Keeping Li-Ion battery on charger vs. plug the charger in and out regularly

One thing worse then overtemperature is overtemperature at charging.

One thing worse then overtemperature at charging is overtemperature at charging with charging >70-9x%.

That speeds up the side reactions of eventual metallic lithium deposits with dialkylcarbonate at high charge.

  The reaction of either DMC or DEC with lithium was initiated at 180 K, a temperature much lower than their bulk melting temperatures, producing lithium methyl carbonate, methyllithium and lithium ethyl carbonate, and ethyllithium, respectively. At temperatures greater than 270−300 K, the lithium alkyl carbonates start to decompose with $\ce{Li2O}$, elemental carbon, and alkyllithium as products on the surface.

Those reactions decrease the cell capacity and increase the cell internal resistance.

Things go worse in summer.

The rule number one is never heavily use the phone ( e.g. at gaming ), while charging it at high percentages.

If it must be charged, then when it is just topping,
or better at low charge percentage
or the best not being charged at all.

Charging scheme is also a question of the cell static aging, that applies more to laptops than to short life cycle of mobile phones.

Another thing is that cells prefer rather  more of shallow cycles than less of deep cycles. That may be unfortunately counteracted by durability of charging connectors.

Fully charged cell is aging several times faster than about half charged one. As at full charge, lithium ions may form a metallic lithium deposit instead of lithium graphite intercalate, which very slowly reacts with dialkylcarbonate solvent.

I frequently charge my phone from my powerbank to avoid immobility of the phone during charging, what gives me the option to charge the phone whenever I want. I do not charge the phone overnight nor keeping  it long time being fully charged.  But the truth is, phone life is typically 2-3 years max, what does  not have much chance to show consequences of not optimal cell maintenance.

My Lenovo T400 laptop from 2009 with 40-80% battery maintenance had its battery ( in late 2017) in better condition than T430 from 2013. (70-75% versus 60% capacity). The latter running RHEL Linux was fully charging for several years, until I found the way to achieve 40-80% mode as well. Both were running mostly with AC charger attached on the office table.


Which ions are accounted for in total and net ionic equations?

After reading into the matter (as a non-native English speaker, I've never heard of total and net ionic equations) here goes my small little guide to ionic equations:

Total Ionic Equations

A total ionic equation is where you start with a balanced equation and write all the ionic compounds in their ionic form. For example, in the following reaction
$$\ce{2 Na3PO4 + 3 CaCl2 -> 6 NaCl + Ca3(PO4)2 v},$$
one of the products precipitates (noted by the downwards pointing arrow). So we will not write it in ionic form for the total ionic equation, which is the following:
$$\ce{6 Na+ + 2PO4^3- + 3Ca^2+ + 6 Cl- -> 6 Na+ + 6 Cl- + Ca3(PO4)2 v}.$$

Net Ionic Equations

Net ionic equations can be derived from total ionic equations: We just eliminate the compounds that appear both on the right and on the left side of the reaction arrow. Thus we get for the above total ionic equation:
$$\ce{2 PO4^3- + 3 Ca^2+ -> Ca3(PO4)2 v}$$

The ions that were eliminated are termed spectator ions, because they do not participate in any notable chemical reaction.

If something is still unclear, maybe have a go at Kristy M. Bailey's
website (Oklahoma City Community College), which explains it pretty neatly.


What happens when we pass sulfur vapors through cold water?

You get more or less coarse sulfur powder-slurry.

  What is the reaction for it?

Sulfur vapor ( S2 mostly) reacts to form S8
depending on speed of cooling, some S6 might
form temporarily

  Which type is colloid is this?

Sometimes this might result in a "colloid",
but I recommend not to use this outdated word,
because so much nonsense was written in the classical
days of "colloids".

  And why doesn't it form in hot or lukewarm water

Who says/writes this?


What is the minimum operating temperature of chemical hand warmers?

They are often made of the crystallic $\ce{CH3COONa.3H2O}$, that dissolves in its own crystal water, when heated up.

As the cooled down liquid state is supercooled solution, it is hard to predict, at what temperature it would spontaneously solidify.

It will provide heat in any case. But as it warms up itself typically by  $\pu{40-50^{\circ}C}$, it would barely warm itself to room temperature, if it was previously cooled down to e.g.  $\pu{-60^{\circ}C}$.

The air activated heaters, based on oxidation of fine metal powders (thanks @MaxW ), may have problem with too slow initial reaction when too cold. So they may need to be kept at pocket temperature for some time before activation.


Energy is not stored in the energy levels themselves but in the states which occupy them.
At normal room temperature the bonding orbitals are occupied according to Maxwell-Boltzmann's statistics $P_i =e^{\epsilon_i /kT}/Z$, and store that chemical energy. At higher temperatures the antibonding orbital will become occupied releasing the chemical energy.


Is concentration polarisation an observable phenomenon over time in a galvanic cell?

There are two different factors that cause a battery to have a lower energy output.

First regardless of the "state" of the battery, there is the open cell voltage and the obtained voltage under load. State meaning here how much of the battery is used up. For the open cell voltage the voltage is measured with a high impedance voltmeter so that essentially no current (meaning a very very small current) is being drawn. Next the battery is put under some load typical for its use and the voltage is measured again while under the load. This second load voltage measurement will always be lower that the first measurement for the open cell voltage. This voltage drop is caused by the diffusion limits of the chemicals in the battery to and from the electrodes.

This behavior can be essentially modeled by assuming that the battery has an internal resistance. So as the current draw increases the voltage drop internal to the battery increases and the output voltage drops.

Now the second problem is related to the Nernst equation. Let's assume that the battery is used up in 10% increments in a series of tests. At the start of each test the voltage and current is measured. Of course during using each 10% of the battery the reactants in the vicinity of the electrodes are used up and both the voltage and current out for a given load drops.

Now if the battery is left to equilateral for some time then the reactants redistribute and the battery regains some power. However the battery will not recovery to the level at the start of the previous load test. Essentially the bulk concentration of the reactants is dropping by about 10% after each test. So even after a recovery period both the voltage and current that the battery can supply for a constant load drops.

Now with all of this being said there are many types of batteries. Some types produce a more constant power output than others.


Reaction mechanism of combustion of hydrogen

This post deals with the mechanism that is observed in the gas phase. It is of course not as simple as the equation might suggest and you did suspect that already.
$$\ce{2H2 + O2 -> 2H2O}$$

This will be divided into many different elementary sub reactions. Any mixture of oxygen and hydrogen is metastable (stable as long as you do not change the conditions). If you provide sufficient energy to overcome the activation barrier, the reaction will take its cause:
$$\ce{H2 <=>[\Delta T] 2 H.}$$

The resulting hydrogen atoms are very unstable and will react with anything in reach, but most importantly with $\ce{O2}$ (which is a triplet biradical), forming hydroxyl radicals. This reaction is endothermic (it requires energy).
$$\ce{H. + O2 -> HO. + O}$$

The resulting oxygen radicals can again react with $\ce{H2}$ to form more hydrogen radicals. This reaction is also endothermic.
$$\ce{O + H2 -> HO. + H.}$$

The hydroxyl radicals can also react with $\ce{H2}$ to form more hydrogen radicals, which is a slightly exothermic (releases energy) reaction.

The net result of those equations leads to a slightly exothermic sum, with high potential:
$$\ce{3H2 + O2 -> 2H2O + 2H.}$$

In principle, the hydrogen radicals react as a catalyst. However, this reaction is a highly branched chain reaction, including a lot of radical reactions. Due to this, more and more hydrogen radicals will be produced.

This scheme will eventually end when the concentrations of $\ce{O2,H2}$ will become lower forcing the excess radicals to react with each other.
$$\ce{HO. + H. -> H2O}$$

Another possibility is forming as a byproduct hydrogen peroxide, which is a very exothermic reaction:
$$\ce{H. + O2 -> HO2.}\\
\ce{2HO2. -> H2O2 +O2}$$

(Also happening but not as important: $\ce{HO. + HO. <=> H2O2}$)

As long as there are hydroxyl and hydrogen radicals present, the peroxide will take part in the chain reaction. However, this reaction also provides the necessary energy to cleave more $\ce{H2}$. peroxide is also easily cleaved again, or reacts with each other:
$$\ce{2H2O2 -> 2H2O + O2}$$

This all results in the main product water.

Please keep in mind, that there are many factors, that influence these reactions. It is strongly dependent on pressure, temperature, and surroundings. Surfaces and/ or catalysts involved in this reaction may change it completely.

Further reading:

How explosive is hydrogen gas?
Eugene P. Dougherty1 and Herschel Rabitz1, J. Chem. Phys. 72, 6571 (1980) most comprehensive (theoretical) study on the $\ce{H2-O2}$ system.
Wiberg, Nils: Lehrbuch der Anorganischen Chemie Kap. 3.2 (german) The mechanism is in part taken from this source.
http://arrow.utias.utoronto.ca/~ogulder/ClassNotes3.pdf Thanks to Silvio Levy


From a scientific standpoint, there is no difference.  Elements=Atoms=the 100 or so basic building blocks of chemistry.

This is purely conjecture, but I suspect that the term/concept "elements" pre-date "atoms" (at least atoms as we know them today).  I could see "elements" referring to the macroscopic description, e.g. a jar of carbon (graphite), but it may not have been realized back-in-the-day that these were aggregations of many carbon atoms.

"Atom" strikes me as quantum scale, and "Element" strikes me as something on the macroscopic scale, perhaps hearkening back to the pre-atomic days.  But really, there is no difference.


All 6 positions on the benzene ring are being protonated (the frequency of protonation at the different sites is related to the stability of the carbocation generated) in acidic media, so all of the ions you suggested are being formed.  Protonation at 5 of these positions can only regenerate starting material when one of the two protons at the initial site of protonation is removed (an identity reaction, we could make this pathway visible by running the reaction in deuterio-acid + $\ce{D2O}$).  Protonation of the benzene ring at the carbon bearing the substituent provides a choice.  Either the proton can again be removed or the stable t-butyl cation can be ejected.  All of these things are happening in the reaction, but only removal of the t-butyl group is visible.


Not knowing what the chemicals in your equation are or what kind of properties they have, I can't tell you how you would measure their concentrations.

Nonetheless, given:

$$\ce{A + B <--> C}$$

If you add an additional, preferably known, quantity of A, then the concentration of B should decrease and the concentration of C should increase.

This is actually just the real meaning behind your statement that "the equilibrium will shift to the right".  In this case, with two reactants, obviously more of the second one (B) must be consumed by the addition of more A in order to produce more C (shift to the right).

Again, how you would measure the concentrations of B and C depends on which compounds B and C actually represent.


Which d-orbital configurations form an octahedral complex?

A metal ion doesn't actually need vacant d orbitals to form a coordination complex according to Crystal field theory. All the ions mentioned in the table form complexes with water (which is a weak ligand) - see this page.

Crystal field theory assumes that all bonding between ligand and metal ion in a complex is electrostatic (ionic). That means there is no exchange of electrons between the metal ion and ligand. Instead, the negatively charged (or polar in case of neutral ligands like $NH_3$) ligand is attracted to the positively charged metal ion by purely electrostatic forces. This is why vacant d orbitals are not required for bonding.


Why do C-F bond has less dipole moment than C-Cl bond?

The bond dipole moment ($\mu$) of a chemical bond is equal to the magnitude of the separation of charge ($\delta$) times the displacement of the charge (or the distance over which the charges are spearated) $d$, or:

$$\vec{\mu}=\delta \vec{d}$$

Thus, there are two factors controlling dipole moment. The charge separation can be estimated from the electronegativity difference, but not perfectly. Disparities in size and polarizability also effect the separation of charge between two atoms. The displacement is the length of the bond. The $\ce{C-Cl}$ bond length (176 pm) is longer than the $\ce{C-F}$ bond length (134 pm) - data from here.

Thus:

For $\ce{C-F}$, the electronegativity difference on the Pauling scale is 1.43, and the bond length is 134 pm: $1.43\times 134 = 191$ (ignoring units at the moment).

For $\ce{C-Cl}$, the electronegativity different on the Pauling scale is 0.61, and the bond length is 174 pm: $0.61\times 174=106$.

So, my quick and dirty method did not work so well. However the disparity between these two bonds by my method $(106/191)=0.55)$ is smaller than by just comparing electronegativities $(0.61/1.43=0.42)$
 and that is because of the other factor mentioned above: size and polarizability. Chlorine is bigger and more polarizable than fluorine (and carbon). The carbon-chlorine bond is polarized toward chlorine not just because of the electronegativity difference. Chlorine's larger size decreases the effectiveness of orbital overlap in bonding, and chlorine's increase polarizability implies that it can support more partial negative charge than suggested by its electronegativity alone.


Complete Combustion of Inorganic, Hydrogen Containing Covalent Compounds in the Atmosphere

The "completeness" of combustion depends on the proportion of the reactants, and on the temperature cycle, not only the peak temperature, but the rate at which the products are quenched. For example, using ammonia and air as fuel in internal combustion engines produces mostly $\ce{H2O}$ and $\ce{N2}$, with less than 0.1% $\ce{NO_x}$ according to Kobayashi, et. al. and Zacharakis-Jutz.

The greatest oxidation state of a substance is often not thermodynamically stable. Perchlorates, chlorine dioxide , $\ce{H2O2}$ and similar compounds are somewhat unstable (i.e., BANG) and are not generally released in great quantities by combustion.


How to RAISE the melting point of water?

source

One way to raise the melting point of water is to increase pressure beyond about 635 MPa.  By raising pressure you could get the melting point to be even greater than the normal boiling point.

A second way is to lower the pressure, but this can only increase the melting point by 0.01 K.

A third way would be to added enough of a high melting point solute that you pass the eutectic point. In other words, generally when you add a solute to water, the melting point decreases, but only until you add enough solute that the eutectic point is reached, beyond which adding more solute increases the melting point.

source: http://www.phasediagram.dk/binary/CaCl2.PNG

For example, in the diagram, above, A is the eutectic point.  Beyond this point, adding more CaCl2 increases freezing point.  Then beyond 38% CaCl2, melting point exceeds ) degrees C.


Cl is slightly more electronegative according to the tables I've seen: http://en.wikipedia.org/wiki/Electronegativity

But I wouldn't say it's an absolute rule that the less electronegative atom is central.  For example, $\ce{NH3}$ has the N central and $\ce{Cl_2O}$ has the O central.


I liked this question because I had never thought much about it. However, it's not such a mystery because the answer is on Wikipedia's “radical” page:

  Historically, the term radical was also used for bound parts of the molecule, especially when they remain unchanged in reactions. These are now called functional groups. For example, methyl alcohol was described as consisting of a methyl "radical" and a hydroxyl "radical". Neither are radicals in the modern chemical sense, as they are permanently bound to each other, and have no unpaired, reactive electrons. However, they can be observed as radicals in mass spectrometry when broken apart by irradiation with energetic electrons.

So, the free in free radical was originally used to distinguish it from a radical bound to a carbon skeleton.

Digging into historical chemistry texts reveals that “free radical” becomes really widely used around 1930:

However, some quotes for earlier works use it, though it does not appear so frequently that it may be considered a common expression. In other words, at that time simply the juxtaposition of the adjective free with radical, became synonymous with group (or today's noun substituent). For example:

  “This hydrocarbon was formerly supposed to be the free radical of the methyl group, but no methyl compounds have in any way been derived from it, whilst by the action, of chlorine upon it, ethyl chloride is obtained as the product” (1870, Lessons in elementary chemistry)

or

  “Although, however, the free radical methylene is unknown, many of its compounds have been prepared, of which some, such as the haloid ethers, &c.” (1884, A Treatise on Chemistry, Henry Enfield Roscoe and Carl Schorlemmer)

Though more quotes could be dug and provided, the ones I've read support this idea that free radical initially meant “the chemical group as a molecule by itself”.


Cation-anion radius ratio of manganese difluoride

You assumed that the cations touch along the $c$ direction.  They don't.  In most ionic compounds you have only anion to cation contact and, therefore, any attempt to determine ionic radii from lattice parameters gives only the sum of the two radii, not the individual values.  There are exceptions such as lithium iodide, in which the lattice parameters enable determination of an ionic radius for iodide, but then they carry no information about the lithium ion.

In principle, you could use the data on lithium iodide, where there is iodide-iodide contact, to get a radius for just the iodide ion.  Then you turn to manganese(II) iodide, where anions contact cations, and get the sum of the manganese and iodide radii.  Then subtract the iodide radius to get radius for manganese ion.  Use that plus the lattice information for manganese(II) fluoride to get the fluoride radius.  It all works perfectly -- if the radius of a given ion is the same in all its ionic compounds.  Maybe it is to sufficient accuracy, maybe it isn't.


At what temperature will proteins and fats boil in a vacuum?

You're talking about the human body being exposed to outer space and asking about what would happen to it's internal structure. Bones, fats and proteins are within the enclosed system that is the human body, they are not subjected to the low pressures of space. The answer to the pressure at which proteins, fats and bones will break down and become vapour is not related to the answer to what would happen to them in the human body if the body is exposed to the vacuum of space. Nevertheless, I'll try to provide both answers.

The answer to the second question (short-term) can be found on NASA's website at "ask an astrophysicist". It's actually quite surprising for someone used to sci-fi movies:

  You do not explode. Your blood does not boil. You do not freeze. You do not instantly lose consciousness.

The question you linked talks about "decay", or the effects of long time exposure. That's hard to predict, but at some point the body would be attracted by a planet and collide and you wouldn't have anything to boil anymore. Now to the chemistry part of it:

There isn't much (if any) data on the boiling point or vapor pressure of the compounds you're asking - it's hard to reproduce and it's more a curiosity than anything else. So let's talk about what would happen if you have fat, bones and proteins directly exposed to outer space pressures and temperatures, which seems to be your point anyway. Temperatures in space go from -270°C (in outer space, unexposed to sunlight) to +120°C (near the earth, exposed to sunlight). Since you're asking about boiling point let's talk about what will happen at the highest temperature.

Let's start with the easier part - the bones. It is safe to assume that the inorganic components will be tougher than the organic ones. The inorganic part of bones is mostly a specific type of apatite, which is a rock. It can be found in moon rocks. It will not boil. So at least the inorganic part of the bone would remain somewhat intact.

Fat is made of triglycerides. It'll probably break down to glycerol and the correspondent acid. Then it would decompose and boil.

Proteins denature at harsh conditions, you'll be left with amino acids at some point. It's hard to tell what would happen since there isn't much data on it, but given the harsh conditions, which include exposure to all sorts of radiations, the amino acids certainly wouldn't last long and at some point you'd be left with more basic molecules that don't resemble a protein anymore. So what can we conclude? Exposed as a whole, the internal structure of the human body would remain somewhat intact for a while (physically speaking). It's hard to predict what happens then, but given that the outer space is such a hostile environment, it's unlikely that the body would last long. Exposed as individual components they would last much, much less. From bones, fat and proteins, it's likely that only the bones stand a chance.


Why is Lanthanum a "fluoride ion sensitive electrode" and hence used for measuring "flouride concentration"?

What is the reason behind choosing "Lanthanum doped with europium"?

Well, you said it yourself, because "fluoride ions bind to the crystal".

Here's a great explanation from Wikipedia:

In the lanthanum fluoride electrode, the sensing element is a crystal of lanthanum fluoride LaF3, doped with europium fluoride EuF2 to create lattice vacancies. Such a crystal is an ionic conductor by virtue of the mobility of fluoride ions which jump between lattice vacancies.

Here's a rough sketch I made to help you visualizing this. This is a simplistic drawing and the actual crystalline structure of the compound is 3-dimensional.

I must stress this: Both the structure and the proportions are probably nowhere near the ones in my drawing, it is only for didactic purposes.

An electrochemical cell may be constructed using such a crystal as a membrane separating two fluoride solutions. This cell acts as a concentration cell with transference where the fluoride transport number is 1. As transference of charge through the crystal is almost exclusively due to fluoride, the electrode is highly specific to fluoride. The only ion which significantly interferes is hydroxide (OH-). Generally such "alkaline error" can be avoided by buffering the sample to a pH below 7.

We had the pH around 5.5. We also used TISAB for the preparation of water samples. The Wikipedia article on TISAB gives an excellent explanation on it's usage in fluoride ion analysis.

Why ASP cannot be used for determining the concentration of anions?

Well, I cannot give you an authoritative answer on that. I'll ask a teacher of mine this week and edit the answer with an update - if someone reading this knows it feel free to improve this answer or add a new one. In the case of fluoride it's probably because the Lanthanum electrode has pretty decent detection limits (it does). It's a lot cheaper to use it instead of AAS.


Why energy of some intermediates of glycolysis increases?

Step 1:

Phosphorylation activates glucose for the following reactions in the pathway. However, the formation
of such a phosphoester is thermodynamically unfavourable and requires
energy input to operate in the forward direction. The energy comes
from ATP, a requirement that at first seems counterproductive.

  $$\ce{-D-Glucose + ATP^4- → -D-glucose-6-phosphate^2- + ADP3  H+}$$
  $\Delta G$ = -16.7 kJ/mol

and energy is released.

Step 3:

Once again, the substrate that provides the phosphoryl group is ATP. Like the hexokinase/
glucokinase reaction, the phosphorylation of fructose-6-phosphate is a priming
reaction and is endergonic(energy absorbed):

  Fructose-6-P + Pi +fructose-1,6-bisphosphate   $\Delta G$ = 16.3
  kJ/mol

When coupled (by phosphofructokinase) with the hydrolysis of ATP, the overall reaction
becomes exergonic:

  Fructose-6-P + ATP → fructose-1,6-bisphosphate + ADP

  $\Delta G$ = -14.2 kJ/mol

Step 4:

Fructose bisphosphate aldolase cleaves fructose-1,6-bisphosphate between the
C-3 and C-4 carbons to yield two triose phosphates. The products are dihydroxyacetone
phosphate (DHAP) and glyceraldehyde-3-phosphate. The reaction has an
equilibrium constant of approximately 10^-4 M, and a corresponding $\Delta G$  of +23.9 kJ/mol.

(the equilibrium is thus greatly influenced by concentration.)

Step 6:

In the first glycolytic reaction to involve oxidation–reduction, glyceraldehyde-3-
phosphate is oxidized to 1,3-bisphosphoglycerate by glyceraldehyde-3-phosphate dehydrogenase.

Although the oxidation of an aldehyde to a carboxylic acid is a highly
exergonic reaction, the overall reaction involves both formation of a carboxylic–
phosphoric anhydride and the reduction of NAD to NADH and is therefore slightly
endergonic at standard state, with a $\Delta G$ of + 6.30 kJ/mol.

(The free energy that might otherwise be released as heat in this reaction is directed into the formation of
a high-energy phosphate compound, 1,3-bisphosphoglycerate, and the reduction of
NAD.)

Step 7:

Here the glycolytic pathway breaks even in terms of ATPs consumed and produced
with this reaction.The enzyme phosphoglycerate kinase transfers a phosphoryl
group from 1,3-bisphosphoglycerate to ADP to form an ATP.

The phosphoglycerate kinase reaction is sufficiently exergonic at
standard state to pull the G-3-P dehydrogenase reaction along. (In fact, the
aldolase and triose phosphate isomerase are also pulled forward by phosphoglycerate
kinase.) The net result of these coupled reactions is

  Glyceraldehyde-3-phosphate + ADP + Pi + NAD → 3-phosphoglycerate + ATP
  + NADH + H+ $\Delta G$ = -12.6 kJ/mol

(Important: Another reflection of the coupling between these reactions lies in their values
of $\Delta G$ under cellular conditions, despite its strongly negative  the phosphoglycerate kinase reaction operates at equilibrium in the erythrocyte ($\Delta G$ =0.1 kJ/mol).

So in this case you'ill notice factors that influence energy expenditure and gain. Step 1 and 3 (exegonic reactions have net -G hence ) will see energy being released (and accroding to your diagram
there is a decrease. Conversely in Step in steps 4-5 (reaction not favoured but due to concentration proceeds and is aided by energy and 6-7( energy is absorbed to form a high energy compound
1,3-bisphosphoglycerate) and 7-8 is mainly determined by reaction environment and relative concentrations of substrates and is slightly endergonic in above case.

Hope this helps

Reference

Biochemistry 4th Ed (Grisham)


About the first two, it's an exception to the rule that resonance dominates hyperconjugation.

About the last inequality, the vinyl cation is sp hybridized which means it is highly electronegative, and a positive charge on it will be very unstable.

About the middle inequality, the benzyl carbocation (with five resonance structures) is more stable than allyl carbocation (two resonance structures). You may verify this here.


Is an SN1 mechanism feasible with allylic or benzylic halides as substrates?

I planned to add this as a comment but I don't have enough reputation ... A definite answer can only come by referring to the literature and/or from the lab. In the context of an undergraduate exam it is reasonable to accept both possible answers ($\mathrm{S_N1}$ and $\mathrm{S_N2}$ ) for allylic and benzylic halides. $\ce{EtOH}$ solvolysis of benzyl bromide is a known reaction and the hypothesis that it goes via $\mathrm{S_N1}$ is the most reasonable one. Is it possible that the solvolysis acually goes via $\mathrm{S_N2}$ instead? The answer lies on performing the appropriate mechanistic studies.

However, in this case, having checked the actual exam question, it specifies acetone as the solvent. This polar aprotic solvent favours $\mathrm{S_N2}$ conditions, which is aprotic (see this question also). Incidentally, aside from the use of $\ce{KI}$ over $\ce{NaI}$, this halide exchange is basically the Finkelstein reaction, which occurs via $\mathrm{S_N2}$. (Credit to Greg E. who wrote this in a now deleted comment.)


In order for a methyl shift to occur, there would need to be an empty orbital resulting from a carbocation intermediate for the methyl group to occupy.

Note that in the bromonium intermediate shown above, neither carbon in the three-membered ring has an entire empty orbital available. Hence methyl shift doesn't occur and the mechanism continues as expected, resulting in halohydrin formation.


Zinc and cadmium are generally considered to be "almost invariably" in the oxidation state +2 (cfr. the introduction to this paper about one of the few exceptions), therefore the indication of the charge is unnecessary.
Zinc is usually listed as a "Type-I cation" (see Monoatomic ion), i.e. a cation that only appears in one oxidation state, or its case is emphasised as an exception among the transition metals:

Several exceptions apply to the Roman numeral assignment: Aluminum, Zinc, and Silver. Although they belong to the transition metal category, these metals do not have Roman numerals written after their names because these metals only exist in one ion.
-- Chemistry LibreTexts

(For the sake of completeness, the IUPAC Red Book section IR-5.4.2.2 "Use of charge and oxidation numbers" neither mentions "Type I/II cations" nor treats the zinc case separately)


What is the intuition and motivation behind the definition of the thermodynamic activity?

So I'm taking Pchem right now so I'm no experts, but here is my two cents.

I'll be honest I essentially followed my books lead on the determination of $a_{i}$, and $\mu_{i}$. When looking at problems like this I try to go backwards. I see that $a_{i}$ is derived from $\mu_{i}$, and I try to find where $\mu_{i}$ comes from then see where that comes from until I get a definition or a well known relation like Gibbs free energy. I'm not savy enough to be like "Oh yea $\mu_{i}$, that's basically $\int dG(T, P_{2})$ I'll just show all that". But, I can slowly go backward. So, I followed the following path.
$$a_{i} \Rightarrow \mu_{i} \Rightarrow (\mu(T,P_{2})-\mu(T,P_{1}) ) \Rightarrow d \mu \Rightarrow d G$$
Then when you have this general path you reverse it to get the derivation.

Also remember that you have to keep track of all the assumptions you make. Its important to remember that. My answer is only applicable for a perfect gas at constant temperature, and a mixture of gas at low pressure.

From deferential of Gibbs free energy the chemical potential of a pure gas is,
$$dG = -SdT +VdP = d\mu$$
at constant temperature,
$$d \mu = V_{m}dP = \frac {RT}{P}dP$$
To get $\mu$ we integrate with constant temperature essentially  expanding the gas on an isotherm for $P_{1}$ to $P_{2}$.
$$\int _{1}^{2} d \mu = RT \int _{P_{1}}^{P_{2}} \frac{1}{P}dP$$
$$\mu(T,P_{2})-\mu(T,P_{1})=RT \ln \Big{(}\frac {P_{2}}{P_{1}}\Big{)}$$
Let $\mu(T,P_{1})=\mu^{\circ} $,  $P_{1}=P^{\circ}=1bar$, $P_{2}=P$, and $\mu(T,P_{2})=\mu_{i} $,

$$\mu = \mu_{i}+ RT \ln \Big{(}\frac {P}{P^{\circ}}\Big{)}$$

Side note your equation 3 comes from this equation. $\mu_{i}=G_{i}$.

For a mix of gas at low pressure, the activity is equal to the ratio of the partial pressure of the gas over the standard pressure,
$$a_{i}=\Big{(}\frac{P}{P^{\circ}}\Big{)}$$

$$\mu_{i} = \mu_{i}^{\circ} +RT \ln a_{i}$$

I think that's how you prove that relationship. I hope that helped.
$$\mu_{i} = \mu_{i}^{\circ} +RT \ln a_{i} \Rightarrow a_{i}= e^{\frac{\mu_{i}-\mu^{\circ}}{RT}} $$
This relation is trivial and just some algebra. When you are trying prove something its important to start as far back as possible.

Citations:

Chemical Activity

Levine, Ira N. Physical Chemistry. 6th ed. pg 175-176 . New York: McGraw-Hill, 1978. Print.

Notes from my professor (sorry).


What do (R) and (Z) mean in chemical names?

The name (R,Z)-chloro-4,6-dimethyloct-3-ene is ambiguous. This could be (R,Z)-1-chloro-4,6-dimethyloct-3-ene, (R,Z)-3-chloro-4,6-dimethyloct-3-ene, (R,Z)-6-chloro-4,6-dimethyloct-3-ene, or (R,Z)-8-chloro-4,6-dimethyloct-3-ene. (2-chloro-4,6-dimethyloct-3-ene, 5-chloro-4,6-dimethyloct-3-ene, and 7-chloro-4,6-dimethyloct-3-ene have one additional stereocenter.) For example:


Does expansion of an ideal gas not always represent an increase in entropy of the system?

Entropy does not depend on just the volume the gas has available.  It also depends on the energy the gas has to exploit that available volume.

Suppose you expand the gas by pushing against a piston.  The gas is doing work to "earn" its extra volume, so even as volume increases the gas is expending energy.  Thus unless heat flow into the gas is allowed the gas will cool.  This loss in energy tends to decrease random motion, and thus tends to decrease entropy, even as having more volume would have the opposite effect.  We have a "tug of war" here.

In an idealized version of such an expansion,  where we aallow no heat flow into or out of the gas (adiabatic) and we maintain equilibrium with no "losses" due to friction ("reversible") then the loss of energy to doing work must effectively balance the gain in volume, leaving zero entropy change.  With no heat flows through the insulation and no heat generation due to friction in this idealized scenario, there must be zero entropy change in both the gas and the surroundings.

The lesson:  Any process (such as expansion  of an ideal gas) that is  reversible (passing through equilibrium states) and adiabatic (no heat flow) has zero entropy change.


Why doesn't hydrazine contain a double bond between two nitrogen atoms?

Very interesting question! Let's start with the nitrogens in hydrazine being approximately sp2-p hybridized, flat. The p orbitals on each nitrogen would overlap in a double bond configuration - but the "double bond" has four electrons, two from each nitrogen. If you consider a molecular orbital, the bonding AND the antibonding orbitals are filled, so the net bonding is zero. Actually worse than that, because the filled p orbitals on each nitrogen are aligned in the worst way - maximum repulsion.

The clever little nitrogens rehybridize to become pyramidal - perhaps not completely sp3, but not at all planar. This rearrangement hardly affects the N-H bonds, but allows the p orbitals to accept some s-character, which increases one lobe at the expense of the other. And of course, the prominent lobes find a way to point away from each other and reduce electrostatic repulsion of the electrons. You could say that the double bond character is reduced/eliminated, but the bonding loss is more than balanced by reducing the antibonding.

A useful comparison can be made with ethylene: planar, sp2-p, because there are only 2 electrons in the double bond.


The highest pressure they can handle is $\pu{1.3 kJ L-1}$, or $\pu{13000 J L-1}$. In the formula you use the ideal gas constant $0.08206$ and the temperature given to find the concentration the maximum pressure it can handle.

This comes out at $\pu{0.53 M}$ or $\pu{530 mM}$ and that's the maximum concentration it can handle. To find the minimum, you subtract $530$ from $630$ giving you $\pu{100 mM}$ or $\pu{0.1 M}$, which is the lowest concentration it can handle.


NMR magnetically equivalent protons for a 1,4-disubstituted benzene ring

when I was in school, I was told that [...] each pair of [equivalent] protons gives a doublet

This is only strictly true for magnetically equivalent protons. If there is magnetic inequivalence, such as in the p-disubstituted benzene ring, then it is no longer true.

It's not as simple as a dd either; when you have magnetically inequivalent protons that see each other, the first-order rules (the so-called $n+1$ rule) doesn't work so well anymore. You can see some real-life examples of how these molecules look like here: https://www.chem.wisc.edu/areas/reich/nmr/05-hmr-15-AABB.htm For the p-disubstituted benzene it sort of looks like a doublet if you zoom out far enough, but if you look closely, there are additional peaks flanking the major peaks.

I say "see each other" because if you think about it, the two protons labelled HA above are also magnetically inequivalent; but since they are so far apart, the two benzene rings are essentially independent of each other, and give rise to the same NMR spectrum.


Which rate is faster: one step with Ea or two steps each with Ea/2?

Instead of looking at $\mathrm{d}[\ce{P}]/\mathrm{d}t$ against time $t$, a better idea might be to plot $[\ce{P}]_t$ against $t$, which you already did. By eyeballing the graphs I'm fairly sure that $[\ce{P}]_t$ in case 2 is always below $[\ce{P}]_t$ in case 1.

In org chem, people probably aren't too interested in what the absolute rate of the reaction at time $t$ is, they are probably more interested in how long the reaction will take to run. Because of the maths, the reaction will never hit 100%, so you can't use that as a benchmark, but you can use e.g. 95% complete (i.e. $[\ce{P}]_t = 0.95$) as a benchmark.


I would recommend this patent which generates AgCl via the electrolytic chlorination of sheet silver, to quote:

  The present invention relates to a method for chlorination of silver materials in the forms of wire, thick film paste and metallic sheet/disc. The method is environmentally friendly since the process does not produce hazardous chemicals or flammable gas. The electrochemical setup includes position the electrode containing silver materials in the anode and a platinum wire at the cathode. A DC voltage supply between 1 to 5 volts oxidizes the silver surface which receives chloride from imidazolium chloride solution to produce silver chloride on silver wire, film or disc. The imidazolium chloride reagent can be in aqueous solution or polar organic solvents or naturally exists in ionic liquid form. The imidazolium radical cation decomposes to produce a stabilized organic radical and an imidazole molecule. The stabilized radical recombines to produce volatile organic compound that can be recovered by simple distillation. The chlorination reagent can be regenerated...

In place of platinum wire perhaps a diamond electrode.

Also, some pertinent comment from atomistry, to quote:

  The rose-red to purple-brown mixtures of silver subchloride and silver chloride are called by Carey Lea " photochlorides." They are formed by the interaction of colloidal silver and a suspension of silver chloride; by the action of organic reducers and of light on silver chloride; and by the electrolytic chlorination of sheet silver. The photochlorides are characterized by their great sensitiveness to light, blue light producing a blue coloration, and red light a red coloration. This action is reversible, the blue coloration being transformed into red by the action of red light, and so on. This phenomenon is inapplicable to the production of colour-photographs, for white light causes a darkening in colour.


How does the ambient temperature affect the rate of redox reactions in a salt lake?

The salt lakes that occur in Western Australia are touted by the article Geochemical characteristics of naturally acid and alkaline saline lakes
in southern Western Australia (Bowen and Benison, 2009) are unique environments, in that they possess

  an extreme and rare
  range of natural hydrogeochemical conditions. Many WA lake
  waters and groundwaters are hypersaline and acidic.

The article discusses observations of effects due to temperature, especially cycles of dry and wet.  The high ambient temperatures would cause increased evaporation, and hence increase local evapoconcentrations of ions, resulting in a lowering of the local pH, storm/inundation events have the opposite effect.  This leads to further observations by the authors that found that

  Field observations of lake pH during storm events
  illustrated that fluids that have been freshened due to meteoric input
  are able to regain acidity rapidly. Part of the acidity may be locally
  sustained by ongoing $\ce{Fe}$ cycling and redox reactions that
  generate $\ce{H+}$ ions. The amount of $\ce{Fe}$ increases significantly in the
  more acid waters, and authigenic $\ce{Fe} mineralization (oxides
  and sulfates, e.g., hematite and jarosite) is abundant throughout
  the acid lake sediments.


Shouldn't the Hofmann product be preferred in this elimination?

Yup, $\ce{t-BuO-}$ is a bulky base.

But is it bulky enough that it can't even abstract a hydrogen from the ring? Nope.

Take a look at this picture of a $\ce{t-BuO-}$ molecules approaching the hydrogen of 1-bromo-1-methyl-cyclopentane.

Does that $\ce{t-BuO-}$ look like it's struggling to pull that hydrogen out? Definitely not. It's doing what it does with ease.


What utility does the Tau bond model of orbital overlap have?

Tau-bonds provide an alternative valid description of electron density in alkenes and alkynes. Even in the modern chemistry
A good example is the conformational preferences in propene (lower energy of the "eclipsed" conformation). You can find more information here: http://pubs.rsc.org/en/Content/ArticleLanding/2011/OB/c1ob05393k#!divAbstract


Mixtures of water with flammable liquids can themselves be flammable, largely due to the fact that the flammable liquids generally are more volatile than water, (ie they evaporate more easily.) As a result, the air above a solution of nail polish will have a lot of molecules of the flammable liquid in it. (That's why you can smell them easily). If you have a flame near the liquid, those molecules in the air above it can combust. The heat generated will evaporate more liquid, which will fuel the fire. Some of the heat is lost to vaporizing water, which doesn't burn, but if the fraction of the total liquid that is flammable is high enough, the combustion continues.

In nail polish, the flammable liquid is usually acetone or ethyl acetate.


Advantages of thioesters over esters in fatty acid catabolism

what is the main aim of the Nature with this choice: faster reactions
  or more extensive reactions?

Nature is frugal, its main aim is to spend as little energy as possible in a chemical reaction.  Nature will always evolve so as to use the lowest energy pathway.

If we compare the resonance structures for an ester and a thioester we see that we can draw the same three resonance structures to describe each structure.

However, the resonance structure on the right, the one with the double bond between the carbonyl carbon and the ester oxygen or the thioester sulfur, is less important in the sulfur case.  Why?  Because a double bond between second and third row atoms is not nearly as strong as a double bond between two second row elements (much better electronic overlap results between atoms of similar size).

As a result the thioester carbonyl is not as stabilized by resonance as the ester carbonyl.  Said differently, the thioester carbonyl is more reactive (lower activation energy to react) than the ester carbonyl.  The thioester motif provides nature with a more reactive carbonyl. This way nature doesn't have to spend as much energy to bring about reaction at this more reactive carbonyl.


Do Grignard reagents deprotonate amines?

If you really wanted to start with the substituted aniline, you could probably oxidize, deuterate, reduce to get the product if you indeed want the amine and not the amide, but the yield would be poor. Various peroxides and peroxybenzoic acids will do this for you. Lui, Li, et al. indicated that oxidation of substituted anilines would work best in dichloroethane or THF with 4-5 equivalents of m-CPBA, with little difference produced by electronic effects. Then you could use your metal reagent, deuterate, and reduce using a method of your choice.

Of course it's probably easier to just nitrate your 1,3,5-tribromobenzene directly with nitric acid.

If this is a question from a book, they're probably trying to make sure you notice that you can't use organometal chemistry when you have acidic protons, so yes, grignard reagents do deprotonate amines.


Finding the valency of transition metals in coordination complexes

For the first example, you are given that there are $6\, \ce{CN}$ ligands which are -1. That means that there is a total of 6- from the ligands. We also know the overall charge of the complex is -3. That means the charge of the central metal must be +3. In the second example, we can look at the charge again. Because the $\ce{CH3-}$ has a  charge of -1, the molybdenum metals must have a +3 oxidation state each to counteract the -6 total charge from the $\ce{CH3-}$ groups. To do these problems, you should look at the charges of the ligands/metal(s) and the total charge of the complex and know that they have to match. After you figure out the oxidation state, you can find the $\mathrm d^n$ configuration by removing electrons from the $4\mathrm s$ subshell first and then 3$\mathrm d$. For example, For $\ce{Fe^{3+}}$, a neutral $\ce{Fe}$ atom has $6$ $3\mathrm d$ electrons. You first remove 2 electrons from the $4\mathrm s$ subshell and then a final one from the $3\mathrm d$ subshell. Finally, you will find that $\ce{Fe^{3+}}$ is $\mathrm{d}^5$. Again, this is based on the oxidation state of the metal.


Difference between Particle in a Box and Harmonic Oscillator approximations

Quantum numbers pop up when you solve Schrödingers wave equation for systems whose solutions are discrete due to the presence of a restraining potential. There is some degree of arbitrariness to what the range of quantum numbers might be, because it depends on how you define the quantum number.

You could just as well define the quantum number and associated energy levels for a 1D  pib as $$\begin{align} E_{n'}&=k(n'+1)^2 \\&=k n'^2 +2kn' + k\end{align} $$ where $$k=\frac{\pi^2 \hbar^2}{2mL^2}$$ Here $n'$ defines a "new" quantum number related to the original one as $n'=n-1$ and therefore ranges from 0 to $\infty$.

For the harmonic oscillator, you could similarly rewrite the energy as

$$E_n = \hbar \omega n'$$

where $n'$ ranges in integer increments from $\frac12$ to $\infty$.

Physically speaking, the non-zero ground-state energy of both the pib and harmonic oscillators is usually  attributed to the Heisenberg Uncertainty Principle, because a particle with no energy would have no velocity and therefore would be completely unrestrained in position. There are of course plane waves, but this would be a strange plane wave, with no amplitude.


Mechanism of reduction by elemental sodium

When alcohol molecules adsorb on the surface of the sodium metal, an electron can be transferred from sodium to the alcohol O-H bond with simultaneous disproportionation  (homolytic cleavage) of the O-H bond thereby generating the alkoxide anion and a hydrogen radical.  The hydrogen radical (probably still on or near the metal surface) can then go on to dimerize and produce hydrogen gas.

EDIT: Support of Mechanism

The mechanism proposed above involves a single electron transfer (SET) from sodium.  This is based on the fact that:

All dissolving (sodium) metal reactions, such as the Birch reduction of alkynes and aromatics, proceed by the SET mechanism
Other reactions involving sodium metal such as the acyloin condensation, the pinacol coupling, the Bouveault–Blanc reduction, etc., proceed through the SET mechanism
The reaction of sodium and water still proceeds in the gas phase where ionization seems less likely, but electron transfer still viable.

Searching on-line I found that water can solvate electrons just like ammonia can.  The chief difference being one of stability; in ammonia solvated electrons are stable for days, whereas in water they are stable for less than 1 msec.  Given the preceding mechanistic evidence and the fact that solvated electrons can exist in water, it seems reasonable to suggest that the mechanism of alkoxide formation, when sodium is reacted with an alcohol, also involves the SET mechanism.


The bond with greater electronegativity difference has greater tendency to break with help of polar $\ce{H2O}$ molecules to hydrated ions, unless it is ionic already. Electrons then stay at the oxygen as the most electronegative atom from all the three.
So if $\ce{M}$ is less electronegative then $\ce{H}$, there will be'
$$\ce{M^{(+)}-O^{(-)}-H(aq) <=> M+(aq) + OH-(aq)}$$
but if $\ce{M}$ is more electronegative than $\ce{H}$ , then there will be:
$$\ce{M-O^{(-)}-H^{(+)}(aq) + H2O <=> MO-(aq) + H3O+(aq)}$$
( The charge in parenthesis is a partial charge.)


basic chemistry alkaline test for flavonoids!!! with reference please

From this reference:

  Treatment of a flavonoid with a strong base will result in ionization of its hydroxyl groups with resulting large bathochromic shifts of both absorption maxima.


How is this mystery chemical reducing carbon emissions?

With reference to the discussion in the question comments, this product does not appear to appreciably reduce carbon emissions in total. It does however appear to reduce carbon monoxide emissions. As carbon monoxide is a product of incomplete combustion, it is plausible that the additive is helping the fuel burn more completely. Broadly speaking, you would hope that the majority of the carbon in your exhaust gas would come out as carbon dioxide, which is 'merely' a greenhouse gas and not truly nasty like CO or PAHs. Prior to the advent of catalytic converters, carbon monoxide poisoning by vehicle exhaust was a much greater risk.

It should be noted that conservation of mass tells us some powerful things about what to expect from an engine - namely that the carbon that goes in to the engine as a component of fuel must come out as exhaust gasses, particulates, polycyclic aromatic hydrocarbons etc., or else be sequestered inside the engine (though the latter is achievable by putting a banana into the exhaust pipe, it is generally done as a prank).

Reduction of carbon emissions in general is achieved through obtaining more work per mass of fuel burnt, which, when coupled with the mass and other design features of the car confers the fuel efficiency. Unfortunately, the data provided don't tell us anything about that. The more complete burning of the fuel, evinced by the reduction of CO concentration in the exhaust gas does imply some increase in efficiency, however CO is a very small component of exhaust gas, indicating alongside the low concentration of unburnt hydrocarbons that the fuel is already burning virtually to completion.

Without knowing the specific claims made by the manufacturer of the additive, I will not make any judgment of good or bad faith, however my distilled conclusions are:

The product does not dramatically impact total carbon emissions
The product appears to appreciably reduce carbon monoxide emissions


What is the mechanism of azulene synthesis?

The @orthocresol link to the Zincke reaction is a good start to the mechanism. You are correct about the first step. Dimethylamine adds to the electrophilic pyridinium salt 1 as illustrated in 1 --> 5.  The reaction is driven by loss of the poor nucleophile (decent leaving group), DNP-NH2. The (Z)-isomer 5 is in equilibrium with the (E)-isomer 6 but the reactions leading to azulene 12 require a cisoid conformation to allow for cyclization.
Cyclopentadienide anion, which is delocalized but for convenience is shown as a localized anion, displaces dimethylamine which is protonated by the methanol in the reaction mixture (see your link to Organic Synthesis). Fulvene 8 must generate the geometry shown to complete the remaining steps to azulene (9 --> 12). ADDENDUM: An added thought. The transformation of 8 ---> 10 may be a 4n+2 disrotatory thermal electrocyclization that forms the new sigma bond of 10 with cis hydrogens.


Why is the enthalpy of vaporization greater than the enthalpy of fusion?

Enthalpies of phase changes are fundamentally connected to the electrostatic potential energies between molecules. The first thing you need to know is:

  There is an attractive force between all molecules at long(ish) distances, and a repelling force at short distances.

If you make a graph of potential energy vs. distance between two molecules, it will look something like this:

Here the y-axis represents electrostatic potential energy, the x-axis is radial separation (distance between the centers), and the spheres are "molecules."

Since this is a potential energy curve, you can imagine the system as if it were the surface of the earth, and gravity was the potential. In other words, the white molecule "wants" to roll down the valley until it sits next to the gray molecule. If it were any closer than just touching, it would have to climb up another very steep hill. If you try to pull them away, again you have to climb a hill (although it isn't as tall or steep). The result is that unless there is enough kinetic energy for the molecules to move apart, they tend to stick together.

Now, the potential energy function between any two types of molecules will be different, but it will always have the same basic shape. What will change is the "steepness," width, and depth of the valley (or "potential energy well"), and the slope of the infinitely long "hill" to the right of the well.

Since we are talking about relative enthalpies of fusion and vaporization for a given system, we don't have to worry about how this changes for different molecules. We just have to think about what it means to vaporize or melt something, in the context of the spatial separation or relativity of molecules, and how that relates to the shape of this surface.

First let's think about what happens when you add heat to a system of molecules (positive enthalpy change). Heat is a transfer of thermal energy between a hot substance and a cold one. It is defined by a change in temperature, which means that when you add heat to something, its temperature increases (this might be common sense, but in thermodynamics it is important to be very specific). The main thing we need to know about this is:

  Temperature is a measure of the average kinetic energy of all molecules in a system

In other words, as the temperature increases, the average kinetic energy (the speed) of the molecules increases.

Let's go back to the potential energy diagram between two molecules. You know that energy is conserved, and so ignoring losses due to friction (there won't be any for molecules) the potential energy that can be gained by a particle is equal to the kinetic energy it started with. In other words, if the particle is at the bottom of the well and has no kinetic energy, it is not going anywhere:

If it literally has no kinetic energy, we are at absolute zero, and this is an ideal crystal (a solid). Real substances in the real world always have some thermal energy, so the molecules are always sort of "wiggling" around at the bottom of their potential energy wells, even in a solid material.

The question is, how much kinetic energy do you need to melt the material?

  In a liquid, molecules are free to move but stay close together

This means you need enough energy to let the molecules climb up the well at least a little bit, so that they can slide around each other.

If we draw a "liquid" line approximating how much energy that would take, it might look something like this:

The red line shows the average kinetic energy needed for the particles to pull apart just a little - enough that they can "slide" around each other - but not so much that there is any significant space between them. The height of this line compared to the bottom of the well (times Avogadro's number) is the enthalpy of fusion.

What if we want to vaporize the substance?

  In a gas, the molecules are free to move and are very far apart

As the kinetic energy increases, eventually there is enough that the molecules can actually fly apart (their radial separation can approach infinity). That line might look something like this:

I have drawn the line a little bit shy of the "zero" point - where the average molecule would get to infinite distance - because kinetic energies follow a statistical distribution, which means that some are higher than average, some are lower, and right around this point is where enough molecules would be able to vaporize that we would call it a phase transition. Depending on the particular substance, the line might be higher or lower.

In any case, the height of this line compared to the bottom of the well (times Avogadro's number) is the enthalpy of vaporization.

As you can see, it's a lot higher up. The reason is that for melting, the molecules just need enough energy to "slide" around each other, while for vaporization, they need enough energy to completely escape the well. This means that the enthalpy of vaporization is always going to be higher than the enthalpy of fusion.


What are the ozonolysis products of phenanthrene?

Phenanthrene reacts quickly with one mole of ozone at the 9,10-double bond according to this paper here and the OrgSyn procedure here referred to above by @Buttonwood. In chloroform a polymeric mono-ozonide is obtained. In MeOH the reaction takes a different course resulting in the formation of a cyclic methoxy-hydroxy-peroxide and a cyclic dimethoxy-peroxide. Reductive work up (Zn/AcOH or Me2S) of these products gives the di-aldehyde.


Is the volume of a protein substrate equal before and after an enzymatic reaction? (pepsin degrading egg white suspension)

The volume of the protein will probably change (although not by much) because it will be denatured as it is degraded and as the pH decreases. As it denatures, the hydrophilic surface area and hydrophobic interior volume will change, leading to a change in the local water liquid structure and density. Denatured, short polypeptides also pack into smaller volumes than native proteins do, and so that would also cause the volume to decrease in most cases. The amount of water in the system will affect the magnitude of the change. I would predict that the volume would decrease slightly, probably on the order of what happens with a few degrees of temperature increase of water (around 0.1%).

So, the short answer is "Yes, the volume will change, but probably not consistently or by enough to be used to reliably measure the extent of the reaction."

This is not my area of expertise, so I don't know what the current standard protocols are for determining reaction rates for pepsin reactions. However, I was able to find a couple of research papers that talk about using ninhydrin to measure reaction extent:

The pH-Dependence of Pepsin-Catalyzed Reactions

Kinetics of Hydrolysis of Egg White Protein by Pepsin

In the second study, they quenched the reaction at a specific time by adding NaOH to bring the pH up, then centrifuged the products before measuring the reaction extent.

If this is not a professional scientific study (meaning it's a school project, or something like that) then you might look at the method used in this report:

The Effect of pH on the Reaction between Pepsin and Egg – White

Basically, they just added pepsin and watched to see how long it took for the solution to become "clear". This won't give you very good accuracy or precision, but it might work for a science project or school report.


How to calculate the emission wavelength of a substance?

There is no general way.

For a particular case, there is need a good quantum chemistry model of energy levels, and a knowledge of character of the emission. It could be an elastic excitation, releasing the same wavelength. It can be fluorescence or phosphorescence with longer wavelength.

For very simple cases, like a hydrogen atom, it is easy and googleable. See the Rydberg constant :

  $$\frac{1}{\lambda} = Ry \cdot {1\over h} \left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)=\frac{m_\text{e} e^4}{8 \varepsilon_0^2 h^3} \left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$$
  where $n_1$  and $n_2$ are any two different positive integers (1, 2, 3, ...), and $\lambda$ is the wavelength (in vacuum) of the emitted or absorbed light.

  $$\frac{1}{\lambda} =  R_M\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$$

  where $R_M = R_\infty/(1+m_{\text{e}}/M)$ and $M$ is the total mass of the nucleus. This formula comes from substituting the  of the electron.

The more complex the real scenario is, the less are usable theoretical models and chemists must rely either on experimental data, either on empirical models not supported by underlying theoretical principle.

There is an example of mercury melting point. Experimentally, it is a piece of cake, some $\pu{-38.8 ^{\circ}C}$. Theoretically, physical chemists waited more then 50 years until there were recently powerful enough computers available. It took combination of extensive relativistic quantum chemistry computations and random calculations to reach result within $\pu{\pm 1 ^{\circ}C}$ error. The prior non-relativistic calculations predicted the melting point with the error more than $\pu{100 ^{\circ}C}$ !!


What would be different about a heavy-water ocean world?

For one thing, it would be hugely more unlikely to exist.  We are more likely to have worlds where there is a little more deuterium in the hydrogen, not 100%.

Most of the hydrogen in our Universe is protium (hydrogen-1) because the vast majority of unfused nuclear particles in stars are protons.  Neutrons must either fuse with something or they decay to make protons with a half-life of ten minutes.  Deuterium and tritium are formed initially when the protons are fused, but they are just intermediate towards the formation of more stable helium-4 nuclei (alpha particles) and so do not accumulate to large concentrations.

If there were a heavy water world, life would have to adapt to it.  Prolonged consumption of pure heavy water is deadly to life on Earth.


How do I remove free biotin after biotin-protein conjugation?

I did some bioconjugation with NHS and maleimide biotins to proteins, peptides, and oligosaccharides. I removed the free biotin using spin filters or dialysis. Just be sure to use a MWCO well below the mass of your protein. I even used prep scale HPLC to remove biotin from the oligosaccharide. The HABA kit works well, I think I used that same one to quantify peptide - avidin linking. But try to keep your biotin protected from light and oxygen, it oxidizes fairly easily, I was getting +16 and +32 masses on LC-MS and couldn't explain it until I found some 50 year old paper on biotin sulfoxide. The oxidized biotin doesn't bind avidin nearly as well as normal biotin.


Nomenclature of the following aromatic compound

Benzalamine seems to be an ocassionally and formerly used general name for compounds of the formula

$$\ce{C6H5-CH=N(R')R}$$

(where C6H5- is phenyl; R′ may be H). This name probably came from the fact that they can be prepared by a reaction of benzaldehyde with an amine:

$$\ce{C6H5-CH=O + R-NH-R' -> C6H5-CH=N(R')R}$$

$\ce{C6H5-CH=NH-CH3}$ could be, in this manner, named methylbenzalamine or N-methylbenzalamine. You can also find name α-methylbenzalamine, which would be a different compound, $\ce{C6H5-CH(CH3)=NH2}$

However, according to the IUPAC nomenclature, the compound class is and imine, or aldimine.

The correct name of $\ce{C6H5-CH=NH-CH3}$ would be

N-methyl-1-phenylmethanimine (preferred IUPAC name, 2013)

Also note that there is cis/trans (Z/E) isomerism possible.


Concentration of a substance in a solution

Using a FULLY MIXED sugar solution (in a closed container to prevent evaporation) as an example, if you could measure out successive 1-ml samples with perfect accuracy, and could count the sugar molecules in each sample with perfect accuracy, you would expect to find that the number of sugar molecules varies and follows a normal distribution with a mean value corresponding to the "nominal" concentration of the solution.

You can observe this phenomenon experimentally in the case of standard solutions containing a given DNA sequence. When you create a series of increasingly diluted samples and then analyze each dilution multiple times using a PCR assay designed to detect that sequence (modern PCR assays can literally detect one molecule of DNA!), you begin to observe the expected proportion of negative sample results as you approach a dilution factor corresponding to one copy of DNA per sample. That is because when you repeatedly sample your solution containing a nominal concentration of one DNA copy per sample, some physical samples will have zero copies, some will have one (highest probability), some will have two, etc. with decreasing probability as you go up.


Cerium (CeO) behavior in Glass Polishing?

My guess is that for the reason why your slurry deteriorates without chilling you should look the other way. Maybe the other slurry components are to blame? I do not see how ceria could impact the slurry performance in your case. Ceria (and yes, it's $\ce{CeO2}$ not $\ce{CeO}$) and some other oxides like zirconia are used in milling and polishing chiefly because of their physical hardness and chemical inertness. In other word, $\ce{CeO2}$ should probably be regarded as purely abrasive component. You need extreme conditions to chemically alter ceria, such as concentrated acids and a lot of prolonged heating, and the chemical interaction between ceria and the other solid materials is highly unlikely unless the temperatures are so high that the whole system starts to glow.


The reducing power of alkali metal such as $\ce{Na}$ and $\ce{K}$ in liquid $\ce{NH3}$ is duee to the solvated electrons ($e^-$), which can reduced certain compounds into unusual oxidation states. One such example is reduction of oxygen ($\ce{O2}$). The the solvated electrons can be reduced $\ce{O2}$ to $\ce{O2^{.-}}$ (superoxide ion) first and then to $\ce{O2^2-}$ (peroxide ion) (Ref.1). However, a solution of $\ce{Na}$ in liquid $\ce{NH3}$ is treated with oxygen, the peroxide ion is generally formed (Ref.2). When a solution of $\ce{Li}$ in liquid $\ce{NH3}$ is rapidly oxidized with $\ce{O2}$ at $\pu{-78 ^\circ C}$, a bright lemon-yellow solution is formed (which is believed to be a solution of $\ce{LiO2}$). When the solution is warmed towards $\pu{-33 ^\circ C}$, the yellow color faded and white suspension is formed (which is believed to be solid $\ce{Li2O2}$). The color change is not reversible (pp. 131-132, Ref.2).

Another example is conversion of $\ce{Ni^{II}}$ to $\ce{Ni^{I}}$ and $\ce{Ni^{II}}$ to $\ce{Ni^{0}}$. Potassium tetracyanonickelate(II) ($\ce{K2[Ni(CN)4]}$) has been reduced by $\ce{K}$ in liquid $\ce{NH3}$ at $\pu{-33 ^\circ C}$ to the red potassium tetracyanonickelate(I) $(\ce{K3[Ni(CN)4]})$ (Ref.3), which is slowly reduced further to yellow potassium tetracyanonickelate(0) $(\ce{K4[Ni(CN)4]})$ at $\pu{0 ^\circ C}$ (Ref.4).

It is known fact that solutions of alkali metal in liquid $\ce{NH3}$ reduce several organic compounds including alkynes to trans-alkenes and aromatics to 1,4-cyclodienes (Birch reduction; e.g., reduction of benzene to cyclohexa-1,4-diene).

Therefore, correct answer to the question is all of (1)-(4).

References:

William H. Schechter, Harry H. Sisler, Jacob Kleinberg, “The Absorption of Oxygen by Sodium in Liquid Ammonia: Evidence for the Existence of Sodium Superoxide,” J. Am. Chem. Soc. 1948, 70(1), 267-269 (https://doi.org/10.1021/ja01181a083).
Nils-Gösta Vannerberg, “Chapter 3: Peroxides, Superoxides, and Ozonides of the metals of groups  Ia, IIa, and IIb,” In Progress in Inorganic Chemistry; Volume 8; F. Albert Cotton, Ed.; John Wiley & Sons, Inc.: Easton, PA, 1962, pp. 125-198 (ISBN:978-04-7017-6733).
John W. Eastes, Wayland M. Burgess, “A Study of the Products Obtained by the Reducing Action of Metals upon Salts in Liquid Ammonia Solutions. VII. The Reduction of Complex Nickel Cyanides: Mono-valent Nickel,” J. Am. Chem. Soc. 1942, 64(5), 1187-1189 (https://doi.org/10.1021/ja01257a053).
George W. Watt, James L. Hall, Gregory R. Choppin, Philip S. Gentile, “Mechanism of the Reduction of Potassium Tetracyanonickelate(II) and Potassium Hexacyanocobaltate(III) with Potassium in Liquid Ammonia,” J. Am. Chem. Soc. 1954, 76(2), 373-375 (https://doi.org/10.1021/ja01631a016).
Additional reference: V. Gutmann, "Chapter III: Coordination Chemistry in Proton-containing Donor Solvents," In Coordination Chemistry in Non-Aqueous Solutions; Springer-Verlag: Wien, Austria, 1968, pp.35-58 (ISBN-13: 978-3-7091-8196-6).


So how are the cells are arranged to form a solid structure a solid body?

Plant cells have cell walls, are filled with liquid, and have strong cell-cell adhesion. This gives a fairly rigid structure that may seem like a solid. Take a carrot for example, it seems solid. But if you put the carrot in a blender, you get carrot juice - not a solid. Also, when the carrot dries out, it loses its rigidity.

Trees

Woody plants contain a lot of cellulose, a water-insoluble polymer, outside of cells. This gives them even more of a solid (and dry) appearance, but the insides of their cells are aqueous as well.

Cells other than plant cells

There are cells that lack a cell wall, they are just surrounded by a lipid bilayer. Tissues made from these cells (muscle, brain, liver, etc.) don't seem solid at all. Animals do have connective tissue (mostly protein-based) and a skeleton that holds everything together unless there is trauma (predators, accidents, etc).

  How do cells form a solid structure?

Let me un-ask the question a bit. Sometimes they do form a solid-like structure, and sometimes they don't. Fibrous biological macromolecules play a large role in making tissues rigid, either directly or in combination with having cells filled with aqueous solution.


The products of the reaction are perfectly miscible with water and are harmless to environment in low concentrations. My guess that the leftovers of ethyl acetate can be turned to ethanol and acetic acid before disposal.

For example here you can see that ethyl acetate is forbidden from sewer disposal while ethanol and acetic acid are allowed.

https://www.vumc.org/safety/waste/chemical-waste-sewer-disposal


Is there a similar characteristic of the lanthanide contraction in the other transition metal groups?

Hope I interpret your question correctly:

A decrease of atomic radii occurs in every period (from left to right). Atomic radii of lower periods are in general larger (Increase from top to bottom). This simple behavior only holds true for elements that only have p-block and s-block electrons.

The first anomaly of this behavior is the d-block contraction. This can be seen in the 4p, 5p, 6p and 7p elements (Elements to the right of Ga, In, Tl). For example the atomic radius increases about 60 % from C to Si. The difference of Si and Ge with the added p-block contraction is just 20 %.

The lanthanoide contraction occurs when the f-block is added. For this reason the 4d and 5d elements are very similar in size. For example Hf (5d, 160 pm) is even smaller than Zr (4d, 159 pm).

Finally the actinoide contraction should have the same effect on 6d elements. But I'm not sure if realiable numbers for the atomic radius exist for those elements.


Can anyone tell me what reaction is this?

Diazomethane under the action of light or heat releases nitrogen leaving a  neutral, divalent carbon species with a lone pair and an empty orbital called a carbene. This carbene then reacts with the double bond of alkene breaking it to form a cyclo- species. in your example, cis-2-butene will react with diazonmethane to form cis-1,2-dimethylcyclopropane with no detectable amount of the trans isomer.

However, do note that if the same reaction is carried out via a triplet carbene, generated via triplet sensitization, then a 1.3:1 mixture of trans to cis dimethylcyclopropane is observed. [3]

Source

https://www.masterorganicchemistry.com/2011/11/05/reagent-friday-diazomethane-ch2n2/
[Source for pic. 1]
http://www.ochempal.org/index.php/alphabetical/c-d/diazomethane/ [Source for pic. 2]
http://reag.paperplane.io/00000832.htm
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alkenes/Reactivity_of_Alkenes/Diazomethane%2C_Carbenes%2C_and_Cyclopropane_Synthesis


Does ethanol "decompose" or "remove" phosgene from chloroform?

It's a weird way of phrasing it, but I think they are trying to lead you to this

Basically, chloroform is stabilized with ethanol or amylene, but if a sample is dried, it will no longer contain stabilizer. To actually remove phosgene, you can wash it with NaHCO3. This would essentially leave you with entirely harmless products.

This is my best guess as to what they are trying to get you to understand. Please let me know if this helps.


How can the solid state structure of cyclopentadienyllithium be explained?

This summary identifies several calculations indicating that covalent interactions between lithium and cyclopentadienyl moieties do occur; the interaction is mostly but not purely ionic.

Draw the occupied $\pi$ orbitals of the cyclopentadienyl groups and, for comparison, draw the $s, p_x, p_y, p_z$ orbitals on lithium with the $z$ axis along the length of the chain.  Observe:

• The lithium $s$ and $p_z$ orbitals overlap the lowest energy cyclopentadienyl $\pi$ orbitals.

• The lithium $p_x$ and $p_y$ orbitals overlap the bonding degenerate pair on the cyclopentadienyl ring.

With these overlaps, we can construct three lithium-cyclopentadienyl bonding combinations per cyclopendienyl group, one corresponding to each cyclopentadienyl $\pi$ bonding orbital.  We then occupy these orbitals with six electrons per ring.  In effect we have extended the aromatic coupling of the cyclopentadienyl ion into the third dimension with the lithium orbitals.

Heavier alkali metals do not work as well because their orbitals are too diffuse, but they could overlap larger rings.


No, because osmosis is not an electrical phenomenon, but rather a chemical one. The concentration of protons would diffuse across the membrane if the fluid, I'm assuming which is water, will be allowed to interact with the other side of the membrane. However, water will not seek to increase the pH of an acidic solution within a membrane the same way it will to dilute concentrations of ions. The resulting balance of pH across the membrane will be as a result of the water to water interaction across the membrane and not because of the pH being imbalanced.


Derivation of the isothermal Joule-Thomson coefficient

When you write a total differential such as
$$dH = \left(\frac{\partial H}{\partial T}\right)_p dT + \left(\frac{\partial H}{\partial p}\right)_T dp \tag{1}$$
you are applying the methods of differential geometry, so in reality the answer to your question lies in the applicability of these methods in thermodynamics (the rest being "math"), something which statements such as "so-and-so is a state function" implicitly justify. One can write a total or exact differential of a state function, as for the enthalpy in the equation above. This equation can be interpreted as follows: small (differential) changes in p and T, which are orthogonal dimensions (in the sense that they can be varied independently), additively cause a linearly proportional differential change in the function H. In the differential limit, the surface of H looks like a plane. The partial derivatives describe the slope of the plane in the orthogonal dimensions.

The cyclic rule can be derived from the above equation by taking the partial derivative wrt one of the independent variables while holding H constant.

$$ 0 = \left(\frac{\partial H}{\partial T}\right)_p \left(\frac{\partial T}{\partial p}\right)_H + \left(\frac{\partial H}{\partial p}\right)_T \tag{2}$$

What does it mean here to hold H constant? It means we are looking for an isenthalpic path on the enthalpy surface, from the initial point at which we computed the partial differentials of the surface wrt T and p, in direction $\left(\left(\frac{\partial T}{\partial p}\right)_H dp, dp\right)$, where the partial differential $\left(\frac{\partial T}{\partial p}\right)_H$ also happens to be given (thanks to the geometry of the problem) by

$$ \left(\frac{\partial T}{\partial p}\right)_H = -\frac{\left(\frac{\partial H}{\partial p}\right)_T }{\left(\frac{\partial H}{\partial T}\right)_p} \tag{3}$$

Alternately, consider the line resulting from intersection of a horizontal isenthalpic plane $c(T,p)=H_0=H(T_0,p_0)$ and the plane $s(T,p)$ tangential to the surface $H$ at the point $(T_0,p_0,H_0)$, the tangential plane given by $$s(T,p) = H_0 + C_p \Delta T + \varphi \Delta p$$

where

$$C_p=\left[\left(\frac{\partial H}{\partial T}\right)_p\right]_{(T_0,p_0)}$$
$$\varphi=\left[\left(\frac{\partial H}{\partial p}\right)_T\right]_{(T_0,p_0)}$$

are the partial derivatives of $H$ evaluated at $(T_0,p_0)$, and $\Delta T = T-T_0,~ \Delta p = p-p_0$. Solving for the intersection line by setting $s(T,p)=c(T,p)$ gives

$$T = -\frac{\varphi}{C_p} p + T_0 + \frac{\varphi}{C_p} p_0 +\frac{c-H_0}{C_p}$$

The slope of the intersection line can be recognized to be the same as $\left(\frac{\partial T}{\partial p}\right)_H$ given by Eq. (3). The geometric nature of the problem and relation between the different derivatives should then be clear.

  So how can we now use this variable when dH≠0?

A Joule-Thompson coefficient $\mu$, like any other thermodynamic (or state) properties, is strictly valid at the conditions under which it is determined (it may have a broader useful range depending on how much it varies with T and p and the tolerated error). This is different from the question of the mathematical accuracy of the relationships used to derive the properties.    $\mu$ is derived at a specific state defined for a pure substance by a specific point (T,p) and as such is a fixed property of the substance at that point.


How do I prepare the following compound from diphenylmethane?

Here's a mechanism that I thought of (It's more theoretical than practical):

Bromination in the first step will occur at the benzyl position due to a stable free radical. Sodium Propynide then adds more carbon atoms via the SN2 mechanism, which are then reduced with Lindlar's catalyst to give the cis-alkene.

The name of this compound should be (Z)-1,1-diphenylbut-2-ene, but as pointed out in the comments, the preferred IUPAC name (PIN) seems to be (Z)-but-2-ene-1,1-diyldibenzene (verified by ChemDraw)


Temperature dependence of reaction enthalpy

All enthalpies are taken to be at $298\mathrm{~K}$ if not specified.

The enthalpy of combustion that is shown in Table 2C.4, $\Delta_\mathrm{c} H^\circ = -890\mathrm{~kJ~mol^{-1}}$, corresponds to $\Delta H^\circ$ for the reaction:

$$\ce{CH4(g) + 2O2(g) -> CO2(g) + 2H2O(l)}$$

which is not the reaction you are interested in (since $\ce{H2O}$ is in the liquid state). Therefore, to calculate the enthalpy of the reaction, you have to use $\Delta_\mathrm{f} H^\circ$ values from Tables 2C.4 and 2C.5:

$$\begin{align}
\Delta H^\circ &= 2\Delta_\mathrm{f} H^\circ(\ce{H2O(g)}) + \Delta_\mathrm{f} H^\circ(\ce{CO2(g)}) - \Delta_\mathrm{f} H^\circ(\ce{CH4(g)}) \\
&= -802.34\mathrm{~kJ~mol^{-1}}
\end{align}$$

as I said in the comments.

Now, even though the question said to use the temperature dependence of heat capacities in Table 2C.1 (in the appendix), the solution manual irritatingly takes the values from Problem 2C.7, on page 106. Here, the heat capacity is expressed in the form $C_{p,\mathrm{m}} = \alpha + \beta T + \gamma T^2$:

$$\begin{array}{c|ccc}
\mathrm{Molecule} & \alpha / (\mathrm{J~K^{-1}~mol^{-1}}) & \beta / (\mathrm{mJ~K^{-2}~mol^{-1}}) & \gamma / (\mathrm{μJ~K^{-3}~mol^{-1}}) \\
\hline
\ce{CH4(g)} & 14.16 & 75.5 & -17.99 \\
\ce{CO2(g)} & 26.86 & 6.97 & -0.82 \\
\ce{O2(g)} & 25.72 & 12.98 & -3.862 \\
\ce{H2O(g)} & 30.36 & 9.61 & 1.184
\end{array}$$

Therefore, we have:

$$\begin{align}
\Delta C_{p,\mathrm{m}} &= 2C_{p,\mathrm{m}}(\ce{H2O(g)}) + C_{p,\mathrm{m}}(\ce{CO2(g)}) - 2C_{p,\mathrm{m}}(\ce{O2(g)}) - C_{p,\mathrm{m}}(\ce{CH4(g)}) \\
&= \left[21.98 - \left(75.27 \times 10^{-3}\right)(T/\mathrm{K}) + \left(27.262 \times 10^{-6}\right)(T/\mathrm{K})^2\right] \mathrm{~J~K^{-1}~mol^{-1}}
\end{align}$$

and integrating from $298\mathrm{~K}$ to $500\mathrm{~K}$ (I got lazy and used my graphing calculator):

$$\int\limits_{298\mathrm{~K}}^{500\mathrm{~K}}\!\!\Delta C_{p,\mathrm{m}}\,\mathrm{d}T = -731 \mathrm{~J~mol^{-1}} $$

and:

$$\begin{align}
\Delta H^\circ (500~\mathrm{K}) &= \Delta H^\circ (298~\mathrm{K}) + \int\limits_{298\mathrm{~K}}^{500\mathrm{~K}}\!\!\Delta C_{p,\mathrm{m}}\,\mathrm{d}T \\
&= (-802.34 - 0.731) \mathrm{~kJ~mol^{-1}} \\
&= -803.07 \mathrm{~kJ~mol^{-1}}
\end{align}$$

Now that's irritating, because if we use whatever data we can find from the appendix, we get the answer $-799.76\mathrm{~kJ~mol^{-1}}$, as mentioned in the comments. However, I suppose there's two things that can be learnt:

Do not assume that the combustion reaction gives $\ce{H2O(g)}$, especially if it is given for $T = 298\mathrm{~K}$.
Check the units in the heading of the table.

Your procedure itself is perfectly sound, which is a good thing.


Why would the pH of a solution not be 7 at the equivalence point of an acid-base titration?

The equivalence point is defined as when you have an equal amount of acid and base. It you have a strong acid and a strong base, the pH at the equivalence point should be pretty close to 7, but if you have a weak acid or a weak base involved, there is not reason why the equivalence point should be at pH 7. You would have a conjugate base or acid that is reasonably basic or acidic at the equivalence point.


Difference between adiabatic and nonadiabatic tunneling for molecule

This is very far from a complete answer, especially w.r.t the part of your question regarding "laser fields". After doing some reading, nonadiabatic tunneling is conceptually the same as adiabatic tunneling except that one allows for coupling to another adiabatic state.

Adiabatic tunneling is the phenomenon where a particle described by a wavefunction is propagating on a potential and begins with less energy than the height of the barrier it approaches. One finds, however, that there is amplitude of the wavefunction on the other side of the barrier despite the particle having insufficient energy to cross the barrier.

Now, in physics and chemistry, nonadiabatic processes typically describe those processes where it is a bad approximation to consider only a single potential energy surface. That is, both surfaces may be adiabatic states, but these states are coupled in some manner. In general, there will always be some small coupling between states that are not too far apart in energy.

Nonadiabatic tunneling just describes the situation that one does not neglect this coupling between states. To a first approximation, nothing will change and one will find the same amplitudes as a function of position and time. If you consider the presence of the higher-energy adiabatic state as a perturbation of the Hamiltonian of the original particle, then the zeroth order picture will just be adiabatic tunneling. The next term will be something like the case where the particle is either reflected or transmitted over the barrier, but it does so by first "visiting" the higher-energy adiabatic state.

Now, with that description in mind, I think it is reasonable to say that nonadiabatic tunneling does not necessarily have to be involved with with conical intersections or curve crossing. Nonadiabatic tunneling is just regular tunneling which allows other states to couple with the original state. Now, of course, conical intersections and curve crossing could very well come into play in nonadiabatic tunneling because this would just be another form of coupling one could introduce.

As an aside, the description I give where the higher-energy state is a perturbation on the original system is probably not a good one in general as one could imagine very strong coupling between the potentials (as would be the case with a conical intersection).


effect of overpressure on lowering viscosity of polysaccharide solution

The pressure is irrelevant, it is only applied so the solution can reach 110°C. Water boils at 100°C at ambient pressure.

Agarose is a polysugar, and at 110°C in water, I expect it breaks down into smaller molecules, disugars, monosugars etc.

Solutions of smaller molecules have lower viscosity. https://en.wikipedia.org/wiki/Mark%E2%80%93Houwink_equation

For typical polymers, the scaling exponent $a$ of Mark and Houwink is 0.5-0.8, but that's for infinitely low concentrations. It goes up drastically for higher concentrations, reaching a typical 3.4 in the melt.

(That's a bit sloppy, Mark-Houwink predicts the intrinsic viscosity, while in an entangled melt the actual viscosity scales with $M^{\approx3.4}$.)

So, if you have a saturated solution, and the average agarose polymer molecule breaks just twice, a factor ten in viscosity decrease is very reasonable.


doesn't explain why electrons in the conduction band are able to conduct electricity while electrons in the valence band can't

Electrons in the valence band can conduct electricity, as long as the valence band is not completely filled. This occurs in semiconductors, when you thermally promote electrons from the valence to the conduction band, you are leaving behind some holes in the valence band. The electrons in the valence band can then conduct electricity as well by "jumping" through the holes. The conductivity of semiconductors arises therefore not only from electrons in the conduction band, but also holes in the valence band.

The origin of conductivity lies in a partially-filled band, and it doesn't matter which band it is. For a more rigorous treatment google for "free electron model". The idea is that each crystal orbital (i.e. MO of the crystal) has an associated value of electron momentum. For each orbital with a positive electron momentum, there is an orbital with an equal but negative momentum. When the band is fully filled, all of the crystal orbitals are filled, and there is no net electron momentum because the positives and negatives cancel out exactly. When the band is incompletely filled, the application of a potential difference in a certain direction across the metal causes the electrons to preferentially populate orbitals with positive momentum in that direction. This leads to a net positive electron momentum in that direction, meaning that electrons "flow".

  The s and p orbitals of the metals combine together to form MOs. [...]

Yes. Bands are simply a collection of crystal orbitals that have such small energy gaps between them such that they are almost continuous. I don't see how the Pauli exclusion principle comes into play except for when you are discussing the population of the orbitals (obviously any one orbital can have a maximum of two electrons). The two ways that I know of to obtain the bands (nearly free electron model, and the LCAO/tight-binding approach) both do not invoke the PEP. However I will add a disclaimer that physicists approach solid state chemistry from a very different direction and there could very well be some explanation that I do not know of.

However, yes, the electrons populate orbitals that are "delocalised" across the entire solid and this is the origin of the "sea of delocalised electrons" description of metals.

  'it is only when these bands become filled with 2p electrons that the elements lose their metallic character'

I honestly don't see how that is relevant or even correct at all. Let's use the example of sodium versus chlorine. The band structure of Na metal is vastly different from that of chlorine since the molecular structures are different to begin with. Sodium is a metal and chlorine is a molecular gas, you wouldn't use a band structure to describe chlorine. You cannot just take a diagram for $\ce{Na}$ and say that it applies to $\ce{Cl2}$ and try to make statements such as "filling of the 3p bands leads to decrease in metallic character" because those bands don't even exist in chlorine. Band structures are just large MO diagrams, it is almost like trying to compare the physical properties of sodium and chlorine using MO diagrams for $\ce{Na2}$ and $\ce{Cl2}$, which is plain absurd.

However, if chlorine were to adopt a structure that has close-packed atoms like in sodium, then you would be right that it would still be metallic due to the partially-filled 3p bands.

If you want to read further I would recommend finding a solid state chemistry textbook, all of them would cover basic band theory. In particular I find this book by Moore and Smart clear and accessible.


You should look on Basis set exchange

After you select your basis set of interest, on the bottom left you click "Get References for selected basis", and in this case you obtain (I did it for elements C,H,O):

  References for the basis set

  H

## VTZ Valence Triple Zeta: 3 Funct.'s/Valence AO for H,He
Krishnan, R., Binkley, J. S., Seeger, R., Pople, J. A.
Self-consistent molecular orbital methods. XX. A basis set for
        correlated wave functions
J. Chem. Phys., 72, 650-654 (1980)
10.1063/1.438955

## 1P Polarization functions associated with 6-311G for H,He
Krishnan, R., Binkley, J. S., Seeger, R., Pople, J. A.
Self-consistent molecular orbital methods. XX. A basis set for
        correlated wave functions
J. Chem. Phys., 72, 650-654 (1980)
10.1063/1.438955

## Diffuse functions asociated with 31G Hydrogen
Clark, Timothy, Chandrasekhar, Jayaraman, Spitznagel, Günther W.,
        Schleyer, Paul Von Ragué
Efficient diffuse function-augmented basis sets for anion
        calculations. III. The 3-21+G basis set for first-row
        elements, Li-F
J. Comput. Chem., 4, 294-301 (1983)
10.1002/jcc.540040303

  C,O

## VTZ Valence Triple Zeta: 3 Funct.'s/Valence AO
Krishnan, R., Binkley, J. S., Seeger, R., Pople, J. A.
Self-consistent molecular orbital methods. XX. A basis set for
        correlated wave functions
J. Chem. Phys., 72, 650-654 (1980)
10.1063/1.438955

## 1P Polarization functions associated with 6-311G
Krishnan, R., Binkley, J. S., Seeger, R., Pople, J. A.
Self-consistent molecular orbital methods. XX. A basis set for
        correlated wave functions
J. Chem. Phys., 72, 650-654 (1980)
10.1063/1.438955

## Diffuse functions asociated with 6-31G
Clark, Timothy, Chandrasekhar, Jayaraman, Spitznagel, Günther W.,
        Schleyer, Paul Von Ragué
Efficient diffuse function-augmented basis sets for anion
        calculations. III. The 3-21+G basis set for first-row
        elements, Li-F
J. Comput. Chem., 4, 294-301 (1983)
10.1002/jcc.540040303


There's an absolute zero, is there an "absolute hot"?

In the actual theories of physics the highest temperature which has a physical meaning is the Planck's temperature.

$$T_\mathrm{P} = \frac{m_\mathrm{P} c^2}{k} = \sqrt{\frac{\hslash c^5}{G k^2}} \approx \pu{1.4e32 K}$$

For the moment no theory predict higher temperature because of the limit of our theories.

There is a Wikipedia article about absolute hot with some references. You must have a look.

  Contemporary models of physical cosmology postulate that the highest possible temperature is the Planck temperature, which has the value $\pu{1.416785(71)e32}$ kelvin [...]. Above about $\pu{10^{32} K}$, particle energies become so large that gravitational forces between them would become as strong as other fundamental forces according to current theories. There is no existing scientific theory for the behavior of matter at these energies. A quantum theory of gravity would be required. The models of the origin of the universe based on the Big Bang theory assume that the universe passed through this temperature about $10^{−42}$ seconds after the Big Bang as a result of enormous entropy expansion.


Is it possible to define the coordinates of an atom using other coordinates while optimizing in gaussian 16?

I just figured out that there is a way to do this. Though its not that clean I think. I can define the initial structure where x3 = (x1+x2)/2 and then freeze the following expression using gic.

x3 - (x1+x2)/2


Using thermodynamics to predict the acid-base character of fluoride ion in water

The textbook walks us through the interpretation of these enthalpies (they also caution us not to use enthalpies as sole criterion for spontaneity of reactions):

So they are asking whether water reacts with the acid HA, and whether water reacts with the conjugate base $\ce{A-}$.

If we try that for the case at hand, we first consider whether HF reacts with water:

$\ce{H2O(l) + HF(aq)->[?] H3O+(aq) + F−(aq)}$

Here, water and fluoride are the bases "fighting for the proton" (reactants: fluoride has the proton; products: water has the proton). Fluoride wins (i.e. HF does not give away the proton) because it has the higher proton affinity, and the reaction does not proceed judged by enthalpy alone.

Second, we consider whether the fluoride ion reacts with water:

$\ce{H2O(l) + F-(aq) ->[?] OH-(aq) + HF(aq)}$

Here, fluoride and hydroxide are the bases "fighting for the proton". Hydroxide wins because it has the higher proton affinity, i.e. this reaction also does not proceed judged by enthalpy alone.

So HF is not a strong acid, and fluoride is not a strong base. Another way of saying this is that hydroxide is a stronger base than fluoride, and hydronium is a stronger acid than HF.

In the textbook walk-through, they conclude that HA is a strong acid and - for a different reaction - $\ce{A-}$ is a strong base. In our case, HF is a weak acid  and fluoride is its conjugate weak base. If you try the same question with HCl, you will see it is a strong acid and chloride is not a strong base. (I don't understand the conclusion of the solution manual - they might have gotten confused because they know fluoride is a weak base, but their calculation together with the textbook argument merely shows that it is not a strong base.)

Now to your questions:

  Why does the text consider the second reaction with H2O? Doesn't F− act as a base in both reactions?

I think they ask "does HF react with water?", and "does fluoride react with water?". I don't know why the answer flips one of the reactions (maybe the solution manual wasn't written by the textbook author, and the original intent of the question got lost).

  If both reactions occur and F− acts as a base, then which one is dominant?

There are three equilibrium reactions, the two shown, and auto-dissociation of water. I don't know the kinetics of the three, but they are all fast and the three equilibrium constants are linked because if you subtract the two reactions you show, you get the auto-dissociation of water. For the equilibrium state, it does not matter which set of two you consider.

  If the first reaction is dominant, then how does it proceed given the low concentrations (1x 10−7 M) of H3O+ in neutral water?

Reaction rate constants vary by many orders of magnitude. The reaction of fluoride and hydronium is nice because the reactants are of opposite charge, so maybe it has a high rate constant. The reaction of fluoride and water is nice because water is the solvent, so it is present at high concentration and already at close proximity. I could not tell you which reaction is faster even if I know the concentrations of all the reactants.

Here is the table of proton affinities as a quick reference:


Is there an efficient and effective way to clean oil from floor?

Assuming you only want chemistry-related answers, I'd recommend any type of detergent.

If you lift that restriction, however, then I'd recommend pressure washers.


What are solvents that can dissolve sebum that is safe for topical use?

Dissolving sebum (the oily substance that protects and waterproofs our skin) sounds like a neat chemical way of removing it - a way to remove skin oiliness. But the solvents that dissolve sebum in a test tube are likely to be too aggressive for extensive use, or to have an oily nature in themselves.

Of the safe and common solvents that might be useful, isopropanol (91% if available, or 70%) would be somewhat invigorating - if not used in excess. It will almost certainly dry the skin, perhaps too much - so that you resort to applying a skin lotion to provide some oiliness.

An alternate way of removing sebum from skin would be to emulsify it with a detergent. Sodium stearate is a common, safe detergent (actually, a soap) which is somewhat soluble in water and available commercially (Ivory Soap is 99.44% pure!). Applied to a washcloth, it will remove much of surface sebum. Rinse, lather, repeat.

Some chemical and cosmetic improvements have been made to other products: added colors, fragrances and acids to reduce the pH from neutral to slightly acidic, for a gentler action on skin; added water for semi-liquid soap (Soft Soap).