The molecular determinants of odor perception is a disputed area in fragrance biochemistry.  There are two basic theories: the "shape" hypothesis and the "vibrations" hypothesis.  Luca Turin is the foremost exponent of the more controversial and less widespread "vibrations" hypothesis, but he does a good job of explaining both theories in his TED talk.

According to the vibrational theory, since different functional groups have different vibrational modes, they should smell quite different.  An example from Luca's talk is a complex alcohol vs. the corresponding thiol.

According to the shape theory, molecules which have similar but not identical functional groups should smell similarly, unless the protein receptors in our nose that sense odors have very exquisite shape sensitivity, enough to perceive the very slight difference in the shape of thiol vs. an alcohol in a large otherwise identical molecule.

  Why might two different chemical structures smell the same?

According to one theory, they would smell the same if they had similar vibrational (i.e. infrared) spectra.  In 2011, PNAS published is a very intriguing result that showed this was true for fruit flies in the case of some odorants.  They changed the vibrations by isotopic substitutions, a method that should leave essentially unchanged molecule's shape and its chemical reactivity.

  So, would that make it so that two different molecules with the same functional groups would smell the same?

In my mind this is true if you believe the vibrational theory as well.  But it is harder to understand why it is true if you believe the shape theory.

Of course, the real answer may be a combination of both theories.  I'm a tepid supporter of the vibration theory myself, but I would be very interested in a chem.se answer from another expert who is less impressed with Dr. Turin's theories than me.


By definition an azetrope is a distillate that boils at a constant temperature, but is a mixture of two or more compounds. So you can't purify an azetrope by using more distillations.

Pure ethanol has a bp of 78.37 °C and pure ethylene glycol has a bp of 197.3 °C. The bp of the azetrope could be (1) below pp of ethanol, or (2) above bp of ethylene glycol. In theory the azetrope cannot be between or equal to either of the boiling points of the two pure chemicals. (The problem with theory is experimental error....)

I can find no evidence of an azetrope between ethanol and ethylene glycol using Google. So I very confident that no such azetrope exists.

You noted that the distillate temperature was about 67 °C as the pot started to boil. You also noted that you read the thermometer when it began to produce the drops of distillate.

Look at the drawing below.

The glassware setup has two thermometers.

Thermometer-1 is in the pot.
Thermometer-2 is in the pot.

Ideally the pot temperature and the distillate temperature would be the same. However when the solution starts to boil the distillate is just starting to contact the thermometer. The thermometers have some thermal mass, so it takes time to get them up to temperature.

The distillate thermometer is relatively slow to equilibrate. So long as it is dripping distillate solution then it is colder than the distillate and will not give an accurate temperature reading.

So the only explanation for this anomalous result is that you made a small mistake taking the reading.


According to Product  Information of Sigma-Aldrich for L-Tyrosine, the solubility is $\mathrm{pH}$-dependent:

Preparation Instructions: This product is soluble in $\pu{1 M} \ \ce{ HCl}$ ($\pu{100 mg/ml}$), with heating. The solubility in water ($\pu{25 ^\circ C}$) is $\pu{0.45 mg/ml}$ in the pH range $3.2-7.5$ (Ref.1). At $\mathrm{pH}$ $1.8$, the solubility is $\pu{2.0 mg/ml}$; at $\mathrm{pH}$ $9.5$, the solubility is $\pu{1.4 mg/ml}$; and at $\mathrm{pH}$ $10$, the solubility is $\pu{3.8 mg/ml}$.

The above claims can be justified by the study of the solubility of tyrosine in the acidic and basic conditions (Ref.2), the summery of which states that:

Measurements have been made of the solubility at $\pu{25 ^\circ C}$ of tyrosine in hydrochloric acid and in sodium hydroxide solutions varying from $0.001$ to $\pu{0.05 M}$, and also in distilled water. The $\mathrm{pH}$ of the saturated solutions was measured with the hydrogen electrode. The following values for the ionization constants of tyrosine have been obtained from the measurements: $k_\mathrm{b} = 1.57 \times 10^{–12}$, $k_\mathrm{a1} = 7.8 \times 10^{–10}$, $k_\mathrm{a2} = 8.5 \times 10^{–11}$. The changes in solubility with $\mathrm{pH}$ can be satisfactorily explained by the use of these ionization constants.

According to plot of solubility versus $\mathrm{pH}$ of their results, it was concluded that the highest solubility of tyrosine can be achieved in the $\mathrm{pH}$ range of $1-2$ or in the $\mathrm{pH}$ range of $9-10$. For example, the concentration of tyrosine in $\pu{0.05 M}$ $\ce{HCl}$ is found to be $\pu{0.0165 M}$ with the $\mathrm{pH}$ of solution being $1.450$. And also, the concentration of tyrosine in $\pu{0.0498 M}$ $\ce{NaOH}$ is found to be $\pu{0.0358 M}$ with the $\mathrm{pH}$ of solution being $9.953$. The least solubility of tyrosine is shown to be with in the $\mathrm{pH}$ range of $3-8.5$ (Ref.2).
Note: The extrapolation of the acidic side of above graph has shown the solubility increases up to about $\pu{0.04 M}$ at $\mathrm{pH} = 1$ (Ref.2).
References:

T. A. Brown, In Molecular Biology: LabFax; 1st Edition; Bios Scientific Publishers: Oxford, England, 1991, p. 29 (ISBN: 1‐872748‐00‐7).
David I. Hitchcock, "The solubility of tyrosine in the acid and in alkali," Journal of General Physiology 1924, 6(6), 747–757 (https://doi.org/10.1085/jgp.6.6.747)(PDF).


$\ce{RCHO}$ is preferred. The problem with $\ce{RCOH}$ is that it gives the misleading impression that the hydrogen is attached to the oxygen. In fact, I don't believe I've ever seen a reference in literature to an aldehyde using the latter representation.


Can glutamic acid and arginine form H-bond at physiological pH?

Yes, glutamate and arginine can form a hydrogen bond at physiological pH. Remember of course, that as soon as you enter aquaeous media, everything will be saturated by hydrogen bonds from the surrounding water molecules. But inside a protein an arginine—glutamate hydrogen bond is perfectly viable. Arginine will donate the hydrogen bond using an $\ce{N-H}$ bond and glutamate will receive towards a carboxylate.

Note, by the way, that your picture gives a wrong structure of protonated arginine. The correct protonation of the guanidine group is given in the image below (taken from Wikipedia).


Why do ionic compounds have such high melting points and boiling points? Do they have IMAFs?

Molecules have IMAFs that determine the melting and boiling points. What about with ionic substances? What keeps the different compounds together? Do they experience IMAFs?

The problem is that you are trying to apply a paradigm that was developed for covalent compounds (the distinction between covalent bonds and intermolecular forces) to ionic compounds. They don't play by the same rules. In an ionic compound, all the ions are held together by electrostatic forces, which mean exactly the same thing as ionic bonds. There is no discrete molecular unit in an ionic compound. The word "molecule" does not apply.

Consequently, there is no such thing as an intermolecular force in an ionic compound. There are only ionic bonds.

  I've heard that it's the ionic bond itself, but doesn't the bond remain after melting

Loosely speaking, it does not remain after melting. The reason why they have such high boiling points is therefore because of the fact that you have to break the ionic bonds for it to begin melting.

  and even if it doesn't then how come covalent bonds don't cause the boiling and melting points in molecules?

Covalent bonds don't break during melting or boiling, with some exceptions (see: giant covalent compounds).


How does electron localization/delocalization work?

Draw the LCAO orbitals.  Delocalization is the summation of discrete electron occupations of orbitals, such as a 1,3 shift (a pi-bond flip).  Remember that LCAO is a terrifically wrong black box. Its only virtue is that it simply, beautifully works (until the Woodward-Hoffmann rules).

Given propene, abstract a methyl proton.  The LCAO anion is half at both ends by a 1,3-shift of the pi bond.  MO shows degenerate orbital occupation still with (nearly) no additional density at the central carbon.  Abstract an allylic proton from 1-butene.  The allylic ends are no longer 50:50 occupied by the anion for the different electronic properties of two attached protons versus a methyl group and a proton.


Most likely, sulfur dioxide. $\ce{SO2}$. was produced. The ingredients sodium hydrosulfite, $\ce{Na2S2O4}$, and sodium metabisulfite, $\ce{Na2S2O5}$, when mixed with acid, release $\ce{SO2}$, which oxidizes to $\ce{SO3}$ eventually.

$\ce{SO2}$ is quite irritating, but not immediately damaging in low doses, <100 ppm, though it can certainly induce a life-threatening asthma attack. "Sulfur dioxide is mildly toxic and can be hazardous in high concentrations. Long-term exposure to low concentrations is also problematic." $\ce{SO3}$ is even more dangerous, since it forms sulfuric acid when it meets water, e.g. in tears in the eye.

Air the room *thoroughly. There is not much else to do: theoretically, ammonia could be used to form a salt, removing the gas from air, but there is no way you could balance the reaction, leaving you with an excess of one irritant or another.


How do I find a concentration from a sample tested on a spectrophotometer?

The question being asked is, how many uL of your solution do you need to withdraw in order for that volume to contain 100 ng of the RNA.  If there are 2 ug/uL in the RNA solution, that is equivalent to 2000 ngm/uL.  So if you want 100 ng of RNA, you will need to withdraw
$$\ce{\frac{2000 ngm}{1 uL}~=~ \frac{100 ngm}{x ~uL}}$$
solving for x yields an answer of 1/20 uL or 0.05 uL.  0.05 uL of the RNA solution will contain 100 ng of the RNA.


Do the letters ol in a substance only indicate that the substance is an alcohol if they came at the end of the name of a substance?

Yes, the ending ol in a name is a good hint that the compound contains an $\ce{OH}$ group.

However, beware of trivial names:

$\alpha$-tocopherol is not an aliphatic alcohol, but a substituted phenol
mannitol does not contain one $\ce{OH}$ group, but six of them

It's getting even more confusing if you consider other languages, such as German. Benzol is the German name of benzene, which has no $\ce{OH}$ group at all. The same is true for styrol, the German word for styrene ;-)


Is there a way to determine if a compound is oxygen or water sensitive?

I think that this will be water sensitive for the reasons you present but also oxygen sensitive for two reasons:

Due to the relatively low oxidation state of $\ce{Co(I)}$ in the compound which will react with $\ce{O2}$ towards formation of $\ce{Co(II)}$.
The $\ce{P^{III}(OPh)3}$ ligand is air sensitive as well and can oxidize to $\ce{OP^V(OPh)3}$ as well as reacting with water: https://www.alfa.com/en/catalog/A18662/.


Chemical compounds responsible for the colors in flowers?

are there other classes of compounds responsible for the red and yellow colors found in flowers?

Interest in colors, specifically dyes, was a real motivator for serious chemical work. Some of the early efforts were aiming to mimic Tyrian purple, also known as Royal Purple, because in antiquity it was worth its weight in gold.

Two classes of molecules come to mind other than anthocyanins in phytochemistry, such as the closely related flavones (its root means yellow) and the distinct carotenes (its root means carrot, but also responsible for flamingo color).

  Can the red colors that give a "normal" anthocyanin pH response arise from pigments held in an acidic environment in the plant?

A real sharp organic chemist by the name of Richard Willstatter proposed this  in the early 1900s. He was working on determining the structure of the pigment responsible for the blue color in cornflower. He extracted the same substance coming from a rose and proposed that all flower colors were due to a different pH in the sap. As your experiment clearly points out, however, alkaline conditions are typically required for the blue anion. A Japanese group argued that alkaline conditions in the sap were unlikely and that instead these different colors can be explained by the formation of metal-organic complexes.

It wasn't until 2005 that a definitive crystal structure was determined from the cornflower pigment called protocyanin. It turns out that the blue color comes from a supermolecule containing one $\ce{Fe^{3+}}$, one $\ce{Mg^{2+}}$, two $\ce{Ca^{2+}}$, six anthocyanin and six flavone molecules packaged together in a complex structure. This color, as I vaguely recall, remains blue from pH 7 until pH 1! This agreed with some earlier evidence which suggested a complex species by the temperature dependence of the pigment.

These are not easy studies to conduct and I'm not familiar with how in-depth red colors have been studied, but it is certainly possible a complex species is actually how the red pigments form in vivo. Keeping in mind that plant vacuoles have a pH on the order of 5, we'll generally notice a loss in red above pH 3 in solution for a pure anthrocyanin in solution as hydration of the cation occurs. Stability of the color, and thus cation, obviously must be obtained either via copigmentation or with metals in a plant.


Why are the densities of europium and ytterbium anomalously low?

Expanding upon @Ivan's comment: Eu and Yb can access the +2 oxidation state instead of +3, due to the +2 ions having a relatively stable half-filled or filled f subshell.
That includes the embedded ions in the structure of the metal. Eu and Yb have $\ce{M^2+}$ instead of $\ce{M^3+}$ ions in the metal. The extra electron and reduced effective nuclear charge on the outer subshells make those $\ce{M^2+}$ ions larger, therefore these metals have lower densities than the surrounding lanthanides. The lower ion charge also means fewer electrons binding the ions together, making Eu and Yb easier to melt.
The difference in ion charge can also show up in other ways. Eu and Yb, for instance, resemble heavy alkaline earth metals more than other lanthanides by dissolving in liquid ammonia.  Europium is also known to appear in some rocks alongside alkaline earth metals rather than alongside other lanthanides (see Wikipedia):

In anaerobic, and particularly geothermal conditions, the divalent form is sufficiently stable that it tends to be incorporated into minerals of calcium and the other alkaline earths. This ion-exchange process is the basis of the "negative europium anomaly", the low europium content in many lanthanide minerals such as monazite, relative to the chondritic abundance.


Whose solubility increases with increase in temperature?

All four substances have higher solubility at higher temperatures in the range from 0 to 100 degrees Celsius.

$\ce{KBr}$ and $\ce{KCl}$ both roughly double their solubility from 0 to 100 degrees Celsius.

To me the question does not make sense.


Two different electronic configurations for cerium

The filling of electrons follows Aufbau principle. However, introduction of $\mathrm{d}$ and $\mathrm{f}$ orbitals and resulting poor shielding affects predicted energy order, and it consists last 2 or three shells incompletely filled. The energy associated with this subshells is nearly same, and one of the subshell of outer shell is bigger in size than subshell of inner shell. So, to perform minimum repulsion between electrons, one or two electrons jump to a bigger subshell (with similar energy) so this exceptions occur.
In case of cerium, energy of $\mathrm{4f}$ and $\mathrm{5d}$ subshells are quite similar and $\mathrm{5d}$ subshell is bigger than $\mathrm{4f}$ subshell. So 1 electron in $\mathrm{4f}$ jump to $\mathrm{5d}$ subshell.


What determines whether a step in a reaction is reversible?

There's no way of figuring that out "from scratch", it is a lot of experience and knowledge. There is no general answer to this but there are some steps which are basically always seen as reversible or irreversible*.

protonation/deprotonation:
unless an extremly strong acid/base (like BuLi) is involved those are always in some kind of equilibrium and are reversible.

exchanging a good leaving group with a bad one:
That can be seen as irreversible.

gas evolution:
as @LinearChristmas pointed out: that can be seen as irreversible (if you are working in solution)

in general: producing a stable structure:
for example aromatization, that's irreversible

*theoretically every reaction is reversible, but if the equilibrium is far on one side we call it irreversible.

For an explanation what makes a reaction reversible see: Is every chemical reaction in equilibrium?


Is FeS2 both an ionic and covalent compound?

Why do you feel a need to classify it under one or another? Chemistry is not about having a list of ionic compounds and another list of covalent compounds.

There are "ionic" bonds (see below) between $\ce{Fe^2+}$ and $\ce{S2^2-}$, and there is a $\ce{S-S}$ covalent bond in $\ce{S2^2-}$. That's all there is to it.

If I absolutely had to pick one, I would say it is an ionic compound. $\ce{S2^2-}$ is a covalently bonded species; $\ce{FeS2}$ is an ionic crystalline solid. The fact that it melts at $1193~^\circ\mathrm{C}$ supports this "classification".

As @Mith notes in the comments, all ionic bonds are not 100% ionic, and a lot of ionic bonds are in fact covalent to a large degree. The fact that $\ce{Fe^2+}$ is a transition metal ion (meaning it has empty 3d and 4s orbitals available for overlap) and also that sulfur is a rather polarisable ion (because of its larger size) means that there is likely to be significant covalency in the $\ce{Fe-S}$ bonds.

What does this mean? It just shows that trying to classify something as "ionic" or "covalent" is futile. $\ce{FeS2}$ is likely somewhere right in the middle of the spectrum.


Is peroxoborate and perborate the same thing?

This is necessarily  not  a chemistry answer but is a continuation of aventurin's answer.

Google ngram viewer is an online search engine h  that charts frequencies of any word or short sentence found in the sources printed between 1800 and 2012 in American English, British English etc.

So, here is the Google ngram for peroxoborate and perborate.

So, you can see that perborate is more often used than peroxoborate. Though, the correct name (IUPAC name) is tetrahydroxidodi-µ-peroxido-diborate, people often mention the compound sodium perborate. Sodium peroxoborate seems to be obsolete.


In this triangular scheme, taking it literally, it is clear that all the population will end up in species C. If $A \rightarrow B$ is mathematically irreversible (rate constant back is zero)  it implies that B is at an energy infinitely below that of A, and the energy of C infinitely below that of B, so clearly C cannot revert to A without the input of an infinite amount of energy. Such an argument hardly makes sense in chemistry, however.

If we now assume that the reaction is reversible but with a very small back rate constant a similar conclusion is obtained, i.e. that the overwhelming (but not total) population ends up in C.

Suppose, without loss of generality and for simplicity, that the activation energy A/B and B/C are the same at $E_A$ and that B is at energy E below A, and C at E below B. Now let the forward rate constant A to B be $k_f=k_0\exp(-E_A/RT) $ and the back rate constant is $k_{BA}=k_0\exp(-E_A/RT)\exp(-E/RT)$ and which is going to be smaller, say it is $10^{-3}$ smaller. The rate constant from C to A is therefore $k_{CA}=k_0\exp(-E_A/RT)\exp(-2E/RT)$ which is now $10^{-6}$ times smaller that the rate constant A to B.


What does 'organic/non-organic molecule' means exactly?

There are many different answers to the question, "What is an organic compound".  Here are a few examples:

Pudue - "compounds that contain both carbon and hydrogen"
Britanica - compounds in which one or more atoms of carbon are covalently linked to atoms of other elements, most commonly hydrogen, oxygen, or nitrogen. The few carbon-containing compounds not classified as organic include carbides, carbonates, and cyanides
Wikipedia - any member of a large class of gaseous, liquid, or solid chemical compounds whose molecules contain carbon. For historical reasons discussed below, a few types of carbon-containing compounds such as carbides, carbonates, simple oxides of carbon (such as CO and CO2), and cyanides are considered inorganic
UC Davis- Any compound containing carbon atom(s) is classified as an organic compound

Trying to pull the best from each of these while still keeping it simple, perhaps a good definition of an organic compound would be - any covalent compound containing, at least, carbon and hydrogen.  Everything else would be inorganic.


What is happening to my chlorine dioxide when I add too much bleach or citric acid?

The "mystery powder" was proposed to be sodium bisulfate, in a recent answer to another question. Sulfuric acid is used in one process to turn sodium chlorite into $\ce{ClO2}$; bisulfite would be slower, but should do the same.
Citric acid will acidify the sodium chlorite and sodium hypochlorite, but will itself be oxidized, perhaps all the way to $\ce{CO2}$. $\ce{HCl}$ is used to convert $\ce{NaClO2}$  and $\ce{NaOCl}$ to $\ce{ClO2}$ and citric acid could well do the same (be oxidized at the same time).
$$\ce{2 HCl + NaClO + 2 NaClO2 -> 2 ClO2 + 3 NaCl + H2O}$$
$$\ce{4HCl + 5NaClO2 -> 4ClO2 + 5NaCl + 2H2O}$$
Chlorite ion and hypochlorite ion are known to react rapidly to form chloride and chlorate (Lister, Can. J. Chem., 30, 879-889, 1952).
The unknown amounts of oxidizing agents and citric acid (a reducing agent) and the complex reactions make the pale yellow color a poor indicator of what might be going on. The presence of $\ce{ClO2}$ does not seem to be positively identified and might not even be needed to explain the observations.


This seems interesting.
The $(CH_3)_3CO^-$ is a strong nucleophile and we observe here methyl bromide is being attacked and there is no point of elimination. There has to substitution in this case.
The bulkier the nucleophile is the more probable is the elimination product due to steric hindrance. In majority of the cases Tert-butoxide undergoes elimination reaction.
The further factor includes temperature at which the reaction is being carried out.

We can show this relation with $$\delta G  = \delta H - T \delta S$$
Hope it was helpful.


ortho/para-Selectivity in the Reimer-Tiemann Reaction

The Reimer–Tiemann reaction will give a mixture of products unless the formation of one product is highly disfavoured. The Kürti-Csakó notes in its introductory paragraph on the Reimer–Tiemann:[1]

the regioselectivity is not high, but ortho-formyl products tend to predominate;

The statement already hints at a product mixture being obtained. To further show that both ortho and para attacks are possible, the 1960 review published by Wynberg contains the following scheme as its very first scheme in the introduction:[2]

So while both products are obtained, a back of the envelope calculation already shows that the distribution is not a perfect $2:1$ ratio. This can apparently be explaned by the effect of positive counterions, as Hine and van der Veen report:[3]

The ratio of ortho to para product was found to be 2.21 [under high base concentrations], showing that the tendency towards o-substitution is indeed increased under conditions where ion-pair formation is encouraged. One factor that would certainly be expected to be present and that would tend to favor o-substitution is an electrostatic effect. When a dichloromethylene molecule attacks the o-position of a sodium phenoxide ion-pair to yield the probable initial product, there is less separation of unlike charges than when the analogous para product is formed.

So while the steric hindrance of a single oxygen atom is not much more than that of a hydrogen atom and thus does not play a huge role, the presence of positive counterions – although increasing steric congestion – may serve to enhance the formation of the ortho-product through favourable electronic effects.

References:
[1]: L. Kürti and B. Czakó: Strategic Applications of Named Reactions in Organic Synthesis. Background and Detailed Mechanisms, Elsevier Academic Press, Burlington, MA, USA, 2005, page 378.
[2]: H. Wynberg, Chem. Rev. 1960, 60, 169–184. DOI: 10.1021/cr60204a003.
[3]: J. Hine, J. M. van der Veen, J. Am. Chem. Soc. 1959, 81, 6446–6449. DOI: 10.1021/ja01533a028.


If you look at the net reaction, it is just combustion of glucose:

$$
\ce{C6H12O6 + 6O2 -> 6CO2 + 6H2O}
$$

So you could just burn a sample of glucose, and it would produce water.


Why carbonyl groups are strong benzene deactivating group for electrophillic aromatic substitution?

A carbonyl group such as an aldehyde is a deactivating group as it withdraws electron density by both the inductive effect (through the sigma bonds) and the resonance effect which involves pi systems.

Here are three resonance forms which show how the aldehyde group takes electron density from the benzene ring in benzaldehyde. There are more resonance forms and some different but related ones account for the reduction in the stability of the cationic intermidate in an aromatic substitution reaction.


From Wikipedia

  The prefix bi in bicarbonate comes from an outdated naming system and is based on the observation that there is twice as much carbonate ($\ce{CO3}$) per sodium in sodium bicarbonate ($\ce{NaHCO3}$) as there is in sodium carbonate ($\ce{Na2CO3}$).

The IUPAC name for it agrees with its chemical formula: sodium hydrogen carbonate.


What happens when acetone reacts with hydrogen cyanide?

The acetone carbonyl is attacked by the CN$^-$ that comes from the NaCN.  This tetrahedral intermediate has a oxygen anion, which then grabs a proton from the HCN, regenerating the CN$^-$.


There might be more recent publications, but an old review by Robert G. Johnson and Robert K. Ingham, Chem. Rev., 1956, 56, 219-269, gives some directions.

Based on early attempts with benzoic acid that date back to the time around 1840, is was long believed that the silver salts of aromatic carboxylic acids do not undergo a Hunsdiecker reaction at all. The authors write:

  In no case was bromobenzene obtained as a result of the
  action of bromine on silver benzoate. The product of Peligot and Bunge was a monobromobenzoic acid; apparently Kekule’s experiment was completely unsuccessful.

However, it later turned out that the inital verdict wasn't quite true.

Benzoic acids bearing electron-withdrawing substituents, such as $\ce{Cl}$ or $\ce{NO2}$ can be converted to their corresponding bromoarenes in a Hunsdiecker reaction. The authors of the review denote:

  Thus, the three isomeric nitrobenzoic acids (as silver salts) were
  converted to the corresponding bromides in excellent yields (ortho, 95 per cent; meta, 89 per cent; para, 79 per cent). The yields from the chlorobenzoic acids were less satisfactory, but sufficiently high to indicate the efficacy of the chloro group.


What will happen if I put this chemical into my water tank?

The product which you have mentioned in your post may have other chemicals like fragrance etc... which may make the water not suitable for washing dishes.

You can purchase sodium hypochlorite (bleach) which is sold separately as powder or solution and add to the water tank. The concentration added to water tank is crucial.

You can use unscented / fragrance free laundry bleach solution also. This will be easier to find.
If you get solution (6% solution of bleach in water) then you can add 40mL to the 8000L full tank first time and mix.

Then, say after a week when your tank has 2400 Liters full (using 800L per day ); you can add 28-30mL again and then fill up tank to full (8000L). (Approximate ratio is 1-1.5mL per 200 Liters of water).

Preventing algal growth:

To prevent the growth of algae it is always recommended to prevent light exposure of stored water. Dark colored/black tanks do well. You can cover the manholes/ crevices through which light may enter the tank.
Additionally, warm water aids in algal growth, so you can try to cover the tank with some tarpaulin etc... to prevent heating.
Regular cleaning is important. Algae already growing on the inner wall of the tank is difficult to to deal with. Proper regular cleaning will help to remove such growth.


What is an effective fabric-friendly solvent for aluminium chlorohydrate?

My undershirts have turned yellow in the armpit area without the benefit of aluminum antiperspirants. I assumed the stains were from carbonaceous exudants from my pores - or the biogenic products of bacterial life on my skin.
So I bleached the undershirts, soaked them in a diluted bleach (sodium hypochlorite) solution (about 1 in 10, I think). BIG improvement. Not perfect, but very good. One of the side effects was that the undershirts turned blue in the bleach because of the whiteners in the surfactant used to wash them. But that faded as the bleach continued to oxidize the stains.
You can go overboard and use more concentrated bleach: this will perk up the entire undershirt! Interestingly, it seemed to not only clean the underarm stains, but also attack/weaken the fabric there, so the stains went away, but small holes started to appear there, which grew larger and larger until the undershirts became nice dust cloths.
Next time, I think I will try an oxygen laundry bleach, even if it takes longer. Or H2O2.
Addendum: the supposition that the brown deposits in the armpit area are due to aluminum compounds is widespread. Even undershirt makers suggest it - but of course, take no blame for providing the substrate that shows the stain. One manufacturer (https://thompsontee.com/blog/the-science-behind-your-pit-stains-and-what-to-do-about-it/) even has a link to fixing it. What do they suggest? Change your antiperspirant; take proper care of your clothing; wear a sweatproof undershirt. (They offer the best sweatproof undershirt: 20% off!) But how to get rid of armpit stains?
#1. Vinegar, baking soda and HYDROGEN PEROXIDE.
#2. OxiClean or Raise Stain Remover (mixed reviews)
Nivea makes deodorants and has a site that recommends ways to remove armpit stains: try an acid. For example: #1: HYDROGEN PEROXIDE #2. Baking soda (?) #3.White vinegar #4.Lemon juice. https://www.nivea.com.au/advice/skin/removing-deodorant-sweat-stains
The desire to use environmentally safe washing agents in certainly admirable, but when you have to do a job, it is quicker to use the powerful tool. Of course, there is a learning curve. I love bleach because its activity is so adjustable and controllable - from an additive at low concentration to eradicate gingivitis by waterpicking to clothes cleaning to mildew removal from house siding, and probably a million more uses.
But I suppose hydrogen peroxide will be adequate. Or OxiClean.


Will selenium react or bond, ionic or covalent, with vanadium?

The ASM Alloy Phase Diagram Database (at asminternational.org if you have access) shows that $V_{2}Se_{9}$, $V_{5}Se_{8}$, $V_{3}Se_{4}$ $V_{5}Se_{4}$, and $VSe$ are all accepted stable phases in the Se-V binary system.


Why does adding salt to water reduce the pH?

This is probably an effect in which sodium ions from the salt are being weakly detected by the glass electrode, giving the appearance of a lower pH. Try add the same concentration of salt to samples of your reference buffers and recalibrate your pH meter to those before measuring the pH of the salt water. I would expect you would see pH 7. Also, keep in mind that the pH of unbuffered water is very sensitive to low concentrations of certain impurities -- from the air or from the added salt. For example, carbon dioxide in the air will lower the pH over time.


Atomic weights in any periodic table are based on the natural abundance of different isotopes of the element. But sometimes the source matters and the results will be different.

The most important thing to note here is that creating new isotopes is hard. You can't just "add" neutrons to an atomic nucleus and often when you do it decays to a different element.

So most periodic tables quote atomic masses using the natural abundance of the elements as they are found in nature.

When there are exceptions, different atomic masses might be quoted for samples know to deviate from the natural abundance (or, more commonly, the specific ratios of the isotopes will be mentioned). Most elements abundance will be determined by the natural processes in stars (or the Big Bang) that created them in the first place. Occasionally natural nuclear processes will alter that ratio. For example, the radioactive isotope of carbon, 14C has a half-life of 5-6 thousand years but is produced naturally in the upper atmosphere. It is well distributed in the air and incorporated into plants. But if those plants die, the amount of the isotope will decay over time lowering the ratio of that isotope in the plant. This is both the basis of carbon dating and the reason why it is possible to tell whether alcohol is produced from fossil fuel sources (old enough that all the 14C has decayed) or from fermentation of sugars (produced from living things so incorporating the atmospheric level of the isotope).

It is even possible for nature to contrive conditions that mimic a nuclear reactor causing havoc with the natural abundance of heavy isotopes of uranium. In the uranium deposits in Oklo (in Gabon) the ratio of 235 was found to be far lower than the normal levels elsewhere because of a natural nuclear reactor that ran 2bn years ago.

But we don't change the numbers in the periodic table because of this. When we talk about specific samples of elements where the ratio is not normal, we specifically mention what the ratio is compared to the natural level.


Why CO2 has 4 vibrational degrees of freedom?

Molecular vibrations can certainly occur by changing the O-C-O bond angle away from 180º. When we call the molecule linear we are saying that the geometric minimum is a linear configuration; however, vibrational modes can (and do) break that symmetry. What's important is that the average nuclear position is a linear molecule (or if you're doing a geometry calculation you would say that it's the 0 K positions).

So the two vibrational modes you're missing are the two O-C-O bends at 90º from each other.

Also as a note: you seem to be suggesting that the two stretches correspond with the two bonds stretching. That's not quite true. One of them corresponds to both C-O bonds stretching outward at the same time (in phase) and the other will correspond one compressing while the other stretches (out of phase).

For a full derivation of the normal modes of CO$_2$ there are many sites, but this one looked good: http://www.colby.edu/chemistry/PChem/notes/NormalModesText.pdf


Crystal structure database search for space groups

You could use COD SQL interface. For example this little Python script returns all entries in space group F-43m that contain Si:

#!/usr/bin/env python
from mysql import connector
con = connector.connect(user='cod_reader', host='sql.crystallography.net',
                        database='cod')
cursor = con.cursor()
cursor.execute("""select file, formula from data
                  where formula LIKE '% Si%' AND sg = 'F -4 3 m'""")
for n, formula in cursor:
    print n, formula

The script prints entry number and formula for each matching COD entry:

$ python cod.py
1010995 - Si C -
1011031 - Si C -
1011254 - Al13 Cl F9 H9 O29 Si5 -
9000281 - Al13 Cl F3.84 H14.16 O34.16 Si5 -
...


There is more to hydrogen bonding than electronegativity.  To get the strongest hydrogen bonds you need a compact atom to concentrate the negative charge for electrostatic attraction and to overlap well with the hydrogen 1s orbital for the molecular orbital/delocalized covalent bond contribution (yes, the latter is a thing.)  Nitrogen has that, chlorine not so much.  So, you get stronger hydrogen bonds with nitrogen than with chlorine.

Did you catch the wording?  Though chlorine has disadvantages it can form (weaker) hydrogen bonds under some conditions.  See here; the abstract indicates that the interaction can exist in some chlorinated organic compounds.


Are there magnetic rocks, and not hematite?

Magnetite is indeed very magnetic. But other minerals are magnetic too, and given a strong enough magnet you will attract more than magnetite. Chromite is one that comes to mind that's also common in some black sands.

You might not always see the magnetite. Magnetite is mixed valence of iron 2 and 3 and as such it isn't stable in our oxygen rich atmosphere. Given enough time it will alter to a variety minerals such as hematite and iron hydroxides. So you might have magnetite in the core of a grain mantles by non magnetic hematite. It will look brown but it will still be attracted to a magnet.

Same thing with your grey rocks: they most likely have magnetite inside, hidden from view.


How do people know HCN smells like almonds?

The odour threshold for hydrogen cyanide $(\ce{HCN})$ is in fact quite a bit lower than the lethal toxicity threshold. Data for $\ce{HCN}$ can be found in many places, but here and here are a couple of good references. That subset of the human population that can detect bitter almonds do so at a threshold of $0.58$ to $\pu{5 ppm}$. The lethal exposure dose is upwards of $\pu{135 ppm}$. That's a whole $\pu{100 ppm}$ range in which to detect and report the fragrant properties.


Writing the overall cell reaction for the Calomel electrode

The calomel electrode is not a cell. It can be part of a cell. It is a half-cell. A cell is made of two electrodes, one calomel electrode and another one. The reaction occurring in the calomel electrode is the reaction you present as "overall cell reaction". What you described as "anode half-cell" has nothing to do with electrochemistry. It is the equation describing how $\ce{Hg_2Cl_2}$ gets dissolved in water.


Why bath sonicators don't come with adjustable power setting?

All the bath sonicators I saw online don't have a power level setting.

This depends on the product lines.

I've mostly used BANDELIN SONOREX SUPER ultrasonic baths, and indeed, the most simple models of this line just have a (mechanical) timer.

However, models of the SONOREX DIGIPLUS line do come with power control.

In the past, the ultrasonic baths were mostly used for cleaning purposes, while the probes, which typically had a much higher power, were used for sample homogenisation and similar purposes. Here, power regulation was apparently more important to avoid degradation of the same due to local overheating.

As usual, exceptions prove the rule:

I've used ultrasonic baths to support oxidations with $\ce{MnO2}$ or PCC on silica
Cleaning of modern (composite) materials seemingly requires power control, e.g. to avoid ablation


What is reactivity really, and can it be quantified?

I think the term "reactivity" is often used to denote kinetic lability, a proclivity to react.  Let's look at the two reaction coordinates shown below.  They both represent the interconversion of molecules A and B through intermediate I.

Note that the potential well that I exists in is much deeper in the drawing on the left than the drawing on the right.  Therefor the activation energy for I to escape will be much larger for the intermediate on the left than the intermediate on the right.  The intermediate on the right would be said to be more reactive (have a shorter lifetime) than the intermediate on the left, because it requires less energy for it to escape.

An experimentalist can infer the depth of the well through experiments.  Through such experiments the experimentalist learns that free radicals, highly strained hydrocarbons, carbenes, etc. exist in potential wells that must be very shallow because they react so readily.  Using this empirical data the experimentalist can then infer how similar compounds are likely to react. A computational chemist can infer the same thing by calculating the profile of the reaction coordinate for the various reactions under investigation by the experimentalist, or as is becoming increasingly common the computational chemist can predict what should be and let the experimentalist confirm those predictions.


The difference between biodegradable polymers and 'regular' polymers is the bonds that make up the backbone. The backbone is the part of the polymer that extents the brackets. Anything attached to that chain is a side group and is less important.

The two examples you show here are very dissimilar. The polypropylene is only made up of Carbon-Carbon bonds. These bonds are very stable: that holds for chemical and 'biological' stability. The PHB how ever bonds monomers with an Ester functional group (Ester). This group is far more reactive that a C-C bond.

The reactivity of the groups the polymer is made up of, is very, very important. That being said, I cannot claim that all poly-esters are biodegradable. Depending on the groups close to the esters and the enzymes (probably from bacteria) involved, the bond might be cleaved, or might be inaccessible.

Hope this helped!


Are all complexes with a polydentate ligand examples of chelation?

I think you should not focus on the ligand to define chelation. You should instead focus on the complex that is formed. In fact, the very same multidentate ligand may participate in binding in more than one way:

by forming more than one bond with the same central atom, therefore forming a chelated complex;
by forming only one bond with that central ion, even though it could form two; this may be the case when a vast excess of this ligand is present with respect to the metal ion;
by bridging between two adjacent central ions.

Only the first of these possibility is referred to as chelation. In other words, forming a chelated complex doesn't depend solely on the fact that the ligand is multidentate. Being multidentate is a necessary, but not sufficient condition for a ligand to form chelated complexes.


Hydrolysis of nitriles: Amide vs Carboxylic acid

Hydrolysis of nitrile is pH dependent. It will lead to formation of different products based on pH.
Theoretically, nitrile hydrolyze to form amide and then ammonium salt.
$$\ce{RCN ->[H2O] RCONH2 ->[H2O] RCOONH4}$$
But in practice, the reaction between nitriles and water is so slow that it is considered negligible. The nitrile is instead heated with either a dilute acid such as dilute hydrochloric acid, or with an alkali such as sodium hydroxide solution to accelerate the hydrolysis reaction. But this pH dependent reaction yields different products.

Acid hydrolysis: yields carboxylic acid

$$\ce{RCN + 2H2O + HCl -> RCOOH + NH4Cl}$$

Alkaline hydrolysis: yields sodium carboxylate salt

$$\ce{RCN + H2O + NaOH-> RCOONa + NH3}$$
To obtain amide, hydrolysis is done selectively and in controlled manner. It can be done in both acidic and alkaline medium. In acidic medium, nitriles are converted into amide in a single step process using TFA-$\ce{H2SO4}$ mixture as reagent.

In alkaline medium, it can done using alkaline solution of hydrogen peroxide (see @Light Yagami's answer).
Reference

Chemguide
Facile and Highly Selective Conversion of Nitriles to Amides via Indirect Acid-Catalyzed Hydration Using TFA or AcOH−H2SO4, Jarugu Narasimha Moorthy and Nidhi Singhal, The Journal of Organic Chemistry 2005 70 (5), 1926-1929
DOI: 10.1021/jo048240a


Is every compound that has a H-bond donating group soluble in water?

Definitely not.
For instance, palmitic acid with a pK$_\mathrm{a}$ of 4.75 has a water solubility of only 0.04 mg/L.

A general rule of thumb is that every oxygen can solubilize four carbon atoms but there are many exceptions.
I addressed a similar question here.

Nitrogen also makes the compound more soluble, but there it depends very much on the pH.


Why is thionyl chloride preferred for preparing alkyl chlorides from alcohols?

Thionyl chloride is preferred for preparing alkyl chlorides from alcohols because the by-products formed in the reaction are $\ce{SO2}$ and $\ce{HCl}$ which are in gaseous form and escape into the atmosphere leaving behind pure alkyl chlorides.

$$\ce{CH3CH2-OH  + SOCl2   →  CH3CH2-Cl  +  SO2 ↑ + HCl ↑}$$


Why is the Standard Entropy of a Proton Zero?

Indeed, it is clear that these values are not... Wait! What "these values"?

You are putting two entirely different things in the same basket. (If your textbook does so, then it does a poor job.) It is much like treating dogs and cupboards similarly, on the basis of both having four legs. Then, however, many generalizations will fail quite miserably. You feed them meat, but the results are not always great. You call in the carpenter to fix them, but...

Same thing here. See, ions and molecules are very, very different things. You totally can have a mole of $\ce{CO2}$. It is a thing. You can work with it, can measure its heat capacity and stuff, and eventually work out the entropy. Indeed, the textbook value of $\ce{CO2}$ entropy is absolute.

On the other hand, you totally can't have a mole of $\ce{Al^3+}$. It is not a thing. Unless you have some counterions, that is, but in that case you can't tell how much of the entropy comes from the counterion and how much from $\ce{Al^3+}$.

How can we assign entropies to ions, then? Why, it is as simple as that: we assign one ion arbitrarily, then work out the entropies of all its counterions from there, and then turn to their counterions. Just what one ion could that be? Why, let's say it is $\ce{H+}$, because so much of our chemistry is dominated by the acid/base equilibria. It is kind of natural to consider its entropy $0$, much like the absolute elevation of the sea level is $0$. Now there you have them, the textbook entropies of ions, which are not quite absolute.

So it goes.


Why does copper react only with nitric acid?

Copper is an unreactive metal and doesn’t react in normal circumstances with dilute acids. However, it does react with nitric acid. Why is this?

Nitric acid is an oxidizing agent and the reaction is not the usual acid + metal reaction. The products are oxides of nitrogen instead of hydrogen. The actual nitrogen oxide formed depends on the concentration and temperature of the acid.

There are actually two equations for the reaction of copper with nitric acid. It depends on whether the nitric acid is concentrated or not. If it is concentrated and in excess then the ratio is 1:4 copper to nitric acid. If it is dilute then the ratio is 3:8.

$$
\begin{align}
\ce{Cu + 4HNO3 &–> Cu(NO3)2 + 2NO2 + 2H2O}\\
\ce{3Cu + 8HNO3 &–> 3Cu(NO3)2 + 2NO + 4H2O}
\end{align}
$$

Copper also reacts with hot conc. $\ce{H2SO4}$ but, here $\ce{H2}$ gas is released as follows :

$$
\begin{align}
\ce{Cu(s) + H2SO4(aq,conc) &-> Cu^2+(aq) + SO4^2-(aq) + H2(g)}\\
\ce{Cu(s) + H2SO4(aq, dil) &-> no ~reaction}\\
\ce{Cu(s) + HCl &-> no ~reaction}
\end{align}$$


Confusion in naming a carboxylic acid with a cyclohexyl group

According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), the senior parent structure has the maximum number of substituents corresponding to the principal characteristic group.

  P-44.1 SENIORITY ORDER FOR PARENT STRUCTURES

  When there is a choice, the senior parent structure is chosen by applying the following criteria, in order, until a decision is reached. These criteria must always be applied before those applicable to rings and ring systems (see P-44.2) and to chains (see P-44.3). Then criteria applicable to both chains and rings or ring systems given in P-44.4 are considered.

  P-44.1.1 The senior parent structure has the maximum number of substituents corresponding to the principal characteristic group (suffix) or senior parent hydride in accord with the seniority of classes (P-41) and the seniority of suffixes (P-43).

  (…)

If there is still a choice (for example, if the ring and the chain each have a substituent corresponding to the principal characteristic group), the ring or the chain can be the senior parent structure; however, for the preferred IUPAC name, the ring is always selected as the senior parent structure.

  P-44.1.2.2 Systems composed of rings and chains (exclusive of linear phanes)

  Two methods are recognized to name systems composed of rings and chains (exclusive of linear phanes).

  (1) Within the same class, a ring or ring system has seniority over a chain. When a ring and a chain contain the same senior element, the ring is chosen as parent. Rings and chains are chosen regardless of their degree of hydrogenation. As a consequence, this approach prefers the choice of a ring over a chain in systems composed of cyclic and acyclic hydrocarbons.

  (2) The context may favor the ring or the chain, so that, for example, substituents may be treated alike or an unsaturated acyclic structure may be recognized, or the one chosen has the greater number of skeletal atoms in the ring or in the principal chain of the acyclic structure.

  (…) For selection of a preferred IUPAC name, see P-52.2.8.

  P-52.2.8 Selection between a ring and a chain as parent hydride

  Within the same heteroatom class and for the same number of characteristic groups cited as the principal characteristic group, a ring is always selected as the parent hydride to construct a preferred IUPAC name. In general nomenclature, a ring or a chain can be the parent hydride (see P-44.1.2.2).

The compound given in the question contains a characteristic group $(\ce{-COOH})$. Since there is only one characteristic group, the seniority order of classes is not relevant in this case; thus, the $\ce{-COOH}$ substituent corresponds to the principal characteristic group.

Since the principal characteristic group is a substituent of the chain, and the ring does not have any substituents corresponding to the principal characteristic group, the senior parent structure of the compound given in the question is the chain and not the ring. The rules P-44.1.2.2 and P-52.2.8 for the selection between a ring and a chain as senior parent structure are not relevant in this case.

Simple carboxylic acids are named substitutively using the suffix ‘oic acid’ as follows.

  P-65.1.2.1 Carboxylic acid groups, $\ce{-COOH}$, that conceptually replace a $\ce{-CH3}$ group of methane or terminate an unbranched hydrocarbon chain are named by replacing the final ‘e’ of the name of the corresponding hydrocarbon by the suffix ‘oic acid’. No locants are necessary to denote the positions of the carboxylic acid groups in a hydrocarbon chain; locants are used when hydrocarbon chains are modified by skeletal replacement, as shown in P-15.4.3.2.3. Except for formic acid, acetic acid, oxalic acid (see P-65.1.1.1), and oxamic acid (see P-65.1.1.1), systematically formed names are preferred IUPAC names; the names given in P-65 .1.1.2 are retained names for use in general nomenclature.

Therefore, the systematic name for the unsubstituted carboxylic acid structure that is given in the question is ethanoic acid (not ‘ethan-1-oic acid’ since no locants are used to denote the positions of the carboxylic acid groups in a hydrocarbon chain). However, the preferred IUPAC name (PIN) of ethanoic acid is the retained name acetic acid according to rule P-65.1.1.1.

The acetic acid chain is substituted using the usual principles of substitutive nomenclature.

However, according to rule P-14.3.4.6, locants are omitted where there is only one kind of substitutable hydrogen.

  P-14.3.4.6 All locants are omitted for parent compounds when all substitutable hydrogen atoms have the same locant.

Therefore, the preferred IUPAC name name for the complete compound given in the question is cyclohexylacetic acid since the locant ‘2’ of ‘2-cyclohexylacetic acid’ is omitted.


Identify the structure of compound with C₉H₁₀ formula using IR and NMR data

The molecular formula of the compound is $\ce{C9H10}$. The number of unsaturations in the compound of $\ce{C_nH_m}$ is $\frac{2n+1-n}{2}$. Thus, the number of unsaturations in $\ce{C9H10}$ is $\frac{2\times 9+1-10}{2}=5$. You already found out that compound is aromatic by FT-IR, and hence $\ce{C6}$ of the formula counted out with 4 unsaturaations (3 double bond and a ring for aromatic nucleus).The remaining number of unsaturarion is $5-4=1$, and remaining number of $\ce{C}$ in the formula is 3. The remaining unsaturation could be a double bond or a ring.
When look at $\ce{^1H}$-$\mathrm{NMR}$ spectrum, chemical shifts of all but aromatic peaks are less than $\pu{3.00 ppm}$. Therefore, it is safe to conclude that the compound is a aromatic bicyclic rather than aromatic alkene. Therefore, the compound is indane:

The protons on 1-$\ce{C}$ and 3-$\ce{C}$ are chemically and magnetically equivalent, and benzylic, hence the their relevant signal would be down-field to that of protons on 2-$\ce{C}$, which is usual acyclic aliphatic carbon. The benzylic protons have two vicinal protons, thus the corresponding peak is a triplet. Meanwhile, protons on 2-$\ce{C}$ have four vicinal protons (benzylic protons), thus the corresponding peak is a quintet (5 peaks).
Note: The integration given is a ratio, not correct number of protons.


Well spotted! I dare say, the book appears to have used a covalent radius for Europium, while using atomic radii for the other lanthanides.

See "atomic radius" vs "covalent radius" values: for Europium, Samarium, and Gadolinium.

Also see this article if you don't like my referencing Wikipedia :P


Solubility of oxygen in water with temperature, why does the change in solubility decrease around 40 °C?

The relatively abrupt changes in slope (do a derivative plot to get an even more distinct picture of what is happening here) are due to changes in the solubility function dominating the solubility behavior at a given temperature and $\ce{O2}$ concentration.  For example, Henry's Law will describe the behavior at low $\ce{O2}$ concentrations.  As Henry's Law applies to the theoretical "infinite dilution" condition, this assumption falls apart at higher concentrations.

The greater the concentration of $\ce{O2}$ the more it departs from the ideal infinite solution assumed for Henry's Law behavior.  At higher concentrations, one or more of the different gas laws then controls the solubility.  Based on your graph, it seems that there are three distinct regions, and thus three different gas laws defining the solubility of $\ce{O2}$ as a function of temperature.

One can also take an empirical approach to predicting the solubility based on experimental measurements.  This website has a live "$\ce{O2}$ solubility in water" calculator based on empirical data.  The authors use two different equations for two different temperature regimes: $\ce{0 ^oC - 30 ^oC and 30 ^oC to 50 ^oC}$.  So their $\ce{30 ^oC}$ cutoff seems to agree with the feature at that temperature in your plot.  I don't know the maximum temperatures / concentrations they used for their second equation.


Diffusion experiment with dialysis membrane bag

You can double-check the area after you're done:

Poke a small hole in the membrane and drain it completely.
Stretch gently by pulling the clips (don't let them slip) until the area between them is rectangular.
Multiply that area (L x W) by two, because the membrane has two sides around the now-flattened middle.

There will always be some small error, because the wet membrane can be stretched somewhat, but that should be insignificant compared to other sources of inaccuracy, such as edge-effects where the clips narrow the membrane, impeding flow.


In thermodynamics, the surroundings are typically treated as an ideal constant temperature reservoir for which $\Delta S$ is always equal to $Q_{surr}/T_{surr}$, irrespective of whether the process that the combined system and surroundings experiences is reversible.  As such, the surroundings is assumed to have an infinite heat capacity and thermal conductivity, and to not change significantly in temperature (its temperature is always $T_{surr}$), no matter how much heat it receives or rejects.  So it always presents the temperature $T_{surr}$ to the system at its interface with the system.  And, all the entropy generated in any process is confined to the system.

In evaluating the entropy change for system and surroundings, the system and the surroundings must first be separated from one another, and then each must be separately subjected to a reversible path.  These separate reversible paths do not have to match up in any way whatsoever.  For the surroundings, the final state is that which was obtained with the irreversible path when $-Q_{irrev}$ was transferred from the system. So the change in entropy for the surroundings is $-Q_{irrev}/T_{surr}$.  And the change in entropy for the combination of system and surroundings is then given by:
$$\Delta S=\Delta S_{sys}-\frac{Q_{irrev}}{T_{surr}}=\left[\int{\frac{dQ_{rev}}{T}}\right]-\frac{Q_{irrev}}{T_{surr}}$$


What Affects Tm and Tr In Chromatography?

Think of it like a solvent extract. Each solvent extraction is a plate.

So at time 1 I do extraction 1  and move top layer to extraction flask 2.

At time 2 I have two extractions to do. I extract flask 2 and move top layer to flask 3 and then I extract 1 again and move top layer to 2 again.

At time 3 I have three extractions to do. and so one.

Another analogy.

Think of the whole thing as binomial trials. Each molecule has a different binomial probability. So the absolute difference in the peak positions between the molecules gets bigger as the number of trials increases.


First, recall that a single bond has the highest free rotation possible (and hence it shows conformational isomerism). A double bond instead does not show any rotation at all (hence, $\ce{abC=Cfg}$ type molecules show geometrical isomerism).

So, it's obvious that, the more double bond character a bond will have, the less likely it is to rotate freely. Here, we're talking about double bond character, because, due to resonance effects, the resonance hybrid in these compounds will only have a partial double bond character.

Thus, you only need to compare the partial double bond character in each compound, and apply the above rule. Bulkiness of the groups isn't really a factor, because the groups actually aren't that bulky, and it isn't a t-butyl group anyway.


What electrodes to use to minimise corrosion during water electrolysis?

It depends on the electrolyte as well as the electrode material. For example, $\ce{NaCl}$ releases atomic chlorine at the positive electrode, quickly attacking most materials except metals of the $\ce{Pt}$ group, as @MaxW mentions.

$\ce{NaOH}$ or $\ce{KOH}$ are a bit gentler on the electrodes. Run the apparatus at a low over-voltage to avoid damaging them. Coated nickel electrodes are suggested for use with alkaline solutions.

Phosphoric acid is also suggested as an electrolyte.


So, shouldn’t the order be the other way round?

Yes, it is the other way around.  Ketones (pKa ~ 20) are more acidic than esters (pKa ~ 25).  The figure below shows the resonance structures for the enolate anion of a ketone and an ester.  Resonance structures II and IV stabilize the enolate anion in the ketone and ester respectively.  In the ester, there is an additional resonance structure, V. Through resonance structure V, the alkoxy group in the ester donates electrons to the carbonyl.  This movement of electron density towards the carbonyl will tend to destabilize the formation of any negative charge on the carbon attached on the other side of the carbonyl (electrostatic repulsion).  This reduces the likelihood of the formation of an enolate anion on that carbon. Therefore, in the ester, to whatever extent structure V contributes, structure IV will contribute somewhat less.  Hence, the ester enolate anion resonance structure IV plays a smaller role and the enolate anion is less stabilized in the ester compared to the ketone.  Consequently, the ester is less acidic than the ketone.


How can I find the symmetry labels of atomic orbitals in a molecule?

First, you have to know the geometry of your compound. The complex $\ce{[PtCl4]^2-}$, for example, is square planar.
The next step is to determine the point group of the compound from this geometry by identifying the symmetry elements that it posesses (a procedure on how to do this can be found here). The point group of the complex $\ce{[PtCl4]^2-}$, for example, is $D_\mathrm{4h}$.
Now you consult the character table of the point group you determined in the previous step. The character tables for many common point groups can be found here. The character table of the point group $D_\mathrm{4h}$ is shown below and can also be found here.

$$\begin{array}{c|cccccccccc|cc} \hline
D_\mathrm{4h} & E & 2C_4 & C_2 & 2C_2' & 2C_2'' & i & 2S_4 & \sigma_\mathrm{h} & 2\sigma_\mathrm{v} & 2\sigma_\mathrm{d} & & \\ \hline
\mathrm{A_{1g}} & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & & x^2+y^2,z^2 \\
\mathrm{A_{2g}} & 1 & 1 & 1 & -1 & -1 & 1 & 1 & 1 & -1 & -1 & R_z & \\
\mathrm{B_{1g}} & 1 & -1 & 1 & 1 & -1 & 1 & -1 & 1 & 1 & -1 & & x^2-y^2 \\
\mathrm{B_{2g}} & 1 & -1 & 1 & -1 & 1 & 1 & -1 & 1 & -1 & 1 & & xy \\
\mathrm{E_g} & 2 & 0 & -2 & 0 & 0 & 2 & 0 & -2 & 0 & 0 & (R_x,R_y) & (xz,yz) \\
\mathrm{A_{1u}} & 1 & 1 & 1 & 1 & 1 & -1 & -1 & -1 & -1 & -1 & & \\
\mathrm{A_{2u}} & 1 & 1 & 1 & -1 & -1 & -1 & -1 & -1 & 1 & 1 & z & \\
\mathrm{B_{1u}} & 1 & -1 & 1 & 1 & -1 & -1 & 1 & -1 & -1 & 1 & & \\
\mathrm{B_{2u}} & 1 & -1 & 1 & -1 & 1 & -1 & 1 & -1 & 1 & -1 & & \\
\mathrm{E_u} & 2 & 0 & -2 & 0 & 0 & -2 & 0 & 2 & 0 & 0 & (x,y) & \\ \hline
\end{array}$$

Finally you need to determine the symmetry labels of the p- and d-orbitals within the point group by identifying the irreducible representations (the labels on the far left of the character table) which match the transformational properties of the basis vectors ($\color{red}{x}$, $\color{red}{y}$, $\color{red}{z}$) and some of their quadratic combinations ($\color{blue}{xy}$, $\color{blue}{xz}$, $\color{blue}{yz}$, $\color{blue}{z^2}$, $\color{blue}{x^2 - y^2}$). These can be found in the two columns on the right columns. Again this will be shown for the example compound $\ce{[PtCl4]^2-}$ where the relevant functions are highlighted.

$$\require{bbox} \begin{array}{c|cccccccccc|cc} \hline
D_\mathrm{4h} & E & 2C_4 & C_2 & 2C_2' & 2C_2'' & i & 2S_4 & \sigma_\mathrm{h} & 2\sigma_\mathrm{v} & 2\sigma_\mathrm{d} & & \\ \hline
\mathrm{A_{1g}} & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & & x^2+y^2,\bbox[cyan]{z^2} \\
\mathrm{A_{2g}} & 1 & 1 & 1 & -1 & -1 & 1 & 1 & 1 & -1 & -1 & R_z & \\
\mathrm{B_{1g}} & 1 & -1 & 1 & 1 & -1 & 1 & -1 & 1 & 1 & -1 & & \bbox[cyan]{x^2-y^2} \\
\mathrm{B_{2g}} & 1 & -1 & 1 & -1 & 1 & 1 & -1 & 1 & -1 & 1 & & \bbox[cyan]{xy} \\
\mathrm{E_g} & 2 & 0 & -2 & 0 & 0 & 2 & 0 & -2 & 0 & 0 & (R_x,R_y) & \bbox[cyan]{(xz,yz)} \\
\mathrm{A_{1u}} & 1 & 1 & 1 & 1 & 1 & -1 & -1 & -1 & -1 & -1 & & \\
\mathrm{A_{2u}} & 1 & 1 & 1 & -1 & -1 & -1 & -1 & -1 & 1 & 1 & \bbox[pink]{z} & \\
\mathrm{B_{1u}} & 1 & -1 & 1 & 1 & -1 & -1 & 1 & -1 & -1 & 1 & & \\
\mathrm{B_{2u}} & 1 & -1 & 1 & -1 & 1 & -1 & 1 & -1 & 1 & -1 & & \\
\mathrm{E_u} & 2 & 0 & -2 & 0 & 0 & -2 & 0 & 2 & 0 & 0 & \bbox[pink]{(x,y)} & \\ \hline
\end{array}$$

Since the three p-orbitals point directly along the $x$, $y$, and $z$ axes, they show the same transformational properties as the basis vectors, and consequently get the same symmetry labels, i.e.

$$\begin{align}
\left.\begin{array}{c}
\mathrm{p}_x \\ \mathrm{p}_y \end{array}\right\} &\rightarrow \mathrm{E_u} \\
\mathrm{p}_z &\rightarrow \mathrm{A_{2u}} \ .
\end{align}$$

The five d-orbitals show the same transformational properties as their quadratic function counterparts and thus they get the symmetry labels, i.e.

$$\begin{align}
\mathrm{d}_{xy} &\rightarrow \mathrm{B_{2g}} \\
\left.\begin{array}{c}
\mathrm{d}_{xz} \\ \mathrm{d}_{yz} \end{array}\right\} &\rightarrow \mathrm{E_g} \\
\mathrm{d}_{z^2} &\rightarrow \mathrm{A_{1g}} \\
\mathrm{d}_{x^2 - y^2} &\rightarrow \mathrm{B_{1g}} \ .
\end{align}$$

Another worked example can be found in this answer of mine.


Is this the correct mechanism of the formation of the nitronium (NO2+) ion from sodium nitrate and sulfuric acid?

The electrons on the two other oxygens are delocalized making them much less basic.


Why does iodine get oxidized to iodic acid and not periodic acid by nitric acid?

Nitric acid is not strongly oxidizing enough to get iodine to +7.  Relativistic effects tend to make the $s$ subshell valence electrons more stable than the others in heavy atoms (http://www.annualreviews.org/doi/abs/10.1146/annurev-physchem-032511-143755), so to some extent iodine acts as if it has only five valence electrons.  To pull out those last two electrons and get periodic acid, electrochemical oxidation is commonly used with a very high oxidation potential (https://en.wikipedia.org/wiki/Periodic_acid).


How do I calculate weight/number average relative molecular mass for polymers?

They perhaps want you to make the calculations relatively, which means that even if you assume by looking at the first graph that the mass of the molecular chain length of $1$ (taken relatively and not as per scale) is $x$ units and thus, the mass of $10$ (again, taken relatively and not as per scale) such chains add up to $10x$ units and finally calculate $M_n$ to be $\approx 68.2273x$ it isn't helping us to get a meaningful answer.

A thing that should be noted is that even if you calculate it by following my example, when you'll finally calculate $\displaystyle \frac{M_w}{M_n}$ which (perhaps) the questioner wants you to calculate it won't make difference because this result is independent of $x$.

It appears that the questioner hasn't explicitly stated the mass of a polymer chain or that of the monomer units. They even haven't stated that all the polymers are considered identical with respect; it is highly probable that they want you to assume these things so as not to make things overly complicated.


There are multiple reactions going on simultaneously. Aluminum metal reacts with the oxygen in surrounding air in order to create a plating of aluminum (III) oxide.

$$\ce{4Al(s) + 3O2(l) -> 2Al2O3(s)}$$

Upon addition of sodium hydroxide, the oxide layer reacts with the base, leaving the aluminum free to attack by water for instance.

$$\ce{2 Al + 6 H2O -> 2 Al(OH)3 + 3 H2}$$
$$\ce{Al(OH)3 + NaOH -> Na+ + [Al(OH)4]-}$$

In summary, we can write the sum of the reactions as:

$$\ce{2Al(s) + 2NaOH(aq) + 6H2O(l) -> 2Na+(aq) + 2[Al(OH)4]- + 3H2(g)}$$

The reason why the reaction proceeded slowly at first was due to the aluminum (III) oxide layer covering the aluminum metal.

From this link, http://www2.uni-siegen.de/~pci/versuche/english/v44-10.html:

  This reaction is used in drain cleaners. They are mostly made out of strong alkalis, to which alumunim or zinc has been added. The alkalis break down organic residues chemically. In addition, the formation of hydrogen leads to a bubbling effect which adds an additional mechanical cleaning mechanism.

Considering the properties of aluminates, one can go to PubChem hosted by the NIH which has a lot of information for almost every single compound necessary as well as literature linked to the specific compound in question:

http://pubchem.ncbi.nlm.nih.gov/summary/summary.cgi?cid=14766&loc=ec_rcs


For thermally grown amorphous $\ce{SiO2}$ films, a bandgap of 9.3 eV was measured by Weinberg.

Measurements on $\alpha$-quartz by Calabrese give $E_g(\Gamma)$ = 6.3 eV, compared to 9.2 eV calculated by Chelikowsky and Schlüter.

Calculations by Ramos on $\beta$-cristobalite with the space groups $Fd3m$ (face-centred cubic, fcc), $I\bar{4}2c$ (body-centred tetragonal, bt), and $P2_13$ (simple cubic, sc) gave direct bandgaps of $E_g(\Gamma)$ = 5.48, 5.79 and 5.73 eV, respectively.


Effect of pressure on adsorption of a gas

Short Answer

The following equilibrium roughly represents what goes on during physisorption of a gases onto a a surface.
$$\ce{Gas + Surface <=>[adsorption][desorption] Gas--Surface}  $$

The effect of pressure can be understood on the basis of Le Chatelier's principle, which states that if a stress (such as a change in pressure) is applied to a system in equilibrium, the equilibrium will shift so as to tend to counteract the effect of the constraint.

In this case, we are increasing the pressure, and the system would respond by wanting to reduce the number of gas phase molecules, and in this case it can do so by favouring the forward reaction wherein the gas phase molecules are immobilised on the surface.

Adsorption is usually described through isotherms, that is, the amount of adsorbate on the adsorbent as a function of its pressure (if gas) or concentration (if liquid) at constant temperature. Several isotherm models exist to describe this phenomenon.

Bonus Round: Additional Commentary

One of the earliest mathematical fits for experimental data was the Freundlich adsorption isotherm. Experimentally it was determined that extent of adsorption varies directly with pressure and then it directly varies with pressure raised to the power 1/n until saturation pressure $P_s$ is reached. Beyond that point rate of adsorption saturates even after applying higher pressure. Thus Freundlich adsorption isotherm failed at higher pressure.

$$\frac{x}{m} = Kp^{\frac{1}{n}}$$

where $x$ is the quantity of adsorbate adsorbed in moles, $m$ is the mass of the adsorbent (this is basically a normalisation to allow for comparison between different materials), $P$ is the pressure of adsorbate and $K$ and $n$ are empirical constants at the given temperature.

The function is not adequate at very high pressure because in reality, as I have already mentioned, because $\frac{x}{m}$ has an asymptotic maximum as pressure increases without bound. As the temperature increases, the constants $k$ and $n$ change to reflect the empirical observation that the quantity adsorbed rises more slowly and higher pressures are required to saturate the surface.

Irving Langmuir was the first to derive a semi-empirical model with a kinetic basis and was derived based on statistical thermodynamics.

It is quite simple, (and elegant in my opinion), and finds extensive use. it is based on the following four assumptions

The surface containing the adsorbing sites is a perfectly flat plane with no corrugations (assume the surface is homogeneous) .
The adsorbing gas adsorbs into an immobile state.
All sites are equivalent.
Each site can hold at most one molecule of A (mono-layer coverage only).
There are no interactions between adsorbate molecules on adjacent sites.
There are no phase transitions.

Unfortunately,these four assumptions are seldom all true: there are always imperfections on the surface, adsorbed molecules are not necessarily inert, and the mechanism is clearly not the same for the very first molecules to adsorb to a surface as for the last. The fourth condition is the most troublesome, as frequently more molecules will adsorb to the monolayer; this problem is addressed by the BET isotherm for relatively flat (non-microporous) surfaces.

Anyway, the surface coverage is given as follows $$\theta = \frac{K_{eq}p}{1+ K_{eq}p}$$


What bond\crystal has the most regular pattern (i.e. least irregularities)?

In nature you need to look for precious gamestones and then do optical selection (diamante and ruby would be good candidates). The same would be true for synthetics but of course as mentioned by Jon you can also take elements (Si, Ge). General rule: they should be hard (smaller chance that pressure will affect structure, by pressure I mean also your tools) and you need to grow them slowly.

btw it's hard to call graphen crystal pattern -- it's 2D.


Most likely this was an artificial "lake" sample and someone gave you distilled water instead. Making artificial water samples is not an unusual practice in teaching laboratories.
Natural water cannot have zero concentration of chloride ion. Sodium and chloride ions are very common contaminants even in pure reagents. There is no "zero" concentration concept in analytical chemistry at all. The proper practice is to write "Not detected" instead of saying 0 ppm. All it means that it was below the detection limit of the ion-selective electrode. Ion selective electrode usually measure samples in parts per million (mg/L) range. Typical limits of detections are in single digit ppm.


Grignard alkoxide intermediate as starting material for ether or ester synthesis

Alkoxide, unless hydrolysed, acts as a very good base and is even used for the synthsis of ethers. However the alkoxide obtained in this case will have a yield that is quite low since the O- anion forms a salt as OMgX, hindering its ability to attack the alkyl group R-X. Thus, it is generally avoided.


Use of the terms "syn" and "anti" to describe the geometry about double bonds is no longer encouraged, but you'll still come across it, particularly in older articles.  Syn and anti are identical to Z(usammen) and E(ntgegen) and were often used to describe the geometry about carbon-nitrogen double bonds. In such cases, the lone pair of electrons is given the lowest priority and the sequence rule applied as usual.  For example, in the following N-methyl imines, the imine on the left has the highest priority group attached to the carbon in the double bond (ethyl) on the same side as the highest priority group attached to the nitrogen in the double bond  (the methyl has higher priority than the lone pair). Therefore the molecule is the syn isomer (but would now be called the Z isomer).

By the same reasoning, the molecule on the right is the anti of E isomer.  Again, the key point is that the lone pair has a lower priority than any atom.


What is the chemical reason talc is used in water-based paint and not in solvent based paint?

Most water based paints are emulsions of plastics particles dispersed in water, it is common to use vinyl acetate to make polyvinyl acetate. To do this an emulsion of vinyl acetate in water is formed, then by adding a free radical initiator.

One purpose which fine clay can furfill in the paint polymer synthesis is to increase the stability of the emulsion of droplets of the liquid monomer. These clay particles will be incorperated into the final paint.

It is also common to add other inorganic substances such as barium sulfate and titanium dioxide. These are used to make the paint more opaque.

One classic organic based paint class is the alkyd paint, these are made by reacting acid anhydrides such as phthalic and maleic anhydride with a polyalcohol such as glycerol or 2,2-dimethylpropane-1,3-diol under acidic conditions. It is normal to distill water out of the reaction, often an unsaturated fatty acid is added in the form of a glycerol triester (vegtable oil).

The idea is that with glycerol the fatty acid ester will react in a transestification reaction to add pedent groups onto the main polymer. Then when the paint is exposed to oxygen a radical cross linking reaction occurs. The other main option I see is to add some maleic acid to the synthesis of the paint binder. The to add some styrene to the paint. This again will cross link with a radical reaction.

For the alkyd paint synthesis there is no need to add talc, these polymerisations are done under homogenous conditions by heating a close to solvent free mixture of monomers. Rather than being a radical polymerisation (example of an addition polymerisation) this reaction is a condensation polymersation where many small chains are converted into a few long chains.


Why does potential energy relate to exothermic reactions in this way?

Energy is a useful concept for rationalising the capability of doing work or heating; other than that, it is purely a theoretical concept.
As Feynman brilliantly pointed out,

  It is important to realise that in physics today, we have no knowledge of what energy is.
  We do not have a picture that energy comes in little blobs of a definite amount.
  It is not that way.
  However, there are formulas for calculating some numerical quantity, and when we add it all together it gives “28”—always the same number.
  It is an abstract thing in that it does not tell us the mechanism or the reasons for the various formulas.

But let's get back to something answerable.

Energy is (a) conserved and (b) convertible into different forms.
If you got any system in a state $q_1$ with potential energy $U(q_1)$ and it changed into state $q_2$ with potential energy $U(q_2)$, then if $\Delta U = U(q_2) - U(q_1) < 0 $, energy has been lost to the surroundings.
This loss could be lots of things, most of the time it is just heat (kinetic motion).
This loss, in the big picture, is just a transfer or conversion.

Let's get to a very simple example, hydrogen combustion:

$$\ce{2 H2 + O2 -> 2 H2O}$$

Since energy can't be produced, we say this reaction releases large amounts of heat.
But, where does it come from?

All that the equation above tells us is the atomic rearrangement that has been taken place: in the left hand side we have two $\ce{H-H}$ bonds and a single $\ce{O=O}$ bond, in the right hand side we got four $\ce{O-H}$ bonds.
So quantifying the bond energies in each side gives us:

     |  Average bond energy (kJ/mol)
-----|-------------------------------
     |        |   Left    |   Right
Bond | Single | hand side | hand side
-----|--------|-----------|----------
H-H  |  -432  | 2 x -432  |    0
O=O  |  -495  |  -495     |    0
O-H  |  -467  |    0      | 4 x -467
--------------|-----------|----------
   Total      |   -1359   |  -1868

This means we get $-1868 - (-1359) = -509$ kJ for two moles of $\ce{H2}$, i.e., $-254.5$ kJ/mol of heat is released during the combustion of hydrogen.

Wikipedia says the correct value is $-242$ kJ/mol, so our calculation has an error of around 5%.

So heat energy has been released and it came from bond breaking and forming.
The energy stored in a bond could be called latent or potential, if you will.

It all boils down to conservation and convertibility.


Why glucose and other similar carbohydrates are oxidised fully by HIO4?

Glucose can react with $\ce{HIO4}$, which can be understood easily by looking at chair conformation of glucose.

Here is example of a diol, for more information looks at this answer https://chemistry.stackexchange.com/a/114836/67254


Number of stereoisomers and symmetry of pentane-2,3,4-triol

It is easier to compare molecules is you rotate them all into a comparable position.  Say 1 with the central C-3 hydroxyl group out of the plane of the screen/page like in 1, 4, 5 and 7.

So when we rotate a molecule by 180 degrees along a vertical axis in the plane of the paper, something going into the plane of the page on one end off the molecule rotates to become something coming out of the plane of the page on the other end of the molecule.  Also central group flips from into to out of the page and vice-versa.

Hence we can tell that actually pairs (1,2), (3,4), (5,8) and (6,7) are the same molecule, just rotated by 180 degrees.

The absolute positions of the three centres doesn't matter, only the relative orientation of the end two to the central site.


Why isn't 4s1 electron of Chromium the last electron?

3d is the highest energy, 0.04 Hartree higher than 4s according to NIST's Atomic Reference Data for Electronic Structure Calculations, Chromium


Thermal and photochemical excitation of electron in photochemistry

But what you say in your last sentence is true, and not specific to polyenes; it is quite general.  The energy of a visible photon at $500$ nm is $20000~ \pu{cm^{-1}}$ but at room temperature the average thermal energy is only $\approx 210 \pu{~cm^{-1}}$. The probability of a molecule with $20000~ \pu{cm^{-1}}$ of thermal energy is $\approx e^{-100} = 3.7\cdot 10^{-44}$ so small! Thus there in effectively zero chance that by thermal excitation an excited state at 500 nm will be produced. (By excited state I mean an electron is transferred from the HOMO orbital into the LUMO or another, higher, unoccupied orbital).

In your calculation you cannot use $C_V\Delta T$ as its not an operator but in a collision transferring energy there has to be an operator so you cannot just remove it from the integral. The properties of the operator determine the selection rules for transitions.

It is possible by collisions between molecules to move up and down in energy, but because thermal energy is small usually this normally involves vibrational and rotational levels. (The effect is immediately obvious by cooling or heating samples in the appearance of hot bands and in relative rotational line intensities.) If there are very low lying electronic states then I can see no reason why these also could not be excited thermally.


What volume of water will dilute an acidic solution back to neutral?

The very idea of diluting a solution back to (almost) neutral with pure water is unreasonable, precisely because it would require insane amounts of water. That being said, your derivation (though impractical) is not meaningless; it is just slightly wrong.

Like J. Ari said in the comments, there is an equilibrium involving $\ce{H+}$ and $\ce{OH-}$, and it will be shifted by the dissolved acid (that is, the concentrations will shift so the equilibrium constant may stay the same). Therefore you can't assume that water will always contain precisely $10^{-7}$M protons from autodissociation (plus maybe some extra from acid). It doesn't work that way.

How does it work, then?

$$\ce{[H+]\cdot[OH-]=10^{-14}}\\
\ce{[H+]-[OH-]=}c(\rm{HCl})$$
(since the acid is strong, and hence fully ionized even in not-so-diluted solutions)
$$\ce{[H+]=10^{-6.995}}\\
\ce{[OH-]=10^{-7.005}}\\
c\ce{(HCl)=10^{-6.995}-10^{-7.005}=10^{-6.995}(1-10^{-0.01})}\\
V_{total}=V_0\cdot{c_0\over c}\\
V_{added}=V_0\cdot\left({c_0\over c}-1\right)
=V_0\cdot\left({10^{6.995-pH_0}\over1-10^{-0.01}}-1\right)$$
So it is only 4.3 million times the original volume, which is about 2 times less than your estimate.

Not that it made much of a difference, really.


Why is ethanol distilled off before adding water in the first step of a malonic synthesis?

I assume this is a real procedure for actual synthesis.

Distilling ethanol off does have an effect on the workup of this reaction. Usually one needs to separate the reaction product and educts or side products afterwards by separation between an aqueous and an organic phase. Having ethanol in the mixture prevents the two phases from separating. In order to isolate your product you will need to distill ethanol off before adding water, because water would also be distilled off, if you added it first.


What is the highest relative abundance for diatomic bromine?

In $\ce{Br2}$, the possible combinations of isotopes are:

$\ce{^{79}Br-^{79}Br} : \mathrm{m/z}~158$
$\ce{^{79}Br-^{81}Br} : \mathrm{m/z}~160$
$\ce{^{81}Br-^{79}Br} : \mathrm{m/z}~160$
$\ce{^{81}Br-^{81}Br} : \mathrm{m/z}~162$

Bromine's isotopic distribution is essentially a $50:50$ ratio of mass $79$ and mass $81$. That means that there's a $25\%$ chance of $\mathrm{m/z}~158$ $(0.5 \cdot 0.5 = 0.25)$ arising from $\ce{^{79}Br-^{79}Br}$. There's a $50\%$ chance of $\mathrm{m/z}~160$ since there are two combinations ($\ce{^{79}Br-^{81}Br}$ and $\ce{^{81}Br-^{79}Br}$) that give that mass $(0.5\cdot0.5 + 0.5\cdot0.5 = 0.25)$. There's a $25\%$ chance of $\mathrm{m/z}~162$  $(0.5\cdot 0.5 = 0.25)$ arising from $\ce{^{81}Br-^{81}Br}$.


This may be what suffices you are looking for :

Periodic Table of Elements w. Selected Phase Transition Temperatures

This may get you started:

MatDL - The Materials Digital Library Synopsis
NIST SRD ( I would write them)
Java Applet: Phase Transition Temperatures (untested)

It is a very good question, which I came across as well. You could just google-image search phase diagram + ELEMENT to copy and past the most appealing charts into a table in your editor of choice.
If the intend is personal use, this might be the fastest route.

wolfram alpha should allow you to look up the state of the element at a given pressure as follows:

sodium phase at 0 bar

Helium itself is a particularly interesting subject.

I hope this helps for a start...


Magnesium powder mixed with CuSO4 solution

Your equation is very close, but you it is not $\ce{Mg2SO4}$. $\ce{Mg^{2+}}$ has a charge of $2+$, and $\ce{SO4^{2-}}$ has a charge of $2-$. They combine in a ratio of 1:1, so the result is

$$\ce{Mg + CuSO4 -> MgSO4 + Cu} ~ .$$

The ionic equation does not include the $\ce{SO4^{2-}}$, and in the case of this equation the net ionic equation is the same as the ionic equation. Ionic equations look at what happens to electrons.
In this case, electrons are being transferred from the magnesium to the copper (since the copper is bonded to $\ce{SO4^{2-}}$ it is in the form $\ce{Cu^2+}$).

$$\ce{Mg + Cu^{2+} -> Mg^{2+} + Cu} ~ .$$

Since the charge on the $\ce{SO4^{2-}}$ doesn't change, it's just a spectator ion.


Why is the oxime tautomer more stable than its corresponding nitroso isomer?

In keto-enol tautomerism generally the keto tautomer is more stable than its enol counterpart because in the former a significant ionic character exists which stabilises it. However, there are certain special cases. For instance if the enol form is aromatic(phenol), or very bulky mesityl groups are attached, etc.

In oxime-nitroso tautomerism however, the electronegativity difference comes into play.

In oxime, nitrogen is attached to carbon by a double bond;

and in nitroso, nitrogen is attached to oxygen by a double bond.

Now it is known that greater the difference in the electronegativity values, stronger is the bond. This property makes the nitrogen carbon double bond stronger hence making oximes more stable than nitroso.

(electronegativity values: N=3.04, O=3.44, C=2.55)


What are the boiling points of DMSO-water concentrations?

The binary mixture of water and DMSO do not form an azeotrope (a constant boiling mixture, where the vapour produced has fixed percentages of each component). Thus, water will boil from the solution first at a bit above 100 ºC, followed by the boiling of DMSO at around 189 ºC.

The reason for the drastic change in freezing point is a little more complicated. Freezing point depression is a common phenomenon; for example, it's the reason sea water freezes at a lower temperature than pure water. In these cases, the reason for the freezing point depression are typically entropic.

However, in the case of DMSO and water, the depression is of particular note—mixtures can remain liquid down to -140 ºC, compared to the normal freezing points of 0 ºC and 19 ºC for water and DMSO, respectively. Quantum mechanical studies have suggested that the formation of DMSO-water clusters is at least partially responsible for the massive freezing point depression. These clusters are strongly hydrogen-bonded, but interact far less strongly with other clusters, making it more difficult to form a solid lattice (and thus a solid).

Kirchner, B.; Reiher, M. J. Am. Chem. Soc., 2002, 124 (21), pp 6206–6215


What does "electrochemically active species" mean in the expression for the mass-transfer limited current?

It is the species that can get oxidized or reduced in the system.

It is specifically written out in order to make sure that it doesn't get confused with supporting electrolyte or any other component in the system.

So, for a reduction reaction, it is the concentration of the oxidized species.


Why are there 3 columns in Group VIII of the Periodic Table?

In the beginning of the $20$th century, the periodic table had $8$ columns, and not $18$ as today. Electrons and of course $spdf$ electrons were unknown. The numerous transition elements were included in this table by doubling the line between Argon and Krypton, then between Krypton and Xenon. It was going this way.
The $4$th period was made of two lines. So each square (each box) contained two elements written on two lines. The first seven atoms $\ce{_{19}K - _{25}Mn}$ occupied the columns $\ce{I - VII}$, and their oxidation state may always have been equal to the number of the column. Then the three atoms $\ce{_{26}Fe}$, $\ce{_{27}Co}$ and $\ce{_{28}Ni}$ did not belong to any column. We will come back to them later on. Then the seven next atoms $\ce{_{29}Cu - _{35}Br}$ had to be put in columns $1$ to $\ce{VII}$, under the corresponding elements : $\ce{_{29}Cu}$ in the same box and under $\ce{_{19}K}$, $\ce{_{30}Zn}$ in the same box and under $\ce{_{20}Ca}$, etc. up to $\ce{_{25}Mn}$ above $\ce{_{35}Br}$. With this system all atoms had at least one oxidation state equal to the number of their column. The same procedure was adopted for $5$th period, with $\ce{_{37}Rb}$ and $\ce{_{47}Ag}$ in the first column,$\ce{_{38}Sr}$ above $\ce{_{48}Cd}$, etc.
This table was useful, except nobody knew what to do with the three atoms  $\ce{_{26}Fe}$, $\ce{_{27}Co}$ and $\ce{_{28}Ni}$ having all valence $2$, and of course of the six corresponding atoms of the platinum group, immediately under them : $\ce{_{44}Ru}$ to $\ce{_{46}Pd}$ and $\ce{_{76}Os}$ to $\ce{_{78}Pt}$. There was plenty of discussion at that time about what to do with them. Usually these elements were "rejected" in the 8th column, with the noble gases. or they were put in a separate column made specially for them. This problem was finally solved with the discovery of the $spdf$ electrons.


Why do NaHCO₃ and KHCO₃ have different crystal structures?

The reason why the crystal structure of $\ce{NaHCO3}$ is face centered and the crystal structure of $\ce{KHCO3}$ is body centered is due to the fact that the cation/anion ratio is different, meaning that certain types of hole are now too small for the cation to fit into them and as a result it must go into a different hole.


Chirality in coordination octahedral compounds

See following image:

Look at the purple loop (en) on upper left hand side structure. Now, if you rotate the whole molecule ${90^o}$ clockwise around an axis going through top $\ce {NH2}$- $\ce {Ni}$-bottom  $\ce {NH2}$, you will get that loop overlapping on similar loop on the mirror image. But, now look what happens to the other two loops? They are not overlapping on other two loops on mirror image. For example, the purple loop on bottom of left hand side structure will move to back side of the same side (it would be ${180^o}$ far from similar loop in the mirror image). Thus this molecule would not superimpose with its mirror image.


It's a modified form of [10] annulene (https://en.wikipedia.org/wiki/Cyclodecapentaene) where internal bridges are used to hold the outer, conjugated ring in a planar geometry and thus with the $4n+2$ rule that outer ring is aromatic.

You may be more familiar with the methylene bridged version of [10] annulene that works the same way, but with a simpler bridging structure.  See the above mentioned Wikipedia link.

Addendum:

As mentioned in comments, it is overly simplistic to use the $4n+2$ rule.  IUPAC justifies it only for simple annulene systems.  When you have internal bonds in the ring like the compound here, you no longer have the symmetry and degeneracy structure on which the rule is based.

When you have saturated interior atoms surrounded by an outer conjugated ring, the $\pi$ orbitals are only slightly altered and that is why we are tempted to still apply the $4n+2$ rule.  But beware: if we unsaturate any internal atoms, in this case by removing the hydrogen from the central carbon, the $\pi$ orbitals are altered more considerably and the $4n+2$ rule is well past its use-by date.

Maybe in the end we can use the rule as a guide, but ultimately we need to back our conclusions up with hard evidence.  In this case the conclusion of aromaticity is supported by https://doi.org/10.1039/C39800000691, in which the authors studied substituted versions of the compound involved here.  They found proton n m r shifts consistent with an aromatic outer ring.


IUPAC nomenclature of protonated aminoacids

Interestingly, Ammonium salts come above acids in the seniority table for functional groups (Reference: Wikipedia). Also, the ion in question has an alkyl group attached to it, which makes the suffix aminium rather than the commonly used ammonium. (Reference: Gold Book, Wikipedia)  This means that the correct name of the compound is 6-carboxyhexan-1-aminium. This was also checked with ChemDraw online, as mentioned in the comments.


How to calculate the solubility product for PbS from galvanic potential?

The galvanic cell must be written like this:

$$\ce{Pb | PbS saturated || Pb(NO3)2 \pu{1 M} | Pb}$$

in order to have a positive  $\mathrm{EMF} = \pu{0.413 V}$.

Your approach is right with a slight error.
This is a concentration cell based on $\ce{Pb^2+}$. Since $\ce{Pb(NO3)2}$ dissociates completely, while $\ce{PbS}$ is a salt with a low solubility, the left semicell will be the anode and the right one the cathode.
So, we have the following semireactions:

$$
\begin{align}
 \ce{Pb^2+ + 2 e^- &-> Pb} &\quad &\text{(cathode)} \\
 \ce{Pb &-> Pb^2+ + 2 e-} &\quad &\text{(anode)}
\end{align}
$$

And for the anode we also have

$$\ce{PbS <=> Pb^2+ + S^2-}$$

where $[\ce{Pb^2+}] = [\ce{S^2-}] = \sqrt{K_\mathrm{sp}}$.
So the semicell potentials are:

$$
\begin{align}
E_\text{cathode} &= E^\circ\\
E_\text{anode}   &= E^\circ + \frac{0.059}{2}\log_{10}[\ce{Pb^2+}]\\
\end{align}
$$

Thus:

$$\pu{0.413 V} = E_\text{cathode} - E_\text{anode} = -\frac{0.059}{2}\log_{10}[\ce{Pb^2+}] \implies [\ce{Pb^2+}] = 10^{-14}$$

And

$$K_\mathrm{sp} = [\ce{Pb^2+}]^2 = 10^{-28}$$


Determining the structure of N2O using hybridization

Usually you would approach this problem from the other direction. First, find the most likely structure with the correct number of bonds, then deduce the hybridisation. This is also how it works in quantum chemistry: hybridisation is deduced from geometry, not the other way around.

If you try to write the structure of $\ce{N2O}$ you should come up with two equally like structures; these are mesomeric structures as given in the spoiler tag below.

   $$\ce{\overset{-}{N}=\overset{+}{N}=O <-> N#\overset{+}{N}-\overset{-}{O}}$$

From this, we can deduce the most likely hybridisation which will result in a linear molecule. The central atom in a linear molecule typically features two $\mathrm{sp}$ hybrid orbitals and two unhybridised $\mathrm p$ orbitals.

So what about the outer atoms? Well, we cannot say for sure because we do not have enough geometric information. We can tell by the type of bonds that we again need two unhybridised $\mathrm p$ orbitals on the end atoms (remember that the orientations of the double bonds are equivalent!) but we don’t know whether it is a better description for the other two orbitals to be seen as $\mathrm s$ and $\mathrm p$ or two $\mathrm{sp}$ hybrids. In these cases, I would tend to go with ‘unhybridised’ until an experimental or calculative result proves me wrong.

After we have done this, we can indeed look back and realise that we could have determined the structure a priori like this from the general rules; we would just have had to ignore the outer atoms and discuss hybridisation only for the central atom. Indeed, ignoring the hybridisation of terminal atoms gets you very far even in the simplified theory that puts hybridisation first (and is practically wrong).


Under what circumstances that Liquid-Liquid extraction would be good to use?

The above reasons to use LLE are focused on the compound thermal similarities and instabilities.

But it does not take into the account the amount of compounds. While distillation separation can be great for major components, it may fail for minor one, and definitely for trace ones.

The LLE is frequently used for separation and content enrichment of trace amounts, in wide range of grams to nanograms per litre.

Using the approach of chemistry of complexes, it may also lead to great separation of non-volatile compounds/ions/elements.

LLE is e.g used for mutual separation of  nuclear fission products and not spent fuel by the PUREX method.


Why is arsenic more chemically similar to phosphorus than is nitrogen?

Every element which is in the first place of the group, is pretty much different from the other elements of the same group. The reason is because the first element has ALWAYS the smallest radius, which leads to some changings also in group properties. Due to the small radius they are more stable than the others and they require more energy for any specific reaction. With some research you can see that also other  first elements of other groups such as Li,O,C etc.  differ from the other elements of that specific group.


Which has higher priority between unsaturated ring and unsaturated chain in the nomenclature of organic compounds?

The name given by ChemDraw is: (E)-1-chloro-3-(prop-1-en-1-yloxy)cycloprop-1-ene.
The ring is determined to be the parent structure.  This is rationalised in the 2013 edition of the IUPAC Blue Book on organic nomenclature by the following rules (pg 451):

P-44.1.2.2 Systems composed of rings and chains (exclusive of linear phanes)
Two methods are recognized to name systems composed of rings and chains (exclusive of linear phanes).
(1) Within the same class, a ring or ring system has seniority over a chain. When a ring and a chain contain the same senior element, the ring is chosen as parent.  Rings and chains are chosen regardless of their degree of hydrogenation.  As a consequence, this approach prefers the choice of a ring over a chain in systems composed of cyclic and acyclic hydrocarbons
(2) The context may favor the ring or the chain, so that, for example, substituents may be treated alike or an unsaturated a cyclic structure may be recognised, or the one chosen has the greater number of skeletal atoms in the ring or in the principal chain of the acyclic structure.

So either the chain or ring may be used as the parent for the name, but the ring will define the preferred IUPAC name.
The preference of ring over chain is further clarified in the following rule (pg 596):

P-52.2.8 Selection between a ring and a chain as parent hydride
Within the same heteroatom class and for tha same number of characteristic groups cited as the principal characteristic group, a ring is always selected as the parent hydride to construct a prefered IUPAC name. In general nomenclature, a ring or a chain can be a parent hydride (see P-44.1.2.2).

All emphasis added.