Why do we use different arguments for determining the strength of hydracids and solubility of ionic compounds?

When pondering solubility, you can't just think about what happens to the compound, you have to think about what happens to the solvent, too.  Solubility is a balance between the loss of interactions between the particles of the compound and the gain of interactions between those particles and the solvent particles.

In this case, the small highly charged F- anion will have much stronger ion-dipole interactions with water, as reflected in its much higher heat of hydration (-524 kJ/mol for F- versus -308 kJ/mol for I-).  That suggests to me that even if the lost Ag+/F- interaction (whether defined as "ionic" or "covalent") is greater than the equivalent lost Ag+/I- interaction when AgI dissolves, the much strong interaction of the F- with water more than compensates, giving AgF the greater solubility.


Does Reverse Osmosis remove elemental ions from water?

The short answer is pressure. What you've described above is basically just osmosis, and without applied pressure to overcome the osmotic pressure of the system, reverse osmosis doesn't happen.

This Wikipedia article gives a good description of the concept, and the difference between filtration, osmosis and reverse osmosis:

  Osmosis is a natural process. When two solutions with different
  concentrations of a solute are separated by a semipermeable membrane,
  the solvent has a tendency to move from low to high solute
  concentrations for chemical potential equilibration.

  Formally, reverse osmosis is the process of forcing a solvent from a
  region of high solute concentration through a semipermeable membrane
  to a region of low solute concentration by applying a pressure in
  excess of the osmotic pressure. The largest and most important
  application of reverse osmosis is the separation of pure water from
  seawater and brackish waters; seawater or brackish water is
  pressurized against one surface of the membrane, causing transport of
  salt-depleted water across the membrane and emergence of potable
  drinking water from the low-pressure side.

  The membranes used for reverse osmosis have a dense layer in the
  polymer matrix—either the skin of an asymmetric membrane or an
  interfacially polymerized layer within a thin-film-composite
  membrane—where the separation occurs. In most cases, the membrane is
  designed to allow only water to pass through this dense layer, while
  preventing the passage of solutes (such as salt ions). This process
  requires that a high pressure be exerted on the high concentration
  side of the membrane, usually 2–17 bar (30–250 psi) for fresh and
  brackish water, and 40–82 bar (600–1200 psi) for seawater, which has
  around 27 bar (390 psi)[6] natural osmotic pressure that must be
  overcome.


Why do glucose, galactose and mannose differ in solubility in water?

The solubility of these sugars will depend on the relative energies of the crystalline form versus the dissolved form of the sugar. Without knowing the specifics, one could surmise that the crystal structures of these three sugars may differ substantially from each other - one may be more tightly or efficiently packed into a crystal lattice than another due to the specific shape of the molecule, resulting in a lower energy state than that of the other sugar's crystal form, and hence giving a lower solubility in water.


First of all, you need to understand the basic difference between the two.

Consider the reaction $$\ce{A + B -> C}$$ Law of mass action states that in a rate equation the powers that the active masses of components raised are always stoichiometric factors.

For a reaction they are always fixed. However, the rate of  reaction differs with the physical conditions. This is applicable to both equilibrium reactions and moderate reactions that are unidirectional.

This makes law of mass action kind of theoretical in nature. In the rate law, the rate equation consists of actual factors that changes with change in experimental conditions. These may or may not be equal to stoichiometric factors. These are generally responsible for determining the order of reaction.

Both the laws are correct because both have a wide area of application. Rate law is although quite experimental in nature and is generally considered to be more appropriate.


Pressure Cooker at High Altitudes and Low Altitudes

Edit - As 1e9dB's answer points out I made a huge mistake in this answer. The inside pressure of the pressure cooker is ambient pressure plus the extra pressure of the valve. The pressure cooker does not cook at an absolute pressure independent of altitude.

You have it all backwards. The pressure cooker allows the pressure inside the cooker to be above the sea-level atmospheric pressure which also increases temperature inside the cooker and thus decreases cooking time. So the critical factor is the increase of the boiling temperature of water as a function of the pressure inside the cooker.

The pressure cooker raises the pressure inside the cooker to the same absolute pressure, regardless of the altitude at which the cooking is being done. A typical value for the internal pressure would be about 2 atmospheres absolute which makes the internal cooking temperature about 120 °C. (The steam is pushing against the pressure regulator (weight) to escape, not the outside atmosphere. It is like the safety valve on an old steam locomotive that keeps the pressure at a certain level so that the pot doesn't blow up.)

The inside temperature of a pressure cooker when steaming (120 °C) is way below the critical temperature of water (about 374 °C) so that factor doesn't enter into the cooking time at all.


Silver's pretty soft—it's possible to just scrape it off with a metal spatula or something. Also, not all of the silver deposits on the glass. You might be able to adjust the conditions of the reaction to make more silver precipitate into the liquid where it's easier to get at. You can, of course, redissolve the silver in nitric acid or the like.
Since the byproducts of the reaction are soluble, it should be pretty pure, depending on how pure the starting materials are. If other easily reducible metal cations are present, you might get some impurities from that.

As a side-note, this isn't a particularly efficient way of getting elemental silver. The typical films produced in a test tube probably contain a couple hundred milligrams at most. Even a whole undergrad lab's worth of tubes is probably not worth recovering.


How to convert an organic carbonate to its formate ester?

First use $\ce{EtO-/EtOH}$, then lithium-tri-tert-butoxy-aluminium hydride. This will convert it to diethyl carbonate and as the reducing agent is hindered you can manipulate reaction conditions to get the desired product.

Or another way is to hydrolyse it in basic medium to get $\ce{H2CO3}$ and then evaporate $\ce{H2O}$ and $\ce{CO2}$ then after separating it will open a large number if ways as you have got ethyl alcohol and methyl alcohol.


Why are some ions mutagenic and others are not?

You are confused about the cause of mutagenesis in alpha particles.

The reason why an alpha particle is mutagenic is not that it is an ion, it is that it is an ion carrying a great deal of kinetic energy. Alpha particles are produced in energetic nuclear reactions and carry a great deal of energy away from the nucleus that emits them (a typical alpha particle has about 5MeV of energy, enough to break a typical chemical bond). It is the kinetic energy that matters not what the ion consists of. Heavier ions with that sort of energy would be just as bad for biological systems as are other forms of energetic radiation (beta particles are just electrons but will still do damage, gamma rays are just powerful photons but will dump enough energy into molecules to break DNA).

Non-energetic helium nuclei would pick up electrons extremely rapidly from the first molecules they encountered (probably from air even before reaching your skin) and would probably be as harmless to you as the helium filling a balloon).


Why do we still use the 'P' subscript in a thermodynamic derivative such as (dH/dT)P?

In thermodynamics, the number of degrees of freedom available to describe a system is given by Gibbs' phase rule:

$$
F = C - P + 2
$$

Here, $C$ is the number of components in the system, $P$ is the number of phases in the system, and $F$ is the number of resulting degrees of freedom.

Consider a pure substance in a closed system.  One can name a wide variety of thermodynamic parameters to describe its state: temperature, pressure, specific volume, internal energy, enthalpy, entropy, etc. However, Gibbs' phase rule says that we can only independently specify a subset of these parameters before we run out of degrees of freedom:

$$
F = 1 - 1 + 2 = 2
$$

So, in any mathematical description of (in your example) the enthalpy, $H$, it can only properly be a function of two independent variables:

$$
H = H\left(X_1, X_2\right)
$$

But: which two variables?

Simply writing the partial derivative $\partial H \over \partial T$ indicates that temperature is to be one of the two available degrees of freedom:

$$
H = H\left(T, X_2\right)
$$

Now, though – what is $X_2$? Well, including the subscript on the differential is a way of explicitly indicating what that other independent variable is:

$$
\left({\partial H \over \partial T}\right)_P \quad \Longrightarrow
\quad H = H\left(T, P\right)
$$

As porphyrin rightly notes in a comment, this applies to multivariate calculus generally, not just the calculus of thermodynamics.


Stoichiometric reaction of hydrochloric acid, potassium nitrate, and copper

My apologizes, I found the answer. The full equation is :
$$\ce{Cu + 4KNO3 + 4HCl -> Cu(NO3)2 + 2NO2 + 4KCl + 2H2O}$$
(I'll detail this later)


Is a time domain spectrum obtainable from a frequency domain spectrum?

Can you do the reverse?

Yes, mathematically the Fourier transform can be reversed. If we define the spectrum $S(\omega)$ as the Fourier transform of a time-domain signal $f(t)$,

$$S(\omega) = \mathcal{F}[f(t)] = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}f(t)\mathrm{e}^{-\mathrm{i}\omega t}\,\mathrm{d}t$$

then the inverse Fourier transform is simply given by

$$f(t) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}S(\omega)\mathrm{e}^{\mathrm{i}\omega t}\,\mathrm{d}\omega$$

Wikipedia has a page on this which is rather technical, but any decent book on Fourier transforms should cover this, so a mathematics book targeted at physicists/chemists may be easier to digest.

  What are the advantages and disadvantages of the frequency and time domain spectra?

The time domain signal (in NMR, "free induction decay") is for the most part, not particularly useful. It is the Fourier transform which makes it valuable for chemists, because the frequencies that appear in the spectrum are related to transitions between different energy states:

$$\Delta E = \hbar\omega$$

In general, probing these energy levels is the main point of spectroscopy, regardless of what range of energies you are using.


In a reversible process, the gas pressure is spatially uniform within the cylinder, and is described globally by the ideal gas law.  However, in an irreversible process, the force per unit area at the piston face is not equal the force per unit area at other locations within the cylinder.  Furthermore, the ideal gas law does not describe the behavior of the gas because viscous stresses contribute to the force per unit area for a rapid irreversible deformation.  So, even though Newton's 3rd law is satisfied at the piston face, unless we specify the force per unit area externally (e.g., manually), we will get the wrong answer if we try to calculate the pressure at the piston face using the ideal gas law.

In applying the equation $W=\int{P_{ext}dV}$ to calculate the work, $P_{ext}$ is supposed to be the force per unit area exerted by the surroundings on your system, at the interface between your system and the surroundings.  So, if the gas is your system, $P_{ext}$ is the force per unit area exerted by the inner face of the piston on your gas (and by your gas on the inner face of the piston).  In your example, if the cylinder is vertical and we do a force balance on the piston, we get:  $$m\frac{dv}{dt}=P_{ext}A-mg$$assuming there is vacuum on the outer face of the piston.  If we multiply this equation by the piston velocity v = dx/dt and integrate, we obtain:$$m\frac{v^2}{2}=\int{P_{ext}dV}-mg\Delta x$$Therefore, the work done by the gas on its surroundings (up until an arbitrary time) is given by:  $$W=\int{P_{ext}dV=mg\Delta x}+m\frac{v^2}{2}$$When the final thermodynamic equilibrium state of the system has been realized, the piston will no longer be moving (including any oscillations of the piston, which will have eventually been damped out by viscous stresses) and the work will then be determined by:$$W=\int{P_{ext}dV=\frac{mg}{A}\Delta V}$$
If, instead of vacuum, there is some constant force external to the piston (say, $P_{atm}$ equal to the atmospheric pressure), the previous results change instead to:$$W=\int{P_{ext}dV=mg\Delta x}+m\frac{v^2}{2}+P_{atm}\Delta V$$when the piston is still moving and $$W=\int{P_{ext}dV=(P_{atm}+\frac{mg}{A})\Delta V}$$at final equilibrium when the piston has been damped to rest.


The gauche effect is commonly explained with LCAO-based bond orbitals. LCAO is short for linear combination of atomic orbitals and implies that we can take two atomic orbitals $\phi_1, \phi_2$ and create two molecular orbitals $\psi_1, \psi_2$ from them by linear combination: $\psi_1 = a_1 \phi_1 + b_1 \phi_2$ and $\psi_2 = a_2 \phi_1 + b_2 \phi_2$.

Considering two neighbouring $\ce{sp^3}$ hybridised carbon atoms, we can create a $\unicode[Times]{x3c3}$-orbital between them by overlapping a hybrid orbital of each; and if $a_1 = a_2$ and $b_1 = - b_2$ (or vice-versa) we can create an antibonding $\unicode[Times]{x3c3}^*$-orbital; both are shown in figure 1 (the bonding in the bottom row, of course). Note that the lobes on each side are the same size.

Figure 1: $\unicode[Times]{x3c3}$ and $\unicode[Times]{x3c3}^*$ orbitals of $\ce{C-C}$, $\ce{C-H}$ and $\ce{C-F}$ bonds (schematically).

If we consider a $\ce{C-H}$ bond rather than a $\ce{C-C}$ bond, we need to take electronegativity into consideration. Compare with the middle column of figure 1: The more electronegative element, carbon, contributes more to the $\unicode[Times]{x3c3}$ orbital, while the less electronegative element hydrogen contributes more to the $\unicode[Times]{x3c3}^*$ orbital. (Note that hydrogen uses a 1s-orbital for binding.)

Similarly for the $\ce{C-F}$ bond, although now fluorine is more electronegative than carbon. And note how fluorine uses a p-orbital for bonding: No energy or anything else would be gained by hybridising fluorine’s orbitals. Neither the size difference between hydrogen’s and carbon’s nor that between fluorine’s and carbon’s orbitals is to any scale and both are grossly exaggerated.

The gauche effect, as a special form of hyperconjugation, relies us on looking at how these molecular orbitals can (or cannot) interact with each other. If we look at the relevant part of the molecule (compare figure 2), we can see that if hydrogen and fluorine are anti to each other, we get a $\unicode[Times]{x3c3}_{\ce{C-H}}$ bond and a $\unicode[Times]{x3c3}^*_{\ce{C-F}}$ bond that have favourable geometries and favourable lobe sizes for overlap.

Figure 2: Potentially overlapping orbitals in the gauche and anti cases of 2-fluoroethanol.

On the other hand, as also depicted in figure 2, the overlap would be bad if oxygen and fluorine be anti due to the small contribution of the less electronegative carbon to the $\unicode[Times]{x3c3}_{\ce{C-O}}$ orbital. (For simplicity, I have rendered oxygen using a plain p-orbital. Oxygen usually has orbitals somewhere between p and $\ce{sp^2}$.)

All this so far tells us why the interaction is geometrically favourable, but we yet did not deduce why it is energetically favourable. And here is where we are, more or less, linear combining the two molecular orbitals we created by linear combination earlier. The combining we are doing is a very general thing, observed not only in the gauche effect here or in hydrogen peroxide, but also is the basis for the anomeric effect in carbohydrates and hyperconjugation. The principle is: If we combine an occupied orbital with an unoccupied one, the occupied orbital gets lowered in energy while the unoccupied gets raised. Because only the occupied orbital contains electrons and their energy is lowered, the entire system gains stability. (You can read the two bolded half-sentences as one.) Compare figure 3.

Figure 3: Stabilisation by combining occupied and unoccupied orbitals

What we have done, is lower the energy level of an already lowered (because bonding is favourable) occupied orbital; the entire system gains. We often state this by saying things such as:

  Electron density is transferred into the antibonding orbital.

But it really boils down to stabilising occupied orbitals. However, that transfer of electron density helps us understand part of the consequences associated with this favourable mixing. If both the bonding and the antibonding orbitals of a certain bond are occupied, you can think of that bond as non-existant; as if both sides had an electron pair. If, on the other hand, neither orbital is occupied, the bond does also not exist, but due to lack of electrons. We can use that analogy to create a mesomeric description of the gauche effect with the two resonance structures depicted in scheme 1.

Scheme 1: A possible depiction of resonance stabilisation corresponding to the gauche effect.

Many chemists out there will vigorously refute this depiction as invalid. Indeed, the resonance structure on the right-hand side of the mesomery arrow shows charge separation and a non-optimal number of bonds. We can therefore assume that it only contributes ever so slightly to the complete picture — which is in line with the electron density only minutely being transferred and also the new bonding orbital (figure 3) being a lot closer in shape and size to the $\unicode[Times]{x3c3}_{\ce{C-H}}$ orbital rather than $\unicode[Times]{x3c3}^*_{\ce{C-F}}$ orbital. But the depiction certainly helps us understand why oxygen cannot do the job: Imagine $\ce{HO+}$ instead of $\ce{H+}$. Looks ugly and unstable from a chemical point of view.


How to prove that entropy is a state function?

I don't understand what you mean by saying "if you don't make the substitution you do not get zero".  The relevant point is that, for ideal gases, $$p = nRT/V,$$ so $p/T$ is constant with respect to $T$.  Therefore $$\frac \partial{\partial T}\left(\frac pT\right)=0$$ for an ideal gas, whether or not you use the substitution to simplify the process.

If you don't use the substitution, you need to use the quotient rule to do the partial derivative properly, and you should still end up getting zero if you do it correctly.  Could that be where your confusion is coming from?


How can water catalyse a reaction between iodine and aluminium?

A naked aluminum surface (freshly sanded) in air immediately grows about 4 nm of adherent aluminum oxide that passivates the surface.  Iodine slightly dissolves in water and disproportionates

$$\ce{I2_{(s)} + H2O_{(l)} -> HOI_{(aq)} + I^{-}_{(aq)} + H+_{(aq)}}$$

Iodine is strongly solubilized by forming triiodide.

$$\ce{I2 + I- -> [I3]-}$$

The acids chew through the alumina surface barrier.  When bare metal and triiodide meet, everything takes off, also igniting the bare metal in air.

http://www2.ucdsb.on.ca/tiss/stretton/database/inorganic_thermo.htm
$$\begin{alignat}{2}
&\ce{AlI3_{(s)}}\quad &&\Delta G_\text{f}^\circ = -300.8\ \mathrm{kJ/mol}\\
&\ce{Al2O3_{(s)}}\quad &&\Delta G_\text{f}^\circ = -1582.4\ \mathrm{kJ/mol}\\
\end{alignat}$$
Absent water the iodine must diffuse through the intact alumina barrier, react with naked aluminum, then swell and disrupt the surface passivation from underneath.


Balanced equation for synthesis of p-nitrophenol from p-nitrodiaznoium chloride salt and sulfuric acid/water?

The sulfuric acid is not consumed in the reaction pathway to product, so there's no reason for it to appear in the equation for the reaction.  Therefor, the overall equation for the reaction is $$\ce{([C6H4N3O2]+ Cl^{-}) + H2O -> C6H5NO3 + N2 + HCl}$$


Why is the Still-Gennari reaction Z-selective?

As you hopefully realise, the stereochemical outcome of the HWE reaction (and variations) is due to the relative orientation of the R-groups in the phosphaoxetane intermediate.
If the R groups are syn as in 5A then the (Z)-olefin is formed, with an anti arrangement 5B favouring the (E)-olefin as the product.

Source: Modern Carbonyl Olefination, Wiley, 2004

The pertinent question when looking at the selectivity is why do we get dominant elimination from the syn phosphaoxetane (leading to the correct olefin), rather than dominant elimination from what is arguably the sterically most favourable anti phosphaoxetane (in which the R groups are pointing away from one another).
As it turns out, the initial 'aldol' addition into the aldehyde is rate determining, with the anti addition being favoured over the syn addition on steric grounds.

Source: Stereoselective alkene synthesis, Springer-Verlag, 2012   (a really useful book, FYI)

Once the aldol addition has taken place, the subsequent steps are rapid due to the high electrophilicity of the Still-Gennari phosphonate (with the additional electron withdrawing groups).
This means that the entire process is essentially controlled by the initial addition to give a syn or an anti adduct. Once formed, the adducts are sufficiently short lived that despite the cis phosphaoxetane being less favoured due to the clash between R-groups, the (Z) olefin is formed via quick and irreversible elimination.
The HWE therefore is (E)-selective, in part, due to the fact that elimination is less rapid and as such there is equilibration to the more favourable phosphaoxetane from which elimination may occur.


Prediction of products in reaction of dicarbonyl with alpha-haloketone

As you've already figured out, cyclohexane-1,3-dione is more acidic than chloroacetaldehyde, the pKa of the diketone in water should be around 5.3.
The enolate of 1 (sorry for the circle in the structure, i'm to lazy to fix it) and 2 undergo an aldol reaction yielding 3. With a bit of charge shifting, we get again a nicely stabilized enolate (4), which dehalogenates (you might call that an O-alkylation as well) to 5. Subsequent elimination under aqueous conditions results in the formation of 6. I'd say this is, more or less, an example of the Feist-Bénary reaction.

Btw, I created the drawings with BKChem.

A conceivable depiction for the "transformation" of 3 to 4 is given below, but please note that this is just a model to please our eyes and minds ;)


$G$ is the Gibbs free energy of a system. It is a conceptual quantity in the sense that there is no reference point that defines $G = 0$ for a substance (unlike entropy). Whenever you see a plot of $G$, notice that the scale is not labeled with absolute values (no indication of where the origin is).
$\Delta G$ is the change in Gibbs free energy, i.e. $G_\text{final} - G_\text{initial}$. Because it is a difference, there is no issue with the lack of reference point.
$\Delta_\mathrm r G$ is the change in $G$ as a reaction proceeds, i.e. $\Delta_\mathrm r G = \frac{\mathrm dG}{\mathrm d\xi}$ with $\xi$ the extent of reaction variable. Because of this definition, it is an intensive quantity with dimensions energy per amount of substance. It is called the Gibbs energy of reaction.
$\Delta_\mathrm r G^\circ$ is the standard Gibbs energy of reaction, for the special case when all reactants and products are at standard state.
$\Delta G_\mathrm f^\circ$ is the standard Gibbs energy of formation of a substance, the $\Delta_\mathrm r G^\circ$ for the reaction that makes the substance from the elements (where all products and reactants are at standard state).
$\Delta G^\circ$ is for any process at standard state (e.g. ice melting at a pressure of 1 bar). It is also sometimes used instead of $\Delta_\mathrm r G^\circ$, but that is confusing because the meaning and the dimensions are different.

  I've seen the first two used interchangably, and seen $G$ specifically referred to as change in Gibbs Free energy. Is this common? Are there cases where $G$ refers to absolute Gibbs Free Energy?

$G$ should always refer to absolute Gibbs energy, and never to a change in Gibbs energy.

  For those with the circle ($^\circ$), I've seen that referred to as "standard", and this means that they are in reference to standard pressure or concentration. Does this make their use noticably different?

Yes, because the Gibbs energy is concentration-dependent. With $Q$ as the reaction quotient,

$$\Delta_\mathrm r G = \Delta_\mathrm r G^\circ + R T \ln Q$$

As concentrations change in a reaction, $\Delta_\mathrm r G$ changes but $\Delta_\mathrm r G^\circ$ does not. When a reaction approaches equilibrium, $\Delta_r G$ approaches zero (but $\Delta_\mathrm r G^\circ$ doesn't change, as just mentioned).

The reason that $\Delta_\mathrm r G$ is concentration-dependent is that the Gibbs energy has an entropy component ($G = H - T S$), and there is an entropy of mixing that depends on concentrations (the $R \ln Q$ part).


An analogy might help. Solubilization of a salt in a solvent (say NaCl in water) is similar to condensation of a gas.
Imagine you have an amount of gas in a cylinder with a piston to regulate the volume and pressure. Beginning at a pressure below the vapor pressure of the substance (which means that all is gas, none is liquid), the pressure is increased (while temperature is constant) until eventually the boiling point is encountered. At this point the pressure is equal to the vapor pressure (at that temperature) and the gas and liquid can coexist (all of this by the way is simpler to illustrate with a T-p phase diagram). Further compression condenses more gas until all of the substance is liquid. Below the boiling point pressure none of the substance is liquid, and above that pressure none is gas.
Next consider solubilization of a substance in a fixed volume of liquid (with T,p constant). One could traverse the solubility diagram by changing the temperature, but it is for the present discussion simpler to consider changing the concentration. At very low concentrations (small amounts of substance) all of the substance solubilizes. Exactly  none (yes, none) is present as the solid. But as you increase the amount of added substance a concentration is reached at which solid remains. That exact point is the solubility limit of the substance at that temperature. It can be thought of as equivalent to the boiling point pressure (coexistence point) of a gas and liquid at a given temperature. If you increase the amount of added substance at that point, no more substance will solubilize (exactly none).
So when a salt is added below its solubility concentration limit no solid remains, all of the substance is in the liquid, and there is no equilibrium only in the sense that there cannot be solid below the solubility limit.

This is also why the equilibrium equation for solubilization of a substance is written as
$$K_{\mathrm{sp}} = \frac{a(\textrm{soln})}{a(\textrm{s})}= a(\textrm{soln})$$
where a's are activities (for small concentrations, the activity of the solubilized substance is equal to concentration ). We ignore the activity of the solid (it is constant and assumed equal to 1) and above the solubility concentration limit the activity of the solubilized substance is also constant (equal to the solubility product).


Mixing water and oil in microgravity outside of a container

An experiment similar to this was carried out using a free-fall drop shaft to provide the microgravity environment in Japan in 2000.

From the abstract of the publication$^1$:

  An experimental study was performed to obtain the detailed information needed to provide a deep understanding of the combustion process and the secondary atomization of an oil-in-water emulsion droplet. The oil-in-water emulsion, which consisted of n-hexadecane as a base fuel, distilled water, and a trace of surfactant was tested. Photographic observation and temperature measurements were made of the suspended emulsion droplet during the heating-up and combustion processes under microgravity. The primary attention was toward the phase separation in the droplet, and the time histories of droplet temperature and the amount of water in the droplet, during the period of time prior to disruptive microexplosion. The results showed that the separation of the base fuel and water as well as their agglomeration and coalescence occurred with the lapse of time. The increase in the droplet temperature resulted in phase separation, and the formation of a single water droplet enveloped by a shell of the base fuel, prior to the microexplosion.

Interestingly, the phase separation was attributed to the increase in droplet temperature. It's my guess that the heating increased the rate of separation, but that given time the final result of "the formation of a single water droplet enveloped by a shell of the base fuel" would have taken place, and I think this would be the result of the situation described in the question; that van der Waals forces between the oil and water would prevent them from separating into two separate balls of liquid. The stronger hydrogen bonds of water were likely the reason for water forming on the inside of the droplet, and I would expect the same to happen in the case described in the question.

Summary, TL;DR:
Based on the results of the drop-shaft microgravity experiment described above, I would expect similar results for an experiment as described in the question. There should be an inner ball of water held together by hydrogen bonding, surrounded by an oil shell held together, and to the surface of the water, by the weaker van der Waals forces.

1) "Water-coalescence in an oil-in-water emulsion droplet burning under microgravity", Proceedings of the Combustion Institute, Volume 28, Issue 1, 2000, Pages 985-990.


Your first misconception is what i and j mean. These are indices of molecular orbitals.

A response matrix is something used in respone theory. It stems from Perturbation Theory.

Response Theory is a mathematical formalism to compute time-dependent molecular properties (in theoretical chemistry, that is).

This pdf (no guarantee for anything) describes it pretty well and concise.
http://www.lct.jussieu.fr/pagesperso/toulouse/enseignement/molecular_properties.pdf


Why is excluded volume in van der Waals equation divided by 2?

Collision is predominantly a bimolecular act. The probability of three gas molecules simultaneously colliding is incredibly low, therefore it quite accurate to say that the excluded molar volume $b$ is overcounted by a factor of 2.


Is there a mathematical formula for electronegativity?

Electronegativity cannot be calculated from the number of subatomic particles in an atom. However, using Ionization Energy and Electron Affinity, one can calculate the Electronegativity of an atom using the Mulliken Electronegativity Equation.

Ionization Energy : The amount of energy required to remove the
electron that is least attracted to the nucleus in a gaseous state.
Electron Affinity : The amount of energy an atom gains when an
electron is added.

To find the Ionization and Electron Affinity values of an atom, one must reference a chart.  In the Mulliken Electronegativity Equation, Ei stands for Ionization Energy and Eea stands for Electron Affinity.  Plug in your values and solve.

Equation for Energies in Kilojoules/Mole :
ENMulliken = (1.97×10^−3)(Ei+Eea) + 0.19
Equation for Energies in Electronvolts :
ENMulliken = 0.187(Ei+Eea) + 0.17

Lithium Example in kJ/mol :

The Ionization Energy for Lithium is 520 kJ/mol
The Electron Affinity for Lithium is 60 kJ mol^-1
ENMulliken = (1.97×10^−3)(Ei+Eea) + 0.19
ENMulliken = (1.97×10^−3)(520 + 60) + 0.19
ENMulliken = 1.143 + 0.19 = 1.333

NOTE : There are multiple scales for measuring Electronegativity, including the Pauling Scale, Sanderson, Allen, and more.  This example measures Electronegativity using the Mulliken scale, and will compute an answer that will differ from other models.


Y-axis of the reaction co-ordinate graph

Single molecule reaction

Some diagrams refer to a single molecule reacting. If that is the case, several quantities are not defined or ill-defined:

pressure
entropy of mixing
temperature

In those diagrams, the best label would be potential energy, and the energy difference between reactant and transition state should be labeled activation energy.

Ensemble reaction

If the diagram refers to an ensemble reacting (reaction mixture with sufficient number of molecules to have macroscopic properties), the y-axis could correspond to Gibbs energy, enthalpy or internal energy.

If the y-axis represents Gibbs energy, the energy difference between reactant and transition state should be labeled $\Delta G^\ddagger$. Those diagrams make sense when discussing how equilibrium constants remain unchanged but reaction rates change when a catalyst is added.

If the y-axis represents enthalpy or internal energy, the energy difference between reactant and transition state should be labeled $E_a$ for activation energy. Those diagrams make sense when discussing reaction rates in the context of collision theory.

What is most popular?

Looking at about 50 images of reaction energy diagrams found in a google image search, I found a wide variety, and lots of mixing and matching (i.e. incompatible quantities in a single diagram). This tells you that the diagram is conceptual, and the concepts are not always clearly articulated in diagrams.


Equivalent conductivity of barium chloride at infinite dilution

One must be really careful with the units while dealing with conductivity, conductance problems.

If you read the question carefully, the equivalent conductivity of $\ce{Ba^{2+}}$ and $\ce{Cl^{-1}}$ are provided to you.

So the molar conductivity of $\ce{Ba^2+}$ is: $2 \times 127 ~\pu{ohm^-1 mole^-1} = 254 ~\pu{ohm^-1 mole^-1}$

and that of $\ce{Cl^{-1}} = 76~\pu{ohm^-1 cm^2 mole^-1} $

Now apply Kohlrausch's law of molar conductivity of solution at infinite dilution:

$\lambda^o_{\ce{BaCl2}}= \lambda^o _{\ce{Ba^{2+}}} + 2\lambda^o_{\ce{Cl^{-1}}} \\ \implies  \lambda^o_{\ce{BaCl2}} = 254 + 2\times 76 = 406 ~\pu{ohm^-1 cm^2 mole^-1}$

Now, as you say, equivalent conductivity is $\dfrac{1}{2}$ times the molar conductivity for $\ce{BaCl_2}$ so equivalent conductivity of $\ce{BaCl_2} = \dfrac{1}{2}\times 406 = 203 ~ \pu{ohm^-1 cm^2 equivalent^-1}$ which is the correct answer.

Alternatively, observe that $203 = 127 + 76$ so Kohlrausch's law for equivalent conductivity of strong electrolyte at infinite dilution may be stated as:

  The equivalent conductivity of a strong electrolyte at infinite dilution is equal to the sum of the equivalent conductivities of the anions and cations.


The sugar will keep dissolving until it reaches it's saturation concentration of around $\pu{900 g/L}$, which is of course horribly sweet, thick and undrinkable as a beverage.  Assuming you don't completely saturate the water with sugar in this way, the undissolved sugar will continue to dissolve so long as there is still any remaining in the glass.

A common way of measuring the dissolved sugar concentration is by refractometry, as described below (note that home refractometers can be purchased for as little as $30 US):

  A refractometer is a laboratory or field device for the measurement of
  an index of refraction (refractometry). The index of refraction is
  calculated from Snell's law while for mixtures, the index of
  refraction can be calculated from the composition of the material
  using several mixing rules such as the Gladstone–Dale relation and
  Lorentz–Lorenz equation.
  ...
  In homebrewing, a refractometer is
  used to measure the specific gravity before fermentation to determine
  the amount of fermentable sugars which will potentially be converted
  to alcohol.

So, in your case a "homebrewing" type of refractometer is probably your best bet. As long as sugar remains in solution, it will continue to dissolve until it reaches saturation.  This also means it's refractive index and specific gravity will continue to increase.  Once this increase is no longer observed, then the sugar is no longer dissolving, meaning you've either reached saturation or all of the sugar has dissolved.  Again, saturation would mean a thick, honey-like solution. Thorough stirring throughout the measurement process is a must.


Conversion from a PPB value to µg/m3 of Isobutylene

This is much simpler than you think, even though the value you've obtained looks reasonable to me.
Part per billion ($\pu{1 ppb} = \pu{1e-9}$) on its own is meaningless.
Based on a context, it can refer to anything: mass fraction, mole fraction, particles etc.
The thing is, for gaseous mixtures $\pu{ppb}$ always refers to the volume fraction $\phi_i$:

$$\varphi_i = \frac{V_i}{V}$$

where $V_i$ is the volume of the $i$th gas (here, isobutylene) and $V$ is the total volume.
In order to convert volume fraction expressed in $\pu{ppb}$ to desired mass concentration $\rho_i$ expressed as

$$ρ_i = \frac{m_i}{V}$$

where $m_i$ is the mass of $i$th component in a mixture, you have to use density $d_i$:

$$ρ_i = \frac{m_i}{V} = \frac{m_i}{V_i}\varphi_i = d_i\varphi_i$$

To conform with your units of choice ($\pu{μg m-3}$) we need to juggle the units a bit as the density is usually given in $\pu{kg m-3}$, e.g. for gaseous isobutylene at NTP $d(\ce{C4H8}) = \pu{2.3959 kg m-3}$ (Source):

$$
\begin{align}
\pu{1 g} &= \pu{1e6 μg} \\
d(\ce{C4H8}) &= \pu{2.3959e6 μg m-3}
\end{align}
$$

Now, using this density value you can multiply your volume fraction in $\pu{ppb}$ or $\pu{ppm}$ and get an answer, e.g. for $\pu{400 ppm}$:

$$ρ_i = d(\ce{C4H8})\varphi (\ce{C4H8}) = \pu{2.3959e6 μg m-3}\cdot\pu{400e-6} = \pu{958 μg m-3}$$

Note that all notations involving "part per something" are deprecated and should be avoided.
It also wouldn't hurt if you ask manufacturer whether $\pu{1 ppb}$ the sensor reports is indeed $\pu{1e-9}$ and not $\pu{1e-12}$ (just in case) if you think the reported values are too high.


Your approach is correct and the result you obtain is right. Checked here.

As you correctly affirm, since the ratio of the base and acid is $1$, log will be zero and therefore $\mathrm{pH}=\mathrm{p}K_\text{a}$, where $\mathrm{p}K_\text{a} = 10.57$ (derived from the value of $K_\text{b}$ you provide).


How to determine pH of food in order to avoid unsuitable storage containers

You can try wetting the tip of an ordinary pH paper, then put it into a sample of your food with that wet side. The capillary forces will suck up the liquid of your food and you should see a colour change in the formerly wetted area and above.

PH paper is not a very reliable method, but it should give you a general idea and it is easily performed at home.


Methanol-water mix freezing points - clarify contradictory online data > 60% w/w?

From Solid-Liquid Phase Diagram of the System Methanol-Water by G. A. Miller and D. K. Carpenter in J. Chem. Eng. Data 1964, 9, 3, 371–373 (July 1, 1964) [https://doi.org/10.1021/je60022a017]

"A solution of methanol and water that is methanol-rich tends to form
a glass when cooled below the melting point. The liquid is very
viscous at such temperatures (below -100° C.) because of the presence
of hydrogen-bonded polymers. Because it is difficult to induce
crystallization, the melting point is determined from a warming curve.
Much of the wide divergence of melting point data for the composition
range of 55 to 95 mole per cent* methanol results from the use of
cooling curves."

* by a back-of-a-napkin calculation (assuming 32.042 g/mol MeOH and 18.015 g/mol H2O) the mole percentage mentioned is approximately 68% to 97% methanol (w/w).
Further - successful - online research was inspired by a comment; it appears the failure to find information with common search terms was due to a language difference. It seems specialists (presumably physical chemists) use "solid-liquid equilibrium" or "solid-liquid phase diagram" in documents which may completely exclude the term "freezing-point" - such as the publication quoted above which answers the original question.


Molecule 1, the cyclopentadienyl cation, is not aromatic. It only has 4 π electrons and does not fit the $4n+2$ rule.

Molecule 2, the cycloheptatrienyl cation (or tropylium cation), is aromatic; it has 6 π electrons and fits the $4n+2$ rule with $n=1$.

Molecule 3, azulene, is generally considered to be aromatic. It fits the $4n+2$ rule with $n=2$, and is often referred to as a "nonbenzenoid" aromatic compound. It can also be viewed as a fusion of two other aromatic units, the tropylium cation and the cyclopentadienyl anion.


I haven't checked your numbers but you're assuming the $\ce{PbI2}$ to be perfectly insoluble. That's a reasonable assumption but in reality its solubility product is $K_{sp}=4.41\times 10^{-9}$ (Wikipedia), so it does have some solubility.

To take this into account it becomes an equilibrium problem. Assume the initial concentrations (just after mixing)  to be $\ce{C(Pb^2+)}$ and $\ce{C(I^-)}$.

After precipitation the lead concentration has been reduced by$X$ and the iodide concentration by $2X$.

With the solubility product after precipation:

$$K_{sp}=[\ce{Pb^2+}]\times [\ce{I^-}]^2$$

So we have:

$$K_{sp}=\Big(\ce{C(Pb^2+)}-X\Big)\times \Big(\ce{C(I^-)}-2X\Big)^2$$

Note that this is a third degree polynomial equation in $X$. Solve for $X$ to find the real equilibrium concentrations $[\ce{Pb^2+}]$ and $[\ce{I^-}]$.

We can also use the equation for the case where $K_{sp}=0$. Then:

$$0=\Big(\ce{C(Pb^2+)}-X\Big)\times \Big(\ce{C(I^-)}-2X\Big)^2$$

First case: $\ce{C(Pb^2+)}-X=0 \implies \ce{C(Pb^2+)}=X \implies [\ce{Pb^2+}]=0$

Also: $[\ce{I^-}]=\ce{C(I^-)}-2\times\ce{C(Pb^2+)}$

In the second case we assume $\Big(\ce{C(I^-)}-2X\Big)^2=0$ and proceed as above.

Which case applies depends on which reagent is the limiting one.


How to balance the equation for the reaction of aluminum bicarbonate and hydrochloric acid?

My question is... what would be the reactants and products in this chemical equation?

You give the equation, you just need to balance it.

  Would HCO3 as a whole compound be a reactant itself?

No, $\ce{HCO3^-}$ is an anion.


Overview of material characterization methods

There is a collection available at Libretexts.org, by students and faculty from Franklin & Marshall College. The intent is to provide a quick overview for readers of the literature (rather than practitioners doing the experiments), so the focus is the type of result each method yields, and how to interpret it.


Does the name 4-nitrodiphenylmethane follow the IUPAC rules?

The preferred IUPAC name for the compound that is given in the question is 1-benzyl-4-nitrobenzene; however, that does not necessarily mean that other names are not in accordance with IUPAC nomenclature.

The name diphenylmethane actually is in accordance with current IUPAC recommendations (see Subsection P-44.1.2.2 in Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book)), although the preferred name is 1,1'-methylenedibenzene (see also this answer).

The name diphenylmethane treats the phenyl groups alike as substituents of methane.

If now one of the simple phenyl substituents in diphenylmethane is replaced by a compound 4-nitrophenyl substituent, the analogous name of the resulting compound is (4-nitrophenyl)(phenyl)methane and not ‘4-nitrodiphenylmethane’.


How to set up an ICE table for adding a weak acid to a strong acid

ICE tables tabulate amounts or concentrations pertaining to an equilibrium reaction: at the initial stage (I), how they change (C), and what they are at the end or at equilibrium (E). They are an alternative to solving multiple simultaneous equations (equilibrium constant expression, mass balance), instead using a single parameter (related to the extent of reaction $\xi$).
For the problem at hand, there is one reaction that goes to completion (strong acid) and one that attains equilibrium (weak acid, both protonated and deprotonated species present at equilibrium).
The ICE table needs to have species from the second reaction only. A good estimate of the hydrogen ion concentration is the concentration of hydrochloric acid (it dissociates completely, and we are ignoring the autodissociation of water for now).
$\begin{array}{|c|c|c|c|} \hline
\ & [\ce{HP}] & [\ce{H+}] & [\ce{P-}]\\ \hline
\text{I} & 0.1 & 0.02 & 0 \\
\text{C} & -x & +x & +x \\
\text{E} & 0.1-x & 0.02 +x & x\\ \hline \end{array}\\\\$
If you do the math, you will find that very little of the propanoic acid will dissociate, as the $\mathrm{p}K_\mathrm{a}$ is far from the initial pH.


Identification of solutions by solubility when mixed with each other

Since It is a Homework Question I would Like to add a hint instead of giving full solution.

You might not be able to Watch the difference Carefully.

For Your information

$\ce{PbCl2}$ = insoluble

$\ce{PbSO4}$ = insoluble

$\ce{AgCl}$ = insoluble

$\ce{Ag2SO4}$ = slightly soluble

Try doing it again and watch the difference carefully afterall now you know which one is soluble.

Hope it helps.


Typically, both two- and three-bond 1H–13C couplings appear in HMBC spectra, with three-bond couplings tending to be more prominent. The presence of other atoms in the path, e.g. oxygen, doesn't affect this. So, the three-bond coupling you have indicated is very likely to show up in the HMBC spectrum, as a cross peak at (f1, f2) ≈ (180 ppm, 4 ppm).

If all you are looking for is structural confirmation + assignment of 1H data (which would be typical if this was a product you expected from an experiment), I'd suggest that you don't necessarily need 2D data. The multiplet structures should contain more than enough information.


I've heard that question before (which is why I asked). So, to head off any potential urban legends...

There is evidence that the regulations against CFC's are making a difference in the loss of ozone. There's a nice little synopsis of the situation here:  The Ozone Hole


Formation of Hydrogen - from a proton and an electron

Both in classical and quantum mechanics you have to get rid of a (rather large) amount of energy, i.e. the difference in potential energy from zero (=infinite distance) to the bound state (= e.g. Bohr radius), also called the binding energy.

Nature has no way to do that without a third particle taking the excess energy with it. In quantum mechanics there is the additional problem with the spin, but also classically that doesn't work if you assume elastic interaction of proton and electron.

So the answer is no.

(Inelastic interaction requires a zoo of additional assumptions, which I let fall victim to Occams razor here. Inverse photodissociation is a possibility, if overall energy and momentum are conserved.)


Asked and answered, but I think one thing that's missing is that allenes are not planar like alkenes or alkynes are. You can refer to this question for an explanation.

The dihedral angle between the two halogens is 90 degrees (ideally). Here's an animation hopefully providing a better view of the 3D structure:

This results in the two mirror images being non-superimposable upon each other, which means the molecule satisfies the criterion for chirality. If you rotate the right-hand molecule by $90^\circ$ such that the blue and white atoms at the top coincide, the green and white atoms will not match each other.

In general, any allene with the formula $\ce{R^1R^2C=C=CR^3R^4}$ will be chiral as long as $\ce{R^1}\neq\ce{R^2}$ and $\ce{R^3}\neq\ce{R^4}$. In this case we have $\ce{H} \neq \ce{Br}$ and $\ce{H} \neq \ce{Cl}$, so the compound in the question is chiral.

If $\ce{R^1} = \ce{R^2}$ or $\ce{R^3} = \ce{R^4}$, then the compound will possess a mirror plane and will therefore be achiral.


Which element should be placed in axial position bulky one or the smaller one?

No, it's the other way around. Axial positions are less spacious, so you'd want to place your smaller substituents there and the bulky ones to the equatorial.


Titration to determine the amount of hydrogen peroxide

I would not use potassium permanganate with sulfuric acid present:

You'll throw off your result due to the competing reaction..and risk a fire depending on the concentration of sulfuric acid.

If you measure the amount of $\ce{O2}$ produced by the decomposition of $\ce{H2O2}$ you can back-calculate how much $\ce{H2O2}$ was present in the sample. I think (with a decent certainty) that none of the species you list here should produce gas besides the peroxide.

That said I would use a catalyst (like $\ce{MnO2}$) for decomposition instead of a potentially reacting species (like potassium permanganate). You will likely get side reactions that consume the catalyst but they don't produce gas. So simply use excess $\ce{MnO2}$.

Measure the volume of $\ce{O2}$ produced (i.e. wait until the gas volume no longer increases) and use the ideal gas law (as a fair approximation) to get moles of $\ce{O2}$ produced. Once you have that, you can stoichiometrically determine moles of $\ce{H2O2}$ in your sample of known volume, thus concentration of $\ce{H2O2}$ (recalling that $2\ce{H2O2} \rightarrow 2\ce{H2O} + \ce{O2}$).

I've done this before with $\ce{MnO2}$ from batteries, washed thoroughly with water. I capture the gas produced in an upside-down grad-cylinder in a tub/beaker of water, e.g.:

P.S. if you manage to / happen to get a pure $\ce{H2O2}$ sample, you can very accurately get concentration (in %) by the following:

$C = \frac{d - 0.9946}{0.00456}$ where $d$ is the density in g/mL at 20deg. C.

That's a standardized function I derived from known and experimental densities of $\ce{H2O2}$


How to calculate the pH of a redox reaction between nickel(IV) oxide and silver?

For the following reaction:
\begin{gather}
\ce{NiO2(s) + 4 H+(aq) + 2 Ag(s) -> Ni^2+(aq) + 2 H2O(l) + 2 Ag^+(aq)}\\
E^\circ = \pu{2.48 V}
\end{gather}
Calculate the $\mathrm{pH}$ of the solution if $E = \pu{2.23 V}$ and $[\ce{Ag^+}] = [\ce{Ni^2+}] = \pu{0.023 M}$.

The important thing to understand that was not immediately obvious to me and a lot of other people was the fact that $n = 2$, and this changes how you calculate your answer.
$\ce{Ni}$ does not change oxidation state in its transition from $\ce{NiO2}$ to $\ce{Ni^2+}$, as its oxidation state in $\ce{NiO2}$ is $+2$.  The 4 $\ce{H^+}$ ions also do not change oxidation state in their transition from $\ce{H^+}$ to $\ce{H2O}$, oxidation state in $\ce{H^+}$ is $+1$ and oxidation state in $\ce{H2O}$ for $\ce{H}$ is $+1$ for 2 $\ce{H}$ atoms.  That leaves us with the only $\ce{e^-}$ transfer occurring to $\ce{Ag}$.  There are 2 moles of $\ce{Ag}$ atoms going to oxidation state $+1$.  Therefore, 2 moles of $\ce{e^-}$ are transferred.  Hence, $n=2$
The Nernst Equation:
$$E = E^\circ  -  (0.0592/n)\log Q$$
Where $E˚$ is the standard cell potential, $E$ is the cell potential for non-standard conditions, and at non-standard pressures or concentrations, $$Q = \frac{[\text{products}]}{[\text{reactants}]}.$$
For this problem, $n(\ce{e-}) = \pu{2 mol}$, and that
$$Q = \frac{[0.023]^3}{[\ce{H^+}]^4}.$$
Following the above calculation with this difference leads to $\mathrm{pH} = 3.34$, which is the correct solution.
\begin{align}
\tag1
2.23 &= 2.48
- \frac{0.0592}{2}\log\left(\frac{0.023^3}{[\ce{H^+}]^4}\right)\\
\tag2
0.25 &= \frac{0.0592}{2} \log\left(\frac{0.023^3}{[\ce{H^+}]^4}\right)\\
\tag3
0.50 &= \frac{0.0592}{1} \log\left(\frac{0.023^3}{[\ce{H^+}]^4}\right)\\
\tag4
0.50 &= 0.0592\log\left(\frac{0.023^3}{[\ce{H^+}]^4}\right)\\
\tag5
\frac{0.50}{0.0592} &= \log\left(\frac{0.023^3}{[\ce{H^+}]^4}\right)\\
\tag6
\frac{0.50}{0.0592}
&= \log\left(0.023^3\right) -\log\left([\ce{H^+}]^4\right)\\
\tag7
\frac{0.50}{0.0592} - \log\left(0.023^3\right)
&= -\log\left([\ce{H^+}]^4\right)\\
\tag8
\frac{0.50}{0.0592} - \log\left(0.023^3\right)
&= -4\log\left([\ce{H^+}]\right)\\
\tag9
\frac{\frac{0.50}{0.0592} - \log\left(0.023^3\right)}{4}
&= -\log\left([\ce{H^+}]\right)\\
\tag{10}
\mathrm{pH} &= -\log\left([\ce{H^+}]\right)\\
\tag{11}
\frac{\frac{0.50}{0.0592} - \log\left(0.023^3\right)}{4} &= 3.34
\end{align}


Glutamine in aqueous solutions degrades slowly when stored in room temperature. Therefore, we can expect that rate of degradation is faster at higher temperatures. It is evident that the hydrolysis product of glutamine is 5-pyrrolidone-2-carboxylic acid (not glutaric acid as shown for enzymatic degradation) and ammonia (Ref.1):

The Ref.1 states that:

The degradation kinetics of ʟ-glutamine (Gln) in aqueous solution was studied as a function of buffer concentration, $\mathrm{pH}$ and temperature. Stability tests were performed using a stability-indicating high-performance liquid chromatographic assay. The degradation product of Gln was 5-pyrrolidone-2-carboxylic acid. The reaction order for Gln in aqueous solution followed pseudo-first-order kinetics under all experimental conditions. The maximum stability of Gln was observed in the $\mathrm{pH}$ range from 5.0 to 7.5. The $\mathrm{pH}$–rate profile described by specific acid–base catalysis and hydrolysis by water molecules agreed with the experimental results. Arrhenius plots showed the temperature dependence of Gln degradation, and the apparent activation energy at $\mathrm{pH}$ 6.41 was determined to be $\pu{9.87 \times 10^4 J mol−1}$.

Factors affecting the stability of ʟ-glutamine in solution has also been studied (Ref.2). They also found that glutamine degradation resulted in the equimolar formation of ammonia and no associated formation of glutamate. At $\pu{22–24 ^\circ C}$, the degradation rate of glutamine was variable depending on the type of solution used. It was 0.23% in water $\mathrm{pH}$ 6.5; 0.22% in dextrose/water (15% $w/v$); 0.8% in mixed total parenteral nutrition (TPN) solution; and on the $\mathrm{pH}$, molarity and type of buffer used. The degradation rate was essentially unaffected by light and $\ce{O2}$.
Nonetheless, the solubility of ʟ-glutamine in water increases with temperature (Ref.3).
References:

Kanji Arii, Hideyuki Kobayashi, Toshiya Kai, Yukifumi Kokuba, "Degradation kinetics of L-glutamine in aqueous solution," European Journal of Pharmaceutical Sciences 1999, 9(1), 75-78 (https://doi.org/10.1016/S0928-0987(99)00047-0).
K. Khan, M. Elia, "Factors affecting the stability of ʟ-glutamine in solution," Clinical Nutrition 1991, 10(4), 186-192 (https://doi.org/10.1016/0261-5614(91)90037-D).
Jan P. Amend, Harold C. Helgeson, “Solubilities of the common ʟ-$\alpha$-amino acids as a function of temperature and solution $\mathrm{pH}$,” Pure and Applied Chemistry 1997, 69(5), 935-942 (https://doi.org/10.1351/pac199769050935).


What Could Formations of Linkage Isomers of [Co(NH₃)₅NO₂]²⁺ be like?

This Co-ordination compound of were studied by Jorgensen and Werner who observed that were two compounds of the same chemical formula but of different colors. (We can see these two geometries in the image).
A red form of low stability converted readily to a yellow form the $\ce{M-NO2}$ nitro isomer, based on comparison with compounds with similar colors. This conclusion was later confirmed, and kinetic and $\ce{^{18}O}$ labeling experiments showed that conversion of one form to the other is strictly intramolecular, not a result of dissociation of the $\ce{NO2^-}$ ion followed by reattachment.

References:
Miessler, Tarr,Inorg. Chem., 2nd Edition, 1998


Why is the radial distribution at the nucleus 0 but not the radial wave function?

You state 'but we already know from the radial wave function that there is a non-zero probability of finding an electron in an infinitesimal volume at the nucleus.'
To add to Poutnik's answer, I think that this sentence states the misunderstanding. The radial part of the wavefunction is $R_{n\ell}(r)$ and when normalised
$$\int_0^\infty R_{n\ell}^*(r)R_{n\ell}(R)r^2 dr=\frac{4}{a_0^3}\int_0^\infty e^{-2r/a_0}r^2 dr=1$$
where the $r^2$ comes from the volume element needed to integrate over all angles which is $r^2\sin(\theta)drd\theta d\phi$. The probability from $r\to r+dr$ is then for the 1s orbital $\displaystyle p=\frac{4}{a_0^3}r^2e^{-2r/a_0}dr$ whereas the 1s wavefunction itself has the form $\sim e^{-r/a_0}$. For other orbitals, e.g $2p_{\pm 1}$ the wavefunction is a function with complex numbers so cannot represent anything physical.


Why are HgO and CdS compounds colourful?

The band gap decreases as either electronegativity of the anion decreases (electronegativity of anion in CdS < CdTe , ZnS < ZnSe) or the overlap decreases (overlap in ZnS > CdS).

Also, note that colored compounds only result when 1.8 < Band gap < 3.0. Larger gaps give white/transparent & smaller gaps give a black colour to compounds.
(The band gap in ZnS is 3.54, the band gap in CdS is 2.45.)


What is the known cheapest chemical solution to extract CO2 from air on industrial scale?

Your intuition is correct, and such compounds are plenty. The cheapest are probably $\ce{NaOH}$, $\ce{Ca(OH)_2}$ and the like, but these are not easily regenerable. Then there are organic amines, like monoethanolamine, which can do the same in a reversible manner. Indeed, they are used in industry.


Why does the reduction of NAD+ to NADH change the absorbance so much, and cause fluorescence?

Essentially NADH has an additional pair of electrons. These additional electrons which are oftein drawn as a lone pair of electrons alters the molecule in a similar way to the way in which many pH indicators change upon protonation.

Here are diagrams of the reduced and oxidized forms of NADH / NAD+

A typical indicator when protonated appears to have a colour, however upon deprotonation the indicator loses a hydrogen and gains a lone pair of electrons now appearing to have a different colour. The addition of the lone pair of electrons actually lowers the transition energy between the pi bonding and the pi anti-bonding states.

For example for methyl red it can be predicted that the unprotonated form has a HOMO/LUMO gap of 2.175 eV (according to extended Huckel calculations). For the protonated form of methyl red has a HOMO/LUMO gap of 3.275 eV.

This lowered energy correlates with a decreased absorbance energy, ultimately reflecting a higher energy wave of light (Frequency of Blue > Red). NADH's absorbance wavelength his greater than NAD+'s absorbance wavelength, because NADH is absorbing lower energy light. This, like pH indicators, is due to the lone pair lowering pi and pi anti-bonding energy state differences.


Whether copper can combine with oxygen in equal proportion?

The Law of Reciprocal Proportions states: "If two different elements combine separately with the same weight of a third element, the ratio of the masses in which they do so are either the same or a simple multiple of the mass ratio in which they combine."

Wikipedia has a different phrasing: "If element A combines with element B and also with C, then, if B and C combine together, the proportion by weight in which they do so will be simply related to the weights of B and C which separately combine with a constant weight of A."

Note this is a statement about the weights of substances, not about the molar fractions or volumes.

For example,

Copper sulfide ($\ce{Cu2S}$) contains 66% copper by moles. It does not, however, contain 66% copper by weight. Instead it contains $(2\cdot63.55)/( 2\cdot63.55+32.06) = 79.9\%$ copper by weight, or 20.1% sulfur by weight.

Likewise, while sulfur trioxide ($\ce{SO3}$) contains 25% sulfur on a molar basis, it contains $(32.06)/( 32.06+3\cdot16.00) = 40.0\%$ sulfur on a mass basis, or 60.0% oxygen by weight.

So when we apply the law of reciprocal proportions to copper oxide, we need to make use of the 20.1% and 40.0% numbers for sulfur. For example, if we had two masses of $\ce{Cu2S}$ and $\ce{SO3}$ each containing 1 g of sulfur, then we would have $1\rm{g}\cdot(79.9/20.1) = 3.98$ grams of copper and $1\rm{g}\cdot(60.0/40.0) = 1.50$ grams of oxygen. The ratio here is $3.98/1.50 = 2.65$.

Now, when we examine copper (II) oxide ($\ce{CuO2}$), for example we see that the ratio of masses for copper and oxygen are $63.55/(2\cdot16.00) = 1.99$ ... and 2.65 is very close to a simple multiple of 1.99 (specifically, 4/3 - the difference is due to rounding.)

The fact that this is always true is trivial, once you know atomic theory. So while the Law of Reciprocal Proportions is important in a historic context - as it helped lead to the development of modern atomic theory - basically no one uses it as such these days.

Say we have the compounds $A_hB_j$, $A_kC_l$ and $B_mC_n$.

In $A_hB_j$ the masses occur in the ratio $(j\cdot M_b) / (h\cdot M_a)$. In $A_kC_l$ the masses are in the ratio $(l\cdot M_c)/(k\cdot M_a)$, so for a fixed mass (say $M_a$g) of A, the ratio of the masses is $(k\cdot j\cdot M_b)/(h\cdot l\cdot M_c)$

Now for the compound $B_mC_n$, the masses occur in the ratio $(m\cdot M_b) / (n\cdot M_c)$. Taking the proportion of this with the previous proportion, we see that the masses cancel, leading to $(k\cdot j\cdot n)/(h\cdot l\cdot m)$. Now because molecules form with small, whole number coefficients in their molecular formulas, this final quantity will typically be a simple, small fraction. (a.k.a. a "simple multiple" of each other.)

So all the Law of Reciprocal Proportions is doing is saying that 1) all atoms of a given element have about the same average weight 2) molecules form from (typically) small, integer numbers of atoms. -- That's it.

This is why you don't see many people discussing it. It's not all that interesting to people who are accustom to thinking atomically.


How to access implicit bond angles that don't appear in Gaussian's z-matrix input?

You can reorder the Z-matrix with a free program like Molden. For example open the molecule in Molden and press the ZMat editor. From there you should see a button that says reorder Z-matrix. Click the atoms one by one, making sure to select the HCH atoms sequentially so they have an explicit angle. Dummy atoms are another good option.


Have you considered a free-energy calculation?

That is, calculate the free-energy of the dissociated species and of the neutral one, and calculate the pKa via the formula for activity.

There was this GROMACS tutorial, but it seems the link is broken.


Roughly speaking, any acid is equally effective if its pKa lies below the pH 7 you want for a neutral solution.  It does not have to be HCl at all, in fact citric acid works for all three of its commonly observed deprotonations (pKa = 6.4 for the third deprotonation and of course lower for the first two; reference).  So you may use citric acid assuming three equivalents per mole and you should get where you want to go.

As a matter of taste I prefer citrate in my tea versus chloride, anyway.


Is there a formula to determine the total number of constitutional isomers?

There seems to be some confusion here : in order to determine the number of stereo-isomers of a molecular formula (say C4H11N) you would need to first know the number of constitutional isomers as you need to know how many stereogenic centers across all the structures.

There are a bunch of existing questions on constitutional isomers (a, b, c, ...) but the summary is : "it is tricky without software for anything more than very small formulae".

There are mathematical formulae for determining the count of constitutional isomers, although the most powerful techniques are non-trivial (such as those based on the Pólya enumeration theorem). To efficiently list a non-redundant set of compounds from an elemental formula is possibly even harder, as it requires a mechanism to avoid duplicates in the output.


Why do the 4f and 5d orbitals have similar energies?

The energy of a hydrogenic orbital with principal quantum number $n$ is proportional to $-(Z/n)^2$, so as $n$ increases the orbitals become more and more bunched together.

In a multi-electron atom the degeneracy of orbitals with different orbital angular momentum $l$ is lifted, i.e. s/p/d/f orbitals are no longer equal in energy. Since the gap between $n = 2$ and $n = 3$ is fairly large, one would not expect the 2p and 3s orbitals to be particularly similar in energy.

But the gap between $n = 4$ and $n = 5$ is smaller, so it is inevitable and not all that surprising that some of those constituent subshells have similar energies when the degeneracy is lifted. This is how it looks like for lithium (diagram from Modern Spectroscopy 4th ed., Hollas, p 214). The diagram will not be the same for the 4f/5d elements – in particular, I would expect the 4p/4d/4f orbitals to have a more significant difference in energy – but the idea is the same.


This source states that MCPBA gives preferential epoxidation of the more substituted alkene. This source explains that due to intra-molecular hydrogen bonding in solution, the high degree of polarisation results in an electrophilic oxygen atom that adds to alkenes. The most substituted alkene will be the most electron-rich alkene and hence the most nucleophilic. This is supported by the data shown here and here 4 on the relative rate of reaction of unsubstituted/mono/di/tri/tetra subsituted alkenes.


The molecule has to be small. trans-1,2-difluorocyclopropane is certainly chiral, and its predicted boiling point is ~ 5 deg C.  trans-1,1,2,3-tetrafluorocyclopropane is another likely candidate.  This one will be easier to prepare (from perfluoropropylene oxide and trans-1,2-difluoroethylene).

Separation of enantiomers is the hard part.


In fact, you need to use two of the fundamental relations.

For an ideal gas, the internal energy $U$ can be considered to be a function of $V$ and $T$. So, the total differential of $U$ would read

$$\mathrm{d}U = \left(\frac{\partial U}{\partial V}\right)_{\!T}\,\mathrm{d}V + \left(\frac{\partial U}{\partial T}\right)_{\!V}\,\mathrm{d}T \tag{1}$$

The first partial derivative is the internal pressure, which we are trying to determine.

The other thing that we know is that (assuming only expansion work and no net reaction)

$$\mathrm{d}U = T\,\mathrm{d}S - p\,\mathrm{d}V \tag{2}$$

The idea here is that we want to change the term involving $\mathrm{d}S$ into a combination of $\mathrm{d}T$ and $\mathrm{d}V$, so that we can compare coefficients with equation $(1)$. So, we just express $S$ as a function of $T$ and $V$:

$$\mathrm{d}S = \left(\frac{\partial S}{\partial V}\right)_{\!T}\,\mathrm{d}V + \left(\frac{\partial S}{\partial T}\right)_{\!V}\,\mathrm{d}T \tag{3}$$

Substituting $(3)$ into $(2)$

$$\begin{align}
\mathrm{d}U &= T\left[\left(\frac{\partial S}{\partial V}\right)_{\!T}\,\mathrm{d}V + \left(\frac{\partial S}{\partial T}\right)_{\!V}\,\mathrm{d}T\right] - p\,\mathrm{d}V \tag{4} \\
&= \left[T\left(\frac{\partial S}{\partial V}\right)_{\!T} - p\right]\mathrm{d}V + \left[T\left(\frac{\partial S}{\partial T}\right)_{\!V}\right]\mathrm{d}T \tag{5}
\end{align}$$

Comparing the coefficient of the $\mathrm{d}V$ term with $(1)$ we find that1

$$\left(\frac{\partial U}{\partial V}\right)_{\!T} = T\left(\frac{\partial S}{\partial V}\right)_{\!T} - p \tag{6}$$

Equation $(6)$ is entirely general and applies to all substances, not just ideal gases. Now, we turn to the Helmholtz free energy:

$$\mathrm{d}A = -S\,\mathrm{d}T - p\,\mathrm{d}V \tag{7}$$

From this we obtain the Maxwell relation

$$\left(\frac{\partial S}{\partial V}\right)_{\!T} = \left(\frac{\partial p}{\partial T}\right)_{\!V} \tag{8}$$

For an ideal gas $p = nRT/V$, so

$$\left(\frac{\partial p}{\partial T}\right)_{\!V} = \frac{nR}{V} \tag{9}$$

Finally, substituting $(8)$ and $(9)$ into $(6)$ gives

$$\begin{align}
\left(\frac{\partial U}{\partial V}\right)_{\!T} &= T\left(\frac{nR}{V}\right) - p \tag{10} \\
&= \frac{nRT}{V} - p \tag{11} \\
&= p - p \tag{12} \\
&= 0 \tag{13}
\end{align}$$

as desired.

Notes

1 Some texts may write that you can obtain $(6)$ from $(2)$ simply by "dividing" by $\mathrm{d}V$ and imposing conditions of constant temperature:

$$\begin{align}
\mathrm{d}U &= T\,\mathrm{d}S - p\,\mathrm{d}V & & \tag{2} \\
\frac{\mathrm{d}U}{\mathrm{d}V} &= T\frac{\mathrm{d}S}{\mathrm{d}V} - p\frac{\mathrm{d}V}{\mathrm{d}V} & &\text{(dividing by }\mathrm{d}V\text{)} \tag{14} \\
\left(\frac{\partial U}{\partial V}\right)_{\!T} &= T\left(\frac{\partial S}{\partial V}\right)_{\!T} - p\left(\frac{\partial V}{\partial V}\right)_{\!T} & &\text{(constant }T\text{)} \tag{15} \\
&= T\left(\frac{\partial S}{\partial V}\right)_{\!T} - p & &\text{(derivative of }V\text{ with respect to itself is }1\text{)} \tag{6}
\end{align}$$

This gives the correct result, and in general this "shortcut" works, but it is mathematically not rigorous.


What is the standard potential of carbon?

To look up the position of carbon in the electrochemical series, you would have to guess what species carbon is oxidised to. This is most likely carbon dioxide; you would then have to look up the value of $E^0$ for the $\ce{CO2}/\ce{C}$ couple in a textbook or some other data source. The standard reduction potential for the $\ce{CO2}/\ce{C}$ couple is given in Shriver & Atkins 5th ed as +0.207 V. Loosely speaking, this means that carbon is harder to oxidise than hydrogen, or in other words, carbon is below hydrogen in the electrochemical series (if you take Li to be at the top).

Just a note: The standard reduction potential for the $\ce{H+} / \ce{H2}$ half-cell is 0 V solely because we decided to define it as such.


How to use this salt analysis chart to determine the cations present in a given solution?

For this question rather than reading the question and trying to understand what is going on all in one go, I suggest just work your way through the flow chart and refer to the question as you go along so that you don't overload your brain with excess information and confuse yourself.
So starting from the top, it asks if a precipitate will form if $\ce{HCl}$ is added. Looking from the information in the question, there is no precipitate formed. Therefore we can rule out $\ce{Pb^2+}$ being in the solution. Now following the steps in the flow chart, it asks if a precipitate forms if sulfuric acid is added to it. Looking from the data in the question, it says that a white precipitate is formed with sulfate. Therefore, we follow the next step which asks if a precipitate is formed when $\ce{NaF}$ is added. In the question it states that a white precipitate forms with fluoride. Therefore we can conclude that $\ce{Ca^2+}$ ions are in the solution.
However, if we look again into the question, it states that a blue precipitate also forms when mixed with hydroxide ions. Therefore, if we look back to the flow chart regarding the section on testing with $\ce{NaOH}$, it can be seen that traces of $\ce{Cu^2+}$ ions also exist in the solution. Therefore, both $\ce{Ca^2+}$ and $\ce{Cu^2+}$ cations are in the solution.
As you can see, once you know the general method, it is really quite simple and self-explanatory.


Where does the potential energy of capillary action come from?

Where does this potential energy of capillary action come from, or is this an exception or something?

First, the laws of thermodynamics still apply. Second, you have to account for the energy balance when you go from dry to wet on the one hand, and continuous flow on the other hand.

Initial wetting of the paper towel or glass tube: Quoting from one of the comments, "the increase in gravitational potential energy is countered by a reduction in the chemical potential of the substances due to a change in the areas of mutual contact". That explains why the capillary fills up in the first place, and in this case it is fine to say "water pulls itself up".
Continuous flow. If you are asking about a tree, where there is continual flow, you are not increasing the water-capillary contact area anymore, it remains constant. The energy has to come from somewhere else.
It takes some energy to remove a water molecule from the top of the
capillary because as you do, you are pulling all the water molecules
up a tiny bit. As explained in the excellent video by Veritasium, water can have negative pressure in the capillary, so trees can grow higher than atmospheric pressure might suggest (thanks, A.K., for the pointer to the video). This is not a problem for the tree because the top of
the tree is the site of photosynthesis, so there is free energy
available to do the "pulling". So you can think of the water-filled
capillary serving both in the role of the rope and the bucket, if
you compare it to fetching water from a deep well. In cases where the water simply evaporates, the energy comes from cooling down the surrounding as the water molecule pulls up another water molecule and evaporates. In the case of continuous flow, water does not "pull itself up", something on the top (or bottom) has to provide the energy. For example, if you put a towel in a bucket of water with the towel reaching over the rim a bit, the towel will get wet. However, if the end of the towel outside of the bucket reaches below the water level, it will continuously siphon water out of the bucket, with water dripping from the towel.

As an aside, it does not take much work to lift a water molecule up a tree (1.8 J/mol for a 10 m tree). Hydrolysis of a single ATP affords about 50 kJ/mol under physiological conditions, so it is sufficient to get 25,000 water molecules up the tree.Gravity is weak compared to electromagnetic interactions, which dominate in chemistry.


Yes, the cis isomer on the right should have the methyl group on C2 in the axial (up) position.  The two rings you've drawn at the bottom of your post are equivalent, they can be interconverted by a 180 degree rotation about an axis that bisects the C1-C2 bond and the C4-C5 bond.


The alkene being reacted with hydrocyanic acid here goes by benzylideneacetone, or 4-phenylbut-3-en-2-one in the language of the IUPACians. But you shouldn't view this molecule as an alkene, you should view it as an "enone", or $\alpha,\beta$-unsaturated carbonyl.

Enones are notable for having "regioselective" reactions where nucleophiles add to either the beta-carbon (via 1,4-addition, vinylogy chemistry) or the carbonyl carbon (via 1,2-addition, standard carbonyl chemistry) regions of the molecule.

In your question, you see we have a 1,4-addition of $\ce{H+}$ and $\ce{CN-}$. (After cyanide attacks the 4-position, the proton adds to O, the "1-position", and the enol intermediate then tautomerizes to a ketone.)

The choice of solvent can have some effect on influencing whether 1,2- or 1,4- addition dominates. Aprotic solvents can encourage 1,4-addition (conjugate addition) to be favored over 1,2-addition (carbonyl addition) (Source).

Your ethanol is a polar, protic solvent because the hydroxyl is capable of hydrogen bonding or donating the proton (its Bronsted-Lowry acidity is about that of water, pKa 15.7 vs 16). DMSO is a polar, aprotic solvent because it does not contain hydrogens that can H-bond nor are (appreciably) acidic. Use of DMSO can encourage 1,4-addition (source).

Simply changing the protic-aprotic nature of the solvent may not always be enough to effectively toggle the dominant regiochemistry between 1,2- and 1,4- addition motifs. It may depend on the other reaction conditions (such as temperature), the structural qualities of the particular enone substrate at hand, and the nature of the nucleophile. In the Hard-Soft Acid-Base Model, "hard" Lewis bases tend to add directly to the carbonyl (1,2) whereas "soft" Lewis bases tend to perform conjugate addition (1,4). For nucleophiles (Lewis bases) not on either extreme end of the hard-soft spectrum, changing the solvent type can sometimes be enough to change the regioselectivity.

I recommend you ask your professor whether there is a rule of thumb he wants you to use to determine the solvent's effect (such as "polar aprotic always gives 1,4"), or whether you should memorize specific examples (because there are many exceptions).


2 electrons with the same spin in the same orbital

It is impossible. Having two electrons with the same spin in the same orbital is a violation of the Pauli exclusion principle.


Do both electrons in a dative bond come from the same atom?

Your textbook’s definition is inconsistent for exactly the reasons you point out. I am going to use slightly different examples than you did but they work equally well.

A prime example for dative bonding is $\ce{H3N\bond{->}BH3}$, amminborane or ‘inorganic ethane’. The latter name because it is isoelectronic to $\ce{H3C-CH3}$, (organic) ethane. Typically, amminborane is generated by ammonia reacting with $\ce{B2H6}$ — or at least it seems very plausible to create it in that way. Ethane can be created in two different ways:

By the recombination of two methyl radicals:
$$\ce{H3C. + .CH3 -> H3C-CH3}\tag{1}$$
By the attack of a formal methide anion with a formal methanium cation:
$$\ce{Li-CH3 + H3C-OTf -> H3C-CH3 + LiOTf}\tag{2}$$

(The second version is absolutely equivalent to your $\mathrm{S_N2}$ reaction to give isopropanol.)

Going by your textbook’s rules, the first reaction might generate a normal covalent bond, but the second should give a dative bond, since both electrons of the $\ce{C-C}$ bond came from methyllithium and none from methyl triflate. And, of course, we cannot distinguish between these electrons and they are delocalised all across the molecule — that goes for both ethene and amminborane.

A much better definition of dative versus normal covalent bonds does not consider their generation but rather what happens when the bond dissociation energy is added, i.e. their dissociation. In amminborane’s case, the dissociation happens at a rather low energy and will yield ammonia and borane again, so the electrons ‘end up where they were before.’ In more technical terms, the compound dissociates heterolyticly. (Equation $(3)$)

$$\ce{H3N\bond{->}BH3 -> H3N + BH3}\tag{3}$$

Ethane, on the other hand will dissociate homolyticly upon being fed with the (much higher) $\ce{C-C}$ bond dissociation energy. It will reform the two methyl radicals used to generate it in equation $(1)$. (Equation $(4)$)

$$\ce{H3C-CH3 -> H3C. + .CH3}\tag{4}$$


Why can some redox half-reactions be combined and some cannot?

In general, you cannot simply subtract electrode potentials like this to find the potential of another half-reaction. They have to be weighted by $n$ (the stoichiometric coefficient of $\ce{e-}$). In cases where your method works, it is only because $n$ is coincidentally the same for both half-reactions you are combining.

The foolproof way is always to convert electrode potentials to Gibbs free energy changes:
$$\Delta G^\circ = -nFE^\circ$$

So $\Delta G^\circ$ for the half-reaction $\ce{MnO4- + 8H+ + 5e- -> Mn^2+ + 4H2O}$ is

$$(-5)(96485 \text{ C mol}^{-1})(1.51 \text{ V}) = -728.4 \text{ kJ mol}^{-1}$$

and $\Delta G^\circ$ for the half-reaction $\ce{MnO4- + 4H+ + 3e- -> MnO2 + 2H2O}$ is

$$(-3)(96485 \text{ C mol}^{-1})(1.7 \text{ V}) = -492.1 \text{ kJ mol}^{-1}$$

Subtracting the second equation from the first, you will find that $\Delta G^\circ$ for the half-reaction of interest ($\ce{MnO2 + 4H+ + 2e- -> Mn2+ + 2H2O}$) is

$$(-728.4) - (-492.1) = -236.3 \text{ kJ mol}^{-1}$$

which gives:

$$E^\circ = -\frac{-236.3 \text{ kJ mol}^{-1}}{(2)(96485 \text{ C mol}^{-1})} = 1.22 \text{ V.}$$

I presume some rounding errors led to the slight discrepancy between my answer and the answer you want. You may leave out the Faraday constant in the calculations once you are more familiar with them; basically that means you work with the quantity $nE^\circ$ instead of $\Delta G^\circ$. The answer should be the same since they are directly proportional.

For more information you can refer to the redox chapters in any inorganic chemistry textbook (Shriver/Atkins, Housecroft, etc.). These calculations are dealt with under the sections discussing Latimer and Frost diagrams.


Why left and right circularly polarized light have different index of refraction in chiral media?

The answer to this is really a combination of the two answers given in the question you linked.

Namely, chiral molecules are unique in that they make a distinction between left and right circularly polarized light.

As porphyrin pointed out, the index of refraction is dependent on the ability of a dipole to be induced in a molecule by electromagnetic radiation. Linearly polarized light is itself achiral, thus any given wavelength of linearly polarized light has only one magnitude of dipole it will induce in a molecule, all else being the same. Circularly polarized light, on the other hand, is itself chiral. Thus, when we bring two chiral objects together, the molecule and the circularly polarized light, there is no reason at all to suspect the dipole induced in the molecule should be the same in both cases. The symmetry necessary to make such a claim simply is not present.

Thus, if you imagined tabulating the index of refraction for some molecule as a function of wavelength using circularly polarized light, you would need to have two columns. One would be for right-polarization and the other for left-polarization.

Also, because it is relevant to the answer, I will reiterate what Ron was saying in the comments. The photon is an inherently quantum object. It is a boson and thus has two possible values of spin angular momentum, $\pm\hbar$. These two values correspond to two allowable states the photon can be in, one which is polarized to the right and the other which is polarized to the left. That is, the electric field vector spins in a circle one way or the other. There also allowable solutions to the state of a photon which are some linear combination of these two polarization states. A 50/50 linear combination gives linearly polarized light. Thus, it seemed like you might have been picturing two photons whose polarizations were cancelling each other out. This is incorrect. Rather, you have to think of a single photon in a superposition of two polarization states.


Sure! A nickel-iron battery uses the oxidation of iron at the negative plate to produce about half the voltage. Though invented by Jungner, it's often called an Edison battery. See http://en.wikipedia.org/wiki/Nickel%E2%80%93iron_battery.

There are very low power semiconductor devices which could be powered by an iron battery. You could make three or four small cells from iron nails and a more electronegative element, e.g. lead, to power a digital watch; even without optimization to remove layers of hydrogen and rust, it should operate for months.

For some other low power ideas, see https://www.enocean.com/en/energy-harvesting/.


I'd suggest to start with writing down the processes occurring during the electrolysis:

\begin{align}
&\text{Cathode (–):} &\ce{2 Ag+ + 2 e- &-> Ag^0} \\
&\text{Anode (+):} &\ce{Zn^0 &-> Zn^2+ + 2e} \\
\hline
&\text{Net} &\ce{2 Ag+ +Zn^0 &-> 2 Ag^0 + Zn^2+}
\end{align}

Applying the unified equation for Faraday's laws of electrolysis:

$$m = \frac{QM}{zF}$$

for each cell, and taking into account that in series circuits current is the same for every element ($I = \text{const}$):

$$\frac{Q}{F} = \frac{z_i m_i}{M_i} = \text{const}$$

$$\frac{z(\ce{Ag}) \cdot m(\ce{Ag})}{M(\ce{Ag})} = \frac{z(\ce{Zn}) \cdot m(\ce{Zn})}{M(\ce{Zn})}$$

The valency numbers of ions are 1 and 2 for silver and zinc, correspondingly. The last thing to do is to express and calculate $m(\ce{Ag})$.

   $$m(\ce{Ag}) = \frac{z(\ce{Zn}) \cdot M(\ce{Ag})}{z(\ce{Ag}) \cdot M(\ce{Zn})} \cdot m(\ce{Zn}) = \frac{2 \cdot \pu{107.87 g mol-1}}{1 \cdot \pu{65.38 g mol-1}} \cdot \pu{7.5 g} = \pu{24.7 g}$$


How does increasing the Hydrogen Concentration of CuSO4 (electrolyte) affect electroplating of copper

In electrolysis, $Cu^{2+}$ is reduced in two steps, first to $Cu^+$ and later on to $ Cu$. Of course $Cu^+$ is automatically disproportioned into $Cu$ and $Cu^{2+}$ in acidic conditions. But in neutral solutions, $Cu^+$ may react with $H_2O$, forming an unwanted red precipitate of $Cu_2O$. This will affect electroplating of copper.


Try writing out the K expression. Remember that in general,

${K = \frac{[product]}{[reactant]}}$

In your case, ${K = \frac{[C]^2}{[A][B]}}$

Or in other words,

$K = \frac{[product]^2}{[reacant][reactant]} = \frac{[product]^2}{[reacant]^2}$

I combined the $[reactant]$ terms because the equation you have, there is 1:1 ratio of both reactants.

We can take the square root of both sides:

$\sqrt{K} = \sqrt\frac{[product]^2}{[reactant]^2} = \frac{[product]}{[reactant]}$

So now just plug in your numbers to determine how far to the right the equilibrium lies.

$\sqrt{K} = \sqrt{4.2} = \frac{[product]}{[reactant]} = 2.0$

Therefore (upon consideration of reaction stoichiometry), there is a 2:1:1 ratio of products to reactants at equilibrium.

We can also plug in the Q (reaction quotient) value; remember that K is just a special Q value (K is the Q value when the system is at equilibrium):

$\sqrt{Q} = \sqrt{1.0} = \frac{[product]}{[reactant]} = 1.0$

The Q value indicates that the system currently lies away from the equilibrium mix of a 2:1:1 ratio between products and reactant gases. At Q = 1.0 we have a 1:1:1 mix of A, B, and C.


The question is indeed poorly worded. What the question seems to be saying is that there are $12.04\times10^{23}$ molecules of hydrogen gas/dihydrogen ($\ce{H2}$).

I'm also noting that the correct answer ends with "... $\pu{44.8L}$ of $\ce{H2O}$ present", while there was absolutely no mention of $\ce{H2O}$. The question is quite misleading.


What is meant by fractional oxidation state?

Nothing too much, really. When you have a compound like sodium tetrathionate $\ce{Na_2S_4O_6}$, if you calculate the oxidation state of sulfur, you get $\frac{5}{2}$. This means that the AVERAGE oxigation state of sulfur is $\frac{5}{2}$. In fact, the sulfurs in the compound are not equal, you have two of them with an oxidation state of $+5$, and two of them with $0$. But when you balance redox equations, you do not really care about this, and you can use the average value of $2.5$ to find the coefficients.


Why does gold react with cyanide and sodium hydroxide

Flotation and treatment of gold with $\ce{NaCN}$ in the presence of air are two distictive steps!

Flotation yields fine particles of gold (and pyrite) at the surface of a flotation tank, attached to foamy bubbles while the rest of the minerals sink to the bottom of the tank.

Flotation does not rely on differences in specific density, but on hydrophobicity!

In the cyanide treatment, the following process - similar to what Uncle Al described for the seawater case in his answer - occurs:

Solid gold is oxidized, the cations are bound by cyanide anions and form a water-soluble complex:

$$\ce{4Au + 8NaCN + O2 + H2O -> 4Na[Au(CN)2] + 4NaOH}$$

In order to reobtain solid gold, the $\ce{Au(I)}$ in the cyano complex $\ce{[Au(CN)2]^-}$ has to be reduced to $\ce{Au(0)}$ again. This can be done either by adding zinc dust, according to

$$\ce{2Na[Au(CN)2] + Zn -> Na2[Zn(CN)4] + 2Au}$$

or electrochemically.


Migratory aptitude in Baeyer–Villiger reaction

In the following image you can see the charge distribution in the transition state of the Baeyer–Villiger reaction (image taken from Clayden Organic Chemistry):

As you can see, there is partial positive charge at the migrating C atom, therefore a group will migrate easier if it contains cation stabilizing moieties. A benzene ring significantly stabilizes positive charge in its proximity through delocalization.


They are all constitutional isomers.  If you are counting enols (your friends #3, it is the enol of your #1), then include the enol of acetone as an additional constitutional isomer.


It's hard to say what exactly you would end up with, but it probably wouldn't much resemble beer. The trick with electrochemistry is that a lot of things are electroactive and what reactions actually occur in a given reaction are not always easy to predict and depend on electrodes, solution composition, pH, applied potential, etc. Depending on your electrodes, ethanol may be significantly more easily oxidized than water. A quick literature search shows that platinum and gold are both decent catalysts for ethanol oxidation (platinum is commonly used in fuel cells for this reason), but even with other materials, I can't imagine any conditions in which one could easily enrich an ethanol/water solution in this way. Many organic compounds, probably including some that impart flavour, are likely to be oxidized as well, even if the ethanol wasn't, so it would certainly affect the taste.

Additionally, the electrolysis of water requires a large amount of energy, far more than distillation, so even if it would work, it would be an expensive way of doing it.


Does tetraoxidovanadium (VO4+) ion exist?

Since the comment by Ivan pointing to a possible misprint got four votes and no answer yet, I'll post it as an answer. In short:  this is a misprint in the book, and the book probably meant $\ce{VO2+}$.

Here's a citation to back it up: Bertrand, G. L.; Stapleton, G. W.; Wulff, C. A.; Hepler, L. G. Thermochemistry of Aqueous Pervanadyl and Vanadyl Ions. Inorg. Chem. 1966, 5 (7), 1283-1284. DOI: 10.1021/ic50041a048.

  ...it does appear that below $\mathrm{pH} \approx2$ the principal species are the pervanadyl and vanadyl ions represented by $\ce{VO2+(aq)}$ and $\ce{VO^2+(aq)}$ (respectively).

Also worth noting that the formation of the pervanadyl ion is more favourable with a free energy of formation of $\pu{-140.3kcal/mol}$, as opposed to $\pu{-106.7kcal/mole}$ for the vanadyl ion (as mentioned in the source above)


Why some cloths become rough after drying?

I believe it is down to water relaxing the fibres in the material. Upon drying in one position the fibres become temporarily fixed in that position. I have also read that the detergents used in washing can cause clothes to dry stiffer. A little experiment such as soaking your clothes in hot water should kill bacteria and then see if they dry stiff. This would give an idea as to whether or not its down to the detergent or just the water.

Interestingly, a recent study suggests that fabric softeners reduce the wettability of materials. So if you use fabric softener on your towels, over time they will become less effective at drying (being towels).


It's oxo-biodegradation* with an asterisk. PET polymers are polyester meaning that the monomer units are link by ester functional groups. This ester group can be hydrolyzed in the presence of water to for the alcohol and acid that formed the ester. For PET the alcohol is ethylene glycol which is readily digestible by bacteria and the acid is terephthalic acid which is not as redily digestable but still digestable.

It is for this same reason that poly(glycolic acid) which is used in dissolvable stitches and poly(lactic acid) which is used in food containers (and 3D printing) are also biodegradable.


hydroxychloroquine (HCQ) Synthesis: Which mechanism?

The 2 and 4 positions of quinolines are similar to the 2 and 4 positions of pyridine in reactivity. 2 and 4 haloquinolines/pyridines are highly reactive to nucleophiles by SNAr. Indeed 4-Cl-pyridine has to be handled as the hydrochloride because of its propensity to self condense.

The 7 position, by contrast, is unactivated and similar in reactivity to benzenes, so the 7-Cl is not going to undergo SNAr without specialist catalysis.


Why do Grignard reagents or Organolithium compounds not displace halides as easily as acetylide ions do?

An acetylide ion proceeds in a nucleophilic substitution pathway (SN2 in the picture due to primary halide). The substitution involves the attack of a highly nucleophilic sp-carbon on a highly electrophilic primary carbon. The enthalpy of the C-C bond formation is relatively low due the pKa of the nucleophile being ~ 25. The ion is treated as a "soft" nucleophile that prefers to substitute alkyl halides instead of acting as a base and conducting an elimination reaction.

A grignard reagent is essentially a highly polarized bond between Carbon and the Bromide(Halide)/Magnesium complex. The electronegative difference between Carbon and Magnesium is 1.3 (http://www.thecatalyst.org/electabl.html <-- Source of values). The Carbon atom in a Gringnard is treated as a carbonanion. This carbonanion characterization is highly unstable in nature and treated as a "hard" nucleophile with pKa of ~ 50. The highly basic character of a Grignard reagent often results in an elimination reaction or no reaction at all. The transition state to substitute the alkyl halide is less stable than the Magnesium/Bromide(Halide) complex. This is due to a ligation formation between the solvent and the Magnesium atom. The solvent (ex. diethyl ether) shares non-bonding electrons with the cation and creates a stable electronic atmosphere with a lower energy than a possible C-C bond with the alkyl halide.

Substitution isn't preferred with a hard nucleophile and therefore, the acetylide ion is more likely to preform a nucleophilic substitution reaction with an alkyl halide. Note that the electronegative difference between Carbon and Magnesium supersedes the hybridization of the sp or "triple" bond complex. The bond is so high in energy that the nucleophilic character of the grignard far surpasses the electronegative character of the sp-hybridization of the acetylide ion.


What is the effect of the temperature on the no of moles in an equilibrium system?

Well, your equation contains the same number of moles either side, so temperature does not matter.

But if the equation had different moles on either side, it should be given if the reaction is endothermic or exothermic.

For eg- if a reaction is exothermic and the temperature is increased, then according to Le Chatelier's principle, the reaction should go in the backward direction (To compensate for the increase in temperature).
So, if the reactant side has more moles than product side, the number of moles will be increased. You can apply similar logic if the reaction is endothermic.


Which is better: iodine number or bromine number?

Bromine number (and Bromine Index) are based on theoretically good chemistry; however, that is where their virtues end. If these tests are run according to the current ASTM procedures, you are virtually guaranteed to get an incomplete reaction and low (perhaps as much as 50%) results. When properly applied, off the shelf equipment can result in exceptionally good accuracy and precision, but it does require a deviation from ASTM-accepted technology.


The sign of enthalpy of formation of magnesium oxide

I guess you've been answered in the comments already, but anyway. You are correct. In thermodynamics, enthalpy is the energy supplied to the system at constant pressure. If a reaction is accompanied by the release of heat, i.e. if it is exothermic, its $\Delta H$ is negative. If $x$ kJ is released into the calorimeter upon dissolving 1 mol of substance, resulting in the rise of calorimeter temperature, then $\Delta H_{diss} = -x$ kJ/mol.

As for the heat, depends on the type of the calorimeter, i.e. on the direction in which the heat flows. For conventional "coffee cup" calorimeter, where $Q=mC\Delta T$, $Q$ is the energy released, so $Q = -\Delta H$. For some power-compensated calorimeter, where you vary the heater power so as to maintain the temperature, $Q$ would be equal to the energy supplied. Since you have to supply excess energy ($Q>0$) to offset the endothermic reaction ($\Delta H>0$), $Q = \Delta H$.

I've heard that a long time ago, at least in some countries, there were two systems of signs for enthalpies, so-called "thermodynamic" and "thermochemical," with reversed signs in the latter. However, I haven't seen "thermochemical" one in real life, and it seems to be long since abandoned. It is not even mentioned in modern textbooks on thermodynamics. So only the thermodynamic sign system, where $\Delta H>0$ for endothermic processes, remains.


Is it possible to determine the age of a building by just looking at its window?

This is but an urban legend. Glass does not get thicker at the bottom. (Sure, glass is fluid to some tiny extent, and it can be demonstrated, but not in this way.) It is just that the glass panes were initially manufactured with uneven thickness. Then the people who stuck the glass into the frames would orient the thicker side down, maybe because it is more stable this way, or because it feels right, or just to mess with us. Sometimes they didn't pay attention and turned it randomly.

Supporting links: 1, 2.


The Ionic Strength is calculated as:

$$I=\dfrac{1}{2}\cdot\sum{c_iz_i^2},$$ where where $c_i$ is the molar concentration of ion i ($M,\,mol/L$), $z_i$ is the charge number of that ion, and the sum is taken over all ions in the solution.

This is your HEPES molecule ($C_8H_{18}N_2O_4S$):

Only one proton at $SO_3H$ group can dissociate, as far as I see. So in your solution you will have two different types of ions with charge equal to 1 (assuming that molecule will dissociates 100%, and than:

$I=0.5\cdot(0.1M\cdot1^2+0.1M\cdot(-1)^2)=0.1M$

So your calculations are wrong.

But also be aware that this molecule is Zwitterion!
READ THIS Do zwitterions contribute to the ionic strength of a solution?
The resume of the articles:

"Hence, electrostatic effects due to zwitterions are important, even if the zwitterions do
not contribute directly to ionic strength."

My calculations are primitive. They are also far from the truth, but they show how the principle works for you.


Enzyme Kinetics - Given Km find substrate concentration at a certain velocity

You can solve for [S]. There are two common ways to write the equation, like you have it:
$$v_0 = k_\text{cat} [E_0] \frac{[S]}{K_\text{m} + [S]}$$

or after dividing both numerator and denominator by [S]:

$$v_0 = k_\text{cat} [E_0] \frac{1}{\frac{K_\text{m}}{[S]} + 1}$$

In this form, it is easier to see that [S] occurs once and you can solve for it.

Get [S] out of the denominator of the "big" fraction and isolate the "little" fraction:

$$\frac{K_\text{m}}{[S]} + 1 =  \frac{k_\text{cat} [E_0]}{v_0}$$

$$\frac{K_\text{m}}{[S]} = \frac{k_\text{cat} [E_0]}{v_0} - 1$$

Finally, take the reciprocal and multiply by $K_\text{m}$:

$${[S]} = \frac{K_\text{m}}{\frac{k_\text{cat} [E_0]}{v_0} - 1}$$