Whats is the difference between Tris and Tris-acetate?

"Tris" simply means "three" and is used in a number of different contexts. However, in biochemistry it's normally used to refer to the common buffering agent tris(hydroxymethyl)aminomethane. You have to be a little careful, though, as there are other uses of "Tris" in biology, which may not refer to tris(hydroxymethyl)aminomethane. (For example, there's two other buffers called "bis-tris methane" and "bis-tris propane", each of which have very different properties from tris(hydroxymethyl)aminomethane.)

Your situation is referring to the standard tris(hydroxymethyl)aminomethane, though. However, when specifying a buffering system, it's often insufficient to specify just the buffering agent used. A buffer consists of a weak acid or weak base in near-equal concentration to its conjugate base or acid. Tris is the weak base being used, but to have a decent buffer you need to also have the conjugate acid form.

As this conjugate acid is charged, there's an obligate counter ion. Different buffering systems may use different counter ions. That's what the acetate is doing here - acting as a counter ion for the conjugate acid of tris(hydroxymethyl)aminomethane.

That's what "Tris-acetate" is specifying. It's using a Tris buffering system with an acetate counter ion. This buffering system is normally obtained by using enough Tris base to get the desired concentration, and then slowly adding enough acetic acid to obtain the desired pH.

Alternatively, you could use something like the Henderson–Hasselbalch equation to calculate the appropriate proportion of base and conjugate acid for your desired pH, and then specifically mix those two forms together. This isn't typically done with Tris-acetate, as the acetate counter ion form of Tris isn't commonly sold. However, Tris hydochloride is, so if you had a system which used a chloride counter ion, you could use this approach. (Though in practice you'll want to aim a little on the basic side, and adjust to the desired pH by titration - the numbers which come out of the Henderson–Hasselbalch equation are theoretically pure, and don't account for non-ideal deviations in behavior and things like weighing errors.)


High errors in frontier orbital energy calculation using DFT

IMHO, this particular comparison is a strange one.

You reference "Comparison of DFT Methods for Molecular Orbital Eigenvalue Calculations" by Zhang and Musgrave, and there are other similar comparisons between density functional orbital methods and ionization potentials and electron affinities.

In general, these Kohn-Sham DFT benchmark papers show that most functionals don't compute orbital energies that directly compare with ionization potentials and electron affinities. That is, in Hartree Fock, Koopmans' theorem suggests that the orbital energy of the HOMO should correspond to the negative of the ionization potential (-IP). In principal, for the exact density functional, a similar property should hold.

What does hold true (e.g., the Zhang and Musgrave paper) is that you can easily calibrate orbital eigenvalues from Kohn-Sham DFT methods to counteract systematic issues. Usually a linear regression is sufficient. (Computing the actual vertical ionization potential using two DFT calculations is generally even more accurate.)

Coming back to your question, I wouldn't be surprised if the absolute errors in orbital eigenvalues are mixed from DFT methods. But that's not how most research uses them - they either perform a calibration (e.g., linear regression) or are comparing multiple different molecules. In the latter case, the relative errors are much smaller (usually ~0.2 eV, IIRC - I'll look for some sources).

I believe the overall argument in the Faber et. al paper is that modern ML methods are very good. In some properties that's probably correct.


tl;dr: The active site catalytic residues (BH+ and B’) are thought to be Lys and a His–Glu dyad, respectively. These facilitate the abstraction of proton from C2 and subsequent protonation on C-1.

Some important points to note:

  "shows histidine protonating c5 oxygen and lysine deprotonating c1
  oxygen to form an open chain aldolase
  ”.

The opening of the ring, is facilitated by Lysine and not by Histidine.

  “Glutamic acid then deprotonates c3 to create an enediol intermediate
  and donates the proton to c1”

It is actually the His–Glu  (B’) dyad (see image below) that abstract this proton from C2. See how this process occurs at the same time as the exchange of H+ (protons) with the solvent or medium. In this step C2 loses a proton while C1 gains a proton (Step 3-4)
Finally a proton is added  from the medium and the ring closes

A proposed reaction mechanism for the Phosphoglucose Isomerase reaction involves general acid–base catalysis by the enzyme:

Step 1

Substrate binding.

Step 2

An acid, presumably the Lys ε-amino group, catalyzes the opening of the pyranose ring.

Step 3

A base, presumably the imidazole portion of the His–Glu dyad, abstracts the acidic proton from C2 to form a cis-enediolate intermediate (this proton is acidic because it is α to a carbonyl group).

Step 4

The proton is replaced on C1 in an overall proton transfer. Protons abstracted by bases are labile and exchange rapidly with solvent protons. This step was confirmed by Irwin Rose by demonstrating that $\ce{[2-^3H]G6P}$ is occasionally converted to $\ce{[1-^3H]F6P}$ by intramolecular proton transfer before the $\ce{^3H}$ has had a chance to exchange with the medium.

Step 5

Ring closure to form the product, which is subsequently released to yield free enzyme, thereby completing the catalytic cycle.

Reaction mechanism of phosphoglucose isomerase


Role of Acetonitrile and Ammonium Acetate buffer in Leucomalachite Green stock solution?

Since they're using ACN/ammonium acetate to elute from solid phase extraction cartridges and that's what they're injecting into their LC analysis, they probably want to use the same mixture to avoid problems with unmatched matrices with their standards. The ammonium acetate is the pH buffer. Malachite green has different protonation states at different pH, so it's important to set the pH properly for elution from the SPE and LC analysis. MG is fairly soluble in water, but still has a lot of organic functionality, so adding an organic solvent would help elute from the SPE column and keep the analyte from being too strongly retained by the LC stationary phase.

As for raman analysis, water has rather weak raman scattering, so it's not usually a problem. Acetonitrile has one very strong peak at 2942 $\mathrm{cm^{-1}}$ and a few moderate peaks, so it might be an issue. Since you're not doing LC, you can likely prepare standards in buffer only. Unfortunately, ammonium acetate is also raman-active, but if you're not doing LC, you might be able to do without as well. Fortunately, one of the great powers of IR and raman spectroscopies is that there is often enough bandwidth to find one or two non-overlapping peaks in a complex mixture and use that for quantitation, so it may still work as written. They're both also very amenable to chemometric techniques like partial least squares when measuring complex samples.

The big problem is going to be sensitivity if you're interested in the concentrations they use in the paper.. Normal raman is not that sensitive and the smallest standard is only ~3 nM, so it's probably going to be hopeless. These guys do surface-enhaced resonance raman and are able to get to about 1 nM, and the raman setup's not actually that hard since MG has a huge absorbance at 620 nm, right where HeNe lasers emit. Unfortunately, MG doesn't fluoresce much in solution or it would be very easy to measure.


Why can't HS- participate in E1 reactions?

As has been pointed out in the comments, the use of the word only is a little misleading. What it should really say is that $E1$ is possible but $S_N1$ is so strongly favoured that to all intents and purposes the only product is the $S_N1$ product.

This can be explained in terms of the relative energies and sizes of the orbitals involved. $\ce{HS-}$ has a large, diffuse, high energy HOMO which can have much better overlap with the empty $2p$ orbital of the carbon than the small, low energy $\ce{C-H}~\sigma^*$. Therefore, the $S_N1$ reaction is strongly thermodynamically favoured and dominates.

If you compare this to something like $\ce{H2O}$, the HOMO is much more compact and lower in energy and so it has much better overlap with the $\ce{C-H}~\sigma^*$ and so $E1$ is comparatively much more favourable.


What temperature is really required to make fused quartz?

Glass is not necessarily made of pure silica $\ce{SiO2}.$ Ordinary glass is a mixture of at least $50\,\%$ $\ce{SiO2},$ plus smaller proportions of $\ce{CaO},$ $\ce{Na2O},$ $\ce{K2O},$ and $\ce{Al2O3},$ with impurities like $\ce{Fe2O3}.$ Pyrex has also some $\ce{B2O3}.$
Glass has no defined melting point. It becomes like a very viscous paste at around 800 °C. Mixtures have melting points that are always much lower than their components.


Finding out the concentration of oxygen in water

This is the old Winkler titration, published as

Die Bestimmung des im Wasser gelösten Sauerstoffes by Ludwig Wilhelm Winkler in Ber. Dtsch. Chem. Ges., 1888, 21, 2843-2854 (SRC)

How does it work?

Add $\ce{MnSO4}$ to your water sample, then $\ce{NaOH}$ and $\ce{KI}$.

Under alkaline conditions, oxygen dissolved in your water sample oxidizes $\ce{Mn(II)}$ to $\ce{Mn(IV)}$.
\[\ce{2Mn^{2+} + O2 + 4 OH- -> MnO(OH)2 v}\]
This is the brownish precipitate mentioned in the question.

Now, the sample is acidified (Winkler used hydrochloric acid) and the solution becomes yellowish due to the iodine formed. Under these conditions, $\ce{Mn(IV)}$ oxidizes iodide:

\[\ce{Mn(IV) + 2 I- -> Mn^{2+} + I2}\]

The amount of iodine formed, which repesents the aount of oxygen in the water sample, can be determined by titration with thiosulfate solution.


When aluminum is the anode (connected to the positive battery terminal), a thin layer of insulating aluminum oxide is produced via the “anodization”. It serves as a dielectric, so basically you have a leaky capacitor: current is low. But when the aluminum is connected to the negative terminal, it is the cathode and then you get reduction of hydrogen ions to hydrogen gas. So considerably more current can flow. This is essentially how tantalum and aluminum electrolytic capacitors have rectifying properties: their oxide films are the cause. Fun link, by the way.


How do I identify the compound(s) that formed from a failed synthesis of another compound?

Theoretical part

Chrome alum $\ce{KCr(SO4)2 * 12 H2O}$ is synthesized according to the following reaction (average oxidation numbers are shown above the elements undergoing redox reaction):

$$\ce{K2\overset{+6}{Cr}_2O7 + 3 \overset{-2}{C}_2H5OH + 4 H2SO4 →[15-40~°C] K2SO4 + \overset{+3}{Cr}_2(SO4)3 + 3 \overset{-1}{C}_2H4O + 7 H2O}$$

after joint crystallization of both $\ce{K2SO4}$ and $\ce{Cr2(SO4)3}$ with water forming a complex which is better described as aquacomplex $\ce{[K(H2O)6][Cr(H2O)6](SO4)2}$ with no sulfate(VI) anions in neither of the first coordination spheres.

Note:

It's absolutely crucial to control the temperature during the addition of alcohol to acidified dichromate. If the temperature below 15 °C, the reaction is too slow, but if it's above 40 °C, ligand exchange starts to take place: $\ce{H2O}$ ligands in $\ce{[Cr(H2O)6]^3+}$ are partially substituted by $\ce{SO4^2-}$, which is reflected in color change from blue-violet to green. Also, evaporation of alcohol also happens quicker at elevated temperatures, depleting the reducing agent at higher rate.
Whereas you need stoichiometric amounts of dichromate and sulfuric acid, alcohol is usually taken in excess to ensure complete reduction of $\ce{Cr^6+}$ to $\ce{Cr^3+}$ and to salt out crystals of chrome alum from the solution.
Working with such small quantities (few mL) of pretty abundant and cheap chemicals seems unreasonable to me as it increases the error of measurement, increases the negative impact of impure chemicals or erroneous concentrations and makes the experiment less demonstrative.
I'd advise to work with larger quantities (e.g. 1-2 g of dichromate) in the future.

Experimental results and discussion

  There were some crystals formed, but the color was orange …

It's extremely likely that the orange crystals are unreacted potassium dichromate(VI), so it looks like the reduction was incomplete.
That's why I doubt that alcohol was ever oxidated further to acetic acid, hence I see the formation of chromium(II) acetate as suggested in the comments as improbable.
Also, chromium(II) acetate is very unstable and would've decomposed under these conditions.

There is also a rudimentary method of checking the oxidized product by its smell.
If it smells like apples, then it's acetaldehyde $\ce{C2H4O}$; if it smells like vinegar, then the oxidation took further and alcohol has been oxidized to acetic acid $\ce{CH3COOH}$.

  … (it should've been purple).

Since you haven't dissolved potassium dichromate in water and just added sulfuric acid to the powder, the concentration of sulfate(VI) was too high.
Plus, you've added all alcohol at once, which might've resulted in temperature spike. Both factors effectively prevent formation of chromium(III) aquacomplex, leaving no chance for alum to crystallize. (see Addendum for the reference)

  From the instruction that we followed, we should heat up the solution, so we did that, and carelessly let the temperature go up past 70 °C. It become greenish in color and doesn't form any crystal after a week.

At this stage you pushed the reduction further and probably evaporated both ethanol and acetaldehyde completely, and the solution was consisting of very concentrated chromium(III) sulfate, still no chance for alum to crystallize due to above mentioned reasons. (see Addendum)

  We resorted to evaporating the liquid part and obtained some leftover greenish yellow powder at the bottom

Green powder is likely a mixture of potassium chromium(III) sulfate $\ce{KCr(SO4)2}$ and also minor amounts of chromium(III) oxide $\ce{Cr2O3}$ formed along the way.
Now yellow powder is with a very high confidence just potassium chromate(VI) $\ce{K2CrO4}$ leftovers from the dichromate(VI) once the acid was used up:

$$\ce{Cr2O7^2- + H2O <=> 2 HCrO4^2- <=> 2 CrO4^2- + 2 H+}$$

One might also claim that it's elemental sulfur formed by reduction of sulfate(VI) after chromium(VI) was completely converted to chromium(III), but I find it rather implausible.

Addendum

From Ullmann’s Encyclopedia of Industrial Chemistry [1, p. 4441]

  Potassium chromium(III) sulfate [10141-00-1], [10279-63-7], [7788-99-0],
  potassium chrome alum, $\ce{KCr(SO4)2 · 12 H2O}$, $M_\mathrm{r}~499.11$, $\pu{1.813 g/cm3}$, crystallizes in the cubic system forming violet regular octahedra
  which decay in air and melt at 89 °C, the color changing to green; the enthalpy of formation is − 5788 kJ/mol. The solubility in water at 25 °C is 11.1 wt %. The solution is violet when cold, but becomes green above 50 °C, this change being accompanied by a decrease in the molar conductance. For 0.125 M solution at 50 °C, the molar conductance is $\pu{221 Ω−1 cm2 mol−1}$ for the violet form and $\pu{202 Ω−1 cm2 mol−1}$ for the green form. This change is reversible and its rate is increased by acids. The green form always occurs as an amorphous solid and crystals
  are unknown.

  […]

  For the preparation of potassium chromium alum, a saturated potassium dichromate solution is reduced with sulfur dioxide in the presence of sulfuric acid.
  During the reaction, the temperature must be kept below 40 °C by cooling to prevent the green modification being produced. Apart from sulfur dioxide, such organic compounds as formaldehyde, methanol, or starch are suitable as reducing agents. Crystallization starts after sulfuric acid has been added and the temperature is kept
  further below 40 °C. If the solution is allowed to settle in vats, large crystals are produced, whereas ﬁne ones result if the solution is stirred.

References

Ullmann’s Encyclopedia of Industrial Chemistry, 40 Volume Set, 7th ed.; Wiley-VCH, Ed.; Wiley-VCH: Weinheim, 2011. ISBN 978-3-527-32943-4.


can i calculate equilibrium constant of this reaction?

$\ce{H2O}$ is a very poor nucleophile and we are dealing with a tertiary substituted carbon, so we can minimize the scope of possible reaction mechanisms to the $\ce{E 1}$ and $\ce{S_{n}1}$ type mechanisms. $\ce{E 1}$ will compete here with $\ce{S_{n}1}$, forming tert-Butyl chloride and isobutylene respectively. That is probably where most of your loss is coming from. Other reasons can be indeed that a bit is left in the equilibrium, but I don't think that would be much since $\ce{H3O+}$ is a much better leaving group than $\ce{Cl-}$. Note that is chemistry papers the yield obtained is almost never a 100% of theory.


Long story short, it goes as follows:

Yes, the freezing water does perform some work. You put a load on top of a bucketful of water, and as it freezes, it will lift the load all right.
Yes, a capsule would prevent the water from freezing, at least for a while (provided that the capsule is immensely strong). Then some of the beer (water?) will freeze. Then... see the next item.
No, the absolute zero would not finally break the capsule. Rather, it would turn water into some alternative kind of ice which is more dense than water. BTW, you won't have to get precisely to absolute zero to achieve that.
True, normally increasing the pressure of a substance will cause it to crystallize ("freeze"). Water is different in this regard, which is quite rare, but by no means unique.


Ventilating combustion byproducts in water heaters

Rather than design something that offers danger of overheating, carbon monoxide poisoning, gas and over-pressure explosion, etc., why not use an off-the-shelf "instant" tankless water heater, available from sources such as Home Depot, Amazon or Lowes?

Find one certified for your area. Making your own is a tankless task!


Is there an analytical chemistry alternative to the Scoville scale for the hotness of chilli?

Matt did find a method that uses HPLC on the peppers alone, not in a dish.

So given that the capsaicinoids is a class of compounds, each different capsaicinoid would have a different perceived hotness. So a chemical test would measure the concentration of each type capsaicinoid and then use a weighted average over the human perceived hotness for each member of the capsaicinoid class.


Never say never! :-) Here is the answer for your question (Ref.1):

  Abstract: We report here that (1) enolate anions of five- to seven-membered cycloalkanones nucleophilically open cyclopentene and cyclohexene oxides in 57−76% yields and with 4−8:1 diastereoselectivity; (2) enolate anions formed regiospecifically via kinetic deprotonation of 2-cyclohexenone and 2-cycloheptenone open cyclohexene oxide in 60−62% yields and with 32−95:1 diastereoselectivity; and (3) an aryl methyl ketone enolate anion opens a monosubstituted epoxide as the key step in a short synthesis of the $\gamma$-hydroxyketone (GHK) aglycon of the natural product curculigine.

Graphical Abstract:

The product(s) and selectivity are depicted in following scheme:

Reference:

G. H. Posner, J. P. Maxwell†, M. Kahraman, "Mild, Fast, and Stereoselective Epoxide Opening by Ketone Enolate Anions. Application to Synthesis of the Norlignan Curculigine," J. Org. Chem. 2003, 68(8), 3049–3054 (DOI: 10.1021/jo020744q).


Migratory preference in a rearrangement involving carbocations

why is the second path not taken?

$\ce{H}$ can migrate, it is just that phenyl migrates better.  The reason that phenyl migrates better is because it is much better able to stabilize the developing carbocation center.  The phenyl group participates by what is termed neighboring group participation (see the "NGP by an aromatic ring" section).  It forms a phenonium ion.  Note the geometry of this ion, the former aromatic ring is perpendicular to the plane of the screen while the former $\ce{C-C}$ bond is contained in the plane of the screen.

(image source)

We can draw several resonance structures to describe the phenonium ion, this leads to it being especially stable.

(image source)

Such extra stabilization is not possible with a migrating hydrogen atom. Therefore the pathway involving the phenonium ion is lower energy (faster) than the pathway involving hydrogen migration.


Why is it not practical for insoluble salts to be prepared from other insoluble salts?

The comment "My textbook says that it is not practically feasible to make insoluble salts (say PbSO4) from other insoluble salts" is not precisely accurate. Limited solubility in water does not necessarily imply a lack of solubility in an acidic medium.

For example, I have prepared Silver acetate (moderately water-soluble, see Table) from the action of dilute acetic acid and 3% H2O2 in the presence of an electrolyte (a small amount of KNO3). Jumpstart the reaction in a microwave. The created Silver acetate is somewhat soluble in acetic acid (see this), hence there is no issue in the creation reaction.

The addition of moderately soluble PbCl2 to Silver acetate/acetic acid likely creates insoluble AgCl and Lead acetate (where the Silver acetate is also moderately water-soluble with respect to water). An explanation of solubility properties, per reference above, relates to a possible creation of an acetic acid salt.


Is there a way to concentrate the vitamin content of a particular food?

You are correct that retinoids are poorly soluble in water. Vitamins A (retinol), D (calciferol), E (tocopherol) and K (phylloquinone) have high hydrophobicity and therefore should be extracted using organic non-polar solvents.

All these vitamins can be identified using UV/Visible spectroscopy.

Since your focus is on retinoids I shall focus on that (but the principles remain the same for other hydrophobic biomolecules).

Note: carrot has beta-carotene and it is a precursor for retinol and not a retinoid itself.

Crush and homogenise the tissue
With some plant tissues you may need to grind the frozen tissue using liquid nitrogen repeatedly. Tissues containing polyphenols and lignins would need additional treatment but not necessary for carrots.
Precipitate proteins (optional; not required in your case).
Extract using solvents such as hexane or chloroform. Isopropanol has also been used for beta-carotene extraction from carrots (Fikselová et al. 2008)[1]. Gundersen (2006)[2] summarizes the methods of extraction and analysis of retinoids from animal tissue.
To purify, you need to perform chromatography. You can differentiate the retinoids using UV/visible spectroscopy. Beta-carotene is easy to detect; it is bright orange in colour and has an absorption maximum at 450nm.

References:

[1] . Fikselová M., Šilhár S., Mareček J., Frančáková H., Czech J. Food Sci., 2008, 26 (4), 268-274.
[2].  Gundersen T. E., J. Neurobiol., 2006, 66 (7), 631–644.


I can't tell you a manufacturer, but I can tell you the purpose of the pointer.

When the height of the column of mercury changes, it changes the height of the mercury exposed to the atmosphere. Since the ruler is (usually) fixed in place, the level of the mercury needs to be adjusted to the zero on the ruler. The pointer shows you the zero point. You adjust the level of mercury so that it barely touches the point.


In coordination chemistry square brackets denote inner coordination sphere; $\ce{[H2O]5}$ makes little sense in this context.
It looks like the correct molecular formula for this compound appears to be $\ce{[FeCl2(H2O)4]+[FeCl4]- * H2O}$ according to the crystal structure determined by Szymański [1].

The IUPAC-compliant name proposed by the author is tetraaqua-cis-dichloroiron(III) tetrachloroferrate(III) monohydrate.

Figure 1. Asymmetric unit $\ce{[FeCl2(H2O)4]+[FeCl4]- * H2O}$ (ICSD-8198) with added hydrogen atoms.
Color code: $\color{#EEEEEE}{\Large\bullet}~\ce{H}$; $\color{#FF0D0D}{\Large\bullet}~\ce{O}$; $\color{#1FF01F}{\Large\bullet}~\ce{Cl}$; $\color{#E06633}{\Large\bullet}~\ce{Fe}$.

References

Szymański, J. T. The Crystal Structure of Tetraaqua-cis-Dichloroiron(III) Tetrachloroferrate(III) Monohydrate (Ferric Chloride $2\frac 1 2$-Hydrate). Acta Cryst. B 1979, B35 (9), 1958–1963. https://doi.org/10.1107/S0567740879008244.


How would you determine the RDS from the rate and order of a reaction

The reaction is 1st-order in A2, so for starters we look for a step with A2 on the left, with a coefficient of 1. In fact, only step II has A2 on the left at all, so really we don't have to look any further.

Comments which may (or may not) make it clearer:

The rate equation says that greater the A2 concentration, the faster the reaction goes, in a linear relationship, and the rate doesn't depend on [B2] at all, since [B2] has an exponent of 0, and anything to the zeroth power is 1. (I suspect that was just put there to make it hard.)

If the whole reaction can be speeded up by adding A2 alone, [A2] must be rate-determining; i.e., [A2] determines the rate, and A2 reacts only in step II.


By burying the copper wire with plenty of sulfur and then heating the mix, the wire becomes completely submerged in liquid sulfur. The liquid sulfur then gets to react with the copper, and any oxygen that could possibly oxidize the copper would have to diffuse into the liquid sulfur before reaching the copper. However, oxygen quickly reacts with heated sulfur forming sulfur oxides, so it never reaches the copper wire. This is all under ideal conditions, of course.

(Originally given as a comment by @Nicolau Saker Neto)


In this experiment, the volumes are totally irrelevant. This is because all that matters (when regarding immiscible liquids) is density.
Density is the mass of a selected volume of matter (in this case olive oil and water). What is important, is that the water is denser, so it sinks to the bottom.
You can think of it, as an interaction between one volume oil and one volume water. Since the water is denser, the same volume of the stuff must weight more. From here, you can use the same logic as before, the heavier one, the water sinks. If you imagine it this way, on the surface of the two fluids there will always be some of both of them, so they'll arrange themselves in the correct order.

Even if the volumes are not the same, we can use Archimedes' principle, which

  indicates that the upward buoyant force that is exerted on a body immersed in a fluid, whether fully or partially submerged, is equal to theweight of the fluid that the body displaces and it acts in the upward direction at the centre of mass of the displaced fluid. Wikipedia

This ask sounds very technical, but what is means, is that the weight of anything submerged in a liquid is decreased by the weight of the liquid it displaced. To put it in other words, it's weight will decrease by the weight of the displaced liquid. Since these two have the same volume, we can compare only their densities.

Think of the experiment as oil droplets submerged in water. When imagining is this way, each droplet will displace the same amount of water (which is denser), so its weight decrease by more, than it ever was, resulting in a negative weight. Since weight is the force an object exerts on other ones "downwards" to keep it in place. We know that the system will at some point reach equilibrium. When this happens, if there was any oil in the water, it would mean, that it has negative wight, so it has to exert a force upwards to keep it in place. Since this is obviously not true, we know, that at equilibrium state all the water will be below all the oil.


Nucleophilic Substitution Reactions help

Yes iodine's good leaving group.  the leaving ability is best described by the Gibbs energy change that occurs.

So since chlorine is a poor  leaving group so that make the reaction slower as less  energy is available.
So I think  that you are right so far.


How to determine if an unknown is potassium chloride or potassium bromate?

Noting the solubilities of potassium chloride and potassium bromate gives a nice point in the identification but apart from it, there are certain chemical test specific for certain anion that gives pinpoint information about the salt.

First is the chromyl chloride test for chloride ions. Perform the test and you will identify potassium chloride (potassium bromate will give -ve result to the test). Second is conc. $\ce{H2SO4}$ test. Potassium chloride will gives vapors of hydrogen chloride. Potassium bromate will give bromine gas.

$$\ce{KCl + H2SO4 -> HCl ^ + KHSO4}$$
$$\ce{4KBrO3 + 2H2SO4 -> 2Br2 ^ + 2O2 ^ + 4K+ + 2SO4^{2-} + 2H2O}$$

Third test is the heat test. Potassium chloride is stable to heat while potassium bromate is prone to decomposition $\ce{\approx 370°C}$.

$$\ce{2KBrO3 ->[\Delta] 2KBr + 3O2 ^}$$

About the silver nitrate test pointed by @Dallon Penney, precipitation will occur in both cases albeit depending on the concentration of respective ions, it seems that the precipitate forming has different properties. Silver chloride exist as a white, curdy precipitate, insoluble in water and in dilute nitric acid while silver bromate exist as a white crystalline precipitate soluble in hot water and sparingly soluble in nitric acid. Both precipitates however are soluble in ammonia solution.


Strength of nucleophiles among SN1 and SN2 reactions

This seems to be true only because SN1 reactions prefer polar protic solvents which stabilize both the leaving group and the carbocation. But such solvents also surround strong nucleophile and hinder it from the carbocation.
Weak nucleophiles on the other hand don't interact with protic solvents as much and thus stand a better chance of getting close to carbocation.


I suggest to use a 1:1 mixture of dichloromethane and 0.2 M NaOH at room temperature.

The protocol was described by Tetsuo Onami and Hitoshi Kanazawa in J. Chem. Educ, 1996, 73, 556-557.

The authors mention that only trace amounts of caffeine are extracted from tea leaves with dichloromethane alone. They reason that due to the hight amount of tannines in the tea leaves, most of the caffeine exist in protonated form and is therefore bound to the tannines.

As an experimentally proven alternative, they suggest to treat the tea leaves with a mixture of dichloromethane and 0.2 M NaOH at room temperature!

According to the article, 20-30 mg of caffeine were extracted from a 2 g tea bag after shaking the above mixture for about ten minutes.

UPDATE
If you insist on the liquid-liquid extraction of previously prepared aqueous solutions and feel the urge to set up a percolator-type apparatus for continuous extraction, the Kutscher-Steudel glassware will not work with solvents that have a higher density than water.

For continuous extractions with solvents like $\ce{CH2Cl2}$, some different glassware has been designed and was described by S. Wehrli in Ein Apparat zur Extraktion von Lösungen mit schweren Lösungsmitteln and was published in Helv. Chim. Acta, 1937, 20, 927–931 (DOI)


How to determine pressure and concentration of carbon dioxide in blood plasma?

Since $p_\ce{CO2}$ is $\pu{30 mmHg}$, and we are given the $0.03$ of $\frac{\pu{mM}~\ce{CO2}}{\pu{mmHg}~\ce{CO2}}$ ratio, we can calculate the concentration of the $\ce{CO2}$ in plasma.

$$\pu{30mmHg}~\ce{CO2}\times 0.03\frac{\pu{mM}~\ce{CO2}}{\pu{mmHg}~\ce{CO2}}= \pu{0.9 mM}~\ce{CO2}$$

$$\ce{CO2 + H2O \xrightarrow{\text{carbonic anhydrase}} H2CO3}$$

$$\text{Thus, } \pu{0.9 mM}~\ce{CO2}= \pu{0.9 mM}~\ce{H2CO3}$$

$$\ce{H2CO3 <=> H+ + HCO3-}$$

$$\frac{\ce{[H+][HCO3-]}}{\ce{[H2CO3]}} = K_{\mathrm{a}_\ce{H2CO3}}$$

$$\log{\frac{\ce{[H+][HCO3-]}}{\ce{[H2CO3]}}}=\log{K_{\mathrm{a}_\ce{H2CO3}}}$$

$$\log{\ce{[H+]}}+\log{\frac{\ce{[HCO3-]}}{\ce{[H2CO3]}}}=\log{K_{\mathrm{a}_\ce{H2CO3}}}$$

$$-\left(\log{\ce{[H+]}}+\log{\frac{\ce{[HCO3-]}}{\ce{[H2CO3]}}}\right)=-\log{K_{\mathrm{a}_\ce{H2CO3}}}$$

$$\mathrm{pH}-\log{\frac{\ce{[HCO3-]}}{\ce{[H2CO3]}}}=\mathrm{p}K_{\mathrm{a}_\ce{H2CO3}}$$

$$\ce{[HCO3-]}=\pu{16 mM}$$

$$\ce{[H2CO3]}=\pu{0.9 mM}$$

$$ \mathrm{p}K_{\mathrm{a}_\ce{H2CO3}}=6.1$$

(PS: It is actually $\mathrm{p}K_\mathrm{a}$ of $\ce{H2CO3}$, not that of $\ce{HCO3-}$. Note that $\mathrm{p}K_\mathrm{a}$ of $\ce{HCO3-}$ is $10.3$.)

$$\mathrm{pH}-\log{\frac{\ce{[HCO3-]}}{\ce{[H2CO3]}}}=\mathrm{p}K_{\mathrm{a}_\ce{H2CO3}}$$
$$\mathrm{pH}-\log{\frac{16}{0.9}}=6.1$$

$$\mathrm{pH}-1.25=6.1$$

$$\mathrm{pH}=7.35$$

  The $\mathrm{pH}$ of blood is maintained within extremely narrow limits of 7.36-7.4 for venous blood and 7.38-7.42 for arterial blood. (Source: Upadhyay - Biophysical Chemistry)

So, we can say the $\mathrm{pH}$ of the blood of the patient is slightly acidic than normal.

I hope I've answered your question.


Pencil leads aren't pure graphite, they're generally graphite powder with various polymers to hold them together and make sure they have the correct hardness (after all, you wouldn't be able to use them to write if they didn't leave graphite on the paper or if they crumbled under slight pressure. This is why the much more rigid glassy carbon is a popular material for carbon electrodes. My guess is that there's no reaction with the electrode happening but gas being produced is causing mechanical damage to the electrode. If you work at less extreme potentials, you might be able to reduce this effect. Otherwise, pure graphite rods should be more durable. Of course, if you're using pencil leads, they're obviously cheap to replace so it might be easier to just use them until they fall apart, then switch to a new one.


Cyclohexane is used for this extraction because it is nonpolar, as is iodine. A more polar solvent like acetone or ethanol will also extract a whole bunch of other stuff like essential oils, proteins, and other polar compounds.

You want Mineral Spirits which you should be able to purchase at any quality hardware store or Walmart. Mineral spirits, mineral turpenine, or whatever you want to call it, is a mixture of saturated hydrocarbons distilled from petroleum. It is a general purpose paint thinner for oil-based paints. Since it is mainly C7-C12 alkanes, it will have a higher boiling point than cyclohexane $(140\ ^\circ\mathrm{C}$ vs $81\ ^\circ\mathrm{C})$., but it will have the same general polarity as cyclohexane.

If you can find products labeled as petroluem ether, light naphtha, or ligroin, all are alkane mixtures with lower boiling points than mineral spirits.

Another common low polarity paint thinner is toluene, which has a lower boiling point at $110\ ^\circ\mathrm{C}$.


How are phosphoglycerides different from phospholipids?

Phospholipids of a lipid bilayer can be divided in to two groups:

Those derived from glycerol -  phosphoglycerides.
Those derived from sphingosine -  sphingomyelin.

Phosphoglycerides are indeed a type of phospholipid.


How do I prepare 2-(3-azido-3-oxopropoxy)ethyl methacrylate from methacrylic acid?

You're going need a protection strategy to make this work. Here is one possible route:

Start with 2-Chloro/bromo ethanol and protect the OH with a silyl
ether, $\ce{SiMe3}$ will do, but requires anhydrous workup.
React the protected haloethanol with the anion of
3-hydroxypropionate-t-butyl ester (commercially available) ($\ce{NaH,
NaI, DMF}$) then deprotect the product (mild acidic work up for $\ce{Me3SiO}$
or $\ce{F-}$ for others).
React this product with the acid chloride of methacrylic acid
(commercially available). Isolate and purify the product.
Remove the t-butyl ester with $\ce{TFA/CH2Cl2}$, concentrate to dry and
form the acyl chloride ($\ce{SOCl2}$, cat DMF).  React this with sodium
azide to give the product following the first part of this prep
(Org. Syn.).


Why is ethyne and not ethene used for welding?

An acetylene molecule is composed of two carbon atoms and two hydrogen atoms. The two carbon atoms are held together by what is known as a triple carbon bond. This bond is useful in that it stores substantial energy that can be released as heat during combustion.The triple bond which makes the oxy-acetylene flame the hottest of all gas flames is also responsible for two rather exceptional properties of acetylene gas.

The first is: If free gaseous acetylene is subjected to severe shock, or a source of ignition, some of the triple bonds may break, releasing enough energy to cause all the other molecules in the enclosed volume to decompose into carbon and hydrogen with explosive force. The force of such an explosion is not so great as that released by the explosion of most mixtures of acetylene and oxygen, or acetylene and air, but it is substantial, and can be withstood only by extra-heavy-wall steel tubing.

The other property of acetylene is that the flammability range of mixtures of air and acetylene is broader than that of any other fuel gas/air mixture. Acetylene/air mixtures can be ignited when they contain anywhere from 2.5 percent acetylene to 80 percent acetylene. Mixtures of methane (the principal component of natural gas) and air are flammable when they contain as little as 5 percent methane and not more than 15 percent methane.

To be suitable for welding operations, a fuel gas, when burned with oxygen, must have the following:

High flame temperature.
High rate of flame propagation.
Adequate heat content.
Minimum chemical reaction of the flame with base and filler metals.

Among the commercially available Oxy Fuel welding fuel gases, acetylene (combination of hydrogen and carbon) most closely meets all these requirements. Other gases, fuel such as MAPP gas, propylene, propane, natural gas, and proprietary gases based on these, have sufficiently high flame temperatures but exhibit low flame propagation rates.


Is melting/boiling point of ionically bonded substance higher than of covalently bound?

The answer relates to the strength of the interactions between the component units that make up a crystal or a solid.

The reason why anything is a solid at a given temperature is, crudely, that the interactions between the units that make up the solid (atoms, ions or molecules) are stronger than the amount of thermal energy available at that temperature.

Diamond, for example, is a solid because each unit carbon is bonded to its neighbours by a strong carbon-carbon bond that takes a great deal of energy to break. Non-polar molecular compounds, like candle wax (in reality a mixture but imagine a pure long chain hydrocarbon to keep it simple) are held together by the intermolecular forces between the molecules (often called dispersion or Van der Waals forces). These are much weaker than chemical bonds and depend on the surface area of the molecules. So small non-polar or not-very polar molecules tend to have low melting points. Hexane is a volatile liquid. The shape and symmetry of the molecule matters a bit to these interactions so toluene is a liquid down to -95°C but benzene, which is very similar to toluene but far more symmetric and slightly flatter only melts at about 6°C as the molecules pack together better encouraging stronger interactions. Polar molecules have stronger interactions because the molecular dipoles interact and this produces stronger forces than the non-polar dispersion forces. So phenol (very similar in size to toluene but far more polar) melts at about 41°C.

Simple ionic solid are made from charged ions. Table salt crystals consist of Na+ and Cl- ions. The key interaction between these is electrostatic and this is a strong force that takes a lot of energy to break. The melting point is about 800°C. Permanent electrostatic interactions are strong especially if the components can get close (remember, it's an inverse square law). This explains why not every ionic compound is a solid. There are ionic compounds that are liquid at room temperature. These tend to consist of molecular ions that are large and sometime irregular so forcing structures where the ionic force has to act over much larger distances and, therefore, be much weaker. An example is [BMIM]PF6 which has a melting point well below room temperature:

These ionic liquids are an interesting area of research in chemistry as they sometimes enable much less wasteful chemical reactions in industry (see wikipedia).


Assay over 100% (L-Cysteine Hydrochloride Monohydrate)

There is a simple reason to have the purity greater than 100% for this compound.

If the substance was exposed to a dry environment for several hours, a small amount of the  water of hydration could be lost, causing the calculation to have a higher purity.

Also, I believe that the purity is calculated from the perchlorate titration of the substance in acetic acid (I will need to check the USP monograph). There are many sources for error in the calculations: the weight of the L-Cysteine Hydrochloride Monohydrate, the volume of the perchlorate solution, the error from the standardization of the perchlorate solution, etc.

Under the right conditions, you can easily get 101% for the purity.


Irreversible process (Isothermal and Adiabatic)

If you have an adiabatic reversible path that takes you from state 1 to state 2, any irreversible path from state 1 to state 2 will require removal of heat, so it won't be adiabatic.  This is a consequence of the 2nd law of thermodynamics, captured by the Clausius Inequality.


Why can't ligands with "locked structures" undergo chelation?

My copy of Concise Inorganic Chemistry, 5th edition (J.D.Lee, Wiley publishers)1 has nothing contrary to say about hydrazine or 1,4-diazabicyclo[2.2. 2]octane (henceforth referred to as DABCO). It is most likely that the author of your adaptation has used his liberty to add these facts in your copy.
This then raises the question: is the author correct? Hydrazine derivatives such as Benzhydrazides and Salicylhydrazides 2 have been proven to show chelation. Pure hydrazine, however, is unlikely to show chelation due to excessive ring strain. Also, no literature has been found which shows hydrazine undergoing chelation. DABCO, on the other hand, forms complexes with Mercury halides3 with a denticity of 1. This ligand also does not show chelation, and fitting a large metal atom even off-center is highly unlikely due to large steric hindrance and the geometry of the lone pair, which is perpendicular to the structure and not favourable for chelation. Most complexes formed by DABCO tend to be linear in nature4 and no rings are formed.
In short, the author is correct in stating that hydrazine and DABCO do not form chelates.
EDIT: You may also want to read more on a class of compounds called Cryptands, specifically Cryptand-222. This compound displays the 'cagelike' ligation you were looking for in DABCO.
References:

Lee, J. D. Concise Inorganic Chemistry. 5th ed, Wiley publishers &
Blackwell Science, 1996.
Issa, R. M., et al. Coordination compounds of hydrazine derivatives
with transition metals. I. Metal Chelates with Benzhydrazide and
Salicylhydrazide. Zeitschrift fur anorganische und allgemeine
Chemie, vol. 354, no. 1–2, Sept. 1967, pp. 90–97.
doi:10.1002/zaac.19673540118.
Shan, Zeng-Mei, et al. Synthesis, Crystal Structures, and
Characterization of Three Mercury(II) Halides Inorganic–Organic
Hybrid Compounds with 1,4-Diazabicyclo[2.2.2]Octane Ligand.
Inorganica Chimica Acta, vol. 366, no. 1, Jan. 2011, pp. 141–46.
doi:10.1016/j.ica.2010.10.023.
Thorp-Greenwood, Flora L., et al. Three-Dimensional Silver-Dabco
Coordination Polymers with Zeolitic or Three-Connected Topology.
Crystal Growth & Design, vol. 14, no. 11, Nov. 2014, pp. 5361–65.
doi:10.1021/cg501231v.


Finding change in enthalpy for a reaction with an atom in elemental state using bond energy

There are two things we need to note with this. For one, you need to know the bond enthalpies if you are going to obtain $\Delta H$, not the bond energies. Second, we can't just use the bond enthalpies in this case because that would ignore that we have $C(s)$ when we need $C(g)$ to make this reaction occur. We have to explicitly include the atomization energy of carbon (the energy needed to break graphite into single carbon atoms).


Molar concentration calculation with absolute uncertainty

Generally, it is preferable to write down the complete quantity equation before plugging in the numbers. If necessary, perform all mathematic operations on quantities symbolically. Do not plug in the numbers until you have only one equation for the desired result. The same applies to the equation for the uncertainty of the desired result.

In this case, the concentration $c$ shall be calculated. The mass $m=0.0933\ \mathrm g$, the molar mass $M=41.989\ \mathrm{g\ mol^{-1}}$, and the volume $V=156.00\ \mathrm{ml}$ are given. The uncertainty of the mass $m$ and the molar mass $M$ are given as $u(m)=0.0004\ \mathrm g$ and $u(M)=0.001\ \mathrm{g\ mol^{-1}}$, respectively. The uncertainty of the volume $V$ is not given; however, when a number is given without any further information, it is generally interpreted so that the last digit is rounded. Therefore, the given value $V=156.00\ \mathrm{ml}$ is assumed to represent a value between $155.995\ \mathrm{ml}$ and $156.005\ \mathrm{ml}$, or $V=\left(156.000\pm0.005\right)\ \mathrm{ml}$. Thus, the assumed uncertainty of $V$ is $u(V)=0.005\ \mathrm{ml}$.

The concentration $c$ is defined as
$$c=\frac nV\tag1$$
where $n$ is amount of substance, which is unknown.
Since the molar mass $M$ is defined as
$$M=\frac mn\tag2$$
the concentration $c$ can be calculated according to
$$\begin{align}c&=\frac m{M\cdot V}\tag3\\[6pt]
&=\frac {0.0933\ \mathrm g}{41.989\ \mathrm{g\ mol^{-1}}\times 156.00\ \mathrm{ml}}\\[6pt]
&=1.42437\times10^{-5}\ \mathrm{mol\ ml^{-1}}\\[6pt]
&=0.0142437\ \mathrm{mol\ l^{-1}}
\end{align}$$

In this case, we may assume that the uncertainties of $m$, $M$ and $V$ are not correlated. Therefore, the relative uncertainty of the concentration $c$ may be estimated as
$$\begin{align}
\left(\frac{u(c)}{c}\right)^2 &= \left(\frac{u(m)}{m}\right)^2 + \left(\frac{u(M)}{M}\right)^2 + \left(\frac{u(V)}{V}\right)^2\tag4 \\[6pt]
\frac{u(c)}{c} &= \sqrt{\left(\frac{u(m)}{m}\right)^2 + \left(\frac{u(M)}{M}\right)^2 + \left(\frac{u(V)}{V}\right)^2}\tag5
\end{align}$$

Thus, the absolute uncertainty of the concentration $c$ is

$$\begin{align}
u(c) &= \sqrt{\left(\frac{u(m)}{m}\right)^2 + \left(\frac{u(M)}{M}\right)^2 + \left(\frac{u(V)}{V}\right)^2} \cdot c\tag6 \\[6pt]
&= \sqrt{\left(\frac{u(m)}{m}\right)^2 + \left(\frac{u(M)}{M}\right)^2 + \left(\frac{u(V)}{V}\right)^2} \cdot \frac m{M\cdot V}\tag7 \\[6pt]
&= \sqrt{\left(\frac{0.0004\ \mathrm g}{0.0933\ \mathrm g}\right)^2 + \left(\frac{0.001\ \mathrm{g\ mol^{-1}}}{41.989\ \mathrm{g\ mol^{-1}}}\right)^2 + \left(\frac{0.005\ \mathrm{ml}}{156.00\ \mathrm{ml}}\right)^2} \cdot \frac {0.0933\ \mathrm g}{41.989\ \mathrm{g\ mol^{-1}}\times 156.00\ \mathrm{ml}} \\[6pt]
&=6.10687\times10^{-8}\ \mathrm{mol\ ml^{-1}}\\[6pt]
&\approx0.00006\ \mathrm{mol\ l^{-1}}
\end{align}$$

Note that the same result may be obtained using the general formula

$$\begin{align}
u(c) &= \sqrt{\left(\frac{\partial c}{\partial m}\right)^2\cdot u^2(m)+\left(\frac{\partial c}{\partial M}\right)^2\cdot u^2(M)+\left(\frac{\partial c}{\partial V}\right)^2\cdot u^2(V)}\tag8 \\[6pt]
&= \sqrt{\left(\frac1{M\cdot V}\right)^2\cdot u^2(m)+\left(-\frac m{M^2\cdot V}\right)^2\cdot u^2(M)+\left(-\frac m{M\cdot V^2}\right)^2\cdot u^2(V)}\tag9 \\[6pt]
&= \sqrt{\left(\frac1{41.989\ \mathrm{g\ mol^{-1}}\times 156.00\ \mathrm{ml}}\right)^2\times \left(0.0004\ \mathrm g\right)^2\\
+\left(-\frac {0.0933\ \mathrm g}{\left(41.989\ \mathrm{g\ mol^{-1}}\right)^2\times 156.00\ \mathrm{ml}}\right)^2\times \left(0.001\ \mathrm{g\ mol^{-1}}\right)^2\\
+\left(-\frac {0.0933\ \mathrm g}{41.989\ \mathrm{g\ mol^{-1}}\times \left(156.00\ \mathrm{ml}\right)^2}\right)^2\times \left(0.005\ \mathrm{ml}\right)^2}\\[6pt]
&=6.10687\times10^{-8}\ \mathrm{mol\ ml^{-1}}\\[6pt]
&\approx0.00006\ \mathrm{mol\ l^{-1}}
\end{align}$$

The numerical values of the concentration $c$ and its uncertainty $u(c)$ should not be given with an excessive number of digits. In this case, it suffices to quote $u(c)=0.00006\ \mathrm{mol\ l^{-1}}$ and accordingly $c=0.01424\ \mathrm{mol\ l^{-1}}$; i.e. $c=\left(0.01424\pm0.00006\right)\ \mathrm{mol\ l^{-1}}$ or $c=0.01424(6)\ \mathrm{mol\ l^{-1}}$. In some cases, however, it may be necessary to retain additional digits to avoid round-off errors in subsequent calculations.

Note

The large equations for the uncertainty $u(c)$ may be confusing and susceptible to errors.

To verify the result, you should check whether the calculated quantity and its absolute uncertainty have the same unit (here: $\left[c\right]=\left[u(c)\right]=\mathrm{mol\ ml^{-1}}$).

Furthermore, you should compare the relative uncertainty of the result with the relative uncertainty of the input data. Usually, the relative uncertainty of the result is greater than or equal to the relative uncertainty of the input data. In this case:

$$\frac{u(m)}{m}=\frac{0.0004\ \mathrm g}{0.0933\ \mathrm g}=0.0042872\approx0.43\ \%$$

$$\frac{u(M)}{M}=\frac{0.001\ \mathrm{g\ mol^{-1}}}{41.989\ \mathrm{g\ mol^{-1}}}=2.38158\times10^{-5}\approx0.0024\ \%$$

$$\frac{u(V)}{V}=\frac{0.005\ \mathrm{ml}}{156.00\ \mathrm{ml}}=3.20513\times10^{-5}\approx0.0032\ \%$$

$$\frac{u(c)}{c}=\frac{6.10687\times10^{-8}\ \mathrm{mol\ ml^{-1}}}{1.42437\times10^{-5}\ \mathrm{mol\ ml^{-1}}}=0.0042874\approx0.43\ \%$$

We find that, in this case, the uncertainty of the concentration $c$ is mainly caused by the uncertainty of the mass $m$.


How to rationalise the basicity order of phenylenediamine?

As Mithoron pointed out in a comment, we had a similar question convering the cases of 2-methylaniline (o-toluidine) and 2-methylbenzoic acid (o-toluic acid) here.

In the case of 1,2-diaminobenzene (o-phenylenediamine), it is conceivable that hydrogen bonding between the two amino groups reduces the electron density on one of the nitrogen atoms, thereby leading to a slight increase on the other. As a result, o-phenylenediamine is significantly less basic than the m- and p-isomer, but a tad more basic than aniline.


Calculating dissociation degree from pH and Ka without given initial acid concentration

Even though the degree of dissociation $α$ depends both on the nature of the dissolved electrolyte (e.g. acid) and the concentration, for the homogeneous medium it is possible to determine $α$ by $\mathrm{pH}$ and $\mathrm{p}K_\mathrm{a}$ only, without any auxiliary information such as initial concentration since $\mathrm{pH}$ is a function of concentration.

I would suggest you not to remember certain formulas, since they have been derived only for particular systems and cannot be universally applied. Instead, use the law of mass action, especially when it comes to multi-component systems and multi-step dissociating electrolytes.

The following is heavily inspired by the German Wikipedia article for the degree of dissociation. Consider all $n$ steps of the dissociation of the $n$-protic acid $\ce{H_nA}$:

$$
\begin{align}
\ce{H_nA + H2O &<=> H_{n-1}A- + H3O+}&\quad &K_{\mathrm{a}1} \tag{1}\\
\ce{H_{n-1}A- + H2O &<=> H_{n-2}A^2- + H3O+}&\quad &K_{\mathrm{a}2} \tag{2}\\
&\ldots &&\ldots\\
\ce{H_{n-(m-1)}A + H2O &<=> H_{n-m}A^m- + H3O+}&\quad &K_{\mathrm{a}m} \tag{$m$}\\
&\ldots &&\ldots\\
\ce{H2A^{(n-2)-} + H2O &<=> HA^{(n-1)-} + H3O+}&\quad &K_{\mathrm{a}(n-1)} \tag{$n-1$}\\
\ce{HA^{(n-1)-} + H2O &<=> A^n- + H3O+}&\quad &K_{\mathrm{a}n} \tag{$n$}
\end{align}
$$

where $n\in\mathbb{N}$, $m\in\mathbb{N}$, $m < n$; $K_{\mathrm{a}m}$ – $m$-th dissociation constant. Dissociation degree $α$ for any component of the composition $\ce{H_{n-m}A^m-}$ formed at the $m$-th step is by definition

$$α_m = \frac{[\ce{H_{n-m}A^m-}]}{c_0}$$

where $c_0$ – initial concentration:

$$
\begin{align}
c_0 &= [\ce{H_nA}] + [\ce{H_{n-1}A-}] + \ldots + [\ce{H_{n-m}A^m-}] + \ldots + [\ce{HA^{(n-1)-}}] + [\ce{A^n-}]\\
  &= \sum_{m=0}^{n}{[\ce{H_{n-m}A^m-}]}
\end{align}
$$

Each concentration in this set can be found recursively from the dissociation constant:

$$K_{\mathrm{a}m} = \frac{[\ce{H_{n-m}A^m-}][\ce{H3O+}]}{[\ce{H_{n-(m-1)}A^{(m-1)-}}]}$$

$$\implies$$

$$
\begin{align}
[\ce{H_{n-m}A^m-}] &= \frac{K_{\mathrm{a}m}[\ce{H_{n-(m-1)}A^{(m-1)-}}]}{[\ce{H3O+}]}\\
  &= \frac{K_{\mathrm{a}m}K_{\mathrm{a}(m-1)}[\ce{H_{n-(m-2)}A^{(m-2)-}}]}{[\ce{H3O+}]^2}\\
  &\ldots\\
  &= \frac{\prod_{i=0}^m{K_{\mathrm{a}i}}[\ce{H_{n-m}A^{m-}}]}{[\ce{H3O+}]^{n-m}}
\end{align}
$$

so that for $α_m$ after rearranging sums of the products the following universal equation  of two arguments ($K_\mathrm{a}$ and $\mathrm{pH}$) emerges:

$$α_m = \frac{[\ce{H3O+}]^{n-m}}{[\ce{H3O+}]^n + K_{\mathrm{a}1}[\ce{H3O+}]^{n-1} + K_{\mathrm{a}1}K_{\mathrm{a}2}[\ce{H3O+}]^{n-2} + K_{\mathrm{a}1}K_{\mathrm{a}2}\cdot\ldots\cdot K_{\mathrm{a}n}}\cdot\prod_{i=0}^{m}{K_{\mathrm{a}i}}$$

For the monoprotic acid $\ce{HA}$ ($n = 1$, $m=0$) the formula becomes much simpler:

$$α_0 = \frac{K_{\mathrm{a}}}{K_{\mathrm{a}} + [\ce{H3O+}]} = \frac{K_{\mathrm{a}}}{K_{\mathrm{a}} + 10^{-\mathrm{pH}}},$$

and if the acid is very weak ($K_{\mathrm{a}} \ll 10^{-\mathrm{pH}}$, e.g. $K_{\mathrm{a}} \ll \text{ca.}~10^{-7}$), only then you can use your formula:

$$α_0 \approx \frac{K_{\mathrm{a}}}{10^{-\mathrm{pH}}}$$


Enthalpy, heat of solution test: Calcium chloride vs magnesium chloride in water

It may be caused by the real content of hydrate water in the used calcium and magnesium chlorides. As there is possible disagreement between declared and true water content, due long term reaching equilibrium with air water vapour. I have once seen a glass bottle with reportedly solid $\ce{LiCl}$, sitting on an shelf for years. The content was a clear liquid of LiCl solution.(*)
Both chlorides form multiple hydrates, stable at different humidity or temperature conditions. Generally, with increasing of crystal/hydrate water content, the dissolution enthalpy raises for both salts from negative to positive values. E.g. anhydrous $\ce{CaCl2}$ warms up water when dissolved, $\ce{CaCl2 . 6 H2O}$ cools water down significantly. Note that both anhydrous chlorides are highly hygroscopic.
I suggest to make repetition of the experiments with salts of different origin and short history. Or, heating them well in an oven to get anhydrous stuff.

(*) This hygroscopicity of $\ce{LiCl(s)}$ is used in some humidity meters, measuring equilibrium electric resistence of a soaked fabric.


Balancing and combining the half equations in redox

Assuming this reaction is taking place in aqueous phase, you can follow the conventional 7 steps for balancing redox reactions although this one does require some extra thoughts.
I will be writing the equation as it should be after each step.

$$\ce{H2 + NO -> NH3 + H2O}$$

Step 1: Ionise the required compounds and remove spectator ions

None of the compounds on either side are ionic other than $\small\ce{H2O}$ also we have no spectator ions as well.*
$$\small\ce{H2 + NO -> NH3 + H+ + OH-}$$
*-Spectator ions are ions which don't participate in the reaction or retain their oxidation state.

Step 2: Split into Oxidation and reduction halves

This is where the thought is required, if you decide of taking the oxidation half as $\ce{H2 -> NH3}$  (you can't simply take $\ce{H3^3-}$ as $\ce{NH3}$ is not ionic rather covalent) you would never be able to balance the nitrogen as the only source of nitrogen is $\ce{NO}$ if you observe the left hand side. Also, since you know that Nitrogen is being reduced (+2 -> -3) the other element, Hydrogen must get oxidised (0 -> +1).
So it is decided that hydrogen cannot convert into ammonia, leaving us with only one option $\ce{H2 -> H2O}$.

$$
\begin{array}{l|r}
\text{Oxidation Half} & \text{Reduction half} \\
\hline
\ce{H2 -> H+ + OH-} & \ce{NO -> NH3}
\end{array}
$$

Step 3: Balance only those atoms undergoing redox

$$
\begin{array}{l|r}
\text{Oxidation Half} & \text{Reduction half} \\
\hline
\ce{H2 -> H+ + OH-} & \ce{NO -> NH3} \\
\text{(because Hydrogen is } & \text{(because only Nitrogen } \\
\text{undergoing redox but} & \text{is undergoing redox,} \\
\text{Oxygen is not.)} & \text{it's already balanced)} \\
\end{array}
$$

Step 4: Balance Oxygen by adding water($\ \small\ce{H2O}\ $)

$$
\begin{array}{l|r}
\text{Oxidation Half} & \text{Reduction half} \\
\hline
\ce{H2 + H2O -> H+ + OH-} & \ce{NO -> NH3 + H2O}
\end{array}
$$

Step 5a: Balance Hydrogen by adding $\ \small\ce{H+}$

$$
\begin{array}{l|r}
\text{Oxidation Half} & \text{Reduction half} \\
\hline
\ce{H2 + H2O -> H+ + OH- + 2H+} & \ce{NO + 5H+ -> NH3 + H2O} \\
\ce{H2 + H2O -> [H+ + OH- ] + 2H+} & \ce{NO + 5H+ -> NH3 + H2O} \\
\ce{H2 + H2O -> \qquad H2O \qquad + 2H+} & \ce{NO + 5H+ -> NH3 + H2O} \\
\hline
\ce{H2 + \qquad-> \qquad \qquad \qquad \quad 2H+} & \ce{NO + 5H+ -> NH3 + H2O}\\
\hline
\end{array}
$$

Step 5b: (Only in case of basic medium) Adding $\ \small\ce{OH-}$ to $\ \small\ce{H+}$

This step must follow Step 5a and means nothing on it's own.
$$
\begin{array}{l|r}
\text{Oxidation Half} & \text{Reduction half} \\
\hline
\ce{H2 \qquad -> 2H+} & \ce{NO + 5H+ \qquad \qquad -> NH3 + H2O \qquad \qquad} \\
\ce{H2 + 2OH- -> 2H+ + 2OH-} & \ce{NO + 5H+ + 5OH- -> NH3 + H2O + 5OH-} \\
\ce{H2 + 2OH- -> [2H+ + 2OH- ]} & \ce{NO + [5H+ + 5OH- ]-> NH3 + H2O + 5OH-} \\
\ce{H2 + 2OH- -> \quad \quad 2H2O} & \ce{NO + \qquad 5H2O \qquad -> NH3 + H2O + 5OH-} \\
\hline
\ce{H2 + 2OH- -> \quad \quad 2H2O} & \ce{NO + \qquad 4H2O \qquad -> NH3 + 5OH- \quad \qquad  }\\
\hline
\end{array}
$$

The medium being acidic or basic merely determines whether $\ \small\ce{H+}$ or $\ \small\ce{OH-}$ will be seen in the final redox equation.

Till the above steps mass has been balanced, now charge balancing.

Step 6: Balance charge in each side of each half by adding electrons

$$
\begin{array}{l|r}
\text{Oxidation Half} & \text{Reduction half}  \\
\hline
(acidic)\quad \ce{H2  -> 2H+ + 2e-} & \ce{NO + 5H+ + 5e- -> NH3 + H2O} \\
\hline
(basic) \quad \ce{H2 + 2OH- -> 2H2O + 2e-} & \ce{NO + 4H2O + 5e- -> NH3 + 5OH-} \\
\hline
\end{array}
$$

→ Verify correctness by ensuring electrons are on right hand side in oxidation half and on left hand side in reduction half.

→ Also verify if net change in oxidation state is equal to the number of electrons appearing in each half.
eg. Hydrogen is going from 0 to +1 oxidation state and there are 2 Hydrogen atoms in the left side of the oxidation half undergoing oxidation(don't count $\small\ce{H}$ from $\small\ce{OH-}$), meaning a net change of '2', which is equal to the number of electrons.
Similarly Nitrogen is going from +2 to -3 and there is just one Nitrogen atom in the left side of the reduction half meaning a net change of '5' which is equal to the number of electrons appearing in the equation. (This is what goes on to be later called as the 'n-factor' or the 'valency-factor')

Step 7: Make 'net electrons produced' equal to 'net electrons consumed'

Multiply oxidation half by 5 and reduction half by 2 (this is basically like taking the LCM and multiplying both halves to make the number of electrons equal to the LCM, here 10.)

$$
\begin{array}{l|r}
\text{Oxidation Half} & \text{Reduction half} \\
\hline
(acidic) \ce{5H2 -> 10H+ + 10e-} & \ce{2NO + 10H+ + 10e- -> 2NH3 + 2H2O}\\
\hline
(basic) \ce{5H2 + 10OH- -> 10H2O + 10e-} & \ce{2NO + 8H2O + 10e- -> 2NH3 + 10OH-} \\
\hline
\end{array}
$$

Finally add the two halves and cancel out the common cmopounds and you get the balanced equation
Acidic Medium
$$\begin{matrix}
\ce{&5H2 + &2NO + &10H+ + &10e- &-> &10H+ + &10e- + &2NH3 + &2H2O} \\
\ce{&5H2 + &2NO & & &-> & & &2NH3 + &2H2O}\\
\end{matrix}$$

Basic Medium
$$\begin{matrix}
\ce{&5H2 + &10OH- + &2NO + &8H2O + &10e- &-> &10H2O + &10e- + &2NH3- + &10OH-} \\
\ce{&5H2 + & &2NO & & &-> &2H2O + & &2NH3}\\
\end{matrix}$$

In this case the oxidation half could have been directly/trivially stated as $\small\ce{H2 -> 2H+}$

You can also carry on $\small\ce{H+}$ ion till the end net reaction and then perform Step 5b which is less tedious.


Is the reaction between a carboxylic acid and alcohol endothermic?

When in doubt, go find data. Here is a typical such reaction - the esterification of ethanol and acetic acid to make ethyl acetate.

$$\ce{CH3CH2OH + CH3CO2H -> CH3CO2CH2CH3 + H2O}$$

The following thermodynamic values are from the NIST Chemistry Webbook.

$$\begin{array}{|c|c|}\hline
\text{compound} & \Delta _\pu{f} H^\circ \ (\pu{kJ mol^-1}) \\ \hline
\text{ethanol} & -276 \\
\text{acetic acid} & -484 \\
\text{ethyl acetate} & -480 \\
\text{water} & -286 \\ \hline \end{array}$$

Now for some Hess's Law:

$$\begin{aligned}
\Delta_\pu{r} H^\circ &= \sum_\mathrm{products} \Delta_\pu{f} H^\circ - \sum_\mathrm{reactants} \Delta_\pu{f} H^\circ \\
&= (-480\ \pu{kJ mol^-1} + -286\ \pu{kJ mol^-1}) - (-276\ \pu{kJ mol^-1} + -484\ \pu{kJ mol^-1}) \\
&= -766\ \pu{kJ mol^-1} - (-760\ \pu{kJ mol^-1}) \\
&= -6\ \pu{kJ mol^-1}\end{aligned}$$

Hey! This one is exothermic (slightly).

Here are the enthalpies of reaction for some other esters (by no means a complete or even representative list):
$$\begin{array}{|c|c|}\hline
\mathrm{ester} & \Delta_\pu{r} H^\circ \ (\pu{kJ mol^-1}) \\ \hline
\text{methyl acetate} & -9 \\
\text{ethyl acetate} & -6\\
\text{methyl benzoate} & -5 \\
\text{ethyl crotonate} & +1 \\ \hline \end{array}$$

I would thus guess that formation of most carboxylic esters from carboxylic acids and alcohols would be slightly exothermic or slightly endothermic depending on structure.

The reasons esterifications are often heated are 1) to increase the rate of reaction and 2) to distill off water as it forms.


The rate of solvolysis is highest in (C) as a 6 membered ring is formed which is more stable than 4 & 5 membered rings.

And (B) more readily formed than (D) because the NGP ability of oxygen is more than that of an alkene(due to high charge density of oxygen).

So the correct order is C>B>D>A.


Find mass of CO₂ and heat released per minute in a combustion reaction

You know the amounts of the four gases burned in a minute.  Just write combustion reactions for each of these and work out the heat produced by each reaction as written. That's your answer to the second part.  Adding up the carbon dioxide produced in each of the four combustion reactions is the answer to the first part.


How do salt additives and impurities affect a critical micelle concentration?

In a solution with two dissociated ionic detergent molecules of equal charge (say two anionic detergent molecules $\ce{A^-}$), their interaction will be dominated by two effects:

Electrostatic headgroup repulsion
Hydrophobic interactions

The first contribution, electrostatic repulsion, will oppose the second and discourage association of the detergent molecules.

If you instead mix two oppositely charged detergents, $\ce{A^-}$ and $\ce{B^+}$, the electrostatic headgroup interaction will now encourage association. All else (e.g. tail length) being equal, this means mixed micelles will have a lower critical micelle concentration (CMC) than pure detergent at the same total concentration.

Adding salt (say $\ce{NaCl}$) to a charged detergent will also lower the CMC, for essentially the same reason as for mixed micelles (electrostatic repulsion is reduced).

Now comes the tricky part. If you add salt to a mixed detergent system, you set up possible competition between electrostatic attraction of salt and headgroups, with electrostatic attraction of oppositely charged headgroups. Individually, either attractive interaction has a tendency to lower the CMC. The question is what effect they will have together. My educated guess is that the effect of salt on mixed micelle systems will be subtle and complicated. The CMC will be further reduced, but interactions between oppositely charged detergent molecules will predominate so the effect of added salt will be minor.

  When you lower a CMC, are the micelles themselves now smaller or bigger? And when the literature describes micelles as smaller, does this mean the circumference is smaller, or the number of detergent molecules is smaller?

I don't think there is a universal answer to the first question.  I will have to wave my hands a little, and say it depends on the detergent, solvent, and temperature. However if you look at a table of aggregation numbers and CMCs for different detergents, you'll usually see that the ones with a lower CMC (for instance, nonionic detergents) also have larger aggregation numbers.

The second question is easier to answer than the first.  The size of a micelle can be measured by a number of different techniques. The aggregation number is often derived by making assumptions about packing in the micelle and starts from estimates of the sizes of the micelle and detergent molecule. So it's pretty safe to generalize that both the dimensions of the micelle (the effective diameter say) and aggregation number are reduced in smaller micelles.


To answer this question at a more relatable level, typically the strength of oxoacids can be determined from the number of "lone" oxygen atoms in the compound--that is, the number of oxygen atoms that do not have a corresponding hydrogen. What usually happens is the central atom in an oxoacid will draw electrons from the bonded oxygen atoms, thereby making them more electronegative. This increases the polarity of the O-H bond, increasing acidity. In your example, H$_2$SO$_4$ has 2 lone oxygen atoms; while in H$_2$S$_2$O$_7$, 5 lone oxygens exist. We therefore suspect that the latter is more acidic. Although there are exceptions to this rule, for a general chemistry course it typically is valid.


Quantum calculation of free energy change in neat conditions

One needs to distinguish between the standard free energy (of reaction) for a process, $\Delta _r G^\circ$, and the free energy in general, $\Delta _r G$. You show how to use quantum methods to compute $\Delta _r G^\circ$. This tells you how much the free energy changes if you take 1 mole of each reagent at 1 M concentration and generate one mole of complex at 1 M concentration. At other concentrations one uses
$$\Delta_r G = \Delta_r G^0 + RT \ln\left(\frac{C_{AB}/C^\circ}{(C_A/C^\circ )(C_B/C^\circ)}\right) \tag{Eq. 1}$$
for the reaction $\ce{A +B<=>AB}$ (assuming all components behave ideally; $\pu{C^\circ= 1 M}$ is the standard concentration). In fact, you use an analogous expression to compute the free energy difference between 1 atm (at $\pu{0^\circ C}$) and 1 M concentrations for each reagent or product:
$$\begin{align}\Delta_r G(\pu{1 atm}\rightarrow\pu{1 M}) &= RT \ln\left(\frac{C_X(\pu{1 M})}{C_X(\pu{1 atm})}\right)\\ &= RT \ln\left(\frac{\pu{1 M}}{(\pu{1e-3 \pu{L/m^3}})\cdot(1/(RT/\pu{1 atm})}\right)\end{align} $$
You use analogous relations (usually in the form of Eq. (1)) if you intend to compute the free energy change for one mole of complex formation when reagents and products are at particular concentrations (information which you then have to provide).
If you are interested in computing the equilibrium concentration of complex given specified concentrations of A and B (assumed constant) then you solve for $C_{AB}$ in Eq. (1) after plugging in values for $C_A$ and $C_B$, remembering again that you are assuming your solution behaves ideally.


I would suggest:

1) Introduction to Chemical Engineering Thermodynamics by Smith, Van Ness, and Abbott;

2) Chemical, Biochemical, and Engineering Thermodynamics by Sandler; and

3) This is more suitable for advanced reading or for a graduate student: Thermodynamics and Its Applications by Tester and Modell.


Electrochemical concentration cells -- how anion levels change

Considering you have chosen a $\ce{Zn}$ concentration cell:
The cell representation is:
$$\ce{Zn(s)/ZnSO4(M1)||ZnSO4(M2)/Zn(s)}$$

Anodic reaction puts more $\ce{Zn^2+}$ ions in the solution:
$$\ce{Zn(s) -> Zn^2+(aq) + 2e-}$$
To balance these extra $\ce{Zn^2+}$ ions salt bridge pulls some anions into anodic compartment.

Cathodic reaction removes $\ce{Zn^2+}$ ions from cathodic compartment:
$$\ce{Zn^2+(aq) + 2e- -> Zn(s)}$$
To make up this equivalent amounts of cations are pulled off from the salt bridge.

So activity of $\ce{SO4^2-}$ is not changed in either of the solutions. It is only those of $\ce{Zn^2+}$ ions in both the compartments, which undergoes a change leading to redox reaction.


On the assignment of axial and equatorial to substiuents of polyhexacyclics

The structure you have drawn where X = Y = Z = H is known as adamantane, C10H16. It has 3 degrees of unsaturation but visually appears to have 4 rings. Once 3 rings have been defined (rings A, B and C), then all C-C bonds have been accounted for. Put another way, you only need to erase three C-C bonds and add hydrogen to obtain a saturated, acyclic hydrocarbon, C10H22.
In structure 1 ring A is highlighted in red with X and Y equatorial to ring A. In structure 2 substituents X and Z are both axial to ring B. Structure 3 has the configuration of the Z substituent changed to equatorial relative to ring B. Ring C bears the same relationship to ring A as ring B does to ring A. Therefore in structure 4, substituent Y is axial to ring C. Now locate ring "D" and define equatorial/axial relationships of the Z substituent to ring "D" in structures 1 - 4.


Selective reduction of nitro group to amine, in benzene ring containing nitrile?

My answer is intended to support Waylander's suggestion elsewhere. I thought it'd be better to give OP some insight of this reduction mentioned (Ref.1). To my knowledge, after doing thorough literature search, the aromatic nitro group reduction by $\ce{SnCl2.2H2O}$ is the best method so far in the presence of other sensitive  groups such as nitrile group on the aromatic nucleus. The abstract of the reference 1 stated that:

  Aromatic nitro compounds are readily reduced by $\ce{SnCl2.2H2O}$ in alcohol or ethyl acetate or by anhydrous $\ce{SnCl2}$ in alcohol where other reducible or acid sensitive groups such as aldehyde, ketone, ester, cyano, halogen, and O-benzyl remain unaffected.

This point is depicted in following diagram as two examples given in Ref.1:

However, as shown in the separate box in the diagram (under the title, Stephen reduction of nitriles), if you use $\ce{HCl}$ (hydrogen chloride gas) with $\ce{SnCl2}$ as other suggestions, the result would be different (Ref.2). $\ce{HCl}$ is added to the nitrile group to give the imidoyl chloride salt first, which is then reduced by the tin(II) chloride. The initial product is the complex with $\ce{SnCl4}$ (which is often crystalline) that is then hydrolyzed to the corresponding aldehyde (see the insert in the diagram).

For the convenience of OP and other readers, I'd include a typical reduction procedure in Ref.1:

  The reduction of p-nitrobenzoic acid: A mixture of $\pu{1.67 g}$ ($\pu{0.01 mol}$) of p-nitrobenzoic acid and $\pu{11.275 g}$ ($\pu{0.05 mol}$) of $\ce{SnCl2.2H2O}$ in $\pu{20 mL}$ of absolute ethanol is heated at $\pu{70 ^\circ C}$ under nitrogen. After $\pu{30 min}$ the starting material has disappeared and the solution is allowed to cool down and then poured into ice. The $\mathrm{pH}$ is made slightly basic ($\mathrm{pH}\ 7-8$) by addition of 5% aqueous sodium bicarbonate before being extracted with ethyl acetate. The organic phase is thoroughly washed with brine, treated with charcoal and dried over sodium sulfate. Evaporation of the solvent leaves $\pu{1.5 g}$ ($94.5\%$) of p-aminobenzoic acid, which gives one spot on TLC and melts over $\pu{300 ^\circ C}$.

References:

F. D. Bellamy, K. Ou, “Selective reduction of aromatic nitro compounds with stannous chloride in non acidic and non aqueous medium,” Tetrahedron Letters 1984, 25(8), 839-842 (https://doi.org/10.1016/S0040-4039(01)80041-1).
James A. Knight, Harry D. Zook, “Reduction of Aliphatic Nitriles by the Stephen Reaction,” J. Am. Chem. Soc. 1952, 74(18), 4560-4562 (https://doi.org/10.1021/ja01138a031).


does Sodium persulfate evaporate at room temperature

Regardless of evaporation, it is an oxidizer and unhealthy so it is recommended to keep it in a closed container.

With that being said, you most likely have a solution of sodium persulfate in water.  Water will evaporate so again, yes, keep it closed.


Conformational analysis determining forces

what is meant with steric interaction
isn't this the same as electron repulsion?

I agree with you Jori, the basis of steric interactions is electron repulsion.  When anyone stops to think about "why are adjacent bulky groups destabilizing" they'll soon arrive at the destabilizing interpenetration of electron clouds.  However, the use of the term "steric" to denote interactions due to size and bulk has been around so long that it has simply become common usage - the way things are said. The following figure shows trans-1,2-di-t-butylethylene to be more stable than its cis counterpart.  It's easier to say and understand that the bulky t-butyl groups produce steric strain in the cis isomer, rather than to say electronic clouds interpenetrate in the cis isomer.  The word "steric" immediately creates a picture of sorts in a chemist's mind.  Tetra-t-butylethylene has yet to be synthesized; attributing this to steric bulk around the double bond provides a clear image of the problem.

Then the last part of bonding and anti bonding orbital interaction
(overlap) stabilizing the staggered conformation of ethane is also not
so clear to me. How does partially putting electrons in a higher
energy molecular orbital (anti bonding) make a molecule's conformation
lower in energy?

First off, let's recognize that the orbital interaction you are referring to is a way to understand the effect of hyperconjugation in various conformations.  Also, we are not putting electrons in a higher energy molecular orbital.
Here are some diagrams that may help explain this.  The first diagram on the left doesn't relate to hyperconjugation, but since it is part of the picture, let's explain it.  The Pauli Exclusion principle says that we can't have electrons with the same quantum numbers in the same orbital. In ethane, each pair of electrons in one C-H bond have the same quantum numbers as the pair electrons in any other C-H bond.  Therefor, the closer we bring 2 C-H bonds, the closer we are to violating the Pauli Exclusion principle.  This argument tells us that the eclipsed conformation should therefor, be higher in energy than the staggered conformation.
The figures in the middle and the right show the effects of hyperconjugation upon the eclipsed and staggered conformations of ethane, respectively.  According to Frontier Molecular Orbital theory we examine the interaction between the highest occupied molecular orbital (HOMO) and the lowest unoccupied molecular orbital (LUMO) to see if stabilization or destabilization results. In the hyperconjugated resonance structure for ethane we have a proton and a pair of electrons in the carbon $\ce{sp^3}$ orbital, that $\ce{sp^3}$ orbital with two electrons is our HOMO.  The only LUMO for it to interact with is the LUMO for the C-H (the $\sigma^{*}$ orbital) bond on the adjacent carbon atom.  We can mix the $\ce{sp^3}$ and $\sigma^{*}$ orbitals to produce 2 new MOs, one bonding (no nodes between these 2 orbitals) and one antibonding (1 node between these two orbitals).  What is pictured below is the bonding (in-phase, no nodes) combination of these two orbitals.  We see that in the eclipsed (middle) conformation the interaction in this orbital is destabilizing, there is no constructive overlap.  However, in the staggered (right) conformation the overlap within this orbital is now constructive - there is positive overlap.  The hyperconjugation argument also favors the staggered conformation being at lower energy.


Why add H2O when calculating the pH of an acid?

Based on what I can make out, you did not use the concentration of $\ce{H2O}$ in either of your calculations.

In the second case, you wrote it as a part of your reaction equation. Now, the reason for doing this becomes clear when you realise that when you are talking about a $\ce{H+}$ ion, you are in fact referring to a "proton"-- a bare positive charge. Moreover, a proton is extremely tiny, and thus consequently the charge density is quite high.

This makes it extremely reactive (in a sense), and thus in a chemical system of any sort would immediately seek out and associate with the electron clouds of a surrounding molecule.

In an aqueous solution an obvious, readily available target is the water molecule, and you end up with hydronium ($\ce{H3O+}$) ions.

In these particular problems, the solvent i.e water doesn't have a significant effect on the equilibria, however, this may not always be the case; which is why I recommend writing your reaction equations as you did in the second case, as it allows you to identify the acid/base, and conjugate acid/conjugate base in the reaction.


Calculating ratio of conjugate base and acid when pH and pKa are given

Recall the fundamental properties of the logarithm
$$+\log \left(\frac{a}{b}\right)$$

is the same as

$$-\log \left(\frac{b}{a}\right)$$

so Henderson Hasselbach can have plus or minus signs. The correct equation is

$$\mathrm{pH} = \mathrm{p}K_\mathrm{a} + \log \left(\frac{\text{conjugate base}}{\text{acid}}\right)$$

or

$$\mathrm{pH}  = \mathrm{p}K_\mathrm{a} - \log \left(\frac{\text{acid}}{\text{conjugate base}}\right)$$

Now sort out whether the textbook solution is right or not.


Is a hue indistinguishable from its equivalent frequency of light?

Certainly not

The problem is not, strictly, one of physics but one of perception. The eye doesn't detect individual wavelengths of light but the relative amount of red, green and blue light (but the sensors in the eye cover a relatively wide and overlapping range of wavelengths centred on what we might measure as "pure" blue, green and red. If a "pure" yellow light (say a wavelength of around 580nm) is shone into the eye it will excite both the green and red sensors and the brain will interpret the result as yellow. So the actual colours we perceive will match a pure light of that colour in that case. But we could get the same effect by shining two lights into the eye of pure green and pure red. The brain can't tell the difference and you would still perceive yellow.

But this system is no perfect and we can perceive colours that don't appear anywhere on the spectrum of "pure" colours. An example of this is magenta which is colour that does not appear in an spectrum of pure, single wavelength, light (or on the rainbow). The brain manufactures this colour when red and blue sensors are activated but green sensors are not. A very simple illustration and explanation of this is given in this short Royal Institution video.

This simple examples shows that colour perception does not relate to or match the perceived colour of a "pure" wavelength of light. Perceived colour is more complex that simply matching the wavelength in a light source and there are colours that we perceive that simply don't have a matching wavelength.


Indeed, most of the names proposed by you correspond to the preferred IUPAC names in accordance with the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book):

(1) 3-bromo-2,5-dimethylhexane

(2) 7,7-dichloro-4-ethyl-3,3,8-trimethyl-4-propyldecane (your edit is correct)

(3) see below

(4) 2-chloropent-1-en-4-yne

(5) 1-bromo-3-ethyl-5-methylbenzene

For compound (3), you have to consider the relevant rules for the numbering of locants for substituent prefixes:

  P-14.4 NUMBERING

  When several structural features appear in cyclic and acyclic compounds, low locants are assigned to them in the following decreasing order of seniority:

  (…)

  (f) detachable alphabetized prefixes, all considered together in a series of increasing numerical order;

  (g) lowest locants for the substituent cited first as a prefix in the name;

  (…)

Note that Rule (f) takes precedence over Rule (g).

In accordance with Rule (f), compound (3) could be named as 5-bromo-1-methylcyclohexa-1,4-diene (which was your proposal) as well as 1-bromo-5-methylcyclohexa-1,4-diene since both names correspond to the locant set ‘1,5’. However, according to Rule (g), this example is named as 1-bromo-5-methylcyclohexa-1,4-diene rather than 5-bromo-1-methylcyclohexa-1,4-diene since bromo is cited first as a prefix in the name.


Techron is a detergent based on PEA, polyeter amine. It is a complex synthetic molecule that will basically 'loosen' and 'solve' deposits in your fuel system. They then become part of the fuel-air mixture and burn in the combustion chamber with it. Be it in the carburettor, the injectors, the valves (for indirect injection vehicles) or the combustion chamber.

The Techron fuel you buy at the Chevron or Texaco stations has only a small amount of Techron added, the rest is just gas/petrol. You can also buy the Techron in seperate high concentrate bottles (typically 12oz.) These bottles contain up to 35% of PEA.

But the working ingrediënt PEA is just as well used in Gum-Out, Amsoil and a bunch of other fuel system cleaners.

So yes, it is a difference to fuel which doesn't have these detergent but every major gas producer (Shell, BP, Chevron, Texaco, ...) will add detergents. They ARE different detergents so they do make a difference.

In fact they are just auxiliary tools to keep your engine working well. It is like a band-aid. It does help tune up your engine, but it helps a lot more to do proper maintenance, change coolant and oil regularly, make sure your radiator works well so your engine operating temp isn't too high, clean your air filter and fuel filter regularly,...

Using these fuels does have an effect eg for emissions testing, but If you ask me. I just buy the cheapest fuel and I will add a bottle of High concentrate Techron or Gumout or another cleaner before I go to an emissions test (make sure you drive about 60 miles before turning up since the deposits have to be burnt before the test). And if you really want to use it, buy a bunch of high concentrate bottles and add them to supermarket gas, that's basically the same


Is it possible for monobasic and dibasic potassium phosphate to complete break down into phosphates?

At basic conditions over pH 10, those phosphorus-containing anions can get further deprotonated, as can easily be seen on the corresponding Wikipedia article on phosphates.

Phosphoric acid $\rightarrow$ Dihydrogen phosphate $\rightarrow$ Hydrogen phosphate $\rightarrow$ Phosphate


I suspect that the issue here is with counting the number of lone pairs.
I think the structure you are referring to looks something like the right hand one shown below.

As you can see, this structure can be described by the two resonance forms shown. The double bond arises because the bromine 'donates' one of its lone pairs to form another bond to the carbon, shifting the positive charge onto the more electronegative bromine and hence stabilising the ione. Since this is the 'more stable' resonance contributor it is usually shown as the structure of the ion in drawings but the actual structure is more accurately represented as a resonance hybrid of the two.


Do Biuret test always means proteins presence in a substance?

Biuret test react with a peptide bond, so nylon can react with.( https://en.wikipedia.org/wiki/Biuret_test, https://twitter.com/jane_essex/status/921359485939134465)


Alkylation of 1-naphthol in trifluoroethanol versus in DMSO

The 1-naphthoxide ion has negative charge on the oxygen and the 2 and 4 carbons. This can be seen in resonance structures such as this:

Since oxygen is significantly more electronegative than carbon we would expect the resonance structure with the negative charge on the oxygen to be the major contributor and so the charge density will be greatest on the oxygen. Therefore in DMSO, where solvation of the anion is poor, we would expect the reaction to proceed through the oxygen nucleophile with an $\mathrm{S_N2}$ mechanism due to the primary nature of the electrophile.

In trifluroethanol the oxygen will be strongly solvated by hydrogen bonding and so its nucleophilicity will be drastically reduced, partly due to steric hindrance of the solvating alcohol molecules. This allows substitution to occur elsewhere. The 4 position is probably preferred over the 2 position again due to the steric hindrance of the solvent coordinating to the adjacent oxygen.

EDIT: Thanks to @jerepierre for pointing this out in the comments: "One of the products (the carbon-alkylated one) requires breaking aromaticity, which is most likely going to be much slower than O-alkylation for either solvent system. A cursory search of SciFinder shows no examples of this reaction. I'm guessing that somebody made up this problem to test solvent effects in alkylation, but came up with a poor example. The best 'predict the products' questions are ones that have actually been performed."


Identification of zinc/aluminium/magnesium with copper sulphate solution

If you mean identification of the metals in their elemental form, then yes, it might be possible to distinguish between them with an aqueous $\ce{CuSO4}$ solution due to the different reactions.

Zinc in $\ce{CuSO4}$ will be oxidized to colorless $\ce{Zn^2+}$, while $\ce{Cu^2+}$ is reduced to elemental copper, because copper is the more noble metal. The $\ce{Cu}$ precipitate is normally not black, although it can appear dark in the background of the blue solution. It can also form a layer on the piece of zinc, and in the course of the reaction, the initially blue solution will become more and more colorless.

$$\ce{Zn + CuSO4 \rightarrow Cu\downarrow + ZnSO4}$$

Aluminium does not react at all because it is protected by its native oxide layer which is stable in this solution. Magnesium, however, is oxidized under evolution of $\ce{H2}$ gas, which fizzes off from the solution because it is poorly soluble in water. The reaction will gradually become slower when insoluble $\ce{Mg(OH)2}$ passivates the metal surface.

$$\ce{Mg + 2 H2O \rightarrow Mg(OH)2 + H2\uparrow}$$


What are the differences between monotropic and enantiotropic polymorphic transformations?

In enantiotropic there exist an transition temperature between two crystalline state in equilibrium but reverese is true for monotropic


What is the Italian ‘Acido Psammico’ (Psammic acid?)

Actually, the meaning is:

used to tan leather.

The acid used traditionally in tanneries is tannic acid, from there its name.

Tannic acid can be found in several plants one of which is the Sicilian Sumac (Rhus coriaria). This plant is also found in Sardegna (as per Flora Italiana), where the novel is placed.
The Italian name for this shrub is Sommacco siciliano.
The description of this plant is:

piccolo albero tipico di luoghi sassosi e rupestri, con fiori giallastri in
pannocchie, le cui foglie e corteccia, ricche di tannino, si adoperano nella
concia delle pelli (fam. Anacardiacee)

[the last part says: the leaves and bark, rich in tannin, are used in the tanning of leather.]

From sommacco to psammicco there is not too much distance.
But I agree that it is speculative.

Tannic acid is toxic, but the oral LD50 in rats is 2.26±0.083 g/kg body weight, so too much to hide in an ostia.
However, that can be a writers license.

Tannic acid is a complex polyphenol with structure:


How do I heat my borosilicate reflux column?

That should pose no issue regarding the glass

Without images to see how unevenly you will apply the heat, I have to modify my instinctual reaction "That cannot possibly be an issue" to "Most likely it will be OK". I'd be much more worried about variation in the distilling if you are at some point drawing a distillate. If you cannot control the heat flux, the yield and temperature profile will be offset from the theoretical and can (if the process is delicate) lead to variation in the results.

But borosilicate glass is sturdy, you can probably heat it with a naked propane flame - it is designed to have a low thermal expansion coefficient. Directly from that, it will have a high thermal shock resistance.


You have the right idea, but somewhere your math is incorrect.
Let the mass of the penny be $M$.
The mass of the copper is $\pu{0.03M}$ and the mass of the zinc is $\pu{0.97M}$.
The volume of copper is $\pu{0.03M}/(\pu{8.93g/cm^3})$ and the volume of the zinc is $\pu{0.97M}/(\pu{7.14g/cm^3})$.
The total volume is
$$\pu{0.03M}/(\pu{8.93g/cm^3}) + \pu{0.97M}/(\pu{7.14g/cm^3})$$
The density of the penny is
$$\frac {M}{\pu{0.03M}/(\pu{8.93g/cm^3}) + \pu{0.97M}/(\pu{7.14g/cm^3})}$$
Factoring out M gives
$$\frac {1}{0.03/(\pu{8.93g/cm^3}) + 0.97/(\pu{7.14g/cm^3})}$$
So the density of a mixture works out to be:
$$D=\frac {1}{(\%_1/D_1)+(\%_2/D_2)}$$


Name of corresponding anion of a hemiacetal

Hemiacetals can actually be named as ‘hemiacetal’ using functional class nomenclature, e.g. pentanal ethyl hemiacetal. However, the preferred IUPAC names are generated using the usual systematic substitutive nomenclature, i.e. as substituted hydroxy compounds, e.g. 1-ethoxypentan-1-ol. The corresponding rules in the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) read as follows.

P-66.6.5.2 Hemiacetals and hemiketals
Compounds with the general structure $\ce{RR'C(OH)(O-R'')}$ have the class name ‘hemiacetals’. They are named substitutively as ‘alkoxy’, ‘alkyloxy’, ‘aryloxy’, etc. derivatives of an appropriate hydroxy parent compound, such as an alcohol; these names are the preferred IUPAC names. Other names are formed by functional class nomenclature using the class name ‘hemiacetal’; similarly derivatives of ketones are denoted by the class name ‘hemiketal’.

According to P-72.2.2.2.2, an anion formed by subtracting $\ce{H+}$ from a hydroxy compound that can be expressed by a suffix ‘ol’ is systematically named by using compound suffix ‘olate’, formed by addition of the ending ‘ate’ to the suffix ‘ol’.

P-72.2.2.2.2 Anions derived from hydroxy compounds
An anion formed by subtracting a hydron from the chalcogen atom of a hydroxy characteristic group, or a chalcogen analogue, that can be expressed by a suffix such as ‘ol’, ‘thiol’, ‘-peroxol’, etc., is preferably named by using suffixes ‘alate’, ‘thiolate’, ‘peroxolate’, etc., formed by addition of the ending ‘ate’ to the suffixes ‘ol’, ‘thiol’, ‘peroxol’, etc. The multiplying prefixes ‘bis’, ‘tris’, etc. are used before these suffixes, to avoid any ambiguity.
(…)

Disappointingly, the current IUPAC recommendations still retain traditional names as preferred IUPAC names for anions of several simple alcohols (such as ‘ethoxide’ instead of the systematic name ‘ethanolate’).
The anion of my example 1-ethoxypentan-1-ol, however, is not affected by such exceptions; the systematic name 1-ethoxypentan-1-olate is also the preferred IUPAC name.


I thought this question was a good idea for a home experiment to do with my 8th grade son.

Into a block of dry ice, we carved a conical mold hole with the pointed end of scissors and poured canola oil into the hole:

We covered this with a second block of dry ice for about a minute and extracted the frozen oil with forceps:

Then we dropped the frozen oil into a jar of liquid canola oil:

And it sank.


Is there any difference between "autoignition temperature" and "ignition temperature"?

The autoignition temperature is the lowest temperature at which a substance can spontaneously ignite in a normal atmosphere without an external source (spark,flame etc).

"Ignition temperature" is what people call autoignition temperature who don't care about details. As Ivan said, there is no such thing as ignition temperature.


The conversion issue you're seeing is because the PDB you're using is not technically well-formatted. If you take a look at the output, you see a bunch of the following warnings.

==============================
*** Open Babel Warning  in parseAtomRecord
  WARNING: Problems reading a PDB file
  Problems reading a HETATM or ATOM record.
  According to the PDB specification,
  columns 77-78 should contain the element symbol of an atom.
  but OpenBabel found '  ' (atom 276)
==============================

This thread from the OpenBabel mailing list goes into more details. Basically, the official PDB specification requires that the atom names be present off to the right-hand side of each ATOM/HETATM line. Older PDBs (and lazy PDB-outputting software) don't do that.

For the most part that works, as the atom name entry normally contains the element designation. Parsing this is tricky, though, as it doesn't always work.

That's what's causing Open Babel to choke. Your PDB contains gold atoms, and Open Babel doesn't know how to handle the "AU" atoms. If you manually edit the ATOM lines to contain the letters "Au" in columns 77-78, then Open Babel is able to make the conversion, but still warns about the C/N/O atoms which don't have their elements correctly designated. - But for C/N/O it's okay, as Open Babel is able to figure out the element from the atom name.

If you don't want to manually edit the lines, you just need to pass the PDB file through a program that can make sense of the AU atoms, and will output the element name in the appropriate colum. VMD doesn't seem to help (it reads things fine, but doesn't add the element annotation), but PyMol is able to read in the file and when re-saving the molecule will put the correct values in columns 77-78.

Regarding the "missing atoms", the PDB file contains a crystallographic symmetry record in the first four lines. (CRYST1, SCALE1, SCALE2 and SCALE3 records). I'm guessing the atoms you're missing are related by crystallographic symmetry to the ones that are present. To generate those atoms, use a program which can use the PDB symmetry records to generate the symmetric partners. (PyMol is one such program.)


Meaning of the term "Activation of alkene"

Activation of an alkene just means that the double bond has a higher electron density than that of a normal isolated double bond. That is, the electron density in the double bond is greater than the one observed in ethene $\ce{CH2=CH2}$.
Activation, in organic chemistry, generally means the compound displays a greater nucleophilic nature than it normally should due to increased electron density. For example, $\ce{-OCH3}$ group activates benzene when it forms toluidine. This makes the benzene ring more electron rich and so make it easier to react in nucleophilic reactions.


Can the concentration of a sodium carbonate solution be determined using volumetric analysis?

The titration of hydrochloric acid with sodium carbonate is a common school practical, used to determine the concentration of the acid (standardisation).

In the first step, as you rightly point out, the sodium carbonate reacts with one equivalent of hydrochloric acid. The phenolphthalein shows the pH change associated with the formation of the sodium hydrogen carbonate (aka bicarb). You could use this value to calculate concentration, however, in order to ensure your answer was correct, you'd need to run this titration at least twice more, averaging your answers and discounting any anomalous titres.

$\ce{Na2CO3(aq) + HCl(aq) → NaHCO3(aq) + NaCl(aq)}$

As the bicarbonate is also reactive with hydrochloric acid, you can do something known as a 'double titration'. After the first titration, you can immediately do a second titration, this time using a different indicator, such as methyl orange. By doing this, you get a second value. If your titration is good, the values for the first and second titrations should be equivalent.

$\ce{NaHCO3(aq) + HCl(aq) → NaCl(aq) + CO2(g) + H2O(l) }$


The H NMR analysis of the hydrogens in salicylic acid

Since you have identified your unknown as salicylic acid (2-hydroxybenzoic acid), I'd help you understand how easily understand the substituents of aromatic nucleus induce shielding-deshielding effects. The following diagram illustrate the $\mathrm{^1H}$-$\mathrm{NMR}$ assignments of salicylic aromatic protons:

A lot of experts do not like explaining shielding-deshielding effects using electron densities around the sought atom. I agree with them, but a student of your capacity could easily related to that explanation. A good rule of thumb is o,p-directing electron donating groups (coursing positive mesomeric or resonance effect) usually shield the relevant protons at o,p-positions (inducing $\delta^-$-charge at the positions). Similarly, m-directing electron withdrawing groups (coursing negative mesomeric or resonance effect) usually deshield the relevant protons at o,p-positions (inducing $\delta^+$-charge at the positions). Accordingly, $\ce{OH}$-group shields $\ce{H_a}$ and $\ce{H_c}$ while $\ce{COOH}$-group deshields $\ce{H_b}$ and $\ce{H_d}$.

Thus, it is safe to say that the most up-field aromatic signals (sheilded) at $\delta \ 6.97$ (apparent $triplet$) and  $\delta \ 7.03$ (apparent $doublet$) are due to $\ce{H_a}$ and $\ce{H_c}$. The assignments of $\delta \ 6.97$ is belong to $\ce{H_c}$ (two vicinal protons) and  $\delta \ 7.03$ is belong to $\ce{H_a}$ (one vicinal proton) are solely due to the number of neighboring (vicinal) protons to each. Keep in mind that chemical shifts of these two signals can be moved to up- or down-field positions depending on the using locking solvent. For example, the locking solvent used to get present $\mathrm{^1H}$-$\mathrm{NMR}$ spectrum here is apparently $\ce{CDCl3}$ (based on solvent residue at $\delta \ 7.28$). However, two signals have been moved to more up-field positions (compared to that in $\ce{CDCl3}$) when the locking solvent is changed to $d_6$-$\ce{DMSO}$ (Ref.1). In that spectrum, two signals at $\delta \ 6.92$ ($t,\: ^3\!\!J=\pu{7.4 Hz}$) and  $\delta \ 6.96$ ($d,\: ^3\!\!J=\pu{7.5 Hz}$) are assigned for $\ce{H_c}$ and $\ce{H_a}$, respectively.

Similarly, the most down-field aromatic signals (desheilded) at $\delta \ 7.55$ (apparent $triplet \ of \ doublet$) and  $\delta \ 7.96$ (apparent $doublet \ of \ doublet$) are due to $\ce{H_b}$ and $\ce{H_d}$ (Note: The splitting of each signal with a small coupling constant is due to meta-or $\omega$-coupling generated by the presence of proton at meta to each; This phenomenon is generally called $^4\!\!J$-coupling). The assignments of $\delta \ 7.55$ is belong to $\ce{H_b}$ (two vicinal protons) and  $\delta \ 7.96$ is belong to $\ce{H_d}$ (one vicinal proton) are also due to the number of neighboring (vicinal) protons to each. Similar to $\ce{H_a}$ and $\ce{H_c}$, these two signals have also been moved to more some what up-field positions (compared to that in $\ce{CDCl3}$) when the locking solvent is changed to $d_6$-$\ce{DMSO}$ (Ref.1). In that spectrum, two signals at $\delta \ 7.52$ ($t,\: ^3\!\!J=\pu{7.7 Hz}$) and  $\delta \ 7.82$ ($d,\: ^3\!\!J=\pu{7.7 Hz}$) are assigned for $\ce{H_b}$ and $\ce{H_d}$, respectively.

References:

Milena Jadrijević-Mladar Takač, Dražen Vikić Topić, “FT-IR and NMR spectroscopic studies of salicylic acid derivatives. II. Comparison of 2-hydroxy- and 2,4- and 2,5-dihydroxy derivatives,” Acta Pharmaceutica 2004, 54(3), 177-191 (PDF).