I can count four reasons that say saliva pH isn't the same among humans:

As people age, the composition of saliva changes in their body; this is easily explainable with the fact that the body isn't performing the actions the same way as you grow up or become old.
People have different eating habits; this will put the body in impression of producing especial kinds of enzymes and reduce\increase the production of certain factors in the saliva.
People aren't the same when it gets to taking care of the insides of their mouth. You can be brushing your teeth twice as much as a close friend of yours does.
Human digestive system isn't always performing its duties consistently and without "ups" and "downs" during the day, which has an important effect on enzymes produced.

And a sea of other reasons....

But, if we are to answer this question:

  Saliva is a watery substance located in the mouths of animals, secreted by the salivary glands. Human saliva is 99.5% water, while the other 0.5% consists of electrolytes, mucus, glycoproteins, enzymes, and antibacterial compounds such as secretory IgA and lysozyme. Saliva, wikipedia

I'll also say that according to here, a healthy range of pH of saliva varies between 6.5 and 7.5 during the day, but is optimally between 6.5 and 6.8.

Source

Thus, saliva is a very weak acid, or sometimes a very weak base. Considering something like that an acid or a base is like saying that the water we drink is a base and the water of the raindrops that fall from the sky of a peaceful country is acidic.

Saliva is really important for it's being a buffer.


MDLCT (Connection Table) Format specification

I know this is an old question but like to provide some detail since this does not seem to be documented officially anywhere.
The MDL connection table format used in MOL files is described in the CT File Formats document referenced in the first answer. This is simple ASCII text with hard linebreaks (LF or CR+LF, depending on platform).

But there is also a special clipboard format MDL_CT created by ISIS/Draw, Accelrys/Draw etc. on Windows. This is not simply a MOL file put as text onto the clipboard. Instead each line of the molfile data is prefixed by a byte holding the number of characters, followed by the ASCII (one byte) characters of the line. The lines are not terminated by line breaks! This is basically the good old TurboPascal string format.

In addition to a single molecule the MDL_CT clipboard format also supports reactions. In that case the data comes from a RXN file, which is converted to the clipboard format the same way a molfile is: prefix each line with a length byte, remove hard linebreaks.


Le Chatelier's principle: what are the effects of decreasing volume?

I think a good place to start is with the idea that that $K_{eq}$ will not change in value unless the temperature changes (see the second answer here for a good explanation). With this in mind we can say that statement III is false because the temperature is not being changed.

Statement II is shown to be false via Le Chatelier's principle, as the equilibrium concentration of reactants should decrease rather than increase due to the decrease in volume because the products have fewer gaseous moles and so would exert less pressure on the container.

Statement I can be shown to be true via Le Chatelier's principle by the same logic as Statement II: for the reaction to shift towards the product, the rate of formation of product must increase.

Therefore, only Statement I is true.

You might wonder how $K_{eq}$ can stay constant, but the rate of forward reaction can increase. Its important to remember that $K_{eq}$ is the ratio of the forward and reverse rate constants, not the forward and reverse rates. The rate constants have no dependence on volume (or pressure) so they won't change.


Infinite Increase in Entropy when Energy added to Absolute Zero

The textbook is referring to the entropy change of the system.  While the textbook is correct that absolute zero can never be attained, its statement that the entropy change is infinite is wrong. The authors' rationale for thinking it is infinite likely stems from a misinterpretation of the definition of entropy change:
$dS = \frac{\text{đ}q_{rev}}{T}$
where $\text{đ}q_{rev}$ is the reversible heat flow into the system.
Let's start with a system at $0 K$, and then warm it to a temperature $T = T'$.  Then:
$\Delta S =\int_{0}^{T'} dS= \int_{0}^{T'}\frac{\text{đ}q_{rev}}{T}$
Let's consider what happens when we flow the first, infinitesimal amount of heat, $\text{đ}q$, into the system.  And to make the calculation easy, let's assume the heat flow is reversible (it doesn't have to be; if it weren't, we'd just need to find a reversible process that gets us to the same final state).  As soon as we do this, the temperature is no longer zero! And once the temperature rises above zero, the singularity disappears and we no longer have a concern about infinite entropy change.
But, you may ask, what about the mathematics right at the very beginning, when the temperature is indeed zero.  Here, to properly calculate the entropy change, we need to use a limit, recognizing that both $T$ and $\text{đ}q$ are approaching zero.  The easiest way to understand this is to consider the Debye $T^3$ law, which models the constant-volume heat capacity of solids as they approach absolute zero:
$C_v = C_v(T) = k T^3$, where $k$ is a constant, and where I've written "$C_v(T)$" as an explicit indicator that $C_v$ is temperature-dependent.
Since, for a constant-volume reversible process, $\text{đ}q_{rev} = C_v(T) dT$, we have:
$\Delta S= \int_{0}^{T'}\frac{\text{đ}q_{rev}}{T} = \int_{0}^{T'}\frac{C_v(T)}{T} dT = \int_{0}^{T'}\frac{k T^3}{T} dT=\int_{0}^{T'}k T^2 dT$
I.e., in evaluating what happens to $dS = \frac{\text{đ}q_{rev}}{T}$ as $T \rightarrow  0$, it is necessary to consider what is happening to both the numerator and the denominator. The textbook authors' mistake was in not understanding this.  Clearly, once we find a reasonable functional form for $\text{đ}q$ in the limit as $T \rightarrow  0$, the singularity disappears.
Think of it this way (borrowing from one of my comments): At constant volume, $dS=\frac{C_v}{T} dT$. As you approach absolute zero, it's not just $T$ that's going to $0$; $C_v$ is going to $0$ as well. And since $C_v$ is going to $0$ faster than $T$ is going to $0$, $\frac{C_v}{T}$ goes to $0$ rather than infinity (for the same reason that $\frac{x^3}{x}$ goes to $0$ rather than infinity as $x$ goes to $0$). I.e., the mistake would be in assuming that $C_v$  is a constant; it's not.
Here's another way we can understand that the textbook's statement is wrong:  Thermodynamics allows us, in principle, to determine absolute entropies for any substance.  The absolute entropy is given by integrating the entropy change from absolute zero to whatever temperature the substance is at:
$\text {Absolute entropy at } T' \equiv S(T') =\Delta S_{0 \rightarrow T'}= \int_{0}^{T'}\frac{\text{đ}q_{rev}}{T}$
If the entropy change between $0 K$ and any higher temperature were infinite, the absolute entropies of all real substances would likewise also be infinite! [Unless the substance were at absolute zero, which is unatainable.]
Here's a link showing how standard molar entropies might be calculated (I don't know if the procedure described here is what is used to determine the official CODATA values):
https://www2.stetson.edu/~wgrubbs/datadriven/entropyaluminumoxide/entropyal2o3wtg.html.
Essentially, it mentions using the Debye extrapolation up to ~$15 K$, and then using a differential scanning calorimeter to determine the reversible heat flow between $15 K$ and $298.15 K$


The similarities and differences between electron transfer and charge transfer

I'm not sure what you are asking but electron 'transfer' is different to 'transport' in the sense that transport ultimately involves many electron transfers events one after the other. Charge transport is the same as electron transport.

There is is difference between electron transfer and charge transfer. Charge transfer is usually associated with charge transfer complexes, such as TTF-TCNQ (tetrathiofulvalene-tetracyanoquinodimethane) but there are very many others. In these complexes the charge transfer from one species to the other occurs as the photon is absorbed so that the lowest excited state is $\ce{D + A + photon -> D+A-}$. This often means that these complexes are highly coloured.

In electron transfer no photon is needed and the process follows Marcus theory whereby a barrier is crossed between $\ce{D + A}$ and $\ce{D+ + A-}$, i.e. this is a thermal process with an activation barrier. The unusual feature of Marcus transfer is the that activation energy has a parabolic dependence on energy thus the rate constant can be slow in both endothermic and very exothermic conditions.


On addition of inorganic coagulants like aluminium sulfate(alum) or Iron salts causes several chemical and physical interaction take place among particles.

This inorganic coagulants neutralize the negative charge on particles. Also metal hydroxide precipitates of the aluminum and iron salts ions begin to form.

This precipitates combine to form larger particles under natural processes such as Brownian motion

Aluminum hydroxides are formed within a fairly narrow pH range, typically: 5.5 to about 7.7. Iron (III) hydroxides can form over a larger pH range including pH levels lower than are effective for alum, typically: 5.0 to 8.5

Source wikipedia


Relative importance of electronegativity and bond length for dipole moment magnitude

I need help understanding, in that trend specifically, as to when the length factor dominates, and when the electronegativity factor does?

There is nothing much to understand here. All you need to understand is that this is basically an experimental observation.

We are trying to use qualitative arguments (higher bond length/more electronegativity) to explain the actually calculated values. When we observed a contradiction in our answer for fluoromethane and chloromethane, we corrected our reasoning, by saying that the bond length dominated over the electronegativity difference in this case. And that is true, because this correction is based on the calculated values only.

So, we first try to predict the order by our own qualitative arguments (logically reasoned, of course). If they work, great! If they don't, we look into the factors that we originally deemed to be minor instead. We calculate their values, and see what effect they had. Then, we justify those minor factors by saying that they dominated over the other in the particular case.

Another such interesting case is the order of the basicity of ammonia derivatives in water. It is $\ce{NEt2>NEt3>NEt1}$ (for the ethyl derivatives) and $\ce{NMe2>NMe1>NMe3}$ (for the methyl derivatives). Here, three factors are operating simultaneously! The solvation effect, the steric hindrance of alkyl groups, and the positive inductive effect of alkyl groups - all either reinforcing or hindering each other. In such edge cases, it becomes extremely tough to predict any order solely based on these factors, and hence, computated values are much preferred.

If you prefer to read the theory of those experiments (be warned it's scary), here is one such paper discussing the detailed calculation of the dipole moments of various fluoromethane derivatives.


What makes trans fats more harmful than saturated ones?

First of all, it's important to distinguish between the types of fatty acids that may form the cell's membrane:

Note that both cis- and trans-fatty acids are unsaturated, the difference being the conformation around the double bond. A trans-fatty acid is more-or-less a straight line, whereas the cis-fatty acid has a noticeable "kink" at the point of the double bond.

Now, the suggested molecular mechanism behind the health risks associated with trans-fatty acids has to do with a property of the cell's membrane called membrane fluidity. This property affects the cell's ability to send and receive signals, and it greatly affects the function of many proteins whose structure and role are dependent on the membrane's state. The membrane has to be fluid enough in order to serve its purpose properly.

(If you don't have time to read a lengthy answer, you can skip to the last paragraph now. Otherwise, keep going...)

One of the main functions of the cell's membrane is to separate the inside of the cell from the outside world, in order to control the cell's internal environment, help it avoid damages from the surrounding environment and deal with harmful substances or unfavorable living conditions. However, the cell needs to communicate with the outside world through different signaling mechanisms which involve sending and receiving chemical signals, as well as adapt itself to different living conditions.
In order to do that, the cell has to have a membrane with a dynamic structure, i.e. one that will be able to adjust itself to the surrounding environment. Evolution decreed that this may be best achieved by a lipid bilayer.
This image shows a portion of a cell membrane, which is made of many fatty acids in a formation that is considered to be the most stable energetically, due to the hydrophobic/hydrophilic interactions of the fatty acids among themselves and with the surrounding water. This image also shows how cholesterol is embedded here and there within the bilayer.
The measure of a membrane's fluidity results from the type of fatty acids that form it and the membrane's cholesterol content.

To put it simply, the more straight fatty acids in the membrane, the more rigid (i.e. less fluid) it will be. The more "interruptions" to the hydrophobic interactions between the fatty acids' tails (by "kinks" in the tails or cholesterol), the more fluid the membrane will be.

The above picture shows two distinct states of the membrane. On the right-hand side the membrane contains mostly straight (i.e. saturated and/or trans) fatty acids, which results in a more rigid and viscous membrane. This state makes it more difficult for the cell to communicate with its surroundings (chemical signals cannot enter the cell easily, proteins in the membrane are less mobile, etc.).
On the left-hand side, the membrane contains more unsaturated fatty acids of the cis conformation. The "kink" that characterizes cis-fatty acids is the main reason this membrane is more fluid - it makes it harder for the fatty acids' tails to stick together due to hydrophobic attraction.
Cholesterol serves the same purpose as those "kinks". by disrupting the attraction between fatty acids' hydrophobic tails. This is a very important role of cholesterol in virtually all of the cells in the body, which makes cholesterol a very important molecule in the body.

To summarize, trans-fatty acids are considered unhealthy because when consumed, they will mostly remain intact (i.e. will not be metabolized; as this form of fatty acids is not found in nature, the body's enzymes have a limited ability to catabolize them), and then they can penetrate cellular membranes, form rigid hydrophobic bonds with the other fatty acids and decrease the membrane's overall fluidity.

See Diet and HD - article from Stanford university.

Other references may be found in most introductory biochemistry texts or online (be sure to read only scientific texts from reliable websites such as the one given above).


Why does the extent of racemisation in an SN1 reaction depend on the solvent used?

Secondary haloalkanes can undergo substitution by both the $S_N 1$ and $S_N 2$ routes. In general these two mechanisms compete with each other. The following diagram is a More-O'Ferrall-Jencks Plot showing the possible mechanism space for a nucleophilic substitution reaction. The reactants are in the upper left. The products are in the lower right. The horizontal axis represents the extent to which the C-LG bond has broken. The vertical axis represents the degree to which the C-Nu bond has formed. The $S_N 1$ pathway is represented by traveling along the perimeter of the plot first to the right and then down. The $S_N 2$ pathway is actually the red diagonal line represented both processes happening simultaneously. There is a third pathway (down and then right), representing nucleophilic attack and then loss of leaving groups. $S_N$ reactions at $sp^2$ centers often follow this route.

Very few reactions stick completely to one of these three extreme paths. For $sp^3$ centers, the mechanism sticks to the upper right of the diagonal depending on a variety of factors like the polarity of the solvent, the nucleophilicity of the nucleophile, the strength of the C-LG bond, steric hindrance at the subsitution center, and how stable the carbocation would be if it formed. The arrows on the plot above suggest the directions that the transition state would move if the nucleophilic attack step became more favored (stronger nucleophile, less polar solvent, less steric hindrance). The end position is not possible for this particular reaction, but such is the result of reusing and image from somewhere else.

In your case, you can tell how close the actual reaction pathway sticks to one of the idealized pathways. Or, you can think of these numbers as the relative likelihood that a substrate will undergo solvolysis by $S_N 1$ or $S_N 2$ mechanisms in that solvent.

In your case, methanol $\ce{CH3OH}$ is less polar than water $\ce{H2O}$, so the $S_N 1$ pathway is less favorable in methanol than in water. However, the $S_N 1$ pathway in methanol is still way more favorable than the $S_N 2$ pathway in methanol (by 73:27).


Molar absorption coefficient - why are the molar absorptivities defined to be the same?

Actually that's not the correct definition of the isosbestic point. From IUPAC goldbook:

Isosbestic point - Wavelength, wavenumber or frequency at which the total absorbance of a sample does not change during a chemical reaction or a physical change of the sample.
A simple example occurs when one molecular entity is converted into another that has the same molar absorption coefficient at a given wavelength. As long as the sum of the concentrations of the two molecular entities in the solution is held constant there will be no change in absorbance at this wavelength as the ratio of the concentrations of the two entities is varied.
Contrary to a widely accepted idea, the existence of an isosbestic point does not prove that the reaction is a quantitative conversion of one species into a unique other species or that an equilibrium exists between only two species. The observation of isosbestic points only indicates that the stoichiometry of the reaction remains unchanged during the chemical reaction or the physical change of the sample, and that no secondary reactions occur during the considered time range, For the reaction A + B → c C + d D + e E, with c, d, and e the percentages of the products C, D, and E, an isosbestic point will be observed at every wavelength where the condition ɛ A + ɛ B = c ɛ C + d ɛ D + e ɛ E, provided that the values of the percentages c, d, and e remain constant during the chemical reaction or the physical change. The use of the obsolete term isoabsorption point is not recommended.


Finding major product when bromonium compound is acted upon by a nucleophile

I'd like to point out that the ketone would actually disfavor an attack at its alpha position:

Placing two positive charges on adjacent carbon atoms is much less favorable than not, especially when there is an oxygen perfectly poised to help stabilize this positive charge, as user55119 pointed out.


Lowering energy in molecular orbital mixing

The orbital $\psi$ is obtained as a mixture of the orbitals {$\phi_i$}

$$\psi= c_1 \phi_1 + c_2 \phi_2$$

choosing the {$c_i$} through energy minimization.

Once obtained $\psi$, if you try to mix it with the initial {$\phi_i$} to obtain a new function, lets say $\mu$, you'll have:

$$\mu = c_3 \psi + c_4 \phi_1 + c_5 \phi_2$$

(notice that the coefficients don't need to be the same)

or equally

$$\mu = c_3 (c_1 \phi_1 + c_2 \phi_2) + c_4 \phi_1 + c_5 \phi_2$$
$$\mu = c_3 c_1 \phi_1 + c_3c_2 \phi_2 + c_4 \phi_1 + c_5 \phi_2$$

$$\mu = (c_3 c_1 + c_4) \phi_1 + (c_3c_2+c_5) \phi_2$$

that has the same functional form that
$$\psi= c_1 \phi_1 + c_2 \phi_2$$

so as the criteria is the energy minimization it should be no surprising that we can not get better energy that making
$(c_3 c_1 + c_4) = c_1$ and $(c_3c_2+c_5)=c_2$.

The idea behind this is that you want to obtain the "better" function $\psi$, but you are restricted to a set of functions of the form: $c_1 \phi_1 + c_2 \phi_2$.  You impose this restriction because it only requires to obtain two coefficients, that is far simpler than obtaining an arbitrary function (defined with infinite images).
One restriction is that the equation $c_i \phi_i = c_k \phi_k$  can only be true if $c_i = c_k = 0$.

You can expand your search to a larger set of functions, for example: $c_1 \phi_1 + c_2 \phi_2 + c_3 \phi_3$, but it requires that the set still be linearly independent, that is, you can not get any of the functions with a linear combination of the others, (i.e. it is not possible to get $c_1 \phi_1  =  c_2 \phi_2 + c_3 \phi_3$ if at least one $c_i$ is different from 0). If not, as in the discussion above, the space of function will contain the same functions so the inclusion of the new function is in some sense redundant or useful.

You can add infinite linearly independent functions, but, it does not mean that the energy will be always get lower.


Does the temperature of water affect its ability to extinguish fires?

A fire needs 3 things as shown in the image:

Water reduces the heat as it turns into steam; and hot water turns into steam faster. But the most important thing water does, is to prevent oxygen getting to the fuel. So as long as there is liquid water, it does a great job.


Does Wurtz reaction follow an ionic mechanism or a free radical mechanism?

I think both mechanisms are valid. As far as I know, both are supported by some experimental evidence. Byproducts, like alkanes (other than the expected one) and alkenes, can be explained by the free radical mechanism. On the other hand, the formation of Organometallic compounds can be verified by the ionic mechanism.

The essential thing to note is that the mechanism involves a $\mathrm{S_N2}$ step. And because of that sterically hindered alkyl halides, like tertiary alkyl halides, instead of giving a substitution product yield an elimination product and you get an alkene.


Is the Nitrogen in pyridine electron withdrawing and what effect does this have on electrophilic substitution?

The orbitals in pyridine are pictured below; 6 pi electrons in a loop of 6 continuous p-orbitals (like benzene) plus a lone pair of electrons in an $\ce{sp^2}$ orbital on nitrogen in the plane of the ring.

Since nitrogen is more electronegative than carbon we would expect the nitrogen to inductively remove electron density from the ring carbons.  Below are the resonance structures we can draw for pyridine.  They show that the nitrogen also removes electron density from the ortho and para carbons via resonance effects.

This removal of electron density from the ring by both inductive and resonance effects suggests that pyridine will undergo electrophilic substitution much more slowly than benzene (the ring is said to be deactivated).

If we look at the resonance structures that we can draw for electrophilic attack at the ortho

meta

and para positions,

we see that in each case we can draw 3 resonance structures, but only in the meta-attack case do we keep the positive charge on carbon and not on the more electronegative nitrogen.  Therefor, meta-substitution is preferred in electrophilic attack on pyridine

So electrophilic attack on pyridine will produce meta-substituted pyridines.  Note too that in the case of nitration, the first thing that will happen is that the lone pair on nitrogen will be protonated by the strong acid and generate the pyridinium cation.  The ring in the pyridinium cation is even more deactivated than the pyridine ring because of the full unit of positive charge on the nitrogen; but again if you look at the possible resonance structures, meta-attack will be preferred.  In this case the ring is so deactivated that extremely vigorous conditions must be used to nitrate the ring.


Are there any strategies to determine bond order and hybridization rankings without drawing the Lewis Structure?

Although it is often taught otherwise, hybridisation is a function of geometry. Therefore you must know (or be able to predict) the geometry. On a basic level, the easiest way to do that are Lewis structures.
A different way is using the VESPR theory, a rudimentary model to predict geometries. Be warned though that this model will fail when it comes to more complex molecules. You need a good imagination of the three dimensions to be able to do it without drawing it.
However you want to spin it, you will have to know which atoms are bonded to other atoms. If you can imagine that, that is good, if not you will have to draw it out. The longer you study, the more present some common molecules are and you will be able to tell certain properties by what we like to call chemical intuition. But that takes time.

Bond order is a bit of a tricky topic, because you have to know the bonding situation. This would include all possible resonance structures for some molecules. While knowing or judging this also comes with experience, it is probably best to draw such structures for learning purposes.

I personally do not like exam questions like these, because they usually just test if a student is good at cramming facts, rather than asking about if these things are understood. But things will get better, if you don't lose interest in the mean time.


Is an SN1 reaction possible in halobenzene?

Yes, that reaction occurs, but generation of the phenyl carbocation was much more difficult than anyone might have guessed. Here is a drawing of the phenyl carbocation.
First note that the 6 p orbitals making up the aromatic pi system are all still intact and overlapping - the aromatic nature of the benzene ring has not been tampered with.  Then notice that the cationic center exists as an empty $\ce{sp^2}$ orbital where the

$\ce{C-H}$ bond used to be. The empty (cationic) $\ce{sp^2}$ orbital is orthogonal to the aromatic pi system; as a consequence there is no resonance interaction or stabilization of any sort between the two systems.  Further, being constrained by the 6-membered ring, the cationic $\ce{sp^2}$ orbital cannot rehybridize.  This results in considerable (low energy) s-character being used to stabilize an empty orbital, and is not available to stabilize other orbitals that do contain electrons! The vinylic carbocation, the ethylenic analogue of the phenyl cation, is much more stable and common (it can even be observed by nmr) because it is not constrained by a ring and can rehybridrize, thereby lowering its energy. It is both this total lack of stabilization and relatively high (higher than need be if rehybridization could occur) energy of the molecule that has made the phenyl carbocation such an elusive target.

To stack the deck in their favor, investigators have run experiments using extremely good leaving groups such as nitrogen and the triflate anion. Magic acid and its variations, a very powerful acid that stabilizes carbocations, was often employed as well.  The nitrogen leaving group route, using thermal decomposition of aryl diazonium cations has been one of the few methods that has allowed access to this transient intermediate.

A few references if anyone would like to read more:

On Attempts at Solvolytic Generation of Aryl Cations - 1976
Attempts at Solvolytic Generation of Phenyl Cations - 2004
A one paragraph review - 2000


Can I reactivate a chemoluminescent reaction?

Short answer: No, you can't!

Longer answer: Still no, and this is why: The reaction between the oxalate and hydrogen peroxide is irreversible. Both compounds are consumed in the reaction and there is no way back!

The only compound that survives is a fluorescent material, that absorbs in the UV, emits in the visible range and gives the colour that you finally observe.

As Ben Norris pointed out in his comment, your only options are:

add fresh diphenyloxalate and hydrogen peroxide (which again will be consumed) or
irradiate the fluorescent compound that has survived with a UV lamp


Why is KI oxidized by H2SO4, whereas NaBr isn't oxidised?

Iodide ions are much more powerful reducing agents than bromide ions:

$$\ce{2I- ->I2 + 2e-} \hspace{1cm}E^{o}=-0.54\ \mathrm{V}$$

$$\ce{2Br- ->Br2 + 2e-} \hspace{1cm}E^{o}=-1.0873\ \mathrm{V}$$

Iodide ions are so powerful, in fact, that they are capable of reducing sulfur from a +6 oxidation state in $\ce{H2SO4}$ all the way to -2 in $\ce{H2S}$ via the following half reaction:

$$\ce{HSO4- +9H+ +8e- -> H2S + 4H2O} \hspace{1cm}E^{o}=+0.80\ \mathrm{V}$$

The full reaction being being:

$$\ce{HSO4- +9H+ +8I- -> H2S + 4I2 +4H2O}\hspace{1cm}E^{o}=+0.26\ \mathrm{V}$$

A quick glance at the reduction potentials tells you that the analogous bromide reaction has a negative reduction potential and is therefore non-spontaneous.

[Source]


The traditional labelling of irreducible representations is the following:

A, when rotation around the principal axis is symmetrical
B, when rotation around the principal axis is asymmetrical

You can see, for example, that $C_n$ groups of even order have both A and B representations, while those of odd order cannot feature B representations.

As others have commented, there are additional rules… E for doubly degenerate representations, T for triply degenerate, u and g for parity under inversion (odd and even, respectively), greek capital letters Σ, Π, Δ for infinite groups (C∞v and D∞h).

Finally, if there are several irreducible representations that are not distinguished under those conventions, they are numbered.


The difference between heat of reaction at constant pressure and that at constant volume

So the question is: Why is the difference between the heat of reaction at constant pressure and that at constant volume equal to $RT$?

First Part: heat of reaction

From the first law of thermodynamics we know

\begin{equation}
  \mathrm{d}U = \delta Q + \delta W = \delta Q - p \mathrm{d} V
\end{equation}

where $U$ is the inner energy, $Q$ is the heat, $W$ is the work, $p$ is the pressure and $V$ is the volume.
So, for a reaction at constant volume you get

\begin{equation}
  \mathrm{d}U = \mathrm{d} Q \ .
\end{equation}

Now, for a reaction at constant pressure you can invoke the relationship

\begin{align}
  \mathrm{d}H &= \mathrm{d}U + p \mathrm{d}V + V \mathrm{d}p \\
              &= \delta Q - p \mathrm{d} V + p \mathrm{d}V + V \mathrm{d}p \\
              &= \delta Q + V \mathrm{d}p \\
              &\overset{\mathrm{d}p = 0}{=} \mathrm{d} Q
\end{align}

where $H$  is the enthalpy.
This gives you the relationships that describe the change in heats released during reactions at constant volume and constant pressure.
The so called heat of reaction is defined as

\begin{equation}
  \Delta Q = \biggl( \frac{\partial Q}{\partial \xi} \biggr)
\end{equation}

where $\xi$ is the extent of reaction.

Second Part: temperature dependence

Now using the equality of mixed partials and the definitions of the heat capacities at constant pressure $C_{p} = \Bigl( \frac{\partial H}{\partial T} \Bigr)_{p}$ and constant volume $C_{V} = \Bigl( \frac{\partial U}{\partial T} \Bigr)_{V}$ you get for constant volume ($\mathrm{d}U = \mathrm{d} Q$):

\begin{align}
  \biggl(\frac{\partial \Delta Q}{\partial T} \biggr)_{V} = \Biggl(\frac{\partial \Delta U}{\partial T} \Biggr)_{V} &= \left( \frac{ \Bigl( \frac{\partial U}{\partial \xi} \Bigr)_{T, V}}{\partial T} \right)_{V} = \Biggl( \frac{\partial^{2} U}{\partial \xi \partial T} \Biggr)_{V} \\
  &= \Biggl( \frac{\partial^{2} U}{\partial T \partial \xi} \Biggr)_{V} = \left( \frac{ \Bigl( \frac{\partial U}{\partial T} \Bigr)_{V} }{\partial \xi} \right)_{T} = \Biggl(\frac{\partial  C_{V}}{\partial \xi} \Biggr)_{T, V} = \Delta C_{V}
\end{align}

and for constant pressure ($\mathrm{d}H = \mathrm{d} Q$):

\begin{align}
  \biggl(\frac{\partial \Delta Q}{\partial T} \biggr)_{V} = \Biggl(\frac{\partial \Delta H}{\partial T} \Biggr)_{p} &= \left( \frac{ \Bigl( \frac{\partial H}{\partial \xi} \Bigr)_{T, p}}{\partial T} \right)_{p} = \Biggl( \frac{\partial^{2} H}{\partial \xi \partial T} \Biggr)_{p} \\
  &= \Biggl( \frac{\partial^{2} H}{\partial T \partial \xi} \Biggr)_{p} = \left( \frac{ \Bigl( \frac{\partial H}{\partial T} \Bigr)_{p} }{\partial \xi} \right)_{T} = \Biggl(\frac{\partial  C_{p}}{\partial \xi} \Biggr)_{T, p} = \Delta C_{p}
\end{align}

where $T$ is the temperature and $\Delta C_{i}$ is change in the molar heat capacities of the reactants and products during the reaction at constant $i$ (where $i$ is either $p$ or $V$):

\begin{align}
  \Delta C_{i} = \sum_{k} \nu_{k} c_{i_{k}}
\end{align}

where  $\nu_{k}$ and $c_{i_{k}}$ are the stochiometric coefficient and the molar heat capacity at constant $i$ of the $k^{\text{th}}$ component in the reaction, respectively.

When you integrate the equations for the temperature dependences of the reaction heats you get:

\begin{align}
  \text{constant volume:  } &\int \biggl(\frac{\partial \Delta Q}{\partial T} \biggr)_{V} \, \mathrm{d} T = \left( \Delta Q \right)_{V} = \int \Delta C_{V} \, \mathrm{d} T \\
  \text{constant pressure:  } &\int \biggl(\frac{\partial \Delta Q}{\partial T} \biggr)_{p} \, \mathrm{d} T = \left( \Delta Q \right)_{p} = \int \Delta C_{p} \, \mathrm{d} T \ .
\end{align}

So, the difference of the reaction heats at constant volume and constant pressure would be:

\begin{align}
  \left( \Delta Q \right)_{p} - \left( \Delta Q \right)_{V} = \int \left( \Delta C_{p} - \Delta C_{V} \right) \, \mathrm{d} T \ .
\end{align}

Third Part: ideal gases

Now, why would this boil down to $RT$? This works when you consider reactions that involve only ideal gases. For ideal gases you have (see here):

\begin{align}
  c_{p} - c_{V} = R \ .
\end{align}

So, the integral simplifies to

\begin{align}
  \int \left( \Delta C_{p} - \Delta C_{V} \right) \, \mathrm{d} T &= \int \sum_{j} \nu_{j} \bigl( \underbrace{ c_{p_{j}} - c_{V_{j}} }_{= \, R} \bigr) \, \mathrm{d} T \\
  &= R \int \sum_{j} \nu_{j} \, \mathrm{d} T \\
  &= R \sum_{j} \nu_{j} \int \mathrm{d} T \\
  &= R T \sum_{j} \nu_{j}
\end{align}

where $\sum_{j} \nu_{j}$ will be some rational number but not necessarily equal to 1 - it depends on the reaction.
This is basically the result you were looking for although the reaction heat difference will only be proportional to $RT$ with the proportionality constant being $\sum_{j} \nu_{j}$:

\begin{align}
  \left( \Delta Q \right)_{p} - \left( \Delta Q \right)_{V} = RT \sum_{j} \nu_{j} \ .
\end{align}


On the activity of liquids and solvents in equilibium constants

I concluded on this question after discussion with the physical chemistry professor at my home university, and reading some physical chemistry books.
Although many references include the concentration of water in the expression of Ka, this is incorrect: the activity of pure water is 1. In dilute solutions, $a(\ce{H2O}) \approx 1$.
Therefore, the second option is correct:
$$K_a=K_c=\frac{[\ce{A-}][\ce{H3O+}]}{[\ce{HA}]}$$
This has a direct consequence on the values of pKa for the two couples of water: $pKa(\ce{H3O+}/\ce{H2O})=0$ and $pKa(\ce{H2O}/\ce{OH-})=14$, instead of the values $-1.7$ and $15.7$ that are often found in textbooks (which result from including $[\ce{H2O}]=55.55$ mol/l).
A good article on this matter can be found here: What is the pKa of water?.
References:

B. Leyh, Department of Chemistry, University of Liège, Belgium.
Atkins, P., & De Paula, J. (2006). Atkins’ physical chemistry (8th ed.). London, England: Oxford University Press.
Reid, P., & Engel, T. (2006). Physical Chemistry. Philadelphia, PA: Pearson Education.


My understanding is that it is an environment in which, when the
  isomers interact with the environment, the two enantiomers can be
  distinguished.

That's exactly right.  Some examples of chiral environments that could be created in a lab would include

Running a reaction in a chiral solvent
Running a reaction in a chiral cavity
Running a reaction with a chiral reagent
Running a photochemical reaction with a polarized light source
Carrying out a reaction on a surface to which chiral molecules have been attached (certain clays have this property)

All of these items are like a glove, and each enantiomer will fit differently.  This different "fit" leads to transition states of different energy as each enantiomer moves along the reaction coordinate towards the product(s).


NMR spectrum of cyclohexanol - hints on integration

To answer your first question about measuring the integrals for the peaks between 1 and 2 ppm: If you seriously need to do a very accurate integral of these peaks without having access to NMR processing software (ie you've just been given the spectrum as a handout), you can resort to one of the methods of choice from the good 'ol days.

Print your spectrum out on a nice big piece of paper, get a nice sharp pair of scissors and cut out your peaks and weight them (a 4 or 5 figure  balance may be required). These method gives incredibly accurate integrals
Print out your spectrum on graph/grid paper and count the grids. Again, very accurate, depending on the scale of grid you choose to print on.

As for 'pairing up' the remainder of the ring protons, this is not a trivial exercise, made even less so with the dispersion of the spectrum you are working with. You are quite right in your assessment. The protons closer to the -OH group will be at higher chemical shifts; equatorial and axial protons will be non-equivalent (equatorial at higher chemical shifts); and the whole fluxional ring dynamic will be temperature, concentration, solvent and field dependent. So, probably it is not worth trying to pair up any of these protons. You could (and hopefully would, if you are a thorough researcher) assign all of your peaks with the assistance of 2D COSY, HSQC/HMBC methods.


This is probably a bit of a simplistic explanation.

Due to the butane being released from the from the container, the pressure in the container decreases, which in turn, decreases the temperature of the container according to the Pressure-Temperature Law (also known as the Admonton Law), that states for an ideal gas:

$$P  \propto  T$$

The boiling point of butane is normally −1 °C at normal pressures – so in typical use at room temperature, it should have already vapourised. But, due to it being under pressure, it remains a liquid, undergoing evaporation in the container. So, butane has a vapour pressure that increases with temperature, as seen below:

Image source: Wikipedia, who sourced it from the CRC Handbook of Chemistry and Physics 44th ed

Even though butane (like all gases) is not an ideal gas, we can still surmise that if you drop the pressure (by releasing the gas), then the temperature of the gas also drops, especially considering the combined pressure due to that from the container and vapour pressure is rapidly released = rapid depressurisation = rapid loss of temperature, which is felt on the outside of the container.


Must a sample of a deuterated compound be isotopically pure for kinetic isotope experiments?

There are two approaches to measuring kinetic isotope effects.

One is to have two completely separate, isotopically pure systems, one heavy, and one light. In each system you can measure the rate of the reaction however you want.  You don't have to be able to measure the rate in an "isotopically resolved" way, just measure the bulk rate.  Each of your systems is isotopically pure, so the rate of each system is 100% reflective of the isotope composition of that system.

The other way is to have a single system with variable/unknown isotopic composition.  You measure the rate of your reaction, but you have to be able to measure (e.g.) the rate of depletion of each substrate isotopomer separately.  (Or the rate of formation of each product isotopomer, but that sounds very difficult in your system.)  For example, if you had a mixture of $\ce{CH4}$ and $\ce{CD4}$ and you were interested in the kinetic isotope effect of some kind of combustion reaction of those gases with oxygen, you could measure the ratio of $\ce{CD4}$ to $\ce{CH4}$ over time, e.g. using mass spectrometry.  It would increase as the $\ce{CH4}$ is preferentially depleted by the combustion reaction.  You could also measure the rates of formation of $\ce{H2O}$ and $\ce{D2O}$.  The rate of formation of $\ce{H2O}$ would be much higher than that of $\ce{D2O}$.  (The exact numerical values here would depend strongly on the nature of your combustion system, e.g. temperature, catalysts used, etc.)

So iff you can measure $\ce{FeOOH}$ and $\ce{FeOOD}$ separately in the same system, then you don't need an isotopically pure material.  I'm not an expert in these types of measurements, but it sounds very hard to me.  Probably if you don't already know how to do it, it will be cheaper just to buy $\ce{KOD}$ and get on with the synthesis of an isotopically pure material so you can use the first approach.


Dependency of Standard Gibbs free energy (using electrode potential) on stoichiometric moles of electron exchanged

It does not depend on how you write the reaction. You were right in saying that, $n$ is the mole number transferred in the reaction. So, it is not the number of electrons you write arbitrarily in the equation that represents the reaction, it is rather proportional to the number of other substances, which, even being all arbitrary always satisfy that proportional relation.
Think of $n$ as the factor needed to relate one intensive property with one extensive property.


This concept of electrolytic solution conductivity is complex and dependent on a variety of factors aside from mere concentration of electrolytes. There are separate laws used to describe the relationship between solution conductance and concentration dependent on the dissociative strength of electrolytes and the extent of overall solution saturation.

In your simple case, just recognize that the net conductance of the final solution may not work out to be the mere sum of conductances in the original solutions

From Wikipedia:

  The specific conductance of a solution containing one electrolyte
  depends on the concentration of the electrolyte. Therefore, it is
  convenient to divide the specific conductance by concentration. This
  quotient, termed molar conductivity, is denoted by Λm
  $$\Lambda _{m}={\frac  {\kappa }{c}}$$

Hence, since conductivity is proportional to concentration, and concentrations aren't additive, you cannot expect conductivities to be additive.


Is there a standard scholarly reference for lattice constants of crystals of the elements?

The CRC Handbook of Chemistry and Physics contains a dedicated compilation by H. W. King, titled «Crystal Structures and Lattice Parameters of Allotropes of the Elements».  In case your research library is closed, you may access some of its editions freely or borrow them with the library card of archive.org.
In case of the 97th edition (by 2016), the section starts by page 12-16.  (Elements liquid or gaseous at ambient conditions are included.)

source


Why does helium seem to exhibit an ionization energy affected by the shielding of electrons?

The premise is problematic as there is no 'shielding' effects in an isolated Helium atom.

However, Phillipp was correct in explaining why the ionization energy of Helium is not exactly twice that of Hydrogen.  This is because the energies of each electron are higher due to their correlated nature (i.e. they repel each other into higher, non-ground-state-like energies).  Therefore, it takes LESS energy to rip one of those electrons out of the system.


Are there any molecules with delta bonds in their ground states?

Chromium(II) acetate has a Cr-Cr quadruple bond - and hence a $\delta$ bond - in its ground state. This sort of "paddlewheel" configuration, with four carboxylic acids coordinating a pair of metal ions, is reasonably common (see also copper(II) acetate, for instance), although the extent to which the metals can be considered covalently bonded varies.


Why is gold sol used for intra-muscular injection?

Will this help?

Toxicity and cellular uptake of gold nanoparticles: what we have learned so far?

Gold has a long tradition as a disease-modifying antirheumatic drug in the treatment of rheumatoid arthritis and it seems to relieve pain and decreases swelling of the joints.


You actually have it backwards. A low $\mathrm{p}K_\mathrm{a}$ is a strongly dissociating acid, and a high $\mathrm{p}K_\mathrm{a}$ is a weakly dissociating acid. It would do you well to remember the origin of $\mathrm{p}K_\mathrm{a}$ \eqref{pKa}, as well as the Henderson–Hasselbalch equation \eqref{HHeq}:

$$K_\mathrm{a}=\ce{\frac{[A-][H3O+]}{[HA]}} \Rightarrow \mathrm{p}K_\mathrm{a} = -\log(K_\mathrm{a})= -\log \left( \ce{\frac{[A-][H3O+]}{[HA]}}\right)\tag1\label{pKa}$$

$${\mathrm  {pH}}={\mathrm  {p}}K_{{\mathrm  {a}}}+\log _{{10}}\left({\frac  {[{\mathrm  {A}}^{-}]}{[{\mathrm  {HA}}]}}\right)\tag2\label{HHeq}$$

From this we can surmise a physical meaning behind what the $\mathrm{p}K_\mathrm{a}$ of an acid is—it is the $\mathrm{pH}$ of the solution when $\ce{[A-]=[HA]}$. This means that for $\mathrm{pH}>\mathrm{p}K_\mathrm{a}$, $\ce{[A-]>[HA]}$. Since ascorbic acid is a diprotic acid, it has two $\mathrm{p}K_\mathrm{a}$ values, one for each deprotonation.


For a reversible process in a closed system (no mass entering or leaving), the general equation for the expansion work done by a real gas on the surroundings is the same for an ideal gas, namely
$$W = \int p\,\mathrm{d}V$$
However, for a real gas, we use the equation of state for that gas $p=p(n,V,T)$ rather than $p=\frac{nRT}{V}$, the equation of state for the ideal gas.


Why does electronegativity increase as effective nuclear charge increases?

The effective nuclear charge can be thought of the charge of the nucleus minus the charge of the core electrons. For an element such as fluorine, the nuclear charge is $+9$ and the core electrons have a charge of $-2$ so the effective nuclear charge is $+7$. Similarly for carbon it would be $+6 - 2 = + 4$.

Now let’s assume a $\ce{C-F}$ bond. There are two electrons in this bond, each charged $-1$. On the left, from the carbon atom, they ‘see’ a charge of $+4$; on their right, from the fluorine atom, they ‘see’ s charge of $+7$.

Electrostatic interactions are greater at shorter distance and at greater charge separation, so the interaction (i.e. force) the bonding electrons experience from fluorine’s side is almost double that from carbon’s side. (The carbon and fluorine atoms can be considered to be roughly of the same size.) It should now be clear why the electrons are drawn to the fluorine.

Electronegativity is just a name humans gave the observed effect of ‘this atom can pull electrons towards it better than others’. It takes a look at the result of the above process and rationalises it. The effective nuclear charge approach is the underlying physical force.

Note: This explanation is extremely simplified, does not take any quantum effects or the like into consideration and should thus only be used at introductory levels.


Looking for raw interferogram data / raw FID data from FT-IR / FT-NIR / FT-NMR

In The Journal of Chemical Education, published by the American Chemical Society, a rather recent article presented a related project, hosted by the University of Minnesota.  Intentionally it is about raw FIDs-as-obtained, out of the NMR spectrometer, aiming to teach correct data processing (including spectra with two orthogonal frequency scales). The article is sponsored by the authors (i.e. freely accessible), the database requires an e-mail to register.  Reference: J. Chem. Educ., 2013, 90 (7), pp 941–943; DOI: 10.1021/ed300787v


How does the rate constant change with the change of the temperature?

Answer B is technically correct although C could also be accepted. Assuming $E_A$ to be constant, at low temperatures because of the inverse $T$ in the exponential $\exp(-E_A/(R\cdot \text{small number}) \equiv \exp(-\text{big})$ which is a small number. At high temperatures $\exp(-E_A/(R\cdot \text{big number})\equiv \exp(-\text{small})$ is a big number, so the rate constant increases with temperature.

At temperature such that the exponential  $\to 1$ the rate constant is $A$ for all practical purposes. Usually this limit is not reached unless $E_A$ is very small and then as $A$ is also a function of temperature this ($A$) term becomes now important. Generally, however, for the vast majority of reactions the exponential term is the most important.


Why are noble metals more electronegative then most metals?

The noble metals are defined by a resistance to oxidation and corrosion, and this should not be interpreted as a lack of reactivity, but instead an aspect of their high EN.  So, there is no contradiction as you seem to be thinking.  Basically, they hold onto their electrons better than other metals, so it is harder for acids and oxygen to steal electrons from these metals.

The noble metals are generally considered to be ruthenium, rhodium, palladium, silver, osmium, iridium, platinum, and gold, so I'll mainly be addressing these particular elements. There are a few reasons why they should have a higher electronegativity:

The lanthanide contraction causes these atoms to have a higher than expected $Z_{\text{eff}}$. This means they hold on to their electrons tightly, so have higher electron affinities and ionisation energies which corresponds to a high EN. This doesn't carry into group 12 because accepting electrons for these elements would result in adding another energy level, so it's not as favorable.
The filling order: By period 6, the 6s, 4f, and 5d orbitals are all so close in energy that the filling order changes, which affects the properties and chemistry of the elements.  Notice that almost all of these metals, apart from Os and Ir, break the typical filling order. Adding electrons to these atoms therefore does not have the same effects that are dictated by normal periodic trends.
Relativistic effects - Somewhat related to filling order. By period 6, nuclei are so heavy that the core electrons are moving close to the speed of light.  This causes a contraction of the s orbitals, the inert s pair effect, and other things that have a big effect on the electronic structure and chemistry.

Also, the noble gases start becoming reactive around Kr, and more so for Xe, because they are so big with such high shielding that electronegative atoms are able to take electrons to form bonds.  Some people get confused about this because they think it's just Kr, and Xe being weird, but it's actually a trend and continues with Rn, but there isn't much data on Rn compounds, or much use for them, since Rn-222 is the longest lived isotope with a half-life of ~3 days.

Periodic Trends of Transition Metals could help explain some of this.


Which will be the most acidic hydrogen in the following organic compounds?

Resonance effect is a more dominating effect than inductive effect. Here I think in both the structures hydrogen attached to carbon 1 will be more acidic. The reason is the negative charge formed on the carbon after removal of hydrogen is resonance stabilised to a greater extent.

In the first structure though after removal of hydrogen from the second carbon, the negative charge formed will be resonance stabilised, but its delocalisation is less compared to the delocalisation of the negative charged which would have been formed from the removal of hydrogen from the first carbon. So, if the negative charge forms from the first carbon ( from the removal of hydrogen), then it is more stabilised. Hence the hydrogen attached to the first carbon is more acidic.

In the second structure also the hydrogen attached to the first carbon is more acidic. This is because if negative charge forms on the second carbon, there is no stabilisation effect. But if the negative charge forms on the first carbon, then again by resonance it will be stabilised. So in the second structure also the hydrogen attached to the first carbon is more acidic.


Will the addition of salt further cool down ice that is already at -30 °C?

If you add salt to icy water, it's the water that gets colder. Pure water under atmospheric pressures will be $0~\mathrm{^\circ C}$ - even when the ice melting in it is colder (for example if it's just out of the fridge). When you then add salt to that water, it lowers the freezing point of your water-salt solution, all the way down to $-21~\mathrm{^\circ C}$ once it contains $5.2\ \mathrm{mol}$ of NaCl per kg of water. That's the maximum amount of salt this solution can contain at that temperature. (Find the thermophysical data at NIST (page 30).

Now, if you add salt to liquid water (containing no ice), you will actually decrease its temperature slightly because of the energy required to solvate the salt in the water. This won't help you in a 'water cooling circuit' because you also lower the heat capacity.


What is the explanation of the structure of H2XeO4?

When I think about such structure, I draw all valence electrons around the atoms and connect them somehow reasonably. There are two ways to connect the hydrogens to xenon. Either directly bonded as H-Xe bonds, which for me does not feel right here, or as part of OH groups, which also make more sense in the "acid" part.
By connecting everything that way, you end up with:

This let you end up with a lone pair at xenon which is the reason for the non-tetrahedral structure.

A quick and dirty quantum calculation[1] shows what I mean. By calculating the ELF (Electron Localization Function), you can easily see where the electrons are located. The lone pair is shown by the yellow volume surrounding no atom.

This structure is not clear from the Lewis formula shown on Wikipedia, but can be seen in the ball and stick representation that is shown there (don't care about the "flipped" hydrogens):

[1] ORCA 4.1.2, best functional of the planet B3LYP def2-QZVP // Crest 2.6, Visualization with Chimera, iso value 0.8 ... whatever ... camels? ... what unit has the ELF?


When you are "containing" a gas in a container then you are by definition restricting its volume. Lets assume you heat the gas. If the volume is confined then the pressure will increase. Similarly, two different gases will have different pressures if confined to the same volume.

If you somehow control the pressure (there is a movable piston) then a real gas will occupy less volume because of its intermolecular interactions.

Also see this: Volume of different gas in a container


Why is O2 (Oxygen) colorless and odorless while O3 (Ozone) is a pale blue gas with a distinctively pungent smell?

Our perception of colour and smell involve different chemistry so there is no reason why they should be related. For example, what ever the chemistry of smell is (and its not certain what this is) it clearly does not involving photons and subsequent isomerisation of a rhodopsin.

Ozone is slightly coloured as it has two low lying electronically excited states (Chappius bands 550 to 600 nm) and a second state (at approx 700 to 1000 nm) both of whose energy gaps from the ground state correspond to a wavelength in the visible spectrum. The combined absorption is rather broad, weak and peaks at approx 600 nm which in the red part of the spectrum, thus ozone appears blue. Oxygen also has two low lying electronic states at around 700 nm and 12000 nm but are very weak and the longer wavelength transition is beyond the range of our eyes to detect it (liquid oxygen appears blue as does liquid ozone). Both O$_2$ and O$_3$ have electronic states in the ultra violet at a wavelength beyond the range of our eyes.

( I have added a moderately detailed description as to why high pressure gas and liquid oxygen molecules are coloured. see Does O2 have a color in the gas phase )


How to determine the minimum pH at which iron(III) hydroxide dissolves completely?

The OP first stated the problem as:

  Knowing $K_\mathrm{s} = 6\cdot10^{-38}$ for $\ce{Fe(OH)3}$ in neutral solutions, calculate the minimum pH of an acidic solution in order to completely dissolve $\pu{10 mg}$ of $\ce{Fe(OH)3}$. Data: $V = \pu{0.1 L}$.

I took that to mean that acid was added to a solution containing 10 mg of iron (iii) hydroxide and the pH of the solution when the iron (iii) hydroxide was totally dissolved was desired.*

Knowing that $K_\mathrm{sp} = 6\cdot10^{-38}$ for $\ce{Fe(OH)3}$ then
\begin{align}
  K_\mathrm{sp} &= \ce{[Fe^{3+}][OH^-]^3} = 6\cdot10^{-38},&
\text{ so }&&
\ce{[OH^-]} &=\sqrt[3]{\frac{6\cdot10^{-38}}{\ce{[Fe^{3+}]}}}
\end{align}

You need to solve for $\ce{[Fe^{3+}]}$ which you can calculating knowing that there is $\pu{10 mg}$ of $\ce{Fe(OH)_3}$ in 0.1 liters of solution. It has a molecular weight of $\pu{107 g/mol}$, so there is
$$0.010/107 = \pu{9.35* 10^-5 mol//L}\text{ of }\ce{Fe(OH)_3}.$$

Going back to the equation for the $K_\mathrm{sp}$:
$$\ce{[OH^-]} =\sqrt[3]{\frac{6\cdot10^{-38}}{\ce{[Fe^{3+}]}}} = \sqrt[3]{\frac{6\cdot10^{-38}}{9.35\cdot10^{-5}}} = 8.63\cdot10^{-12}$$

Knowing $\ce{[OH^-]}$ you can calculate $\ce{[H^+]}$ via

$$\ce{[H^+]} = \frac{1\cdot10^{-14}}{\ce{[OH^-]}} = \frac{1\cdot10^{-14}}{8.63\cdot10^{-12}} = 1.16\cdot10^{-3}$$

and then pH:

$$\mathrm{pH} = -\log\left(1.16\cdot10^{-3}\right) = 2.94.$$

Now the OP has modified the problem statement to be:

  Knowing $K_\mathrm{s} = 6\cdot10^{-38}$ for $\ce{Fe(OH)3}$ in neutral solutions, calculate the minimum pH of an acidic solution in order to completely dissolve $10\ \mathrm{mg}$ of $\ce{Fe(OH)3}$. Data: $V = \pu{0.1 L}$.

We again end up with the final pH being just acidic enough to dissolve $\pu{10 mg}$ of iron (iii) hydroxide in $\pu{100 ml}$ of water, but obviously the solution must start out more acidic. The final solution will again have a pH of $2.94$ or a $\ce{[H^+]} = 1.16\cdot10^{-3}$.

In $\pu{100 mL}$ of a solution of pH $2.94$ there are $\pu{0.116 mmol}$ of acid.

We calculated before that there were $\pu{9.35*10^{-5} mol}$ of $\ce{Fe^{3+}}$, but there are three $\ce{OH^-}$ anions for every cation of $\ce{Fe^{3+}}$, so there are $3 \times 9.35\cdot10^{-5} = \pu{2.81*10^{-4} mol}$ of $\ce{OH^-}$.

The total amount of substance (in millimoles) of acid needed is
$$0.116 + 0.281 = 0.397.$$

For $\pu{100 ml}$ to contain $\pu{0.397 mmol}$ the solution would need to be 3.97 millimolar in $\ce{[H^+]}$, or the pH = $2.40$.

If you start with $\pu{100 ml}$ of solution with pH $2.40$ and dump in $\pu{10 mg}$ of $\ce{Fe(OH)3}$, then the final pH will be $2.94$ and all of the $\ce{Fe(OH)3}$ will be just dissolved.


When was oil discovered to be of biological origin?

If you look at a gas or liquid chromatogram of crude oil, you see a remarkable preponderance of (unbranched) even-numbered carbon compounds (re acetyl CoA homologation).  You also see traces of polycyclics that are net chiral - terpenoids, terpenes, pristane, phytane, cholestane, chlorins, and porphyrins.  Fischer-Tropsch gives a very different distribution and no net chirality.

Mikhailo V. Lomonosov, "Slovo o reshdenii metallov ot tryaseniya zemli"
Proceedings of the Imperial Academy of Sciences, St. Petersburg, 1757.

(Mikhail Tsvet was born in 1872)


What products will the following reactions yield?

No, I hesitate to give the correct products since this is homework, but maybe some of what's below will help
Yes, the two reactions are both esterifications; converting an acid, an acid anhydride, an acid chloride, etc., to the corresponding ester
The proton is a catalyst, it is not consumed in the reaction.  It protonates the carbonyl oxygen, making the carbonyl carbon more susceptible to nucleophilic attack
benzoic acid


How is molecular orbital theory used in drug research?

Common drug design approach does not involve much QC. There are two general approaches - QSAR and docking

QSAR

The idea is that a family of chemicals (training set) with some having desired biological activity, is used to train a pattern recognition algorithm. The trained algorithm is then used on a library of candidates and once recognised as active are then synthesised and tested in a lab.

Docking

The idea is to construct a molecule strongly interacting with a biological important target (a receptor or an enzyme). Since intermolecular interaction are weak, their calculation from first principles is very computationally expensive and is not practical in most cases. Since biologically relevant molecules are usually fairly trivial, even if complex molecules (For example, there is usually no need to consider interactions involving continuous conjugated systems), the intermolecular interactions are easily described using approach of molecular mechanics. So, in this approach a set of molecules is generated and then tested for interaction with a protein target. The tight spot is to find a way two molecules can bind, typically involving search over extremely large conformer space. Billions of potential candidates for ligand-protein complex must be tested, so the computationally cheapest solution (molecular mechanics) is used.

========

As you can see, neither approach normally uses orbital theory or any complex QC, and even if such complex QC is involved, it is clearly subservient. Instead machine learning algorithms and ways to efficiently search conformation space for low-energy conformers are the focus of research.


What is the major product of the reaction between 4-chlorobut-1-ene?

In solvolysis of simple primary cyclopropylmethyl systems the rate is
  enhanced because of participation by the $\sigma $ bonds of the ring. The ion
  that forms initially is an unrearranged cyclopropylmethyl cation that is
  symmetrically stabilized, that is, both the 2,3 and 2,4 $\sigma $ bonds help
  stabilize the positive charge(page 464 ,JerryMarch ,Advanced Organic Chemistry
  Reactions, Mechanisms,and Structure,Sixth Edition).

The reaction of various cyclopropylmethyl systems are given below.

In the question 4-chlorobut-1-ene  (1) undergoes rearrangement as shown in the scheme 3 and scheme 4 to give 1a, 1b and 1c.Assuming solvolysis with acetic acid ,1a, 1b and 1c would give 2, 3 and 4.


Why is the boiling point of m-dichlorobenzene less than that of p-dichlorobenzene?

The difference between the melting points is small: ortho 180 °C, meta 172 °C, para 174 °C, so they are basically the same. It is just not possible to argue bigger or smaller with hand-waving arguments about dipoles or polarity etc. with something so complicated. However, when the thermal energy is increased, whatever intermolecular interactions exist become less important relative to the average thermal energy, and which now has a dominating and disruptive effect. Thus I would expect all isomers to have a similar boiling point.


Your question covers two different topics:

Redox properties can be well described with redox potentials attributed to a given reaction. It is essentially the same concept that you are talking about: we assign a number (conveniently it is an actual measurable potential) to an electrode reaction which includes an oxidized and reduced form of the species we are talking about. This number, the redox potential, can show you how strong oxidizer/reducer is the corresponding corresponding form, and just as you guess, strong oxidizers generally have a weak reducer pair.
The $\ce{HCl}$ formation reaction is generally a radical chain reaction, triggered e.g. by light, and not a redox reaction. The key here is to split the $\ce{Cl2}$ (e.g photo-chemically), then the obtained free radicals induce a chain-reaction with elementary steps like
$\ce{Cl + H2 -> HCl + H}$ or
$\ce{H + Cl2 -> HCl + Cl}$
The two gas mixed is more or less inert in total darkness.


Which one really determines how people feel. Heat or temperature?

Regarding the last part "Do we feel heat or the temp?" - this is more physiological then chemical, but I want to share this anyway.

We feel heat flux, i.e. how fast heat is removed or supplied to our skin. Since heat flux can be both advection in the surrounding medium (e.g. air) or conduction in the skin it depends on the situation which is dominant.

In general you can write this as
$$ Q=k A \Delta T$$
where $Q$ is the heat flux, $k$ is some transfer coefficient, $A$ is the exchange area and $\Delta T$ is the temperature difference. The value of $k$ depends on the mechanism of transfer e.g. for advection it scales with $\rho C_p v$ i.e. it depends on the heat capacity and flow velocity of the surrounding medium.

The consequence of feeling heat flux is that we can only feel temperature differences not absolute temperature and that advection gives rise to a phenomenon called 'wind chill' that causes us to feel colder when there is more wind, because the higher wind velocity increases the heat transfer coefficient.


If the temperature of a solution decreases, what is the sign of the enthalpy change?

When I was doing enthalpy, this was one of the tricky concepts that I still sometimes get wrong. Here is how you are supposed to remember it.

  Identify the System

  The system is where the reaction happens. You cannot physically monitor the change in heat here. The change you always measure happens in the surroundings.

  Identify the Surroundings

  This is the surroundings where usually the change is expressed as.

Now that we got the basics cleared away, let us look at your problem.

  My reasoning: if the temperature of the solution decreases, it means
  heat was released into the surroundings, so the enthalpy change is
  negative.

This is what I usually think of, but now you must understand that as I've said before the change always happens in the surroundings and is associated with the surroundings. So the temperature of the solution decreases, which means that the surroundings temperature decreased. The solution is part of the surroundings. This means that the temperature of the system increases. Thus making the $\Delta H$ positive and the reaction endothermic.

Good Question, always remember that you cannot, in most situations, physically measure the change in the system.

Thanks.


Sulfur is sometimes described as a hypervalent atom, or one that has more than eight valence electrons. This is possible to some extent; sulfur can indeed utilize its d-orbitals for bonding. However, recent quantum mechanical calculations suggest that the extent of this utilization is negligible. Therefore, your book is most likely representing the $\ce{Cl_2SO}$ molecule as a "charge-separated" molecule.

EDIT: Upon further thought this still makes no sense. Charge is conserved and thionyl chloride should always have 26 electrons (no matter how they are distributed). So perhaps, due to the high degree of ionic character in the $\ce{S-Cl}$ and $\ce{S-O}$ bonds, the missing two electrons are "dispersed" among these ligands. (This is speculation). Perhaps your book is genuinely in error. Perhaps they did not count a lone pair on the central sulfur atom or otherwise miscounted the number of electrons?

EDIT 2: Here's a picture of two possible Lewis structures of thionyl chloride; as we can see, the book agrees that the thionyl chloride molecule is better represented as a "charge-separated" molecule. Nonetheless, all the electrons are present.

EDIT 3: Here's a picture of thionyl chloride which makes it seem as if the molecule only has 24 valence electrons. NB: the lone pair is still there and its presence is implied through the lack of any indication of a non-zero formal charge on the sulfur atom.


Yes, it can be done by mimicking the conditions in the earth's mantle.

It was shown by Alexander Goncharov and coworkers [1] by placing methane in a diamond anvil cell, applying a pressure of more than 2 GPa and heating everything to a temperature around 1000-1500 K with a laser.

Is it economical? Hardly, but very interesting anyway!

Kolesnikov, A.; Kutcherov, V. G.; Goncharov, A. F. Nature Geosci 2009, 2 (8), 566–570. DOI 10.1038/ngeo591.


Why is Cr²⁺ is a good reducing agent but Mn³⁺ is a good oxidising agent?

Why can't the $\mathrm{t_{2g}}$ argument be used for $\ce{Mn}$ ?

Because the number of $\mathrm{t_{2g}}$ electrons is not changing upon going from $\ce{Mn^3+}$ to $\ce{Mn^2+}$.  In both $\ce{Mn^2+}$ and $\ce{Mn^3+}$ there are 3 $\mathrm{t_{2g}}$ electrons.  I think the point they are trying to make with $\ce{Cr^3+}$ being half filled is that there are 3 $\mathrm{t_{2g}}$ electrons and no $\mathrm{e_g}$ electrons.  $\ce{Mn^3+}$ and $\ce{Mn^2+}$ both have $\mathrm{e_g}$ electrons.

You could make a similar argument for $\ce{Mn^3+}$ being oxidized to $\ce{Mn^4+}$, which would have half filled $\mathrm{t_{2g}}$ and empty $\mathrm{e_g}$.

  Why can't the half-filled d-orbital argument be used for Cr?

Because neither $\ce{Cr^3+}$ nor $\ce{Cr^2+}$ has a half filled d subshell.  $\ce{Cr^3+}$ has 3 electrons while $\ce{Cr^2+}$ has 4 electrons.


Orbitals are defined by a set of wavefunctions. Hence, when two orbitals overlap they can either interfere constructively (bonding) or destructively(anti bonding).

Lets take two pz orbitals orbitals overlapping with each other. You must understand that the p orbitals look like a dumbbell. One side of the dumbbell is + and the other side is -.

If - meets - we have constructive overlap. If - meets + we have destructive overlap.

The former gives a bonding orbital ($\sigma$) and the latter an antibonding orbital ($\sigma*$).

Depending on the molecule it is often energetically favourable to be in a bonding state hence bonding orbitals are lower in energy to their anti-bonding counterparts.


If aliens lived in a hydrogen (or any flammable gas) based atmosphere, would they perceive oxygen to be flammable?

Absolutely yes.

Lighting a torch in such an environment would simply be the reverse physical process (and same chemical process) of what is done in our oxygen-containing atmosphere. In the chamber or alien world of hydrogen gas, providing an ignition source to a stream of oxygen would give a flame. The chemical reaction would actually be the same as if you had lit a hydrogen stream in an oxygen-containing atmosphere:

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

The only thing we've done is reversed the sources of the $\ce{H2}$ and $\ce{O2}$, so the hydrogen-breathing aliens would perceive oxygen to be explosively flammable, just as we perceive hydrogen gas to be so.  The bottom line is that it is the two of the gases together that are flammable.


Predicting the products of the radical chlorination of propane

Let me give you two helpful hints.

  Propane

If you use an internet search tool of your choice, you should quickly find out that propane is $\ce{C3H8}$, which has the structure $\ce{CH3CH2CH3}$. The formula and structure you are using is octane. The product you drew, 1-chlorooctane $\ce{CH3CH2CH2CH2CH2CH2CH2CH2Cl}$, is one of the monochlorination products of octane, so you should feel good about being able to correctly apply the reaction type and produce a correct product.

  What is the question asking?

A careful read of the intro part of the question When propane reacts with chlorine, two different monochloropropanes are formed indicates that we should be expecting two different organic products with chlorine atoms attached to the carbon skeleton. In other words this reaction should give two products with the formula $\ce{C3H7Cl}$.

However, part a) is not worded well to indicate that you mostly likely are to draw the structures of the two monochloropropane products. You are correct in noting that a radical halogenation reaction will have two products, though the inorganic byproduct is usually considered to be $\ce{HCl}$ and not $\ce{H2}$ or $\ce{H2O}$.

$$\ce{C3H8 + Cl2 -> C3H7Cl + HCl}$$

So let me ask your question in a clearer way:

  Propane reacts with chlorine to produce two different monochlorination products with the formula $\ce{C3H7Cl}$. (a) Draw the structures of  both monochlorination products and provide their names. (b) Identify the type of isomeric relationship that exists between the two structures that you drew in (a).


Why aren't Boron and Aluminium assigned to group 3 of periodic table? What determines the group?

Back in 1933, John Drury Clark published 1 his "race track" periodic table that subsequently became famous as the cover of Life magazine in 1949:

Looks like B and Al are nicely positioned. Just need to add an outer track for elements 105 - 118, element names and symbols for elements 97 - 103 and element 41 is niobium, Nb.

1 Clark, John D. (November 1933). "A new periodic chart". Journal of Chemical Education. 10 (11): 675–677. Bibcode:1933JChEd..10..675C. doi:10.1021/ed010p675.

Clark is also the author of Ignition! An Informal History of Liquid Rocket Propellants, Rutgers University Press Classics, New Brunswick, NJ, ©1972 and 2017.


Preparation of an aromatic Grignard reagent

I have prepared many aryl Grignards in my career. Generally the aryl bromide or iodide is easier than the chloride, but it can be done. THF is certainly fine for this, run under inert atmosphere, of course. Magnesium powder may prove to have more of an oxide film than turnings and hence be less reactive. Most Grignard preps use turnings. It might be helpful to give your $\ce{Mg}$ turnings a brief wash with $\pu{2N}$ $\ce{HCl}$, then rinse with $\ce{EtOH}$ and $\ce{Et2O}$ just before you use them.

To initiate the reaction - do not stir the reaction but either hotspot it by using a heatgun to heat the region immediately around where the turnings sit at the bottom of the flask or put the whole flask in an ultrasound bath. A crystal of iodine in the reaction mixture will often serve to initiate the reaction, when it decolourises you know the reaction is proceeding. Depending on scale the heat of the reaction may bring the solution to reflux so you'll need a condenser on it.

Finally if all else fails at couple of mls of Red-$\ce{Al}$-$\ce{H}$ solution should start it.

If all fails with $\ce{Mg}$ metal, consider running a Knochel exchange with $\ce{iPrMgCl}$.


Work done by an expanding gas, how to solve this particular example?

You have system where 1 mole of solid (0 moles of gas) decomposes into two moles of gas at a constant temperature (given) and likely a constant pressure (not given).

Your formula for work is not correct for this problem. You cannot have $n$ outside of the integral since $n$ is not constant.

Let's write the total differential of the ideal gas equation:

$$\begin{align}
\mathrm{d}(PV)&=\mathrm{d}(nRT) \\
P\mathrm{d}V + V\mathrm{d}P &= nR\mathrm{d}T + RT\mathrm{d}n
\end{align}$$

Since $T$ is constant and $P$ is constant, we know $\mathrm{d}T=0$ and $\mathrm{d}P=0$, so this simplifies to:

$$P\mathrm{d}V = RT\mathrm{d}n$$

Can you do something with this relationship that avoids integrating with respect to (unknown) volumes?


Relative stabilities of tropylium and cyclopropenyl carbocations

Both ions are indeed strongly stabilized by aromatic coupling, and salts containing both ions are known.  But, of course, aromaticity is not the only thing that determines stability.  Tropylium ion, with its lesser ring strain and more atoms over which to distribute the positive charge, forms ionic salts with more anions (e.g., ionic $\ce{C_7H_7Br}$ versis covalent $\ce{C_3H_3Br}$) and in that sense is more stable.
The ionic structure of some tropylium salts is documented in Wikipedia[1][2][3][4]

In 1891 G. Merling obtained a water-soluble bromine containing compound from a reaction of cycloheptatriene and bromine.[1] Unlike most hydrocarbyl bromides, this compound, later named tropylium bromide, is a water-soluble solid and is insoluble in hydrocarbons, chloroform, and ether. It is purified by crystallization from hot ethanol. Reaction with aqueous silver nitrate immediately gave a precipitate of silver bromide. The structure of tropylium bromide was deduced to be a salt, C7H7+Br–, by Doering and Knox in 1954 by analysis of its infrared and ultraviolet spectra.[2][3] The ionic structures of tropylium perchlorate (C7H7+ClO4–) and tropylium iodide (C7H7+I–) in the solid state have been confirmed by X-ray crystallography.[4] The bond length of the carbon-carbon bonds were found to be longer (147 pm) than those of benzene (140 pm) but still shorter than those of a typical single-bonded species like ethane (154 pm).

References:

Merling, G. Ueber Tropin. Ber. Dtsch. Chem. Ges. 1891, 24 (2), 3108–3126. DOI: 10.1002/cber.189102402151.

von E.doering, W.; Knox, L. H. The Cycloheptatrienylium (Tropylium) Ion. J. Am. Chem. Soc. 1954, 76 (12), 3203–3206. DOI: 10.1021/ja01641a027.

Balaban, A. T.; Oniciu, D. C.; Katritzky, A. R. Aromaticity as a Cornerstone of Heterocyclic Chemistry. Chem. Rev. 2004, 104 (5), 2777–2812. DOI: 10.1021/cr0306790.

Kitaigorodskii, A. I.; Struchkov, Y. T.; Khotsyanova, T. L.; Vol'pin, M. E.; Kursanov, D. N. Crystal structures of tropylium perchlorate and iodide. Russ Chem Bull 1960, 9 (1), 32–36. DOI: 10.1007/BF01178699.


The difference between the different types of rhodamine

Rhodamine green
CAS: 189200-71-3

Rhodamine Red
CAS: --


$$\begin{align}
\text{AFR (Air to Fuel ratio)} &= \frac{n_{\text{air}}}{n_{\text{fuel}}} \\
\text{Equivalence ratio} &= \frac{\text{AFR}}{\text{AFR}_{\text{stoich}}} \\
\end{align}$$

According to the problem statement, if you did everything right,
$$\begin{align}
\text{AFR} &= 37.54 \\
\text{AFR}_{\text{stoich}} &= \frac{\nu_{\text{oxygen}}}{\nu_{\text{fuel}}}\times 4.76 \\
\ce{C4H10 + $6.5$\,O2 &-> 4 CO2 + 5 H2O} \\
\text{AFR}_{\text{stoich}} &= 4.76 \times 6.5 = 30.94 \\
\text{Equivalence ratio} &= \frac {37.54}{30.94}=1.21
\end{align}$$
Now there are two types of equivalence ratio. One is air to fuel and another is fuel to air. I suspect in the answer they meant fuel to air equivalence ratio:
$$\text{FAR equivalence ratio} = \frac{1}{\text{AFR equivalence ratio}} = \frac{1}{1.21}=0.82$$


Clearly, it cannot be an $\ce{O+}$, as that would require a triple bond to the oxygen, making the phosphorus in the middle quite unhappy.

Of the other two alternatives, both are correct -- however, it matters what environment you're considering the molecule. In neutral or slightly basic conditions, it will be an $\ce{O-},$ while in slightly acidic conditions, it will be an $\ce{OH}.$

Let me know if you need any clarification on why this is the case -- I have no idea what your level of knowledge is.


Cleaning stains with hydrogen peroxide and bleach

Peroxide already is a "bleach", and you don't say what the other one is. The gas you see evolving is most likely oxygen (from the peroxide).

Generally (most importantly) bleach oxidises double bonds (which make the colour in many organic compounds). The remaining molecule is also more polar afterwards, because an oxygen-containing group was inserted, and therefore easier to wash out with water.


How do I calculate the mass percent in the original mixture?

Looks close. Your math looks sound, but the reasoning is messy. $\ce{O2}$ is the only oxygen here.

$$\ce{2KClO3 -> 2KCl + 3O2}$$

You might have skipped writing out that reaction — it helps a lot.

If $0.700~\mathrm{g}$ of residue remained, that means $0.950\ \mathrm{g} - 0.700\ \mathrm{g} = 0.250\ \mathrm{g}$ of $\ce{O2}$ left the container (which means that $0.250\ \mathrm{g}$ of $\ce{O2}$ was made!)

We can calculate the number of moles, similar to what you did.

$0.250\ \mathrm{g} \times \frac{1\ \mathrm{mol}}{32\ \mathrm{g}} = 0.00781\ \mathrm{mol}\ \ce{O2}$

However, we have no reason to divide by 2 — $\ce{O2}$ is the only way (well, at least for this purpose) oxygen can exist in nature. Treat it as "oxygen".

Let's apply a mole ratio, now. There were $\ce{2 KClO3}$ molecules used for every $\ce{3 O2}$ molecules we got.

$0.00781\ \mathrm{mol}\ \ce{O2} \times \frac{2~\mathrm{mol}\ \ce{KClO3}}{3~\mathrm{mol}~\ce{O2}} = 0.00521\ \mathrm{mol}\ \ce{KClO3}$ which we had at the beginning. Note that we can do this because it says all of it reacted completely. If only $95~\%$ of it reacted, we would multiply by $0.95$, for example.

Find the grams of $\ce{KClO3}$ expended in reaction.

$0.00521\ \mathrm{mol} \cdot \frac{122.55\ \mathrm{g}}{1\ \mathrm{mol}} = x~\mathrm{g}\ \ce{KClO3}$

Your percent is pretty easy from there.

$\frac{x\ \mathrm{g}}{0.950\ \mathrm{g}} \times 100~\% = y~\%$

Where I'm sure you can figure out $x$ and $y$. (Remember to find $x$ to 3 significant digits).

Leave a comment if I wasn't clear on something!


As you might have guessed: It depends!

What is the educational level of the study (High School, College, etc.)?
What's the time frame - how many man-hours?
What's the equipment of the lab?
If equipment or material has to be bought: What is the budget?
Is it a plain academic study or is it intended to go public with the results in order to change/improve consumer awareness?

"Something unique" might mean that

there are no reference data
it is too "academic" to have social implications (or at least that it will be difficult to explain to the public that it actually has)

What about something as "boring" as sugar content in local soft drink brands as compared to similar products on the international market? I'm sure you'll find lots of references, such as this.

Go for something that is relevant for the average Joe Bloggs. Go for something that can be bought thoughout the while time of the study, just in can you have to repeat measurements or need to cover seasonal variations. If it is a fresh, seasonal product, you might have to plan ahead and stockpile and deep-freeze material.

In the end, it is up to you to decide - planning and outlining the study is part of the project.

Speaking of relevance for the average consumer: Everybody needs excrete - but you do not want to examine that. Fortunately, everybody needs to eat too and seemingly, a lot of people tend to eat too much of the wrong stuff. The recent report of the US Dietary Guidelines Advisory Committee might give some directions what to look for.


Relation between activation energy and temperature sensitivity

In chemicals reactions that have more than one product, being one of them the kinetic product and the other the thermodynamic product, the temperature dependency is a relevant factor. For example, the kinetic vs. thermodynamic control can manipulate the regiochemistry of the alkylation of enolates. In other words, you have a competition between reactivity and stability.

Reactions are usefully to describe it in terms of potential energy diagrams, $\Delta G$ vs. reaction coordinate ($\xi$), where a one-step reaction have a global maximum corresponding to a transition state leading to the product and the free energy of this transition state, relative to the reactant, determine only the rate of reaction. The enthalpy part of $\Delta G$ deal with the stability of the molecule or intermediate but this doesn't give information about the rate of reaction.

Absolute rate theory (or transition state theory), by Eyring, have an expression for the rate constant $k$:

$$k=\frac{\kappa k_{B}T}{h}e^{\Delta S^{\ddagger}/R}e^{-\Delta H^{\ddagger}/RT}$$

where $\kappa$ is the transmission coefficient, $k_{B}$ the Boltzmann's constant, $h$ the Planck's constant, $R$ is the gas constant and $T$ is the absolute temperature. Here, the term $\frac{\kappa k_{B}T}{h}e^{\Delta S^{\ddagger}/R}$ varies slightly with temperature compared to the enthalpic term. Arrhenius equation also reflects the temperature dependence of reactions:
$$k=Ae^{-E_{a}/RT}$$
For remark, both relationships are temperature dependent.

So, if you have a reaction under kinetic control, the ratio of the products is determined by the relative energies of the transitions states. Whereas under thermodynamic control, the ratio of the products is determined by the relative energies of the products. This leads to the question if there are any relationship between $\Delta G$ or $\Delta G^{\ddagger}$, where you might search about Bell-Evans-Polanyi formula, Hammond's postulate and the Marcus equation.

Suppose you have a reaction
$$\text{A}\leftarrow\text{R}\rightarrow\text{B}$$
with the energy profile:

The product A have a lower $E_{a}$ or $\Delta G^{\ddagger}$ (kinetic product), whereas B have a higher $E_{a}$ but a lower $\Delta G$. The potential energy profile generally is temperature independent, i.e. the barrier and the heat of reaction are determined by the molecular structure. When you raise the temperature, the kinetic energy of the molecules is raised and this helps the molecules to pass the potential energy barrier more easily (you can check it by a Maxwell-Boltzmann distribution for a molecule changing the temperature and fixing an energy corresponding to $E_{a}$), being obvious that if the barrier is smaller will be more easy to pass.

If the temperature is high enough to make the molecules pass over all the barriers on the energy profile, the molecules will choose the pathway with the most stable product. All of this is determined by whether the products are or aren't able to equilibrate under the reaction conditions (search for the principle of microscopic reversibility). So, at lowers temperature you will disfavor the equilibrium and only get the kinetic product.


React benzene with chlorine in the presence of $\ce{FeCl3}$ to yield chlorobenzene.

According to Ullmann's Encyclopedia of Industrial Chemistry:

Most monochlorobenzene is now produced from benzene and chlorine in continuously operated plants. Depending on the ratio of benzene to chlorine chosen, one can achieve either a low rate of benzene conversion and little dichlorobenzene
formation, or almost complete conversion of the benzene with a higher degree of dichlorobenzene formation. [...] The composition of a chlorination mixture containing the highest possible proportion of monochlorobenzene has been given as
4–5% unreacted benzene, 73% monochlorobenzene, and 22–23% dichlorobenzenes.

Nitrate chlorobenzene to obtain a mixture of 2-nitrochlorobenzene and 4-nitrochlorobenzene. Separate the isomers.

In Ullmann's Encyclopedia of Industrial Chemistry, we find:

Nitration of chlorobenzene with mixed acid (30/56/14) typically gives an isomer
mix in 98 % yield consisting of 34–36% 2-chloronitrobenzene, 63–65 % 4-chloronitrobenzene, and only ca. 1% 3-chloronitrobenzene.

"Mixed acid" refers to a mixture of nitric acid, sulfuric acid and water.

React 4-nitrochlorobenzene with ammonia (Nucleophilic aromatic substitution) to obtain 4-nitroaniline.
According to Ullmann's Encyclopedia of Industrial Chemistry, the reaction conditions as as follows:

4-Nitroaniline is produced in an agitated titanium reactor in 99.3 % yield by reacting 4-chloro nitrobenzene with a tenfold excess of aqueous ammonia at 175 °C and 4.2 MPa (42 bar) for 10 h.

Ullmann's Encyclopedia of Industrial Chemistry cites a patent by MONSANTO (DE1768518) as the relevant reference.

Reduce the nitro group with iron.


How can you tell if a sugar is D/L in its chair form?

D-glucose stays D-glucose regardless of whether it is in chair (hemiacetal) form, open chain form, or any other minor form. These different forms (if you could isolate them) would likely have different specific rotation values (interaction with plane polarized light). There are two chair forms-one with an axial OH and the other equatorial (probably favored)--but both are forms of D-glucose.


Bridged ring system reduction through Wolf Kisher: Is it possible?

If you are thinking of reducing the ketone with hydrazine under alkaline conditions then I see a problem. The hydrazine will also act as a nucelophile on the alkyl bromide in the molecule. I think that you will have a SN2 reaction on the alkyl bromide to form an alkyl hydrazine at the same time as the hydrazine is reacting with the ketone. If you then heat it up with base you will remove the ketone but assuming that you use a large excess of hydrazine you will end up with an alkyl hydrazine in the top left of the molecule rather than an alkyl bromide.

Why not react the original compound with benzyl alcohol using a mild base to form a benzyl ether at the start. Then do the reduction with hydrazine, then hydrogenate with hydrogen gas and palladium on carbon. This hydrogenation will remove the benzyl group and give you a alcohol. Then you can react the alcohol with tosyl chloride/pyridine and use the tosylate as an electrophile. If you very much want an alkyl bromide then react the tosylate with sodium bromide in DMF.


This is an old question, but in case anyone comes down this road...

  Aren't alpha halo carbonyl compounds extremely fast and reactive via $\ce{S_{N}2}$
  (the fastest i guess)?

I don't think so. In an $\ce{S_{N}2}$ reaction a nucleophile is attacking a positively polarized carbon atom.  The more positive charge on the carbon, the more the nucleophile is attracted.  Look at structure I below, it shows one of the resonance structures for a carbonyl that leads to a polarized carbonyl with positive charge on the carbonyl carbon.  Now to whatever degree the adjacent C-Cl bond is polarized with positive charge on carbon, we wind up with an unfavorable electrostatic situation of adjacent positive charges.  Said differently, the presence of a carbonyl group suppresses polarization of the adjacent C-Cl bond, which will make the C-Cl carbon less prone to nucleophilic attack.

  And if the alpha carbon(to which the halide is attached) is tertiary,
  there should be a high chance of an elimination product via E 2  (leading to the
  formation of alpha beta unsaturated carbonyl), right?

No, you want to remove hydrogen D in the diagram below.  Hydrogen D is much less acidic than hyrdrogen A because removal of the latter results in a resonance stabilized carbanion.  The difference in acidity between hydrogens A and D is so large that effectively only hydrogen A will be removed by a nucleophile.

These two arguments effectively eliminate pathways C and D from consideration.  Pathway B is simply an equilibrium that goes nowhere, so it has no real effect on the outcome of the reaction. This leaves pathway A, the Favorskii reaction, as the only viable alternative.  Pathway A, besides being the only game in town, is actually favored by two factors, 1) as noted above, hydrogen A is acidic and it's removal produces a resonance stabilized carbanion, and 2) completion of the Favorskii reaction is strongly favored by entropic factors since the carbanion and the C-Cl carbon are contained in the same molecule and held relatively close together.


Would there be a solid residue left behind from a reaction between NaK and H2O?

According to this site:-

  NaK reacts explosively with water to form hydrogen, potassium hydroxide
  and sodium hydroxide.

This site says:-

  NaK is a strong reducing agent. It is corrosive and will react violently with water. The reaction with water will liberate extremely flammable gases potentially causing a fire or explosion.

This youtube video confirms the above fact.


Why do polar aprotic solvents favour SN2 over E2?

Competition between substitution and elimination is one of the many points of confusion for introductory organic students.
In general, they are taught that there are four factors that will determine the outcome of a reaction where there may be competition between substitution or elimination.
They are:

Type alkyl halide (methyl, $1^o$, $2^o$ or $3^o$)
The strength/basicity of the nucleophile
The solvent
The temperature

I think that many students over-emphasize the impact of any one of these factors.  It is more useful to look at the overall impact of the factors and evaluate the reaction as a whole.

Let's examine the case in question:

Primary alkyl halide - generally favours substitution
The nucleophile is $\ce{^-CN}$ which is a good nucleophile but a weak base.
Polar aprotic solvent
Low temp.

The big factor here is not the solvent.  All of the factors point toward substitution, especially the alkyl halide and the nucleophile.  Remember that elimination reactions are essentially acid-base in nature.  If the nucleophile is not a strong base, elimination reactions are less likely.


Investigating a factor affecting buffer capacity

As far as I can tell, something like temperature won't affect buffer capacity.

In this particular case, I wouldn't be too sure about it. Isn't it conceivable that your buffer will lose ammonia upon heating?

Did you already decide about the pH indicators? For the lower end, thymol blue might be a good choice. The supplementary data
to J. Mater. Chem. B, 2015,3, 2801-2807 contains a UV/VIS spectrum of this indicator.


Is there always a need for a salt bridge to let a galvanic cell work continuously?

It depends where and how the electrodes are positioned. The UC Davis ChemWiki page Electrochemistry 2: Galvanic cells and electrodes describes the possible arrangements.

For example, if the electrodes are placed in the same vessel, a salt bridge is not required (as shown below). Ions can pass through the porous barrier.

If, however, the electrodes are are placed in separate vessels (such as in the diagram below), then a salt bridge is required to facilitate ion movement.


Why doesn't open-chain glucose react with water readily in aqueous solution?

Why doesn't water react with the aldehyde to form a linear geminal diol?

It does, and depending on the nature of the aldehyde, this will be the major species. You are right in saying that the high concentration of water helps to form the diol. In other solvents, the diol form will be present at lower fraction, even when they contain a bit of water. The Wikipedia article on geminal diols has some equilibrium constants for various aldehydes and ketones in aqueous solution: https://en.wikipedia.org/wiki/Geminal_diol#Reactions

  Glucose's overwhelming preference as a cyclic pyranose is commonly explained by the cyclization being intramolecular; while the aldehyde group in its linear form can react with other molecules like protein and lipids during rare encounters, it far more readily reacts with an alcohol group that is always around by virtue of being on the same molecule.

Yes, that is the special thing about glucose: the aldehyde can form an intramolecular hemiacetal without strain. It can also form a geminal diol, but that is no different from other aldehydes, so it is usually not mentioned in that context.

Is the geminal diol ever important for the reactivity?

There is an enzyme called aryl sulfatase which has an aldehyde (a cysteine modified post-translationally to a formyl glycine) in its active site [review]. X-ray crystal structures show that the main form is the hydrated species, i.e. the geminal diol. In the suggested mechanism, one hydroxyl attacks a sulfate ester, displacing the alcohol. Then, the second hydryoxyl group reforms the aldehyde and releases the sulfate in an elimination reaction.

Are there other instances where we don't explicitly mention reaction with water?

The hydrogen ion occurs in aqueous solution in various hydrated states. Carbon dioxide in aqueous solution occurs either hydrated (carbonic acid) or not. In either case, it is easier to lump the species into one than to talk about them separately.

What is the equilibrium constant for the hydration of the aldehyde in glucose

I did not find data for glucose, but did find some for glyeraldehyde 3-phosphate (Petersson and Petersson 1999). [In the HTML version of the paper, some equations and schemes don't show, but the PDF is an improvement.] In aqueous solution, only about 3% of this phosphorylated sugar is present in the aldehyde form. In the reaction they were studying, only the aldehyde form reacts in the enzymatic reaction, so there is a slow non-enzymatic first step to dehydrate the diol and a fast second step.

Wikipedia has an article on glycoaldehyde, with 4% of it in the aldehyde form.
If glucose behaves in a similar manner, most of the linear form would be in the diol form. Equilibrium constants given in the literature probably use the total concentration of linear glucose (hydrated or not) in the equilibrium constant expression rather than the actual concentration of aldehyde.


Polypyrrole is quite a rigid polymer (one of the reasons it's a conductive polymer) and it's generally not considered particularly soluble without being derivatized. It seems to be possible to dissolve it somewhat in polar organic solvents if it's been polymerized with a large anion like dodecylbenzenesulfonate, basically reducing the interactions of the chains of polymer, but it loses some of its conductivity when cast into a film. One group has apparently managed to make it water soluble by polymerizing oxidatively with horseradish peroxidase in the presence of a sulfonated polystyrene counterion, which keeps the chain growing linearly along the length of the polystyrene, apparently allowing it to be soluble.


What happens when an ionic compound is boiled?

Interesting question with no generic answer. Boiling the salts, or ionic compounds, makes them a plasma or freely moving ions is nothing but a misconception. The temperature is too low to convert a substance into a plasma.

Let us start with ionic liquids which are room temperature cations and anions but made up of organic molecules. You heat them, to the state of boiling in the absence of oxygen. They decompose into various products. So nothing special, this is as expected for organic compounds.
Heating of non-decomposable salts at high temperatures, for example, NaCl, KCl: What will happen when we heat them to the point of boiling? Answer is nothing. They remain in vapor phase as NaCl "molecules". The vapors behave as gases. People have developed equations of state. Interestingly enough $\ce{Na2Cl2}$ (dimers) are found in gas phase.

See: Critical point and vapor pressure of ionic fluids including NaCl and KCl, Chemical Physics Letters, 105, 1984, 484-489

Ionic materials, KCl or NaCl, when introduced into flames or plasma convert into gas phase "atoms". For example $\ce{Na+}$ would be converted into a neutral Na atom. This is utilized in a completely different field in chemistry, called flame atomic absorption or flame atomic absorption spectroscopy.

In short, these complexities make science very interesting.


Yes it can happen.  This paper offers an example, in which $\ce{BF3}$ or $\ce{B(C6F5)3}$ adds to a low oxidation state iridium complex.  There is really not a lot of mystery behind it.  The iridium in its low oxidation state has electrons to offer and the boron with its vacant orbital can take them up.