Calculating pH of reaction mixture of silver nitrate and ascorbic acid

For solutions of weak acids and with simplifying assumptions [H+]=[A-], [A-] << [HA ], there is formula:

$$\mathrm{pH}=\frac12(\mathrm{p}K_\mathrm{a} - \log {c})\tag{1}$$

For our case, if the stoichiometric ascorbic acid concentration $0.368/2=\pu{0.184 mol/L}$ is used:

$$\mathrm{pH}=\frac12(4.1 - \log {0.184}) \simeq 2.42\tag{2}$$

If reaction is completed, $\mathrm{pH}$ would be given by $\ce{HNO3}$ only.

Situation would get complicated, if nitric acid starts to react with metallic silver or residual ascorbic acid. These side reactions would decrease mineral and eventually also organic acidity.

$$\ce{3 Ag + 4 HNO3 -> 3 AgNO3 + 2 H2O + NO}$$
$$\ce{3 C6H8O6 + 2 HNO3 -> 3 C6H6O6 + 4 H2O + 2 NO}$$

If more of the ascorbic acid is used, initial $\mathrm{pH}$ is again given by the above equation.

The ending $\mathrm{pH}$ is given by equation:

$$[\ce{H+}]= c_1 + [\ce{A-}]\simeq K_\mathrm{a}\frac{c_2}{[\ce{A-}]}\tag{3}$$

where $c_1$ is the nitric acid concentration, $c_2$ is the residual ascorbic acid concentration. and the right equation side is taken from $(1)$

Concentration of $\ce{H+}$ is equal to sum of $\ce{HNO3}$ concentration $c_1$ + $\ce{H+} from weak acid, that is equal to A- concentration ( HA -> H+ + A-). The latter is then approx equal to simplified extression derived from the acid dissociation constant.

$$c_1 \cdot  [\ce{A-}]  + {[\ce{A-}]}^2 - K_\mathrm{a}\cdot c_2  \simeq 0 \tag{4} $$

$$[\ce{A-}]=\frac{-c_1 + \sqrt{{c_1}^2+4\cdot c_1\cdot c_2 \cdot K_\mathrm{a}}}{2} \tag{5}   $$

From (3):

$$[\ce{H+}]=\frac{c_1 + \sqrt{{c_1}^2+4\cdot c_1\cdot c_2 \cdot K_\mathrm{a}}}{2} \tag{6}  $$

$$[\ce{H+}]=c_1 \cdot \frac{1 + \sqrt{1+ \frac{4\cdot c_2 \cdot K_\mathrm{a}}{c_1}}}{2} \tag{7}  $$

In the case by we use acid concentration twice as much as stoichiometric one, then $c_1 = 2c_2$. As $K_\mathrm{a}=\pu{7.94e-5}$:

$$[\ce{H+}]=c_1 \cdot \frac{1 + \sqrt{1+ 2 \cdot K_\mathrm{a}}}{2} \simeq c_1 \cdot  1.00004 \tag{8}  $$

You can see the dissociation of the weak acid is almost suppressed by $\ce{H+}$ from nitric acid, so we can neglect the weak acid contribution to $\mathrm{pH}$.


Atoms or molecules with spin 1 in the ground state?

Atoms

Atomic carbon with its $\mathrm{1s^2 2s^2 2p^2}$ configuration has a triplet ground state ($S = 1$), precisely because of Hund's first rule.

However, in the context of the Stern–Gerlach experiment, you might run into a problem with orbital angular momentum, as carbon's ground state also has nonzero orbital angular momentum ($^3\mathrm{P}$ ground state, $L = 1$). The behaviour in a magnetic field will be rather more complex and you probably need to take into account spin-orbit coupling.

At the moment I can't think of any atoms with a ground state term symbol of $\mathrm{^3S}$. I actually suspect that it's impossible, but I'm not really up to proving it right now.

Molecules

Triplet dioxygen (as Zhe mentioned) has no orbital angular momentum ($^3\Sigma_\mathrm{g}^-$ ground state, $\Lambda = 0$), but I'm not sure if the inhomogeneity of the electron density would have any impact. (as in, I'm genuinely not sure.)

Ions

The $\ce{^2H}$ nucleus (which us chemists usually refer to as $\ce{D+}$) is lightest stable nucleus that has a spin of 1.

The only issue is that you then have a moving charge in a magnetic field. If you could ignore it, or take it out of the equation, somehow...


Is a temperature decrease required for hydrogen bonds to form in water?

You are right, in a sense.

A more accurate statement would be that as the temperature decreases, the average lifetime of a hydrogen bond increases. Thus, on average the liquid/solid/molecule will be more stable. This is because, as you point out, hydrogen bonds are weak enough to be broken by collisions due to thermal motions (translation and rotation). They are almost always immediately formed again (in liquid), but the more frequently they are broken, the less hydrogen bonding there will be on average.

One indicator of the degree to which hydrogen bonds stay formed is the average number of hydrogen bonds over time per water. In water, this number comes out to be around 3.4 hydrogen bonds, where it could have a maximum of 4.0. Two donors and two acceptor molecules.

Thus, you can rationalize the transition from liquid water to ice in this way by considering that in ice you come quite close to the number of 4.0 hydrogen bonds per water. As you say, then, when temperature decreases, more hydrogen bonds are formed in general. I would hesitate to call it a requirement though because hydrogen bonding can take place at any temperature, but is just rarer when the temperature increases.


Calculate molarity given density and mass percentage

Solution
By definition, molarity $c$ is the amount of substance $n$ per volume of solution $V$:
$$c = \frac{n}{V}\label{eqn:1}\tag{1}$$
The volume $V$ can be found using total mass of solution $m_\mathrm{tot}$ and its density $\rho$:
$$V = \frac{m_\mathrm{tot}}{\rho}\label{eqn:2}\tag{2}$$
The amount of substance $n$ can be found using mass of solute $m$ and molar mass of solute $M$.
Knowing mass fraction $\omega,$ one can switch to the mass of solution $m_\mathrm{tot}$ already used in \eqref{eqn:2}:
$$n = \frac{m}{M} = \frac{\omega m_\mathrm{tot}}{M}\label{eqn:3}\tag{3}$$
By plugging both \eqref{eqn:2} and \eqref{eqn:3} into \eqref{eqn:1}, one receives the final expression:
$$
\require{cancel}
\begin{align}
c &= \frac{\omega \cancel{m_\mathrm{tot}}\rho}{M\cancel{m_\mathrm{tot}}} \\
  &= \frac{\omega\rho}{M} \\
  &= \frac{0.27\times\pu{1.198 g cm-3}}{\pu{98.08 g mol-1}} \\
  &= \pu{3.30E-3 mol cm-3}~\text{or}~\pu{3.30 mol L-1}\tag{4}
\end{align}
$$
Notes

There is little to no point of converting anything prior to calculations as if the dimensional analysis is done right, you'll get the answer anyways.
Density is an intensive property: you cannot use mass fraction to determine the density of solute alone.
Regarding "no of moles": first, don't use non-standardized abbreviations like "no", and second, it's amount of substance, not "number of moles".
"Number of moles" is an appropriate term in a dialogue with your neighbor about the population of talpids in your garden, but not in science.
Don't forget to define each physical quantity properly and always try to solve the problem algebraically first, and plug in the values at the end when you have the ready-to-use equation for the quantity you were asked to find.
Keep the track of significant figures (slang: "sigfigs"), e.g. you are given the density with four sigfigs, so it would make sense to also use molar mass with four sigfigs as well.
Yet due to two sigfigs of mass fraction, the accuracy of the final answer is reduced.


Positive charge on a zwitter ion at pH=7?

To sum up: of the side chains listed, only the aminoalkyl ones will be positively charged at pH 7, so four peptides -- 4, 6, 8 and 9 -- will have a net positive charge at pH 7.


Why doesn't a table sublimate, while ice does?

This relates to the difference between how ice, iron spoons and wooden tables stick together.

In wood, most or all of the bonds between the individual atoms and fibre units are covalent; making them very strong. Similarly, a spoon is an alloy of iron, nickel, chrome and carbon, also a very coherent structure.

In ice, on the other hand, the units of the crystal is held together with comparatively weak hydrogen bonds, meaning that not a lot of energy is required for a surface molecule of water to escape, in a process called sublimation.

Usually, once a water molecule escapes the ice, it is, given a relatively constant temperature, very unlikely to rejoin the body of ice; meaning that especially in open air and windy conditions, ice will sublimate at sub-zero temperatures, although as the temperature of the ice drops, this effect becomes steadily less noticeable.

Edit: It is worth noting that given enough time, a table or a spoon can indeed sublimate as well, at any temperature, but the odds against it are utterly staggering.


Regioselectivity for electrophilic aromatic substitution with multiple strongly electron withdrawing groups

As Mithoron suggests, the correct answer is probably "no reaction." There are no examples of electrophilic aromatic substitution of p-cyanonitrobenzene in SciFinder, by the introductory organic chemistry mechanism at least.

There is one report of halogenation (with N-halosuccinimide) of this compound with palladium catalysis. The cyano group serves as a directing group to place a halogen ortho to the cyano. Chlorination and bromination are both reported.

Reference: Palladium-Catalyzed Highly Selective ortho-Halogenation (I, Br, Cl) of Arylnitriles via sp2 C–H Bond Activation Using Cyano as Directing Group


The method consists
of slowly adding a dilute solution of barium chloride to a hot solution of the
sulphate slightly acidified with hydrochloric acid

It is  customary to carry out the precipitation in weakly acid solution in order
to prevent the possible formation of the barium salts of such anions as chromate,
carbonate, and phosphate, which are insoluble in neutral solutions; moreover,
the precipitate so obtained consists of large crystals, and is therefore more easily filtered

Reference: Vogel's Textbook of 'Quantitative Chemical
Analysis 11.72 Sulphate: Determination of sulphate as barium sulphate


$\ce{NaOH}$ and $\ce{LiOH}$ are both classified as ionic compounds, but as with most things in chemistry, this is always on a continuum scale, in this case between ionic and covalent bonds.  $\ce{NaOH}$ is 'more' ionic in character which means its $\ce{OH^-}$ is more frequently unaccompanied by its counter cation, $\ce{Na+}$ as compared to $\ce{LiOH}$ and its corresponding $\ce{Li+}$.

Without the cation to balance the character, $\ce{OH^-}$ will balance its charge via deprotonation, and hence it is a stronger base.


We all learn right away (Chem I) that most silver salts are insoluble - except nitrate - in water, but how soluble are they in isopropanol?  That lonely alcohol functional group may not give the same "umph" that water would for the dissolution of the Mighty Silver Cation.

Normally, you'd start with a ZnOAc dihydrate salt, so there's gonna be some water in your mix to dissolve the nitrate, but if you're going for an {aq.} soluble salt in ethanol, did you consider just adding DI to the mix?

What about changing your starting components to include the silver and zinc salts together in dissolution before adding the alcohol?  Add them both gravimetrically ratioed as salt to your alcohol solvent and use the DEA to assist in solvation.  There's docs on this, let me know if you can't find them.

Update:(moved from a second answer, ed.)

So we actually just tried using water to help dissolve the ZnOAc and AgNO3 in the ethanol solution and we found that adding DI water dropwise did help to dissolve the ZnOAc, but after aging overnight a cloudy precipitate had developed, after googling around a bit, I saw that water in ethanol could react with ZnOAc to form ZnOH. In order to test this, I added ~3M Nitric Acid dropwise, and it cleared up the solution. Unfortunately, the solution then failed to stick to our slides when spin-coating.

The good news is, the solution with the cloudy ZnOH actually produced similar XRD results to the solution without water added. We have been using monoethanolamine as a stabilizer to help dissolve the ZnOAc. MEA acts as a bidentate ligand to hold the Zn+ ions in solution. I should have a parametric analysis on a ZnO sol-gel process published by the end of semester - will post it when available.

Additionally, note from Ahmed's comment below.  Silver salts tend to be extremely photosensitive, so the smallest amount of exposure of their solutions may cause discoloration.  We have had improved success by performing as many steps as possible in darkness, and/or by covering glassware with aluminum foil.


To have conductuing valence zone requires to have infinite amount of ovarlapping valence orbitals with lesser amount of electrons. This is not usually possible for oxides, because they contain direct contacts O-O, but not M-M. O-atoms in oxides has completely filled electronic shell. Pure metals has electronic shells with much less electrons then available orbitals so they can allow electrons to move freely. This is the difference.


Why did a former emulsion of oil and water precipitate after heating and cooling when heat made it originally?

The simple explanation is that the speed of de-emulsification depends strongly on physical state and temperature.

If you have created a solid emulsion by cooling the initial blend, the de-emulsification will be very slow as there is little scope for the components to move around at all never mind separate. warming this solid up so it liquefies creates two factors that make separation easier: the mixture liquefies and the temperature rises, creating many easier ways for separation to occur. Once separation has occurred only re-blending will restore the emulsion.

Emulsions often separate slowly (milk, for example, slowly forms a layer of creamier fat on top but it doesn't happen quickly after thorough mixing). In your case the initial cooling "froze" the emulsion but the microwave heating enabled de-emulsification to occur by heating the mix. Cooling the result without blending is not going to restore the emulsion without further blending.


I see what went wrong. Such questions are often confused with aldol or cannizarro as they have a carbonyl group and a base is given.
Actually this is a completely different name reaction called Favorskii rearrangement. The key to identify it is that its applicable to α-halo ketones.
In Favorskii rearrangement, the base takes α-hydrogen from that side of ketone which is opposite to the side having the halogen. It then forms a 3 member cyclic ring as an intermediate product. The product formed is a carboxylic acid derivative, mostly ester.
You can learn more at organic-chemistry.org or wikipedia


Calculating entropy change: reversible vs irreversible process

You are confused about how to calculate the entropy change for a system and surroundings that have experienced an irreversible process.  For the irreversible path, you correctly calculated the final state using the first law of thermodynamics.  From this point on, you must totally forget about the irreversible path and, instead, focus exclusively on the initial and final end states.  You must separate the system from the surroundings, and then devise an alternative  reversible path for each of them separately that takes each from its initial state to its final state.  You did that correctly for the system, and determined its correct entropy change (-7.62 J/K).  This is the entropy change for the system both for the reversible path you devised as well as for the actual irreversible path.

In the case of the surroundings, during the irreversible process, its internal energy increased by 3.741 kJ at the constant temperature of 300 K.  So, for its alternative  reversible path, to achieve this same internal energy change at 300 K, the amount of reversible heat it would have to absorb would be the same 3.741 kJ.  So, the change in entropy of the surroundings in the irreversible process was 12.47 J/K.

I know that this might all sound a little confusing.  So here is a more detailed explanation of the fundamentals as well as a cookbook (foolproof) recipe for determining the change in entropy of the system and surroundings for any arbitrary irreversible process on a closed system (including worked examples of the methodology):  https://www.physicsforums.com/insights/grandpa-chets-entropy-recipe/


Specific gravity of H2O2 bleached viscose

You can assume that the density remains unchanged.

Hydrogen peroxide simply oxidizes dyes, thereby destroying chromophores (groups that move the absorption of a molecule into the visible spectrum). The viscose structure should remain largely unscathed, and the change in density is so minuscule I would argue it is within the experimental error.


Is there an edible, colour-less, taste-less, odour-less, water-soluble, cheaply-producable gel, solid or other substance?

Are you talking about maltodextrin? It's a solid powder assembled from glucose units with no taste smell or color. It's commonly used as a carrier; it's used to dilute the actual sweet -flavored molecule in Splenda.

If you're dealing with Hot oatmeal, you could add methyl cellulose to gel it up. Upon mixing with cold milk it will melt, acting like reverse jello; solid when hot and liquid when cold.

You asked a rather strange question, I'd recommend instant oatmeal to prevent a possibility of bacterial/fungus growth when storing oatmeal for extended periods of time.


Reaction of 1,1-diphenylethanol on Montmorillonite K10 under microwave radiation

Is my answer correct?

No! Your "answer" is just a random collection of facts and does not address the question.

Actually, Ben Norris has already kindly provided all the necessary information in his comment:

Montmorillonite K10 is a solid, acidic catalyst.
Microwave irradiation means that you are applying heat to the reaction.

Under the given conditions, 1,1-diphenylethanol (1) is protonated, eliminates water (which might be driven out since heat is applied) and yields 1,1-diphenylethene (2).

Given that you're still applying heat in the presence of an acidic catalyst, the reaction will not stop here.

Protonation of 2 again results in the known bisbenzylic cation, which will add to a second molecule of 1,1-diphenylethene in Markovnikov fashion.

The resulting intermediate is will undergo an intramolecular Friedel-Crafts alkylation which furnishes the indane derivative 3.


Why is water instead of hydroxide ions used to balance oxygen atoms in redox half equations?

It's not that we really have a choice in this case between water or hydroxide or pick one or another arbitrarily to fit the stoichiometry of the whole redox reaction. The redox reaction is obviously taking place in quite a strong acidic medium (makes sense, e.g. if $\ce{Zn}$ is used as reducing agent) as you supply protons $\ce{H+}$ to the nitrate according to your reaction.

In fact, 10 protons for each reduced nitrate group is needed; and even if you propose hydroxide ion formation, you'll still need 7 protons (second reaction). Either way, there is no chance for $\ce{OH-}$ to coexist with $\ce{H+}$ in acidic medium and equilibrium will be shifted towards water formation instead.

Note that it doesn't mean that reduction of nitrate is forbidden if hydroxide ions are formed. For example, you can use another reducing agent such as $\ce{Al}$ and do perform the reduction in basic media instead. In this case you are going to end up with hydroxide ions all right:

$$
\begin{align}
\ce{Al + 6 OH- &→ [Al(OH)6]^3- + 3 e-} &|\cdot 8 \tag{ox}\\
\ce{NO3- + 8 e- + 6 H2O &→ NH3 + 9 OH-}&|\cdot 3 \tag{red}\\
\hline
\ce{8 Al + 3 NO3- + 21 OH- + 18 H2O &→ 8 [Al(OH)6]^3- + 3 NH3} \tag{redox}
\end{align}
$$

Just as with the formation of human personality, environment also plays gigantic role in chemistry.


What is the physical significance of the universal gas constant R?

It may be helpful to look at a related value $k_\mathrm{B}$, the Boltzmann constant, which is widely used in thermodynamics.
These two are related by $R = k_\mathrm{B}N_\mathrm{A}$, allowing the ideal gas law to also be written:
$$PV = Nk_\mathrm{B}T$$ where $N$ is the number of particles, as opposed to the number of moles.
The units are $\pu{J\cdot K^{-1}}$.
It's a proportionality between energy and temperature. In the ideal gas law, it relates temperature on the right to the product of pressure and volume, which is an energy.
This general idea is frequently used in thermodynamics, as you will see factors of the form $\exp{(-E/k_\mathrm{B}T)}$, where the $k_\mathrm{B}$ allows the exponent here to be unitless.
As examples:

Planck's law, where the energy is in the form of quantum energy level spacing:
$$B_{\nu}(\nu, T) = \frac{2h\nu^{3}}{c^{2}}\frac{1}{\exp\left(\frac{h\nu}{k_{\mathrm{B}}T}\right) - 1}$$
Maxwell-Boltzmann distribution, where the energy refers to the kinetic energy of gas molecules:
$$f(v) = \left(\frac{m}{2\pi k_{\mathrm{B}}T}\right)^{\frac{3}{2}}4\pi v^{2}\exp\left(-\frac{mv^{2}}{2k_{\mathrm{B}}T}\right)$$


To start off, here is the equilibrium we are talking about so we can refer to it later:

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

Since the pH of the solution is $\mathrm{-log[H+]}$, the pH of pure water will decrease upon the addition of carbonic acid.

Where I think you are getting confused is regarding the action of the bicarbonate that is produced. As this is an equilibrium reaction, bicarbonate will indeed act as a base, which is the reaction from right to left above. But lets say a molecule of bicarbonate reacts with hydrogen ion to reform carbonic acid. That just means another molecule of carbonic acid will dissociate to take it's place. This is all just a way of saying that the reaction shown above is an equilibrium, such that both the left to right and right to left reactions are occurring. But in the end there is never any net hydroxide produced or hydrogen ions consumed by this reaction, as there is always plenty carbonic acid available to counteract the basic action of the bicarbonate.

I hope I understood your question correctly. Feel free to use the comments below to ask for any clarifications.


You might want to raise the temperature a little bit as the reaction proceeds at 100-110°C to form formic acid. If the temperature is further raised to 260-280°C, then the reaction will proceed to form allyl alcohol.

Formic acid has almost unlimited shelf life if unopened and is protected from heat and light sources. Allyl alcohol is stable in normal storing conditions but it may polymerize into a thick syrup if left untouched for several years.

Discussing about safety concerns is pretty much off-topic here but you might want to check the safety data-sheets: formic acid and allyl alcohol.

Here is a research paper for further reading.

Reaction sources: Google Images


Is the effective nuclear charge only defined for valence electrons?

$Z=Z_{eff}+\sigma$, where $\sigma$ is shielding constant calculated according to Slater's rule. Since Slater's rule can be applied to any electron regardless of whether or not it's a valence electron, I think $Z_{eff}$ and $Z$ can be defined for any electron too, $Z$ in particular remaining constant.
The effect of outer electrons not being taken is also explained by Slater's rules empirically.


Moist Chlorine & Hydrogen Psychrometric or Mollier Chart

I have searched hard but without proper psychrometric data am unable to even estimate preliminary values.

You have to google for 'vapor pressure' of water and ideal gas model. In cases ideal gas model is close enough to true, the total pressure is divided into partial pressures, attributed to different gases, and the partial pressures are proportional to amount of molecules in the gas mixture. So, in the case the equilibrium partial pressure of water vapors above the liquid can be found in any table of vapor pressures, the rest of the pressure would be other gases. In the case of vapor above saline water, the equilibrium pressure is reduced by the fact that the salt is present. The exact reduction is guarded by Raoult's Law and is linearly proportional to the concentration of the salt.

  Two coolers (shell/tube exchanger with chilled water on shell side) are planned to dehumidify these two gases.

To dehumidify these gases effectively, if this is really required, consider bubbling them through layer of concentrated $\ce{H2SO4}$, it is a quite common and easy way to dehumidify gases.


Identifying the molecular structure of compound with sum formula C8H14O3 formula using NMR and IR spectrometry data

Compare the quartet at $\pu{4.2ppm}$ and the singlet at $\pu{3.5ppm}$. These quite obviously would, in your friend’s structure, correspond to the protons on either side of the ether-oxygen. However, we should expect the methylene singlet to be more deshielded than the ethyl group’s methylene: It is squashed between an electron-withdrawing $\ce{OR}$ and an electron-withdrawing $\ce{C=O}$. The ethyl group’s methylene is between an electron-withdrawing $\ce{OR}$ and a mostly electron-neutral $\ce{CH3}$ group. Therefore, the order of peaks we observe in the spectrum does not make sense when we consider your friend’s proposal.
Thereafter, one might also argue that a quartet at $\pu{4.2ppm}$ is so obviously an ethyl ester as likely every organic chemist has had way to many NMR spectra after a column that contained a peak exactly there. An ethyl-ether is more shielded (diethyl ether’s methylene proton has a chemical shift of $\pu{3.5ppm}$.


Is the mole fraction used in Raoult's law calculated initially or at equilibrium?

Assume the system to be in equilibrium, otherwise you can't use the standard equations of thermodynamics. As for your second question, it's impossible to answer because it depends on the solvent and details of the experimental setup (size of container relative to volume of solvent, temperature, etc)


Coordination geometry: why is of Cu(acac)2 square planar?

With bidentate ligands of the form $\ce{LL'}$, it is difficult to force triple coordination to a copper(II) centre. I don't have data for acac offhand, but I have some data for en binding to Cu. (Source: some past exam question somewhere.)

$$\begin{array}{c|ccc}
n & 1 & 2 & 3 \\ \hline
\log K_n & 10.55 & 9.05 & -1.0
\end{array}$$

The exact numbers don't matter; only the trend does. (Which is why I'm lazy to find a better source.) The binding of the third ligand, which would necessitate coordination at the axial position, is very weak. To get to a $\ce{[Cu(acac)3]}$ complex, you would have to either 1) adopt a perfect octahedral environment and sacrifice the electronic stabilisation from the Jahn-Teller distortion; or 2) stretch two of the acac ligands to a longer length than they would like to be.

As for why axial elongation is favoured over compression, it is perhaps due to electronic reasons: under elongation, the $\mathrm{d}_{z^2}$ orbital is doubly occupied and the $\mathrm{d}_{x^2-y^2}$ orbital is singly occupied. On the other hand, under compression, $\mathrm{d}_{z^2}$ orbital is singly occupied.

Now, there is a second-order mixing which allows the $\mathrm{4s}$ orbital to mix with the $\mathrm{d}_{z^2}$, since they possess the same symmetry ($A_\mathrm{1g}$) under $D_\mathrm{4h}$ symmetry; in the process the $\mathrm{d}_{z^2}$ orbital is stabilised, and the $\mathrm{4s}$ destabilised. The destabilisation of the $\mathrm{4s}$ orbital doesn't matter, since it's empty. However, the stabilisation of the $\mathrm{d}_{z^2}$ orbital is, of course, more profound when there are more electrons in the $\mathrm{d}_{z^2}$ orbital, which corresponds to elongation.

Here's a diagram from my notes.


As per classic literature, PF5 shows identical bond lengths of all P-F bonds which are supported by data from NMR and IR spectroscopy which led to the development of berry-pseudo-rotation. But data's from electron diffraction shows that there are eventually two types of bond lengths (P-F (axial)> P-F(equatorial)) which supports the fact that a naive might predict the hybridisation to be sp²d and hence T.B.P. Even in JD lee, it has been clearly stated that this fluctuation may be called the "Paradox" of Chemistry


Derivation of the Heisenberg uncertainty principle

The proof I will use is taken from Griffiths, Introduction to Quantum Mechanics, 2nd ed., pp 110-111.

Defining "uncertainty"

Let's assume that the normalised state $|\psi\rangle$ of a particle can be expanded as a linear combination of energy eigenstates $|n\rangle$, with $\hat{H}|n\rangle = E_n |n\rangle$.

$$| \psi \rangle = \sum_n c_n |n\rangle \tag{1}$$

The expectation value (the "mean") of a quantity, such as energy, is given by

$$\begin{align}
\langle E\rangle &= \langle \psi | H | \psi \rangle \tag{2}
\end{align}$$

and the variance of the energy can be defined analogously to that used in statistics, which for a continuous variable $x$ is simply the expectation value of $(x - \bar{x})^2$:

$$\sigma_E^2 = \left\langle (E - \langle E\rangle)^2 \right\rangle \tag{3}$$

The standard deviation is the square root of the variance, and the "uncertainty" refers to the standard deviation. It's more proper to use $\sigma$ as the symbol, instead of $\Delta$, and this is what you will see in most "proper" texts.

$$\sigma_E = \sqrt{\left\langle (E - \langle E\rangle)^2 \right\rangle} \tag{4}$$

However, it's much easier to stick to the variance in the proof. Let's generalise this now to any generic observable, $A$, which is necessarily represented by a hermitian operator, $\hat{A}$. The expectation value of $A$ is merely a number, so let's use the small letter $a$ to refer to it. With that, we have

$$\begin{align}
\sigma_A^2 &= \left\langle (A - a)^2 \right\rangle  \tag{5} \\
&= \left\langle \psi \middle| (\hat{A} - a)^2 \middle| \psi \right\rangle \tag{6} \\
&= \left\langle \psi \middle| (\hat{A} - a) \middle| (\hat{A} - a)\psi \right\rangle \tag{7} \\
&= \left\langle (\hat{A} - a)\psi \middle| (\hat{A} - a) \middle| \psi \right\rangle \tag{8} \\
&= \left\langle (\hat{A} - a)\psi \middle| (\hat{A} - a)\psi \right\rangle \tag{9}
\end{align}$$

where, in going from $(7)$ to $(8)$, I have invoked the hermiticity of $(\hat{A} - a)$ (since $\hat{A}$ is hermitian and $a$ is only a constant). Likewise, for a second observable $B$ with $\langle B \rangle = b$,

$$\sigma_B^2 = \left\langle (\hat{B} - b)\psi \middle| (\hat{B} - b)\psi \right\rangle \tag{10}$$

The Cauchy-Schwarz inequality

...states that, for all vectors $f$ and $g$ belonging to an inner product space (suffice it to say that functions in quantum mechanics satisfy this condition),

$$\langle f | f \rangle \langle g | g \rangle \geq |\langle f | g \rangle|^2 \tag{11}$$

In general, $\langle f | g \rangle$ is a complex number, which is why we need to take the modulus. By the definition of the inner product,

$$\langle f | g \rangle = \langle g | f \rangle^* \tag{12}$$

For a generic complex number $z = x + \mathrm{i}y$, we have

$$|z|^2 = x^2 + y^2 \geq y^2 \qquad \qquad \text{(since }x^2 \geq 0\text{)} \tag{13}$$

But $z^* = x - \mathrm{i}y$ means that

$$\begin{align}
y &= \frac{z - z^*}{2\mathrm{i}} \tag{14} \\
|z|^2 &\geq \left(\frac{z - z^*}{2\mathrm{i}}\right)^2 \tag{15}
\end{align}$$

and plugging $z = \langle f | g \rangle$ into equation $(15)$, we get

$$|\langle f | g \rangle|^2 \geq \left[\frac{1}{2\mathrm{i}}(\langle f | g \rangle - \langle g | f \rangle) \right]^2 \tag{16}$$

Now, if we let $| f \rangle = | (\hat{A} - a)\psi \rangle$ and $| g \rangle = | (\hat{B} - B)\psi \rangle$, we can combine equations $(9)$, $(10)$, $(11)$, and $(16)$ to get:

$$\begin{align}
\sigma_A^2 \sigma_B^2 &= \langle f | f \rangle \langle g | g \rangle \tag{17} \\
&\geq |\langle f | g \rangle|^2 \tag{18} \\
&\geq \left[\frac{1}{2\mathrm{i}}(\langle f | g \rangle - \langle g | f \rangle) \right]^2 \tag{19}
\end{align}$$

Expanding the brackets

If you've made it this far - great job - take a breather before you continue, because there's more maths coming.

We have1

$$\begin{align}
\langle f | g \rangle &= \left\langle (\hat{A} - a)\psi \middle| (\hat{B} - b)\psi \right\rangle \tag{20} \\
&= \langle \hat{A}\psi |\hat{B}\psi \rangle - \langle a\psi |\hat{B}\psi \rangle - \langle \hat{A}\psi | b\psi \rangle + \langle a\psi |b\psi \rangle \tag{21} \\
&= \langle \psi |\hat{A}\hat{B}|\psi \rangle - a\langle \psi |\hat{B}\psi \rangle - b\langle \hat{A}\psi | \psi \rangle + ab\langle \psi |\psi \rangle \tag{22} \\
&= \langle \psi |\hat{A}\hat{B}|\psi \rangle - ab - ab + ab \tag{23} \\
&= \langle \psi |\hat{A}\hat{B}|\psi \rangle - ab \tag{24}
\end{align}$$

Likewise,

$$\langle g | f \rangle = \langle \psi |\hat{B}\hat{A}|\psi \rangle - ab \tag{25}$$

So, substituting $(24)$ and $(25)$ into $(19)$,

$$\begin{align}
\sigma_A^2 \sigma_B^2 &\geq \left[\frac{1}{2\mathrm{i}}(\langle\psi |\hat{A}\hat{B}|\psi \rangle - \langle \psi |\hat{B}\hat{A}|\psi\rangle) \right]^2 \tag{26} \\
&= \left[\frac{1}{2\mathrm{i}}(\langle\psi |\hat{A}\hat{B} - \hat{B}\hat{A}|\psi \rangle ) \right]^2 \tag{27}
\end{align}$$

The commutator of two operators is defined as

$$[\hat{A},\hat{B}] = \hat{A}\hat{B} - \hat{B}\hat{A} \tag{28}$$

So, the term in parentheses in equation $(27)$ is simply the expectation value of the commutator, and we have reached the Robertson uncertainty relation:

$$\sigma_A^2 \sigma_B^2 \geq \left(\frac{1}{2\mathrm{i}}\langle[\hat{A},\hat{B} ]\rangle \right)^2 \tag{29}$$

This inequality can be applied to any pair of observables $A$ and $B$.2

The Heisenberg uncertainty principle

Simply substituting in $A = x$ and $B = p$ gives us

$$\sigma_x^2 \sigma_p^2 \geq \left(\frac{1}{2\mathrm{i}}\langle[\hat{x},\hat{p} ]\rangle \right)^2 \tag{30}$$

The commutator of $\hat{x}$ and $\hat{p}$ is famously $\mathrm{i}\hbar$,3 and the expectation value of $\mathrm{i}\hbar$ is of course none other than $\mathrm{i}\hbar$. This completes the proof:

$$\begin{align}
\sigma_x^2 \sigma_p^2 &\geq \left(\frac{1}{2\mathrm{i}}\cdot\mathrm{i}\hbar \right)^2 \tag{31} \\
&= \left(\frac{\hbar}{2}\right)^2 \tag{32} \\
\sigma_x \sigma_p &\geq \frac{\hbar}{2} \tag{33}
\end{align}$$

where we have simply "removed the square" on both sides because as standard deviations, $\sigma_x$ and $\sigma_p$ are always positive.

Notes

1 I have skipped some stuff. Namely, $\langle \hat{A}\psi |\hat{B}\psi \rangle = \langle \psi |\hat{A}\hat{B}|\psi \rangle$ which is quite straightforward to prove using the hermiticity of both operators; $\langle \psi |\hat{A}|\psi \rangle = a$; $\langle \psi |\hat{B}|\psi \rangle = b$; and $a = a^*$ since it is the expectation value of a physical observable, which must be real.

2 This does not apply to, and cannot be used to derive, the energy-time uncertainty principle. There is no time operator in quantum mechanics, and time is not a measurable observable, it is only a parameter.

3 Technically, it is a postulate of quantum mechanics. (If I am not wrong, it derives from the Schrodinger equation, which is itself a postulate.)


Why is the volume in this compress gas cylinder constant throughout?

It is about the volume of the cylinder, which of course does not change. You do not also consider the volume of the escaped nitrogen gas.


Why don't hydride shifts occur with hydrohalogenation of 4-methyl-cyclohexene?

The reason for not having consecutive hydride shifts is that generally, a new carbocation is formed through shift only when it is more stable than the immediately previous one.

To form a tertiary carbocation, 3 consecutive hydride shifts would have to occur consecutively (when + is at c-4 from the alkyl group, your first product) in which more than one intermediates have same stability (same number of alpha-H). Hydride shifts will not occur for your second product to form tertiary carbocation as the next intermediate is infact less stable (3 alpha-H as compared to 4 alpha-H previously).

Since the ring already has 6 members (least strain) so ring expansion/contraction will not occur either.


Which equilibrium constant is appropriate to use?

As noted in this previous question, the correct definition of the equilibrium constant $K$ depends on activities. If you are interested in the derivation of the equation $\Delta G^\circ = -RT \ln K$ (which requires "proper" thermodynamics), read Philipp's answer to that question. For a reaction

$$0 \longrightarrow \sum_i \nu_i \ce{J}_i$$

(this is a fancy way of writing the reaction which makes sure that the stoichiometric coefficients $\nu_i$ of the reactants are negative), the equilibrium constant is defined as follows:

$$K = \prod_i a_i^{\nu_i}$$

where $a_i$ is the activity of the species $\ce{J}_i$ at equilibrium. (Note that since $\nu_i$ for reactants is negative, the terms for the reactants will appear in the denominator of a fraction.)

The question is, why are $K_c$ and $K_p$ taught at an introductory level despite them not actually being the equilibrium constant $K$? The answer is likely partly because it is difficult to introduce the idea of an activity, but also because $K_c$ and $K_p$ are good approximations to the actual equilibrium constant, $K$. To see why, we have to look at the mysterious quantity called the activity and see how it is related to the concentration of a solution, or the partial pressure of a gas.

If we assume certain ideality conditions, then we can write the activity of a gas as:

$$a_i = \frac{p_i}{p^\circ}$$

where $p_i$ is the partial pressure of the gas and $p^\circ$ is the standard pressure, defined by IUPAC to be equal to $1~\mathrm{bar}$. Therefore, $a_i$ is nothing but the partial pressure of the gas in bars, but without the units. If we therefore define $K_p$ as follows:

$$K_p = \prod_i p_i^{\nu_i}$$

then, as long as the pressures you use are in bars, the numerical value of $K_p$ will be the same as that of $K$, the true equilibrium constant. As such, if you ignore the fact that you are taking the natural logarithm of a quantity with units, the value of $\Delta G^\circ$ that you obtain will be "correct".

Similarly, for a solution, we can write:

$$a_i = \frac{c_i}{c^\circ}$$

where $c_i$ is the concentration of the species and $c^\circ$ is the standard concentration, again defined by IUPAC to be equal to $1~\mathrm{mol~dm^{-3}}$. So, the same idea applies. If you keep all concentrations in units of $\mathrm{mol~dm^{-3}}$, then the numerical value of $K_c$ will match that of $K$.

An example

Consider the Haber process (all species gaseous):

$$\ce{N2 + 3H2 <=> 2NH3}$$

We would define the "true" equilibrium constant as:

$$K = \frac{(a_{\ce{NH3}})^2}{(a_{\ce{N2}})(a_{\ce{H2}})^3}$$

and you would define $K_p$ as:

$$K = \frac{(p_{\ce{NH3}})^2}{(p_{\ce{N2}})(p_{\ce{H2}})^3}$$

Let's say (just as an example - the numbers are not correct!) that at a certain temperature $T$, we have the equilibrium partial pressures:

$$\begin{array}{c|c|c}
p_{\ce{NH3}} & p_{\ce{N2}} & p_{\ce{H2}} \\ \hline
20~\mathrm{bar} & 50~\mathrm{bar} & 50~\mathrm{bar}
\end{array}$$

Then you would have $K_p = 6.4 \times 10^{-5}~\mathrm{bar^{-2}}$. Now you can see why it is not correct to use $K_p$ in the equation above: you would have

$$\Delta G^\circ = -RT \ln(6.4 \times 10^{-5}~\mathrm{bar^{-2}})$$

which makes absolutely no sense since you can't take the logarithm of a unit. In a simplistic treatment you would just ignore the fact that a unit existed and just take the logarithm of the number.

However, you will not run into that problem if you use the "true" equilibrium constant, as you really should. If we assume ideality, i.e. $a_i = p_i/p^\circ$, then we find that $K = 6.4 \times 10^{-5}$. Since activities are dimensionless, you will find that $K$ is dimensionless and you can now take the logarithm. As I mentioned earlier, the numerical quantity of $K$ is exactly equivalent to $K_p$ as long as you keep your pressures in units of bars.

I have seen some questions where students are requested to determine $K_c$ for a gas-phase reaction using the ideal gas law and hence $\Delta G$:

$$\begin{align}
pV &= nRT \\
c = \frac{n}{V} &= \frac{p}{RT}
\end{align}$$

The value of $c_i$ and $p_i$ will therefore differ by a factor of $RT$. Therefore, the numerical values of $K_c$ and $K_p$ in gas-phase reactions will, in general, differ. (If you have the same number of moles of gas on both sides, then the factors of $RT$ will cancel out and numerically $K_c = K_p$, but that is not always the case.) As established earlier, because of the equation for the activity of an ideal gas, only the use of $K_p$ will lead to the correct numerical value for $\Delta G^\circ$.

If a question requires you to do so, then do it. Just bear in mind that it is not only wrong in terms of taking the natural logarithm of a quantity with units; it will also give you a wrong numerical result for $\Delta G^\circ$.


Activation energy in physical adsorption

The definition given above would more closely describe chemisorption. At large separation from the surface the interaction is zero. In physisorption the interaction becomes attractive and forms a potential well as the surface is approached. There is no activation energy needed to do this. At shorter separations than the potential minimum, the potential becomes repulsive. The minimum is usually at larger separation than the minimum for equivalent species exhibiting chemisorption. Species may physisorp before chemisorption occurs

Comment by porphyrin converted to answer


Partial and total concentration in solutions

It's true and false.

For example:
1 Solution with 1mol/L of NaCl and 1mol/L of KBr.
The total concentration of salts is 2mol/L.

But the ions concentration are:

Na+ 1mol/L
K+ 1mol/L
Cl- 1mol/L
Br- 1mol/L

Is it clear?


Your dilution approach can be heavily improved.

Rule no. 1: 2+2 $\neq$4 in solution preparation from an analytical chemist's perspective. An extreme example is that you add 50 mL of water+50 mL of ethanol $\neq$ 100 mL solution. If you add a solid NaCl to 1 L of water, the volume may not remain 1 L. It may decrease!

Analytical dilutions are always prepared in highest quality volumetric glassware (Called class A glassware), which comes in fixed volumes. You might have 2 mL, 5 mL, 10 mL, 25 mL, 50 mL, 1000 mL flasks in the university. Similarly, pipettes comes in fixed volumes. The graded ones (so called Mohr's) are not that great.

With these limitations, you design dilutions, in such a way that volumes are not fractional so a volume of 24.75 mL for dilution is a no-no.

The dilution relation is handy $$\mathrm{C_iV_i=C_fV_f}$$, where i and f indicate initial and final concentrations (C) and volumes (V).

So think about it now in order to prepare 1000 mg/mL you would like to dilute it 1000-fold in steps to make it a 1 ppm solution or 1000 ppb $\ce{Pb(NO3)2}$. From there you would like to dilute it 100-fold to make it 10 ppb.

The first question you would like to think now, is what volumetric glassware is available.


Negative ions affecting galvanic cell voltage

Well, assuming that you're talking about an aqueous cell, in order to get the anion into solution, you need a salt: eg $\ce{CuSO4}$ or $\ce{Cu(NO3)2}$. Yes, the local electric field around a single $\ce{SO4^2-}$ anion would be stronger than around a single $\ce{NO3^-}$. However, at the same concentration of $\ce{Cu^2+}$ you would have twice as many $\ce{NO3^-}$ ions so the overall solution would still be neutral.

In a microscopic level, each ion is surrounded by a solvation shell of water molecules that all have their partial positive charge on their hydrogens pointing at the anion (or oxygen towards the cation). For this reason they are also shielded from the electrode. Speculation: double charged anions would probably have a thicker solvation shell.

I don't see how this explains the difference in potential of $\ce{Cu^2+/Cu}$ vs $\ce{Zn^2+/Zn}$ since both have the same charge.


This is an area of corrosion science. Chloride ions do not aid general corrosion  of stainless but rather the special processes called crevice corrosion and pitting corrosion. General corrosion is of large areas. Crevice corrosion is like　a hole in the metal and local corrosion effects occur. Pitting corrosion is corrosive fluid forms its own hole and similar local corrosion effects occur. The local corrosion effects for chloride crevice and pitting corrosion corrosion are complex and dependant on the fluid but a major one is the increase in acidity from  ferric ions (which normal complex with water to be acid) as the chloride ions displace a water molecule and the new complex is more acidic. This acidity damages the complex chromium iron oxide layers. Another effect is that the chloride penetrates the oxide layer and opens it up to allow corrosive species mainly oxygen to penetrate to base metal.

Nickel gives some benefit to corrosion protection of  stainless steels by substituting for iron in the iron-chromium-oxide but that's not main reason its included but rather allows the stainless steel to be austenitic- basically more ductile and easier to weld.
Molybdenum is main element that is added to improve pitting/corrosion resistance. 304 is the most common basic number stainless steel 16-18% Chromium and 8-10% Nickel  (they add letters to give more information) . 316 is second as it contain  16-18% Chromium, 8-10% Nickel and 2-3% Molybdenum- greatly reduces pitting corrosion- conditions need to be more severe to have pitting and crevice corrosion.


How do point mutations affect complementary base-pairing and widths of the nucleotide pairs?

When Point mutations occur in one strand of DNA, irregular base-pair emerge. One of the main features of this irregular state is that the width of the double-strand differs from the standard form(usually larger width as the new, irregular base pairings are weak) in that specific area. In other words, the portion of DNA which has the defective base-pair bulges out or protrudes leading to the asymmetry of DNA.(analogous but not the same as Thymine dimers) Most of the times the DNA repair system finds this and deals with it in the correct way. I do not know about the exact number of hydrogen bonds that appear in these situations but perhaps it obeys either Watson-Crick, Hoogsteen, or Wobble rules.


Electrolysis of Anhydrous Sodium Hypochlorite

Obviously, sodium does not have a choice other than getting reduced. Oxidising it to sodium(II) is not an easy feat considering that the electron would have to be extracted from a core orbital. That means, the hypochlorite must be oxidised.

Usually, an oxidation of hypochlorite will yield chlorate, $\ce{ClO3-}$. If we attempt to form a redox half-reaction out of that, we can only use $\ce{ClO-}$ as a charge balancing agent:

$$\begin{align}\ce{ClO- \phantom{\ce{+ 4 ClO-}} &-> ClO3- + 4e-}\tag{Ox1}\\
\ce{ClO- + 4 ClO- &-> ClO3- + 4 e-}\tag{Ox2}\end{align}$$

If we do this, we have four extraneous chlorine atoms and two extraneous oxygen atoms. Thus, it makes most sense to do the mass balance by adding $\ce{Cl2O}$:

$$\ce{ClO- + 4 ClO- -> ClO3- + 4 e- + 2 Cl2O}\tag{Ox3}$$

However, it will be difficult confirming this. Sodium hypochlorite is not stable, especially not at elevated temperatures, especially not in pure form, especially not anhydrous sodium hypochlorite (sensing a pattern?). You would need a certain amount of recklessness to attempt this and sufficient luck to be able to report the result. Which is possibly why I cannot find any scientific publication on the topic; so take the equation with two grains (or more) of salt.


As a short answer, Hydrogen bonds (for elementary chemistry) are formed with donation of lone pair of a lewis base into LUMO of H attached to a highly electronegative atom. A better donor makes a better bond and more electronegative atom makes a better bond. From my experience, the second factor overrides the first one in most cases.

N is a better donor than O and H in $\ce{O-H}$ is a better acceptor than H in $\ce{N-H}$.


More effective fire extinguisher liquid than water

You could consider an aqueous film forming foam (AFFF) system, this is water with some additives to form a foam. Originally foam was something intended for class B fires, but there are now foams which are intended for both class A and class B fires. They are even considered to be acceptable for use on gas fueled fires where there is no risk of explosion (class C). Thus there are ABC foams.

A word of warning some ABC foams such as those based on PFOS are horrible and harmful. It might be best if you stick to a water spray system. The PFOS foams are considered to be problem when they enter the environment, if I recall correctly they are very harmful to fish and the perfluoro compound PFOS only breaks down very slowly in the environment.


Boiling points of v. similar organic molecules

Solids and liquids are hold together by cohesion. Intermolecular interactions contributing to cohesion depend on the geometry of the molecules in interaction,too.  Here, n-pentane with an elongated molecular structure exhibits a much larger molecular surface as compared to neo-pentane of a more sphere-like structure*).  As a result, other criteria kept constant (like the molecular mass), the lesser the molecular surface area of molecules of a substance, the lower the boiling temperature of this substance.

*)True, neo-pentane's methyl groups point towards the vertices of a tetrahedron. Actually I recommend to build a molecular model of both compounds and place them side-by-side to see it.


Can all NMR spectrometers perform special techniques such as COSY, NOESY, HETCOR,..?

It depends on a lot of things, software-wise

some of these techniques are simply a software option
you can implement pulse sequences yourself, if that´s possible with your control software
you need to have the right software to transform and evaluate the data
... or write it yourself (not exactly rocket science if you know your way with Matlab/Octave etc.)

and also regarding your hardware

you can have several probes for one spectrometer

they can be single channel, or dual ("HX" usually), triple ... channel
the "X" channel can be tuneable over a wide(r) range, or you are limited to e.g. carbon

of course you need a transceiver with enough channels, which can be used at the wanted frequencies
and a suitable preamp for every channel
some techniques need more special hardware, like CPMAS, pulsed gradients, cryo etc.
low field spectrometers on the market (afaik) have no user exchangable parts, so you are stuck with the hardware you bought

But to answer your immediate question: I guess with basically any living (as in "not undead") spectrometer, all those you named above should be possible, provided you can tune the X channel to the het nucleus you are after. They are the basic standards.

If you want to know a lot more, 200 and More NMR Experiments by Berger and Braun is probably your book.


What is the bond energy of H-O? Wikipedia disagrees with itself

According to the 97th edition of the CRC Handbook of Chemistry and Physics, table 1 Bond Dissociation Energies in Diatomic Molecules [1, p. 9-76]:
$$D^\circ_{298}(\ce{H-O}) = \pu{(429.74\pm 0.03) kJ mol-1}.$$
The value has been taken from the article by Ruscic et al. [2]. Note that this value is strictly for the diatomic molecule $\ce{OH}$ in gas phase and is determined as the standard enthalpy change for the fission process
$$\ce{O-H → O + H}.$$
References

Haynes, W. M.; Lide, D. R.; Bruno, T. J. CRC Handbook of Chemistry and Physics: A Ready-Reference Book of Chemical and Physical Data, 97th ed.; Taylor & Francis Group (CRC Press): Boca Raton, FL, 2016. ISBN 978-1-4987-5429-3.
Ruscic, B.; Feller, D.; Peterson, K. A. Active Thermochemical Tables: Dissociation Energies of Several Homonuclear First-Row Diatomics and Related Thermochemical Values. Theor Chem Acc 2014, 133 (1), 1415. DOI: 10.1007/s00214-013-1415-z.


Shift in the position of equilibrium by a catalyst to favor the forward reaction

Wikipedia says

  Catalyzed reactions have a lower activation energy (rate-limiting free energy of activation) than the corresponding uncatalyzed reaction, resulting in a higher reaction rate at the same temperature and for the same reactant concentrations.

This means that a catalyst will reduce activation energy as seen here

We clearly see that the activation energy of both forward and backward reaction equally. Hence by Arrhenius equation both the rate constants change by same factor thereby there is no change in equilibrium constant.


Calculating the pH given number of moles of HI and 500. ml of water

Simple questions are often interesting.

(1) Most importantly, volumes are rarely additive. So the notion of adding the volume of the HI and the volume of water just doesn't work.

(2) In an answer to a question about hydroiodic acid,  Curt F. showed a chart of concentration and density. For the $\pu{20 ^\circ C}$ values the liters of water in a liter of the acid solution can easily be calculated. So for the OP to get an error of 1% or less, the molarity would have to be about 0.2 molar or less.

\begin{array}{|c|c|c|c|}
\hline
\text{Concentration, %}(w/w) & \text{Density, }\pu{kg/L}(\text{@ }\pu{20 ^\circ C}) &  \text{Molarity} & \pu{L(water)/L(acid)}  \\ \hline
5.2 & 1.0342 & 0.420 & 0.980 \\ \hline
10.8 & 1.0812 & 0.922 & 0.964\\ \hline
16.4 & 1.1226 & 1.44 & 0.938 \\ \hline
22.4 & 1.1765 & 2.06 & 0.913 \\ \hline
27.2 & 1.2333 & 2.62 & 0.898 \\ \hline
33.1 & 1.2918 & 3.34 & 0.864 \\ \hline
38.7 & 1.3605 & 4.12  & 0.834\\ \hline
42.9 & 1.4208 & 4.77 & 0.811 \\ \hline
48.7 & 1.5072 & 5.74 & 0.773 \\ \hline
53.0 & 1.5913 & 6.59 & 0.748 \\ \hline
57.0 & 1.6933 & 7.55 & 0.728 \\ \hline
\hline
\end{array}

You can see from the red line that the data is not linear.


What is a kg.cal per gram, as a heat of combustion?

$\mathrm{kg~cal}$ probably represents the kilogram calorie (especially given that it is an old book). Nowadays it is usually just known as the Calorie, usually abbreviated $\mathrm{kcal}$. It is roughly the amount of energy required to raise the temperature of $1~\mathrm{kg}$ of water by $1~\mathrm{^\circ C}$. Since this is temperature and pressure dependent, there are slight variations in the actual definition depending on the context in which it is used.


I agree.  This is a dangerous reaction and should not be tried without proper equipment, including a lab hood and fire extinguishing equipment.

The fumes could be "just" droplets of sodium hydroxide solution.  I've seen similar fumes involving hydrochloric acid, just by adding water to rinse out traces of concentrated acid (in a hood, of course).  They may be carried by increased evaporation even without boiling and without need for a further "chemical reaction".


OK, I was too lazy to look up LD50 and LC50 of $\ce{HCN}$ for you ;)
But if you have a chance, have a look at

G. Padmaja & Keith H. Steinkraus,
Cyanide detoxification in cassava for food and feed uses,
Critical Reviews in Food Science and Nutrition, 1995, 35(4), 299-339 [DOI]

Seemingly, the processing of cassava (manioc) involves several steps, such as drying in the sun, grating, watering and cooking, each of which either facilitates the enzymatic or thermal cleavage of the linamarin (i.e. the toxic compound) to glucose or supports the release of $\ce{HCN}$.
In total, the processing minimizes the risk of exposition through inhalation. None of the steps is perfomed in confined space, watering helps to wash out the $\ce{HCN}$ which in water dissociates a bit ($\ce{HCN}$ is a weak acid).
Thus, digestion of the untreated roots is likely to be fatal, while the slow processing is apparently not.

Ok, I digged a bit and maybe you have you have access to this one:

Bala Nambisan
Strategies for elimination of cyanogens from cassava for reducing toxicity and improving food safety Food and Chemical Toxicology, 2011, 49, 690–693 [DOI]


Is there a difference between negative/positive controls and control/experimental groups?

Consider a chemical test, such as a test for the presence of lead in drinking water. You would use both experimental and control samples. Whenever you run the test on experimental samples, you also need to run it on both positive and negative control samples--positive samples are prepared deliberately with a certain known concentration of lead, and negative samples are known not to contain any lead (you have to be careful here!). If you get the incorrect results for the control samples, then you know that the results for the experimental samples may be off, so you have to repeat the test. You also might want to periodically have the analyst run a proficiency test with the standard positive and negative controls as well as samples containing various known amounts of lead.

Experimental and control groups might more typically refer to situations such as drug trials, where you are comparing a new drug to another drug that is already approved and used to treat an illness. The experimental group would be the group of subjects having the illness that are given the new drug and the control group would also have the illness but would be given the approved drug (preferably in a double-blind study design--where neither the subjects nor the people administering the drug know which treatment is which). Note that both the experimental group and control group are "experimented on" - otherwise they could tell whether they received the treatment or not, and the results could be erroneous due to the placebo effect.


How many peaks of HCl are found in a mass spectrum?

Collision between an electron and the molecule during a MS-experiment usually results in two phenomena:

\begin{align}
\begin{cases}
\ce{AB + e &-> [AB]+ + 2e} &\text{electron ionization}\\
\ce{AB + e &-> [AB]-} &\text{resonance capture}
\end{cases}
\end{align}

Resonance capture is less frequent and usually doesn't add much to the spectra. For $\ce{HCl}$ considering electron ionization as the primary process the following process takes place:

$$\ce{HCl + e -> [HCl]+ + 2e}$$

Based on ionization method (e.g. applied energy) several fragmentation processes may accompany the formation of cation/cation-radical ($\ce{[HCl]+}$/$\ce{[HCl]^{+.}}$):

\begin{align}
\ce{[HCl]+ &-> H+ + Cl^. }\\
\ce{[HCl]+ &-> H^. + Cl+ }\\
\ce{[HCl]+ + e &-> H+ + Cl-}
\end{align}

Fragmentation is often followed by recombination acts of different degrees of probability, e.g.:
\begin{align}
\ce{H+ + Cl- &-> HCl}\\
\ce{H+ + HCl &-> [H2Cl]+}\\
\ce{H^. + [H2Cl]+ &-> [H3Cl]^.+}
\end{align}

One can possibly observe peaks of $\ce{[H]+}$ and $\ce{[H2]+}$, but those are usually have negligible intensities due to recombination processes.

Regarding the spectra OP provided, it looks somewhat strange as the base peak @ $m/z = 36$ is not $100\%$ (approx. $90\%$ instead). I would recommend to check for the experimental data in more reliable sources, e.g. in NIST database:

Considering the fragmentation processes and the fact that $\ce{^{35}Cl} : \ce{^{37}Cl} = 3 : 1$, one can perform the following peak assignment:

\begin{array}{rrr}
m/z & \text{r.i.}, \% & \text{fragment}\\
\hline
35 & 18  & \ce{[^{35}Cl]+}\\
36 & 100 & \ce{[^1H^{35}Cl]+}\\
37 & 6   & \ce{[^{37}Cl]+}\\
38 & 33  & \ce{[^1H^{37}Cl]+}\\
\hline
\end{array}

Alternatively, for $m/z = 37$ and $m/z = 38$ fragments $\ce{[^1H2^{35}Cl]+}$ and $\ce{[^1H3^{35}Cl]+}$, respectively, could be proposed, but it wouldn't be correct due to the following:

recombination acts leading to the formation of large molecules as detected end-products are less possible and usually result in much smaller peaks;
observed relative intensities are not in agreement with isotope distribution for chlorine ($\ce{^{35}Cl} : \ce{^{37}Cl} = 3 : 1$, which is in agreement with r.i. ratios for $\ce{[^{35}Cl]+} : \ce{[^{37}Cl]+} = 18 : 6$ and $\ce{[^1H^{35}Cl]+} : \ce{[^1H^{37}Cl]+} = 100 : 33$, but not with alternatively proposed fragments);
the problem itself states that

  hydrogen chloride molecules will split into atoms.


A minimal basis set is a set that employs just one function to describe each orbital. For $\ce{H}$ you take into account one orbital ($\mathrm{1s}$), so you use one function, and for $\ce{Li}$ you consider five orbitals ($\mathrm{1s}$, $\mathrm{2s}$, $\mathrm{2p_x}$, $\mathrm{2p_y}$, $\mathrm{2p_z}$) and you use five functions, one for each.

What you described ("the minimal number of orbital/spatial wavefunctions needed for a specific number of electrons") would use a function for each occupied orbital - leaving unoccupied orbitals, even in a partially occupied subshell, out of the picture. This is a very bad idea - shells are a set of degenerate solutions for the hydrogen atom, so reducing the number of functions to describe them will seriously distort the depiction of the atom.

The usual approach is, instead, to always provide functions at least for the entire valence shell of the atom (and its inner shells). So for all 2nd period elements, you would use all 2nd shell orbitals ($\mathrm{2s}$, $\mathrm{2p_x}$, $\mathrm{2p_y}$ and $\mathrm{2p_z}$), even in a minimal basis set, no matter how many electrons actually occupy that shell. What still makes that basis set a minimal one is that you are only using one function for each orbital.

Similarly, a minimal basis set calculation for $\ce{Na}$ would include one function for each orbital in the 1st, 2nd and 3rd shells, and so on.


How do I extract cyanide from apple seeds?

Forget about the apple seeds, they contain about 1 to 4 mg amygdalin per gramm seeds (DOI).
Instead, collect apricot seeds during the right season, the amygdalin content varies though the year and can be as high as 5% of the dry weight of the seed (DOI).
It is probably advantagenous to break the husk with a nut cracker, a plier, etc. and cut the softer inner material to smaller pieces.
Extraction of amygdalin can be performed by

immersing the material in methanol and subsequent ultrasonification
Soxhlet extraction with methanol
reflux extraction in water in the presence of citric acid

A comparison of the extraction methods is given here.

Removal of the solvent in vacuum will yield a crude material with significant amounts of amygdalin. You might want to have a look at this article from the Western Journal of Medicine on the toxicity. Here, an $\mathrm{LD_{50}}$ of 522 mg amygdalin per kg body weight was estimated for the oral application to rats. The online resource of the U.S. National Library of Medicine gives a value of 405 mg/kg.

Further information on the health risk of apricot kernels are provided by of the German Bundesinstitut für Risikobewertung (Federal Institute for Risk Assessment) and the British Committee on Toxicity.

A note in the a German medical journal, Deutsche Ärzteblatt, (PDF) describes a case where boy of four years (110 cm, 18 kg body weight) was given apricot kernels during an alternative cancer treatment. Upon additional treatment with a single dose of 500 mg amygdalin, the kid showed agitation, spasms and the eyes started to roll.

I'll leave it up to your fantasy as a writer on how to apply the poison, but spicing some marzipan with it might help ;)


I have discussed this problem once in the past with an Italian colleague, teaching in Palermo University, and who was in charge of training the Italian team at the International Chemistry Olympiad. I remember he told me that this order can be established by plotting the square root of the successive ionization potentials of any atom versus its atomic number. Of course there is one point for H, two for He, three for Li, etc. If you report these √I values versus the atomic number Z, you will find that all highest values (relative to last $1$s electron) are steadily increasing, of course. But these points remain alined across the page. Likewise, all next to highest values (also relative to $1$s) are alined, and surprisingly the slopes of both lines are equal. The first line starts at H, and the second at He. But they are parallel, after my Italian colleague.
Now if you consider the next but two highest √I values, the curve starts at Li, and the corresponding points ($2$s) are also alined through the diagram, with a slope which is nearly the same as the first two curves. Likewise the points corresponding to the next but three values (starting at Be) are alined and the line is parallel to the preceding line.
Something new happens with the next points. If you consider the fifth √I values by decreasing numerical order, these points start with Boron. And they are alined across the diagram. They correspond to $2$p electrons. New surprise : the points are still alined. But the slope of this line is much smaller than before. The slope seems to be typical of p electrons.
If you join the sixth √I values, stqrting with Carbon, the curve is still a line, with the same slope typical for p electrons. Same thing for the $7$th, $8$th, $9$th and $10$th electron. All lines are parallel.
First difficulty occurs with the $11$th √I values, starting with Na. The curve joining these points is much steeper than for the preceding electrons. It has the same slope, typical of the $1$s and $2$s electrons. So, this line cuts preceeding lines. As a consequence, when studying the points through Mg, Al, Si, etc., it is not easy to decide which √I value belongs to what.
I have not done it myself, but my colleague told me that this job may be carried out through all the periodic table, and that it is possible to recognize a bundle of steep lines for all s electrons, another bundle of not so steep lines for p electrons, another bundle of still less inclined lines for d electrons, etc. and the same with f electrons.
Apparently, this job explains why there are irregularities in the electronic configurations through the periodic table.
If some reader will try to check this statement, I would be pleased to hear of his job.


Mechanism of formation of peak of mass spectrum of isobutane at m/z = 27

Some of the first mass spectrometrists to see this ion were Stevenson and Hipple in 1942.1  They were puzzled by it.
Fifteen years later, Rosenstock and Melton explained the source of this ion in an exhaustive study.2  First, the molecular ion $\ce{C4H10+}$ dissociates into fragments via five possible pathways, each with its own activation energy.  See Table VIII in their paper, which I reproduce below.

None of these primary dissociation pathways gives rise to $\ce{C2H3+}$.  Instead, it arises from a variety of processes from a subset (the metastable subset) of these five primary transitions.  These processes are shown in Tables X and XI of their paper.

All of the reactions that give rise to $\ce{C2H3+}$ are probably happening to some extent, but the lowest-energy route there for isobutane (right column) appears to be via $\ce{C2H5+ -> C2H3+ + H2}$ (only 1.9 eV).  However, $\ce{C2H5+}$ itself requires more energy to form than $\ce{C3H7+}$ ($m/z=43$), so the route $\ce{C3H7+ -> C2H3+ + CH4}$ may also occur as well.
The general conclusions of their paper are worth quoting:

The basic concept of the quasi-equilibrium theory is
that mass spectra are formed by a series of competing
consecutive unimolecular decomposition of excited ions.
The results discussed above lend strong support to this
concept. It was found that of several possible reactions
of a given ion, the metastable transition was associated
with the reaction having the least activation energy.
The general rate equation appropriate for these systems
is the product of a frequency factor and a function of
the excitation energy $E$ and activation energy $\epsilon$. When
the value of $E$ is such that the rate constant is of the
order of $10^6~\mathrm{sec^{-1}}$, the rate constant is extremely
sensitive to the value of $\epsilon$. The result is that a second
reaction having an $\epsilon$ higher by more than a few tenths of
an eV will have a considerably smaller rate constant,
even if the frequency factor is larger by a factor of a
thousand. Thus, in general, the correlation can be
carried out in terms of activation energies alone.

References

D. P. Stevenson, J. A. Hipple, Jr., J. Am. Chem. Soc. 1942, 64 (7),  1588–1594.
H. M. Rosenstock, C. E. Melton, J. Chem. Phys. 1957, 26 (2), 314-322.


Hill equation with inhibition and activation

Mechanism of cooperative enzyme activity

The easiest way to explain cooperativity is the example of hemoglobin, where four nearly identical subunits switch cooperatively between two states T and R of different binding affinity. When no ligand (L) is bound, hemoglobin is in the T state; once at least one ligand is bound, it is predominately in the R state.

$$\ce{T4 + 4 L <=> R_4L + 3 L <=> R4L + 3 L <=> R4L2 + 2 L <=> R_4L3 + L <=> R4L4}$$

The states with 1-3 ligands are less populated (because the R state has higher affinity for the ligand L thant the T state), and we will ignore them:

$$\ce{T4 + 4 L <=> R4L4}$$

From this, you can write the law of mass action and the binding isotherms.

If we apply this to an enzyme with a Hill coefficient of $n$, and call the active state of the enzyme E and the inactive state F, we would have the equilibrium:

$$\ce{F_n + nS<=> E_nS_n}$$

For sufficiently high concentration of substrate S, all enzyme would be in complex with substrate, leading to a rate of $v_\text{max}$.

Mechanism of cooperative inhibition

In the OP's question, the Hill coefficient for inhibition $m$ is different from that for enzyme activity ($n$). So we could have a regulatory protein with $m$ subunits attached to the enzyme. It binds cooperatively to inhibitor, which allosterically affects enzyme activity. The regulatory protein would also have two states, T and R, either free or bound to $m$ inhibitors:

$$\ce{ T_m + m I <=> R_mI_m}$$

When the allosteric inhibitor is bound to ligand (i.e. in the R state), the enzyme lacks activity.

Combination of the two

Only a certain fraction of enzyme is bound to substrate, and only a certain fraction of enzyme-bound substrate is not inhibited by the regulatory protein. For this scenario, the enzyme activity is $v_\text{max}$ times those two fractions.

$$\frac{\mathrm d[\ce{P}]}{\mathrm dt} = v_\mathrm{max}\frac{[\ce{S}]^n}{K_S + [\ce{S}]^n} \cdot \frac{K_I}{K_I + [\ce{I}]^m}\tag{2}$$

What happens when there is competitive inhibition?

It gets more complicated. Will the inhibitor also cause a switch in the enzymes conformation? In which direction? For the simpler (non-cooperative) Michaelis Menten model, a competitive inhibitor changes the apparent affinity of enzyme to ligand in a concentration-dependent manner (the enzyme partitions between free, ligand-bound and inhibitor-bound state). For the cooperative case, you would have to decide whether mixed states (such as $\ce{ES2I2}$) are populated; if they are, they would contribute to product formation, but less than $\ce{ES4}$ because not all active sites are turning over substrate.


Resonance stabilization and size of ligand atoms

Yes, this is true. The $\pi$ overlap between a valence $\ce{p}$ orbital of a second row element and one of a higher row element is much smaller than the $\pi$ overlap between two second row elements. One reason is that the bond length is generally higher when higher row elements are involved (due to their size). Another reason is, that the $3 \ce{p}$ or $4 \ce{p}$ orbitals are much more diffuse than the $2 \ce{p}$ orbitals. If you want to read deeper into the topic have a look at:

[1] W. Kutzelnigg, Angew. Chem. Int. Ed. Engl., 23, 272 (1984).

[2] N. C. Norman, Polyhedron, 12, 2431 (1993).


What is the the full name of TAPD? Structure provided

TAPD = Tetraalkyl-p-phenylenediamine

This is because you have a phenyl (benzenoid) ring, with two amino groups (diamine) opposite (para) to each other, and each of those amino groups has two alkyl substituents, giving four (tetra) in total.


Acetaldoxime Synthesis: Hydroxylamine vs Hydroxylamine HCl

Hydroxylamine is a low-boiling liquid readily oxidized to explosive nitrogen-oxygen compounds.  It is stored as the hydroxylammonium hydrochloride salt for convenience and safety, since the salts do not oxidize as easily.  Additionally, hydroxylamine itself may explode when heated.

The sodium hydroxide is to render the freebase which then may participate in the reaction.  In principle, the hydroxylammonium bisulphate would work as well.

It is somewhat important that an exact stoichiometry be used for base to hydroxylamine to prevent aldol side-reactions if relevant (which for acetaldehyde, they are).  Additionally, this reaction is reversible in the presence of acid and water, so all the acid must be consumed, since water is a byproduct of the condensation.  Given the explosive nature of hydroxylamine, azeotropic distillation via Dean-Stark trap is not viable for dehydration.  Chemical dehydration may be effective.


Software for chemical reaction simulation in ICE

I am not sure about the phenomena you are describing but simulators are just tools with built-in models and thermodynamic data together.

What you need is, for example, to find some kinetic (combustion kinetic) that it has in its equation a parameter regarding the acetone concentration.

Try to find at the literature some of these kinetic equations, if you are student you can access with your university credentials into sciencedirect.com papers.

If not, check arxiv.org or google ;)

Once you have these, then you can use simulators to reproduce your "special" kinetic.

Another possibility is to describe better your system behaviour (including graphs) and try to fit them suing a correlation or a polynomial fit. Or ask the mathematicians, they could recognise models just seeing the patterns and give you an equation type that you can use.

Check this paper as an example;

http://www.combustion.org.uk/ECM_2007/ecm2007_papers/1-20.pdf


Why don't we use an alkyl halide directly in Gabriel Phthalimide Synthesis?

It is used to make the conjugate base of the imide which is far more nucleophilic than the imide itself, but not too much to attack another molecule of the alkyl halide.


Link between centrosymmetry, dipole moment and chirality

A molecule can't be chiral if it has an $S_n$ improper rotation (note that $S_1$ is equivalent to a mirror plane and $S_2$ is equivalent to inversion). Chirality is only possible in the $C_n$ and $D_n$, as well the less common $T$, $O$, and $I$, point groups, which only have rotation axes as symmetry elements.

A dipole moment can only occur along the primary $C_n$ rotation axis of a molecule (if it has one) and if there are any perpendicular mirror planes or rotation axes, it will have to be zero by symmetry. A $S_n$ improper rotation with $n>1$ also precludes a dipole. This limits the point groups that can have dipoles to $C_n$, $C_{nv}$, and $C_s$.

Centrosymmetry is equivalent to having the $S_2$ symmetry operation. As mentioned above, this eliminates the possibility of chirality or a dipole. However, centrosymmetry is just a special case and  molecule won't be chiral and/or won't have a dipole under other conditions as well. That is to say, centrosymmetry is sufficient, but not necessary for eliminating chirality and/or a dipole.

To summarize:
$$\begin{array}{|c|c|c|} \hline
\text{Condition} & \text{Exact Point Group} \\ \hline
\text{Chiral/Nonpolar} & I,O,T,D_{n} \\
\text{Achiral/Polar} & C_{nv}, C_s  \\
\text{Chiral/Polar} & C_n \\
\hline
\end{array}$$

Where I'm taking "exact point group" to mean the highest symmetry group the molecule is contained in. This delineation has some minor exceptions. Technically, molecules of the appropriate symmetry still might be nonpolar (or at least too small to measure) by an accidental cancellation. For chirality, while all molecules of these symmetries are formally chiral, they can appear achiral in practice if the enantiomers can readily interconvert, as occurs with nitrogen inversion.


If the same atoms (cells) that now compose my body get arranged in N years away, am I that person?

Let me start from the bottom up: If we assume the earth as a closed system, i.e. exchange of energy with the surrounding space is possible, but exchange of matter impossible, than naturally our body contains atoms previously present in other form of matter.  Already our breakfast is matter we incorporate, subject to chemical reaction (digestion), and eventually excrete in one form, or an other.  Hence it is likely that we contain some atoms that once constituted a mammoth, or a dinosaur, or by your tomato in the back yard.

Digestion often splits the larger molecules present in our food into smaller ones.  This eases transport of matter across membranes, for example.  Think of proteins de-constructed into amino acids; our organism uses the smaller to build new proteins in need.

Beside chemical degradation as in aforementioned example of the proteins, digestion includes physical degradation, too -- chewing food, for example increases the surface area the chemicals responsible for digestion may access.  By  both means of action, physical and chemical degradation, entire cells face only a tiny chance to remain intact "as such".  It is more likely that they are dismantled, are split into their constituents, eventually -- as molecules -- harvested in the ways briefly mentioned earlier.

To sum up: so far there is no evidence a synapse of yours will be recycled, maintaining the original form and function, to find its way back as synapse in an other brain, similar to a spare tire.  Atoms of yours however, as easily they entered your body, eventually will leave you; at their pace.


In a battery what would happen if you had a perfectly uniform and pure cathode?

I don't think that's accurate. Even the universe doesn't have an "unlimited supply of electrons".
Impurities in the cathode likely give rise to internal resistance inside the battery (among other things), leading to deviation from an ideal cell. I would guess that as your cathode approached purity, your battery would approach the behavior of 'ideal cells', and that discharge rates would approach those of capacitors, and efficiency would approach 100%. To get an ideal cell though, the anode and electrolyte would also have to be 'pure' and some real-world effects would have to be ignored.

Because of internal resistance, the terminal voltage of a cell that is discharging is smaller in magnitude than the open-circuit voltage and the terminal voltage of a cell that is charging exceeds the open-circuit voltage. An ideal cell has negligible internal resistance, so it would maintain a constant terminal voltage of $ \epsilon $ until exhausted, then dropping to zero. If such a cell maintained 1.5 volts and stored a charge of one coulomb then on complete discharge it would perform 1.5 joule of work. In actual cells, the internal resistance increases under discharge
Wikipedia

What do you mean by pure, exactly? Complete homogeneity and lack of any other elements? That sounds like a physics problem where you "ignore air resistence" and other real world effects for simplicity. If this is what you're after, search for things like 'Ideal Battery' and 'Ideal Electrochemical Cell'.


During the decomposition of sodium bicarbonate lab, the mass of the final solid I received was less than expected. Errors?

I understand that the correct equation is supposed to be the first one where sodium carbonate was produced. However, through calculations using the ratio between reactant and solid product, my experiment showed that equation 2 where sodium oxide was produced is supposed to be correct.

So now you're faced with a decision. You can either:

Massage the result of your experiment until it comes up with the answer you think is "supposed" to be correct.
Report the result of your experiment as it actually happened.

If you take the first route, I'm sure we can come up with some convincing explanations. Maybe your starting sodium bicarbonate contained water (so it wasn't 100% pure) and that water was driven off in the heating. Maybe you spilled some as you put it in the crucible. Maybe some of the mixture in the crucible splashed out during the reaction. Maybe you read the scale wrong. Maybe there is another reaction that also happens, perhaps involving the air, or the crucible, or the gas from the bunsen burner.

But I think it would be much more honest to report your result as it actually happened. Your data supports the hypothesis that the sodium bicarbonate decomposed to sodium oxide. In fact, the agreement with that hypothesis is quite strong, and the agreement with the supposedly correct hypothesis is terrible. It could be that this is due to an error you made... or maybe the stupid sodium bicarbonate didn't get the memo that, according to the teacher (or the text or Wikipedia or whatever), it was supposed to decompose to sodium carbonate, so it just went ahead and decomposed all the way to sodium oxide.

I realize you're probably worried about being marked down on your lab report because you came to the "wrong" conclusion. Any decent teacher, though, would give you much more credit for correctly interpreting your data and coming to a conclusion based on the data, even if it's wrong, than for massaging your data to support a preconceived hypothesis.


What is the momentum of a particle described by the following equation?

The idea of superposition of eigenstates is quantum 101, so you should probably go and read a proper book on QM. I personally recommend Griffiths.

In quantum mechanics, observables such as $p$ are represented by Hermitian operators such as $\hat{p} = -\mathrm{i}\hbar(\mathrm d/\mathrm dx)$.

The momentum of a particle is not well-defined unless it is in an eigenstate of the momentum operator $\hat{p}$. If it is, then a measurement of the momentum will always return the corresponding eigenvalue. For example, if the wavefunction of the particle is simply $\exp(ikx)$ , then a measurement of the momentum will always return the value $\hbar k$, and the particle is said to exist in a state of well-defined momentum.

Now, if the wavefunction of the particle is $A\exp(ikx) + B\exp(-ikx)$, then a measurement of the momentum will sometimes yield $\hbar k$ and sometimes yield $-\hbar k$. So, the actual momentum of the particle is not well-defined, as it could be either of the two values (before you observe it).

The expectation value of the momentum, though, is well-defined. It is simply

$$\langle p \rangle = P_+(\hbar k) + P_-(-\hbar k)$$

where $P_+$ and $P_-$ are the probabilities of measuring the momentum to be $\hbar k$ and $-\hbar k$ respectively. Assuming that your wavefunction is normalised, i.e. $|A|^2 + |B|^2 = 1$, basic quantum mechanics tells us

$$P_+ = |A|^2;\quad  P_- = |B|^2$$

and so the expectation value in this case is found to be

$$\langle p \rangle = \hbar k (|A|^2 - |B|^2)$$

However, that simply means that if you prepare many particles in the state above, measure their momenta, and average their momenta, the average will tend toward this value. Each individual measurement is guaranteed to give either the value $\hbar k$ or $-\hbar k$.

And yes, each individual particle does not have one single well-defined momentum. That's quantum mechanics for you.


Nucleophilic aromatic substitution on pyrimidines: C2 vs C4 selectivity

An explanation can be provided by frontier molecular orbital theory.  The LUMO coefficient is higher on the C4 vs C2.  This is a general effect meaning that an EWG para to the leaving group is more reactive than ortho.  See 4-halopyridines vs 2-halo and 4-nitrobenznes vs 2-nitro.  There could also be an effect from the repulsion between the nitrogen lone pairs and the approaching electron rich nucleophile, which would be greater in proximity to 2 nitrogens rather than 1.


Can it be said that on our planet we have roughly the same number of electrons as we have protons?

Yes

We can see that the charges in the observable world are close to balance by considering what we would observe if they were not.

Big imbalances of charges cause strong electrical fields. Those are measurable and their effects are observable. We can create such imbalances locally (that's what a Van de Graaff generator does) and the effects are spectacular: really big sparks as the strong fields ionise air and cause big electrical discharges. These separations also occur in nature (that's what thunderstorms are). In both cases, the consequence of significant temporary separations of charge are spectacular natural phenomena that act to restore the balance.

It could be true, however, that the earth as a whole had a net charge. We wouldn't notice this while on the surface as the net charge would be relative to interplanetary space. On the other hand any such charge would probably be neutralised by selective interaction with the charged particles in the solar wind which we could observe. Also, if the net charge were substantial, we would observe significant effects on planetary orbits (the electrical field is a lot stronger than the gravitational field.

These observations together suggest that both locally and globally we have approximately the same number of protons as electrons on the earth.


Octahedral Crystal Field Splitting Orbital Degeneracy

It is because perfect octahedral symmetry is normally assumed; the two $\mathrm{e_g}$ levels and three $\mathrm{t_{2g}}$  are degenerate.  If there was a distortion, say by lengthening both $z$-axis ligand positions then the $\mathrm{e_g}$ degeneracy would be removed as the $\mathrm d_{z^2}$ becomes more stable than $\mathrm d_{x^2-y^2}$. This happens because the $z$-axis ligand has more effect on $\mathrm d_{z^2}$ than on $\mathrm d_{x^2-y^2}$ orbitals simply due to its position. The $\mathrm d_{xy}$ orbital also increases in energy removing the degeneracy of the $\mathrm{t_{2g}}$ and the  $\mathrm d_{yz},$ $\mathrm d_{zx}$ lowered.


The term "normality" stems from the older German term for equivalent concentration (die Äquivalentkonzentration), namely die Normalität, which was coined by Karl Friedrich Mohr.
According to German Wikipedia which also refers to Straube's Der historische Weg der Chemie [1, p. 220], Mohr introduced the concept of normal solution and normality (as it's known in the modern world) in the first edition of his Lehrbuch der chemisch-analytischen Titrirmethode [2], e.g. the preface contained the following definitions among others:

Die Normaltemperatur der Flüssigkeiten ist 14° R. = 17,5° Cent.
[…]
Normalflüssigkeit bedeutet eine Lösung, welche 1 Atom Substanz, in Grammen ausgedrückt, in 1 Litre enthält.
Zehend-Normalflüssigkeit bedeutet eine Lösung, welche $1/10$ Atom Substanz, in Grammen ausgedrückt, auf 1 Litre enthält oder mit $1/10$ Atom eines darin enthaltenen Körpers (z. B. Sauerstoff) wirkt.

The book also lists numerous "normal" definitions not really in a sense of "adequacy", but rather in a a sense of "standard": Mohr basically invented standardized quantitative analysis and proposed numerous analytical tools and techniques.
Among these definitions there are:

Normalwägung — standard weighing [procedure];
Normalbürette — standard burette (a glass tube $\pu{500 mm}$ long, $\pu{13.5 mm}$ inner diameter, holds $\pu{60 cm3}$ of liquid divided into five parts, each $\pu{cm^3}$ is $\pu{7 mm}$ long);
Normale Stärke — strength of a normal solution of acid/base;
Normalsäure and Normalnatron — standard/normal solutions of acid and soda, which, interestingly enough, Mohr suggested to label with the pieces of red and blue paper, respectively;
Normalkleesäure, Normalsalpetersäure, Normalchlorbaryum, etc. — normal solutions of oxalic, nitric acids and barium(II) chloride and so on.

At the same time, normality was used to describe the normal ratio between stoichiometry of the elements (esp. C, O, N) and the number of found equivalents in complex compounds such as proteins or drugs [3].
References

Wilhelm Strube: Der historische Weg der Chemie. Aulis Verlag Deubner & Co KG, Köln 1989, ISBN 3-7614-1180-4. (In German)
Friedrich Mohr. Lehrbuch der chemisch-analytischen Titrirmethode; Vieweg und Sohn: Braunschweig, 1855. (In German. Available for free at Google Play (accessed 2019-05-25))
Zeising, A. Das Normalverhältnis der chemischen und morphologischen Proportionen; Weigel, 1856. (In German. Available for free at Archive.org (accessed 2019-05-25))


Order of Reactivity of Halogens in Electrophilic Addition

The rate-determining step of halogenation of alkenes is the formation of cyclic intermediate.
The cyclohalonium intermediate of bromine will be more stable than that of chlorine owing to its lower electronegativity.
The following paragraph is taken from Peter Sykes [$1$, p.$181$-$182$]:

It is not normally possible to add fluorine directly to alkenes as the reaction is so exothermic that bond fission occurs. Many alkenes will not add iodine directly either and when the reaction does occur it is usually readily reversible.

As is clear $\ce{I2}$ will be more reactive than $\ce{F2}$.
Hence, the correct answer is D.
References:

Peter Sykes; A Guidebook to Mechanism in Organic Chemistry 6th ed.; 2003 ISBN-10:8177584332


cross exchanging the coefficients and valences of cations and anions with each other in chemical formulas

The formulas for ionic compounds can be predicted from the charges of their ions. Cations and anions combine to form ionic compounds with a ratio of cations to anions that will produce a neutral (uncharged) unit. Generally, compounds are uncharged species.

Most binary ionic compounds are comprised of a metal and a nonmetal. In the most generic case, a metal cation with a charge of $+m$ and a nonmetal anion with a charge of $-n$ will combine in a $n:m$ ratio to balance charge. The example below, I am using $\ce{M}$ to represent a generic metal and $X$ to represent a generic nonmetal.

$$\ce{nM^{m+} + mX^{n-} -> M_n N_m}$$

This ratio guarantees charge balance. In the generic case the total charge of all cations in the formula unit is $+m\cdot n$, while the total charge of all the anions in the unit is $-n\cdot m$. These two values are equal and we get a compound. It is possible that $n:m$ is not the simplest whole number ratio and we can simplify. See some examples:

Calcium $\ce{Ca^2+}$ chloride $\ce{Cl-}$ gives $\ce{CaCl2}$.
Iron(III) $\ce{Fe^2+}$ sulfide $\ce{S^2-}$ gives $\ce{Fe2S3}$.
Magnesium $\ce{Mg^2+}$ oxide $\ce{O^2-}$ gives $\ce{Mg2O2}$, which can be simplified to $\ce{MgO}$.

We need to be careful to only think this way for ionic compounds. Covalent compounds can be more complex, and their are often multiple binary compounds based on varying oxidation state. Additionally, for covalent compounds, we do not simiplify the coefficients to the lowest whole number ratio, as the formula unit describes discrete molecules with fixed composition.

For example, $\ce{C4H8}$ (butene) is a different compound from $\ce{C2H4}$ (ethylene).


The anode is where oxidations happen (the cathode is where reduction takes place).

In the vicinity of your anode, you have the following compounds: $\ce{H2O}$, $\ce{Cu}$, $\ce{Cu^2+}$, $\ce{SO4^2-}$. If you check oxidation states, you will realise that oxygen is always $\mathrm{-II}$, hydrogen is $\mathrm{+I}$, sulphur is $\mathrm{+VI}$ and copper is present in both $\mathrm{\pm 0}$ and $\mathrm{+II}$. Thus, the only species available for oxidation are copper($0$) and oxygen($\mathrm{-II}$).

Checking standard potentials:
\begin{align}
\ce{O2 + 4 H+ + 4 e- &<=> 2 H2O} \qquad &E^\circ = \pu{+1.23 V}\tag{1}\\
\ce{Cu^2+ + 2 e- &<=> Cu} \qquad &E^\circ = \pu{+0.35 V}\tag{2}
\end{align}

It should be clear why copper is oxidised.

On the reduction side, hydrogen, copper and sulphate all could accept electrons. Copper is again the happiest of the three to accept electrons.


Reduction of copper oxide by hydrogen in the lab

Burning the hydrogen gas at the exit stream of the copper oxide reduction tube is simply one way to assure that it does not accumulate to explosive levels in the lab. Although hydrogen gas is not toxic, it is flammable in air at concentrations as low as $\pu{4\%}$. As the combustion product of hydrogen and oxygen is mainly water vapor, this technique makes for a simple, safe means for disposing of the excess hydrogen from the copper oxide reduction process.


Let the anhydrous acid be $\ce{HA}$ and assume its Molar Mass to be $X$. There is no a priori reason to assume $\ce{A}$ is a mere element.

The copper salt in question is $\ce{CuA2.3H2O}$.

Its Molar Mass is:

$$63.55+2\times(X-1)+3\times 18=115.55+2X$$

Assume reasonably and without losses, that:

$$\ce{CuA2.3H2O(s) -> CuO(s)}$$

So that $1\ \mathrm{mol}$ of the hydrated salt decomposes to $1\ \mathrm{mol}$ of cupric oxide.

The Molar Mass of $\ce{CuO}$ is $79.545$, so $0.3306\ \mathrm{g}$ of it is:

$$\frac{0.3306}{79.545}=4.156\times 10^{-3}\ \mathrm {mol}$$

So now we know that $1\ \mathrm{g}$ of the hydrated salt is $4.156\times 10^{-3}\ \mathrm {mol}$, so its Molar Mass is:

$$\frac{1}{4.156\times 10^{-3}}=240.6\ \mathrm{g/mol}$$

And with the above:

$$115.55+2X=240.6$$

Solve for $X$.

As the acid is monobasic, its Molar Mass is also its Equivalent Mass.


Is the electrophile in sulfonation of benzene SO3 or SO3H+?

From Clayden, under the heading "Sulfonation of Benzene" (electrophilic aromatic substitution):

  The cationic intermediate can
  also be formed by the protonation
  of sulfur trioxide, $\ce{SO3}$, and
  another way to do sulfonations is
  to use concentrated sulfuric acid
  with $\ce{SO3}$ added. These solutions
  have the industrial name oleum.
  It is possible that the sulfonating
  agent in all these reactions is not
  protonated $\ce{SO3}$ but $\ce{SO3}$ itself.

From March's $7^{th}$ edition, under Sulfur electrophiles:

  Sulfonation is often accomplished with concentrated sulfuric acid, but it can also be
  done with fuming sulfuric acid, $\ce{SO3}, \ce{ClSO2OH}, \ce{ClSO2NMe2/In(OTf)3}$ or other reagents.

  Sulfur trioxide reacts much more rapidly
  than sulfuric acid with benzene it is nearly instantaneous

  A great deal of work has been done on the mechanism, chiefly by Cerfontain
  and co-workers. Mechanistic study is made difficult by the complicated nature of
  the solutions. Indications are that the electrophile varies with the reagent, though
  $\ce{SO3}$ is involved in all cases, either free or combined with a carrier. In aqueous
  $\ce{H2SO4}$ solutions, the electrophile is thought to be $\ce{H3SO4+}$
  (or a combination of
  $\ce{H2SO4}$ and $\ce{H3O+}$) at concentrations below $\sim 80–85%$ $\ce{H2SO4}$, and $\ce{H2S2O7}$ (or a
  combination of $\ce{H2SO4}$ and $\ce{SO3}$) at concentrations higher than this. (the changeover
  point varies with the substrate)

  In fuming sulfuric acid ($\ce{H2SO4}$ containing
  excess $\ce{SO3}$), the electrophile is thought to be $\ce{H3S2O7+}$
  (protonated $\ce{H2S2O7}$) up to $\sim 104% \ce{H2SO4}$ and $\ce{H2S4O13} (\ce{H2SO4+ 3SO3})$ beyond this concentration.

  Finally, when pure $\ce{SO3}$ is the reagent in aprotic solvents, $\ce{SO3}$ itself is the actual
  electrophile. Free $\ce{SO3}$ is the most reactive of all these species, so that attack
  here is generally fast and a subsequent step is usually rate determining, at least
  in some solvents.


This is a good question: Tin is above copper in the galvanic series, so logically, copper should displace tin, not the other way around.

Electroless (autolytic) tin plating can be accomplished by disproportionation of a tin compound. $\ce{2Sn(OH)3- -> Sn + Sn(OH)2-6}$, where the less stable Sn(II) yields Sn(IV) + Sn(0), or metallic tin. The divalent tin is usually $\ce{SnCl2}$.

Electroless tin can also be deposited by reduction of a tin compound, e.g.  thiourea reducing $\ce{SnCl2}$ to metallic tin. In a similar fashion, silver nitrate can be reduced to metallic silver to make a mirror on a glass surface. Note that the silver is deposited without  any effect on the glass (other than a nice, shiny coating).


What is a material more easily cut than shattered?

The crystalline property that would allow to cut the "diamond creature" (or whatever those things are in your universe) you are looking at is cleavage (a writer might also use a term in a wordplay).
In crystallography, cleavage is the ability of crystals to crack or split in certain crystallographic directions.
This mechanical property of crystalline media is associated with their internal structure (the ratio of adhesion forces in the crystal lattice) and does not depend on the external shape of the crystals.
This property is also often utilized in the beginning of the process of fashioning the gemstones.
There is a fascinating demonstration with explanation: Crystals - Alan Holden 1958 (YouTube).
With diamonds you can do that too, just not always as easily as with mica: Diamond Cleavage (YouTube).
Diamond (as well as ﬂuorite, galena, topaz and some other minerals) possesses a so-called "perfect" cleavage, and the cleavage planes in diamond are parallel to the octahedral crystal faces [1, p. 30]:

Figure 4.1 Two examples of directional-dependant properties in diamond. The cleavage planes are parallel to the octahedral faces, which are also planes of near
maximum hardness. The easiest direction for sawing is at right-angles to the dodecahedral plane, which is also the plane of minimum hardness.

References

Read, P. G. Gemmology, 3rd ed.; Elsevier Butterworth Heinemann: Amsterdam; Boston, 2005. ISBN 978-0-7506-6449-3.


What are the other end products in sulfur oxoacid formation?

I am not one hundred percent certain, but I will try to give you some guidelines. There is the well known reaction of sulfur dioxide with calcium oxide.
$$\ce{SO2 + CaO -> CaSO3}$$

Under aerobic conditions, this would further react with oxygen to form sulfate, since we have excess sulfur here I think it is fair to assume it will react to form thiosulfate.
$$\ce{1/8S8 + CaSO3 -> CaS2O3}$$

Adding sodium sulfate is probably just a way to control/ buffer the pH during the reaction, I fail to see the actual relevance. But judging from the side reaction
$$\ce{SO2 + 1/8S8 (s) + 3H2O <=> S2O3^{2−} + 2 H3+O}$$
it is actually necessary to have a buffer system based on sulfate.
$$\ce{Na2SO4 + H3+O <=> 2Na+ + HSO4- + H2O}$$

I am not aware of any reaction of sulfur with sulfate.

In total I would write your equation like this:
$$\begin{multline}\ce{SO2 (g) + S (s) + CaO  (s) + Na2SO4 (s) ->[\ce{H2O}]\\ 2Na+ (aq) + Ca^{2+} (aq) + S2O3^{2-} (aq) +  SO4^{2-} (aq)}\end{multline}$$