilize it. I think colloidal silver generator would be for anyone who makes colloidal solution a dream I assure you is a very complex and tedious process.
Moreover both $Ag^{+}$ and colloidal silver are very difficult to store, the first may  undergo photo-reduction to the metal the second is unstable as many colloidal solution because colloidal solution have grater energy due their larger surface (More energy many times means more instability). Without any stabilizer (such as t-BuONa or citric acid) I don't think the colloid will last long.
However I don't see why you should drink something like this. In fact $Ag^{+}$ has antiseptic properties such as  Mercurochrome,HydrogenPeroxide or Lysol(R) but this don't mean drinking Lysol, Hydrogenperoxide or mercurochorme every morning will make you feel better. I strongly suggest you to do not drink such thing ever.


Can I use sodium hydroxide (NaOH) to remove oxides from metals?

Your second reaction isn't a redox reaction. When carbon dioxide is dissolved in water, two equilibriums are forms:

\begin{eqnarray}
    \ce{H_2CO_3 + H_2O &<=> HCO_3^- + H_3O^+} \\
    \ce{HCO_3^- + H_2O &<=> CO_3^{2-} + H_3O^+}
  \end{eqnarray}

Remember that water also is in equilibrium:

$$\ce{2H_2O <=> OH^- + H_3O^+}$$

Adding $\ce{NaOH}$ will — according to Le Chatelier's principle, the equilibriums are disturbed, and more carbonate ($\ce{CO_3^{2-}}$) is formed.

If you're interested in a nice redox reaction involving iron(III) oxide, aluminium is much better choice.


Why is chloride oxidised instead of water in copper chloride electrolysis?

The standard redox potential of chlorine is $E^\circ = +1.358\ \mathrm{V}$; the actual potential depends on the concentration of $\ce{Cl-}$:

$$\ce{Cl2 + 2e- <=> 2 Cl- }\quad\quad E^\circ = +1.358\ \mathrm{V}$$
$$\begin{aligned}
E&=E^\circ+\frac{RT}{2F}\cdot\ln\frac{1}{\left[\ce{Cl-}\right]^2}\\
&=1.358\ \mathrm{V}-0.05916\ \mathrm{V}\times\log{\left[\ce{Cl-}\right]}
\end{aligned}$$

The standard redox potential of oxygen is $E^\circ = +1.229\ \mathrm{V}$; the actual potential depends on $\mathrm{pH}$:

$$\ce{O2 + 4H+ + 4e- <=> 2H2O}\quad\quad E^\circ = +1.229\ \mathrm{V}$$
$$\begin{aligned}
E&=E^\circ+\frac{RT}{4F}\cdot\ln\left[\ce{H+}\right]^4\\
&= 1.229\ \mathrm{V}-0.05916\ \mathrm{V}\times\mathrm{pH}
\end{aligned}$$

The redox potentials $(E_{\ce{O2}} < E_{\ce{Cl2}})$ suggest that oxidation of $\ce{H2O}$ to $\ce{O2}$ at the anode should be preferred over oxidation of $\ce{Cl-}$ to $\ce{Cl2}$.

However, depending on the material and shape of the anode, the overpotential of oxygen at the anode can be very large. Therefore, the oxidation of $\ce{Cl-}$ to $\ce{Cl2}$ at the anode is feasible.

Nevertheless, at high $\mathrm{pH}$ and at a low concentration of $\ce{Cl-}$, the oxidation of $\ce{H2O}$ to $\ce{O2}$ is still preferred over oxidation of $\ce{Cl-}$ to $\ce{Cl2}$. Hence, if also $\ce{H2}$ and $\ce{OH-}$ are produced during the electrolysis of aqueous solutions (this is less relevant for the reduction of $\ce{Cu^2+}$ to $\ce{Cu}$ which is considered in the question, but for example important for the chloralkali process)
$$\ce{2H2O + 2e- -> H2 + 2OH-}$$
the electrolysis of chloride cannot be made complete in a single step since $\mathrm{pH}$ is increased and the concentration of $\ce{Cl-}$ is decreased during the electrolysis:
$$\ce{2 H2O + 2 Cl- -> H2 + 2 OH- + Cl2}$$


Why are substituted biphenyls optically active?

In general, compounds without improper axes of rotations are chiral. More concretely (and less correctly), compounds with less symmetry are more likely to be chiral.

Substituted biphenyl compounds generally have little symmetry. Why is this? Steric hindrance causes the two phenyl rings to be tilted relative to each other and serves as a barrier to rotation. This introduces dimensionality to the compound and breaks symmetry in the plane of the ring. If, in addition, there are substituents located asymmetrically on the rings, then the two rings can be differentiated from one another and thus lead to chirality. A similar situation is observed for spiro compounds.

BINAP is an example:

Remark. Most textbooks mention chiral centers as a sufficient (but not necessary) condition for chirality, and fail to emphasize other possibilities. Substituted biphenyl compounds, for example, may be chiral even without chiral centers.


Variation of angle in GeX2 molecules where X is an halogen

As for all atoms of period 3 and higher, we can consider germanium to be essentially unhybridised and the bonds being formed entirely by $\unicode[Times]{x3c3}_\mathrm{p-p}$ overlaps. Therefore, we expect a perfect $90^\circ$ angle for all three compounds.

However, while fluorine is small, the size of the halogens increases from fluorine to iodine. The larger the halogen the more space it requires. Since the central atom remains the same size (and, let’s approximate, is roughly the size of bromine) it cannot offer more space for a strict $90^\circ$ angle without lengthening (= weakening) the bonds which would be unfavourable. It is less unfavourable to widen the angle to generate more space. This requires the s-orbital to hybridise slightly (less favourable) but keeps the desired shorter $\ce{Ge-X}$ bond length (more favourable than anything else in this consideration). Not the bonding electrons repulse each other, but the steric stress of the halogen’s lone pairs causes them to move apart. The same effect can be observed in $\ce{NH3}$ versus $\ce{PH3}$ or $\ce{PH3}$ versus $\ce{PF3}$.

Note the following: If you have two bonds from a central atom which are each described by a bonding and an antibonding orbital, then these two sets of orbitals will always be orthogonal to each other (the integral of overlap is $0$). Thus they will be indifferent to each other, not repulsive.


Is a person warmed by fire an example of endothermic reaction?

I wouldn't call it an endothermic chemical reaction. The fire is an example of an exothermic reaction. The man constitutes part of the surroundings for the fire (the system), and heat is being transferred between the system and its surroundings. Looking at it the opposite way, the fire could be considered the surroundings (a heat reservoir) for the man (the system), and there is heat being transferred from the surroundings to the system. But the man is not undergoing a chemical reaction.


Synthesis of hydrazine with bleach and ammonia

I sympathize with your desire to launch an action figure. However, the chemistry you propose is really dangerous. You would be better off with black-powder based rockets.

  Is there any way to synthesize hydrazine with bleach and ammonia?

Do not do this.

My answer is not "yes" or "no". My answer is Don't. Do not try this without proper safety equipment and training, and then think twice or three times before doing so, and then have an ambulance standing by. Certainly don't do it at home, or by yourself.

This particular reaction is capable of producing hydrazine. However, it also producing a whole host of additional stuff, like chloramine, hydroxylamine, nitrogen trichloride, hydrogen chloride, and chlorine gas, all of which are either gases or volatile liquids, and all of which are poisonous. Hydrazine is poisonous. Nitrogen trichloride is a shock-sensitive explosive on the order of a sneeze can set it off. The reactions are as follows.

Chloramine:
$$\ce{NH3 + HOCl -> Cl-NH2 + H2O}$$

Dichloramine:
$$\ce{Cl-NH2 + HOCl -> NHCl2 + H2O}$$

Nitrogen trichloride:
$$\ce{NHCl2 + HOCl -> NCl3 + H2O}$$

Hydrazine and HCl:
$$\ce{NH3 + NH2Cl -> H2N-NH2 + HCl}$$

Hydroxylamine and HCl
$$\ce{NH2Cl + H2O -> NH2OH + HCl}$$

Chlorine gas:
$$\ce{HCl + HOCl -> H2O + Cl2}$$

Any organic impurities in the mix can lead to other deadly stuff being produced.

This reaction is not controllable to produce a single specific product. Everything it produces is poisonous and/or explosive. I will recommend against, and I will not help you find a recipe.


What is the formula for evaporation rate of water?

The temperature of water is implicitly covered by actual and saturated vapour content, or equivalent vapour pressures.

There is deterministic relation between relative humidity, air temperature and temperature of water, supposed to be in dynamic equilibrium.

In such a state, the cooling effect of water evaporation is balanced by the thermal transfer from warmer air.

This relation is used in meteorology/climatology as the standard method to determine humidity via the psychrometric tables of temperature of "dry" and "wet" precise thermometers.

Effectively, the wet bulb temperature is  somewhere between air temperature and the air dew point.

They are standard items in the white meteorological wooden box 2m above the lawn.

See psychrometrics particularly the sections dry and wet bulb temperature.

The main article Wet bulb temperature.


Best ingredients for a homemade desinfectant

H2O2 is caustic and reactive, but of the list above, it would be the most complete germicide. H2O2 is an oxidizing agent. H2O2 compares well with bleach (bleach is arguably, the king of disinfectants). However, typically the usage level of H2O2 is higher (3%) than other disinfectants. You cannot mix bleach with H2O2 (H2O2 reacts rapidly with bleach). Also, you cannot mix bleach with acid, although many people will mix vinegar and bleach in tiny amounts to give it an extra kick (which it will)... this is DANGEROUS. H2O2 can generally be mixed with (most) acids- especially organic acids like citric acid or vinegar.

Essential oils (especially phenolics like clove, cinnamon, and/or thyme oil) are the most germicidal e-oils, but e-oils are kind of like gasoline (plastic dissolving-staining etc.). Other, terpenic e-oils (like peppermint, pine, citronella, lemongrass, etc.) are somewhat germicidal. Limonene and pine oil are often used as a germicide, although it is comparitively weak to most essential oils. However, they is not as caustic or toxic as clove (for example)- although they do have some toxicity. Pine oil, thyme oil, peppermint, and limonene are popular choices for germicidal use. I have seen thymol and limonene mixtures before. Peppermint is common for school mop-products (becuase in schools, its common to clean up vomit, and peppermint helps quell nausea). Cinnamon, thyme and peppermint are common in mouthwash. Usage levels for e-oils is usually much less than 1%... they are not as effective (but still helpful) as other disinfectants at such levels, but the lower usage is balanced between the noxious-toxicity of the oil and the efficacy of the germicidal effect. Since antibiotic e-oil products would need to contain a noxious and cost prohibitive amount of e-oil to be effective (as a stand-alone antimicrobial), most antibiotic e-oil products (pine-sol for example) contain ammonia (high pH), and simply list pine as a scent (it would help kill bacteria- but to claim antimicrobial on a bottle, the product must be EPA registered and there are lots of problems with e-oils... again, cost, toxicity, and efficacy to name a few). You can mix e-oils with soap and alcohol. Other chemicals (acids, bases, and oxidizers) can/will break e-oils down, but not very rapidly. And the break down/byproducts would probably still have some degree of antimicrobial activity.

Vinegar is used by many DIY people, but it is not a popular commercial product because it smells kind of bad, it's not very effective, and if it is mixed with bleach (WARNING) it can cause a serious health hazard. Also, 5% acetic acid is not extremely germicidal (it will work to kill bacteria in food after a few days); many spores will survive for hours in acetic acid. Vinegar is most effective against non-spore forming (gram negative) bacteria- which are often the more nasty kinds of bacteria (Salmonella, E. coli, Serratia, Pseudomona, etc.)... but it's less effective against spore forming (gram positive) bacteria (Candida, Bacillus, Streptococcus, Micrococcus, etc.). Again, it is an effective preservative in food... as is 5% ethanol (ie., beer). 50% acetic acid is much more effective and it works faster... as does 50% ethanol. pH is a factor, so vinegar is a bit more effective than ethanol, but in general, a high pH is much more effective than a low pH. So, ammonia or lye solutions are more effective than organic acids.

70% isopropanol and 60% ethanol are a common hand and wound disinfectants. They are fairly antimicrobial and antiviral, but again, spores (and other gram positive bacteria) can survive in dilute alcohols (for a little while).

Borax is not germicidal (I don't care what the popular opinion on the internet is). The insecticidal efficacy of borates is also relatively poor, but since it doesn't leave or break down, it is effective over time (weeks, years, etc.)... persistence is why it is used as an effecitive termiticide for wood-framed houses. And although borax is claimed to be effective for roaches... mixed with bait and such- I have tested it thouroughly and it's not very good. But more at the subject here- I would not consider borates to be antimicrobial... unless you consider pH to be the antimicrobial agent. In which case, ammonia or lye are extremely antimicrobial (they really are). But you can't mix lye and ammonia for cleaning (DANGER). Also, bleach has a relatively high pH (about 11).

Notably, but not mentioned, SOAP is the originial antimicrobial and insecticide. It works great; pound for pound- it works better than vinegar. It's been used for centuries, and it has very low toxicity. But, since it has very low usage levels (generally less than 1% in water) and it is more for removing bacteria- so it is not purely germicidal. Generally speaking, soap more strongly inhibits gram positive, spore forming bacteria (at say 0.1% concentration) and inhibits most gram negative bacteria at around 4% concentration. Also, soap manufacturers don't need to pay hefty tolls for registering all their products as antimicrobial because people buy soap anyway. But adding a strong antibiotic (usually triclosan, which tends to work better on gram negative bacteria) they can easily skate through EPA with an affordable antimicrobial product (affordable especially compared with the cost of the claim which costs MILLIONS of dollars).

To claim "antibiotic" with EPA, a product must kill most bacteria (I think it's something like 99.9%, minimum, within 5 minutes against several standard/representative microbial species). You really can't do that without something like 60% ethanol, or 0.15% bleach (ie., 3.33% of a 5% bleach solution), or 3% H2O2, or 2% clove oil (ooh that stings!) or 0.1% triclosan in soap (which gets diluted 100x to about 0.001% in common use) or some antibiotic molecule in high enough concentrations to kill.

Some antimicrobial combinations are synergistic, some are antagonistic, some are additive, and some are deleterious. Most of the reliable combinations of over-the-counter antibiotic materials have already been tried and the best products (soap, hydrogen peroxide, alcohol, bleach, ammonia, and yes, even fragrance) have stood the test of capitalism and time.

Here is some good reading about this: Guideline for Disinfection and Sterilization in Healthcare Facilities, 2008


Books for a very advanced 3rd grader (elementary school)

One of the best is The Golden Book of Chemistry Experiments.  By today's standards, this book is considered "dangerous", but keep in mind that it was written for teenagers in the 1960s.  In any case, parental supervision should always be required in chemistry, and if you carefully research the experiments before doing them you should be fine.  It takes you all the way through technique, building apparatus, and safety.  The only problem is obtaining chemicals: the sources cited therein are hopelessly outdated (i.e. your will NOT find most of this stuff in drugstores anymore), and you will need to use educational supply companies to obtain small quantities of chemicals.  Have fun!


Dipole moment of molecule (isolated vs in solution)

I've actually been investigating this in my comp chem coursework. A large dielectric can support more charge separation and therefore stronger dipoles, but the effect diminishes. Going from an $\epsilon$ of 2 (cyclohexane) to 9 (DCM) gives a much bigger change in dipole than going from 9 to 80+ (water).


1.) So is iron the only element that can do that? If not, what is it about iron that makes it the only element found in hemoglobin?

No, iron is not the only metal that will fit inside the porphyrin structure. But its ability to be an electron acceptor and donor makes it very useful for oxygen and carbon dioxide to bind to the hemoglobin. Since iron is also very abundant, it was easy to incorporate it and utilize its properties.

Other macrocyclic structures include chlorophyll with magnesium in the center and vitamin B12 (cobalamin) with cobalt in the center.

  2.) What is it about being able to gain and lose electrons make it essential to life? I think it has something to do with oxygen being able to bind with iron. ... What is special about iron?

Iron is abundant. The iron in the hemoglobin can weakly bind both oxygen and carbon dioxide.

You can find more information on the Wikipedia page.


Why is the effect of ring activators more dominant than ring deactivators?

$\ce{-NH2}$ activates the ring due to its +M effect, increasing electron density in the ring at (b) and (d). On the other hand, $\ce{-NO2^{-}}$ causes de-activation because of its -M and -I effect and the electron density would be higher at (c) relatively.

Now since in activated rings, the rate of electrophilic aromatic substitution is higher than those in deactivated rings, the effect of $\ce{-NH2}$ as ortho-para director is seen more during product formation and not of $\ce{NO2^{-}}$ majorly. The electrophile will attack at (c), but since its rate will be quite slow, it would not form the major product.


I've seen similar demonstrations with magnesium, and in both cases the explanation is the same.  The metal reacts to produce the hydroxide plus hydrogen; with zinc this would be
$\ce{Zn + 2 H2O -> Zn(OH)2 + H2}$
But, the zinc hydroxide is essentially insoluble in cold water and seals up the metal before the reaction gets very far.
Boiling the water speeds things up in two ways.  First, the metal hydroxide, with either zinc or magnesium, is inherently more soluble.  Second, the turbulent action created by the boiling (the bubbles you see are actually steam, not hydrogen) acts as a stirring mechanism, speeding up the mass transfer so the slight concentration of dissolved hydroxide gets to the litmus paper faster.
Magnesium differs from zinc because the former hydroxide is somewhat more soluble.  Therefore, the basic nature of the product with magnesium can be demonstrated with phenolphthalein, which requires a slightly higher pH to react than litmus.  Some demonstrations with magnesium show a rose coloring of the magnesium metal in cold water with the phenolphthalein showing that the magnesium hydroxide remains essentially on the surface.


Mendeleev was the first to realize that properties recur among the elements. To highlight this, he devised the early periodic table based on atomic masses. Moseley came along and stated that the periodic properties of elements was a function of their atomic numbers rather than their atomic masses.

It was Niels Bohr who came up with the idea of shells and subshells. He developed his own take on the periodic table, which looked somewhat like this:

Finally, coming to your question, A lot of derivatives of the periodic table were made by various scientists during the 1920's - 1940's. The version we use today was made by Glenn T. Seaborg and his colleagues.

  Seaborg and collaborators had synthetically produced several new elements with atomic numbers beyond uranium, the last naturally occurring element in the table. Seaborg saw that these elements, the transuranics (plus the three elements preceding uranium) demanded a new row in the table, something Mendeleev had not foreseen. Seaborg’s table added the row for those elements beneath a similar row for the rare earth elements, whose proper place had never been quite clear, either.

Further reading:

How the periodic table went from a sketch to an enduring masterpiece
Further development of the Periodic Table - Wikipedia


We'll assume the reaction loses only $\ce {CO2}$ from the system (although it is a exothermic reaction, we'll assume $\ce{H2O}$ produced would not be lost as vapors):
$$\ce{CaCO3 + 2 HCl -> CaCl2 + H2O + CO2}$$

Mass of $30\%~\ce{HCl}$ needed to react with $\pu {605.5 g}$ of $\ce{CaCO3}$ $\pu {= 605.5 g \ce{CaCO3} \times \frac{\pu{1mol}~\ce{CaCO3}}{\pu{100.09g}~ \ce{CaCO3}}\times \frac{\pu{2mol}~\ce{HCl}}{\pu{1 mol}~\ce{CaCO3}}\times \frac{\pu{36.45g}~\ce{HCl}}{\pu{1mol}~\ce{HCl}}\times \frac{\pu{100g}~\text{solution of 30%}~\ce{HCl}}{\pu{30 g}~\ce{HCl}}= 1470.04 g of 30\% \ce{HCl} solution}$.

Similarly, mass of $\ce{CO2}$ released when $\ce{HCl}$ reacted with $\pu{605.5 g}$ of $\ce{CaCO3}$ $\pu{= 605.5 g \ce{CaCO3} \times \frac{\pu{1mol}~\ce{CaCO3}}{\pu{100.09g}~\ce{CaCO3}}\times \frac{\pu{1mol}~\ce{CO2}}{\pu{1mol}~\ce{CaCO3}}\times \frac{\pu{44.0g}~\ce{CO2}}{\pu{1mol}~\ce{CO2}} = 266.18 g \ce{CO2}}$.

According to the law of conservation of mass,

$\pu {mass of reactant = mass of products}$

Assuming no solvent left the system,

$\pu {mass of reactant + solvent = mass of products + solvent}$

$\pu{mass of reactant + solvent = 1470.04 g + 605.5 g = 2075.54 g}$

$\pu{mass of products + solvent remained in the flask = mass of reactant + solvent - mass of \ce{CO2} released = 2075.54 g - 266.18 g = 1809.41 g}$

Now, suppose extra $\pu{A g}$ of $30\%~\ce{HCl}$ added to the solution, so that final concentration is $3\%~\ce {HCl}$.

$\pu {mass of \ce{HCl} in A g of 30\% solution = A g of 30\% \ce{HCl} solution \times \frac {\pu{30g}~\ce {HCl}}{\pu{100g}~ \text{of 30% $\ce{HCl}$ solution}} = 0.3A g of 30\% \ce{HCl}}$

Thus, since final concentration of solution is $\mathrm {3\%}$,

$$\pu {\frac {(0.3 A) g}{(1809.41 + A) g} = 0.03}$$
$$\pu {0.3 A = 0.03 \times 1809.41 + 0.03A}$$
Thus, $\mathrm {A} = \frac {0.03 \times 1809.41}{0.3-0.03} = \pu{201.04 g}$

Finally, the mass of $30\%~\ce{HCl}$ initially added is $\pu {(1470.04 + 201.04) g = 1671.1 g}$.


Why for compounds of elements in a group with larger anions have nearly constant lattice enthalpy?

If we assume that the lattices have the same structure, the lattice enthalpy is dominated by the charge of the ions and the distance between them. Since we are dealing here with cations from the same group, the charges are the same, so the remaining variable is the distance between them.

With moderately sized or small anions, a smaller cation can get closer, resulting in a shorter overall distance and a larger magnitude lattice enthalpy. In the case of very large anions, however, the anions are already packed as closely together as possible, so if a smaller cation moves closer to one sulfate, it moves farther from another. The equilbrium position of the cation is thus determined solely by the distance between the sulfates and does not vary with cation size.

For the second part of your question, I'm guessing that your statement that "larger cations stablise larger anions and small cations stabilise small anions" might be made with respect to solubility?

Solubility is a function of both the lattice enthalpy and the solvation enthalpy, which are in competition with each other. Generally speaking, small ions have higher solvation enthalpy in water. However, they also can have greater magnitude lattice enthalpy if they are able to pack more closely together.

Empirically, it has been observed that if the cation and anion are both small, the lattice enthalpy "wins" over the solvation enthalpy, and the salts tend to have low solubility in water.

If both ions are large, both the lattice enthalpy and the solvation enthalpy have smaller magnitude, and the empirical observation is again that the lattice enthalpy "wins".

When the sizes are mismatched, however, we find that the solvation enthalpy "wins" and the salts are generally soluble. We can see why in the extreme example from the first part of your question, where a smaller cation does not increase the magnitude of the lattice enthalpy at all, but will definitely increase the magnitude of the solvation enthalpy relative to a similar salt with a larger cation.


Electrolysis of water: Which equations to use? (IB Chem)

For the acidic electrolysis, use the reactions where $\ce{H+}$ occurs.

As $\ce{OH-}$ is not available in considerable amount there as a reagent, neither it is created as a product.

Generally, for a reaction choice, apply the principle of availability and stability, allowing for a reagent to exist in (relative) abundance.
$\ce{OH-}$ or anions of weak acids like $\ce{ClO-}$ do not survive in acids. Acids do not survive in hydroxides.

But note that using reactions with half of a molecule is not necessery.

$$\begin{align}
\ce{O2(g) + 4H+(aq) + 4e-  &<=> 2 H2O(l)}\\
\ce{2H+(aq) + 2e- &<=> H2(g)}
\end{align}$$

For the alkaline electrolysis, similarly, use the reactions where $\ce{OH}$- occurs.

$$\begin{align}
\ce{2 H2O(l) + 2e- &<=> H2(g) + 2 OH^-(aq)}\\
\ce{O2(g) + 2 H2O(l) + 4e- &<=> 4 OH^-(aq)}
\end{align}$$


How to optimise battery usage on modern electronics?

There is always some internal resistance in a battery. All other things being kept constant, the larger the battery, the less internal resistance, so less energy is wasted inside a larger battery due to internal heating and energy loss.

This is particularly important in high-drain devices such as phones and laptops, where batteries are discharged in just a few hours. By having the power bank in parallel, the load is shared and batteries are being drained over a longer time.

For example, a once-cell penlight with an incandescent lamp might burn for an hour. Two cells, one replacing the other, would then last two hours. But two cells in parallel might last 2.2 hours, because of the lower internal resistance and decreased energy waste in the cells.


producing ozone through electrolysis of water

First of all $\ce{H2SO4}$ produces $\ce{H2SO5}$ which is a peroxyacid during electrolysis  which most electrolytes will not do, so I suspect that $\ce{H2SO4}$ will be mandatory unless phosphoric acid ($\ce{H3PO4}$) can.  This peroxyacid give you a potential for a mechanism of ozone creation.
$$\ce{H2SO5 + O2 -> H2SO4 + O3}$$
Note: I do not claim this to be the mechanism but only an explaination for why the $\ce{H2SO4}$ may be important.

The carbon and platinum electrodes are not only chemically resistant, but may likely have a catalytic effect in producing ozone.  You can get platinum wire at a reasonable price if you look around.

In regard to testing if you have ozone, you can use paper strips that have potassium iodide impregnated in them.  The ozone will oxidize the iodide ion and turn the paper strip brown.


You're right that IF the metal calorimeter and the styrofoam calorimeter were both perfect and behaved in the way that we usually pretend calorimeters do (they don't absorb or lose heat) then they should produce identical results, for the enthalpy reason you provide.

However, since we're asking about specific materials, metal vs styrofoam, we're considering real calorimeters instead of ideal.

There are a few things that calorimeters can do that will affect how large a temperature change you measure.

Heat loss - As you note, metal conducts heat much more effectively than styrofoam.  This means that the metal calorimeter will carry the heat of the reaction away from the reaction vessel to the outside world faster than styrofoam will.  If that is the case, which reaction vessel will get hotter, styrofoam or metal?
Heat absorption - The calorimeter (and thermometer) gets heated up along with the reaction mixture.  A calorimeter that absorbs large amounts of energy will have a relatively lower final temperature, since more of the reaction's heat went into heating up the calorimeter than went into heating up the reaction mixture.  This quantity is known as the calorimeter constant and varies depending on the materials and mass of the calorimeter.  Styrofoam absorbs more heat per gram than steel, but the steel calorimeter is probably heavier.  With a fast reaction there is the additional fact that the styrofoam calorimeter probably never comes up to the full temperature of the reaction before you quit recording temperatures.  This winds up leading to a small source of lost heat for the metal calorimeter as compared to the styrofoam.
The quality of the equipment -  Metal calorimeters are often made specifically for the purpose of calorimetry.  Styrofoam calorimeters are often a couple of coffee cups jammed together.  If the calorimeter allows a significant amount of heat loss to the atmosphere due to its construction, then it will result in a cooler reaction mixture and a lower calculated enthalpy for the reaction.

You've got to take all three of these into consideration in order to determine the particulars for your experiment.


What would be the effect if someone were to drink ultra-pure, 18 M-ohm water?

Not much, although if they drank many gallons of the water it could be a problem.

  Would they immediately die?

No.

  Would they just need to pee more?

Probably.

  Would $\ce{CO2}$ from the air (after the bottle is opened) and whatever's in saliva dissolve into the water making it much less pure before it gets into the important parts of their digestive tract?

It would become orders of magnitude less pure as soon as it mixed with their saliva and then their gastric juices.  Same as regular water.

  What would be the threshold for bad things to happen -- a teaspoon, a liter, 12 gallons in 10 minutes (at which point I assume regular water would kill you)?

I don't think the effect would be that different from drinking regular water.  All over the internet there is this idea that super pure water is somehow too pure or dangerous, no good evidence or mechanism is ever presented.

Water is toxic primarily because it dilutes the sodium and potassium ions in your body, leading to Hyponatremia or possibly the far more dangerous Hypokalemia.  The normal level of $\ce{Na+}$ in blood plasma is about 135 millimolar (mM), and the normal level of $\ce{K+}$ is about 5 mM.

A very rough approximation to understanding how toxic water can be would then be to compare the concentration difference of sodium (or potassium) between the ingested water and the blood.  A 25 micromolar -- i.e. 25 μM or 0.000025 molar solution of sodium hydroxide has a conductivity of about 6.7 microsiemens per cm, more than 100-fold higher than "pure" 18.2 megohm water (0.055 microsiemens per cm).  According to this rough difference-based model, the ability of the very dilute sodium ion solution to cause problems is proportional to (135 - 0.025) mM, and the ability of the "pure" water to cause problems is (135 - 0) mM.  For nearly biological or medical conditions of interest, 135 mM is not really different from 134.975 mM, and so ultra-pure water is not really more dangerous than everyday drinking water.


Reason for endothermic nature of NaHCO3 + C2H4O2 reaction

Your "all of the above" answer is correct, with emphasis on number 2.  The phase change from a condensed phase (solid or liquid) to gas is almost always endothermic.  It's the same thing as wetting your hand then blowing on it.  The liquid water undergoes and endothermic phase transition which takes in heat from its surroundings, i.e. it evaporates and your hand gets cold.


There's no further need to comment on the coffee maker, but you might give the following cleaning method for the condenser and the rotation tube a try:

Note that you might want to first check whether this is in accordance with local safety regulations!

Take a small or medium-sized (max. 500 mL) 2-neck flask (NS29 socket connects to the rotavap, NS14.5 socket is for your thumb) and fill it with max. 250 mL acetone
Hook it up to the rotavap, support it with your hand
No spinning :D
Start the vacuum pump
Close the NS14.5 socket with your thumb
Watch the acetone starting to boil under reduce pressure
Rapidly remove your thumb from from the socket and let the acetone "spit" all through the rotavap


How to determine whether a organic compound is diamagnetic or not?

If all the electrons are spin paired then the molecule will be in a singlet state$^*$ and will be diamagnetic, if they are not so paired then it will be  will be paramagnetic. The rules are the same whether or not the molecule is 'organic'.

Most 'normal' organic compounds, e.g. methane, benzene, etc. have paired electrons in each of the orbitals from the lowest one to the highest occupied one (HOMO), and thus are diamagnetic. The orbitals including the LUMO and up are unoccupied.

Radicals such as $\ce{CH3\cdot}$ have an unpaired electron (doublet state) and so are paramagnetic.

If a biradical is formed e.g. carbenes (see IUPAC Gold Book 'biradical' for an example structure) then it will depend upon whether the spins are paired in a singlet state (diamagnetic) or parallel as in a triplet state which will be paramagnetic. Often singlet and triplets are close in energy with the triplet being lowest. Similarly, dioxygen has two unpaired electrons and forms a triplet ground state so is paramagnetic. Nitric oxide  has one unpaired electron and is paramagnetic.

If a molecule is put into an electronically excited state by absorbing a photon then both singlet and triplet states can be produced, however, theses are short lived and decay back to the ground state.

$^*$(singlet /doublet/triplet refer to spin multiplicity, spin quantum number  $\pm 1/2$ and total spin e.g. $S=s_1+s_2$ gives multiplicity g  as $g=2S+1$)

example of biradical from IUPAC Gold Book.


Enthalpy change of gas expansion against constant external pressure

Assume an ideal gas
I think we have to assume an ideal gas, otherwise it is not clear how to get the final temperature of the gas. In the OP's comments, the final temperature is determined as 600 K.
The difference between change in energy and change in enthalpy
Enthalpy is defined as:
$$H = U + PV$$
If both P and V change from state one to state two:
$$H_1 = U_1 + P_1 V_1$$
$$H_2 = U_2 + P_2 V_2$$
we can relate the change in enthalpy to the change in energy by subtracting the two equations:
$$\Delta H = \Delta U + P_2 V_2 - P_1 V_1$$
We could write $\Delta(PV)$ for $P_2 V_2 - P_1 V_1$, but maybe that is confusing.
Change in energy of the gas
In the question, the temperature, the volume and the pressure of the gas all change as we are going from initial to final state. Because energy and enthalpy are state functions, we can decide in which order these changes happen. It makes sense to use steps from initial to final that are easy to calculate. I will change temperature first, then volume (the pressure will change as well, and we will see how that plays into the calculation).
Initial state: $V = \pu{1 L}, T = \pu{300 K}$
Intermediate state: $V = \pu{1 L}, T = \pu{600 K}$
Final state: $V = \pu{4 L}, T = \pu{600 K}$
The change in energy from the initial state to the intermediate state is given by the OP:

$$\Delta U = C_V \Delta T = \pu{1.5 kJ}$$

Taking that intermediate state (V = 1L) to the final state (V = 5L) does not involve any change in internal energy if the gas is ideal, so $\Delta U = 0$ for this step. We could expand into a vacuum, or against pressure. In the latter case, the temperature would drop and we would have to transfer heat into the system to the same amount as the work it did on the surrounding.
So for the entire process, $\Delta U = \pu{1.5 kJ}$.
Change in enthalpy of the gas
$$\Delta H = \Delta U + P_2 V_2 - P_1 V_1 = $$
$$ = \pu{1.5 kJ + 5 atm 4 L - 1 atm 10 L} = $$
$$ = \pu{1.5 kJ + 10 atm L = 1.5 kJ - 0.101 kJ = 1.4 kJ}$$
What about the external pressure?
It does not matter for the calculation because the internal energy of a gas (assuming ideality) does not depend on it. In other problems, where the volume of the gas is not constrained, it might be important, but here it is not.
You could solve this problem going along another path, and keeping track of volume work and heat transferred to get at the changes in energy (using the first law of thermodynamics). This would be more complicated, though.


What is the name and formula of this structure?

That's Atorvastatin, a cholesterol lowering drug (works by inhibiting the enzyme HMG-CoA reductase).
https://en.wikipedia.org/wiki/Atorvastatin
For more data, see e.g. https://www.trc-canada.com/product-detail/?CatNum=A791725


Which is the most acidic proton of (3Z)-3-ethylidenecyclopent-1-ene-1-carbaldehyde?

If you draw curly arrows for the deprotonations, it becomes clear that only the pathway in green is productive due to the ability to form an extended enolate.

If you deprotonated the green protons, the resulting negative charge is able to be delocalised through the alkene and onto the carbonyl, the product of which is essentially an extended enolate. Due to this stabilisation, the pka of the green protons is relatively low (the resulting conjugate base is stabilised).

Deprotonation of the red proton however doesnt yield a charge which may be delocalised onto the carbonyl ( you would end up with a 5 valent carbon if you draw the arrows ). What you essentially end up with in this case then is an allyl anion (which itself is conjugated to the exocyclic alkene), which although stabilised by resonance isn't quite as favourable as in the green case (the pka of the red protons is higher).


The said lesson tends to oversimplify things to the point of getting them downright wrong. Atom dipoles do not line up with each other. They are way too weak for that. Should it be according to your description, they would indeed line up all in one direction and stay locked forever, much like magnetic particles in a magnet.

Now, I brought up the magnetic analogy because it is much better known than the similar electric phenomenon, called ferroelectricity. That's where your explanation is precisely right. Compounds that exhibit ferroelectric properties, once polarized, never lose their dipoles unless you force them to. But they constitute just a tiniest minority of all compounds.

What happens in the rest of them? Well, the molecules turn this way and that, atoms bounce and wiggle, spontaneous dipoles arise every now and then, only to vanish almost instantly. But the position where two neighboring dipoles are aligned in the same direction is energetically favorable, so they spend a little more time in this position than in the opposite. And that's what we call induced dipoles, dispersion forces, etc.

So it goes.


If ionic compounds are strongly bonded, why are acids with a more polar central atom the stronger types of acids?

Ionic compounds make a network of electrostatic attraction. Molecules have covalent intramolecular bonds which are very strong, but each molecular unit is weakly attracted to one another (relatively) due to lack of ionic attraction.  So, essentially, in the solid form, each $\ce{NaCl}$ has more attraction to another $\ce{NaCl}$ than does each molecule to another in a molecular system. That does not, however, translate to each $\ce{Na+}$ ion of an $\ce{NaCl}$ unit being more tightly bound to a $\ce{Cl-}$ ion than two atoms of a covalent bond are to each other.

The idea that ionic bonds are "stronger" than covalent bonds is confusing to early chemistry student (and maybe more so to chemists, but it's tough to convey concisely what is actually occuring) but, for some reason, seems to be nearly ubiquitous in general chemistry courses. It is much easier for compounds like water to stabilize $\ce{Na+}$ and $\ce{Cl-}$ than to dissociate something like paraffin wax into ions in order to solvate the molecules as ions.

So I guess what I'm trying to say is: the ions of ionic compounds aren't usually held more tightly together than two atoms of a covalent bond.

As for the idea that it is a higher electronegativity difference in $\ce{H-A}$ bonds that allow for higher acidity, that can be disproven by looking at $\ce{HF}$ vs $\ce{HI}$. $\ce{HF}$ has a higher electronegativity difference but is a weaker acid. This is because $\ce{I-}$ is larger, so it can spread the negative charge out, essentially stabilizing the negative charge on its own. $\ce{HF}$ produces $\ce{F-}$, and because fluorine is more electronegative than iodine, that may be expected to be a stronger acid than $\ce{HI}$. However, because $\ce{F-}$ is much more of a point-charge rather than a diffuse charge like $\ce{I-}$, it has to be stabilized by solvation effects to a larger extent.

So what causes an acid to be more acidic is the thermodynamic stability of the resultant conjugate base. Because the charge in $\ce{I-}$ is more stabilized by the ability to spread out its own charge, it more readily donates $\ce{H+}$ than does $\ce{HF}$.


The compound you are presenting is a natural product, compactin. The numbers of the carbon atoms are arbitrary. The digraph of C9 shows how the configuration is determined. Sphere 2 contains the atoms C8, C10, C14 and H. Clearly, hydrogen has the lowest priority. Sphere 3 has the locants C10{15,5,H}, C8{7,(7),H} and C14{H,H,H}. In a one-to-one comparison of the locants, C10 and C8 have priority over C14 because C>H. So C14 has the next lowest priority. C(7), which is a duplicate atom of atomic number 6, is attached to three atoms in sphere 4 of atomic number zero. Zero is as low as you can go. Therefore, the priorities are C10>C8>C14>H. C9 has the (S)-configuration.


According to an 1894 tract with the title Merck's Market Report and Pharmaceutical Journal: An Independent Monthly Magazine Devoted to the Professional and Commercial Interests of the Druggist, Volume 3, "cresol soap" was also called "crelium" or "crelium soap".
The google n-gram viewer suggests that "cresol soap" is a historical term, whose usage began in the 1890s, peaked dramatically during World War I and its immediate aftermath, and declined substantially afterwards.
A 1915 British medical report argues for using cresol soap as a vermin control agent in troop populations.  It gives a recipe as:

10 gallons of water
1.5 pounds of soft soap
Jeyes' fluid 1.5 ounces

This raises the question of what "Jeyes' fluid" is. Amazingly, Jeyes' Fluid is apparently still a commercial product, and contains a mixture of 4-chloro-m-cresol, isopropanol, terpineol, and "tar acids".
"Cresol soaps" are apparently still an item of commerce somewhere in the world, as a 2011 applications report form an analytical equipment vendor illustrates the analysis of a "cresol soap" sample, although a specific source is sadly not mentioned.  The primary detected constituents were m-cresol and p-cresol.

I suspect there are many other recipes that involve mixing soap, water, and crude cresol preparations of various kinds.  In the era of cresol soap's invention and popularity, coal was far more ubiquitous than now, and the cresol sources were probably crude fractions of distilled coal tar that were enriched in various isomers of cresol.  In those days awareness of potential toxicity from cresols was also not as developed as now.  The "tar acids" in Jeyes' fluid probably include a variety of cresol isomers.


How to explain two NMR spectra of the (allegedly) same compound with slightly different shifts?

That's a pretty small chemical shift difference.  Keep in mind that chemical shifts depend on temperature, concentration, pH, and other factors.  The concentration difference is the most likely, but I wouldn't rule out T effects.  If you're running it in a different batch of CDCl3 (or even the same batch, but it's aged), the pH could be different as CDCl3 becomes acidic over time.

If you used identical settings when collecting the spectra (e.g. using a default Proton) then you can measure the signal-to-noise ratio of two identical peaks and get an estimate for how different the concentrations are.  You can also simply test how sensitive your chemical shifts are to concentration by making up a more concentrated sample (or diluting your current sample).  Or try changing the temperature with the VTC and ramp it up to 25.5 or 26.0 C and see how much the chemical shifts change (referencing everything to CHCl3 for consistency).

If your compound were reacting with something, there would be a much more dramatic change in the chemical shifts than 0.07 ppm.


Why is dinitrogen tetraoxide less stable at high temperature?

I figured out the answer using Van't Hoff Equation:
$$\ln\left(\frac{\ce{K_2}}{\ce{K_1}}\right)=-\frac{\ce{\Delta H}}{\ce{R}}\left({\frac{1}{\ce{T_2}}-\frac{1}{\ce{T_1}}}\right)$$

Let's assume that $\ce{\Delta H}$ is constant for a limited range of temperature.

Since $\ce{\Delta H}$ is negative for the reaction $(=\pu{-13.87 kcal mol^{-1}})$ if we increase the temperature i.e. $\ce{T_2>T_1} \to\ce{K_1> K_2}$. Thus, equilibrium shifts towards the left as the temperature increases implying the decreasing stability of $\ce{N2O4}$ with increasing temperature.


What would be the product (B) in the following sequence of reaction?

This is a fun one, and a common question in this course.  First, as you say, H and Br are lost by anti elimination, since they helpfully tell you base and E2 are involved.  That means the H and Br are lost from opposite faces of the future double bond, leaving us with a trans intermediate that has a mirror plane (the plane of the molecule)

Now we need to add Br+ to make a bromonium intermediate.  Notice that both carbons end up as (R) or else both end up as (S), so two enantiomers are possible here.

But, whichever carbon is attacked by Br- ends up with an inversion of symmetry, so you now have two carbons of opposite chirality.  The result, as shown in your answer, is a molecule with a mirror plane.  In other words, adding bromine to opposite sides of a trans double bond makes both sides come out as mirror images, once you turn the bromines around to line up.  This compound is a meso compound.  But ... what about the other isomer if the Br+ originally goes to the other side of the double bond?  Well, just turn your Fisher structure upside down. :)


How to cancel once-mentioned element in thermochemical reactions

We can make use of this formula: $$\Delta H = \Sigma ~\Delta H_f~(\text{products}) - \Sigma ~\Delta H_f~(\text{reactants}) $$

Take $\ce{O3 \rightarrow \frac{3}{2} O2}$ into consideration. Ozone molecule is the less stable form of O. If you know that how much the $\Delta H$ for that reaction is, you can in fact predict the $\Delta H_f$ for $\ce{O3}$. If you're interested, wikipedia has documented some of the enthalpies of formation. Notice that the most thermodynamically stable form of the element is considered to have $\Delta H_f = 0$.

Now that the $\Delta H_f$ for ozone is determined, you can apply the formula above for the first reaction: ($x$ is $\Delta H$) $$x = (\Delta H_f \ce{(NO2)} + 0) - (\Delta H_f (\ce{NO}) + \Delta H_f (\ce{O3}))$$

So, from the equation above we can conclude that $$x + (\Delta H_f (\ce{NO}) + \Delta H_f (\ce{O3})) = (\Delta H_f \ce{(NO2)}) $$

Now write the master formula for the last reaction: $$(2 * \Delta H (\ce{NO}) + 0) - 2 * \Delta H_f (\ce{NO2})$$

What did we conclude from the last reaction? $x + (\Delta H_f (\ce{NO}) + \Delta H_f (\ce{O3})) = (\Delta H_f \ce{(NO2)}) $ Assuming that $x$ is at hand, do you really need to know the $\Delta H_f$ for $\ce{NO}$ & $\ce{NO2}$?


It's probably considerably notorious historical trade name. In many older publications, e.g. "Skellysolve B (petroleum ether)"[1,2] appears for solvent used.

It was a product or trade name of Skelly Oil Company[3]
 founded in 1919.

It's not a pure hexane, rather narrow-boiling hexane oil fraction.[3]

I could find analysis of similar product Skellysolve C, which is C7 heptane fraction:[3]

24.9% heptane
22.9%

2,3-dimethylpentane
2-methylhexane
3-methylhexane

19.3%

trans-1,2-dimethylcyclopentane
trans-1,3-dimethylcyclopentane

15.0% methylcyclohexane
5.5% cyclohexane
5.2% heavy nonaromatics
2.7% toluene
2.5% hexane
…

Other ones:[4]

  Skellysolve A is essentially n-pentane, b 28-30°/atm,
  Skellysolve B is essentially n-hexane, b 60-68°/atm,
  Skellysolve C is essentially n-heptane, b 90-100°/atm,
  Skellysolve D is mixed heptanes, b 75-115°/atm,
  Skellysolve E is mixed octanes, b 100-140°/atm,
  Skellysolve F is petroleum ether, b 30-60°/atm,
  Skellysolve G is petroleum ether, b 40-75°/atm,
  Skellysolve H is hexanes and heptanes, b 69-96°/atm,
  Skellysolve L is essentially octanes, b 95-127°/atm.

Unfortunately, Chemical Abstracts Service, mixes names for different substances, including this tradename: “n-hexane”, “hexane (all isomers)”, “Skellysolve B” under single CAS RN 110-54-3.[5]

References:

Newman, H.; Zilversmit, D. Quantitative Aspects of Cholesterol Flux in Rabbit Atheromatous Lesions. Journal of Biological Chemistry 1962, 237 (7), 2078–2084.
Holick, M. F.; DeLUCA, H. F. A New Chromatographic System for Vitamin D3, and Its Metabolites: Resolution of a New Vitamin D3, Metabolite. J. Lipid Res. 1971, 12 (4), 460–465.
Griswold, J.; Andres, D.; Berg, C. F. V.; Kasch, J. E. Pure Hydrocarbons from Petroleum - Separation of Straight-Run Fractions by Distex Process. Ind. Eng. Chem. 1946, 38 (1), 65–70.
Armarego, W. L. Purification of Laboratory Chemicals; Butterworth-Heinemann, 2017.
CommonChemistry.org – CAS Registry Number: 110-54-3


The law of mass action only rigorously applies to elementary reactions (in that otherwise the exponents do not match stoichiometric coefficients). These are reactions whose equations match which particles are colliding on the microscopic scale. Consider the reaction

$$\ce{$r_1$R1 + $r_2$R2 + \ldots + $r_n$R_n <=> $p_1$P1 + $p_2$P2 + \ldots + $p_m$P_m}\tag1.$$

Reaction $(1)$ is elementary if and only if it proceeds in a single step. At some time $t = t_c$ all reactant particles $\ce{R1, R2, \ldots, R_n}$ collide. Quantum mechanically this collision is not too well defined, so view it to be Newtonian in essence. Not to mention: the probability $P$ of the simultaneous collision goes to zero as $n \to \infty$. This is why in actuality most elementary steps involve at most three individual constituents.

We therefore sacrifice some generality and instead consider the easier to handle

$$\ce{R1(g) + R2(g) <=> P1(g) + P2(g)} \tag{1'}.$$

Energies

Collision theory postulates that particles only initiate a transformation when they have enough energy. This is called activation energy and denoted by $E_a$. Maxwell-Boltzmann distribution

$$\frac{\mathrm{d}N_v}{N} = 4\pi\left(\frac{m}{2\pi k_bT}\right)^{3/2}v^2\exp\left(-\frac{mv^2}{2k_bT}\right)\mathrm{d}v\tag2$$

allows us to estimate the fraction of particles that have the required amount of movement. Note we are technically assuming $\mathrm{total\ energy} = \mathrm{kinetic\ energy}$. Similarly relativistic effects are ignored, and no cap is placed on maximum velocity. With these simplifications, the fraction of interesting particles is

$$f = \frac{N^*}{N} = \frac{1}{k_bT}\int_{E_a}^\infty\exp\left(-\frac{E}{k_bT}\right)\mathrm{d}E\tag3$$

which evaluates to

$$f = \exp\left(-\frac{E_a}{k_bT}\right)\tag{3'}.$$

Number of interesting collisions

Say $Z^*$ is the number of collisions that matter. We will look for a function $Z$ such that

$$Z^* = Z\cdot f.\tag4$$

The Maxwell-Boltzmann distribution $(2)$ gives for a single kind of particle (say two $\ce{R1}$'s) that there are

$$Z = 2N^2\left(\sigma_\ce{R1}\right)^2\sqrt{\frac{\pi RT}{M_{\ce{R1}}}}\tag5$$

collisions in $1$ second per $1\ \mathrm{cm^3}$ if $N$ is the number of particles per square centimeter. (This is the traditional starting point unit.) Hence the rate of such a reaction would be

$$v_r = \frac{2Z^*}{N_A} \cdot 10^3\ \ \ \left(\frac{\mathrm{mol}}{\mathrm{s\cdot dm^3}}\tag6\right).$$

Reaction rate $v_r$

If you like, you are now able to substitute $f$ from $(3')$ and $Z$ via $(5)$ into $(4)$. Then insert $(4)$ into $(6)$.  After necessary manipulation, this yields $(7)$

$$v_r = \underbrace{4 \cdot 10^{-3} N_A\left(\sigma_\ce{R1}\right)^2\sqrt{\frac{\pi RT}{M_{\ce{R1}}}}\exp\left(-\frac{E_a}{k_bT}\right)}_k \cdot  \overbrace{\left(\frac{N}{N_A}\cdot 10^3\right)^2}^{c^2}.$$

Coming back to our actual reaction $(1')$ involves more relative quantities. For example, $\sigma_\ce{R1}$ becomes $$\sigma_\text{aver} = 0.5\left(d_\ce{R1} + d_\ce{R2}\right).$$

So it is a bit more difficult. But the result is analogous to $(7)$.

$$v_r = \underbrace{2\sqrt{2} \cdot 10^{-3} N_A\left(\sigma_\text{aver}\right)^2\sqrt{\pi RT\left(\frac{1}{M_{\ce{R1}}} + \frac{1}{M_{\ce{R2}}}\right)}\exp\left(-\frac{E_a}{k_bT}\right)}_k \cdot c_{\ce{R1}} \cdot c_{\ce{R2}}$$

More compactly, if $\ce{R1} = A$ and $\ce{R2} = B$

$$v_r = k[A][B]\tag8$$

which is what we set out to prove.

This derivation assumes that every active collision leads to a reaction. When theoretically computed $k$ were compared to experimental, an extra factor was introduced, $P$. This is called the steric factor, and in classical collision theory $P$ remains empirical in essence.

Collision theory is for elementary steps

As your quote suggests, collision theory is a first theoretical explanation for the proportionality to products of concentrations. It does not generally explain various exponentiation. It does not have to either. The higher (or non-integer or negative) exponents usually derive from the mechanism itself. In other words, the fact that common reactions are not elementary comes into play.

For instance, the transition

$$\ce{2Br- + H2O2 + 2H+ -> Br2 + H2O}$$

is experimentally found to follow

$$v_r = k\ce{[H2O2][H+][Br-]}\tag{a}.$$

Mathematically, we can verify that one possible mechanism is

$$\ce{H+ + H2O2 <=>[K] H2O+-OH,} \tag{fast equilibrium}$$ $$\ce{H2O+-OH + Br- ->[k_2] HOBr + H2O,}\tag{slow}$$ $$\ce{HOBr + H+ + Br- ->[k_3] Br2 + H2O.}\tag{fast}$$

Indeed,

$$K_c = \frac{\ce{[H2O+-OH]}}{\ce{[H+]}\ce{[H2O2]}} \implies \ce{[H2O+-OH]} = K_c\ce{[H+]}\ce{[H2O2]}\tag{b}.$$

Applying the method of stationary concentration gives

$$v\left(\ce{HOBr}\right) = 0 \implies k_2\ce{[H2O+-OH]}\ce{[Br-]} = k_3\ce{[H2O2]}\ce{[H+]}\ce{[Br-]}\tag{c}.$$

The overall rate of the reaction is characterised by the rate of formation of bromine $\ce{Br2}$. So,

$$v_r = v\left(\ce{Br2}\right) = k_3\ce{[H2O2]}\ce{[H+]}\ce{[Br-]} \overset{(c)}{=} k_2\ce{[H2O+-OH]}\ce{[Br-]} \overset{(b)}{=} k_2K_c\ce{[H+][H2O2]}\ce{[Br-]}$$

or more briefly using $k_2K_c = k$

$$v_r = k\ce{[H2O2][H+]}\ce{[Br-]}\tag{d}.$$

I repeat: this only shows that it might be a valid mechanism, not that the reaction actually follows such a pathway. Therefore, while we can use collision theory to derive the law of mass action as a first approximation, the exponents for non-elementary reactions are determined via experiment.


I have some additional comments. As mentioned previously, by the most straightforward measure of polarity for small molecules (net molecular dipole moment), $\ce{HCl}$ is undoubtedly more polar than $\ce{H2S}$. If some other measure of polarity is being considered, then it needs to be stated for clarity.

I suspect the authors of the book and some of its readers are taking the higher boiling point of $\ce{H2S}$ compared to $\ce{HCl}$ to mean the former is a more polar molecule. In fact, this behaviour holds for all of the group 15/16/17 molecular hydrides. However, this is only an indirect measure of polarity, and there are other confounding factors. In this particular case, I believe the reason $\ce{H2S}$ has a higher boiling point does not have to do with the strength of each intermolecular interaction, but the average number of intermolecular interactions which can be held at once, compared to its neighbouring hydrides ($\ce{PH3}$ and $\ce{HCl}$).

A hand-waving explanation is that the group 16 hydrides have a better "balance" of partially charged atoms compared to the group 15 and 17 hydrides in the same period. For example, looking at the hydrogen bonded molecules of the second period and considering an optimal hydrogen bonded network, $\ce{NH3}$ is capable of being the donor atom for three H-bonds and being the acceptor atom for only one H-bond. In $\ce{HF}$, the situation is inverted; now there are three H-bond acceptor sites, and only one H-bond donor. $\ce{H2O}$, however, is a convenient match of being a H-bond double donor and double acceptor. Thus, water performs more hydrogen bonds per molecule on average than either ammonia or hydrogen fluoride, and has a significantly higher boiling point, even though the actual hydrogen bonds are arguably strongest for $\ce{HF}$. The logic would be similar for the group 15/16/17 hydrides of the third period.

Lastly, I'd like to clarify the role of non-bonding pairs in the dipole moment of a molecule. Electrons in pure atomic orbitals are symmetric with respect to the nucleus and produce no net dipole. In a hybridization picture, one might say that non-hybridized $s$ and $p$ orbitals possess no dipole moment, while hybrid orbitals are not symmetric and do possess a dipole moment. In the case of the second period element hydrides, the lone pairs in hybridized $sp^3$ orbitals would provide a net dipole moment. However, for the hydrides in the third period and below, the molecular geometries are consistent with no hybridization (bond angles close to 90°), and therefore the lone pairs would be present in effectively $s$ and $p$ atomic orbitals, with no effect on a molecule's polarity.

In a more accurate picture without invoking hybridization, the molecular orbitals which contain the lone pairs are almost the same as the central atom's atomic orbitals in the heavier group 15/16/17 hydrides (and hence produce no net dipole), while for the second period hydrides the molecular orbitals are not as symmetric and hence actually produce a dipole.


Chemical reaction and shift in magnetic susceptibility sign

Seems to me any reaction involving the recombination of two radicals leads to a closed-shell diamagnetic species (which oxygen is definitely not!). I can't think of any other examples immediately. You could make singlet oxygen, which is diamagnetic, from paramagnetic triplet oxygen by irradiation, but then there is only one reactant.

If you heat a paramagnetic substance enough so that the thermal energy can overcome the energy that is gained by the alignment of your magnetic moments, I believe the substance will not exhibit paramagnetism in a magnetic field above that temperature. So that seems doable.

Edit: Actually, I found an example: ozone is diamagnetic and can be made from paramagnetic oxygen and an oxygen radical. So that could be a possible reaction where you have two paramagnetic reagents combining into a diamagnetic product.


This questions has two parts if you look closely:

  Which of the following elements is important in the semiconductor industry to improve the conductivity of Ge?

All of the above will improve the conductivity of $\ce{Ge}$.  This is because germanium is a semiconductor which conducts electricity via the movement of holes and electrons.  Since germanium is tetravalent and gallium is trivalent, doping germanium with gallium creates a p-type semiconductor which is exponentially rich in holes, thus increasing the conductivity. This mechanism holds true for aluminum, potassium and magnesium meaning all of them can increase the conductivity of germanium

  Which of the following elements is important in the semiconductor industry to improve the conductivity of Ge?

This part is what makes gallium special.  Gallium and germanium are sequential elements in atomic number.  They have very similar atomic radii.  Gallium is also very much less reactive than the other elements mentioned. These characteristics allow gallium to dope the germanium to a higher concentration without forming a new  undesired solid phase.  Sodium and magnesium form germanides with germanium and will not dope as shown by the phase diagram.  Aluminum will dope, but its smaller atomic radius can alter and strain the crystal structure of germanium much more than gallium which can affect band gap.


Oxyphosphorus compounds, all of which contain phosphorus-oxygen linkages, are the most dominated subset in Phosphorus Chemistry. You may find good review of oxyphosphorus compounds in Ref.1. In particular, most of these commonly known as phosphates are described in Chapter 3 of Ref. 1 (Pages 169-305) which states that:

  Oxyphosphorus compounds may be defined as compounds which contain phosphorus-oxygen linkages. They may contain up to six oxygen atoms linked to a central phosphorus atom. Pyramidal derivatives are represented by phosphite esters (la), tetrahedral compounds by orthophosphate salts (lb) and esters (lc), trigonal bipyramidal compounds by pentaoxyphosphoranes (Id), and octahedral compounds by a few hexaoxyphosphorides (hexaphosphates) of type (le). In addition, there are the comparatively rare 2-connected angular phosphenites (If) and 3-connected planar phosphenates (lg).

The term normal phosphates is used to described the compounds, in which only $\ce{P—O}$ linkages are present (e.g., structures 1a-g). Compounds such as 1b containing discrete $\ce{PO4^{3-}}$ anions are known as orthophosphates. If the neutralizing cation is $\ce{H+}$, then it is called phosphoric acid or orthophosphoric acid.

On the other hand, if some of the oxygen linkages are replaced by other atoms or groups (e.g., $\ce{H, NR2, CR3}$, etc.), the compounds can be termed substituted phosphates. For example, if one $\ce{P—O}$ linkage is replaced by a $\ce{P-H}$ linkage, the compound series are called phosphites (2a) while when the two $\ce{P—O}$ linkages are replaced by two $\ce{P-H}$ linkages, that series is known as hypophosphites (2b).  Similar to phosphates, if the neutralizing cation is $\ce{H+}$ in 2a and 2b, then they are traditionally called phosphorus acid (Wikipedia1) or hypophosphorus acid (Wikipedia2), respectively. The IUPAC names of these acids are phosphonic acid and phosphinic acid, respectively. Meanwhile, substituted phosphates containing a single $\ce{P=O}$ linkage (e.g., 2c) are usually referred to as phosphoryl compounds.

When $\ce{PO4^{3-}}$ anions are linked together with sharing oxygen atom in common, they are called condensed phosphates (e.g., pyrophosphate, 3a). If the neutralizing cation is, again, $\ce{H+}$, then it is called pyrophosphoric acid or diphosphoric acid (IUPAC name). On the other hand, one can expect when the two anions linked together with sharing oxygen atom in common are phosphites, they would be called pyrophosphites, 4a (Wikipedia3).  However, recent research showed that there are some controversies on the structures, not only on pyrophosphites, but also on phosphite itself.

Phosphorous acid, $\ce{H3PO3}$ (systematic IUPAC name is phosphonic acid), exists in two tautomeric forms: An one with tricoordinate about phosphorous $(\ce{P(OH)3})$; and another with tetracoordinate about phosphorous $(\ce{HP(O)(OH)2})$ (See Scheme below), which is the favored tautomer between two (Ref.2).

The $\ce{P(OH)3}$ form of two tautomer is rarely isolated or directly obtained, yet its presence in the equilibrium has been repeatedly postulated from kinetics studies (Ref.3,4) and thermodynamic studies (Ref.5,6).

According to most literature, the $\ce{P(OH)3}$ form of tautomers is possible only in triesters such as 1a $(\ce{P(OR)3})$. Even in such cases, there is a strong tendency to rearrange to the tetracoordinate species, $\ce{RP(O)(OR)2}$ (Ref.5). Hence, it is safe to say that pyrophosphites has high tendency to have a structure resembling of 4a (See first scheme).

References:

D. E. C. Corbridge, In Studies in Inorganic Chemistry, Volume 20: Phosphorus – An Outline of its Chemistry, Biochemistry and Uses; Elsevier B.V.: New York, NY, 1995, Pages 1-1208 (https://www.sciencedirect.com/bookseries/studies-in-inorganic-chemistry/vol/20).
F.A. Cotton, G. Wilkinson, C.A. Murillo, M. Bochmann, Advanced Inorganic Chemistry, 6th Edn.; Wiley-Interscience: New York, NY, 1999, “Chapter 10: The Group 15 Elements: $\ce{P, As, Sb, Bi}$,” pp. 380-443.
G.A. Haight Jr., M. Rose, J. Preer, “Reactions of chromium(VI) with phosphorus(III) and phosphorus(I). I. Dihydrogen phosphite, phosphorous acid, and hypophosphorous acid,” J. Am. Chem. Soc. 1968, 90(18), 4809-4814 (DOI: 10.1021/ja01020a011).
R.O. Griffith, A. Mckeown, “Kinetics of the reaction of iodine with phosphorous acid and with phosphites,” Trans. Faraday Soc. 1940, 36, 766-779 (DOI: 10.1039/TF9403600766).
J. P. Guthrie, “Tautomerization equilibria for phosphorous acid and its ethyl esters, free energies of formation of phosphorous and phosphonic acids and their ethyl esters, and $\mathrm{p}K_a$ values for ionization of the $\ce{P—H}$ bond in phosphonic acid and phosphonic esters,” Canadian Journal of Chemistry 1979, 57(2), 236-239 (https://doi.org/10.1139/v79-039).
J. P. Guthrie, “Carbonyl Addition Reactions: Factors Affecting the Hydrate–Hemiacetal and Hemiacetal–Acetal Equilibrium Constants,” Canadian Journal of Chemistry 1975, 53(68), 898-906 (https://doi.org/10.1139/v75-125).


Conversion of 1,2-dione to a 7 membered oxepane?

By this late date Coherent has probably resolved the structure of his product (9). From the limited information provided, it appears that a photolysis of benzil (1) was conducted in hydroxylic solvent (methanol?). Let's assume for the sake of argument that the product has the structure that Coherent has proposed. Coherent asks, how can it (9) be formed? First of all, hydroxyl radicals need not be invoked. Structure 2 represents an excited state of benzil that may cyclize to diradical 3 whose resonance structure is diradical 4. Rearomatization of 4 provides enol 5, tautomerization of which affords 2-phenylbenzofuran-3-one (6). Now the reorganization 4 --> 5 need not be intramolecular as indicated (red hydrogen) but intermolecular or, less likely, via the solvent acting as a source of hydrogen atoms (methanol C-H bond DHo = 96 kcal/mol. Probably too strong a C-H bond to deliver a hydrogen atom to stabilized radicals).
Benzofuranone 6 is a known compound. If benzofuranone 6 is truly formed in the reaction sequence,  it should independently afford the final product 9 upon irradiation conditions. Cleavage of benzofuranone 6 by the Norrish type I mechanism leads to diradical 7, which may recyclize to 8 and subsequently rearomatize to 9. [I am not overly excited about going from a 5-membered to a 7-membered ring, but then again who knows.]
Finally, compound 9 should be able to be synthesized from O-benzylsalicylic acid by intramolecular acylation.


How can I derive whether the methyl cation or anion is planar?

The VSEPR model is a rather simple model that actually helps in this case. It assumes that anything ‘on’ a central atom — atoms that it is bound to and lone pairs — requests almost equal amounts of space, and therefore the compound’s final geometry is dictated by the number (and type; lone pairs are deemed ‘invisible’) of substituents only.

It is important to recognise that only lone pairs (and radicals) count towards the VSEPR model; empty orbitals do not. The reasoning is that something cannot take up space if it is empty.

Applying the theory to the methyl cation should give you three substituents: the three hydrogens. These three will, according to the theory, align themselves in a trigonal planar fashion around the central carbon, giving a perfect bond angle of $120^\circ$ between them. Thus, the methyl cation is planar.

Applying it to the methyl anion should give you four substituents: the three hydrogens we had previously plus the lone pair. The lone pair takes up space, so including it into our visualisation would give us a tetrahedral geometry. However, it is also invisible (we can only ‘see’ full atoms) hence the final structure is predicted to be trigonal pyramidal with the four atoms not occupying the same plane.

Therefore, assuming you copied accurately, the answer key should have given A as the correct answer.


How do indicators work when they mixed with each other?

In general, the chemistry of one indicator does not usually affect another as far as I know, but the colors will mix as you would expect. It gets tricky here because the question shows the colored ranges but leaves a blank in the transition pH ranges, but I would guess the answer they're looking for is from 3-7.5


Liquids that keeps the lowest temperature for the longer time after being frozen

To maintain a stable temperature while heat is entering a system implies some phase change: that is, the substance is either melting or boiling, keeping the temperature the same, while some part is changed form solid to liquid or form liquid to gas.

Pure water freezes at 0 °C, but if another substance is added, the freezing point (f.p.) may be lowered. Ordinary table salt, $\ce{NaCl}$, can lower the f.p. to -20 °C. Calcium chloride, $\ce{CaCl2}$, is even more effective. Both are used for melting ice from roads, with $\ce{CaCl2}$ preferred in cold weather.

You might test various concentration to see what best meets your needs. Ideally, the solution f.p. should be a bit above the temperature of your freezer.


Adsorption (with a d) is caused by intermolecular Van der Waals Forces allowing molecules to stick to a solid surface.  The solid surface can be described using a Langmiur model.  In this model a surface can bind with a number of gas molecules at any number of active sites. These active sites may be packed much closer together than molecules dispersed in a gaseous state. When dispersed gaseous molecules in a closed container interact with a surface packed with active sites, the surface may appear to "suck the gas molecules" as the gas molecules bind to the densely packed active sites.

This suction is particularly true of several activated carbons because the porous surface causes a large number of active sites to pack together.

The Langmuir model is only one model. In this model once all of the active sites are filled, the gas around the surface forms a monolayer. The monolayer is expanded upon in more sophisticated models such as the BET model.  The BET model is used to analytically determine BET surface area of a material like activated carbon.

Sometimes the behavior of the surface is described empirically such as in the Freundlich adsorption isotherm.  In this case, as with all empirical models, there is no explanation for the attraction sucking molecules to the surface.  It is simply accepted the model works.


Your books statement that "carboxylic acid is a functional group" doesn't seem quite right to me. I think it's better to say that a carboxylic acid is an organic molecule containing a carboxyl functional group.

Similarly, I wouldn't say that an alcohol is a functional group, but that an organic molecule containing a hydroxyl group is an alcohol.

Of course you can have single large molecules with multiple functional groups that could be called, at once, for example, an alcohol, a ketone and a carboxylic acid, because it contains a hydroxyl functionality, a carbonyl functionality and a carboxyl functionality.

I hope this addresses what you were asking, don't hesitate to ask if I missed something.


The diameter of human hair is typically 100 microns (from 17 to 181 microns: https://hypertextbook.com/facts/1999/BrianLey.shtml ). The diameter of a cellulose fiber is about 10 microns ( https://en.wikipedia.org/wiki/Paper ), and a cellulose molecule is much smaller (up to about 10,000 glucose units long: https://en.wikipedia.org/wiki/Cellulose_fiber ).

While it should be possible to make a coarse papery product from hair, something is needed to make the hair fibers stick together without being sticky in the final product. Think how obnoxious it would be if you wash your hair, then dry it and it turns into cardboard!

Paper from cellulose fibers sticks together when it dries because there are so many hydroxyl groups on every cellulose molecule and fiber - no extra adhesive needed. It can be made quite thin - commercial paper is available as thin as 60 microns ( https://hypertextbook.com/facts/2001/JuliaSherlis.shtml ).

Hair, however, is long enough, and frequently curly enough, to be able to be woven into yarn or thread, and then into thick 2-dimensional products. The thickness of the yarn or thread gives the product more strength.


Cationic vacancies created when non-stoichiometric iron oxide is doped with silicon(IV) ions

Your approach is correct, however, you forgot to account for the vacancies created when Fe3+ replaces Fe2+. First, calculate the cationic vacancies formed due to the replacement of Fe2+ by Fe3+.
Assuming  to be the fraction of Fe2+ in the compound, the fraction of Fe3+ in the compound is (0.93 - x).The total charge on x Fe2+ and (0.93 – x) Fe3+ should be equal to the charge due to O2- ions.
Hence, 2x + 3(0.93 – x) = 2
=> 2x + 2.79 – 3x = 2.00 or x = 0.79. So, fraction of Fe3+ ions in the compound is 0.14
Now, 2 Fe3+ ions replace 3 Fe2+ ions, so there's one vacancy created for every 2 Fe3+ ions. That means for 0.1 mole of F0.93O, the number of vacancies will be 0.007A due to Fe3+ replacing Fe2+ in FeO. This is the part you missed.
Now, as you mentioned, each Si4+ ion will replace two Fe2+ ions in the compound, so again, one vacancy per replacement. And, as you already solved, The number of replacements would be (0.079/2)A = 0.0395A. So, the total number of vacancies are 0.0395NA + 0.007NA = 0.0465NA.
Hope this helps!


SN1 reaction of ethanol at tertiary alkyl halide

Just to leave this question not unanswered. As ron already pointed out, you reasoning is correct. Based on the information given, the ethanol may perform a nucleophilic attack at the carbocation. To a very minor extent you might find the elimination product, but I guess you will have to cook it quite well. The elimination pathway is more likely to occur, when there is a stronger base than ethanol.

Just to get your terminology right. The carbocation is a reactive intermediate, not a transition state. The $\ce{S_{N}1/E1}$ reaction have always (at least) two transition states. The first one corresponds to the carbon leaving group bond break and the second one corresponds to the nucleophile carbon bond formation, proton abstraction, respectively.


Nomenclature of methyl groups in an alkanol

Could someone please explain why I am wrong?

Find the longest chain! This determines the base name: hexane.
Find the position of the $\ce{OH}$ substituent. Lower number wins.


Why does zinc react while its outermost suborbital is full?

You could ask the same question about the alkali earth metals with their $\rm{s}^2$ valence shell. While there is an energy cost to removing electrons, in a spontaneous reaction (e.g. $\ce{Cu^2+_{(aq)} + Zn_{(s)} -> Cu_{(s)} + Zn^2+_{(aq)}}$) there will be a compensatory transfer of energy elsewhere (like the reduction and deposition of $\ce{Cu^2+}$) which leads to a more stable overall system.


Is this the correct equation for the mean free path?

Your derivation is correct. The $\sqrt{2}$ comes from $v_{rel}=\sqrt{2}v_{mean}$, where $v_{rel}$ is the mean relative speed (relative to the other molecules in the gas) and $v_{mean}$ is just the mean speed of the molecules in the gas.

In conclusion: note what the subscript says about the variable!


What's the difference between a tetrahedron and a trigonal pyramid?

In a rigorous geometrical sense, there is no difference between tetrahedron and trigonal pyramid--the terms both mean the same thing. In colloquial and chemical use, however, 'tetrahedron' typically implies the 'regular tetrahedron', where all four faces are equilateral triangles.

Chemically speaking, when referring to these two shapes as descriptors of molecular geometries, there is (usually) a central atom in addition to the four atoms at the vertices.  In the symmetrically-proper regular-tetrahedral geometry, the central atom is inside the solid volume of the tetrahedron and equidistant from all of the four vertex atoms.  In the trigonal pyramidal geometry, the central atom can be located inside the solid volume, contained within a plane of the trigonal pyramid/tetrahedron, or I believe even outside the tetrahedral volume.


Fajans' rules (note the difference) were formulated in 1923 by Kazimierz Fajans.

They are a method for predicting ionic vs. covalent that predates electronegativity (by three decades) and make use of ionic and atomic radius data that was becoming available through x-ray crystallography.

To use Fajans' Rules, assume your binary compound is ionic and identify the potential cation and anion.

By Fajans' Rules, compounds are more likely to be ionic if: there is a small positive charge on the cation, the cation is large, and the anion is small. For example, $\ce{NaCl}$ is correctly predicted to be ionic since $\ce{Na+}$ is a larger ion with a low charge and $\ce{Cl-}$ is a smaller anion.

Compounds are more likely to be covalent if: there would be a large positive charge on the cation, the cation would be small, and the anion would be large. For example, $\ce{AlI3}$ is correctly predicted to be covalent since it would have a small cation with a high charge and a large anion.

Note that Fajans' Rules have been largely displaced by Pauling's approach using electronegtivites. However, the remnants of Fajans' Rules are found in Hard-Soft Acid-Base Theory, which predicts bonding properties based on polarizability (which is based on size and charge). Binary compounds having a soft acid and/or a soft base are often covalent.

$$\begin{array}{|c|c|c|c|}\hline
\mathrm{compound} & \mathrm{Fajans} & \mathrm{Pauling} & \mathrm{HSAB}\\ \hline
\ce{NaCl} & \mathrm{low + charge \\ larger\ cation \\ smaller\ anion \\ionic} & \mathrm{3.16-0.93 = 2.19\\ ionic} & \mathrm{hard\ acid \\ borderline\ base \\ ionic}  \\ \hline
\ce{AlI3} & \mathrm{high + charge\\ smaller\ cation \\ larger anion \\covalent} & \mathrm{2.66 - 1.61 = 1.05 \\covalent} & \mathrm{hard\ acid\\ soft\ base \\covalent} \\ \hline
\end{array}$$


Why is the standard atomic weight of chlorine not a whole number?

The process is explained (for silicon) here:

  The calculation is exemplified for silicon, whose relative atomic mass is especially important in metrology. Silicon exists in nature as a mixture of three isotopes: $\ce{^28Si}$, $\ce{^29Si}$ and $\ce{^30Si}$. The atomic masses of these nuclides are known to a precision of one part in 14 billion for $\ce{^28Si}$ and about one part in one billion for the others. However the range of natural abundance for the isotopes is such that the standard abundance can only be given to about ±0.001% (see table). The calculation is

  $$A_r(\ce{Si}) = (27.97693 \times 0.922297) + (28.97649 \times 0.046832) + (29.97377 \times 0.030872) = 28.0854$$

For chlorine, the idea is the same, but you only have two isotopes, $\ce{^35Cl}$ (76%) and $\ce{^37Cl}$ (24%).


What is the expected number of rings obtained in the following ring formation process?

There is no simple answer here because the outcome will be heavily dependent on the concentration and on the rate of addition.

The most salient point in your question to me is: "excess ammonia."

You can ask what the competing reactions are.

1,5-dibromopentane reacts with ammonia to form 1-amino-5-bromopentane
1-amino-5-bromopentane reacts with itself to form piperidine
1-amino-5-bromopentane reacts with ammonia to form 1,5-diaminopentane
1-amino-5-bromopentane reacts with another straight chain alkyl amine to elongate the chain
1,5-dibromopentane reacts with piperidine to form N-(5-bromopentyl)piperidine
and others...

The main problem I see is that in excess ammonia, unless you perform a slow addition of ammonia, the concentration may be such that the ammonia will over add via (3). Then you're going to get any product. The intermediates are also reactive, so the reactant concentration has to be relatively low as well, otherwise, you may see (4) or (5) or other side reactions. You may also need to run the reaction at lower temperature to disfavor reaction of the product with any alkyl bromide intermediate or reactant.

These issues need to resolved before you even begin to think about distribution of ring sizes because you might not even form any rings.


Nitrobenzene and benzene would NOT have the same rate towards any type of electrophilic substitution reaction as strength of nucleophile - (nitro)benzene does matter.

Nitro group as in nitrobenzene is highly deactivating group due its electron withdrawing nature and therefore decreases the nucleophilicity of benzene ring.

IJOART_document


Calculating base dissociation constant given the molarity and pH

$\ce{C6H5NH2}$ (aniline) is a weak base, and therefore, in its aqueous solution, it takes a proton from water and forms $\ce{C6H5NH3^+}$ and $\ce{OH-}$, which are correct as you have written.

Suppose the extent of protonation is $\alpha$. After taking up the proton, $[\ce{C6H5NH2}] = c (1 - \alpha)$ and $[\ce{C6H5NH3^+}]= c \alpha = [\ce{OH-}]$.

So, we can write the following, from the above mentioned reaction:

$$
K_{\mathrm{b}} = \frac{[\ce{C6H5NH3^+}][\ce{OH-}]}{[\ce{C6H5NH2}]} = \frac{c^{2}\alpha^{2}}{c(1 - \alpha)} \approx c \alpha^{2}
$$

As the degree of protonation is very less than 1 ($\alpha << 1$), we get $\alpha = \sqrt{K_{\mathrm{b}} / c}$. Now,

\begin{align*}
\mathrm{p}\ce{OH} &= -\log[\ce{OH-}] \\
& = -\log(c\alpha)=-\log(c\sqrt{K_{\mathrm{b}}/ c})=-\log(\sqrt{cK_{\mathrm{b}}}) \\
&= \frac{\mathrm{p}K_{\mathrm{b}}}{2} - \frac{\log c}{2}
\end{align*}

(as $-\log K_{\mathrm{b}} = \mathrm{p}K_{\mathrm{b}}$).

In the question, $\mathrm{pH} = 8.8$, so, $\mathrm{pOH} = 14 - 8.8 = 5.2$, and $c = 0.1$. Putting these values in the equation we will get

$$
\mathrm{p}K_{\mathrm{b}} = 2.(5.2 + \frac{\log(0.1)}{2}) = 9.4 \implies K_{\mathrm{b}} = 10^{-9.4} = 3.98 \times 10^{-10}
$$

Thus we get the $K_{\mathrm{b}}$ of aniline as $3.98 \times 10^{-10}$.

Note that, if you have any aqueous solution consisting of only weak base, you can apply the direct formula as $\mathrm{pOH} = \frac{\mathrm{p}K_{\mathrm{b}}-\log c}{2}$, and easily get the $K_{\mathrm{b}}$.


What would be the aldol condensation products in B

B is the correct answer, but not for the reasons you have given.

4-Me-benzaldehyde cannot self-condense, and is a far more reactive electrophile than acetophenone.  Acetophenone has a pKa of around 16 so will readily undergo base-catalysed aldol reaction with the aldehyde. This is the only reaction that will occur. All the other sets of reactants shown are likely to undergo self-condensation as well as cross-condensation giving a mixture of products.


What is the triple point pressure for Gallium?

At the very least your pressure has to be consistent with a whole number of molecules or atoms in your vapor phase.  Suppose you have a pressure of $7×10^{-34}$ atmosphere or (roughly) $7×10^{-29}$ Pa at $300$ Kelvins.  The average volume per atom/molecule is computed by the molecular level version of the Ideal Gas Law, in which the gas constant $R$ per mole is replaced by Boltzmann's constant $k=1.38×20^{-23}$ J/K per unit entity.  Thus

$V=kT/P=(1.38×10^{-23})×300/(7×10^{-29})=6.9×10^7$

cubic meters per vaporized entity.

So unless you are equilibrating your gallium over that much vapor volume, with high probability you do not have in practice even one atom or molecule of vaporized gallium.  In realistic laboratory or even industrual settings the triple point content of gallium vapor is most probably zero.


Reason for inability to find the absolute value of internal energy

In order to answer this question, one needs to define what the absolute energy of a system is. Energy can be trapped in a system in ways not yet discovered or fully understood. Think of the energy associated with mass ($E=mc^2$), which is a result that was not known to the early founders of thermodynamics.

We need to define reference points, which we can use to compare our products with. Usually this "zero" point is defined as the energy of the individual atoms in their most energetically favorable state (phosphorus is an exception, since red phosphorus is much more available), such as $\ce{H2}$ and $\ce{He}$.


Does increasing the number of interferometers increase signal in FT-IR?

I use FTIR in Physics, pretty much top-of the-range commercial instruments (actually more than just IR - ours go down to the UV and would do THz if we had the detectors; this isn't uncommon so I'll mainly refer to FT rather than FTIR).  There's definitely only one interferometer in each of our systems (and the ones used by our friends in Chemistry).  In astronomy FT spectroscopy has been used on the ground, but the best IR work these days is done from space using dispersive instruments (e.g. Spitzer's IRS).
The only common commercial use of three series-connected spectral devices (trying to coin a deliberately vague term) is dispersive instruments, which use three diffraction gratings for high resolution spectra and, for Raman spectroscopy, laser rejection with low stray light. A good benchtop FT instrument (with one interferometer) can give just as good resolution, though slowly and without the dynamic range. I mention this because if you were giving a lecture introducing a range of systems and techniques, and skipped a slide or two, you could jump from FTIR to triple dispersive systems and cause a lot of confusion.