Cleaning a round-bottom flask, especially after distillation

@Permeakra suggests some good cleaning solutions. I have rearranged them (adding a few) in a suggested order from mildest to most evil. The specific thing you want to use will vary based on what you want to clean. If an initial rinse isn't good enough, you may want to let it soak for a few hours. Always take safety precautions (don't add acid washes to basic compounds, for example) and where the appropriate personal protection equipment (PPE).

Detergent + water + "elbow grease" - there is a lab grade soap out there called "microwash" that is pretty good, and I also tell my students that things are more soluble in elbow grease than water.
Acetone - Acetone will dissolve most organics, and it is less harmful to you and the environment than other organics.
Dilute acid (HCl or H2SO4) - good for some "baked"-on stuff or for inorganic compounds. Watch out for vigorous acid-base reactions.
Dilute nitric acid (HNO3) - since most nitrate salts are soluble, nitric acid does a better job with resistant inorganics. It is also an oxidant and will convert resistance organics to more soluble compounds. Also watch for acid-base reactions.
Other organic solvents - Use whatever will dissolve your crud, but use proper personal protective equipment (PPE), fume hoods, and waste disposal.
"Base bath" (saturated KOH in isopropanol) - This solution eats a fine layer off the glass, which also removes the crud stuck to it. Rinse first with water or neutralize any acid present to avoid vigorous reaction with the bath. Immerse the flask into the bath for a few hours to overnight and it should come out clean. Base bath will dissolve silicone grease. Leave it in too long and it will come out etched. Use heavy butyl gloves.
Concentrated HCl - more powerful than dilute acids, but more dangerous. Neutralize any bases present to avoid vigorous reaction. Use proper PPE.
KOH + KMnO4 + H2O - this solution is basic (eats a fine layer of glass) and oxidative (converts many things, like organics, to a more soluble form). Neutralize any acids present and wear appropriate PPE.
Concentrated H2SO4 - has the added bonus of reacting with some insoluble organics, like polyaromatics, to form soluble sulfonates. H2SO4 is more dangerous than HCl. Follow the same precautions, and be more careful. I would use use butyl gloves instead of nitrile or latex. It will cause serious chemical burns.
"No-Chromix" - a non-chromium alternative to chromerge (below), no-chromix is a proprietary cleaner that is added to sufluric acid. The main ingredient is ammonium persulfate (NH4)2S2O8. You can make a reasonable knock-off by dissolving 2 g of (NH4)2S2O8 in 100 mL of water and adding to 2.5 L of sulfuric acid. Feel free to scale down, as this stuff has no shelf-life to speak of.

  The following five items have dangerous reputations, and for good reasons. Use them as last resorts.

Concentrated HNO3 - strongly acidic and powerfully oxidizing. Dissolves many resistant things, including burned-on crud, but may ignite low molecular weight organics (like solvents). Rinse the flask first with water to remove bases, then with acetone to remove organics, and then with water to remove acetone before adding HNO3. Use butyl gloves.
Dilute HF (no more than 5%) - HF eats glass also, and it does so more quickly than base baths. However, the the hazards associated with HF are severe. HF burns are not like other acid burns. The HF continues to bore into you and the damage might not be evident until later. DO NOT USE UNLESS YOU HAVE THE BURN ANTIDOTE HANDY. The burn antidote
is calcium gluconate, which neutralizes the acid and more important sequesters the fluoride ions. It is available as a gel or cream. You should still go overboard on PPE.
Aqua Regia: 1 part Concentrated HNO3 to 3 parts concentrated HCl. This is more powerful than just concentrated HNO3, since HCl is a stronger acid. It also decomposes to form NO2 and Cl2, both more powerful oxidants than Concentrated HNO3 alone. Aqua regia is famous as the only substance that can dissolve gold. It will also dissolve all manner of inorganics and organics, including silicone and hydrocarbon grease. Treat aqua regia with the respect deserved of its two components and keep it in the fume hood. Do not seal the vessels since it gives off gas. Aqua regia has no shelf life, but it remains a strong acid and will need neutralized before disposal. Note that aqua regia will not dissolve metallic rhodium, iridium, osmium, and ruthenium (all less active than gold).
Piranha solution: H2SO4 + H2O2 - this strongly acidic strongly oxidizing solution will eat off a lot of metal, organic, and inorganic residues by oxidizing then to more soluble forms. As a strong acid oxidant, it shares the same incompatibility warnings that concentrated nitric acid has. Piranha solution is good for glassware that needs to remain calibrated (volumetrics) or uniform (NMR tubes) since it does not eat glass. Piranha solution does not have a long shelf life.
Chromerge H2SO4 + K2Cr2O7 - forms chromic acid, a powerful oxidant. This stuff is considered toxic waste in most jurisdictions, meaning it will be expensive to get rid of. You can store it, however, if you store it in teflon. Do not use on NMR tubes, since chromium is paramagnetic. Treat chromerge with a respect due to the sum of everything above it on this list.


Does the term "Alkaline" necessarily indicate the presence of an actual alkali?

Most every English language dictionary definition of alkaline that I've found defines it as the adjective form of alkali and includes something like "having a pH greater than 7".  As pointed out in the comments there is no IUPAC definition, so it's going to be difficult to come up with an authoritative scientific answer.  Here are some interesting items I've found on the word's origin, usage in different contexts, etc.:

Online Enemology Dictionary:
alkali: late 14c., "soda ash," from Medieval Latin alkali, from Arabic al-qaliy "the ashes, burnt ashes" (of saltwort, which abounds in soda due to growing in alkaline soils), from qala "to roast in a pan." Later extended to similar substances, natural or manufactured. The modern chemistry sense is from 1813.

Cambridge Dictionary :
alkali: chemistry - a substance that has the opposite effect or chemical behavior of an acid. adj. alkaline

Biology Online:
alkaline: (Science: chemistry) Having the reactions of an alkali. Origin: L. Alkalinus Relating to or containing an alkali; having a ph greater than 7; alkaline soils derived from chalk or limestone.A substance that has a ph higher than 7.

Merriam-Webster:
alkaline: :  of, relating to, containing, or having the properties of an alkali or alkali metal :  basic; especially, of a solution :  having a pH of more than 7

The almighty Wikipedia, fairly representative of a half-dozen other online definitions:
In chemistry, an alkali (/ˈælkəlaɪ/; from Arabic: al-qaly القلي, القالي , “ashes of the saltwort”) is a basic, ionic salt of an alkali metal or alkaline earth metal chemical element. An alkali also can be defined as a base that dissolves in water. A solution of a soluble base has a pH greater than 7.0. The adjective alkaline is commonly, and alkalescent less often, used in English as a synonym for basic, especially for bases soluble in water. This broad use of the term is likely to have come about because alkalis were the first bases known to obey the Arrhenius definition of a base, and they are still among the most common bases.

I think in the end its usage is simply tied to the jargon and vocabulary of the context in which it is used. And with so many definitions, most of the ways that we hear it used are probably at least partially correct.  And as is so often the case, I think the Wikipedia definition probably comes as close to the best definition as any.


ASTM D2270 Viscosity Index Formula/Table derivation

back when it was invented they measured all the oils they could find.  They got the worst (most viscosity change) and the best in each group and ranked these as 0 to 100.  All others were plotted on this "linear" scale.

The unit was used for dome time until the best was bettered, then they invented VIE (E is a subscript, and stands for extension, strictly all VI's above 100 are VIE's).

It is a horrid unit because it is far from having any useful meaning, unfortunately it has some significance.


Equilibrium rate constants forward and backwards

In simple cases, the rate is equal to the rate constant "k" times the concentration.  The forward rate constant and the backward rate constant don't change, but the rate of change (rate constant times concentration) will change as the concentration of a reactant or product changes.


You should look into what is meant by Gaussian statistics or normal distribution. In the case of a normally distributed random variable, approximately 68% of the set should be within one standard deviation of the mean. Particularly look at the section entitled "Standard deviation and coverage".


Creation of ester from acetylene derivative

No, your approach doesn't give the right product. There's a carbon missing between the benzene ring and the carbonyl carbon. See the difference in the products below.

I don't know exactly what your solutions manual suggested, but I've included a conservative approach through phenylacetaldehyde. As you suggest, it's not absolutely necessary to go through an acid chloride to make an ester from a carboxylic acid. The Fischer esterification will work. Also, for many carboxylic acid to ester conversions, you can simply react the carboxylic acid with thionyl chloride and the desired alcohol all in the reaction together, rendering it a one-step transformation.


The composition of dry air is about $78\%$ $\ce{N2}$, $21\%$ $\ce{O2}$ and $1\%$ $\ce{Ar}$.  The molecular weights of these compounds are:
$\ce{N2} = \pu{28 g/mol}$
$\ce{O2} = \pu{32 g/mol}$
$\ce{Ar} = \pu{40 g/mol}$

So, the average molecular weight of dry air is given as:

$\pu{(0.78 * 28 g/mol) + ( 0.21 * 32 g/mol) + ( 0.01 * 40 g/mol) g/mol = 29 g/mol}$

The molecular weight of water is only $\pu{18 g/mol}$.

So, a molecule of air being replaced by a molecule of water results in an air molecule with an average molecular weight of $\pu{29 g/mol}$ being replaced by a water molecule having a molecular weight of only $\pu{18 g/mol}$.

This scenario assumes a constant temperature and pressure when humidifying the air.


Why replace c with v in Broglie hypothesis

C is for radioactive rays while v is for the matter waves..here c is for a particle travelling with light speed, i.e., radioactive rays, so we replace it with v velocity of any particle for its matter wave

$E=mc^2$ and $E=hv$.
Therefore,  $$mc^2=hv$$
Or $$mc^2=\frac{hc}{k}$$, where k=wavelength
$$mc=\frac{h}{k}$$
Or, $$k=\frac{h}{mc}$$
   Replacing c by v we get
$$k=\frac{h}{mv}$$
Note the difference between matter waves and electromagnetic waves. E.W. are associated with electrical and magnetic field while matter waves are not so E.w. travel at 3×10^2 m/s. While matter waves travel with different velocity. The equation of de broglie with c is valid, it can be treated for finding the wavelength of electromagnetic waves, or electrons. But the velocity of matter waves are quite smaller as compared to light speed. So he used a general v in place of c V can be equal to c when the particle is travelling at light speed


Why are carbocations with sp3 hybridisation highly unstable?

When you hybridize towards $sp^3$, you use one $p$ orbital more compared to the $sp^2$ hybridization.

The additional $p$ orbital raises the energy of the 4 $sp^3$ orbitals higher with respect to the energy level of $sp^2$-hybridized orbitals ($p$ orbitals are higher in energy than $s$ orbitals), which in the energy-minimizing world of chemistry is not favoured by the system.

The energy is minimized by hybridizing in a $sp^2$ fashion (in which all the $\sigma$-bonding electrons go) and adopting a planar geometry.


How pure gold is obtained back from dissolved state in Aqua regia?

By electrolysis. In general electrolysis is a very common way of purifying metals if you have them dissolved, since you can selectively control the reaction by adjusting the potential current. Al and Cu are also obtained through electrolysis of their salt solutions.
Electrolysis is a redox reaction that occurs in the non spontaneous direction driven by a external potential. In electrolysis we have two electrodes (a cathode and an anode) submerged in a solution with the salt of the metal we want to purify. As it always happens in electrochemistry, the reduction reaction will occur in the cathode, while the oxidation in the anode.
Since in a gold solution obtained through Aqua Regia we have the gold forming a tetrachloride complex (AuCl4), the oxidation reaction we will have will be the production of chlorine gas from the chloride. So the global redox reaction that will be occurring will be:
$$\ce{2[AuCl4]- + 6e- -> 2Au + 8Cl-}$$
$$\ce{6Cl- -> 3Cl2 + 6e-}$$
And if you look at the standard potentials of the two reactions in this Wikipedia webpage you can see that you just need a 0.43 V external potential for the reaction to occur. The negative electrode should be made out of gold and it will grow during the process by the deposit.
Industrially this process is done by having hundreds of electrodes connected in series and submerged into the gold salt solution. In this image you can see a factory for Cupper purification.

You can see from the electrodes lifted by the crate that they have copper lamina that grow during the process. Electrolysis is normally used when you want very high purity in your product, like gold and copper destined for electric applications in which any oxygen impurities greatly decrease conductivity.
Another great thing about electrolysis is that it can be used to easily apply a thin layer of the metal over a piece, if the piece is conductive itself. This is how gold coated audio jacks are produced, by using the audio jack adapter as the cathode.


All three would appear accepted by IUPAC, though A is preferred generally.

Quoting from the red book (the IUPAC guide for nomenclature of inorganic compounds):

IR.5.4.2.2:

  It is possible to provide information on the proportions of the constituents in names by using one of two other devices: the charge number, which designates ionic charge, and the oxidation number, which designates oxidation state. In nomenclature, the use of the charge number is preferred as the determination of the oxidation number is sometimes ambiguous and subjective. It is advisable to use oxidation numbers only when there is no uncertainty about their assignment.
  The charge number is a number whose magnitude is the ionic charge. It is written in parentheses immediately after the name of an ion, without a space. The charge is written in arabic numerals, followed by the sign of the charge. Note that unity is always indicated, unlike in superscript charge designations (which are used in formulae). No charge number is used after the name of a neutral species.

  The oxidation number (see Sections IR-4.6.1 and IR-9.1.2.8) of an element is indicated by a Roman numeral placed in parentheses immediately following the name (modified by the ending ‘ate’ if necessary) of the element to which it refers. The oxidation number may be positive, negative or zero (represented by the numeral 0). An oxidation number is always non-negative unless the minus sign is explicitly used (the positive sign is never used). Non-integral oxidation numbers are not used for nomenclature purposes.

Potassium hexacyanidoferrate(II) and potassium hexacyanidoferrate(4–) are specifically given as examples based on this rule, making A and B both valid IUPAC names.

Section IR.5.2.1:

  The proportions of the constituents, be they monoatomic or polyatomic, may be indicated in generalized stoichiometric names by multiplicative prefixes, as was the case for the constituents of binary

Tetrapotassium hexacyanidoferrate is given as an example based on this rule, making C also a valid IUPAC name.

As an aside, the red book specifically makes provision for the fact that, at present, there are multiple allowed IUPAC names:

  A planned future project on inorganic PINs will need to face the problem of choice between the equally valid nomenclature systems currently in use.

This differs from organic nomenclature when one preferred IUPAC name is given (known as the PIN), even where multiple valid names may exist (and are often preferred in everyday use for historical reasons).


Why don't ionic compounds show stereoisomerism?

They can, and they do.  The secret is to have a chiral cation and separately a chiral anion.  Letting $D^+$ be the dextrorotatory form of the cation, $L^+$ be the levorotatory form, and anologously for the anion, we then have four isomeric salts:

$D^+D^-$

$D^+L^-$

$L^+D^-$

$L^+L^-$

The first and second are diastereomers because only the anion is mirror-reflected; the cation is not mirror-reflected.  Similarly for three of the other five possible pairs; only two of the six possible pairs are enantiomeric.

Being diastereomers instead of enantiomers, the first and second salts, for instance, may have different physical properties such as solubility in a given solvent.  This may be used for separating different enantiomers of the anion (or, with a different pair such as first and third, separating different enantiomers of the cation).  And it's actually done.  See here for a brief summary and here for an example of diastereomeric salt formation.


Conversion of potassium dichromate solution to a less toxic solution

That looks like one way to reduce the dichromate ion which is known to be toxic. MSDS sheets indicate that potassium dichromate has category 1B carcinogenicity.

Another way I believe is to just start from an aldehyde and convert it into a carboxylic acid:

$$\ce{3RCHO + 2Cr_2O_7^- + BH^+ -> 3RCOOH + 2Cr^{3+} + 4H_2O} $$

Alternatively, one can put dichromate in acidic solution with iron:

$$\ce{Cr2O7^{2-} + 6Fe^{2+} + 14 H^+ -> 2 Cr^{3+} + 6 Fe^{3+} + 7H_2O }$$

Dichromate oxidizes $ \ce{Fe^{2+}}$ ions and produces $\ce{Cr^{3+}}$ and $\ce{Fe^{3+}}$ ions.


Shouldn't the pressure of gas above a liquid slightly decrease when some of its molecules dissolves in the liquid?

1) If you induce a sudden chance in pressure you will indeed take the system out of equilibrium. Lets consider a closed recipient containing some liquid and some gas above. If you suddenly inject more gas into the recipient, the gas pressure is initially going to raise fast and then going to decrease slowly while part of the gas is dissolving, until the system reaches equilibrium again. However, this new equilibrium will have a higher gas pressure AND greater amount of gas dissolved in the liquid when compared to the initial equilibrium. Of the amount of gas you injected part will be dissolved and part will stay above the liquid, contributing to the higher gas pressure. It makes no sense to have pressure under the initial value, this would only be possible if the system hadn't been at equilibrium at the start.

2) Shaking doesn't increase the pressure, unless you shake the bottle long enough to cause a considerable increase in the temperature inside, but that would be quite an impressive amount of shaking. If you could take the system out of equilibrium by releasing some of the dissolved gas from the liquid phase, then that would increase the pressure, but shaking doesn't accomplish that, and I believe it wouldn't be possible to shift a system out of equilibrium like that without transference of matter or heat and without a change in volume or temperature.

There is a very interesting video by Veritasium on youtube about this bottle issue, I recommend you to watch it:
https://www.youtube.com/watch?v=K-Fc08X56R0&t=3s


Would concentration of a compound in equilibrium increase even if it's not favored by a volume decrease?

Suppose you have the following generic reaction:

$\ce{xA_{(g)} <=> yB_{(g)}}$

With the condition, $x > y$. The reaction quotient, written in terms of concentration would be:

$Q = \frac{(\frac{n_B}{V})^y}{(\frac{n_A}{V})^x}$

Where $n_A$ and $n_B$ are moles of the respective species and $V$ is volume. With some algebraic manipulation, this can be rewritten as follows (note here I'm explicitly treating $Q$ as dimensionless, q.v., the relevant Wikipedia page on equilibrium constant units, which applies to the reaction quotient as well):

$Q = \frac{n_B^y}{n_A^x} \cdot V^{x-y}$

Written in that form, it should be apparent that, if $x > y$, then $x - y > 0$, and therefore $Q$, the ratio of products to reactants, varies directly with $V$. In other words, as volume grows, $Q$ grows, and as volume decreases, $Q$ decreases. Suppose that, prior to some volume change, the reaction is at equilibrium, i.e., $Q = K$. If volume instantaneously begins to increase, $Q$ becomes instantaneously greater than $K$, and therefore the rate of the reverse reaction becomes greater than the rate of the forward reaction, favoring generation of $A$ until equilibrium is reestablished. Conversely, if volume instantaneously begins to decrease, the opposite happens and production of $B$ is favored. This is precisely what one would expect based on an intuitive understanding of Le Chatelier's principle: if volume decreases, equilibrium shifts to favor production of fewer moles of gas to counteract the increasing pressure; if volume increases, equilibrium shifts to favor production of more moles of gas to counteract the decrease in pressure.

Edit: I misread part of your question initially. In short, if the volume decreases, generation of product $B$ becomes favored, so the concentration of reactant $A$ must decrease relative to $B$ (since $x$ moles of $A$ are consumed in the production of $y$ moles of $B$, and the reaction quotient approaches the equilibrium constant over time).

Just to clarify this further, it's true that, in an absolute sense, if volume decreases in an instantaneous fashion, then the concentrations of the reactant and product, considered separately, would both have to increase. However, in relative terms, the concentration of $B$ must increase relative to $A$ as the reaction proceeds. The ratio between the two must grow to compensate for the diminished volume, otherwise $Q$ will remain perpetually smaller than $K$, when in fact $Q$ always converges to $K$, for a given set of conditions, over time.


Is the compound CM-7 in Red Comet fire grenades azeotropic dichloromethane?

Passive fire suppression systems by Red Comet included a glass bulb with colorless transparent liquid, and had a sticker:

  MC-7 This fire extinguisher contains azeotropic chlormethanes®

Registered trademark symbol already signifies that marketing already prevailed at that time. Indeed, among chlormethanes $\ce{CH_xCl_{4-x}}$ chloromethane $\ce{CH3Cl}$ is flammable. Dichloromethane $\ce{CH2Cl2}$ is not combustible itself, but supports burning and flashes. Both chloroform $\ce{CHCl3}$ and $\ce{CCl4}$ are not flammable, but don't form azeotrope and still would produce phosgene $\ce{COCl2}$ when react with atmospheric oxygen. Also, the company denied presence of $\ce{CCl4}$ in MC-7 (Kiplinger's Personal Finance, Oct 1956, p. 48):

  CM-7, which has replaced carbon tetrachloride in all our grenade extinguishers. CM-7 does not produce toxic gases. It converts toxic gases created by the fire itself into additional fire-extinguishing chemicals.

As it seems like Red Comet doesn't hold any patents for inventions in chemistry, most likely it's a hydrochlorofluorocarbon that's been known at that time.

These days MC-7 extinguishers are sold as vintage items, and uniformly it's claimed (e.g. here and there) that the liquid component is 1,1,2-Trichloro-1,2,2-trifluoroethane, which at that time indeed has been studied as a very unreactive (even upon heating) vaporizable liquid agent (e.g. Fire Research Abstracts and Reviews, Vol. 3
1961, p. 148).

Starting 1960s, several acts in Montana, Chicago, and later across US no longer permitted usage of both tetrachloromethane and MC-7 as fire-extinguishing chemicals. Halons took over.


Measuring mole fraction in lab using Raoult's law

The mole fractions in the gas phase also depend on the relative volatility of the two liquids, so they will not in general be equal to the mole fractions in the liquid mixture. For instance, if you have components A and B, then their vapor pressures will be

$$ \begin{align} p_A &= \chi_A p^\circ _A \\ p_B &= \chi_B p^\circ _B \end{align}$$

where $\chi_i$ is the mole fraction of component i in the liquid and $p^\circ_i$ is the vapor pressure of the component when pure. According to Dalton's law the mole fraction of a given component in the ideal gas is proportional to its contribution to the total pressure so that

$$\begin{align}  p&=p_A+p_B \tag{1}\\ &= \chi_A p^\circ _A + \chi_B p^\circ _B \\ &= \chi_A (p^\circ _A -p^\circ _B) + p^\circ _B \end{align}$$

From this last expression you can conclude that the total pressure above the mixture will vary linearly with the mole fraction of either component in the liquid.

You can also show that the mole fraction of either component in the liquid is not equal to that in the gas. For instance for component A you can write by combining Raoult's and Dalton's laws that

$$\chi_{A,g}=\frac{p_A}{p}=\chi_{A,l}\frac{p_A^\circ}{p} \tag{2}$$

where the mole fractions in the liquid and gas are labelled explicitly with subscripts $l$ and $g$. It should be clear that $\chi_{A,g} \neq \chi_{A,l}$ unless $p=p_A^\circ$, but the latter condition is only satisfied for the pure substance A.


Is it possible for the kinetic energy integral to be negative?

Generally, the off-diagonal elements of matrices, especially in physical systems, are interpreted as the coupling between whatever the $i$ and $j$ elements correspond to. So, in this case, a negative element of $T_{ij}$ corresponds to a negative kinetic coupling between atomic orbital basis functions. In plain language, this means that basis functions $i$ and $j$ tend to mutually lower the kinetic energy of an electron placed in one of those orbitals. I am not sure it really makes sense to give much of an interpretation to this, however, because what one is really interested in is diagonalizing the Fock matrix, of which the kinetic energy is only one part.

Also, it is always possible to choose a basis where the kinetic energy matrix is diagonal, but again, this would not get you anywhere as in solving the HF problem, you will diagonalize the Fock Matrix which will surely put you in a basis in which the kinetic energy matrix is non-diagonal. What might be more interesting would be to take the actual molecular orbitals which are part of the solutions of the Roothan-Hall equations and re-compute the kinetic energy matrix over these orbitals. Looking at the coupling between these orbitals may be more easily interpretable for e.g. aromatic $\pi$-systems.


Michaelis menten constant motivation behind 1/2?

To see why the $K_m$ is equivalent to the substrate concentration at which the rate is equal to $V_{max}/2$, we need to look at the derivation of the Michaelis-Menten, which is usually included in detail in any introductory biochemistry book. Here is a brief version.
We will consider an enzyme-catalyzed reaction described by the following scheme:
$$\ce{E + S <=>[k_1][k_{-1}] ES ->[k_2] E + P},$$
where "E" is enzyme, "S" is substrate and "P" is product.
To get an expression for the rate, we start by noting that the rate of formation of product is $k_2[\ce{ES}]$, so we need to find an expression for $[\ce{ES}]$ using known quantities. Next, we make a steady-state approximation, setting $\frac{d[\ce{ES}]}{dt}=0$, so we have
$\displaystyle k_1[\ce{E}][\ce{S}]-(k_{-1}+k_2)[\ce{ES}]=0$
$\displaystyle \implies [\ce{ES}]=\frac{k_1}{k_{-1}+k_2}[\ce{E}][\ce{S}].$
To finish, we note that when the reaction has not consumed very much substrate, we can approximate $[\ce{S}]$ as equal to the starting substrate concentration $[\ce{S}]_0$. We also know that the total amount of enzyme in the reaction must be equal to the sum of the concentrations of E and ES, ie $[\ce{E}]+[\ce{ES}]=[\ce{E}]_{Total}$. We already have an equation for $[\ce{ES}]$ in terms of $[\ce{E}]$, so with some algebra, we can find that
$\displaystyle [\ce{ES}]=\frac{k_1[\ce{E}]_{Tot}[\ce{S}]}{k_{-1}+k_2+k_1[\ce{S}]}$.
Combining everything, we get
rate of formation of $\displaystyle \mathrm{P} =k_2[\ce{ES}]=\frac{k_2k_1[\ce{E}]_{Tot}[\ce{S}]_0}{k_{-1}+k_2+k_1[\ce{S}]_0}$.
In order to isolate $[\ce{S}]_0$ in the denominator, we divide the numerator and denominator by $k_1$:
rate $\displaystyle =\frac{k_2[\ce{E}]_{Tot}[\ce{S}]_0}{\frac{k_{-1}+k_2}{k_1}+[\ce{S}]_0}$,
and our final simplification is to define the unwieldy $\frac{k_{-1}+k_2}{k_1}$ as $K_m$ and $k_2[\ce{E}]_{Tot}$ as $V_{max}$:
rate $\displaystyle =\frac{V_{max}[\ce{S}]_0}{K_m+[\ce{S}]_0}$.
Getting back to your original question: Note that we never explicitly defined $K_m$ to be the substrate concentration at half-maximal rate. It simply resulted from our choice to isolate the substrate term in the denominator. However, that was not a completely arbitrary choice.
If we look at $K_m=\frac{k_{-1}+k_2}{k_1}$, we can see that if $k_{-1}$ is large and $k_2$ is small, $K_m$ approaches $\frac{k_{-1}}{k_1}$, which is the equilibrium constant for dissociation of the substrate from ES. For that reason, $K_m$ is often treated as an apparent measure of the affinity of the enzyme for the substrate, and that is an additional reason why we choose to define $K_m$ this way. Always remember that it is not actually the dissociation constant, and in some reactions will be quite far off from the actual $K_d$.


You may estimate the contribution of thermal radiation by using the Stefan–Boltzmann law, which describes the power $P$ radiated from a black body in terms of its temperature $T$:

$$P=\sigma\cdot A\cdot T^4$$

where $\sigma=5.670\,367(13)\times10^{-8}\ \mathrm{W\ m^{-2}\ K^{-4}}$ is the Stefan–Boltzmann constant[reference] and $A$ is the surface area.

The formula can be corrected for real materials by introducing the emissivity $\varepsilon$:

$$P=\varepsilon\cdot\sigma\cdot A\cdot T^4$$

For an ideal black body, the emissivity is $\varepsilon=1$. For polished metal surfaces, the values are significantly lower.

For example, for a hot black frying pan with $\varepsilon\approx1$, $A\approx600\ \mathrm{cm^2}=0.06\ \mathrm{m^2}$, and $T\approx230\ \mathrm{^\circ C}\approx500\ \mathrm K$, the power may be estimated as

$$\begin{align}
P&=\varepsilon\cdot\sigma\cdot A\cdot T^4\\
&=1\times5.670\,367\times10^{-8}\ \mathrm{W\ m^{-2}\ K^{-4}}\times0.06\ \mathrm{m^2}\times\left(500\ \mathrm K\right)^4\\
&\approx200\ \mathrm W
\end{align}$$

However, in addition to thermal radiation, heat transfer also occurs via thermal convection and thermal conduction. You could exclude these modes by placing the frying pan in a vacuum.

You could also estimate the total heat transfer due to all three modes from the electrical power of the stove that is required to keep the frying pan at a constant temperature.


Green kerosene in copper container goes colorless

The brass is a metal mixture of cooper and zinc. Copper compounds are known to form blue and green salts.

Examples: Copper Chloride

I believe that green color it's a copper compound and the copper reacted with oxygen (air) and the preciptate is copper oxide.

Also, the oxygen solubility in hidrocarbons it's around 60/70 ppm¹.
¹Corrosion for everybody - Alec Groissman


Pt[BrCl(PR₃)₂] cis, trans isomer nmr - Which is cis and which is trans?

Look at the two structures drawn below

In the trans-isomer each phosphorous atom is in the exact same chemical environment.  Each one has a chlorine on one side, a bromine on the other side and a phosphorous diagonally across.  Therefore only one signal is observed for the phosphorous.

In the cis-isomer, each phosphorous is in a different chemical environment.  One phosphorous has a chlorine diagonally across from it, the other has a bromine diagonally across from it.  In this case we observe two phosphorous signals, one for each of the two different phosphorous environments.


What is the purpose of balancing a redox half equation using H+ and H2O and electrons?

One reason to track solvated hydrogen (or hydroxide) ions and water in a redox reaction is that the hydrogen ions often get transferred too, and you need to know how that takes place to render the products.  We should expect hydrogen-ion transfers as part of the process in a protocol solvent such as water or a concentrated protic acid, as the electrons are charged and some compensating charge-transfer process must take place to keep the charges in balance.
Copper in concentrated nitric acid, where nitrogen is reduced (primarily) to the +4 oxidation state, is a well-worn example.  The oxidation half-reaction is, of course, copper being oxidized to $\ce{Cu(II)}$.  So we look at the reduction half-reaction.  In its "raw" form we would have
$\ce{HNO_3 -> NO2}$
We balance the elements other than hydrogen and oxygen, which is already done, and then compare the combined oxidation states of those elements in the reactants and products to find that the nitric acid must have taken one elecron:
$\ce{HNO_3 + e^- -> NO2}$
Now the charges are not balanced, and so we invoke that compensating hydrogen ion transfer.  In acid this is set up by adding (solvated) hydrogen ions to balance the charges:
$\ce{HNO_3 + \color{blue}{H^+} + e^- -> NO2}$
Water is used to balance the hydrogen which, if we did things correctly, balances the oxygen as well giving our familiar result
$\ce{HNO_3 + H^+ + e^- -> NO2 +H2O}$
and upon combining with the copper oxidation:
$\ce{Cu + 2 HNO_3 + 2 H^+ -> Cu^{2+} + 2 NO2 + 2 H2O}$
in which we see that in the final reaction, hydrogen ions are indeed transferred.  So we needed to account for them.


Your problem seems to have no solution without further information. You may be interested in the the general method for any acid /base calculation is summarised below.

When there is solution of a weak acid and it salt, or just the weak acid or just the salt, i.e. pure HA, NaA + HA or pure NaA, then there is no distinction between these types of solution because of the equilibria involved. ( NaA just represents any salt ).

The following calculations apply for each or any of these solutions.

The full calculation for problems of this type starts by defining the equilibrium constants for the reactions. A molecule dissociates as $\mathrm{HA \leftrightharpoons H^+ + A^-}$, but to be general we let the base be [B] instead of $\mathrm{[A^-]}$ and so write $\mathrm{HA \leftrightharpoons H^+ + B}$

$$K_a=\mathrm{[H^+]_e[B]_e/[HA]_e} \tag{1}$$

where the concentrations are those at equilibrium.

There is also the equilibrium $\mathrm{H_2O \leftrightharpoons H^+ + OH^-}$ to consider and

$$K_w=\mathrm{[H^+]_e[OH^-]_e}$$

(The activity of water is 1 so is ignored in this equilibrium ).

We know the amounts of acid $c_a$ and base $c_b$ added at the start of the reaction. To obtain the pH at equilibrium the amount $\mathrm{[H^+]}$ has to be calculated. To do this work out the concentration of acid and base in terms of the initial amount and the ionized species.

The total concentration of HA is

$$c_a=\mathrm{[HA]_e+[H^+]_e - [OH^-]_e}\tag{1a}$$

and for the base

$$c_b=\mathrm{[B]_e -[H^+]_e + [OH^-]_e\tag{1b}}$$

The e subscripts are now dropped for clarity. (Note that some authors use mass and charge balance instead to work out the concentrations.)

These values can be substituted into the equilibrium constant equation (1) and using $K_w=\mathrm{[H^+][OH^-]} $. Substitution leads to the general equation that can be used to solve all problems of this type;

$$K_A=\mathrm{[H^+]}\frac{c_b + \mathrm{[H^+]} - K_w/\mathrm{[H^+]} }{c_a - \mathrm{[H^+]} + K_w/\mathrm{[H^+]}  }\tag{2}$$

which is a cubic equation in $\mathrm{[H^+]}$ that is best solved numerically in the most general case. This equation can be simplified under different conditions.

Rewriting eqn 2 gives

$$K_A=\mathrm{[H^+]}\frac{c_b + \mathrm{[H^+]} - \mathrm{[OH^-]} }{c_a - \mathrm{[H^+]} + \mathrm{[OH^-]}  }$$

assuming that $\mathrm{[OH^-]} \gg \mathrm{[H^+]}$ then

$$K_A=\mathrm{[H^+]}\frac{c_b  - \mathrm{[OH^-]} }{ \mathrm{[OH^-]}  }$$

As $K_w=K_AK_B$ then

$$K_B = \frac{\mathrm{[OH^-]^2}}{c_B-\mathrm{[OH^-]}}$$

and making a further approximation, if $c_b >\mathrm{[OH]^-}$ then $K_Bc_B\approx \mathrm{[OH^-]^2}$, which is the equation you quote. Therefore there seems that your problem cannot be solved without more information.


How is hair tissue mineral analysis performed?

The analytical technique is called inductively coupled plasma optical emission spectroscopy. It was invented by a chemist V. A. Fassel, although it is pure physics. The specimen is digested in an acid, and a fine spray is created. That "spray" is heated to a temperature, which is higher than the surface of the Sun. Most of the elements of the periodic table emit specific wavelengths in that hot plasma- the characteristic emission of each element is just like a fingerprint. Not all elements can be detected easily but it is very good for metals. This is the mineral analysis. Organic analysis is even more complicated and it is a separate story.


How can I remove the bitartrate anion from an organic bitartrate salt?

I see on Wikipedia: "Dimethylethanolamine (DMAE or DMEA) is an organic compound with the formula (CH3)2NCH2CH2OH. It is bifunctional, containing both a tertiary amine and primary alcohol functional groups. It is a colorless viscous liquid. It is used in skin care products."

Now the fact that it is used in skin care products does not make it safe: the SDS  (https://www.fishersci.com/store/msds?partNumber=AC116180010&productDescription=N%2CN-DIMETHYLETHANOLAMINE+1LT&vendorId=VN00032119&countryCode=US&language=en) shows these pictograms:

Uh oh! Oh, well.

Well, let's forget that for a minute. DMAE is a liquid, boiling at 134C. Bitartrate anion is very insoluble if attached to potassium (potassium bitartrate is cream of tartar). If you added a stoichiometric amount of potassium bicarbonate (KHCO3, quite safe) to an aqueous solution of DMAE, you would precipitate almost all of the bitartrate, leaving the DMAE in water. Filter off the ppt and distill off the water and then distill and collect the DMAE. It will probably be a little wet and you will hate the overwhelming fishy smell.

Oh, and use a respirator (from the SDS: "Follow the OSHA respirator regulations found in 29 CFR 1910.134 or European Standard EN 149. Use a NIOSH/MSHA or European Standard EN 149 approved respirator if exposure limits are exceeded or if irritation or other symptoms are experienced.")

Good luck. Actually, I think you will have better luck if you do something else.


Determination of the iodine value of a fatty oil: Does the concentration of Na2SO3 effect the iodine value?

The reaction of $\ce{I2}$ or $\ce{IBr}$ with double bonds is slow. You have to wait half an hour for the reaction to be finished. But of course, you may wait a longer time, maybe $\pu{3 h}$ as you suggest. Or one night. Usually we are in a hurry, and we are looking for ways of avoiding to spend too much time for a given operation. That is why the procedure recommends half an hour.

And of course you may carry out the titration with a solution 5 times more diluted than recommended. The only drawback is that you may have to refill many times your burette before going to the end of the titration. Suppose $\pu{42 mL}$ of the $\pu{0.1 M}$ solution is required for performing the titration. This could be done with a usual $50$-$\pu{mL}$ burette. With a $\pu{0.02 M}$ solution, you will use $5 \times \pu{42 mL} = \pu{210 mL}$. In this case you will have to refill four times your burette. Four times! Bad luck!


Is there a piece of software that can help me see the (hkl) planes in a crystal lattice?

Mercury (available on Windows, Linux, MacOS) has a free version and is arguably the best crystallographic viewer for beginners. It also allows to add as many $(hkl)$ planes as you want. I used the structure of calcite from COD (1010962.cif).

Open your CIF file; by default Mercury loads an asymmetric unit. Once the CIF is loaded, go to Calculate → Packing/Slicing... and tick Pack option. Alternatively, tweak the volume populated by the atoms if needed. (This also automatically populates the list of planes with the planes denoting BFDH crystal morphology.)

Navigate to Calculate → Planes... → New Plane... → hkl and add the planes by Miller indices:

and, if needed, BFDH planes one by one:

If you have CSD-Materials (paid option), then you'll be able to display BFDH theoretical crystal morphologies at once by going to CSD-Materials → Calculations → BFDH Morphology...; on export you'll get something like this:

Alternatively, there is KrystalShaper (Windows only), a free lightweight program made specifically for displaying crystal morphologies:

P. S. Thanks to Buttonwood for useful comments regarding graphical details and reminding about BFDH crystal morphology.


Synthesis of phenethyl alcohol from benzene

$\ce{AlCl3}$ is a Lewis acid and will bind to the oxygen of the ethylene oxide giving a species that functions as $\ce{[R2O-AlCl2]+}.$

This cationic species will be attacked at carbon by the electron-rich aromatic ring in much the same way as a standard Friedel-Crafts reaction intermediate generated from an alkyl halide to give $\ce{[Ph-CH2CH2-OAlCl2]+}.$ This then loses $\ce{HCl}$ to give $\ce{PhCH2CHOAlCl2}$ which is stable until aqueous workup which gives phenethyl alcohol.


A cereal sample was analysed for Calcium content as follows:

The absorbance of this solution was found to be 0.351.
A 7ppm calcium solution gave an absorbance of 0.254

You have slightly (~1.5 times) higher absorbance (.35) than a solution of 7 ppm calcium (0.24)

  step 1: 7 ppm/ (0.351 x 0.254) = 78.515 ppm ca in final dilute sample.

This is not correct. You should have proportion
7/0.254 = x/0.351   leading to x = 7*0.351/0.254

You get 10 times higher value. This is the first error. You should have done 7*(0.351/0.254) = 9.67 ppm.

In 50 mL it is A, but when it was in 5 mL it was 50/5 = 10 times higher.
7*(0.351/0.254)*(50/5)=96.7 ppm.

This means you have 96.7 mg calcium per 1 kG of cereal (ppm is one millionth, and thus, mg per kG). Then one tenth of that amount of cereal (0.1 kg/1 kg) contains 1/10 of the calcium. that is 9.67 mg.

Once again, you should have had these steps:
So, you should have these steps:

(1) Since the final sample absorbs .35/.254 ~1.38 times more it is 1.38 times more concentrated than 7 ppm. That is 9.67 ppm.

(2) It was diluted 50/5= 10 times, so it was 10 times more concentrated before. That is 96.7 ppm.

(3) For 100 g of cereal: 100 000 mg * 96.7 /1 000 000 = 96.7 mg /10 = 9.67 mg.


Distinguish between two possible configurations for angular momentum in carbon atom

You are separating the spatial and spin part of your two wavefunctions. I don't think that is a good idea here, because it does only work for one of the two configurations.

The wavefunction of a configuration is given by a Slater determinant build up from spin orbitals. In the following I will neglected the normalization constant of $\frac{1}{\sqrt2}$ for simplicity. The wavefunction for the first configuration ($\uparrow - \downarrow$) is

\begin{align}
   \Psi_1 &= \begin{vmatrix}
     p_-(1)\alpha(1) & p_+(1)\beta(1)\\
     p_-(2)\alpha(2) & p_+(2)\beta(2)\\
     \end{vmatrix}\\
    &= p_-(1)\alpha(1)p_+(2)\beta(2) - p_+(1)\beta(1)p_-(2)\alpha(2)
\end{align}

Note that this differs from your suggested wavefunction by the spin part of the orbitals, and there is no way to separate them out.

And for the other configuration ($-\uparrow\downarrow -$) we have
\begin{align}
   \Psi_2 &= \begin{vmatrix}
     p_0(1)\alpha(1) & p_0(1)\beta(1)\\
     p_0(2)\alpha(2) & p_0(2)\beta(2)\\
     \end{vmatrix}\\
    &= p_0(1)\alpha(1)p_0(2)\beta(2) - p_0(1)\beta(1)p_0(2)\alpha(2)
\end{align}

Both, $\Psi_1$ and $\Psi_2$ look quite similar now. However, since the spatial part of both orbitals in $\Psi_2$ is the same here, we can factor them out

\begin{equation}
  \Psi_2 = p_0(1)p_0(2)\underbrace{[\alpha(1)\beta(2)-\beta(1)\alpha(2)]}_{\chi}
\end{equation}

And we get what you have suggested for this configuration. But this does only work for $\Psi_1$!

Now, why is $\Psi_1$ assigned to $L=2$, while $\Psi_2$ is $L=0$?

I assume your confusion is because if you add up the orbital quantum numbers $m_l$ you get $0$ in both cases ($1-1=0$ and $0+0=0$). However, those sums give you $M_L$, not $L$. The many-electron quantum number $L$ has $2L+1$ components with quantum numbers $M_L=-L,-L+1,\dots 0, \dots L-1,L$. So for $L=2$ we have $M_L=-2,-1,0,1,2$, much like there are three orbitals $p_-$, $p0$ and $p+$ for $l=1$

So $\Psi_1$ would be the $M_L=0$ component of $L=2$, while $\Psi_2$ is the $M_L=0$ component of $L=0$.

But there is one issue with such assignments: You cannot really do it, as the configurations get mixed together. The point is, that the atom as a whole needs to be spherical symmetric (unless there is some external electric or magnetic field). But the individual $p$ orbitals are not. For example $p_0$ is (in the usual convention) aligned along the $z$-axis and zero in the $xy$ plane.

Mathematically speaking the configurations form a basis (Hilbert space) in which the actual electronic states are expanded. This is known as the configuration interaction method.

So for example the electronic state with $L=0$ would be
\begin{equation}
\Psi_2 = \frac{1}{\sqrt3}(|\uparrow\downarrow - -\rangle + |-\uparrow\downarrow - \rangle + |--\uparrow\downarrow \rangle)
\end{equation}

where each $|\dots\rangle$ denotes a Slater determinant.


Apparent contradiction in the type of deviation of an ethanol-water mixture

As suggested in the comments, there is an entropic contribution to the excess free energy which should not be overlooked, as it is the excess Gibbs free energy that determines whether the deviation from Raoult's law is positive or negative. All of the statements made by the OP are correct, except that negative $\Delta H$ (exothermicity) and $\Delta V$ (volume contraction) do not necessarily imply a reduced volatility (negative deviation from Raoult's law), that is they are not sufficient conditions for a negative deviation, a negative excess Gibbs free energy instead being the sufficient condition. Again, as alluded to in the comments, a simple analysis based entirely on $\Delta H$ is not appropriate here, largely because of the hydrogen bonding (associative) nature of the liquids. In fact, the negative mixing volume and enthalpy might partly explain the negative excess entropy (and vice-verse: they are correlated). The excess entropy of mixing of water and ethanol is highly negative over the entire range of composition at room temperature, leading to a net positive excess Gibbs free energy.

  Does the mixture actually exhibit positive deviation? Which interactions are stronger - unlike or like?

Therefore the mixture does exhibit a positive deviation from Raoult's law, even as interaction enthalpies between unlike molecules are stronger.


As stated in user208322's answer, $\ce{CH3}$ and $\ce{H3C}$ are exactly the same, they're only written differently for ease of reading. As per your question about how much truth there is to having 4 groups, it's 100% truth. You cannot have a chiral carbon that doesn't have 4 groups attached to it (i.e. must be sp3 hybridized). If you think you've found a chiral carbon that only has 3 groups attached to it, you're probably forgetting the unwritten hydrogen.

In this example specifically, there is only one chiral carbon, the one attached to the secondary alcohol. Although it may look like the carbon is attached to the same group twice, you need to look further than just the first atom. Yes it is attached to two carbons, but one of those carbons is attached to a nitrogen and the other is attached to an oxygen, so the groups are different. Two groups are  only the same if you can follow it down every single atom and they match exactly.

Edit: In regards to your question about following the chain all the way down, realistically you only need to follow it until you see that it differs from the other chains. So for chains 1 and 3 once you get to the nitrogen and oxygen atoms you see that they're different and there's no point continuing. And the unwritten hydrogen has to do with the fact that organic chemists use shorthand in their molecular structures. Basically, carbons and hydrogens are extremely common in organic molecules so they just don't write them. All the points are carbons unless otherwise labelled, and they must have 4 bonds, so where you see that the carbon is only bonded to 3 things, it is also bonded to a hydrogen as well. See this wikipedia article for more info (specifically, this section).


What does a square mean in a chemical formula?

That square refers to a vacancy (or framework vacancy) in the crystal structure. Here's one reference from Advances in Electronic Ceramics. Ceramic Engineering and Science Proceedings. Volume 28, Issue 8. Ed. C. Randall ${\it et\;al.}$

In short, a framework vacancy is a feature of some inorganic materials in which a "guest atom" (or more than one) can fit into the structure or denotes a structure that is missing interstitial or packing atoms (as outlined here).

Here's the info on your mineral where you can see that the square is instead denoted by square brackets [].


Greater bond distance for silicon diamond structure than for carbon diamond structure?

Yes, both structures belong to the same diamond cubic type, and the interatomic distances for diamond and silicon (you can use "Measurements" tool from the JSmol menu)

$$
\begin{align}
d(\ce{C-C}) &= \pu{1.54 Å} \\
d(\ce{Si-Si}) &= \pu{2.35 Å} \\
\frac{d(\ce{Si-Si})}{d(\ce{C-C})} &\approx 1.53
\end{align}
$$

are in a good agreement with the corresponding covalent radii

$$
\begin{align}
r_\mathrm{cov}(\ce{C}) &= \pu{0.76 Å} \\
r_\mathrm{cov}(\ce{Si}) &= \pu{1.11 Å} \\
\frac{r_\mathrm{cov}(\ce{Si})}{r_\mathrm{cov}(\ce{C})} &\approx 1.46
\end{align}
$$

which would set the silicon atoms further apart from each other.


what contributes to the temperature of a vacuum

It really depends on how the vacuum is defined. If you, for instance, pump all the air out of a steel container, you will have a vacuum in there. Yet, there will be photons constantly radiated off of the walls. This interplay of photons will be in thermal equilibrium with the walls and therefore will have a contribution to the temperature. For instance, if you would place a thermometer inside this container, the material of the thermometer would be hit by the photons of the container wall and therefore the thermometer would give a temperature equal to the temperature of the wall. As a matter of fact, even in the deepest of deep space, there is some radiation to be found which gives a temperature of about 3K. (This radiation is left over from the Big Bang)

Nevertheless, if you make a true vacuum inside some container of which the walls are at 0 K that cannot be penetrated by any kind of radiation or matter, then the temperature therein is 0 K.


Understand Hexabromocyclododecane CAS numbers

The (−)- in the chemical name refers to the optical activity of the compound, that is to tilt the plane of linerarly polarized light to the left (levo-rotatory), or to the right (dextro-rotatory) as determined with a polarimeter.  It may be either (−)- (minus sign) or (+)- (plus sign), as seen, for example, for two of the three forms of tartaric acid.  Compounds with stereogenic centers which do not tilt the plane of the polarization (e.g., meso-compounds) may be labeled by (±)-, too.
Similar to this, and equally used for molecules with stereogenic centers − especially carbohydrates  − is the the use of (D)-, (L)- or (DL)- which however relates to the spatial arrangement around the sterogenic atom of highest priority (there are rules how to determine this just from reading structure formula).  It may be used independently as well as simultaneously to the former system of (−)- and (+)-; but note the knowledge what to write within former system does not translate 1:1 into what to use in the later system.
As for the CAS number, this is a book-keeping number (see here); contrasting to a chemical name, it does not convey structural or even chemical meaning.  CAS for example systematically attributes more than one registry number to the same molecule if there is evidence that it crystallizes in different polymorphs.  This «already» happens with inorganic molecules like $\ce{TiO2}$ which may crystallize in the form of the mineral anatase (CAS 1317-70-0), or rutile (CAS 1317-80-2).  Thus, if your literature search is about the isolated molecule alone (say, molecular mass), it does not matter which entry you follow further.  Yet if you are interested in say the melting point of your compound or crystal structure, you have to treat the CAS numbers separate, like referring to different compounds.  Thus, the relationship between CAS number and a structure is not bijective.
(Occasionally, e.g. for some of the more substituted fullerenes, you may encounter many CAS numbers for one and the same compound in a database like Scifinder or Reaxys.  Possibly the difficulty to draw their formulae contributed to create whole sets of CAS numbers.)


Thin Layer Chromatography & Stationary Phase

Aluminum Oxide is more polar than Silica Gel and polar compounds do not migrate far from the starting point.

Thin Layer Chromatographic Analyses

7. Thin-Layer Chromatography


CHARMM parameter files for NBD (nitrobenzoxadiazole)

Converting between force fields is always an iffy affair, as different force fields were parametrized with different reference sets, and possibly even with different intent in mind.

You have two options here:

You can extract the raw parameters from the particular GROMACS or AMBER force field used here, and try to translate them into the parameter formalism CHARMM uses. Generally speaking, force fields consist of a set of parameters for each atom type, and a set of equations these parameters are used in to calculate forces and geometries. Those equations represent aspects of the atom's or molecule's properties that have an influence on the calculated forces or the molecule geometry, like bonding interactions, non-bonding interactions, angle contraints, bond length constraints, and so on. You need to find the corresponding equations in CHARMM's force field framework, and then translate the parameters from AMBER or GROMACS so that the force field is the same in CHARMM.
Alternatively, you can look at the various CHARMM force fields and try to find one that works with NBD and any other molecules you got in your simulation. The first article you cited details where the various general parameters they used were from and how their simulation in GROMACS was set up. You should be able to define the same simulation parameters in CHARMM and rerun their simulation there. You can then use their published data to benchark your own simulation against theirs.


How do I combine z-matrices of multiple molecules into one input stack?

The Setup

The first thing you need to do is to get the difference between a Z-matrix (sometimes also called internal coordinates) and cartesian coordinates straight. They are truly different things.

Secondly, take the Z-matrices of a sufficiently large molecule. The basic structure is:

C1 .  .  .
C2 r2 .  .
C3 r3 a3 .
C4 r4 a4 d4
C5 r5 a5 d5
(and so on from this point)

where in this simplified notation, I have omitted the anchor centers for the coordinates. One notices that there are 6 coordinates left out, marked by dots. They can be omitted because three degrees of freedom for translation and three degrees of freedom for rotation of the entire molecule are irrelevant for the calculation of properties.

Now imagine a second molecule tacked onto the other one:

C1 .  .  .
C2 r2 .  .
C3 r3 a3 .
C4 r4 a4 d4
C5 r5 a5 d5
C6 *  *  *
C7 r7 *  *
C8 r8 a8 *
C9 r9 a9 d9

This Z-matrix is invalid. One needs to fill the missing coordinates marked by the asterisks. In a sense, the second molecule does not have the 6 degrees of freedom mentioned above, because any change in them would change the orientation relative to the first one, impacting all properties.

So, what can you do?

Filling in the missing coordinates is hard, at the very least by hand or some other automated fashion. Doing it in an editor is basically the same amount of work as the solution I actually suggest: Create the system of both molecules in one Z-matrix and split that Z-matrix. Remove the superfluous coordinates in the resulting matrices and run calculations as necessary.


Why is pyridine used when making tosyl esters from alcohols?

The function of pyridine is actually not so simple and not so easy to notice at first glance. There is a fundamental reason why pyridine is used to promote the acylation reaction, which is that it can act as a catalyst.

Despite its basicity and the subsequent formation of its chlorhydrate salt after the tosylation reaction, pyridine is also an excellent nucleophile. Pyridine is, in fact, more nucleophilic than the alcohol, and it attacks the acyl chloride rapidly, forming a highly electrophilic (because of the positive charge) intermediate: N-tosylpyridinium chloride. It is indeed this intermediate the actual tosylating agent which reacts with the alcohol to give the ester. Because pyridine is acting as a nucleophile to speed up the reaction, yet is unchanged by the reaction, it is a "nucleophilic" catalyst.


Water solubility of different types of Iron III

Iron(III) oxide is insoluble.
Iron(III) oxalate is, in practical situations, insoluble.
Iron(III) chloride-hexahydrate is soluble.

Aditionally, Iron(III) oxalate is destroyed by visible light, and I intend to complete my experiment during the day, especially considering that involves colour changes.


Why doesn't calcium carbonate dissolve in water even though it is an ionic compound?

As someone said here, this:

  The teacher stated that the ionic compounds dissolve in water except
  some carbonates.

Is indeed an oversimplification. First of all, the distinction between an "ionic compound" to other compounds isn't too defined. What your teacher probably said, or didn't say but wanted to, is that some ionic compounds easily dissolve in water. Salt (halite - NaCl) is the best example.

Calcium carbonate, in nature, also commonly dissolves. It's just not as immediate as dissolution of the more soluble ionic compounds. You are probably familiar with this phenomenon:

This forms because calcium carbonate dissolves. The rock is limestone, which is usually composed of pure calcium carbonate. Acidic water greatly enhances the solubility of calcium carbonate, and it doesn't even need to be highly acidic. Rain or river water that come into contact with the atmosphere absorb the $\ce{CO2}$ as $$\ce{H2O + CO2 <=> H2CO3},$$ which then facilitates calcium carbonate dissolution with $$\ce{CaCO3 + H2CO3 <=> Ca^2+ + 2HCO3-}.$$


Extraction of spilled methylamine from mud

The wording of the question (and confirmed in a comment) suggest a major problem with the scene in question. If mixing dirt with methylamine made mud, then it was either some solution of dissolved methylamine, or liquid methylamine, but not a solid methylamine salt like methylamine hydrochloride. The first scenario wouldn't make sense for the synthesis of methamphetamine, and the second case is the real problem: methylamine boils at -6oC.

Assuming they had a solution containing methylamine that had become contaminated with dirt, the recovery would actually be pretty easy. Because methylamine has such a low boiling point, simply basifying the solution during distillation would readily separate the methylamine from anything else that could be present in the contamination.


What can create black stains on metal after putting HCl on a tap full of limestale?

The red is copper; the knob is likely brass , possibly bronze depending on age. Your knob is not zinc or aluminum or the HCl would have done more damage. Difficult to determine what is black and what is shadow. It is less likely to cause damage in the future if you use vinegar and steel wool to clean-off carbonates.


The formation of dihalocarbenes (at least cloro- and bromo-counterparts) in reactions of haloforms with bases has been well established (e.g., Ref.1,2). In 1959, Closs and Closs introduced a novel reaction to prepare chlorocarbene as a reaction intermediate in situ in order to synthesize chlorocyclopropanes derivatives (Ref.3). They have reported that when n-butyllithium is added to methylene chloride in the presence of olefins at $\pu{—25 °C}$, chlorocyclopropanes are formed in fair to good yields depending on the structure of the olefin. They further claimed that the novel procedure strongly suggests the existence of chlorocarbene as a reaction intermediate.

They further developed the method and published a full paper as a general method (Ref.4) where abstract states:

  Alkyllithium compounds react with methylene chloride in the presence of olefins to form chlorocyclopropanes. This reaction is believed to proceed through the intermediate chlorocarbene. Seven alkyl-substituted chlorocyclopropanes have been prepared by this method. Configurations are assigned to the products by considerations of predictible non-bonded interactions in the transition states of the addition of chlorocarbene to the olefins. The configurations are confirmed by n.m.r. spectra.

Later, Billups and Haley have used the same method to synthesize spiropentadiene (Ref.5).

References:

W. von E. Doering, A. Kentaro Hoffmann, “The Addition of Dichlorocarbene to Olefins,” J. Am. Chem. Soc. 1954, 76(23), 6162–6165 (https://doi.org/10.1021/ja01652a087).
W. von E. Doering, Wm. A. Henderson, Jr., “The Electron-seeking Demands of Dichlorocarbene in its Addition to Olefins,” J. Am. Chem. Soc. 1958, 80(19), 5274–5277 (https://doi.org/10.1021/ja01552a065).
Gerhard L. Closs, Lisellote E. Closs, “Synthesis of Chlorocyclopropanes from Methylene Chloride and Olefins,” J. Am. Chem. Soc. 1959, 81(18), 4996–4997 (https://doi.org/10.1021/ja01527a057).
Gerhard L. Closs, Lisellote E. Closs, “Carbenes from Alkyl Halides and Organolithium Compounds. I. Synthesis of Chlorocyclopropanes,” J. Am. Chem. Soc. 1960, 82(21), 5723–5728 (https://doi.org/10.1021/ja01506a041).
W. E. Billups, Michael M. Haley, “Spiropentadiene,” J. Am. Chem. Soc. 1991, 113(13), 5084–5085 (https://doi.org/10.1021/ja00013a067).


Calculating reaction products programmatically?

You are looking at the most simple form of double displacement reactions of salts with

\[\ce{M^1X^1 + M^2X^2 -> M^1X^2 + M^2X^1}\]

Dissolving these salts leads to complete dissociation $\ce{MX -> M+ + X-}$. From the solution, pairs of anions and cations recombine.

Using a lookup table with elements, you could identify the cation from each string that represents a starting material (=salt). The remaining string is the anion.
Build three lists:

a list of cations cations = ['Na', 'Li']
a list of anions   anions = ['Br', 'F']
a list of the starting materials

With Python, itertools.product(cations, anions) will give a generator object that yields tuples of all possible cation-anion combinations
These tuples can be joined to strings for a possible salts
Put these strings in a list and subtract the list of the starting materials to obtain the possible products

import itertools as it
reactants = ['NaBr', 'LiF']
# splitting the string via lookup table is something I leave up to you
cations = ['Na', 'Li']
anions = ['Br', 'F']

pr = it.product(cations, anions)
pairs = list(pr)
# yields list of tuples [('Na', 'F'), ('Na', Br'), ...]

salts = [''.join(pair) for pair in pairs]
# yields a list of strings by joining the tuples:
# ['NaBr', 'NaF', 'LiBr', 'LiF']

products = [salt for salt in salts if not salt in reactants]
# only gives those salts that are not in the list of reactants


If the atomic number is # of protons, why does emission of a beta (electron) particle increase the atomic number?

During beta particle emission one neutron is converted into proton and electron, to balance charge and electron can't remain inside the nucleus hence emitted from the nucleus.


I think that, this acidity order can be justified by considering a combined effect of +R effect, -I effect and intermolecular hydrogen bonding (where it is dominant).

For 4-fluorophenol, the +R effect of $\ce F$ is significant but weaker than -I effect. The electron pair on the $\ce{2p}$ orbitals of $\ce F$ can easily delocalise with the benzene ring because of the stronger $\ce{2p\pi -2p\pi}$ overlap, which makes $\ce {+R} $ effect significant . That's why it increases the electron density on the carbon near the $\ce{-OH}$ group to such an extent which decreases its ability to donate protons, greatly decreasing
its acidity.
Considering intermolecular hydrogen bonding also, the acidity of the compound decreases further due to strongest intermolecular hydrogen bonding between the halogens.

For 4-chlorophenol, the +R effect is negligible due to much weaker $\ce{3p\pi -2p\pi}$ overlap and thus the $\ce{-I}$ effect becomes more prominent. But $\ce {Cl}$ is electronegative enough to form strong intermolecular hydrogen bonding($\ce{HO-C6H4-Cl...H-O-C6H4-Cl...)}$. That's why, it becomes difficult to remove hydrogen atoms from $\ce{-OH}$ group, decreasing its acidity.

For 4-bromophenol and 4-iodophenol, there is no strong hydrogen bonding due to much less electronegative $\ce{Br}$ and $\ce{I}$ atoms, and also +R effect is ineffective due to very weak $\ce{4p\pi -2p\pi}$ and $\ce{5p\pi -2p\pi}$ overlap. Only -I effect is acing for these two compounds in favour of acidity. As -I effect of $\ce{Br}$ is larger than $\ce{I}$, the acidity of p-bromophenol is greater than the p-iodophenol.


How does the degree of saturation influence the colour of the flame?

The flames that you see when you burn carbon-containing compounds have colour for a completely different reason that if you add ionic compounds (e.g. sodium chloride) to a flame. Ionic compounds won’t combust, their bonds will merely be broken and the metallic atom then has an electron that can be excited and that can re-relax under photon emission.

For carbon, we are actually talking about proper combustion, i.e. a reaction of the carbon compound with oxygen to give carbon dioxide and water. This comes in two flavours: complete and incomplete.

In complete combustion, all carbon atoms are fully oxidised to carbon dioxide due to sufficient oxygen amounts. We can write a nice equation e.g. for methane:

$$\ce{CH4 + 2 O2 -> CO2 + 2 H2O}\tag{1}$$

Complete combustion is often shown with methane as it often tends to combust completely.
Incomplete combustion, as the name says, is incomplete, i.e. some carbon atoms won’t make it all the way to carbon dioxide. It is impossible to give an exact equation for incomplete combustion, but one possible one could be the following:

$$\ce{2 CH4 + 3 O2 -> CO2 + C + 4 H2O}\tag{2}$$

Incomplete combustion is commonly shown with unsaturated compounds because they have a greater tendency to display it.

The key factor in incomplete combustion is the fact that elemental carbon (soot) is generated; it is that soot whose electronic transitions cause the yellow colour of a flame. This is not a feature of unsaturated versus saturated hydrocarbons: as the equation above shows, methane can burn both in a complete and in an incomplete manner depending on how much oxygen is added.

Likewise, the most common example for incomplete combustion, ethyne, can be combusted completely if only enough oxygen is added, see this question.


Dissolving to Create a Solution with Ethanol

Knowing that the material (solute)  has a solubility of 1 mg/ml in 95% ethanol does not help you predict its solubility in 75% ethanol. It could be more soluble or less soluble, depending on the properties of the solute.


I would start by looking at the cis rather than trans isomer as the starting point for interpreting the NMR spectrum. It is much more likely to be the product you got from this reaction. Also, be sure to identify all the chemically distinct hydrogens in your structures.


translation of a sentence involving antigen/buffer/solution

You actually need to prepare a stock solution of 10,000 pg/ml antigen in 7.5 % BSA-TSA buffer.


Why don't atoms with expanded valence shells have a formal charge?

Formal charge is the difference between the number of valence electrons that a neutral unbonded atom has and the number of valence electrons assigned to that atom when after it forms bonds in a molecule. It looks like you are making your mistake in assigning the proper number of electrons to the sulfur atom.

The electrons assigned to an atom are:

All nonbonding (lone pair) electrons on that atom in the Lewis Structure.
One half of all bonding electrons (one electron for each single bond, two electrons for each double bond, etc.).

Consider water, $\ce{H2O}$. The oxygen atom has two bonds and two lone pairs. Oxygen originally had 6 valence electrons. We assign 6 electrons to it (4 nonbonding and 2 bonding). The formal charge is $6-6=0$.

Now consider the hydroxide ion, $\ce{OH-}$. The oxygen in hydroxide has 7 electron assigned to it: 6 nonbonding and 1 bonding. The formal charge is $6-7=-1$.

Now consider the nitrogen atom in the nitrate ion, $\ce{NO3-}$. Nitrogen originally has 5 electrons. In the nitrate ion we assign 4 electrons to that nitrogen (1 for each single bond and 2 for the double bond. There are no lone pairs). The formal charge is $5-4=+1$.

As a final example, let's look at the P in phosphate, $\ce{PO4^3-} $. Phosphorous original had 5 electrons, like nitrogen. We assign 5 electrons to phosphorous in phosphate (1 for each P-O bond and two for the P=O bond). The formal charge is $5-5-=0$

How many electrons are assigned to the sulfur atom in $\ce{SO3}$? From the question it sounds like you have the correct Lewis Structure. It looks like you are making your mistake in assigning the proper number of electrons based on the number of bonds.


How to arrange substances in order of increasing vapor pressure?

To solve this problem, use Raoult's law. In Raoult's law, the vapor pressure of a fluid is lowered proportionally with the mole fraction of solute particles in the fluid.
You know the formula, but you only need it if you want to calculate the actual resulting vapor pressure. We can know which is bigger like this:
A. $\pu{0.5M }$ sugar. Sugar does not ionize, thus there are $\pu{0.5 M }$ particles in the solvent.
B. $\pu{0.5M \ce{NaCl}}$. Sodium chloride dissociates to two ions. Thus there are $0.5 \times \pu{2 M}$ particles in the solvent.
C. $\pu{2M}$ ethylene glycol. Again, this doesn't dissociate, so there are $\pu{2M}$ particles.
D. $\pu{2M \ce{LiBr}}$. Dissociates, resulting in $2 \times \pu{2 M}$ particles.
Thus, listing the concentration of particles in the solvent, you see where the answer comes from:
D: $\pu{4 M}$
C: $\pu{2 M}$
B: $\pu{1 M}$
A: $\pu{0.5 M}$


Cloudiness indicates light scattering by small particles, with size on the order of the wavelength of light and larger. If those small particles are denser than the liquid, they may eventually sink and become the settled precipitate.

However, some small particles are so small and so nearly equivalent in density to the liquid that they do not sink, but remain suspended. Brownian motion helps. Some clays can remain suspended indefinitely; in water treatment plants, FeCl3 is used as a flocculant to bring such turbidity down.

Sometimes intermolecular interactions keep particles suspended. For instance, many surfactants form micelles comparable in size to the wavelength of light and may, in pure form, be considerably more, or less, dense than the liquid, and yet remain suspended indefinitely because they are at least partially hydrophillic.

Centrifugation may help bring down suspended particles (or float them if they are lighter than water).


Why does this compound have two benzene rings?

Chemists are already accustomed to this type of formula, even though it is not absolutely correct. More exact would be drawing all bonds with double solid/dashed bonds to express that the bond order is between 1 and 2, approximately 1.5, but is almost never done

Rarely can be seen formula with the two circles merged (and maybe dashed also), like

But the electrons are in fact delocalized across all bonds of the polycycle structure in "electron cloud" that looks somewhat like the infinity sign ∞ in this case (naphthalene) above and below the planar molecule. What matters is that the count of π (those from the "second bond" of the double bond) electrons is 4n+2 for n is some natural number (here it is n=2). It is a special feature called aromaticity, which has something to do with advanced topic of quantum chemistry; and it greatly stabilizes the molecule. Any of these formulas is unable to describe reality.


As I already mentioned in my comment, phosphines such as $\ce{PEt3}$ can bind to transition metals with partially filled d orbitals as a σ-donor as well as a π-acceptor.

The σ interaction is simply a donation of electrons from the lone pair on phosphorus to empty metal d orbitals.

The π interaction is backdonation of electrons from filled metal d orbitals to an empty orbital on the phosphine. Traditionally this was thought to be the 3d orbitals on phosphorus, hence "dπ–dπ". However, as Martin pointed out in the comments this is a rather outdated mode of thinking. The current accepted view is that the acceptor orbitals are $\ce{P-R}$ σ* orbitals (to be precise, a linear combination of them that has appropriate symmetry to overlap with metal d orbitals).


Determining percentage of copper in brass

Well the question would had more to it. The question would have probably gone something like this. A sample of brass was dissolved with excess nitric acid. The $\ce{Cu^{2+}}$ ions are then reduced with excess iodide ions. Iodine is then titrated with thiosulfate (giving its concentration and volume used). Find the percentage of copper in brass.

To do this question, calculate the amount of moles of thiosulfate used, using its concentration and volume. Then using equation 3, you can see the number of moles of iodine is half the number moles of thiosulfate. Then using equation 2, you can see that the number of moles of $\ce{Cu^{2+}}$ ions are twice as the number of moles of iodine and hence are equal to the number of moles of iodine. Finally, using equation 1, the number of moles of copper is equal to the number of moles of $\ce{Cu^{2+}}$ ions and hence are equal to the number of moles of thiosulfate.

Now that we know the number of moles of copper, you can multiply by its molar mass to get its mass and then find the percentage of copper in brass.


Why doesn't phosphorus form as many oxides as nitrogen?

It is generally due to the larger size of phosphorus. Nitrogen and oxygen both are members of 2nd period.Hence the p$\pi$ - p$\pi$ bond formed between nitrogen and oxygen has larger area of overlap resulting in shorter bond length and increased stability. Phosphorus is a member of 3rd period so when 3p orbital of phosphorus overlaps with 2p orbital of oxygen, due to difference in energy and size the area of overlap decreases which leads to lesser stability.Hence it forms less oxides.


How to calculate the remaining mass in the mother solution after multiple extractions?

In old lecture notes of mine I found this formula

$ g_i = g\left(\frac{V_1}{V_1 + V_2K_2}\right)^n$

where $g_i$ is the remnant mass of a product initially dissolved in volume $V_1$, to be $n$-times extracted by volume $V_2$ with a given Nernst dissipation coefficient $K = \frac{a_2}{a_1}$.  Here, $g$ refers to the mass of the product initially dissolved in $V_1$.

Applied to your case, assuming a molar mass of 180 g/mol, this would read like 0.54 g initially dissolved in you 150 mL solution 1 (3 mmol).  After a single extraction with 50 mL of 2, only 90 mg were remaining (0.5 mmol) in $V_1$, after the third time 2.5 mg (about 14 umol).

I speculate a textbook like P. W. Atkins' Physical Chemistry may contain an example to confirm such a calculation.


Why do the f-block elements ( inner transition metals) have an oxidation number of +3?

First of all, lets define which elements we're discussing as f-block.  Generally the lanthanides are treated as f-block elements, though this introduces some ambiguity as La (which starts the lanthanides) has 1 d electron and no f electrons while Lu (which ends the series in some definitions) has 1 d-electron and all 14 4f electrons.  So one of these two is a d-block element but its not entirely clear which one.  Recall that in the f-block there can only be 14 elements due to the 14 electrons that fill f orbitals (just as the d-block can only have 10 elements).

Not all of the lanthanides have 5d electrons, only La, Ce, Gd and Lu do!  The rest fill the 6s and 4f only.  The 6s are heavily shielded and easily lost.  Only one 4f electron is lost because the 4f closer to the nucleus and more effected by losing electrons, ie once one is removed its difficult to remove another because the rest are far more attracted to the nucleus.

So the real question should be why are La, Ce and Gd exceptions to the filling order?  For La and Ce this is because they have a lower Z effective, they're not splitting the energy of the 4f and 5d as effectively.  For these two the 5d is lower energy and fills first, then the 4f fills.  For the later Ln with a higher Z effective the 4f (being in a lower shell) is pulled to significantly lower energies and fills first.

The reasons for Gd having a different filling order is slightly more complex but can be related to whats seen in the transition metals, recall that Rh and Ru don't actually fill the 5s despite not being at a half-shell closure, they fill one 5s and the rest in 4d.  Lu is not an exception.


Reason for same frequency of spectral line in infrared and visible region

According to Bohr's theory of the structure of the hydrogen atom, the atom is defined as a system with a positively charged nucleus that has electrons revolving around it in discrete orbits of fixed size and energy and stated that the orbits were held stationary using electrostatic forces.
Upon calculation of the various parameters of the model, we get the energy of the $\mathrm{n}^{\text{th}}$ orbital in the hydrogen atom to be:
$$E=\frac{\mathrm{-13.6}}{\mathrm{n^2}} \mathrm{eV} \tag{1}$$
If we state that the atom to be considered were to be made mono-electronic and had an atomic number of Z, satisfying the conditions for the Bohr's model to work, we get that the energy of the $\mathrm{n}^{\text{th}}$ orbital to be:
$$E= -13.6 \cdot\frac{\mathrm{Z^2}}{\mathrm{n}^2} \mathrm{eV} \tag{2}$$
A spectral line is observed when the electron de-excites itself from a higher energy state to a lower energy state that is more stable. This is done by emitting a photon whose energy is the same as that of the difference of energy between the two levels. Applying $\frac{\mathrm{hc}}{\lambda} =E$, and $\frac{\mathrm{c}}{\lambda} = f$ we get
$$f = \frac{\mathrm{c}}{\lambda} = \mathrm{c} R_H\big[\mathrm{\frac{1}{n_1^2}-\frac{1}{n_2^2}\big]Z^2} \tag{3}$$
Here $R_H$ is the Rydberg constant whose value is $\approx\pu{109,678 cm^-1}$
In the question, you have been asked to figure out which line in the Balmer series($\mathrm{n_1} = 2$) of hydrogen would have the same frequency as that of the $3^\text{rd}$ line in the Brackett series($\mathrm{n_1} = 4$). I believe you would know how to solve the question and if not would now be able to via equating the frequencies of the two spectral lines.
Moving on to your doubt which is based on a misconception.

How is it that the frequency of a spectral line belonging to infrared region be same as that of line belonging to visible region knowing that frequency is characteristic property unique to an electromagnetic radiation?

The issue with this statement here is that you are assuming that the Balmer series is always visible for all atoms. This is only true for hydrogen as the the wavelengths provided for the different series are based on measurements taken on the hydrogen atom. That is, the Paschen series is infrared for hydrogen and this may change if the atom changes. As the atomic number increases, the frequency becomes greater as you can see from equation ($3$). This implies that at higher atomic numbers, there is a greater shift towards the UV region.
TL;DR: What you say is correct, frequency is a characteristic property unique for a certain electromagnetic radiation. However, the Balmer series is defined to be visible specifically for the hydrogen atom and not for any other mono-electronic atom/ion. As the atomic number becomes greater, a shift towards the UV region takes place.


Which of the following is most likely to undergo hydride shift?

Besides Avinsh's comment, there is another factor.  When you have a carbocation center next to a cyclopropane ring or cyclobutane ring, the "bent bonds" of the ring partially overlap the "vacant" orbital on the cation center.  To the extent that this occurs, the formally primary ion (c) is something like a more stable tertiary one.  Ion (a) does not have this factor because the cation center is already inside the ring instead of next to it.


Reaction intermediates of MnO2 catalyzed H2O2 decomposition reaction

Reaction conditions

As Watts et al. have shown, the decomposition products of this Fenton-like reaction strongly depend on the $\mathrm{pH}$ of the solution.1

If performed in acidic conditions, the reaction generates mostly hydroxy radicals, but no reductants (which would be the hydroperoxide and superoxide anions). If, conversely, the reaction is held in neutral conditions, Watts et al. have shown that the reaction produces significantly more of the aforementioned reductants.

Do et al. have conducted further research into the reaction mechanism at $\mathrm{pH}=7$ and concluded that the observed reaction order may be approximated as a pseudo 1st order reaction, where the ratio $\ce{[H2O2]}/\ce{[#MnO2]}$ is vital to the description of the decomposition rate.2 They have further shown that reactive intermediates, such as superoxide and hydroperoxide anions are generated by the reaction. After slightly modifying the pH towards alkaline conditions, the production rates for the reactive anions increased drastically.

Reaction mechanism

Do et al. have meticulously assembled a table with a proposed reaction pathway:

And finally, I quote one of their closing statements:2

  The existence of hydroperoxide/superoxide anion implies that the suggested reaction mechanism could be explained by hydrogen peroxide being decomposed, not only directly on the surface of manganese oxide, but also through a propagation reaction involving intermediates such as hydroperoxide/superoxide anion in solution.

References

Watts, R., Sarasa, J., Loge, F., and Teel, A. Oxidative and Reductive Pathways in Manganese-Catalyzed Fenton’s Reactions. J. Environ. Eng., 131(1), 2005, 158–164, DOI link.
Si-Hyun Do, Bill Batchelor, Hong-Kyun Lee, Sung-Ho Kong. Hydrogen peroxide decomposition on manganese oxide (pyrolusite): Kinetics, intermediates, and mechanism. Chemosphere, Volume 75, Issue 1, March 2009, Pages 8-12, DOI link.


Shouldn't mineral NPK fertilizer ingredients add up to 100%?

There is a specific convention how the nitrogen, potassium and phosphate concentration is given, and it might be surprising because it mentions substances that are not in the fertilizer, namely nitrogen atoms, $\ce{P2O5}$, $\ce{K2O}$ and $\ce{MgO}$. You have to imagine that all of the nitrogen, phosphorous, potassium and magnesium in your sample reacts to form these species, and then you express the masses of those fictitious products as percentage of the original mass.

$$\text{fertilizer}\ce{-> x N + y P2O5 + z K2O + q MgO}$$

For an example, look at this answer: https://chemistry.stackexchange.com/a/64282 which links to this reference https://en.m.wikipedia.org/wiki/Labeling_of_fertilizer


Kinetic Energy Evaluation Integral Evaluation Program

The equation for the Kinetic Energy matrix element that you quote is for two unnormalised 1s Gaussians. The d factors contain the normalisation factors and the contraction coefficients - look more carefully at the code, you have got it slightly wrong what the d's represent.

Talking of the code please, Please don't use that as a model for your own Fortran - that style is getting on for 50 years out of date!


Using MO theory, give an explanation for the C-C bond length in cyanogen

See my answer to your question on methyl vinyl ether for some background on molecular orbital (MO) theory.

In this case, your molecule is analogous to the all-carbon system, butadiene.  In cyanogen you have 4 p orbitals that can align, overlap and mix together to generate 4 molecular orbitals and you have 4 electrons (1 from each p orbital involved in a pi bond) to place in your molecular orbitals.  Since the carbons in cyanogen are $\ce{sp}$ hybridized you actually have a second set of molecular orbitals that are identical to the first in all respects except that the 4 p orbitals in the second set are rotated 90 degrees with respect to the other set of p orbitals.

Let's count electrons. We have 4 electrons in each of these orthogonal molecular orbitals that each contains 4 p orbitals, so we have 8 electrons in our two orthogonal sets of molecular orbitals.  We also have 2 lone pairs remaining on each nitrogen for a total of 4 electrons.  Finally we have the sigma system; we have 2 C-N sigma bonds and 1 C-C sigma bond each containing 2 electrons for a total of 6 sigma electrons.  For the whole molecule we have 8+4+6=18 electrons as expected.

As I mentioned in my other answer, we typically focus on the delocalized molecular orbitals formed by combining the p orbitals and disregard the sigma system (usually) when we create our MOs. Disregarding the sigma system usually simplifies matters.

There are a number of ways to create the MO diagram for butadiene or cyanogen, but all of them will have the common theme of mixing 4 p atomic orbitals and generating 4 molecular orbitals.  Here is one way to do it, let's start by mixing 2 p orbital to create the MO's for ethylene (or nitrile).

image source

Now we can mix 2 ethylenes (or nitriles) to generate butadiene (or cyanogen)

image source

We label molecular orbitals $\ce{\Psi_1 ~- ~\Psi_4}$ in order of increasing energy (so $\ce{\Psi_1}$ is the molecular orbital at the bottom of the figure and $\ce{\Psi_4}$ is at the top).  Don't worry about the "S"'s and "A"'s in the diagram, they are just telling us about the symmetry properties of the molecular orbitals which we don't need to focus on right now.

Notice how $\ce{\Psi_1}$ has overlap between the orbitals on the two middle carbons.  This indicates overlap and a higher bond order between these 2 carbons, just as your resonance structures II and II indicate.  Just like in butadiene, this carbon-carbon bond has some double bond character and, therefore there is a higher barrier to rotation about it then you might otherwise expect.


Mechanism of hydrolysis of acyl chlorides

This reaction is mostly ignored in basic organic texts and the mechanism you give is straight out of Wikipedia with some modification. I am sure some serious study exists some where but not on the web. Please bear with me because I can't[or just don't know how to] do drawings.  The first step is probably as written and is most likely slow because it involves charge separations. You are right there is a + charge on the O of water. [btw there is no homolysis involved, that generates radicals, and there is no reasonable path to generate hydride ions] The second step seems to require the participation of a base this may be loosely described as an E2 reaction where a proton is removed from the protonated O atom by another water molecule and the developing negative charge on the other oxygen expels the chloride ion. This mechanism[if plausible] requires enough water for active solvation. This mechanism possibly explains the success of the Schotten-Bauman method where base is added to the reaction as a "catalyst" [tho it is consumed]. If water is limited the base may be the attached chlorine and references do state that dry reactions evolve HCl gas. These are my thoughts on the reaction; library research is in order.


Is Neutron radiation considered electromagnetic wave?

Electromagnetic radiation consists of electromagnetic waves (oscillations of the electric and magnetic field), whereas corpuscular radiation consists of actual particles. Certainly, wave–particle duality can make the distinction between waves and particles fuzzy. However, electromagnetic radiation (whether described as wave or as photons) is massless (its invariant mass is zero) and it travels at the speed of light. In contrast, the particles of corpuscular radiation have a non-zero mass and therefore a velocity below the speed of light. The particles of corpuscular radiation can be charged positively (e.g. α particles) or negatively (e.g. β− particles), or they can have no charge at all.

Neutron radiation is a corpuscular radiation because it consists of particles (i.e. neutrons). Neutrons are subatomic particles with no charge and a mass of $m=1.674\,927\,471(21) \times 10^{-27}\ \mathrm{kg}$ [source], which is slightly larger than the mass of a proton.

The kinetic energy of free neutrons can vary significantly. For example, the so-called fast neutrons, which are released during nuclear fission, have energies above $E_\text{kin} = 1\ \mathrm{MeV} \approx 1.6 \times 10^{-12}\ \mathrm{J}$. The thermal neutrons, which are used to initiate nuclear fission in moderated reactors, have energies below $E_\text{kin} = 0.1\ \mathrm{eV} \approx 1.6 \times 10^{-20}\ \mathrm{J}$.

Since, in classical mechanics, kinetic energy $E_\text{kin}$ is
$$E_\text{kin} = \tfrac{1}{2}mv^2$$
where $m$ is mass and $v$ is velocity,
the velocity of a neutron with a kinetic energy of $E_\text{kin} = 0.1\ \mathrm{eV} \approx 1.6 \times 10^{-20}\ \mathrm{J}$ may be estimated as
$$\begin{align}
v &= \sqrt {2\frac{E}{m}}\\[6pt]
&= \sqrt {2\frac{1.6 \times 10^{-20}\ \mathrm{J}}{1.674\,927\,471 \times 10^{-27}\ \mathrm{kg}}}\\[6pt]
&= 4.4\times 10^3\ \mathrm{m\ s^{-1}}
\end{align}$$
which is clearly less than the speed of light.


Is hot air or hot water more effective at melting a frozen pipe?

Specific heat relates to the heat capacity; how much temperature change you get for a specific mass and energy once the heat has finished ‘moving’.

But you also have to consider how fast the heat is moving.

For melting a frozen pipe with a fluid you also need to consider the heat resistivity of the fluid which is the inverse of heat conductivity. Some fluids like air are good heat insulators - much more so than water. The resistivity will limit the rate at which energy can be transferred to the mass.

Since water has a much lower resistivity than air ($\sim 1.7\ \mathrm{m\cdot K/W}$ vs $\sim 50\ \mathrm{m\cdot K/W}$) you can expect a heat transfer rate into the cold pipe about 25 times faster. This all assumes the air and water are at the same temperature.


Is 4-methylchlorobenzene an acceptable IUPAC name for this compound?

In the generation of a preferred IUPAC name (PIN), the locant '1' may only be omitted under certain circumstances, as stipulated by Rule P-14.3.4.2 in Nomenclature of Organic Chemistry: IUPAC Recommendations and Preferred Names 2013.

  (c) in monosubstituted homogeneous monocyclic rings, e.g. cyclohexanethiol (PIN) or bromobenzene (PIN) ...

Since 1-chloro-4-methylbenzene is not a monosubstituted ring, it does not fall under this category. Therefore, in the preferred IUPAC name, the locant '1' must be cited to avoid any ambiguity.

The omission of the locant '1' when there is no ambiguity is widespread, and is permissible within general nomenclature (refer to Loong's answer here). In this case, the omission of the locant '1' from the name "4-methyl-1-chlorobenzene" is likely to be unambiguous. However, note that the name "4-methyl-1-chlorobenzene" is not in accord with IUPAC nomenclature to begin with. Therefore, the abbreviated name "4-methylchlorobenzene" is probably not acceptable, either.

  P-14.5.1 Simple prefixes (i.e. those describing atoms and unsubstituted substituents) are arranged alphabetically...

Since 'chloro' comes before 'methyl' in alphabetical order, the order of citation of substituents must be chloro followed by methyl; hence the correct name must be of the form x-chloro-y-methylbenzene.


Draw each ring with one double bond and the two atoms having no double bond with two electron pairs instead of one.  Verify that the total electron count agrees with two positive charges on $\ce{S4}$ and zero on $\ce{S2N2}$.  The double bond and the extra electron pairs on the two atoms are the six ring electrons in the Huckel rule.


Can equilibrium be reached in a non-reversible chemical reaction?

In any chemical reaction there is a generally a potential energy barrier between reactants and products. How easily this is crossed from reactants to products or vice versa depends primarily on the temperature. This is because to react, molecules have to gain energy that is more that the average to be able to cross the barrier. (This is called the activation energy). The energy is accumulated randomly from solvent in solution and by collisions in gas phase, which is why some chemical reactions can be slow. Barriers are often high compared to thermal energy so it can take ages to get enough energy by chance to cross a barrier. The chance of having a certain energy E is determined by the Boltzmann distribution $exp(-E/k_bT)$ where $k_B$ is the Boltzmann constant, T temperature. As Boltzmann distribution also depends on temperature reactions are generally observed to be faster at higher temperatures.
If a reaction is called 'reversible' then one expects to see reactants and products in the mixture at the same time at amounts depending on the equilibrium constant. The product is reverting to reactant and reactant to product, assuming that all product is not continuously being removed, say by precipitation or otherwise.  In an 'irreversible' reaction the barrier from products to reactants is so high that in normal experience enough of the product never returns to reactant to be measurable. The effect can be very dramatic (orders of magnitude) with a relatively small change in activation energy because of the exponential dependance of the Boltzmann distribution.  All this means that there is no fundamental difference in 'reversible' and 'irreversible' reactions, just a matter of time-scale. However, it is obviously practically useful to distinguish these cases and so we can look at equilibrium constants to guide us. Note (a) That thermodynamics does not inform us about time-scales but only the relative free energy of reactants and products. (b) The words ' reversible' when used as 'reversible process' has a special meaning in thermodynamics and needs to be used with care. (c) I have excluded the special case of 'barrierless' reactions.


How does a "sea of electrons" hold metal atom together?

The metal atoms lose electrons to form positive metal ions and a sea of delocalised electrons. The positive ions are attracted to the oppositely charged electrons and form an electrostatic attraction.


Ligand substitution reaction in coordination compounds

The reaction you wrote down is wrong on two counts. The reactant is not a hypothetical tetraaqua complex and the product is not a hypothetical tetraammin complex. The correct reaction is as shown below:

$$\ce{[Cu(H2O)6]^2+ (aq) + 4 NH3 (aq) <=> [Cu(NH3)4(H2O)2]^2+ (aq) + 4 H2O (l)}\tag{1}$$

Note that I have used an equilibrium arrow here: the reaction can proceed in both directions depending on the concentration of ammonia in your sample (in reality, there is also the competing precipitation of $\ce{Cu(OH)2}$ at slightly basic pH levels but I will ignore that here).

As an equilibrium, this can be pushed either way. So let’s take a look at what nitric acid can do. It is both an acid and an oxidising agent – but as you noted, copper(II) is already very highly oxidised and it is generally not possible to sustain copper(III) or higher copper oxidation states in aquaeous solution (it does exist as an intermediate in various copper-catalysed organic reactions). On the other hand, $\ce{HNO3}$ can very much act as an acid in your solution as there is a basic compound able to receive protons (see spoiler after figuring out the reaction yourself).

   $$\ce{HNO3 (aq) + NH3 (aq) -> NH4+ (aq) + NO3- (aq)}\tag{2}$$

Note that while $(2)$ is also a reversible reaction, the equilibrium is far to the product side so we can essentially treat it as completed. Obviously, this has an effect on the original reaction – and with a simple application of Le Chatelier’s principle the resulting answer should be obvious.

   Yes, since ammonia will be protonated to give ammonium, equilibrium $(1)$ will be shifted back to the reactant side, regenerating hexaaquacopper(II) $\ce{[Cu(H2O)6]^2+}$.


Solvent for copper(I) iodide, copper(I) thiocyanate, silver iodide for HTM deposition; safety for methylammonium lead iodide

Tell your mentor that to be concerned about chlorobenzene toxicity on perovskites is riddiculous. The most toxic thing in perovskites is the lead iodide...chlorobenzene´s toxicity is much lower than that. If he is so worried about toxicity, he should not work on perovskites.


To understand the paramagnetic nature of $\ce{O2}$, we must first understand how atomic orbitals mix together to form molecular orbitals. In the diatomic molecules of the elements in the second period, a phenomenon known as $\mathrm{s}$-$\mathrm{p}$ mixing  results in an increase in the energy of the $\sigma_\mathrm{2p_z}$ molecular orbital, and a decrease in the energy of the $\sigma_\mathrm{2s}$ orbital. This is also observed in the $\sigma^{*}_\mathrm{2p_z}$ and $\sigma^{*}_\mathrm{2s}$ orbitals.

The degree of $\mathrm{s}$-$\mathrm{p}$ mixing is determined by the energy gap between the $\mathrm{s}$ and $\mathrm{p}$ orbitals. The higher the energy gap, the less the orbitals mix. Because the $\mathrm{s}$-$\mathrm{p}$ gap increases across a period, $\mathrm{s}$-$\mathrm{p}$ mixing in $\ce{O2}$ and $\ce{F2}$ is actually so low that the $\sigma^{*}_\mathrm{2p_z}$ orbital is lower in energy than the $\pi_\mathrm{2p_{x,y}}$ orbital, unlike the preceding elements.

This can be seen here:

[source]

By constructing the molecular orbital diagram for $\ce{O2}$ and filling each  orbital according to Hund's rule, it becomes evident that $\ce{O2}$ is a diradical, with two unpaired electrons of the same spin. This is what gives oxygen its paramagnetism.