Can caustic soda (sodium hydroxide) be used to dissolve urine odour?

Well, it would actually make things a little awkward.
Why do Vinegar and baking soda work?
The reason for the usage of vinegar or sodium bicarbonate must first be understood.

Vinegar ($\ce{CH3COOH}$) is an acid, and hence protonates the carbonyl oxygen present in urea[1].

$\ce{NaHCO3}$ (baking soda) is amphoteric, meaning it can react with both acids and bases. Urea is basic, and hence forms a salt. This answer on our site talks exclusively about the deodorizing property of baking soda.

Both actions are quite much the same, and prevent the hydrolysis of urea (which happens to release ammonia ($\ce{NH3}$) which is the "foul" smelling substance)

Why wouldn't sodium hydroxide work?
Not that $\ce{NaOH}$ wouldn't react, but because it would accelerate the hydrolysis of urea[2].

So if you're looking to stop the hydrolysis of urea (and the inevitable generation of ammonia) $\ce{NaOH}$ might not be the right guy.
What I wouldn't advise
But, hey, if you're looking to remove all the urea once and for all, you could pour a few buckets of $\ce{NaOH}$ and let the ammonia create a riot in your unsuspecting neighborhood.
What I would advise
Urea hydrolysis is catalysed by the enzyme urease (so it is dead slow without it)
From reference [3] (emphasis mine)

Urea is a stable compound with a decomposition half-life in aqueous media of 3.6 years. However, hydrolysis of urea is 104 times faster when the urease enzyme is present.

If we could inhibhit its action, we'd have got what we needed.
If you're wondering where urease comes from and why the dog urine stench is greatly enhanced in summer months, you're looking at the same answer: microorganisms.
It has been published that Metals and fluorides could be used to inhibhit urease. I quote further from source [3].

2.3. Inhibiting chemicals
Seven different chemicals, silver nitrate, zinc nitrate, sodium fluoride, glacial acetic acid, vinegar, citric acid, and sulfuric acid, were used to inhibit urea hydrolysis.

The list is by no means exhaustive.
From source 4, different classes of compounds which might be used:

The classes of urease inhibitors include: amides and esters of phosphoric acid, thiols, hydroxamic acids, phosphinic and thiophosphinic acids, boric and boronic acids, phosphate, heavy metal ions, bismuth compounds, quinones, and fluoride.

If you're not keen on purchasing baking soda, I'm sure you'll find any one of the above an interesting buy.

References:

Interaction of urea with weak acids and water, Prabhat K. Das Gupta and S. P. Moulik, J. Phys. Chem. 1987, 91, 5826-5832 DOI:10.1021/j100306a061

What is the reaction between NaOH and urea?

Characterization of urea hydrolysis in fresh human urine and inhibition by chemical addition, Hannah Ray, Daniella Saetta, Treavor H. Boyer, Environ. Sci.: Water Res. Technol., 2018, 4, 87-98 DOI: 10.1039/C7EW00271H

A combined temperature-pH study of urease kinetics. Assigning pKa values to ionizable groups of the active site involved in the catalytic reaction, Barbara Krajewska, J. Mol. Cat. B: Enzymatic Volume 124, Feb. 2016, 70-76 DOI: 10.1016/j.molcatb.2015.11.021

EDIT
There has been some discussion on if the stench is truly ammoniacal or is just bacterial odor. In that case, I would suggest sprinkling of significantly higher concentration of acid. Extreme pH is known to disrupt bacterial life forms and the acid will also prevent to a great extent the possibility of urea hydrolysis, thus preventing the odor anyway.


Comparing the mesomeric effects of amino and mercapto groups

For a mesomeric effect to occur, an orbital has to align with the π system it is supposed to resonate with. Since π systems have a defined $z$ axis — the orientation of the $\mathrm p_z$ orbitals that make up the π system — only orbitals that also feature such an alignment an interact. This limits any mesomeric interaction of any heteroatom with a π system to that heteroatom’s $\mathrm p_z$ orbital. Therefore, both nitrogen and sulphur can only have one orbital at their disposal for $+M$ effects.

The reason why the $+M$ effect of nitrogen is stronger despite its higher electronegativity is the period these elements are in. Nitrogen, being in the same period as carbon, has orbitals that fit together well size-wise with carbon orbitals. This is also reflected by the ease of forming $\ce{C=N}$ or $\ce{N=O}$ double bonds. Sulphur is one period down. Its orbitals are more diffuse than carbon’s or nitrogen’s; therefore, any π overlap is much weaker. This is also the reason why $\ce{S=O}$ double bonds are rare and $\ce{S=C}$ double bonds are typically more reactive than $\ce{O=C}$ double bonds.


To understand the confusion here you need to take two different points of view about what is happening in a reaction: the molecular view and the statistical view.

The statistical view ("Gibbs free energy is a minimum") is an average of all the molecular activity in the reaction. It describes the net, bulk behaviour of the reaction. At equilibrium, there is no net direction of change. But this tell us little about what is happening to the individual molecules or atoms in the reaction.

The molecular (or atomic) view is about what is happening with the individual things that make up the reaction. At any temperature above absolute zero there will be plenty of things happening at the molecular level. There is plenty of energy around and the molecules will bang into each other a lot. Moreover, not all the components will have the same energy (some molecules are slow moving, others fast: the distribution of energies will follow a Boltzmann distribution). Sometimes there is enough energy for the products of the reaction to fall apart; sometimes there is enough energy for the reactions to react to give the product. Equilibrium is reached when, at a given temperature, the rates of both reactions are the same not when everything stops at the molecular level. Only when we look at the overall averages in the bulk reaction do we see no net change: at the molecular level things are always happening.

It is worth noting that the molecular view gives more insight than the purely thermodynamic view which only considered bulk averages. Some reactions at the molecular level happen fast, others slowly. Often in chemistry it is the kinetics that determine the product not the thermodynamics and thinking about the molecular point of view often explains the observed situation in the bulk.

It is quite important and insightful in chemistry to be able to switch from one view to the other as you will get a much richer interpretation of what happens in real reactions if you can.


Overview of mechanisms

There have been recent excellent summaries of the reaction profiles of the Wittig1 and the Corey–Chaykovsky2 reactions. For the purposes of answering this question, there's no need to go into exquisite detail about each individual step (a lot of the research has been poured into understanding stereoselectivities which isn't relevant here).

The general mechanisms are shown below, which should suffice; however, the interested reader is encouraged to consult the articles above.

Both reactions are considered to be essentially irreversible. As such, an argument based on the thermodynamic stability of the products cannot be valid.

For a kinetically controlled reaction, one should look at the activation energies of betaine vs. oxethietane or oxaphosphetane formation from the ylid and carbonyl.

Phosphorus ylids

The mechanism of the Wittig reaction has been studied very thoroughly, and the general consensus is that betaines are not intermediates. See: Which is the currently accepted mechanism of a Wittig reaction?

As described in the answers there, true betaines have never been observed as intermediates in "salt-free" Wittig reactions, and have been rigorously disproven in some cases. Betaines can be obtained by deprotonation of their conjugate acids, called β-hydroxyphosphonium salts. However, the only observed products of this deprotonation are either oxaphosphetanes3 or the alkene + phosphine oxide products4. There is a large body of evidence for a four-membered transition state leading directly to an oxaphosphetane.1

As for computational studies, Harvey and Aggarwal5 found a betaine intermediate at 14.6 kcal/mol higher than the cis oxaphosphetane for the reaction of $\ce{Me3P=CHMe}$ with acetaldehyde, but could not find a corresponding transition state. For the reaction with $\ce{Ph3P=CHMe}$, the betaine was found to be 16.2 kcal/mol higher than the oxaphosphetane, and the TS leading to the betaine 3 kcal/mol higher than the TS leading to the oxaphosphetane.

If the phosphorus betaine is never formed, the next step to form an epoxide (intramolecular nucleophilic substitution) cannot occur. Hence, the Corey–Chaykovsky reaction will not occur with phosphorus ylids.

Sulfur ylids

Very few cases of oxathietanes are known. A rare case was provided by Okazaki in 1996,6 characterised by NMR and crystallography:

However, the oxathietanes possibly formed in a typical Corey–Chaykovsky reaction differ from this in structure. For the model reaction of $\ce{H2S=CH2}$ with formaldehyde, an early computational study found a local minimum corresponding to an oxathietane with S–O bond length of 2.10 Å,7 and postulated that a Wittig outcome (alkene formation) did not occur because of the high activation energy to pseudorotation about sulfur as well as [2+2]-cycloreversion. However, these studies also did not take solvation effects into account, which will stabilise the highly charged betaine.

In a more recent DFT study, Harvey and Aggarwal2 were able to reproduce the molecular geometry of Okazaki's oxathietane, proving that there was no inherent computational bias against such structures. However, attempts to find a stable oxathietane from the addition of $\ce{Me2S=CH2}$ to formaldehyde, or from the addition of $\ce{Me2S=CHPh}$ to benzaldehyde, failed. In particular, the S–O distances in the optimised intermediates were 2.81 and 2.48 Å respectively (the full geometries are in the supporting information). Geometry optimisation starting from an oxathietane structure also yielded nothing but betaines.

Likewise, if an oxathietane cannot be formed, then the decomposition to alkene and sulfoxide cannot occur. Hence, sulfur ylids do not undergo the Wittig reaction.

Why the difference?

Harvey and Aggarwal2 wrote of their inability to find oxathietanes:

  Clearly, the energetic gain associated with bonding between these atoms [sulfur and oxygen] is more than offset by the ring strain it would entail.

Evidently, the key to understanding the different behaviour lies in bond strengths: not the double bonds in the product, but the single bonds in the oxetane intermediate. Kutzelnigg8 has found binding energies for $\ce{H2P-OH}$ (relative to $\ce{PH3 + O}$) and $\ce{HS-OH}$ (relative to $\ce{SH2 + O}$) to be 114 and 59 kcal/mol respectively. A qualitative explanation is not easy, and I am not really inclined to try.

Extra: Phosphorus betaines

As mentioned above, phosphorus betaines can be independently generated by deprotonation of β-hydroxyphosphonium salts. If we ignore the immediate cyclisation to form oxaphosphetanes, it is theoretically possible that these betaines could undergo intramolecular nucleophilic substitution to yield epoxides:

In a 2005 study, Aggarwal and Harvey again showed that such a process was much more energetically unfavourable for phosphorus than for sulfur.9 This was mainly rationalised by the poor leaving group ability of $\ce{PR3}$ compared to $\ce{SR2}$. Firstly, the intrinsic leaving group ability of phosphorus is lower (this is a purely kinetic measure of leaving group ability, based on the activation energy of the reaction $\ce{X + CH3X+ <=> X+CH3 + X}$). Secondly, the C–P bond is stronger than the C–S bond, which adds a further thermodynamic cost to epoxide formation. The difference in the energy profiles is striking:

For $\ce{X = PMe3}$, epoxide formation from the betaine is associated with an activation energy of 28.3 kcal/mol. Compared to an essentially barrierless cyclisation to the oxaphosphetane, it is easy to see why such a process cannot compete.

For $\ce{X = SMe2}$, on the other hand, the activation energy is only 5.4 kcal/mol. In fact, as the authors had noted in their earlier paper,2 the major challenge in the Corey–Chaykovsky reaction is getting the betaine into the correct conformation for intramolecular $\mathrm{S_N2}$ (i.e. going from 4-syn to 4-anti in the scheme above)!

References

Byrne, P. A.; Gilheany, D. G. Unequivocal Experimental Evidence for a Unified Lithium Salt-Free Wittig Reaction Mechanism for All Phosphonium Ylide Types: Reactions with β-Heteroatom-Substituted Aldehydes Are Consistently Selective for cis-Oxaphosphetane-Derived Products. J. Am. Chem. Soc. 2012, 134 (22), 9225–9239. DOI: 10.1021/ja300943z.
Aggarwal, V. K.; Harvey, J. N.; Richardson, J. Unraveling the Mechanism of Epoxide Formation from Sulfur Ylides and Aldehydes. J. Am. Chem. Soc. 2002, 124 (20), 5747–5756. DOI: 10.1021/ja025633n.
Vedejs, E.; Marth, C. F. Mechanism of Wittig reaction: evidence against betaine intermediates. J. Am. Chem. Soc. 1990, 112 (10), 3905–3909. DOI: 10.1021/ja00166a026.
Maryanoff, B.; Reitz, A. B.; Mutter, M. S.; Whittle, R. R.; Olofson, R. A. Stereochemistry and mechanism of the Wittig reaction. Diasteromeric reaction intermediates and analysis of the reaction course. J. Am. Chem. Soc. 1986, 108 (24), 7664–7678. DOI: 10.1021/ja00284a034.
Robiette, R.; Richardson, J.; Aggarwal, V. K.; Harvey, J. N. Reactivity and Selectivity in the Wittig Reaction: A Computational Study. J. Am. Chem. Soc. 2006, 128 (7), 2394–2409. DOI: 10.1021/ja056650q.
Ohno, F.; Kawashima, T.; Okazaki, R. Synthesis, Crystal Structure, and Thermolysis of a Pentacoordinate 1,2λ6-Oxathietane: An Intermediate of the Corey−Chaykovsky Reaction of Oxosulfonium Ylides? J. Am. Chem. Soc. 1996, 118 (3), 697–698. DOI: 10.1021/ja953533u.
Volatron, F.; Eisenstein, O. Wittig versus Corey-Chaykovsky Reaction. Theoretical study of the reactivity of phosphonium methylide and sulfonium methylide with formaldehyde. J. Am. Chem. Soc. 1987, 109 (1), 1–14. DOI: 10.1021/ja00235a001.
Wallmeier, H.; Kutzelnigg, W. Nature of the semipolar XO bond. Comparative ab initio study of H3NO, H2NOH, H3PO, H2POH, H2P(O)F, H2SO, HSOH, HClO, ArO, and related molecules. J. Am. Chem. Soc. 1979, 101 (11), 2804–2814. DOI: 10.1021/ja00505a002.
Aggarwal, V. K.; Harvey, J. N.; Robiette, R. On the Importance of Leaving Group Ability in Reactions of Ammonium, Oxonium, Phosphonium, and Sulfonium Ylides. Angew. Chem. Int. Ed. 2005, 44 (34), 5468–5471. DOI: 10.1002/anie.200501526.


Why do we say there's no carbon dioxide present?

You're doing a titration of the carbonic acid $\ce{H_2CO_3}$ which is the product of the dissolution of $\ce{CO_2}$ in water  by a base here the sodium hydroxide. According to wikipedia for carbonic acid, $\ce{pKa_1=3.6}$, $\ce{pKa_2=6.3}$ and $\ce{pKa_3=10.32}$.

When the carbon dioxyde is in water, the solution is acid then your phenolphtalein is colorless. During your titration while not all the carbonic acid as react then the solution will still be colorless because phenolphtalein's color change appears at around $\ce{pH=8.2}$ (Ref). Then when you finish you're titration the color of the solution will change.

You need to make some prediction to find the theorical pH at the equivalence.
In this case at the equivalence $\ce{[H^+]=\sqrt{Ka_1Ka_2}}=1.26 \times 10^{-10}$ then $$\ce{pH_{eq}=-\log_{10}(1.26 \times 10^{-10})=9.9}$$

So at this point the phenolphtalein must be pink and then no more acid (so carbone dioxyde) will be in your solution. You'll have only carbonate ion which will not react with sodium hydroxyde.


Why do single elements in a molecule have an oxidation state of zero?

Oxidation states are something of an "accounting fiction", used to describe where electrons are spending most of their time in a molecule.  This is turn is dictated by electronegativity: the more electronegative, the more the element attracts electrons.

So when you say oxygen's oxidation state is -2, this is only with respect to other elements of lower electronegativity. Now in the case of oxygen, the only other element with higher electonegativity is fluorine. For example, the molecule Oxygen Difluoride ($\ce{OF2}$) has oxygen in the +2 oxidation state because fluorine is more electronegative!  For any other element in combination with oxygen, however, we assume the electrons spend their time with the oxygen filling its outer valence shell, thus the -2 oxidation state.

But two oxygens in a molecule of gas have the same electronegativity, and there is no way to say that the electrons want to spend more time with one or the other oxygen.  Therefore, the "average" electron state for an oxygen molecule is as for the individual atoms, and so the oxidation state is 0.


If glucose can react with phenyl hydrazine to form osazone why can't it react with 2,4-DNP to give positive test?

Although glucose 1a exists principally in the hemiacetal pyranose forms (α and β), nonetheless it reacts with one equivalent of phenylhydrazine to form the hydrazone 1b via the aldehydic open form. Two additional equivalents serve to form the osazone and the products of reduction of phenylhydrazine, ammonia and aniline.The glucose 2,4-dinitrophenylhydrazone 1c has been reported1 but not the corresponding osazone. In their studies on the mechanism of osazone formation, Shemyakin and coworkers2 utilized p-nitrophenylhydrazine labeled with nitrogen-15 at the terminal nitrogen (the red N in the structures). Although glucose was not examined, the p-nitrophenylhydrazones  of fructose (forms the same osazone as glucose), benzoin and 2-hydroxycylohexanone were studied. The labelled nitrogen was located in the ammonia that was liberated during osazone formation.
Structures 2-7 represent the essentials of the mechanism.  Enolization of the hydrazone 2 produces the enol 3, which, via the zwitterion 4, undergoes oxidation of the carbon framework and reduction of the phenylhydrazine moiety. The transformation 3 --> 4 is essential. The basicity of the electron pair on the nitrogen in the 2,4-dinitrophenyl case is sufficiently reduced by delocalization into the aromatic ring, thereby repressing the proton transfer. This step must be completed before osazone formation can be accomplished.

1: Lloyd, E. A. and Doherty, D. G., J. Am. Chem. Soc., 1952, 74, 4214.
  2: Shemyakin, M. M., Maimind, V. I., Ermolaev, K. M. and Bamdas, E. M., Tetrahedron, 1965, 21, 2771.


Library / function for computation of molecular mass / molecular weight in C


Which base accepts the proton abstracted from water in propyne hydration?

Protons can basically jump on and off at any time if there is sufficient incentive for them to do so. In your reaction, you not only have the hydrogensulphate anion $\ce{HSO4-}$ that can accept a proton to go back to sulphuric acid but also water itself which can be protonated to $\ce{H3O+}$. The only important aspect is to identify the least weak base and use that in your mechanism.

Furthermore, general chemistry should have taught you that a solution containing sulphuric acid and water will contain $\ce{H3O+}$ ions by deprotonation immediately.


If you take "paint" in the very broad meaning of substances used to treat surfaces:

wood can be waxed, and waxes melt at not too elevated temperature and can then behave like oils.

I'd expect the behaviour of waxes here as very prone to surprises, since they can also become sticky when some compounds are liquid while others are solid.


As permeakra mentions, glucose can be purchased pretty easily. It might be labeled as "dextrose".

Another commonly available is maltose (or malt sugar).

Sucrose (or table sugar) is not a reducing sugar, but one of its constituents (glucose) is. Glucose can be made from sucrose by heating with a little bit of acid to make invert sugar syrup. Invert sugar is approximately a 1:1 ratio of glucose to fructose, and only the glucose in this mixture is reducing. An example procedure can be found here:How does adding lemon juice to sugar make better caramel?. Honey is a naturally occurring invert sugar.


Is sodium percarbonate the safest and by weight the most efficient oxygen generator?

Careful, as sodium percarbonate (basically $\ce{3Na2CO3.2H2O2}$) is usually marketed as a chlorine-free bleach. The latter usually has an additive, TAED, which acts as a key activator. See, for example, Wikipedia on OxiClean, where not even listed on the label is TetraAcetylEthyleneDiamine, which is apparently commonly employed as a bleach activator in many laundry products, see this, for example.

TAED reacts with H2O2 in alkaline conditions (referred to as perhydrolysis) creating, in part, Peracetic Acid. The latter PAA is actually preferred over sluggish H2O2 for bleaching. In other words, it is not the H2O2 as the active bleaching agent, but PAA created in situ.

So, other than possibly misleadingly implying that eco-friendly hydrogen peroxide is the active agent, it is, in reality, PAA, a probematic compound with associated health concerns (see Marquand, E. C.; et al. , 2007. "Asthma Caused by Peracetic Acid-Hydrogen Peroxide Mixture". J. Occup. Health. 49 (2): 155–158.). Here is an MSDS on an actual PAA mix, ( 76-61% water, PAA 20-35%, 3% Acetic acid and 1% H2O2), which is far from friendly but does eventually totally decompose into harmless O2, CO2, H2O,...

As such, anyone using a commercial percarbonate product should at least be aware of potential safety issues that one may likely be employing a Peracetic acid mix.

Note, there is an indicator that you do not have pure sodium percarbonate. It is the presence of some dire warning labels.


Butler-Volmer equation with concentration dependence?

I just figure it out that both are right (of course!). It's just that the definition of over-potential $\eta_s$ is different. In the case of Newman's, positive $\eta_s$ will guarantee positive (anodic) current. Here $\eta_s=V-U=\phi_{metal}-\phi_{solution}-U$ is positive in the sense that $V>U$, where $U$ is the equilibrium potential corresponding to the surface concentration. Whereas in the case of Bard's, sometimes positive $\eta_s$ will lead to positive (cathodic) current if $C_O$ is sufficiently large (possibly with convention or precipitation). But here $\eta_s=V-E_{eq}$, and $E_{eq}$ is defined as equilibrium potential corresponding to bulk (initial) condition. So the value of $V$ here will result in negative $\eta_s$ in the case of Newman's

In short, it’s just that the sign of over-potential is different with their corresponding definition of equilibrium potential.


What is the difference between a radical and a neutral alone atom?

A chlorine radical and a chlorine atom are effectively the same thing. This does not mean that "radical" and "atom" are interchangeable as jargons. They are used in fairly particular contexts, namely

Saying an "atom" is a reference to an instance of a particular element. They are often discussed abstractly, in terms of their constituent parts (nuclei and electrons) without coming to discuss their  surroundings or conditions within a system.
It is natural to say molecules are made of atoms, because they conserve the total number of nuclei and electrons with respect to the individual atoms, so we can rationalize them as bonded atoms. But saying molecules are made of radicals is weird.
Radicals can (very often) be polyatomic, like a methyl radical $\ce{CH3.}$
A radical has an odd number of electrons at its outer shell, it is chemically unstable, although electrically neutral. Atoms can be neutral and chemically stable, as in the noble gases, which fully comply to the octet rule.

Thus an atom being a radical and vice-verse is more of a "chemical coincidence" than anything else, kind of like a hydrogen cation is a proton. But they are definitely not equivalent.


The exact nature of the mechanism is unknown

Be aware that some organometallic chemists consider arrow-pushing mechanisms inadequate/inappropriate for a variety of reasons (including the depiction of single electron transfer). For example, the arrow-pushing mechanisms we draw with alkyllithiums are incomplete in that they do not account for the oligomeric nature of the reagent.[2]

The elements which are known:

The aldehyde or α-hydroxyketone must be enolizable. The enolate tautomerization of reducing sugars is discussed in McMurray[3]. This is common to the Benedict’s test mechanism as well.
The rate of enolization determines the rate of oxidation J. Am. Chem. Soc. 1970, 92, 537
Cu(II) has long been known to mediate oxidative coupling of enolates by a radical mechanism J. Am. Chem. Soc. 1971, 93, 4605. Mechanistic studies point to a metal-chelated single-electron-transfer process.5
The active species is the metastable [Cu(tartrate)2]6- ( Eur. J. Inorg. Chem. 2016, 1798).

Drawing one plausible mechanism

Convince yourself that the Cu(II) complex  is a d9 17-electon complex.  That suggests you could plausibly draw an inner-sphere or outer-sphere mechanism.
For the inner-sphere mechanism: coordinate the aldehyde, and generate the enolate (considering C-vs-O coordination and necessity of ligand-slip).
Form the hydrate of the C-bound enolate, then homolyze the bond to form an organic radical and Cu(I).
A second molecule of Cu(II) oxidizes the radical to the cation, which gets quenched the elimination of H from the hydrate to generate the enol form of the carboxylic acid.

For the basics of an outer-sphere single electron transfer see Mark Vander Wal’s Basics of Electron Transfer

[2]: Advanced Organic Chemistry: Part A: Structure and Mechanisms, Carey & Sundberg, p. 413.
[3]: Organic Chemistry: With Biological Applications,  McMurry, p. 758


$\ce{BF3}$ has trigonal plannar structure all the three $\ce{B-F}$ bonds lie in plane and thus p-orbitals of boron and fluorine become parallel.

Boron has empty p-orbital and p-orbital of fluorine contains lone pair, and hence  boron acts as Lewis acid and fluorine as Lewis base. Fluorine donates its lone pair to boron and this bonding is called backbonding.


Why does CH3CH(OH)CH2CH3 give iodoform test?

The substrate you have provided (2-butanol) would indeed give a positive test.

The iodoform test is performed in presence of iodine and potassium or sodium hydroxide, which first oxidizes the aforesaid compound to 2-Butanone.

Mechanism (taken from this Wikipedia page) which shows $\ce{BrO-}$ instead of $\ce{IO-}$, but would be executed in a similar fashion:

Thus 2-Butanone responds to iodoform test successfully.

References:

Ueber Entstehung von Jodoform und Anwendung dieser Reaction in der chemischen Analyse, Lieben, Ann. Spl. Bd. 7, 218 1870


Baking Soda + Citric Acid + Water = CO2, how much though?

I'm not sure what you mean by 90% and 10%. If you mean 90% yield (i.e. 90% or more of the bicarbonate ions in sodium bicarbonate will become carbon dioxide), then yes, that will happen if there is enough acid. How much is enough acid? We'll get to that in a minute. If you mean something else, like 90% of the mass of the components will become carbon dioxide, then that probably will not happen.

Let's look at why 90% of the mass of products is not carbon dioxide.

Bicarbonate anions react with protons from acid to produce carbonic acid, which decomposes to form carbon dioxide:
$$\ce{H+ + HCO3- -> H2CO3 -> H2O + CO2}$$
Let's start with a gram of sodium bicarbonate. How many bicarbonate ions are present? The molar mass of sodium bicarbonate is $84.0066\mathrm{\ g/mol}$, where a mole (mol) is a quantity equal to the number of carbon-12 atoms in 12 grams of carbon-12 or an Avogadro's number $(6.022\times 10^{23})$ of things. Atoms and molecules are very small, and the numbers of them in macro measurements are large.
So we have
$$\mathrm{1.00\ g \times \dfrac{mol}{84.0066\ g}= 0.0119\ mol}$$
$$\mathrm{0.0119\ mol \times \dfrac{6.022\times 10^{23}}{mol}=7.17\times 10^{21}\ molecules}$$
Let's stick with the mole unit since it gives us easier to conceptualize numbers.
The molar mass of carbon dioxide is $\mathrm{44.01\ g/mol}$, so if all of the bicarbonate ion in sodium bicarbonate is converted into carbon dioxide, we will get $\mathrm{0.0119\ mol}$ of carbon dioxide which is
$$\mathrm{0.0119\ mol \times \dfrac{44.01\ g}{mol}=0.524 \ g}$$
So 1.00 gram of sodium bicarbonate will product 0.524 grams of carbon dioxide. We have not even accounted for the citric acid yet. Citric acid is $\ce{H3C6H5O7}$, and it is tempting to think that some of those carbon and oxygen atoms in that molecule might become molecules of carbon dioxide. They can, but not in a reaction with sodium bicarbonate (acid-base). If you react citric acid with oxygen (combustion), then you can produce carbon dioxide.

How much acid?

Citric acid serves as a proton source - in fact it has three acidic protons. That means in theory, each molecule of citric acid can donate three protons to convert three bicarbonate anions into carbon dioxide:
$$\ce{H3C6H5O7 + 3HCO3- -> C6H5O7^3- + 3H2O + 3CO2}$$
However, those three acidic protons are not equal (and the reason for that would make a great follow-up question). Two of the protons are more acidic than carbonic acid, and the third one is slightly less acidic than carbonic acid. In theory, since carbonic acid decomposes into carbon dioxide, which escapes from solution, we can count on Le Châtelier's principle to pull even that third proton. However, in practice, we might not want to wait around. Let's use one equivalent of citric acid per two equivalents of sodium bicarbonate:
$$\ce{H3C6H5O7 + 2HCO3- -> HC6H5O7^2- + 2H2O + 2CO2}$$
So, to figure out how much citric acid (molar mass $\mathrm{192.12\ g/mol}$) we need, we have to do some calculations. Remember that 1 gram of sodium bicarboncate is 0.0119 moles, and we need 0.5 moles of citric acid for every mole of bicarbonate:
$$\mathrm{0.0119 \ mol\ bicarb\times \dfrac{1\ mol\ CA}{2\ mol\ bicarb}\times\dfrac{192.12\ g\ CA}{1\ mol\ CA}=1.14\ g\ citric\ acid}$$


Assuming you have equal moles of two different perfect gases, according to the Sackur–Tetrode equation the difference in their molar translational entropy is:

$$\frac{\Delta S_\mathrm{trans}}{R} = \frac{3}{2}\ln\left(\frac{M_2}{M_1}\right)$$

where the $M_i$ are the molecular masses. For instance, since $\ce{D_2}$ and $\ce{He}$ have identical masses this term would vanish, i.e. $ \Delta S_\mathrm{trans} = 0$.

If the atoms are in their electronic ground states then the electronic state makes no contribution to the entropy. On the other hand, a diatomic molecule has rotational and vibrational degrees of freedom. Assuming it's a rigid rotor, then in the high temperature limit (in which many rotational quantum levels are significantly occupied),

$$\frac{S_\mathrm{rot}}{R} = 1 + \ln\left(\frac{8\pi^2IkT}{h^2\sigma}\right)$$

where $I$ is the moment of inertia, $\sigma$ is a symmetry number.

Assuming the diatomic molecule is a harmonic oscillator, then

$$\frac{S_\mathrm{vib}}{R} =  x(e^x - 1)^{-1} - \ln(1 - e^{-x})$$

where $x = hν/kT$ and $T$ is around room temperature. However, the contribution of the vibrational entropy is tiny compared to the rotational component, so it can be ignored.

That would wrap up the story, in any case, if nuclear spin didn't have something to say about it. Nuclear spin contributes additionally to the entropy of $\ce{H2}$ and $\ce{D2}$ as explained in this post, resulting in an altered rotational entropy term. This rotational/spin contribution lifts the entropy of these molecules above those of Ne and He.

Therefore at room temperature and $\pu{1 atm}$ pressure, the molar entropy of $\ce{D2}$ is expected to be larger than that of helium due to the rotational contribution of $\pu{31.3 J mol-1 K-1}$.

Similarly, the translational entropy of neon is $\pu{3.5 J mol-1 K-1}$ higher than that of hydrogen gas, but $\ce{H2}$ has a rotational entropy of $\pu{18.7 J mol-1 K-1}$ so the entropy of neon is expected to be smaller by about $\pu{15.2 J mol-1 K-1}$.


Why is standard reduction potential an intensive property?

Look at the unit volt. 1 volt = 1 joule per coulomb. If the number of moles is doubled, the coulombs are doubled, as is the number of joules. Think of the volt as the driving force behind an individual electron in an electrochemical system.


How can I increase the current on a voltaic cell for high school lab?

Increase the area of the electrodes (e.g more ribbon).
Decrease the resistance of the electrolyte (e.g. reduce the spacing between electrodes, increase concentration of salts).
Reduce polarization of electrodes (bubbles on surface, etc.). Scrape or lightly sandpaper the electrodes and test immediately after immersion, before bubbles form.


Why do we add copper when testing a salt for nitrate ion?

This is a preliminary test, which is neither particularly selective nor sensitive.
With nitrate, brown vapours of $\ce{NO2}$ are released.
However, bromine or iodine could also release brownish or violet vapours.
Copper (or silver) is used to remove bromide and iodide by precipitation.


How are imaginary frequencies and force constant calculated by DFT?

Within DFT the vicinity of the solution to the Kohn-Sham equations is explored treating small changes of nuclei positions as a perturbation. The restoring forces that act in the system upon such perturbations give us information about the force constant matrix, from which we can get the harmonic vibrational frequencies as explained elsewhere, e.g. in the book of Wilson and Decius. Basically, you need the second derivatives.

A key point is to do it efficiently. Doing it numerically is very costly.
Since the basis functions are known and they have some nice properties one can do it analytically. Thus, researchers have devised a set of coupled equations to compute the effect of perturbations on the density matrix, which is the core of the calculation.

You will find a detailed explanation in this article by F. Neese.


The methyl epoxy ether is chiral, with the chiral center being the middle carbon in the chain, so it has two enantiomers.


Nucleation and Phase Equilibrium at the Melting Temperature

You probably know that water can be supercooled down to well below -30°C. And it has probably been suggested (if not outright stated) that at those temperatures, the water is in a "metastable" liquid state and that "eventually" it will spontaneously solidify - even if "eventually" may require trillions of years. Since thermodynamics is empirical, you could wonder how that has been empirically observed...but let's not go there. You've been taught that ΔG is a measure of the spontaneity of a process at constant temperature and pressure and at equilibrium. (How a "process" happens when the system is at equilibrium (steady-state) is often ignored. It's fair to say that nothing "happens" at equilibrium, rather when we say "equilibrium process" we mean "a process occurring sufficiently close to equilibrium". But I think I'm getting off topic, again. In considering nucleation, it is vital you distinguish between homogeneous and heterogeneous nucleation. This is the crux of the issue. Water is almost never without contaminates which provide nucleation sites, even if those sites are only on the surface of the container holding the liquid (getting the water this pure is an enormously difficult feat, and how you create a liquid phase without a container I leave to your imagination (hint:rain)). So, when reading about nucleation, you need to distinguish between the situations where there is preexisting centers (contaminants) or whether the water is assumed to be absolutely pure and centers must form spontaneously. Perhaps to answer your question (it would have been most useful if you had copy/pasted the exact text from your book). Once centers (seeds?) exist, the continued growth is pretty much identical whether their origin was homo- or hetero-. Obviously there is no equilibrium between a dust particle (solid) and liquid water. So how can nuclei exist at the MP? Obvious, right? They're hetero-, no formation is required. (Here I avoid the issue of whether FP is (or is not) identical to MP). Wikipedia has a short article on Classical Nucleation Theory which may help. Consider a solid. At its surface, the atoms (or molecules) are different than in the bulk (for most solids, bulk properties aren't reached for top 2 or 3 (or more) layers) but just consider an ideal crystal. Obviously, the surface has fewer bonds and hence has higher energy (is less stable). We call that surface tension (or exactly equivalently, surface energy). The classical theory states it's always positive. This is the energy which needs to be "overcome" to get nuclei, and it's proportional to the surface area. You should note here that our discussion has gone down to the microscopic level, where classical thermodynamic quantities become less and less useful. In other words, the macroscopic classical world thermodynamics needs to be treated carefully here. Statistical thermodynamics is most suitable transition from the classical to the quantum world. These types of processes can be studied using either, but fundamentally the world behaves according to the rules of (statistical) quantum mechanics when you're concerned with atoms depositing on or leaving a surface. The perspective of Classical Thermodynamics is of limited use. (In this case, I think it is probably just as useful, if not more useful, than the quantum mechanical approach. But you're unlikely to get many "fundamental" insights using it.) I'll point out one other apparent flaw in the "activation energy" approach: if the energy barrier is symmetric, then at its peak there should be no preferential direction: the odds of the system moving towards the solid or towards the liquid should be exactly the same (i.e. 50:50), yet the Classical Nucleation Theory requires a fudge factor for this probability - it is an indication that there is more going on here than this approach can handle, that the nucleation activation energy is, at best an approximation. Meaning don't take it too seriously; unless it can be derived from the quantum mechanics, it has no fundamental basis. I don't know much about this area of chemistry. I don't know whether or not it can be derived from the QM, but given that the Classical Nucleation Theory article in wikipedia is speaking about radius as if it is continuous, then I have my doubts. In some systems you begin to see bulk properties at N=100 to N=1000 (N=number of particles), so you'd have to be above this threshold for "classical" behavior to emerge.


Magnetization as a process condition (during synthesis procedure) can affect the crystal structure and nanoparticle size and morphology which in turn can affect accessibility of reactive sites. See the following papers:

Hu, Dawei, Yanmin Wang, and Qi Song. "Weakly magnetic field-assisted synthesis of magnetite nano-particles in oxidative co-precipitation." Particuology 7.5 (2009): 363-367.

Xiao, W., et al. "Magnetic-field-assisted synthesis of magnetite nanoparticles via thermal decomposition and their hyperthermia properties." CrystEngComm 17.19 (2015): 3652-3658.

Li, Yang, et al. "Fe 3 O 4@ PSC nanoparticle clusters with enhanced magnetic properties prepared by alternating-current magnetic field assisted co-precipitation." Colloids and Surfaces A: Physicochemical and Engineering Aspects 520 (2017): 348-354.


This lab equipment is called 'Fermentation Tube'.

If you want to see how an experiment is carried out with this equipment, please watch the video below.

Mark Garcia: Lab Protocol - Fermentation in Yeast (Unit 9 Respiration); YouTube


All the normal thermodynamic properties of a pair of enantiomers are the same.
Unless there is some interaction with a chiral environment. The only way any properties will differ is if the environment in which they are measured is chiral.
For example, the heat of combustion of D glucose is the same as for L glucose (as combustion doesn't care about the stereochemistry). But only D glucose is usable by higher organisms in metabolism as all the enzymes involved are chiral (despite the thermodynamics of the overall reaction being exactly the same).


What is the hybridisation of carbon in transition state in SN2 mechanism?

In transition state of SN2 mechanism, the carbon atom is $\ce{sp^2}$ hybridised due to its planar structure. At this point, carbon almost acquires 'pentavalency' with three full bonds and two 'partial' bonds, and a planar complex is formed, which is $\ce{sp^2}$ hybridised.


How to make 1,1-diphenyl-1-butene from benzophenone and 1-bromopropane?

Two steps:

Form the Grignard reagent with 1-bromopropane and magnesium metal. This can be down in a variety of ethereal solvents, THF or $\ce{Et2O}$ are most commonly used. This species is nucleophilic through carbon and will add to the carbonyl group.

Cool the Grignard solution in an ice bath under nitrogen with stirring, slowly add a solution of 0.9 eq of benzophenone. Allow it to stir until TLC shows consumption of the benzophenone starting material.

Add a solution of aqueous ammonium chloride to quench residual Grignard. Add organic solvent $(\ce{Et2O}$ or $\ce{EtOAc})$ and collect the organic phase, dry and concentrate. This will form 1,1-diphenyl-butan-1-ol.
Take the crude product from step 1 and dissolve in toluene. Add a catalytic amount of p-toluene sulfonic acid (PTSA) and heat under Dean-Stark reflux to remove the water. The PTSA will protonate the tertiary alcohol you have formed in step 1. This gives a cation stabilised by the two phenyl groups so it forms readily. This eliminates water to give the required product cleanly.

When the system has ceased to generate water allow it to cool, wash with diluted aqueous sodium carbonate solution, dry and concentrate. This should result in fairly clean 1,1-diphenyl-1-butene.

To produce cleaner material use chromatography on silica with hexane/ethyl acetate (0 to 10%).

Further reading on Grignard reagents: Master Organic Chemistry.


Why HF is said to be formed from H-1s and F-2pz overlap?

You fell into a wrong analogy with the picture of energetically far apart. Let me break down your picture into something visible:

Imagine the orbitals of fluorine at different heights of a mountain (I’ll use the Zugspitze, because it feels like home). 1s is at the very peak of the mountain (yeah, I’m turning this around just for simplicity). The Eibsee (c. $1000~\mathrm{m}$ over sea level) would correspond to the 2s-orbital. Garmisch-Partenkirchen, the town in the valley before the mountain, ($700~\mathrm{m}$) would be the 2p orbital.

We need to place the 1s-orbital of hydrogen somewhere so that the height difference to the Eibsee is too large for an interaction. One possibility would be the Höllentalanger hut at around $1300~\mathrm{m}$. That would render the energy difference to both fluorine’s 2s and 2p too large. But in fact, hydrogen’s 1s is better placed at the height of Oberau, $650~\mathrm{m}$. That is way too far away from the 2s but rather close to the 2p.

This is confirmed by energy diagrams of $\ce{HF}$ like that presented in this lecture (in German). (Just search for HF on the page, it’s not far down. Ignore the text around it and look at the scheme.)

Now you may be asking yourself why the hydrogen’s 1s energy is so high compared to fluorine’s orbitals. Compare ionisation energies, which are typically understood as the energy difference of the highest orbital to ‘an electron in vacuum’. For hydrogen, this energy is $1312~\mathrm{kJ/mol}$ while for fluorine this is $1681~\mathrm{kJ/mol}$. The reasoning behind it is fluorine’s very high electronegativity, attracting the electrons towards its nucleus. Fluorine’s nucleus has a $+9$ charge which is much better at drawing electrons in than hydrogen’s $+1$ (even though it needs to draw nine electrons in rather than one). This stabilises all of fluorine’s orbitals; and those which are ‘closer to the nucleus’ (s-orbitals) most.

I am not well-educated in modern physical chemistry methods, but I heard that orbitals were visualised. If that is the case, it won’t be long until we can visualise the $\ce{HF}$ orbitals, and indeed show that the bonding orbital (highest energy σ type) will have a nodal plane through fluorine.


How to calculate the mass of magnesium necessary to produce a certain volume of hydrogen gas?

It is very helpful when you start with using proper notation, i.e. $V(\ce{H2})=44.8~\mathrm{L}$. Proper notation like $M$ for molecular mass, $m$ for mass, and $n$ for amount of substance will help you follow through your thinking.
When do the calculations, always carry the units with every quantity you use.

It is necessary to know a few principle formulae, for example that molecular mass and mass are connected through the amount of substance: $$M = \frac{m}{n}$$
There are a few more you will need and with their repeated use you will get to know them easily.

When attacking this problem, find out how the quantities relate to each other. In this example it is: If I want to produce one mole of hydrogen gas, how many moles of magnesium do I need?

   one, because $n(\ce{Mg})\ce{->}n(\ce{H2})$

A key principle (and asumption) in this exercise is the ideal gas law,
$$pV=n\mathcal{R}T.$$
You need it to calculate the amount of substance of hydrogen that should be produced.
(You have indirectly used it by dividing the given value by the molar volume of an ideal gas at STP.)

   In this case about two moles of hydrogen gas shall be produced, i.e.
 $\displaystyle n(\ce{H2})=\frac{pV(\ce{H2})}{\mathcal{R}T}\approx2~\mathrm{mol}$
 (Old definition of STP: $p=1~\mathrm{atm}$ and $T=298.15~\mathrm{K}$)

Now you also know the amount of substance of magnesium you need.

   two moles, i.e. $n(\ce{Mg}=2~\mathrm{mol}$

You just have to use the molecular mass of magnesium to find the mass of magnesium you need with the above connection.


As with many chemical compounds (see the large class of terpenes, for instance) "croton" derivatives take their name from the natural source of their "father" compound. They are a family descending from crotonic acid, which was found in the oil from the seeds of the Croton plant.

"Croton" compounds are derivatives of the father compound "crotonic acid" (2-butenoic acid), so you can guess the names of the derived compounds by the logical chemical modifications of such compounds. For instance, as you might guess, crotyl alcohol is But-2-enol, from the reduction of crotonic acid; crotonylene's structure can be imagined with an analogy between acetic acid and acetylene, and so on.

I don't think that nowadays new names for the derivatives could be accepted, but for historical reasons many derivatives have these funny names.

And, knowing the stucture of the father compound, it's easy to imagine the structure of the derived ones.


What will be the final product when the above bridged compound is heated with concentrated sulfuric acid?

Protonation of both $\ce{OH}$ groups and subsequent formation of the corresponding cations is possible.

The tertiary cation will yield an exomethylene group
The "secondary" non-classical cation leads to a $\ce{C=C}$ double bond in the ring. With other words, a norbornene is formed.

The combined eliminations will give 7-methylene bicyclo[2.2.1]hept-2-ene as the product.

In J. Am. Chem. Soc., 1973, 95, 7801-7813, the same product was obtained by elimination from the tosylate of 7-hydroxy-7-methylnorborn-2-ene.


It is possible that the white material is calcium carbonate or magnesium carbonate from your water source precipitating.

See if vinegar dissolves the white residue.

If the white residue dissolves in vinegar, it is probable calcium carbonate and/or magnesium carbonate.


In Simonini reaction how does the $2:1$ ratio of reagents enable formation of ester as major product?

If you have the extra equivalent of the carboxylate, then this carboxylate will never react with iodine to form the alkyl iodide. Instead, it will do the next best thing, which is $S_{\mathrm{N}}2$ reaction with the alkyl iodide that you did make.

Note that this probably only works if R is primary.


Calculation of Gibbs free energy of a chemical reaction


The acetic acid in "Cleaning vinegar" may come from an industrial process. One way to produce acetic acid industrially, is carbonylation of methanol. And methanol is toxic!
So unless the product has been analyzed for residues of methanol and approved for food consumption, it is probably unwise to use it for consumption.


How can nitrogen achieve +5 oxidation state?

The concept of oxidation states is just that: A theoretical concept. You assign electronegativity values based on one of the numerous scales available, split the bonds heterolytically according to those numbers, and then count electrons.

As such, it is very easy to make a cation like @permeakra stated in the comments that satisfies the octet rule and still gets to state +5 (the molecular cation in question is $\ce{NF4+}$).

There are no $d$ orbitals necessary (which are also theoretical, but slightly closer to reality) to explain how a +5 state is possible.


What is the difference between a ylide and a betaine?

The more definitive source for chemistry terminology is the IUPAC Gold Book.

Betaine

  Originally, the compound betaine, $\ce{(CH3)3N+–CH2C(=O)O−}$ (N,N,N-trimethylammonioacetate), and similar zwitterionic compounds derived from other amino acids. By extension, neutral molecules having charge-separated forms with an onium atom which bears no hydrogen atoms and that is not adjacent to the anionic atom. Betaines cannot be represented without formal charges.

Ylides

  Compounds in which an anionic site $\ce{Y−}$ (originally on carbon, but now including other atoms) is attached directly to a heteroatom $\ce{X+}$ (usually nitrogen, phosphorus or sulfur) carrying a formal positive charge. They are thus 1,2-dipolar species of the type $\ce{R_mX+–Y−R_n}$.

So, ylides are not betaines and betaines are not ylides. Primarily due to the atoms in a ylide that contain the charges must be adjacent, and in betaines they must not be adjacent.

An example of a ylid would be the phosphonium ylid used in Wittig chemistry:

Since it has adjacent positive and negative charges, it is classified as a ylid.

The addition of this ylid to a carbonyl compound would form a betaine (note that the mechanism of the Wittig reaction is now thought to be a concerted [2+2] cycloaddition leading to the oxaphosphetane immediately, without the intermediacy of the betaine):

Note that the ylid has an alternate resonance form (the one with a P=C double bond) which has no formal charges, whereas the betaine cannot be drawn without formal charges.


If accuracy is not essential, weigh a small amount drained for a fixed time but still wet, and then weigh again dried, to find the ratio of wet-to-dry weight (carefully disposing the dried, sensitized, explosive without destroying the scale or personnel).

Though this could give a rough idea of the actual dried weight, it would vary from batch to batch and with temperature.


Is the absolute temperature concept 100% viable?

The concept of absolute temperature is real. Temperature can be thought of as thee measure of the concentration of heat energy in a substance and isn't effected by different elements or compounds. Absolute temperature which is −273.15 degrees Celsius (0 kelvin) is the temperature where there is no heat energy in the substance at all. That means that all the molecules are existing in their lowest possible energy state. So conventionally is thought that it is impossible to get lower than absolute zero as to reduce the temperature, you have to remove heat energy but you can't remove heat energy in a substance which doesn't have any heat energy in the first place.

However theoretically, the Boltzmann equation does allow for substance to go minus Kelvin and in fact it actually lies above infinity Kelvin. Recently there have been several claims that this has been achieved. If you search this up on Google I am sure that you will be able to find more information on this.


In the Wittig reaction, why do stabilised ylides produce E-alkenes and unstabilised ylides produce Z-alkenes?

I have given some justification on why unstabilized ylids give Z-alkenes in this answer of mine.

A possible explanation why stabilized ylids show E-selectivity is given in the organic chemistry textbook by Clayden, Warren, Wothers and Greeves on p. 817:


Is it possible to curve fit GC, Mass-spectrometry and NMR peaks?

You should not assume any given peak shape as in practice this may not be what it should be from theory.

You have a set of data points so once you have determined the baseline a numerical integration should work well. There are many of these in the literature. When fitting use the chi squared statistic as a measure of fit, the Pearson R is just useless and should never be used in fitting. (This is because a poor fit has, say, an R = .99 and a good fit .999 so its not anywhere near discriminating enough). You should also look at the shape of the residuals, autocorrelation function, runs tests etc to make sure you have a good fit. A one parameter good/bad test is often not good enough.

If you have overlapping peaks then you will have to assume some shape simply to use non-linear least squares to estimate one peak from another. This will lead to some more uncertainty but depending on what you are trying to determine this may not matter.

In gc data the peak shape is not necessarily gaussian, often it is lopsided so a numerical method is better here.

It may not be necessary to fit your data to try to get such high precision, the data simply may not allow it due to instrumental factors. But what is important is results not data, i.e. its what the data tells you that is important; this may not need such high precision as you seek. I'm absolutely not being patronising, but I would go as far as to say that if you need 0.1% accuracy you should perhaps think of another way of trying to learn what you want to know.


The open circuit voltage of a battery: Does it decrease with increase in pH of electrolyte while the voltage across a load increase?

Your aluminum half cell has a standard OCV (i.e., at 1 M $H^+$) of 1.676 volts. Ref. 1
You don't have standard conditions, but that doesn't matter; your conditions do not change for this half cell.
The other half cell is a reduction of water, in alkaline solution, with the electrons from the aluminum to produce hydroxide ions.

The CRC Handbook confirms -0.8277 volts for this reaction. I had intended to use a picture from Ref 1 showing the OCV, but it was incorrect by a factor of 10! Just goes to show you that you can't trust it just because it's in print, or online!
The point is, this reaction wants to consume hydroxide ions and generate electrons, but you are running it in reverse by pumping aluminum electrons into it.  When you increase the pH of the half cell, you increase its voltage by giving the half cell more reactant - but then you are fighting it because it is yielding to the aluminum voltage, so your total OCV out decreases. That's why your cell output is so low (that, and the non-standard conditions). But the good thing is that the extra ionic concentration decreases the cell resistance, so it can supply more current to your 1k resistor even with the lower OCV.
Ref 1. https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Electrochemistry/Voltaic_Cells/The_Cell_Potential


Is there any difference between negative electron gain enthalpy and electron affinity?

Well, energy is not always "released" when an electron is added to an atom in the ground state.
Noble gases for example, have a stable configuration and a high amount of energy is required to force the electron into a higher energy state and a less stable configuration, thus making the electron affinity negative.


Why does a smaller bandgap correlate to increased absorption of visible light?

In your case, absorption takes an electron from a valence band and in to a conduction band. At energies below the narrowest gap between the valence and conduction band, the photon can't give enough energy to an electron to get it up from the valence to the conduction band - there is no absorption there.

Right at the band gap, you can now just get a small population of electrons from the valence to the conduction band, those right at the top of the valence band. Absorption starts to kick in.

As the photon energy increases, there is a larger and larger number of valence band states that can now be excited in to the conduction band.

Depending on details of the shapes of the conduction and valence bands (density of states of initial and final vs energy), and on the exact absorption process (if you need a phonon for momentum matching it is harder), you get various expressions relating the absorption to the band gap ranging from $\sqrt E$ to $E^{2}$ or higher powers.


Why is water's IR spectra indistinguishable from the IR spectra of a solution of glucose in water?

I see the spectra in the following way:

The sample and measurement: glucose in water and IR absorption measurement

Considering free glucose molecule, we should expect the vibrations from $C-H$, $O-H$, $C-O$ and $C-C$ bonds and other associated vibrations.
The approximate vibrational frequencies are :
a) $C-H$ = ~2800 - 3300 $cm^{-1}$ (stretching)
b) $O-H$ = ~3800 - 3200 $cm^{-1}$ (stretching)
c) $C-O$ = ~1200 - 1030 $cm^{-1}$ (stretching)
d) $C-C$ =  usually not good for interpretation
e) $C-H$, $CH_{2}$ and $O-H$ bend(in plane) = 1500-1200 $cm^{-1}$
and now considering water,
the $O-H$ stretching band  spectrally overlaps with the bands of glucose (~3750 - 3200). The combination band (~2300) and scissor mode (~1600) are also relatively broad and strong. Due to this spectral overlap and also larger infrared  absorption cross-section of water, glucose is not clearly identifiable in solution.

Instrument and spectral quality : For obtaining a high resolution spectra with good s/n ratio the instrument should be good (FT-IR) and long exposure is needed.
The spectra shown by @user1155386 has the $x-axis$ as nanometers (not wavenumbers  ! ), indicating that the spectral resolution is low. It is noisy showing that exposure time can be improved (the exposure time shown there is only 10 milliseconds).

However, still it would be difficult (but not impossible) to differentiate glucose solution and water. The main hurdle would be get a good spectra and the reason 1. mentioned above.
Additionally, I should mention Raman spectroscopy is better for aqueous solution measurements.


The most common glass-etching powders are solid ammonium bifluoride and solid potassium bifluoride. See for example the Material Safety Data Sheet for “EtchON Acid-Frost Glass Frosting Powder”, or this one. This is not a definite identification, but it fits your description, and those two are the most commonly available such chemicals.

Disposal is a bit tricky. Definitely don't put it down the drain! The first MSDS linked above says:

  Ecological Information:
  Noxious for the water inhabitants. Avoid disposal into rivers, sewer system and land penetration.

The second gives a hint at disposing of it:

  Disposal Instructions:
  All wastes must be handled in accordance with local, state and federal regulations. Material can be converted to a less hazardous material by weak reducing agents followed by neutralization.

So, contact your local waste management authorities, and ask them how to proceed. Do not attempt it neutralize it by yourself unless (i) you know what you are doing, and (ii) you have the appropriate hardware (fume hood, e.g.).


Would gas come out of solution in an infinite volume?

According to this abstract from a paper titled "Why Is It Much Easier To Nucleate Gas Bubbles than Theory Predicts?":

  Bubble nucleation from supersaturated gas solutions generally takes
  place at much lower supersaturations than are expected from the
  theory. Furthermore, the same theory predicts that the threshold
  concentration of gas needed to cause nucleation should be essentially
  independent of the gas species used, a finding contradicted by
  experiment. There are two general explanations:  first, that the
  theory is wrong, or second, that there is a previously unidentified
  factor which is influencing the results of the experiments. Given the
  success of the fundamental theory in other areas, the second
  explanation is preferred. The previously unrecognized factor is
  identified here as being the surface activity of the gases which form
  the bubbles.

Although this is discussing conditions of low supersaturation and not subsaturation, I assume the same concepts apply, though it would cost me $40 to get the full article text and find out for sure.

I'm pretty sure what's happening here is that through the random motion of dissolved gas molecules, occasionally enough of them adhere to one another to form the nucleus of a bubble.  Particularly at very small radii, the high surface tension of water dominates otherwise destructive forces, and a micro-bubble begins to form.  But by the definition of saturation with respect to bubble formation, it would seem that you cannot have "macro-scale" bubble formation and growth persisting under conditions of subsaturation as equilibrium is reached.

If I can paraphrase your question as:

  "If I lower the gas pressure to below its saturation pressure, does
  the gas come out of solution if there are no interfaces (nuclei) for
  bubbles to form?"

then the answer is yes, regardless of the presence or absence of bubbles.  The dissolved gas molecules will partition at the gas-liquid interface until equilibrium concentrations are achieved in each phase.  If some gas bubble began to form, it would quickly redissolve under conditions of subsaturation with respect to the gas solubility.

Unless I'm missing something about the importance of bubbles in your question, this would pretty clearly be the bottom line answer.  Please comment if I've missed the crux of your question.


The obvious Star Wars puns aside, I have never heard about the dianion of a carbon dioxide dianion and/or that it was named carbonite.
However, the carbon dioxide radical anion, $\ce{CO2^{.-}}$, is experimentally available.
Typically, it is generated by hydrogen atom abstraction from formate, $\ce{HCOO-}$, using $\ce{HO^.}$ radicals. The $\ce{HO^.}$ radicals, in turn, are obtained from vacuum ultraviolet (VUV) irradiation of water at a wavelength of $\lambda = 175~\mathrm{nm}$.

UPDATE
Apparently, I was wrong about the dianion! As Mithoron pointed out in his comment, species like $\ce{(M^+)_2CO2^{2-}}$, formed in the reaction carbon dioxide with alkali metals in argon or nitrogen matrixes, have been detected by IR spectroscopy. See Inorganic Chemistry, 1984, 23, 177-184 for details.


Energy of orbitals under a crystal field

They are equivalent in energy because symmetry says so, and because the calculations give that as an answer. Okay, that wasn’t exactly satisfactory, was it?

The next complicated answer on the road from over-simplified to reality is that the $\mathrm{d_{z^2}}$ extends to both ligands, too: Do not forget the ‘ring’ in the $xy$ plane. By extending towards the ligands on the $z$ axis further than the $\mathrm{d_{x^2 - y^2}}$ orbitals do to their ligands, it is less stabilised than it looks, and the interaction with the equatorial ligands also destabilises.

The really sophisticated and probably most correct answer is that we generally have a wrong view of those two orbitals. They are mathematically equivalent and would actually form a group of three: $\mathrm{d_{x^2 - y^2}}$, $\mathrm{d_{z^2 - x^2}}$ and $\mathrm{d_{y^2 - z^2}}$ — but because we can only have two, wee need to add two together and commonly take the two that give ‘$\mathrm{d_{z^2}}$’. In some images of molecular orbitals you can see that they are much more identical than they initially look.


Your question implies that hydrogen bonds give rise to discrete peaks in the vibrational spectrum. It's better to say that hydrogen bonding affects the position of already existing peaks, in particular those vibrational modes coming from the $\ce{X-H}$. A hydrogen bonding network may also broaden low-frequency peaks coming from librations and hindered rotations.

For example, take the $\ce{O-H}$ stretching region in water.

The vapor spectrum consists of water molecules effectively in the gas phase that aren't interacting with each other, or are doing so rarely enough that the molecular configurations are "countable", shown by the discrete peaks between ~3600-3900 wavenumbers. A lack of water molecules interacting means hydrogen bonds aren't forming.

Moving to a bulk phase, where the water molecules are now interacting with each other all the time, hydrogen bonds are being formed and broken much more often. This results in a continuous peak, as more configurations are being sampled. In the case of water, charge transfer into the $\ce{O-H}$ bond causes a red-shift (lowering) of its frequency, moving the center of the distribution to 3400 wavenumbers. The distribution begins to narrow again in the solid phase, as the number of configurations is still large but dynamical switching between them slows.

It is also possible to experimentally measure spectra of small clusters through more advanced techniques such as mass spectrometry-detected rare gas tagging. Note the appearance of peaks around 3400 wavenumbers as the size of the cluster grows. (The presence of a proton doesn't affect these spectra in the hydrogen bonding region.)

A similar effect can be seen in molecular dynamics simulations. See how the peaks broaden and shift to lower energies when going from 1 water to the dimer, then 3 in a line/triangle, and finally 4 in a square.

First image taken from http://www1.lsbu.ac.uk/water/water_vibrational_spectrum.html. Experimental cluster spectra taken from http://pubs.acs.org/doi/abs/10.1021/acs.jpca.5b04355. BOMD plot my own creation (unnormalized dipole ACF spectra from HF/STO-3G, 0.36 fs time step, 10000 steps).


Oxygen is not flammable, as it does not burn, it supports burning. For burning to happen, a strong oxidizer (for example, oxygen) and a strong reducer (for example, carbon) must be present.

No, the problem with oxygen is, that while not being flammable, it is much more efficient in supporting burning. In air oxygen is mixed with nitrogen, meaning that for the same amount of reducer a quadruple amount of oxidizer is required and a lot more heat is lost with inert gases (nitrogen). For example, iron wire can burn in oxygen, but not in air. It is also a lot easier to lit things up in oxygen. A classic experiment is to move a smoldering splinter into a vial with pure oxygen. It immediately starts to burn with a bright flame

Now, liquid oxygen is an entirely new level of crazy. In addition to constantly releasing gaseous oxygen... If an organic material (for example, sawdust) is mixed with liquid oxygen, it becomes an explosive.


Determining the mass of sodium sulfate that forms when reacting sulfuric acid with sodium hydroxide

Rather than working with funny fractions, let’s use decimals.

$$m(\ce{H2SO4}) = 4.9~\mathrm{g}\\
m(\ce{NaOH}) = 3~\mathrm{g}\\
M(\ce{H2SO4}) = 98.09~\mathrm{g \cdot mol^{-1}}\\
M(\ce{NaOH}) = 40.00~\mathrm{g \cdot mol^{-1}}\\
M(\ce{Na2SO4}) = 142.05~\mathrm{g \cdot mol^{-1}}\\
~%
\\
n(\ce{H2SO4}) = 50~\mathrm{mmol}\\
n(\ce{NaOH}) = 75~\mathrm{mmol}$$

Remember when doing stoichiometry to correctly consider your reaction equations. In this case:
$$\ce{H2SO4 + 2NaOH -> Na2SO4 + 2 H2O}$$

If we want to do this, we see that we don’t have enough $\ce{NaOH}$ — it is present only in substoichiometric amounts. We would need $100~\mathrm{mmol}~\ce{NaOH}$ for a full reaction with $\ce{H2SO4}$:

$$\frac{n(\ce{NaOH})}{n(\ce{H2SO4})} = \frac{2}{1}\\
n(\ce{NaOH}) = 2 n (\ce{H2SO4})\\
n(\ce{NaOH}) = 2 \times 50~\mathrm{mmol}= 100~\mathrm{mmol}$$

If we now assumed that all $\ce{NaOH}$ would participate in the formation of $\ce{Na2SO4}$, we would arrive at $37.5~\mathrm{mmol}~\ce{Na2SO4}$ or $5.4~\mathrm{g}\ \ce{Na2SO4}$. But that would leave free $\ce{H2SO4}$ lying around. Instead, consider a stepwise process:

$$\ce{H2SO4 + NaOH -> NaHSO4 + H2O}\\
\ce{NaHSO4 + NaOH -> Na2SO4 + H2O}$$

We see that we need to use the first $50~\mathrm{mmol}\ \ce{NaOH}$ to generate $50~\mathrm{mmol}\ \ce{NaHSO4}$; with $25~\mathrm{mmol}\ \ce{NaOH}$ remaining. These $25~\mathrm{mmol}\ \ce{NaOH}$ can react with the $\ce{NaHSO4}$ to generate $25~\mathrm{mmol}\ \ce{Na2SO4}$ and leave $25~\mathrm{mmol}$ unreacted $\ce{NaHSO4}$.

$$n(\ce{Na2SO4}) = 25~\mathrm{mmol}\\
m(\ce{Na2SO4}) = n \cdot M = 25~\mathrm{mmol} \times 142.05~\mathrm{g \cdot mol^{-1}} = 3.55~\mathrm{g}$$


Your formulation us correct for an acid with one dissociable proton ($x=1$) below pH 7, and if you assume ideal solution behavior it even works beyond your upper bound.  Typically you should take $x=1$ because, mostly, even most strong acids dissociate one proton.  However...

One limitation, however, is when you have an acid with two or more dissociable protons.  If the second proton dissociates before you get to pH 7, then you need to add that into your calculation (and then it gets a lot more complicated).  Among common acids sulfuric acid is the one that will do this; in the limit of extremely dilute solutions (0.001 molar or less, roughly), the pH of a sulfuric acid solution will be about 0.3 unit lower than your formula with $x=1$ because of that second dissociable proton.


Is this problem about acid-base titration wrong?

Your calculations are correct. I also did a second sanity check: Adding $200~\mathrm{ml}\ \ce{NaOH}$ to a neutral solution does indeed give a $\mathrm{pH}\ 12$ solution, so they also wanted you to assume that the neutral point is the equivalence point.

Which answer to select? Well, considering that:

$$-\log(2 \cdot 10^{-2}) \approx 1.699 \approx 2$$

I would probably go with $2$. Along with telling the pupils that this is a badly worded question and in their exam there will be clearly solveable ones.


What substances form crystal structures?

Maybe to focus more on the question: what kinds of things can you make crystals from:

Typically a pure element or an inorganic or organic compound can be made in crystalline form. It might be a fine powder or small granular crystals. If grown under controlled conditions, you could get large crystals. If it is a gas or liquid at room temperature, it could perhaps solidify to a crystalline form at lower temperatures.

On the other hand, an impure, indeterminate, or polymeric material, or a mixture of compounds, very seldom will form crystals. Still, there are exceptions. Inorganic salts often crystallize from water solution as "hydrates" - with extra water molecules in the crystal lattice of the salt. Other solvents also sometimes co-crystallize with inorganic or organic substances. Neutral compounds can co-crystallize together as clathrates, such as methane hydrate - a crystalline structure containing both methane and water that forms at high pressures and low temperatures.

Even a complex or high molecular weight substance, such as an enzyme or virus particle can form crystals. In the realm of nanotechnology, new materials are being engineered specifically to aggregate into crystal-type structures.


Which species gives electron in the following reaction?

What you have here is a reduction half-equation as the $\ce{Fe(IV)}$ is accepting electrons to get reduced to $\ce{Fe(III)}$. As the name suggests, this only shows half of the actual reaction as if something is getting reduced, something else must be getting oxidised. So what you need is the oxidation half-equation which shows the species that is getting oxidised and donating electrons. When you add these two half equations the electrons will cancel each other out and you will get your equation for the reaction.

For example, one possible oxidation half-equation is the following:
$$\ce{Mg -> Mg^{2+} + 2e-}$$
Now if you times the co-efficients of the reduction half-equation by 2 and times the co-efficients of the oxidation half-equation by 3 and then add them together you will get the following equation:
$$\ce{2FeO4^{2-} 3Mg + 16H+ -> 2Fe^{3+} 3Mg^{2+} + 8H2O}$$
As you can see all the electrons balance out and that is your complete equation for the reaction.

Sorry, I don't think I really understand your second question about basic medium. Could you please expand on it to clarify what you mean.


Why is cyclopentadiene more stable than its aromatic counterpart cyclopentadienyl anion?

This is a nice question because it forces you to step back and think rather than fall into the obvious trap.

You are entirely correct that aromatic compounds are more stable than non aromatic compounds which are more stable than anti aromatic compounds assuming that all other things are equal. In this case, all other things are not equal because one of the species is neutral while the other two are charged. Having a positive or negative charge on carbon is not very favourable, even if the negative charge does make the species aromatic. The $\text{p}K_\text{a}$ of cyclopentadiene in DMSO is 18[1] which means the equilibrium constant for dissociation of cyclopentadiene into the cyclopentadienyl anion and a hydrogen ion (which will actually be attached to a DMSO molecule) is $10^{-18}$! In other words, cyclopentadiene is much more stable than the cyclopentadienyl anion.


You have to assume it is first order in each reactant.

Zero-order kinetics is always an artifact of the conditions under which the reaction is carried out.

The reaction can't really be zero-order in a reactant, otherwise it would imply the reaction continues after there is no reactant remaining.


How does solubility of corrosion product affects rate of corrosion?

The process in general is called passivation. It means, that the surface of a metal is corroded and the corrosion product forms a protective layer. If the corrosion products are hardly soluble, it therefore blocks the metal beneath it from further corrosion. If the corrosion products are soluble, this protective layer can be washed away and the material is exposed to further oxidation.

In iron, for example, the initially formed iron hydroxides are somewhat soluble in water. The formed rust is flaky and friable, which basically means it is removable by little or no stress, i.e. rain can break the surface and expose more metal to the oxidising agent. It therefore usually always decomposes completely.

In another example, copper does form such a protective layer of copper oxides. This is also called patina. These layers are hard(er) to remove and protect the underlying metal from further oxidation. This is also the reason, why for example many church roofs and the Statue of Liberty are green.

This method is often used to protect metals from corrosion, i.e. the formation of the protective layer is induced chemically.


How is mean square displacement defined in Brownian motion?

The mean squared distance is a measure of deviation: how far away two things are.

Let's say we want to find the MSD of a particle from it's origin. What we do is find the square distance at specific time intervals, say every second. We then make a table of data:

$$\begin{array}{c|c|}
 \text{time} & \text{square distance} \\ \hline
 1 & 3.6 \\ \hline
  2 & 3.8 \\ \hline
 3 & 3.4 \\ \hline
 4 & 3.2 \\ \hline
\end{array}$$

The formula for the average is adding everything up and dividing by the number of things, so in this example:

$$MSD = \frac{3.6+3.8+3.4+3.2}{4}$$

What I've done above is stated more generally, and much more elegantly, as the formula you stated,

$$MSD = 1/T * \sum_\text{t=1}^\text{T} (x(t)-x_\text{0})^2$$

Your cumulative sum formula doesn't make intuitive sense, it instead finds the sum of the mean square distance at every time increment (there is no reason why you'd want to do that)


Solubility product when you add an acid to a solid?

Are not you complicating the problem?

The solution is simple. You have $\pu{0.015M}$ of $\ce{Me(OH)2}$ and according to the reaction you will need twice as much acid (based on stoichiometry): $\pu{0.030M}$. Knowing that $\pu{pH = - \log [H^+]}$, the dissolution will be complete at $\pu{pH = 1.52}$.


Derivation of relationship between equilibrium constant and Gibbs free energy change

Using the fundamental equations for the state function (and its natural variables):
\begin{align}
  \mathrm{d}G &= -S\mathrm{d}T + V\mathrm{d}P\\
  V &= \left(\frac{\partial G}{\partial P}\right)_T\\
  \bar{G}(T,P_2) &= \bar{G}(T,P_1) + \int_{P_1}^{P_2}\bar{V} \mathrm{d}p
\end{align}
Here $\bar{x}$ represents molar $x$, i.e. $x$ per mole
\begin{align}
  \bar{V} &= \frac{RT}{P}\\
  \bar{G}(T,P_2) &= \bar{G}(T,P_1) + RT \ln\frac{P_2}{P_1}
\end{align}
Defining standard state as $P = \pu{1 bar}$ and $\bar{G}=\mu$
$$\mu(T,P)=\mu^\circ (T) + RT\ln \frac{P}{P_o}$$
consider the general gaseous reaction $\ce{a A + b B -> c C + d D}$
$$\Delta G=(c\mu_\ce{C} + d\mu_\ce{D} - a\mu_\ce{A} - b\mu_\ce{B})$$
for "unit progress" in reaction. Using $\mu_i = \mu^\circ_i + RT\ln \frac{P_i}{\pu{1 bar}}$
\begin{align}
\Delta G &= (c\mu^\circ_\ce{C} + d\mu^\circ_\ce{D} - a\mu^\circ_\ce{A} - b\mu^\circ_\ce{B}) + RT \ln\frac{P_\ce{C}^c P_\ce{D}^d}{P_\ce{A}^a P_\ce{B}^b}\\
\Delta G &= \Delta G^\circ + RT\ln Q
\end{align}


What is the energy change of ethene reacting with hydrogen gas?

Your reasoning is correct, but your numbers are wrong.

You are not breaking a C-C bond, but changing a C=C bond to C-C. So you will also need the energy for C=C which is 614 kJ/mol.


How to convert a linear dicarboxylate species to the dimethyl ester?

You have a couple of options.

Your acid-catalyzed Fisher esterification is one of them. The downside is that water, which is liberated in the reaction, has a higher boiling point than methanol. You can't just destill it off to drive the reaction to completion.
Zhe and  Hexacoordinate-C mentioned diazomethane and methyl iodide which might work well. You might want to make sure that your DNA does not get methylated.
The Steglich esterification, using dicyclohexylcarbodiimide (DCC) and N,N'-dimethylaminopyridine (DMAP) might be an alternative to the Fisher esterifiction.
You might want to try something different, using dimethyl carbonate as a non-cancerogenic alkylation reagent. The procedure is described in Green Chemistry Letters and Reviews, 2013, 61, 89-93 and uses potassium carbonate as a base, dimethyl carbonate as a reagent (and solvent), and tetrabutylammonium chloride as a catalyst.


Please explain the following graph for a quantum mechanical harmonic oscillator

Background

Let's start by considering the case of 2 hydrogen atoms separated by a large distance.  If they happen to be on a trajectory that brings them closer together they can interact with one another.  If they get close enough together they can form a bond and create a hydrogen molecule.  This process is pictured in the following graph.

image source

This "potential well" is often referred to as a "Morse potential".  Out on the right edge of the curve (say 5 angstroms), we have our 2 separated hydrogen atoms.  As the distance between them lessens, stabilization occurs, they can fall into the well and form a hydrogen molecule.

The red lines in the Morse potential represent different vibrational states of the hydrogen molecule.  Initially, the hydrogen atom is formed in one of the higher vibrational states.  It can either absorb more energy through collisions and go back to 2 dissociated atoms or it can give away some excess energy through collisions and fall to a lower vibrational level.

The classical view of the hydrogen atom is two weights connected by a spring.  The more energy in the system, the wider the oscillation of the spring.  With lower energy the oscillations are smaller.  The energy of the spring system (y-axis) as a function of separation (x-axis) is given by Hooke's Law as

$$E(x)={1\over2}kx^2$$

This is really an equation that describes a parabola, so the lower part of a Morse potential can be accurately described by a parabola.  Your top drawing represents the lower part of the Morse potential diagram.

While we have used hydrogen as our model in the above explanation, the same reasoning applies to any bond formation (and subsequently bond stretch) process.  For example, we could describe a $\ce{C-H}$ bond stretch in a completely analogous manner.

Your Questions

  What does each line represent

Different vibrational levels, as explained above

  why are they waves

Each energy state (vibrational level) is described by its own wavefunction ($\Psi_n$).  The square of the wavefunction (${\Psi_{n}^2}$) describes the probability of finding the electron at a certain point on the x-coordinate for that specific vibrational energy level. From the solution to the wavefunction we find that as we go to higher and higher energy levels, the number of nodes (a point where the wavefunction = 0) in our wave continually increases.  The square of the wavefunction produces positive values (the "upside-down" parabolas). We always have the same number of nodes in the wavefunction and the probability distribution.  For example, in the lowest vibrational energy level (n=0) there is no node in either the wavefunction or the probability distribution.  In the n=1 vibrational level there is one node in the wavefunction and probability distribution.

  what is the parabola?

Explained in the background section


What is wrong with storing deionized water in glass containers?

It is very difficult to make a container that does not introduce impurities to a liquid you place into it. As mentioned in the comments to the question, storing in a glass bottle will result in ions in your formerly pure water.

This is thought to proceed by a mechanism called ion exchange (source):

$$\ce{ Glass.Na+ + H3O+ -> Glass.H3O+ + Na+}$$

The Wikipedia article on dealkalisation states:

  Routine tests for surface dealkalization in the glass container industry all generally aim to evaluate the amount of alkali extracted from the glass when it is rinsed with or exposed to purified water. For example, dealkalization can be quickly checked by introducing a small volume of distilled water to a freshly made bottle and rolling the bottle gently to pass the water completely over its inside surface. The pH of the rinse water is then measured; untreated containers will tend to yield a slightly alkaline pH in the 8-9 range due to extracted alkali, while dealkalized containers tend to yield a pH that remains approximately neutral.

The article also mentions a technique called secondary ion mass spectrometry that can probe the composition of solid glass up to 1-2 nm depth below the surface. Leaching has been studied quantitatively in the context of storing nuclear waste in the glass state, see for example this PhD thesis (abstract in English, main text in French).

So storing in a glass bottle is not ideal because of contamination with ions. If you store it in a plastic bottle, on the other hand, you will find organic substance in your formerly pure water. Depending what kind of analysis you want to use your water for, different types of impurities will be more relevant (i.e. give spurious signals in the analysis). If you are analyzing samples with a GC/MS, glass bottles would be better (well, you would use an organic solvent, not water). If you are analyzing samples with atomic emission or absorption spectroscopy, plastic bottles would be better.

A first pointer would be the materials used in sample handling by the instrument itself.


Why is zinc not considered to have a noble gas configuration?

The difference hinges on the word "valence" in "their outer shell of valence electrons is considered to be full".

That's the difference between a $\ce{1s^2 2s^2 2p^6 3s^2 3p^6}$ configuration and a $\ce{1s^2 2s^2 2p^6 3s^2 3p^6 3d^{10} 4s^2}$ one. It's not just filled principle quantum numbers or filled subshells, but (un)filled valence subshells.

The definition of valence is a bit circular here. A valence subshell is one which can react to form a bond. For argon, all the valence subshells are filled because there's a substantial stability gap in adding that last $\ce{3p}$ electron and the first $\ce{4s}$ or $\ce{3d}$ electron. (If you think about potassium, the $\ce{3d}$ orbital is even less easily filled than the $\ce{4s}$ one.) That's why the $\ce{3d}$ orbital doesn't count as a "valence" orbital for Argon -- it takes too much energy to fill. Even though it has a principle quantum number of 3, it (re)acts like it has one of 4.

Contrariwise for zinc, there's a relatively small energy gap between the $\ce{4s}$, $\ce{3d}$ and $\ce{4p}$ orbitals. You probably already know this for $\ce{4s}$ and $\ce{3d}$, as they trade back and forth during the transition metals, but $\ce{4s}$ and $\ce{4d}$ are also relatively close in energy. This is part of the reason why atoms like carbon can "borrow" a $\ce{2s}$ electron to make four one-electron-donated bonds, rather than being forced to make two one-electron-donated bonds and one two-electrons-borrowed one.

So the definition of valence orbitals doesn't depend on their quantum numbers, but on the energy required to fill them. That's why zinc is not a noble gas - the $\ce{4p}$ orbitals count as valence (reactive) orbitals for zinc even while the $\ce{4d}$ don't.


Why are antiaromatic compounds unstable?

Antiaromatic compounds are not necessarily unstable - they are just less stable than a bunch of ethenes connected by sigma bonds. If the pi electrons in an antiaromatic compound were to delocalize across a large degenerate orbital, then the result would be a half filled shell. While a half filled shell is not the end of the world, it is a lot more stable to have a full shell of electrons. As a result, compounds such as cyclobutadiene undergo a Jahn Teller distortion (in this case, becoming a rectangle rather than a square) in order to form two separate (and non degenerate) orbitals. This creates a collection of separate, full, pi orbitals. The stability gained by having full rather than half full orbitals is more than enough to compensate for the energy required to break the symmetry in the first place. Cyclobutadiene is stabilized due to stretching not because it prevents antiaromaticity, but because it causes all the pi electrons to be part of complete shells. While it might seem weird that a physical distortion such as stretching can consume less energy than is released by going from a half to a full shell, let me remind you that this is not the case. Higher energy light (UV) is often required to stimulate electronic transitions while lower energy light (IR) can cause bond stretching in compounds.

I was able to figure this out thanks to the help of orthocresol.


Does increasing concentration of copper sulfate decrease or increase its conductivity?

I'm thinking maybe at the higher concentrations your tap water is making a film of $\ce{Cu(OH)2}$ on the surface of your electrode, and that is what's shutting down your electroplating. Your tap water might be mildly alkaline (tap water pH's can range from about $6.5$ to as high as $9.5$).

Try adding a little sulfuric acid to your solutions. If alkaline films are the problem, that will fix it.

It will also increase the conductivity of your solutions tremendously, because hydrogen ions are extremely mobile.

The solubility of copper sulfate decreases with increasing sulfuric acid concentration, so don't add too much acid.


Disclaimer:

Here I take the molar masses of carbon, hydrogen and oxygen to be $\pu{12g, 1g, 16g}$ respectively, which is what would be expeccted of you at school.

For the sake of simplicity, let's assume you have $\pu{100g}$ of the said compound. Which would mean it contains $\pu{21.62g}$ of oxygen.

Employing a bit of basic math in the analysis of that fact, leads us to similar (but more useful) re-statement:

  $\pu{74g}$ of the said compound will contain $\pu{16g}$ of oxygen (the number also being the same as the molar mass of oxygen).

Now our substance possess a general formula of $\ce{C_{x}H_{y}O_{z}}$.

So in $\pu{74g}$ of it (of which $\pu{16g}$ is oxygen), the molar masses of $\pu{C}$ ( times the number of moles of it, which is $x$ ) and $\pu{H}$ ( times the number of moles of it, which is $y$ ) must add up to $\pu{74-16=58}$. Which gives us the equation:

$$\ce{x(12) + y(1) = 58}$$

Now neither $x$ nor $y$ can carry the value $\pu{0}$ (else we won't be dealing with a hydrocarbon anymore). This gives us three mathematically (not necessarily physically) plausible options:

$\pu{x= 2; y= 34}$
$\pu{x= 3; y= 22}$
$\pu{x= 4; y= 10}$

Even if you've only just started studying organic chem. you should be able to tell that the first two options are absolutely ridiculous (that's way too many hydrogen atoms that can be accommodated around 2 or 3 carbon atoms). So we go with the third option: $\pu{x= 4; y= 10}$, which sounds promising.

Which gives us the formula  $\ce{C_{4}H_{10}O}$

You may now work out what alcohols or ethers that could give you ;-)


I think this question is impossible to answer. There are dozens of chemical types for label adhesives. Most are thermoplastic, meaning they don't form new covalent bonds with the surface. But some are thermosets, meaning they react and cross-link.

In general, it is quite difficult to dissolve a crosslinked polymer since the dissolution process requires destruction (scission) of covalent bonds. The ways to do that are obvious: heat, oxidation, reduction, acids, or bases.

One thing that needs to be said here is consideration of both the opportunity cost as well as the environmental cost of your method generally makes all but the easiest methods inadvisable. When you say the adhesive "does not respond" to X, do you soak the label in X? (I typically use a slightly larger container (say part of a PET bottle) and a shallow layer of the solvent and allow it to soak overnight.

It's always a good idea, in my opinion, to start with water with a drop or two of surfactant in it. If hot water doesn't work, I move on to 70% IsoPrOH and give that a few hrs. Next, I use an alkaline (basic) detergent solution and give that another overnight soak. If it's worth my time, I end with an acetone soak (I have acetone which I reuse for this purpose in its own container.) One of these works most of the time. When none of these work, I toss (recycle) the jar. You may or may not know that not all polymers are soluble. So, your assumption that there is a solvent for the label you want to remove isn't necessarily true.

The most aggressive solvent I am aware of is mixtures of methylene chloride, dodecylbenzene sulfonic acid, and N-methyl pyrrolidone. That was used to de-pot Soviet microelectronic back in the Cold War days to reverse-engineer them. Anyway, glass is mostly Si-O-Si bonds and alkali bases should loosen any organic's adhesion to it.

Be sure to remove any surface layer on the label - abrasion is generally quickest. Oh, speaking of dangerous chemicals (methylene chloride is a carcinogen, and ddbsa is corrosive to the skin), I've also used aqueous HF to clean glass. The HF actually removes some of the silica, but if done carefully enough doesn't visibly damage the glass surface (never would try this with glass jars; why bother? but with expensive glass flasks, and other apparatuses, it was worth the hassle. The disposal of the waste solvent was a pain, though.)


Chemical potential and it's implications in an open system

Too long for a comment, so here goes.

It's quite subtle and I'm really not sure how to explain in words, but the gist is that the chemical potential does not reflect a real, physical process of adding a substance that can be carried out. If that were the case, then the chemical potential would be the difference in the Gibbs free energies of the two systems:

System 1: $n$ moles of substance and $\mathrm{d}n$ moles of substance, separated by some imaginary barrier
System 2: $n + \mathrm{d}n$ moles of substance

and measuring the chemical potential would simply mean removing the imaginary barrier, allowing intermolecular attractions to form, etc. But that's not the case.

The chemical potential is the difference between these two systems:

System 1: $n$ moles of substance
System 2: $n + \mathrm{d}n$ moles of substance

That's all there is to it. Where do the extra $\mathrm{d}n$ moles come from? It doesn't come from anywhere. There's no real physical process going from System 1 to 2 here, so don't think of it as one.

Since Gibbs free energy is an extensive property, the $\mathrm{d}n$ moles of substance carry their own Gibbs free energy, just like they carry their own internal energy. That is why the chemical potential is positive - yes, the formation of intermolecular forces, etc. will affect the exact value of the chemical potential, and in a sense that is why the chemical potential is a partial derivative - but remember just simply having more of the substance means that there is more Gibbs free energy.


What decides whether a reaction releases light or heat

The amount of energy released by a reaction is what most often creates light. Burning, the energetic reaction usually caused by oxidation of something, creates light because the amount of heat released is large enough to heat the reactants or products to temperatures where the radiation emitted contains a lot of visible light (coals in a fire glow red because they are hot enough that black-body radiation has a significant component in the visible).

Most exothermic chemical reactions don't generate this amount of heat and, although they release heat, you don't see it. Hand warming pads, for example, often contain supercooled sodium acetate which can be made to crystallise by popping a small metal insert. The exothermic reaction (crystallisation) releases heat to warm your hands, but not enough to see visible light.

A very small class of chemical reactions produce light directly as the reaction leaves molecules in an excited state which decays with the emission of visible light. Glow sticks are an example of this.

Overall, though, most reactions produce heat because the products have lower energy than the reactants and the released energy has to go somewhere. And it usually just ends up as heat. No photons are usually involved as the energy passes directly into the vibrational and translational energy of the products. So the excess energy in the reactants' bonds ends up as vibrational or translational energy in the products.


How to separate benzyl alcohol and meta-cresol

Your observation on the boiling points is correct.

Take a look at the melting points instead: the m.p. of benzyl alcohol is -15 °C, while the m.p. of m-cresol is 11 °C. You might want to cool down the mixture and see whether you can see some enrichment.


Boiling aluminum in high-alkalinity water

It appears your creating calcium aluminate by reacting aluminum with alkaline water. Since that is probably not wanted in the brew, you might use glass or anodized aluminum, which is more resistant to chemical attack, and a water softener, if the calcium carbonate isn't wanted, either.


How to determine the specific heat of methanol from calorimeter data?

You should have included the heat capacitance of the calorimeter. Think about it, if the calorimeter was at $\pu{24.7 ^\circ C}$ before the methanol was added and then it rose $\pu{3.8 ^\circ C}$ along with the liquids, where did the energy come from to increase its temperature? Since you know that the heat capacitance of the calorimeter is $\pu{19.3 J/^\circ C}$ then you can find out how much energy the calorimeter absorbed to increase in temperature by multiplying its heat capacitance by its temperature increase. I.e. $\pu{19.3J/^\circ C} \cdot \pu{3.8 ^\circ C} = $ "Joules absorbed by the calorimeter".
Then we simply apply the heat lost equals heat gained concept:
$$\begin{multline}
\pu{25.95 g} \cdot (\pu{35.6 ^\circ C} - \pu{28.5 ^\circ C}) \cdot S \\
  = \pu{38.65 g} \cdot (\pu{28.5 ^\circ C} - \pu{24.7 ^\circ C})
    \cdot \pu{2.46 J/(g* ^\circ C)} + \pu{19.3J/^\circ C}
    \cdot \pu{3.8 ^\circ C}
\end{multline}$$
After rearranging, we get $S = \pu{2.36 J/(g * ^\circ C)}$.


Note: I'm not a modeler myself, but tested several recipes of rocket candy and read rocket modeler forum for a while.

First of all, you have to think if you will be satisfied with solid propellant or want a liquid fuel engine.

Building home-made liquid fuel rocket engine is very hard. One need to find pumps, make combustion chamber from something very heat-proof and reasonably stable chemically. Anyway, some successful reports may be found on the internetz. Note, however, that if you can choose safe liquid fuel, like kerosene or some alcohol, there is no safe oxidant for liquid-fueled rocket engine. The closest thing to safe oxidant I can imagine is nitric acid, that require work on fresh air or in specially designed room and tends to form explosives on contact with many things, like milk protein (yep!), some of which are volatile

Solid-fuel model propellants are much easier to prepare and use, there are numerous articles in the internetz about such fuels. The most common one is rocket candy, prepared from potassium nitrate and sugar. I tried it to see how it works (I didn't build any rockets), but instead of sugar I used sorbitol. Rocket candy based on sorbitol can be melted and then can be casted in any form. wiki article contains both info and links to sites with well-testes recipes by successful modelers.

Another possible option is making missile model based on some classical jet. There is quite simple pulse jet and there are some books about home-made turbine jet designs.

Anyway, I recommend to search nearby rocket modeling club and find supervisor.