Ethanol and water hygroscopic equilibrium concentration

You have a closed system with (say) 100 gallons of (say) pure ethanol and 100 gallons of moist air.  There is not enough water in the air to go into the ethanol and reach the azeotropic ratio -- not by 2 or 3 orders of magnitude (see below).

If you know the relative humidity of the air at the time the tote was opened (and the temperature), you can find the absolute humidity (the mass % of water vapor in the air) and so estimate the maximum amount of water that went into the tote.  Since 100 gallons is about 20 times the volume of a mole of gas, assuming an absolute humidity of 10% (it's very likely way less) the answer is about 2 moles, or about 40 grams. In equilibrium (in the closed system) almost all of the water vapor will very soon go into the ethanol, so you have 100 gallons of ethanol (with a bit of toluene), plus 40 g of water.  That's 0.01%.  Whether this much water ruins your alcohol or not, you will know better than anyone.

Every time you open the tote, you're potentially letting in 40g of water.
I say "potentially" because, since the gas content is not instanty exchanged, much less water than that might go in.  But still, if you open it often, the percentage will keep going up -- asymptotically to the azeotropic ratio, but since the alcohol will also evaporate, as has been pointed out, you probably won't ever get near that ratio.


Creating an exothermic reaction through exposure to oxygen

There are products like this already on the market.  A product called Hand Warmers are gloves that make use of exothermic reaction to warm your hands.  And there are cold packs (that make use of endothermic reactions) to put on sports injuries.


What are the byproducts in a Suzuki reaction?

It is postulated that the trifluroborates are not the transmetallating specis, rather than to be the source of slow hydrolysis to yield boronic acids:

(source 1, slide 41; citing Molander and Biolatto, 2003 (Ref.1))
which then enter the catalytic cycle:

(preview in Kürti & Czakó, page 448 bottom (Ref.2)
Organic Reactions (vide infra) explicitly states that complete hydrolysis of the trifluoroborates is necessary to allow successful transmetallation.
In addition, the fluoride anion is said to interact in the following way with the palladium catalyst:

(source 2)

References:

Gary A. Molander, Betina Biolatto, "Palladium-Catalyzed Suzuki−Miyaura Cross-Coupling Reactions of Potassium Aryl- and Heteroaryltrifluoroborates," J. Org. Chem. 2003, 68(11), 4302-4314 (https://doi.org/10.1021/jo0342368).
László Kürti, Barbara Czakó, In Strategic Applications of Named Reactions in Organic Synthesis: Background and Detailed Mechanisms; Forwarded by E. J. Corey, Introduction by K. C. Nicolaou; Elsevier Academic Press: Burlington, MA, 2005 (ISBN-13: 978-0-12-429785-2).


You are correct. It is unnecessary to include the 2 in order to provide an unambiguous name. This compound is also often called isobutane which is an older name that is still in common usage.


Molecular Orbital Mixing to get the Molecular Orbitals of a large molecule

Running the calculation without imposing symmetry constraints should give the same ground energy as one with them because the true electronic ground state of the system (within the Born-Oppenheimer approximation) always has the same symmetry as the nuclei. The variational principle always puts an upper bound on your ground state energy, so in principle it will work just as well. Using symmetry, however, could speed up your calculation significantly because the symmetry operators commute with you Hamiltonian. This gives you an additional set of constraints on the coefficients. It is worth mentioning that single reference methods like Hartree-Fock and DFT will sometimes give lower ground state energies without symmetry constraints because the true ground state is a symmetric superposition of several configurations, but the single reference methods can only examine a single one.

For polymer molecules and other 1D infinite systems you would use periodic boundary conditions. Just make the box really large in the two dimensions you don't want to be periodic (so the system doesn't 'see' itself) and only evaluate non-gamma k-points along the periodic vector.


Calculating conductivity of a solution given molar conductivities and concentration

Just be explicit with all the units, including for the molar concentration, and be sure they cancel out properly.  You have $\pu{0.1 mol L-1}$ for the solution concentration, but your $K$ values are in $\pu{S cm2 mol-1}$.

So convert your molar concentration to cubic centimeters (there are $\pu{1e3 cm3}$ in one liter) and you get $\pu{1e-4 mol cm-3}$.

Then when you multiply both values and your units cancel nicely to give you

$$\pu{1e-4 mol cm-3} \times \pu{247.5 S cm2 mol-1} = \pu{2.475e-2 S cm-1}$$


For starters, note that you will probably be using the hexahydrate $\ce{CrCl3 * 6 H2O}$. In fact, there is not the hexahydrate, but three distinct hexahydrate crystal structures known. These can be described as $\ce{[CrCl2(H2O)4]Cl . 2 H2O}$, $\ce{[CrCl(H2O)5]Cl2 . H2O}$ and $\ce{[Cr(H2O)6]Cl3}$. In solution, these ions will dissociate, giving you the corresponding free cation, one to three free chloride anions and potentially additional water molecules. It is safe to say that these three anions will take part in an equilibrium reaction $(1)$.

$$\ce{[CrCl2(H2O)4]+ + 2 H2O <=> [CrCl(H2O)5]^2+ + H2O + Cl- <=> [Cr(H2O)6]^3+ + 2 Cl-}\tag{1}$$

It doesn’t really matter which one is the most prevalent species, the argument is generally the same for all three. In all three cases, we have a positively charged cation to which water forms a dative bond. In neutral solution, water is known to self-ionise according to equation $(2)$

$$\ce{H2O <<=> H+ + OH-}\tag{2}$$

The displacement of a proton is facilitated by neighbouring electron withdrawing groups: They reduce the charge density making an anion more stable. And for all intents and purposes, cations are electron withdrawing, displaying a $-I$ effect (negative inductive effect). Therefore, forming a coordinate bond to a chromium(III) ion — even if the overall complex only has a $1+$ charge — will increase the tendency of a water molecule to deprotonate. Or, in equation terms, the equilibrium of equation $(3)$ is right-shifted when compared to equation $(2)$’s equilibrium. (I’ll only discuss this for the monochloridocomplex for simplicity.)
$$\ce{[CrCl(H2O)5]^2+ <=> [CrCl(OH)(H2O)4]+ + H+}\tag{3}$$

Since displacing a proton is easier, the solution is more acidic, the $\mathrm{pH}$ is lower.


Generally available fluid with good heat transport properties

400 C is beyond long term use of any organic heat transfer fluid.  Dow Corning boasts 427 C for one of its silicone fluids.  Don't bet on that.  Inorganic eutectic salts will long term decompose into corrosive volatiles.  Best bet for a red hot closed system is fusible alloys of indium, gallium, and bismuth - beware of metal boiling points.  Open systems will air oxidize.   Accommodate sealed inert gas fill (argon) with a metal bellows add-on, preferably on a pipe extension away from the hot zone.  Fill composition must be compatible with container composition.  Mercury is out for toxicity and boiling point, 357 C.  Thermal coefficient of linear expansion, solid vs. liquid alloy, must be looked up.  The increment is cubed for volume change.

500 C (red heat) will change the metallurgy of stainless steels over time.  They will harden (carbide precipitation), among other things.  Don't try to drill a hole or tap a thread in stainless that has been soaked at red heat.


Why is Oxygen compulsory for free radical reactions?

There is no such condition that oxygen must be present to generate free radicals. Chlorine forms free radicals, and so do many organic molecules. In fact the presence of oxygen is bad for free radical reactions.


How to calculate the concentration of NO at equilibrium?


Explain how to filter Ca and Ba ions separately out of solution using insoluble compounds

You could carry out the standard qualitative analysis procedure. Dissolve the mixture in  hot acetic acid solution.

Then add pottasium chromate ,barium is precipitated as its chromate salt(pale yellow ppt).

Filter it and you could then precipitate Calcium from the resultant filtrate by adding ammonium oxalate.(precipitated as calcium oxalate, white ppt)


Slater determinant as an unperturbed atomic wave function

However if the single or linear combinations of Slater determinants in
  a same configuration are the zeroth order functions (eigenfunctions of
  the unperturbed atomic Hamiltonian), how the fact that S2 and L2
  commute with the atomic Hamiltonian makes the Slater determinants be
  the eigenfunctions of the atomic Hamiltonian??

Actually, you go it right, but let us spell this out step-by-step. First, forget about angular momentum operators, since you do not need them to understand that a single Slater determinant is indeed not an eigenfunction of the atomic Hamiltonian (as you called it). That is it, you are right. A Slater determinant is an eigenfunction of the unperturbed Hamiltonian which describes a system of independent electrons, but not of the exact one.

Nevertheless, a Slater determinant can be used as a trial wave function in a variational procedure. That is the whole point of the Hartree-Fock method: a Slater determinant $\Phi$ is not an eigenfunction of the atomic Hamiltonian, but we could evaluate it energy $\left\langle \Phi \mid H \mid \Phi \right\rangle$ using Slater rules, and consequnetly, we could minimize it to find an upper bound to the ground state energy. The resulting Slater determinant obtained by minimizing the energy is indeed only an approximation to the ground state wave function.

Note also that the exact wave function (which is an eigenfunction of atomic Hamiltonian) can be expressed as a linear combination of Slater determinants for the various possible electronic configurations, not just a single Slater determinant. And this is the theoretical basis for configuration interaction (CI) method.

Now back to all this business with angular momentum operators. A single Slater determinant is not necessarily an eigenfunction of neither $\hat{L}^2$ nor $\hat{S}^2$. However, as you mentioned, the Hamiltonian indeed commutes with all these operators (in non-relativistic approximation), and thus, the exact wave function is also an eigenfunction of $\hat{L}^2$ and $\hat{S}^2$ (as well as $\hat{L}_z$ and $\hat{S}_z$). To be more precise, the exact wave function can be chosen to be a simultaneous eigenfunction of all these commuting operators $\hat{H}, \hat{L}^2, \hat{S}^2, \hat{L}_z, \hat{S}_z$. And since it is really desirable, we require the same from out trial wave function: we would like it to be an eigenfunction of these angular momentum operators.

So we want to construct our trial wave function being an eigenfunction of angular momentum operators $\hat{L}^2, \hat{S}^2, \hat{L}_z, \hat{S}_z$, but a single Slater determinant would not necessarily do it, so we form what is called a spin-adapted configuration state function (CSF) - a linear combination of Slater determinants which is an eigenfunction of angular momentum operators.

Note, however, that this is not always possible.
For instance, for the case of unrestricted determinants using a linear combination would not help in making the trial wave function an eigenfunction of $\hat{S}^{2}$.


Can Ethanol be used to carry out SN1 reaction?

Be careful!  Ethanol is not a strong acid ($\mathrm{p}K_\mathrm{a} \approx 16$). So its conversion to ethoxide anion is not so easy. Rather, it would attack the $\ce{C}$ center by its lone pair and then it would release proton to form the product.

Actually here what I think is, the hetero atom $\ce{O}$ just near to the $\ce{C}$ undergoing substitution will assist by its lone pair to kick out chloride and then being planar ethanol can attack from either side. And finally, you get the product of $\mathrm{S_N1}$ type.

But don't mess up between ethanol and ethoxide ion. They are totally different in nucleophilicity. Treat them as two different nucleophiles.


How can the formula of WD-40 remain a trade secret?

"How can it remain secret?". Trivial answer is - it can't (it isn't).

WD40 is a Fast Moving Consumer Good (FMCG). It is also a reasonably cheap product - as an industrial chemist this suggests at least two things to me:

It doesn't contain anything rare or expensive
It isn't difficult to manufacture

As an ex-petroleum chemical analyst, with access to a reasonably complete lab, checks to do :

Get the Safety Data Sheet (below is composite information from a number of WD-40 Penetrant SDS)

cas        %
64742-47-8      50-70    isoparaffins petroleum hydrotreated HFP

64742-56-9      30-35    paraffinic distillate, light, solvent-dewaxed

61789-86-4      <5       calcium petroleum sulfonate

111-76-2        1-3      ethylene glycol monobutyl ether

Not Available    1       fragrance
Check the SDS info using physical/chemical testing
If you have the gear available :
Density
Distillation
FTIR
GC (GC-MS if you have it)
AAS/ICP - metals (ICP multielement would be best)
CHOSNP analysis for hetero atoms
Rebuild the product from data. Check to see if it performs the same.
Improve on the product (believe me - WD-40 is NOT the be all and end all of penetrating oils - technology has improved since its formulation).


Why are enthalpy of combustion and molar mass of organic compounds proportional?

The biggest contributor to the heat of combustion is the difference in covalent bond energy of reactants and products. The intermolecular forces that have to be overcome for combustion of liquids (with products all gases) are an order of magnitude lower (adapted from https://chemistry.stackexchange.com/a/125013):
$$
\begin{array}{lrr}
\hline
n\text{-Alkane} & T_\mathrm{b}/\pu{K} & \Delta_\mathrm{vap}H(T_\mathrm{b})/\pu{J mol-1}\\\hline
\text{Methane} & 112 & 8176\\
\text{Ethane} & 184 & 14640\\
\text{Propane} & 231 & 18832\\
n\text{-Butane} & 273 & 22390\\
n\text{-Pentane} & 309 & 26352\\
n\text{-Hexane} & 342 & 28850\\
n\text{-Heptane} & 372 & 31800\\
n\text{-Octane} & 399 & 33972\\
n\text{-Nonane} & 424 & 36910\\
n\text{-Decane} & 447 & 38750\\
\hline
\end{array}
$$

[OP] I know that the longer the chain is, the stronger the LDF get, so it is expected to see an increase in heat of combustion if molar mass increases.

Some of these alkanes are gases, so there are no dispersion forces to overcome. Also, you got the sign of the contribution wrong. If a reactant has higher LDF than the products (which have none), this would make the reaction less exothermic. So you would expect lower magnitudes for the larger alkanes, opposite of what is observed.
Instead, the heat of combustion scales with the number of carbons (see e.g. this table and graph) because the number of $\ce{C=O}$ and $\ce{O-H}$ bonds made (and the number of $\ce{C-C}$ and $\ce{C-H}$ bonds broken) also scales with the number of carbons.


Choosing hyperconjugation when resonance doesn't give a reason for stability

My teacher says that while comparing the stability of the species concerned, we should remember that stability provided by Resonance>hyper conjugation>inductive effect. Of course there are exceptions(like in the case of chloro benzene where according to resonance the ring should be activated but inductive effect dominates it and the ring gets deactivated) to these but I think we can follow it in majority of these cases and there is no reason to not be confident about it.So likewise in your case resonance is not enough so you look upon the next one i.e hyper conjugation


Short Answer(as it applies to humans):

Not in short exposures.

Longer Answer:

Polyvinyl acetate (PVA or PVAc) is the substance that makes up the active portion of Elmer's / school / wood glue -- the rest being water. Sodium borate (borax) can be used to cross-link the polymers present in the emulsion to form a slimy substance.

Some polymer cross-linking processes can be exothermic, but I'm going to set that aside for now because... extended exposure to borax alone can cause the symptoms described. Some of the articles relating to the incident linked to in the question state that the child had been making and handling this material for weeks.

There are a few studies in which volunteers were given lotion/shampoo containing >1% sodium borate. After about a week, some experienced the described symptoms.

DOI:10.3109/10915818309142004

  A cumulative irritancy test was used to study the effect of repeated exposures
  of a hair preparation containing 3.2% Sodium Borate and a cleansing cream containing
  1.7% Sodium Borate on the skin of 12 and 14 subjects, respectively (Table
  2). The test material was applied under an occlusive patch to the backs of subjects
  daily for 21 consecutive days. Sites were scored one hour after patch
  removal. Applications 4 to 21 of the hair preparation produced erythema and
  papules in most subjects; the total cumulative irritancy score was 571 (maximum
  score = 630). The cleansing cream caused slight erythema in two subjects only,
  resulting in a total irritancy score of 6.4. The investigators concluded that, under
  the conditions of the study, the hair preparation was a "mild to moderate"
  cumulative irritant, whereas the cleansing cream was practically non irritating.

  In a similar study, the contact-sensitizing potential of a cleansing cream containing
  1.7% Sodium Borate was tested in 22 subjects (Table 2). Preliminary irritancy
  testing revealed no irritation to a 48 h patch containing this product;
  therefore, skin sites were pretreated with 5% SLS for 24 h prior to application of
  the initial induction patch. The product was applied under occlusion for 48 h,
  every other day for 10 days (five applications). Following a 10- to 14-day rest,
  48-hour occlusive patches containing the test material were applied to fresh sites
  (with and without SLS pretreatment). SLS controls were also applied. Sites were
  scored at 48 and 72 h. More than half of the subjects reacted to SLS treatment
  (with or without application of test material). No significant irritation was observed
  at sites tested with the cleansing cream alone. The product containing 1.7%
  Sodium Borate was nonirritating and nonsensitizing.

In summary, this material should be pretty safe for short term handling with unbroken skin. Continuous use increases the risk of mishandling (broken/compromised skin contact) leading to boron absorption which may result in skin irritation (redness, bumps).

Note: Abstract does not mention the specific study referenced.


Which quantities change when choosing a different standard state?

Quantities that are dependent on the standard state

When calculating the equilibrium constant K or the reaction quotient Q from a set of observed concentrations, these concentrations are given relative to the standard concentration of the species. For example, in aqueous solutions the solvent water has a different standard concentration (that of pure water) than solute (roughly 1 M). Therefore, the value of K and Q depend on choice of standard state.

Whenever a thermodynamic quantity has a Plimsoll symbol appended, it refers to a system at standard state, e.g. $\Delta_r G^\circ, \Delta_r H^\circ, S_f^\circ, \Delta H_f^\circ$ or to the substance at standard state. The choice of standard state changes the value of these quantities. For the biochemical standard state, the quantity is appended with a prime, e.g.  $\Delta_r G^\circ {'}$ to indicate that difference.

Quantities that have special values at standard state

At standard state, activities are equal to one. Consequently, the reaction quotient Q is also equal to one.

$$Q = 1 \ \ \ \ \text{(at standard state)}$$Also, and this is a bit of a tautology but useful in deriving some relationships, when concentrations are at standard state (or more generally if $Q = 1$), quantities are equal to the standard quantities. Take for example the Gibbs energy of reaction:

$$\Delta_r G =  \Delta_r G^\circ + R T \ln Q$$

When concentrations are at standard state, this reduces to:

$$\Delta_r G =  \Delta_r G^\circ \ \ \ \ \text{(at standard state)}$$

Of course, that is the idea in calling $\Delta_r G^\circ$ the standard Gibbs energy of reaction.


What is the accuracy of medical syringes?

The requirements for ordinary sterile single-use syringes made of plastic materials are standardized in the international standard ISO 7886-1:1993 Sterile hypodermic syringes for single use – Part 1: Syringes for manual use.

According to this standard, the tolerance on the graduated capacity depends on the nominal capacity of the syringe $V$ and the expelled volume $V_\text{ex}$.

For syringes with a nominal capacity of $V\lt5\ \mathrm{ml}$, the tolerance amounts to

$\pm5\ \%\ \text{of}\ V_\text{ex}$ for volumes equal to or greater than half nominal capacity $V$
$\pm{\left(1.5\ \%\ \text{of}\ V +2\ \%\ \text{of}\ V_\text{ex}\right)}$ for volumes less than half nominal capacity $V$.

For syringes with a nominal capacity of $V\geqslant5\ \mathrm{ml}$, the tolerance amounts to

$\pm4\ \%\ \text{of}\ V_\text{ex}$ for volumes equal to or greater than half nominal capacity $V$
$\pm{\left(1.5\ \%\ \text{of}\ V +1\ \%\ \text{of}\ V_\text{ex}\right)}$ for volumes less than half nominal capacity $V$.

The values apply to the volume of water at a temperature of $T=\left(20\pm5\right)\ \mathrm{^\circ C}$ expelled from the syringe when the line circumscribing the end of the piston traverses a given scale interval.


Mercury assisted deprotection of dithiane

$\ce{Hg^2+}$ has a high affintity to thiols and other sulfur functional groups, and therefore $\ce{HgCl2}$ forms a Lewis acid-base adduct with the dithiane. The cyclic sulfonium cation undergoes ring opening, and after nucleophilic attack by water, a second Hg-S bond under ring closure is formed. Deprotonation finally yields the deprotected carbonyl compund and the cyclic mercury dithiolate.


Does 1,3-dichloroallene possess a C2 symmetry axis?

A $C_2$ axis is indeed here—although not the one that you are thinking of: your suggestion of a $C_4$ then $C_2$ rotation does not qualify as a $C_2$ symmetry operation. There is a true $C_2$ axis which bisects the allene and is tilted by $45^\circ$ from both the $\ce{Cl-C-Cl}$ and $\ce{H-C-H}$ planes.
One can visualize symmetry elements for small molecules using, for example, Jmol.
Launch the program GUI and do the following:

File → Get MOL
Type in an identifier, e.g. the name, 1,3-dichloropropa-1,2-diene, or CAS number, 83682-32-0.
Open console (File → Console) and enter show pointgroup to list symmetry elements and draw pointgroup to add them to the picture:


What are the priorities for substituent groups on a cyclohexane?

If the numerical sequence of locants is equivalent, the numbering should be assigned to correspond with alphabetical order. Hence, in your example, correct would be 1-ethyl-2-methylcyclohexane. Quoting the IUPAC Blue Book, Rule A-2.4:

  If two or more side chains are in equivalent positions, the one to be assigned the lower number is that cited first in the name.


In halogenation of ethene using bromine for example, why does the lone pair of bromine attack the C atom?

There are too many bond lines to explain the reaction; the question suggests that we have to explain the bond lines!

Let's start from scratch, with a different viewpoint. Ethylene isn't the attacker; it doesn't burst into flame in oxygen or do anything aggressive. It may be reactive, but in a "normal" range. Bromine, on the other hand, is a potent oxidizer and does go around interacting with relatively inert materials - it's a powerful bleach.

So, when bromine comes near the ethylene pi bond, it attacks the electron pair and grabs it, but the second bromine steals it away and departs as Br-. After all, bromine is more electronegative than carbon; the first bromine is just a link in the chain of events that allows the second bromine to escape with an electron. The first bromine winds up stuck to the C-C double bond; the two carbons and the bromine share the positive charge. The bromine certainly gets the lion's share of electron density, leaving the carbons with most of the positive charge. That's why another bromine can attack the carbons from the rear to give a dibromide. (If a Br- attacked the bromine attached to the double bond, it would just regenerate the reactants.) I bet that the second bromine to bond to the ethylene is not from the same bromine molecule as the first bromine. (But since all bromine molecules look the same, I can't prove it.) Doing the bromination in a solution with an overwhelming abundance of another nucleophile would probably give a mixed product.

A C-C double bond is about 1.34 Angstroms long; a bromine atom (in Br2) is about 2.29 Angstroms in diameter. Rather than envisioning the cation as a triangle of three grapefruit, I see it as a cantaloupe sitting on two large oranges. Just consider the empty p orbital of bromine and the pi lobe of the carbons; no need or benefit from making a three-membered ring.


Enantioselective organic reaction not working

The original "Rutgers' reference" deploys $2\,\mbox{mmol}$ aldehyde in $20\,\mbox{mL}$ solvent (i.e. starting at $0.1\,\mbox{mol/L}$, or $0.1\,\mbox{M}$) and recommends a stirring of $3\,\mbox{h}$.

Subsequently, you mention $1\,\mbox{mmol}$ per $20\,\mbox{mL}$ solvent (i.e. starting at $0.05\,\mbox{M}$), but without an indication of reaction time.

But if the starting concentration is lowered further ($0.01\,\mbox{mmol}$ in $5\,\mbox{mL}$ equals to $0.002\,\mbox{M}$ or $2\,\mbox{mM}$), not only the detection of the product becomes difficult. Did you take into consideration that by diluting the reaction you over-proportionally slowed down the advancement of the intended reaction?  Compared to the initially mentioned starting concentration, your reaction starts at a dilution by a factor of 50 (fifty).  This is more than a big factor for process involving a bimolecular reaction.

By the way, $0.01\,\mbox{mmol}$ of proline, molar weight of $115.13\,\mbox{g/mol}$ as stated here would correspond to little more than $1.1\,\mbox{mg}$.


Energy per atom in a parallel radioactivity equation

You may state that, in $60$ s, one A atom has liberated $20$ MeV by the first reaction. In the same time, three A atoms have liberated $3·40$ MeV by the $2$nd reaction. The total energy liberated in $60$ seconds, is $120 + 20 = 140 $ MeV.- As a consequence, the average energy liberated by one A atom is $140$ Mev$  /4 = 35$ MeV


Why is the inside of a cell both more acidic and more negative than its environment?

The key is that $\ce{H+}$ and $\ce{OH−}$ are not the only ions at play here.

The selective permeability of different cellular membranes for different anions (i.e. $\ce{Cl−}$) and cations (i.e. $\ce{K+, Na+, Ca^{2+}, Mg^{2+},}$ etc.) results in charge separation. Cellular membrane potentials are thereby created, with or without an intercellular / intracellular pH gradient.


food chemistry maltodextrin vs organic maltodextrin

"Food additive 1405" refers to the process for the modification of starch; in this case the process uses enzymes to break up the starch molecules into oligosaccharides between 3 and 20 saccharides.

The difference between regular maltodextrin and 'organic maltodextrin' would be that the source was grown without synthetic pesticides and synthetic fertilizers. 'Organic maltodextrin' may also need to use natural enzymes.


Term for compound of carbon, hydrogen, and oxygen

There is no term for this group as it is too broad and diverse.  Even the narrower term hydrocarbon is usually qualified e.g. aliphatic, olefinic, aromatic.

With thanks to Karl.


Why do all medicines require storage temperatures below 25 °C?

People do not realize that the quality requirements for a medicine is rather high. We are talking about a complex mixture of more or less reactive organic compounds that are potentially sensitive to air, moister and light. Any reaction of the original ingredient is potentially biological active and harmful, and even simple things like recrystallization or coagulation can destroy the effect of the original mixture.

The question "will they really spoil" makes no sense in the given context: if 99.999% of the cases not spoil, but there is a handful cases when things goes bad then the public goes banana. It took only one (purposefully false) paper to convince tens of thousands of people to go anti-vaccine, and decades till it dies out.

If it happens with a medicine, it potentially kills the product, so easier to err (by huge margin) on the safe side.


The symbols, which are standardized in ISO/IEC 80000 Quantities and units (14 parts) and recommended by IUPAC in Quantities, Units and Symbols in Physical Chemistry (Green Book), distinguish the various related quantities:

$$\small
\begin{array}{llll}
\hline
\text{Quantity} & \text{Quantity} & \text{Unit} & \text{Unit} \\
\text{name} & \text{symbol} & \text{name} & \text{symbol} \\
\hline
\text{heat capacity} & C & \text{joule per kelvin} & \mathrm{J/K} \\
\text{heat capacity at constant pressure} & C_p & \text{joule per kelvin} & \mathrm{J/K} \\
\text{heat capacity at constant volume} & C_V & \text{joule per kelvin} & \mathrm{J/K} \\
\text{specific heat capacity} & c & \text{joule per kilogram kelvin} & \mathrm{J/( kg \cdot K )} \\
\text{specific heat capacity at constant pressure} & c_p & \text{joule per kilogram kelvin} & \mathrm{J/( kg \cdot K )} \\
\text{specific heat capacity at constant volume} & c_V & \text{joule per kilogram kelvin} & \mathrm{J/( kg \cdot K )} \\
\text{molar heat capacity} & C_\mathrm m & \text{joule per mole kelvin} & \mathrm{J/( mol \cdot K )} \\
\text{molar heat capacity at constant pressure} & C_{\mathrm m, p} & \text{joule per mole kelvin} & \mathrm{J/( mol \cdot K )} \\
\text{molar heat capacity at constant volume} & C_{\mathrm m, V} & \text{joule per mole kelvin} & \mathrm{J/( mol \cdot K )} \\
\hline
\end{array}$$

However, in contrast to symbols for units, symbols for quantities are only recommendations. Unfortunately, quantity symbols are used inconsistently in the literature, sometimes in a sloppy way. In principle, a given symbol can indicate different quantities.


How do I analyze 17 different gases from a sample?

You could use a permanent gas analyzer that has an FID channel for hydrocarbons and CO2, with a TCD detector for H2 and HF and a third MS channel if needed. Basically, this would mean you split your sample into three different columns in the oven to analyze them for different things. I agree with Karl and don't reinvent the wheel. I've done the hydrocarbons (C1-12) and CO2 and permanent gases using a Thermo Trace 1310 with an FID and TCD detector and would recommend that for ease of use. Then I would probably add an MS detector to it in case you need it.
How are you hoping to get the battery gases into the GC? I think you could fit a battery into a headspace vial and heat it up to release the gases then directly inject the headspace into the GC.


What is the hybridization of the nitrogen in the azide ion?

Here are the resonance structures we can draw for the azide anion.

Resonance structure A has an octet on each nitrogen and just one unit of formal
charge on each nitrogen
Resonance structures B and C also have an octet around each nitrogen, but there is a rather large formal charge of -2 on one of the nitrogens.  Such a large charge on a small atom like nitrogen makes structures B and C less favorable and therefore we would expect them to count less to the "true" representation of the azide anion, than resonance structure A
Resonance structures D and E do not have an octet around the singly-bound, terminal nitrogen, these resonance structures will count even less.  In fact, since the azide ion is linear and these two resonance structures are bent, we can say that they don't really contribute to the true picture of the azide ion.

So if we consider only resonance structures A and B (C is equivalent to B) we can see that the central nitrogen is $\ce{sp}$ hybridized in both cases (like in allene for A and like in acetylene for B).  The $\ce{sp}$ hybridization of the central nitrogen is consistent with the linear structure of the azide ion.

In A, the terminal nitrogens may both be $\ce{sp^2}$ hybridized; after all the drawing looks similar to the double bond in ethylene. However, there are other options.  For example, both nitrogens could be $\ce{sp}$ hybridized, with an $\ce{sp}$ single bond, one lone pair in a $\ce{sp}$ orbital and the other lone pair in a $\ce{p}$ orbital (just like the 2 lone pairs on oxygen in water). Without more data there is no way to tell for sure, but in an introductory class $\ce{sp^2}$ hybridization would be a reasonable answer.

In B, the terminal nitrogen with the triple bond appears to be $\ce{sp}$ hybridized just like in acetylene.  The hybridization of the other terminal nitrogen in resonance structure B is anybody's guess; there are many possibilities and since there is only one ligand attached (the central nitrogen), we don't know what direction the 2 electron lone pairs are pointing in.  This nitrogen might be unhybridized, $\ce{sp}$ hybridized, $\ce{sp^2}$ hybridized - again, we just don't know.  Since the true hybridization of the terminal nitrogen is an equal mixture of these two hybridizations (e.g. a blend of B and C), we really don't know how the terminal nitrogens in B and C are hybridized.

In an introductory class, the answer would be that the central nitrogen is $\ce{sp}$ hybridized, and, since resonance structure A is the major resonance contributor, we might say that the terminal nitrogen is $\ce{sp^2}$ hybridized (but, from the above discussion, we know that is an oversimplification).


Why does H+ attack alkene first in this reaction?

It's better to check the resonance hybrid whenever we have a confusion, so, in this case resonance takes place between lone pair of oxygen atom and the double bond and the electrons shift from oxygen towards the alkene. This creates a partial positive character on the oxygen atom and partial negative character on the alkene carbon which is farthest from the oxygen atom. Now we can easily decide that in between ether oxygen atom and alkene proton will tend to attack alkene carbon and that to on the carbon, which is farthest from the oxygen which results in the positive charge on the carbon next to ether oxygen (this cation is also stable because of octet completion due to resonance of oxygen (may also be called as backbonding)). Now deciding between hydroxy oxygen and alkene is a bit tougher task. Let us assume that proton attacked hydroxy oxygen and it formed oxonium ion, now it can be removed as water and a cation can be formed there. After formation of cation there is nothing to do and it will be waiting for the hydroxy component of water to get attacked by. But if the cation is formed at the alkene carbon it can go for rearrangement which increases the size of 4-membered ring to 5, which increases entropy of the system and also it will lead to formation of $\ce{C=O}$. That will tend the reaction to move forward (like pinacole pinacolone rearrangement). So,it's better to form cation at alkene site rather than forming oxonium ions.

Following is an arrow-pushing diagram of the above description:


Best Cheminformatics API for drawing SMARTS reactions?

OpenEye is probably the most powerful and fast depiction SDK, but it costs money. Also OpenEye is a low-level, cumbersome, complicated library - so I'd consider it only if you have very sophisticated needs. It's written in C++, but it has Python/Java wrappers.
CDK (Java) has an OS-agnostic depiction API, works great for simple needs, haven't tried anything super complicated.
I tried Indigo several years ago and it worked fine, maybe you misconfigured it? Though it did require an additional OS-level dependency for fonts (Cargo?) on Linux. Don't remember if there were problems with other OS's.


Explain the structure of HNC (Hydrogen Isocyanide)

In such type of questions, we can't directly determine Oxidation state by the normal algebra method.

First you have to draw the structure.

1.In a covalent bond, the more electronegative element gets a contribution of -1and the other element gets a contribution of +1 charge(valid for both sigma and pi bonds. Treats double double bond as two single bonds and triple bonds as three singles bonds).

Bonds between same element have contribution 0 to each element.
In a coordinate bond, if the bond is directed from less electronegative element to more electrononegative element, the more elecronegative element gets a contribution of -2 and the less electrononegative element gets a contribution of +2.

4.in case the coordinate bond is directed from more electrononegative to less elecronegative elements, the contribution is 0 to each element.

5.To these 4 steps for each and every bond. The value of Oxidation state of an element is sum of all contributions due to all other elements/bonds.

In your question, due to C-H bond, carbon gets -1 and hydrogen +1. Due to the triple bond between C and N, N gets -3 and C gets +3.

Summing all,

Oxidation state of hydrogen is +1

Oxidation state of carbon is -1+3=+2

And oxidation state of nitrogen is -3


What organizational methods pre-date the Periodic Table?

One attempt to order chemical elements was Döbereiner's system of triades, published in Annalen der Physik und Chemie, back in 1829 (doi 10.1002/andp.18290910217 with Wiley); or (open access with Gallica).

Although an actual view into his paper permits the speculation he was not using the atom masses we know today; rather than using specific weight and molecular weight as known / defined at his time of oxides of Ca, Ba, Sr, for example:

(The dots on top of Ca, Ba, Sr seem to adhere a nomenclature, as on top of S, Se, Te three of them; Cl, Br, I are topped by five.)


Which is more basic: 1,2‐dihydropyridine or 1,4‐dihydropyridine?

In compound $\ce{A}$ , lone pair on $\ce{N}$ is in conjugation with two $\ce pi$ bonds (as shown below). Therefore this lone pair is delocalized (resonance structures in the figure below) to a greater extent.

In  compound $\ce{B}$ , lone pair on $\ce{N}$ is in conjugation with a single $\ce pi$ bond , while the other pi bond is not in conjugation or is cross conjugated.

Since lone pair on $\ce{N}$ in compound $\ce{B}$ is less delocalized then in compound $\ce{A}$ , compound $\ce{B}$ is more basic then compound $\ce{A}$.


Using results of current experiments in silanization, how to achieve a higher contact angle?

It's always good to clean glass surfaces with oxygen plasma or UV ozone treatment before silianizing, but 108° is pretty good with an alkylsilane.
The best you will be able to do with this type of surface treatment is using fluorosilanes.
Silanization agents are basically made of three parts, a functional group, a linker and a coupling group:

The coupling group is what allows the silane to attach to a surface and is found bound to the silicon atom.
Typical couplers are hydrolyzable groups like alkoxy, acyloxy, or halogen.
These don't generally influence the hydrophobicity of the resulting surface directly, though they will affect how well the silane adheres to a given surface, the manner and speed with which it can be applied, and the deposition density.
Alkoxy and acyloxy groups can be applied from aqueous solutions and are generally less reactive, but halogen groups can be deposited from anhydrous solvents or the vapour face quite quickly.
The linker is typically just a short alkyl chain between the silicon atom and the functional group to reduce steric interactions between adjacent silanes.
The functional group is what has the greatest influence on hydrophobicity.
Many different functional groups can be added, some even conferring hydrophilicity instead, if desired.
In general, longer alkyl chains are more hydrophobic, but long fluoroalkane chains are the most hydrophobic and will also not interact easily with organic solvents either so fluorosilane coated surfaces are often known as omniphobic by virtue of being both hydrophobic and oleophobic.

So if you want the maximum contact angle possible from a simple surface treatment, you're looking for something like 1H,1H,2H,2H-Perfluorododecyltrichlorosilane.
It may also be possible to simply spin or spray coat a fluoropolymer like Teflon AF or Fluoropel, depending on your application and surface. All these types of coatings will give contact angles up to about 120° in the best case and to get any higher will require surface patterning to achieve superhydrophobicity.


What is the best way to detect the presence of tear gas?

Owlstone make an electronic multi-agent detector - I don't know how much it costs but I would guess somewhere in the range \$10K - \$100K. There might be cheaper single-agent electronic sensors or chemical tests.


Would decomposing an object into molecules be an environmental problem?

Decomposing something into 'molecules' is immediately indicative of a non-chemistry background! To try to address your question a bit more directly, I can see a couple of ways of looking at the thought experiment:

1) It decomposes everything to simplest possible component stable molecular structures.

In the case of organic material, plastics etc, you will probably end up with a preponderance of various gases and water, as largely this disassembly is equivalent to complete combustion (CO₂, H₂O, NH₃) likely being the main resulting candidates. These would float out into the atmosphere unless captured.

In the case of inorganic material, e.g. a computer motherboard you will end up with many molecules of various metals, most of which would rapidly oxidise due to their very high surface area (being unimolecular), and various silicates and ceramic unit cells. All of these would be incredibly fine aerosols or dissolved in atmospheric water, similarly rapidly dispersing unless contained.

Results would likely include enhancing the greenhouse effect (on balance as some compounds will diminish it), inducing acid rain (sulfates, nitrates), causing algal blooms/biome damage.

If you had storage for different outputs then you would be producing several industrially useful compounds and if you partnered it with metal reclamation facilities to deal with all the metal oxides it might be a useful way of recycling, energy no object.

2) It decomposes and sorts everything by chemical element.

Assuming you have a nice little storage container for each element (reasonable given you are sorting into the simplest molecule of a pure element), you would end up with various cylinders of oxygen, hydrogen, nitrogen, lumps of metal, lumps/powders of inorganic materials like phosphorus, sulfur, silicon. This would be potentially useful, but incredibly energetically expensive even beyond the separation system outlined by Uncle Al, so would be considerable only in the case of having essentially limitless free energy (many nuclear fusion plants/orbital solar collectors/dyson sphere type energy availability I suspect).

Issues here would relate to various elements being highly reactive on contact with air (or rather the oxygen/water in it). Obvious examples include allotropes of phosphorus and the rare earth metals.

Other issues unaddressed include fractional distillation and cracking of small molecule gases into smallest molecule gases, how to prevent reaction inside the vessel i.e. actually separate things rather than just provide them with huge amounts of energy to react with each other etc


The conductivity of molten lithium carbonate has been determined for vast range of temperatures (Ref.1). The abstract of which states that:

  The present communication reports the results of an investigation of the properties of surface tension, density, and electrical conductance for molten $\ce{Li2CO3}$, $\ce{Na2CO3}$, and $\ce{K2CO3}$ and some mixtures in the temperature range of $750^\circ –\pu{1000 ^\circ C}$. The surface tensions are approximately twice the values for the corresponding chlorides; the densities and electrical conductance are quite comparable to those of the chlorides. The ionic nature of the molten carbonates is examined from the physicochemical criteria based on these properties; and the mechanism of electrical transport is considered in the light of current theoretical concepts. Relative to $\ce{Na2CO3/K2CO3}$ mixtures, the surface tensions and partial molal volumes indicate but little deviation from the predictions for thermodynamically ideal mixtures.

The conductivity of molten lithium carbonate/metal carbonate mixture has also been determined for range of temperatures (Ref.2), the abstract of which states that:

  The electrical conductivities of the molten binary salt mixtures $\ce{Li2CO3–X2CO3}$ ($\ce{X: Na, K, Rb,}$ and $\ce{Cs}$) and $\ce{Na2CO3–Z2CO3}$ ($\ce{Z: K, Rb,}$ and $\ce{Cs}$) were measured using an ac two-probe technique and electrochemical impedance spectroscopy. The electrical conductivities of the eutectic compositions of $\ce{(Li_{0.52}Na_{0.48})2CO3}$ and $\ce{(Li_{0.62}K_{0.38})2CO3}$ at $\pu{923K}$ were $2.06$ and $\pu{1.31Scm−1}$, respectively. The conductivity of molten $\ce{Li2CO3–X2CO3}$ ($\ce{X: Na, K, Rb,}$ and $\ce{Cs}$) and $\ce{Na2CO3–Z2CO3}$ ($\ce{Z: K, Rb,}$ and $\ce{Cs}$) decreases with an increase in the ionic radius of $\ce{X}$ or $\ce{Z}$. We have succeeded in correlating the equivalent conductivity of a binary carbonate mixture $\Lambda$, with the composition ($x_\mathrm{A}$ and $x_\mathrm{B}$), the equivalent conductivities of the pure components $\ce{A}$ and $\ce{B}$ ($\Lambda_\mathrm{A}$ and $\Lambda_\mathrm{B}$), and two parameters ($\lambda_\mathrm{A}$ and $\lambda_\mathrm{B}$) as $\Lambda = x^2_\mathrm{A}\Lambda_\mathrm{A} + x^2_\mathrm{B}\Lambda_\mathrm{B} + x_\mathrm{A}x_\mathrm{B}(\lambda_\mathrm{A}x_\mathrm{A} + \lambda_\mathrm{B}x_\mathrm{B})$. The pair of interaction parameters ($\lambda_\mathrm{A}$ and $\lambda_\mathrm{B}$) can be estimated for a binary system of molten carbonates using the relationship between the cationic radii and the equivalent conductivities of the pure carbonates.

The table 107 in page 85 of government technical report (Ref.3) gives some insight to the conductivity (equivalent conductance ($\Lambda$)  and specific conductance ($\kappa$)) of molten lithium carbonate (m.p.: $\pu{618 ^\circ C}$ or $\pu{891 K}$; eq. wt.: $\pu{36.95 g/eq.}$) in different temperatures together with density ($\rho$) and viscosity ($\eta$) at the same temperature:

$$
\begin{array}{c|ccc}
\text{Temperature in }\pu{ K}  & \Lambda & \kappa & \rho & \eta \\
\hline
1010 & 82.88 & 4.097 & 1.8260 & - \\
1020 & 84.58 & 4.172 & 1.8222 & - \\
1030 & 86.31 & 4.249 & 1.8185 & - \\
1040 & 88.06 & 4.326 & 1.8148 & - \\
1050 & 89.83 & 4.404 & 1.8111 & 4.64 \\
1060 & 91.63 & 4.483 &  1.8073 & 4.34 \\
1070 & 93.46 & 4.563 &  1.8036 & 4.01 \\
1080 & 95.31 & 4.644 & 1.7999 & 3.67 \\
1090 & 97.19 & 4.726 &  1.7961 & 3.36 \\
1100 & 99.09 & 4.808 & 1.7924 & 3.10 \\
1110 & 101.02 & 4.892 & 1.7887 & 2.91 \\
1120 & 102.98 & 4.976 & 1.7850 & 2.83 \\
\hline
\end{array}
$$

The tabulated values of specific conductance ($\kappa$) for molten $\ce{Li2CO3}$ in this table were calculated from the data of Ref.1, Using the following quadractic equation for specific conductance (the precision is $3.00 \times 10^{-4}\; (0.0066\%))$:

$$\kappa = 0.9877 - 1.3529 \times 10^{-3} T + 4.3873 \times 10^{-6} T^2 $$

The uncertainty of the specific conductance values is estimated to be about $1.5\%$. The ref.3 listed this as the best equation for temperature range $\pu{1018-1118 K}$. However, the ref.3 did not provide the relevant units for these calculations and, unfortunately, I do not have access to the Ref.1.

The Ref.3 also provided three other equations for their calculations (I mention this only for interesting viewers):

$$\rho = 2.2026 - 0.3729 \times 10^{-3} T$$
The ref.3 listed this as the best equation for temperature range $\pu{1011-1115 K}$ (The uncertainty of the values is estimated to be about $0.2\%$).

$$\eta = - 5259.12 + 14.8091 T - 1.38581 \times  10^{-2 }T^2 + 4.31294 \times  10^{-6} T^3$$
The ref.3 also listed this as the best equation for the temperature range $\pu{1046-1122 K}$ exracted from Ref.4 (The uncertainty of the values is estimated to be about $3.0\%$).

$$\Lambda = 754.5 e^{-\frac{4438}{RT}}$$
The ref.3 listed this as the best equation for equivalent conductance (The uncertainty of the values is estimated to be about $0.15\%$). Here, the units are given as $\pu{ohm-1 cm2 equiv-1}$. The equation is derived from Arrhenius type equation, $\Lambda = A_\Lambda e^{-\frac{-E_\Lambda}{RT}}$. For $\ce{Li2CO3}$, $A_\Lambda = \pu{754.5 ohm-1 cm2 equiv-1}$ and $E_\Lambda = \pu{4438 cal mol-1}$.

References:

George J. Janz, Max R. Lorenz, “Molten Carbonate Electrolytes: Physical Properties, Structure, and Mechanism of Electrical Conductance,” J. Electrochem. Soc. 1961, 108(11), 1052-1058 (doi: 10.1149/1.2427946).
Toshikatsu Kojimaa,b,z, Yoshinori Miyazakia, Katsuhiro Nomuraa,* and Kazumi Tanimoto, “Electrical Conductivity of Molten $\ce{Li2CO3–X2CO3}$ ($\ce{X: Na, K, Rb,}$ and $\ce{Cs}$) and $\ce{Na2CO3–Z2CO3}$ ($\ce{Z: K, Rb,}$ and $\ce{Cs}$),” J. Electrochem. Soc. 2007, 154(12), F222-F230 (doi: 10.1149/1.2789389).
G. J. Janz, F. W. Dampier, G. R. Lakshminarayanan, P. K. Lorenz, R. P. T. Tomkins, In Molten Salts: Volume 1, Electical Conductance, Density, and Viscosity Data; National Standard Reference Data Series: National Bureau of Standards 15, US Department of Commerce, National Bureau of Standards, US Government Printing Office: Washington DC, 1968.
George J. Janz, Fumihiko Saegusa, “Molten Carbonates as Electrolytes: Viscosity and Transport Properties,” J. Electrochem. Soc. 1963, 110(2), 452-456 (doi: 10.1149/1.2425785).


Unlike the other factors affecting chemical equilibrium, (namely concentration, pressure), temperature follows the Le Chatelier principle via a different mechanism.

Temperature alters the equilibrium constant $K$, and this change in $K$ also changes the relative concentrations of the reactants and the products. The way in which the equilibrium constant depends on temperature is given by Van't Hoff equation:
$$\frac{d \ln K}{d T}=\frac{\Delta H}{RT^2}$$
Therefore, for exothermic reactions, the change in $K$ acts as if heat is one of the products, since increasing the temperature causes the $K$ to decrease, making the equilibrium at lower temperatures more favorable for Haber's process for making ammonia. (While this only talks of the equilibrium concentrations, lower temperatures in general result in slower rates, hence a balance is always sought)

While for your other question, as the temperature increases, the equilibrium is reached faster by both increasing ammonia (the cause of the increasing temperatures) and the decreasing $K$, until at a certain temperature, the reaction reaches equilibrium. I am not sure of the practical constraints employed in temperature regulation in the industry.


Your textbook contains an error. Either the formula is written incorrectly or the name is. Since according to comments the name is correct, it must be an incorrect formula. Let’s break it down:

If the formula is correct

We have two complex ions that together create a doubly-complex salt. According to the rules of (pseudo-) binary ionic compounds, one ion must be positively charged and the other negatively. By convention, the cation is written first so it is safe to assume that $\ce{[Pt(NH3)4]^{?}}$ be the cation — which fits in well with chemical intuition because $\ce{NH3}$ is a neutral ligand and platinum is a metal.

Therefore, $\ce{[Pt(NH3)Cl3]^{?}}$ must be the anion — very nice, because whe have chlorido ligands that are commonly negative. Since every chlorido ligand adds $-1$ charge to the complex, the anion can be anything from $-1$ to $-3$ with decreasing probability.

The complex needs to be a $1\,:\,1$ binary complex, so the negative charge needs to be equivalent to the positive charge. From here onwards it gets difficult. We have the following possibilities:

$\ce{[Pt^{I}(NH3)4]^{+} [Pt^{II}(NH3)Cl3]^-}$
$\ce{[Pt^{II}(NH3)4]^2+ [Pt^{I}(NH3)Cl3]^2-}$
$\ce{[Pt^{III}(NH3)4]^3+ [Pt^{0}(NH3)Cl3]^3-}$

They all differ in the assumed oxidative states of platinum. We cannot decide a priori which one of these three is correct, although the last one is by far less probable. Technically, we can even assume further diverging oxidation states but they get proportionally less probable. Only if the charge were clearly stated could we distinguish between the cases. For the sake of consistency, here are the names:

tetraammineplatinum(I) amminetrichloridoplatinate(II)
tetraammineplatinum(II) amminetrichloridoplatinate(I)
tetraammineplatinum(III) amminetrichloridoplatinate(0)

If the name is correct

Then we can logically build up the resulting salt from the name. Let’s start with the cation:

It is platinum(II), so we get $\ce{Pt^2+}$
adding four ammine ligands gives us $\ce{[Pt(NH3)4]^2+}$

For the anion:

It is still platinum(II) (although rendered as platinate, so we need to end up with an anion): $\ce{Pt^2+}$
We have one ammine ligand: $\ce{[Pt(NH3)]^2+}$
and three chlorido ($\ce{Cl-}$) ligands: $\ce{[Pt(NH3)Cl3]-}$

Since we have a cation with $+2$ charge and an anion with $-1$ charge, we need two anions per cation, giving us the final formula:

$$\ce{[Pt(NH3)4][Pt(NH3)Cl3]2}$$

So what is correct?

Going by chemical intuition (platinum(II) being a well-known $\mathrm{d}^8$ system that forms square-planar complexes), I would say that the name is correct and the formula isn’t. This also has the implication that they messed up the question by missing a subscripted 2 (rather likely) rather than messing up the answer by getting a Roman numeral wrong (not quite as likely).

Hence there is nothing miraculous or confusing about assigning oxidation states to metals in complexes — and the IUPAC even formally defines oxidation states for promiscuous ligands such as nitrosyl (which is always considered to be a radical neutral ligand, never $\ce{NO^+}$ or $\ce{NO-}$) allowing for a clean-cut naming of complex compounds without knowing the chemistry involved.

The charge does not have to be mentioned in doubly-complex salts for any component. However, it always helps students to give them and in non-trivial cases (like the assumed correct formula) it might be vital to know the overall charges. If there are no charges written, though, the final product has to be assumed neutral.


Can I add the molarities of multiple solutes together?

It doesn't make sense to add the total number of moles then divide by volume to find molarity. By doing that you are assuming that all the solutes have the same concentration which is not true(you are finding an average concentration of the solution but it's not necessarily correct). The solution contains different solutes at different concentrations. I think what you're doing only works if you know that the concentrations of solutes in the solution are the same.


The phosphate ester groups which connect the nucleotides contain one acidic proton at their OH group, and two at the end of each strand. As you can see here, here and in the picture below (with the groups marked turquoise), most of these groups are deprotonated depending on pH. They act as Brønsted-Lowry acids.


First, in order to compare mp's you need to know the mp's. You have two choices: 1. run melting point measurements or 2. find literature written by someone who has run the experiments. Perhaps this is a translation problem from your native language? Perhaps you meant predict the ranking of mp's for those 4 compounds.

You state that CaCO3 melts at 811 °C. As far as I know, that is incorrect. First, the thermodynamically stable polymorph is calcite, which according to wikipedia has a mp of 1339 °C but which decomposes at temperatures above ~840 °C (also according to Wikipedia!). I hope you see this is clearly contradictory. My first hand experience is that it does indeed decompose at around 820–840 °C, so I am unable to explain how they were able to measure its melting point 500 °C above that. 500 °C is an enormous gap, but perhaps they measured it under some enormous pressure (far from STP) - I really don't know.

The other major problem I'm having is understanding what you (or your teacher) means by "counterbalance the effect of electropositive field of cations ... and electronegative field of an anion." It seems to me to be meaningless, at least I don't recognize this terminology in reference to the electrostatic forces usually invoked to account for ionic solids. BeSO4 is an obvious problem since Be is known to for its tendency to form covalent compounds. Indeed, according to Wikipedia, anhydrous Be sulfate has Be-S (covalent) bonding. While you could predict that the covalent bonding would reduce the mp, what you'd do with the polymeric nature of the solid is beyond me - I've no idea how you or your teacher can explain that, except by hand waving.

This leaves the Ba and Sr salts. Simple logic of ion size would predict the order Ba < Sr and ignoring the ridiculous assertion on Wikipedia that the Ba salt's melting point is 811 °C while at the same time (you did READ the article, right?) stating that that is the temperature at which it a "polymorphic transformation" occurs. A little further digging provides a safety data sheet which claims a mp of 1400 °C "close to its decomposition temperature", but such data should be suspect.

Your teacher's choice of picking carbonate as the anion was a poor one, in my honest opinion. Because of the tendency of these salts to decompose, rather than melt, you need to be very careful where you obtain your data from. But perhaps that is the lesson your teacher hopes some of you will learn - to not just accept a number because it is readily available (without subjecting it to critical analysis (who, what, where, why, when, how - which in this case means how reliable is the reference? does it reference a primary (academic) source? if not, an authoritative secondary source? if not, then you should seek independent confirmation. Such confirmation is hard to find on the internet, since one lazy site steals from another until the noise drowns out the signal.

I really don't know if density functional theory (DFT) quantum mechanical modeling by supercomputers is able to accurately predict the mp's of these material. My guess is that because of the change in crystal structure(s) that it remains beyond our current capabilities (for some things, calculating the value has become reasonably accurate – as far as I know (and I'm no expert) mp isn't one of the successes. If it was accurate it would be a third way to "compare" the mp's.


Determining Molecule Based on Proton NMR (HNMR)

Your assignment seems reasonable. One thing to keep in mind is that even though  phenyl protons are in different chemical environments, they may be too close to resolve and with a low-field instrument, some may appear as a "blob" of superimposed signals. In this case, the peak from (B) is simply unresolved from that of the (C) protons, giving a single signal that integrates to 3 protons. See here with a 90 MHz instrument. That spectrum is a bit different than the one you have, but the one you've been given may be a simulated spectrum.


The answer appears to be forthcoming in the wikipedia article on silicon monoxide and references therein — to quote:

  $\ce{(SiO)_{n}}$ irreversibly disproportionates into $\ce{SiO2}$ and $\ce{Si}$ in a few hours between 400 and 800 °C and very rapidly between 1000 and 1440 °C, although the reaction does not go to completion.

The article implies that $\ce{(SiO)_{n}}$ generates a passivating layer of $\ce{SiO2}$ in contact with dioxygen, so presumably the slag would be encapsulated in a thin layer of that.

Additionally, silicon is not generally considered to be a metal, but rather a semimetal or metalloid.


How do I prepare citric acid solution (from anhydrous powder) of accurate concentration, given that the solution volume expands when water is added?

The "expanding solution" problem is irrelevant if you use standard techniques
First, you need to realise that a standard solution is a known amount of solid in a known volume of solution not a known amount of solid and a known amount of the solvent. You don't need to know how much solvent you use. So you need to target getting a result that occupies, say, 100mL not a solution made from 100mL of water.
There is an easy way to achieve this. Use a volumetric flask with an accurate volume marking. Add the solid to the flask. Fill the flask slowly (while shaking enough to ensure the solid dissolves before the flask is full). Keep adding small increments of the solvent until the total volume reaches the graduation. You now have an exact volume of a solution containing an exact amount of the solid. At no point do you need to worry about any volume expansion or contraction.


Reason for the greater energy of the eclipsed conformer of alkanes?

Just take two electrons. If you want to push them closer together you need to input some energy and the system overall will have higher energy. There is a preference for the two electrons to be farther apart.

This situation is not any different if we look at pairs of electrons in bonds instead of individual electrons.


Both citrulline and arginine are nitrogen-rich, basic amino acids. A basic acid sounds confusing, but that's the way it is.

The colour change indicates a change in the pH of the solution (the tea mix).

The pH indicator that changes the colour is in the hibiscus flower and belongs to the class of the anthocyanes.

Under acidic conditions, the dye exists as a deeply-red flavylium cation. At a slightly higher pH, the dye is reversibly converted to a colourless form. Here, the colour of the mixture is determined by the other ingredients. When more base (read: more amino acid) comes to action, further colour changes are conceivable and the double deprotonated form of some anthocyanes are known to be bluish-green.


The primary reference for this type of data is the article by Li [1]. Snyder cites this paper as the primary source for UV cutoff values ("Ref 7" on the screenshot below) [2, p. 299]:

References

Li, J. B. Signal-to-Noise Optimization in HPLC UV Detection. LC GC 1992, 10 (11), 856–864.
Snyder, L. R.; Kirkland, J. J.; Glajch, J. L. Practical HPLC Method Development, 2nd ed.; Wiley: New York, 1997. ISBN 978-0-471-00703-6.


There are two parts to the answer to your question. (a)  When solving a differential equation the boundary conditions must be specified (this is not always emphasised enough in textbooks), and (b) these conditions are determined by the physics of the problem via the postulates of quantum mechanics, including the nature of the wavefunction. Quantum mechanics (QM) has no derivation; we accept its postulates and QM survives only because experiment has so far always confirmed it.
As an example of quantisation consider the particle in a box. Here there are all sorts of solutions to the differential equation, i.e Schroedinger's equation. However, as the walls are infinitely high the wavefunction must have zero amplitude here and this means that the only solutions that remain and are physically realistic are when quantum number $n$ is quantised. The fact that the wavefunction must be zero at each wall are the boundary conditions.
A similar situation occurs on a particle on a ring where the angular momentum quantum number $l$ is quantised only because we insist, by applying the postulates of QM that the wavefunction repeats itself exactly each $2\pi$ round the circle. This is in effect the boundary condition.
In the H atom the equation is far more complicated but the general conditions imposed on the wavefunction remains the same; the imposition of boundary conditions, as determined by QM, results in the generation of integer quantum numbers so that theory matches experimental data.


Confusion about boiling point and vapor pressure

Shouldn’t we say that the liquid boils when its vapor pressure is equal to the greatest fluid pressure in the liquid?

No.  If you heat up water at the ocean's surface to 100 °C, it will boil, even though the pressure at the bottom of the ocean is far higher.

As you note, pressure is not guaranteed to be homogeneous in space (i.e. the same everywhere), especially for deep bodies of water.  But neither is nearly any other fluid property.  Temperature can vary with position in poorly-mixed waters.  So can whether the water is boiling or not.

A given small-ish region of water with nearly constant temperature and pressure will boil if and only if the (P, T) condition of the water is at or above the water's vapor pressure at the same temperature.


Does mercury fulminate really look like crystal methamphetamine?

Mercury fulminate $\ce{[Hg(CNO)2]}$ can have a white, gray or greenish-brown color depending upon purity and dryness (it is often kept moistened to prevent accidental decomposition).

  Does fulminated mercury really have the translucent white/blue crystal
  appearance as was shown in this scene?

I've never seen a picture where it appears translucent or where it has a bluish color.  The pictures I've seen always show it as an opaque powder.  I suspect that it could be crystallized, but that's probably never done as it would further increase its sensitivity to detonation (e.g. if a crystal cracked, or rubbed against another crystal, the friction, triboelectrification, etc. could cause the material to explode).  Here are some pictures.

  Is the compound that sensitive that throwing it against the wall would
  cause detonation?

Here is a link to an examination of the Breaking Bad "detonation by throwing" experiment (with video).  The conclusion is that it can detonate when thrown, but 1) a larger amount of material than what Walt had, and 2) a much harder throw (greater force, greater shock) would have been required to cause detonation.


How to prove at what pH the color of an indicator will change based on its pKa?

Thank you guys for your answers! I am fairly confident I have found the correct answer, or at least one of the possible correct answers. Here's it goes:

The indicator reacts with water according to the following reaction equation:
$$\ce{HA + H2O <=> H3O+ + A-}$$
Then using the Henderson-Hasselbach equation, I arrive at the following point:
 $$ K_s = \frac{[\ce{H3O+}]*[\ce{A-}]}{[\ce{HA}]} $$
Now, at the equivalence point, the concentration of the indicator an it's conjugate base are equal, hence:
 $$\require{cancel} K_s = \frac{[\ce{H3O+}]*\cancel{[\ce{A-}]}}{\cancel{[\ce{HA}]}} $$
$K_s = [\ce{H3O+}]$. After taking the negative log of both sides I arrive at the equation: $pK_s = pH$ (which was given to be 6).

Now, when the solution is acidic(pH lower than 6 and high [$\ce{H3O+}$] concentration), Le Chatelier sais then the balance will shift to the left of the top most equation. Favoring the indicator $\ce{HA}$ as opposed to it's conjugate base. Naturally, if the solution contains more indicator molecules, then it will also take on the indicators color.

 Analogue to the previous reasoning, when the pH reaches values above those of 6, Le Chatleier sais that the reaction will favor the formation of the indicators conjugate base [$\ce{A-}$](the right side) and the solution takes on the color accordingly.


For the record, I am not a biochemist, a nutritional expert, or any sort of person who is required to know this for my profession.

That said, as far as I know, to get the vast majority of nutritional food substances, you'll need to be able to produce at least:

sugars and starches
lipids
amino acids
cellulose
various other non-polymeric organic molecules of various levels of complexity
salts
proteins

These together will give you a reasonable approximation of food which will be most of what you need to stay alive, without necessarily going into the complexity of actually producing a better analogue of dead tissue "under the hood".

Sugars and starches are entirely composed of carbon, hydrogen, and oxygen. The same goes for lipids, cellulose, and the majority of the organic components you'll need. For amino acids, and the other organic bits, you'll need to add nitrogen and sulfur.

You also need phosphorus for DNA, RNA, ADP, and ATP, as well as a whole bunch of other signalling and respiration-based molecules, and bone material.

You need sodium and potassium ions which are needed in living tissue for nerve signalling, and you also need calcium ions to make bones.

Many plant materials seem to contain small amounts of boron, though the biological role of it isn't concrete as far as I know.

As far as I know, the proteins are where the metallic awkwardness comes in. A lot of proteins have a reaction site containing a metal, metal-sulfur, metal-porphyrin, or another metal-organic molecule complex, which can come in various metallic flavours. For example, haemoglobin requires an iron-porphyrin complex, while chlorophyll requires magnesium-porphyrin, and a lot of respiration-based proteins seem to involve iron-sulfur complexes.

I'm fairly confident you could make reasonable analogues of most food with C, H, O, N, S, P, Na, K, Cl, and Ca, with relatively small additional quantities of B, Fe, and Mg. You might suffer after a while from micronutrient deficiencies, though.

(I'll probably find that I've forgotten something important somewhere.)


I think the two steps that are missing from you approach are "carding", where, after immersion in the boiling water, the steel is rubbed/brushed to knock off the larger rust structures and repetition. The process is repeated several times before the piece builds up a stable blued surface.

Check out this video of Larry Potterfield applying a rust blue surface to a rifle action and barrel: https://www.youtube.com/watch?v=VPD3GWqk-yE


Is it wrong to allege that lemons and limes are alkaline?

Grouping of food items into various classes is a very old idea of Ancient Greek, Alchemists & Ancient Indian School of Medicine (Aryuveda). These ideas are perhaps thousands of years old. From a scientific perspective these classifications don't mean anything today. For example, in those schools of thought, fish has a very "hot" effect on the body but yogurt is "cold". Similarly, citrus fruits are "cold" but dried dates are "hot". Now nobody can define what hot or cold means and what is the criterion?

Similarly, the chart shown in the original posts are all wrong. It is just a modern version of these ancient schools of medicine. I think the major source of confusion in calling limes and lemons is that citrates (salt of citric acid) are indeed alkaline. Many antacid concoctions use solid sodium bicarbonate and citric acid. The moment you add a spoon to water, it starts to fizz vigorously. The pH of such solutions are slightly alkaline. Perhaps whoever made those charts had these ideas in mind.


Is pentamine or pentylamine the correct name?

The following IUPAC guidelines suggest the correct systematic nomenclature for simple primary amines.

P-62.2.1.2
Primary amines are systematically named in the following ways:
(1) by adding the suffix ‘amine’ to the name of the parent hydride;
(2) by adding the name of the substituent group R - to the parent hydride ‘azane;
(3) by adding the name of the substituent group R - to the term ‘amine’ used as a preselected parent hydride name for NH3; this method is used only with monoamines.

Method (1) gives the preferred IUPAC name (PIN). So in your example this would give pentan-1-amine.
Method (3) gives what is often the more commonly used name, in your example this would give pentylamine.
Both names are 'fine' and in reality method (3) is the one that you will most often see used/written on bottles of reagent etc. But for the correct IUPAC identifier method (1) should be used.
You should also note that pentamine is a wholly different compound:

Other example:

By method (1). Methanamine
By method (2). Methylamine


The iron oxide from rust is not soluble in high pH (basic) solutions such as bleach. If you want to remove the the rust, you need an acid such as muriatic acid (wear PPE, follow label instructions). I would recommend an oxalic acid based product such as Bar Keepers Best Friend (I have no affiliation).  In my (pseudoscience) experience, oxalic acid its more effective at sequestering rust (when scrubbed) than anything else I've used including nitric, muriatic, and sulfuric acid. I would be cautious if the white rocks are marble as acid can disslove marble into oblivion.


All quoted stuff copied from IUPAC Red Book, IR-4 unless otherwise specified

It's not only $\ce{H3AsO3}$ . What about $\ce{H3BO3,H3PO4,H2SO4,HClO4}$? Or indeed, any other oxyacid?

Basically, we have multiple ways of writing inline formulae of covalent compounds. One is the boring way (molecular formula), and one is the fun way (structural formula) where the formula reveals a bit of the structure. $\ce{As(OH)3}$ reveals the structure, $\ce{H3AsO3}$ doesn't.

  For compounds consisting of discrete molecules, the molecular formula, as opposed to the
  empirical formula, may be used to indicate the actual composition of the molecules. For the
  order of citation of symbols in molecular formulae, see Section IR-4.4.

  A structural formula gives partial or complete information about the way in which the atoms
  in a molecule are connected and arranged in space. In simple cases, a line formula that is just
  a sequence of atomic symbols gives structural information provided the reader knows that
  the formula represents the order of the atoms in the linear structure.

So, what are the rules for writing the molecular formula?

Well, initially these rules were basically that we just follow electronegativity--the most electropositive element is listed first and so on. Looks like they slightly changed it.

For one, there are actually two ordering standards--one is the extremely boring alphabetical one, which I've never seen used (except in emperical formulae), and one is via electronegativity.

Now, instead of following the absolute electronegativities, the IUPAC has assigned a "pseudo-electronegativity" order (my terminology). The rules are as follows (source):

  for two elements in different groups - then the element in the higher numbered group has higher "electronegativity"
  for two elements within the same group the element with the lower the atomic number has the higher "electronegativity"
  Hydrogen is fitted in to be less electronegative than polonium and more electronegative than nitrogen. Hence the formulae of water and ammonia can be written H2O and NH3 respectively.

This leads to the following "pseudo-electronegativity" order of elements:

(click to enlarge)

The motivation behind this must probably be the erratic and hard-to-quantify nature of "real" enectronegativity.

Now, we can apply these rules in the following way:

If the compound is a binary compound (only two elements, not two atoms), then we just follow the pseudo-electronegativity thingy.

  In accordance with established practice, the electronegativity criterion (Section IR-4.4.2.1)
  is most often used in binary species.

If the compound is a coordination compound, we order central atoms by pseudo-electronegativity, and ligands by alphabetical order, as written (eg $\ce{(en)}$ before $\ce{(dmg)}$, but $\ce{C2H4(NH2)2}$ after $\ce{(dmg)}$, even though $\ce{(en)}$ and $\ce{C2H4(NH2)2}$ are the same).

  The order of citation of central atoms is based on electronegativity as described in
  Section IR-4.4.2.1. Ligands are cited alphabetically (Section IR-4.4.2.2) according to the
  first symbol of the ligand formula or ligand abbreviation (see Section IR-4.4.4) as written.
  Where possible, the ligand formula should be written in such a way that a/the donor atom
  symbol is closest to the symbol of the central atom to which it is attached.

Finally (barring annoying extra rules for isotopes and hydration), we have what are called "generalized salts". This is for compounds which can be treated as being made up of ions. The ions are ordered by electronegativity, and the elements within each ion are ordered alphabetically{*}.

  If the formula of a compound containing three or more elements is not naturally assigned
  using the preceding two sections, the compound can be treated as a generalized salt. This
  term is taken to mean any compound in which it is possible to identify at least one
  constituent which is a positive ion or can be classified as electropositive or more
  electropositive than the other constituents, and at least one constituent which is a negative
  ion or can be classified as electronegative or more electronegative than the rest of the
  constitutents. The ordering principle is then:

  (i) all electropositive constituents precede all electronegative constituents;

  (ii) within each of the two groups of constituents, alphabetical order is used.

This is the rule we need to write the formula for $\ce{H3AsO3}$. First, we need to clear up what is meant by "treated as being made up of ions". Well, for this, as @Terry noted above, we look for the weaker bond. In $\ce{NaOH}$, we have relative bond strengths $\ce{Na...O-H}$, whereas $\ce{H3AsO3}$ is $\ce{As-(O...H)3}$. Thus, in $\ce{NaOH}$, the "ion" breakup is $\ce{Na+, OH-}$, while in $\ce{H3AsO3}$, the ions are $\ce{H+,AsO3-}$.

So, arsenous acid is broken up into ions, of which $\ce{H+}$ is written first. Within each ion, we order alphabetically, so we get $\ce{AsO3}$, not $\ce{O3As}$. Putting them together, we get $\ce{H3AsO3}$.

*Despite being the IUPAC rule, it seems that, within ions, most people just follow the electronegativity rules again. Selenous acid seems to be commonly named $\ce{H2SeO3}$, when it should be $\ce{H2O3Se}$. But this seems to be a minor offence:

  Deviation from alphabetical order of constituents in the same class is allowed to
  emphasize similarities between compounds.


Good to see classical analytical chemistry queries. These graduated pipettes or Mohr pipettes they are called, confuse even the relatively advanced users. First you have to check where the graduations end. If the pipette was meant to deliver 10 mL or 50 mL you should not see 10 or 50 mL mark near the tip of the pipette. That is one hint. Add a picture of the tip as well. Secondly, a double ring indicates that all liquid has to be expelled.
Notice that when you are doing very high precision work, one has to know the full drainage time as well after the liquid has been delivered. For example, after delivering 10 mL you need to wait for 5 s before blowing out the last drop. Very high quality glassware use to mention this time on the glassware.
In your blurred picture one can se Ex 5s, which means wait for 5 s before blowing out the last drop.
Volumetric pipettes should never ever be blown out, but they also have a drainage time, you have to wait for a certain time before dislodging the last drop purely from surface tension effects.


Why does Nitrogen form N₂ (and not N₄) while but Phosphorus forms P₄ (and not P₂)?

Due to larger atomic size $\ce{P}$ is unable to form pi bonds and so it is tetra-atomic in which each $\ce{P}$ atom is linked with 3 other $\ce{P}$ atoms by 3 sigma bond. But, due to smaller atomic size $\ce{N}$ forms 1 sigma and 2 pi bonds i.e. triple bonds with other $\ce{N}$ atom and exists as diatomic molecule.

Reference: http://philmintz.tripod.com/Chemistry/page4.html


How energy is released when chemical bond is formed?

Let us assume the simplest case to form a bond from two atoms (assumed to be in their ground state) that approach each other from an infinite distance. For convenience, we put the potential energy of the system at zero here. If the atoms come closer, they start to interact via long-range forces (such as van der Waals forces) and the potential energy is slightly lowered. If the atoms come even closer, their wave functions start to overlap and the potential is reduced rather steeply as compared to the long-range forces. Moving the atoms even closer increases the potential as the electrons will start to repel each other. The potential energy as a function of the interatomic distance can be approximately described by a Morse potential (the blue curve in the figure below (taken from Wikipedia))

In the figure you also see horizontal lines in the Morse potential which represent the energies of the vibrational states that are supported by the potential. Now that we know what the potential looks like we can take our two atoms at infinite distance and move them towards each other, that is, the total energy $E_\text{tot}(r)=E_\text{Morse}(r)+E_\text{kin}(r)>0$. As the atoms get closer, the potential energy gets lower and the velocity and kinetic energy increase, the total energy is - of course - conserved. When the atoms come so close that they start to repel each other (when they hit the repulsive "wall" on the left), they fly away in opposite direction and end up at infinity again! This is the same result that you would find with an ideal marble track (no friction) that is similarly shaped to the Morse potential: the marble will roll down, first accelerate, then climbs up the wall on the left and rolls back to where it started. In order to from a bond (or to stop the marble in the middle) we need to remove part of the energy of the system. As Ivan Neretin pointed out, this can be done by emitting a photon. When the atoms are separated at $r_e$, the release of a photon with energy $h\nu=D_0+E_\text{kin}$ will result in a loss of energy and the atoms will be trapped in the potential. (In our marble track, friction can play a similar role)

This process is called radiative association and it is believed that it played an important role in the chemistry of the early Universe. However, not all molecules can emit a photon. There are some rules - selection rules - which have to be fulfilled. The molecule needs to have a dipole moment for instance. Therefore $\text{H}_2$ molecules are unlikely to be formed via radiative association.

Another, more likely, method to remove the energy is to have a third atom (or molecule) colliding with the other two atoms. This atom can then take away the excess energy in the from of kinetic energy in a so-called three-body collision.

When you have more complex molecules that collide to form bonds, the excess energy can be distributed over the different degrees of freedom that the molecule possesses. For instance, the molecule's vibrational and/or rotational energy is increased which results in an increase in temperature.


Is associative ligand exchange possible with octahedral complexes?

It could — and there isn’t even a general restriction on the electron count.

Any pentacoordinated species can easily be expanded to a hexacoordinated one. It is just a question of the ligand approaching from the correct angle. In fact, I would even want to prefer associative ligand exchange for pentacoordinated species over dissociative since there is less reorganisation involved (only two ligands need to shift positions compared to at least three in dissociative).

Also, it should be mentioned that the number of electrons in each structure does not have to be 18 or less. Many (pseudo)octahedral complexes such as $\ce{[Co(H2O)6]^2+}$, $\ce{[Cu(NH3)4(H2O)2]^2+}$ or $\ce{[Ni(H2O)6]^2+}$ have more than 18 electrons and are fully comfortable with it.


Suggestions for solvents capable of efficiently performing alpha-bromination and nucleophilic substitutions

As a general comment published procedures have been optimised. I don't have access to the paper you link so I don't know if the procedure is from process chemistry (where you expect a high degree of optimisation) or from med chem (where the priority is to get the compound out fast but there is still some optimisation).

But your questions are still valid. So I agree that doing the bromination - amination in one pot using the same solvent is a reasonable experiment to try and actually acetonitrile is a good choice. Why do you think acetonitrile is nucleophilic? it is not (unless you deprotonate it with a strong base). So it is very possible that the first researcher tried your approach initially and possibly found that the bromination was not very selective or the amination step was too slow, or low yielding and switched to the current procedure.

I also agree that you can use a base instead of excess of amine, for example an amine base like triethylamine or inorganic bases like $\ce{K2CO3}$ are standard in this case and this is what they would have used if they had to do the reaction with a different amine (as the nucleophile) of higher cost.


Is elemental lead as toxic as lead compounds?

Pb(IV), in all likelihood, will quickly get reduced to Pb(II) once it gets into a human body. So as far as poisonings go, these two can be considered equal.
Elemental lead will partially dissolve in stomach, but mostly go out undigested. So it is way less harmful than any compound of Pb(II). There were people in the old times who survived bullet wounds and used to carry bullets in their bodies for decades, with little or no apparent ill effects.
Then there is a wholly different story of the covalent compounds of Pb. They are more toxic than ionic Pb(II), and some are capable of penetrating through the skin. In particular, the notorious $\ce{Pb(C2H5)4}$ has caused our civilization a great deal of harm.
So it goes.


Electrophilic substitution of benzene with conc. HNO₃ and HNO₂

Nitrosyl cation, $\ce{NO^+}$, is a better electrophile than molecular $\ce{HNO3}$ (you don't get $\ce{NO_2^+}$ in this system without sulfuric acid or other strong auxiliary acid) and so you get nitrosobenzene in the substitution reaction.  But then the nitrosyl group with its nonbonding electron pair is an attractive target for oxidation by the nitric acid and thereby is converted to $\ce{NO_2}$.  Thus nitrobenzene.  Then, the nitrobenzene, deactivated by the nitro group, will not react with the relatively mild electrophile $\ce{NO^+}$ anymore, meaning products C and D are never reached (and their nitrosyl groups would not have survived anyway).


Detection of sulphide in solution also containing sulphate

I will add one thing that may be the reason behind why (C) is an acceptable answer choice. Yes, both $\ce{PbS}$ and $\ce{PbSO_4}$ are insoluble. However, if you look at the $\ce{K_{sp}}$ for both compounds, $\ce{PbS}$ is about $\pu{3.2 x10^-28}$ while the $\ce{K_{sp}}$ for $\ce{PbSO_4}$ is $\pu{1.3 x 10^-8}$. When you have a mixture of ions, the resulting molecule with the lowest $\ce{K_{sp}}$ will precipitate first. In this case, since $\ce{PbS}$ has a lower  $\ce{K_{sp}}$ than $\ce{PbSO_4}$, it will precipitate first.
This technique is called fractional precipitation where you separate two ions, in our case $\ce{S^{2-}}$ and $\ce{SO_4^{2-}}$, by precipitating one of the ions selectively through taking advantage of their differing $\ce{K_{sp}}$ values.
This is the likely reason behind why choosing (A) itself is acceptable or choosing (A) and (C) is also acceptable.


What is the difference between non-adiabatic coupling and diabatic coupling?

The difference is in the representation: adiabatic representation, being exact, allows you to make adiabatic approximation. If you do not want to have adiabatic approximation, there will be off-diagonal matrix elements in kinetic energy, so the coupling goes through kinetic energy. In diabatic representation, the kinetic energy is diagonal, and the couplings come from potential.
As was pointed out above, diabatic coupling allows you to apply Landau-Zener rule and estimate probabilities of transition.


Does acetylene give a positive Tollens' test? If not, why is this so?

Tollen's reagent (ammoniacal silver nitrate) reacts with terminal acetylenes to form the silver acetylide which precipitates out of solution

$$\ce{RC#CH + AgNO3 + NH4OH -> RC#CAg v + NH4NO3 + H2O}$$

In the case of acetylene itself, since both ends of the molecule are (if you will) terminal acetylenes, a bis-silver acetylide salt is formed and precipitates out of solution (see page 3, item 6b in a PDF from www.sakshieducation.com/ or via the Internet Archive)

$$\ce{HC#CH + 2AgNO3 + 2NH4OH -> AgC#CAg v + 2NH4NO3 + 2H2O}$$

Tollen's reagent is not very stable, therefore it needs to be freshly prepared before use.  The reagent has a number of other uses in chemistry (test for aldehydes and α-hydroxy ketones), anatomical staining, and silver mirroring.  This Wikipedia article provides a nice, concise overview.


Assuming that the fields marked by x are solid and y is liquid, these are the points:

a: eutectic point. Cooling will turn the single liquid into two solids. The point is between the composition of the two solids involved.
b: eutectoid point. Cooling will turn the single solid into two solids.
c: peritectic point. Cooling from immediately to the left will turn the solid into a liquid and a solid. The point is outside the composition of the two solids involved.
d: congruent point: A solid will melt at this point to a liquid of the same composition.
e: incongruent melting points. A solid will melt to a liquid and a solid. Note that the left one is intimately associated with the peritectic point.

There are no peritectoid points in this diagram. A peritectoid point is similar to a peritectic point, but with all phases solid.


Microwaving a glass of water, what happens?

Heating water on a hot plate is safe, because the hottest point is at the bottom of the pot. A lot of relatively small bubbles appear there without much overheating of the water, because there is a lot of nucleation at the uneven phase boundary steel-water.

In a microwave, the hottest place is IN the water. The glass does not get heated by microwave (at least not much), and radiates off some heat to the surrounding.

Problem: In clean water, there are few good nucleation points to form bubbles, only some dust particles perhaps. So the water gets overheated rather strongly, and a first bubble that appears can grow a lot before it has cooled its surroundings down to 100 °C. That one huge bubble can throw most of the water out of the glass. It boils over, violently.

Btw. a microwave does not heat its content uniformly. It forms a standing electromagnetic wave (that's not radiation, strictly speaking) in the oven, like a rope swung quickly between two people, or a guitar string. The wave pattern has knots at a distance of $0.5c/f\approx6\ \mathrm{cm}$ (with $f=2.45\ \mathrm{GHz}$ and speed of light $c$ which of course is a bit less in your chicken), where there is very little heating. That's why the microwave oven has the rotating plate, to generate some uniform heating. For pure water, that doesn't matter, because convection sets in anyway and distributes the heat.


Volume Element in the Integral for Variational Method

If you are working in spherical coordinates, and if the integrand has no dependence on the angular coordinates, then that volume element would be correct ($4\pi$ from each integration over $\theta_i \phi_i$ and $r_i^2$ for each spherical shell); however, the interelectronic potential $\frac{1}{r_{12}}$ depends not only on the radial coordinates, but also on the angular coordinates of each electron, so you can't just integrate those out.


What happens to the first ionization potential when a hydrogen-like atom captures a particle?

When we solve the Schrodinger equation for the hydrogen atom we general make the simplifying assumption that the proton stays fixed and the electron moves in the potential of the fixed positive charge. So when we write, for example, the $1s$ orbital as:

$$ \psi_{1s} = \frac{2}{a_o^{3/2}} e^{-r/a_0} \tag{1} $$

the variable $r$ is the distance from the proton, and in the equation for the Bohr radius:

$$ a_0 = \frac{\hbar^2}{me^2} \tag{2} $$

The $m$ is the mass of the electron. If you're interested in some detail this is discussed on the Physics SE in Reduced mass in quantum physics (Hydrogen Atom) but taking into account the motion of the hydrogen atom turns out to be surprisingly simple. We simply define $r$ to be the distance to the centre of mass of the atom, and the mass $m$ then becomes the reduced mass of the electron-proton system:

$$ m = \frac{m_e m_p}{m_e + m_p} \tag{3} $$

If we take the limit of $m_p \to \infty$ then the reduced mass just becomes the electron mass $m_e$, but for finite $m_p$ the reduced mass is less than $m_e$.

Given all this you can now see what the question is getting at. If the proton captures a neutral particle of mass $207m_e$ the effect is to increase the mass of the proton. This increases the mass $m_p$ we have to put in to equation (3) so it increases the reduced mass in equation (2) and hence the wavefunction (1). The end result is that in the equation for the ionisation energy:

$$ I = \frac{me^4}{8 \epsilon_0 ^2 n^2h^2} \tag{4} $$

the reduced mass $m$ is increased slightly so the ionisation energy increases slightly.

A good example of this is to compare the ionisation energies of hydrogen, deuterium and tritium, where the increased nuclear mass of deuterium and tritium increase the ionisation energy by the mechanism discussed above. In fact there is an existing quesion discussing exactly this: How can difference in neutron number cause a difference in ionisation enthalpies?


Distance requirement for hydrogen bonding to occur

You will probably find all this information in biochemistry textbook. To give you an idea in proteins, typically; alpha helices H..O distance is 2.06 Angstrom, and N..O, 2.99. In beta sheets (parallel) H--O 1.97, N--O 2.29 and in anti parallel sheets, H--O 1.96 and N--OI 2.91. These H bonds are not linear, e.g C=O..H-N typically 150 to 160 degree for COH or NHO angle.

In Watson-Crick base pairing in DNA, (C-G pairs) the linear H-bond distances  are  $\mathrm{N_4-(H)-O_6}$ 2.91, $\mathrm{N_3-(H)-N_1}$ 2.95, and $\mathrm{O_2-(H)-N_2}$ 2.86.


Equilibrium constant expression for heterogeneous equilibrium

First, as kaliaden commented, the true equilibrium constant

$$K(T,p) = a(\textrm{CaO,s}) a(\textrm{CO$_2$,g}) / a(\textrm{CaCO$_3$,s}) \approx
p_\textrm{CO$_2$} / p^\circ \equiv K_p(T)$$

where $a$ are the activities, $p$ the partial pressure and $p^\circ$ is one bar.
Thus $K$ basically only depends on temperature (unless you are interested in brutally large pressures, and we safely disregard the sublimation of CaCO$_3$ and the effect of the air). Obviously you can obtain

$$K_c(T) \approx K_p(T)/ RT = [\textrm{CO$_2$}]$$

Now your question is: how is it that this constant does not depend on the amount of matter that we have?

The fact is that as long as there is enough matter to reach the equilibrium (to saturate the air with carbon dioxide vapor) it doesn't matter how much CaCO$_3$ you start with. The reaction will proceed until reaching an equilibrium characterized by $K(T) \approx p_\textrm{CO$_2$} / p^\circ$. Obviously if you don't have enough CaCO$_3$ (for instance, a few micrograms) then everything will react and you will not reach the equilibrium.


Why is it important to use a salt bridge in a voltaic cell? Can a wire be used?

There's another question related to salt bridges on this site.

The purpose of a salt bridge is not to move electrons from the electrolyte, rather it's to maintain charge balance because the electrons are moving from one-half cell to the other.

The electrons flow from the anode to the cathode.  The oxidation reaction that occurs at the anode generates electrons and positively charged ions.  The electrons move through the wire (and your device, which I haven't included in the diagram), leaving the unbalanced positive charge in this vessel.  In order to maintain neutrality, the negatively charged ions in the salt bridge will migrate into the anodic half cell.  A similar (but reversed) situation is found in the cathodic cell, where $\ce{Cu^{2+}}$ ions are being consumed, and therefore electroneutrality is maintained by the migration of $\ce{K+}$ ions from the salt bridge into this half cell.

Regarding the second part of your question, it is important to use a salt with inert ions in your salt bridge. In your case, you probably won't notice a difference between $\ce{NaCl}$ and $\ce{KNO3}$ since the $\ce{Cu^{2+}}$ and $\ce{Zn^{2+}}$ salts of $\ce{Cl-}$ and $\ce{NO3-}$ are soluble. There will be a difference in the liquid junction potential, but that topic is a bit advanced for someone just starting out with voltaic/galvanic cells.


Both sources are correct. However, they are referring to two different $\mathrm{p}K_\mathrm{a}$ values.

$$\ce{H3C-NH3+ <=>[$K_\mathrm{a1}$] H3C-NH2 + H+ <=>[$K_\mathrm{a2}$] H3C-NH- + 2 H+}$$

The second value is the one that your marking scheme refers to when it asks for the $\mathrm{p}K_\mathrm{a}$ of $\ce{H3C-NH2}$. However, it is of little practical relevance. Most of the time, the question is not how acidic methyl amine is but how basic. And for that, the $\mathrm{p}K_\mathrm{a1}$ value of the conjugate acid, methyl ammonium, is used. $\ce{H3C-NH3+}$ has a $\mathrm{p}K_\mathrm{a1}$ value of approximately $10$, and it is the value you found online.

Formally correct usage would only name $\mathrm{p}K_\mathrm{a2}$ as the value of methyl amine. Colloquial chemists’ usage mainly uses $\mathrm{p}K_\mathrm{a1}$.