What's the difference between a nucleophile and a base?

The two are related, in that most nucleophiles are (Lewis) bases and vice versa. Some good nucleophiles are also strong bases, e.g. $\ce{HO-}$. However, a species can be a good nucleophile and a weak base, e.g. $\ce{I-}$; or a species can be a weak nucleophile and a strong base, e.g. $\ce{t-BuO-}$. How can we separate this behavior?

  Nucleophilicity is a kinetic phenomenon.

Nucleophilicity is most often defined based on the relative rate of the reactions of nucleophiles with a standard substrate in a standard solvent.

For example, a standard reaction might look like:

$$\ce{CH3I ->[Nu-][H2O] CH3Nu}$$

The nucleophilicity will be related to the relative rate constant of reaction with the nucleophile (relative to the rate constant of the reaction with water $\equiv 1$).

  Basicity is a thermodynamic phenomenon.

Basicity is based on the position of equilibrium:

$$\ce{B + HSol <=> BH+ + Sol-}$$


The reaction proceeds via a SFR (Substitution Free Radical) mechanism and yields an allylic or benzylic halogenated alkene. This result is specific for low concentration of halogen and high temperature.

The results of the same reaction of halogens with alkenes but at different reaction conditions, i.e. high halogen concentration and room temperature result into anti addition of the halogen atoms on the double bond. Thus the reaction takes place faster in presence of $\ce{CCl4}$, i.e. non polar solvents. The reaction proceeds via formation of cyclic intermediate. The result of the same reaction in water instead of $\ce{CCl4}$ will result in to anti addition of halide ion and $\ce{OH-}$ ion on the double bond.


The book that you're reading is measuring by mass.

If you have pure water then you would expect oxygen to make up $\frac{16}{16 + 2}\times 100\% \approx 89 \% $ by mass. Likewise, hydrogen would make up $\frac{2}{16 + 2}\times 100\% \approx 11 \% $ by mass.


I do agree with your point and Philipp explanation. Getting more formal, your $K_p$ at 35°C is:

$$ K^\circ_p(308.15\,\mathrm{K}) = \frac{\left(\frac{p_\mathrm{tot}}{p_\circ}\right)^2\chi^2_\mathrm{NO_2}}{\left(\frac{p_\mathrm{tot}}{p_\circ}\right)\chi_\mathrm{N_2O_4}} $$

Where $p_\mathrm{tot}$ is the pressure in your flask and $p_\circ$ the standard pressure. The $\chi$ quantities stand for molar ratio. For the reaction given, those ratios are:

$$ \chi_\mathrm{NO_2} = \frac{n_{NO_2,0} + 2\xi}{n_{N_2O_4,0} + n_{NO_2,0} + \xi} $$

and:

$$ \chi_\mathrm{N_2O_4} = \frac{n_{N_2O_4,0}  -\xi}{n_{N_2O_4,0} + n_{NO_2,0} + \xi} $$

Those identities come from amount balance, where $\xi$ is the molar coordinate of the reaction and zero-indexed quantities amount of matter when starting experience.

Assuming that initially there was only $\mathrm{N_2O_4}$, and you are using the following definition of dissociated ratio:

$$ \tau = \frac{\xi}{n_\mathrm{N_2O_4},0} = 0.15$$

You will obtain by substitution, molar ratio values:

$$ \chi_\mathrm{N_2O_4} = \frac{0.85}{1.15},  \chi_\mathrm{NO_2} = \frac{0.3}{1.15}$$

Then you inject those values in the $K^\circ_p$ expression above. Do not forget to express pressures in bars - this is a common mistake, you must remeber that $K_p$ are dimensionless. Therefore, $p_\mathrm{tot} = 1.2 \times 1.01325 \,\mathrm{bar}$ and $p_\circ = 1.0 \, \mathrm{bar}$. This is also an approximation because we are assimilating pressure as activities. But in your case you can consider fugacities close to unity and assimilate your gas mixture as an ideal gas.

Yes, you were right about molar ratio. Do not forget that equilibrium constants are dimensionless: they are ratio of activities!


Regarding Question 1:

NMR measures the energy difference between the field-aligned energy of a nuclei and the field opposed energy of the nuclei.

The biggest field in the situation is the external magnet.

The next biggest field is the magnetic moments of the electrons.  Electron density leads to what we refer to as chemical shift.

Lastly, there are the magnetic moments of other nuclei.  But, the other nuclei are also undergoing their own spin flipping and such like.  So if a hydrogen has its molecular neighbor with a spin aligned with the external magnetic field while relaxing, the total field it experiences is stronger.  Otherwise it is weaker.  The difference between stronger and weaker is essentially the distance between the two peaks on the NMR spectrum.  It is important to remember the external magnetic field is the strongest here, so when the nuclear spins oppose and weaken (or align and strengthen) the magnetic field, it is only a tiny amount.

Regarding question 3:

Carbon and Oxygen have complete Nuclear quantum shells, and as a result have no net magnetic moment. The concept of 'spin up' or 'spin down' means nothing to these nuclei.

Since splitting is the result of a magnetic interaction between spins, there will be no splitting.

Regarding chlorine, The coupling constants are small, and generally not observed due to limitations in the machines making the measurements.


I give a very general answer: it depends on the solid and the dopant.

One of the main issues is best viewed from a thermodynamic point of view: you want to create something similar to a solution. The dopant should randomly occupy some positions in the solid, and this should be comparably stable or more stable than the pure phases themselves. The dopant can substitute an atom (then it is often chosen to be with similar size and chemical properties), but it can go originally unoccupied places, too.
Some examples:

You can introduce dopants with co-crystallization. You have a method to prepare the host solid through crystallization from its melt or from a solution, then you can mix the dopant during the synthesis, and if you are lucky it works. With molecular materials or simple inorganic salt, it can be a relatively easy process.
Chemical vapor deposition techniques are often used to create (doped) thin films.
Some dopants can be introduced by simple immersion. Hydrogen can diffuse inside of the host material, and act as a dopant in metals and other materials. The preparation can be just letting the sample stay in hydrogen gas for a longer period. Similarly, there are many oxides where oxygen acts as a dopant, and its level can be controlled by heat treatment in oxygen (or in a vacuum, to remove the extra oxygen). My understanding is that the boron doping in silica using diborane gas is also based on similar principles.


Validating a method to meet accuracy criteria using precision and accuracy?

Possibly, but the rational is a bit convoluted.

First, as you noted there is a fundamental difference between precision and accuracy. You explained the difference correctly.

Now let's think of a hypothetical example. Suppose that you have a pan balance in the lab and an analytical balance down the hall in a balance room.
So you are going to use the analytical balance (which is accurate and precise) to validate the pan balance.

You weigh a rock specimen on the pan balance and the weights have a precision of 10%. This means +/- 10%. So to get a better precision, you weigh the rock multiple times. Now the average weight will have a better precision than a single weighing. The average will be smaller by a factor of 1/sqrt(n). So 4 weighings will give you +/- 5%.

So you get a bunch of rocks that span whatever weight range that you would use. You weigh the rocks on the pan balance and on the analytical balance (which is assume to give the "true" weight). Now you can establish a calibration curve for the pan balance.

The calibration curve is going to have some error too, so you'll probably have to adjust how many repeat measurements and how many different samples you need to get the desired accuracy. But it should be possible in this example. The real twist here is how you "validate" the calibration.


Arranging different oxygen-containing organic compounds by decreasing acidity

And this is the type of question for which Professor Mayr made us chemistry students rote memorise typical $\mathrm{p}K_\mathrm{a}$ values. Here are some.

\begin{array}{lc}\hline
\text{compound type} & \text{typical }\mathrm{p}K_\mathrm{a}\\ \hline
\text{alcohol} & \approx 15 \\
\text{ketone} & \approx 20\\
\text{carboxylic acid} & \approx 5\phantom{0}\\
\text{secondary }\ce{C-H} & \approx 45\\ \hline\end{array}

Therefore, the order is $\mathrm{Q > P > R > S}$


What does the symbol ϕ mean in skeletal formulae?

According to the Graphical Representation Standards for Chemical Structure Diagrams (IUPAC Recommendations 2008), the Greek lowercase letter phi $(\phi)$ may be used to represent a phenyl group.

  It is acceptable to use the Greek lowercase letter phi $(\phi)$ to represent a phenyl group. That abbreviation has a long history, and ambiguity is unlikely since that letter is rarely used for other purposes in chemical structure diagrams.

Note that this traditional use is only considered “acceptable”, i.e. not “preferred”.

According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), the preferred name for the hydrocarbon parent ring component is the retained name chrysene. The indicated numbering of this polycyclic hydrocarbon is is the result of the application of the specific criteria used to number fused ring systems:

Therefore, the complete name for the entire compound given in the question is 1,3,7,9-tetraphenylchrysene.


IMO this might be tungsten(VI) oxide, $\ce{WO3}$, which is fairly non-toxic, unless you consume it spoonwise.

From the look of this DIY setup I would doubt the quality of the light bulb itself was superb, and most likely one wouldn't use a fancy lamp anyway. So, the glass enclosure could be breached and some point, and the inner inert atmosphere (Argon/Nitrogen) was substituted with air. When hot tungsten filament oxidizes, it produces tungsten oxides which evaporates/scatters from the filament as white smoke:

$$\ce{2W + 3O2 -> 2WO3}$$

Actually, bulky $\ce{WO3}$ is yellow, but being dispersed in a form of tiny crystals, tugsten(IV) oxide has pale-yellow or white color depending on the fraction.

Anyway, I would get rid of this can enclosure and buy a decent light source, mainly from an aesthetic point of view.

P. S. Here is a funky experiment showing what could happen in your case: https://www.thenakedscientists.com/get-naked/experiments/light-bulbs-liquid-nitrogen


How does pH of soda drink vary with time

I will make few simplifications:

I will neglect presence of other acids like citric or phosphoric acid, considering just water and carbon dioxide.
I will use the simplified exquation of $\mathrm{pH}$ of a weak acid (  :

$$\mathrm{pH}=\frac 12 (\mathrm{p}K_\mathrm{a} - \log {[\ce{CO2}]}) \tag{1}$$

I will  consider $\ce{CO2}$ escaping as the exponential process of the 1st order kinetics.

$$[\ce{CO2}]=A + B \cdot \exp{(-Ct)} \tag{1a}$$

I will  neglect the final equilibrium $\ce{CO2}$ concentration, compared to initial one:

$$[\ce{CO2}]=B \cdot \exp{(-Ct)} \tag{1b}$$

We can realize that the exponential function for $\ce{CO2}$ escaping kind of nullifies the logarithm function of $\mathrm{pH}$ definition:

$$\mathrm{pH}=\frac 12 \left(\mathrm{p}K_\mathrm{a} - \log {\left(B \cdot \exp{(-Ct)}\right)}\right) \tag{2}$$

$$\mathrm{pH}
\simeq \frac 12 \left(\mathrm{p}K_\mathrm{a} - \log {B} + \frac1{2.303}Ct\right) \tag{3}$$

Therefore, $\mathrm{pH}=f(t)$ is approximately linear.

It is not linear if we consider all factors, especially if other acids are present.

Additional requested clarification:

The equation (1) is quite notoriously known simplified equation for $\mathrm{pH}$ of weak acid. It can be directly derived from definition of a dissociation constant of an acid:

$$K_\mathrm{a}=\frac{\ce{[H+][A-]}}{\ce{[HA]}} \tag{4}$$

involving 2 simplifications:

solution is acidic enough to ignore water auto-dissociation:

$$K_\mathrm{a} \simeq \frac{\ce{[H+]}^2}{\ce{[HA]}} \tag{5}$$

solution contains enough of a weak acid, so it's majority is not dissociated:

$$K_\mathrm{a} \simeq \frac{\ce{[H+]}^2}{c} \tag{6}$$

where

$$c=\ce{[HA]}+\ce{[A-]} \tag{7}$$

Then

$$\ce{[H+]}=\sqrt{ K_\mathrm{a} \cdot c} \tag{8}$$

After logaritmization of (8), we get (1).


Distinguish between symmetric and asymmetric molecules with IR

Whether the stretch of a bond in a molecule is asymmetric or symmetric depends on the relative movement of the atoms connected by the bond. Lets take a $\ce{CH2}$ group as an example. If the two hydrogen atoms connected to the carbon move in the same direction during stretching, it s a symmetric stretch, if they are going in opposite directions, the stretch is asymmetric (see also here):

Left: symmetric stretch, right: asymmetric stretch

A bond vibration like stretching will only be IR-active (i.e. give a band in the IR spectrum) if it is accompanied by a change of dipole moment. In the case of the $\ce{C-H}$ bond, carbon is slightly negatively and hydrogen slightly positively polarized. A symmetric stretch increases and decreases the distance between these partial charges periodically, and therefore also changes the dipole moment, because the latter is dependent from the distance between charges (source). During an asymmetric stretch, the elongation of one $\ce{C-H}$ bond is compensated by the shortening of the other, so the overall change of dipole moment should be weaker.

In the case of 2-(4-tert-butylphenyl)ethanol (the correct name for your compond), the band at 3006 $\mathrm{cm}^{-1}$ can either be assigned to the $\ce{CH2}$ groups of the substituted ethanol or to $\ce{C-H}$ bonds in the benzene ring. In the first case, you can distinguish between symmetrical and asymmetrical stretch (symmetrical seems more likely to me), while in the latter case, a single hydrogen is attached to a carbon, and a distinction of the stretching, like in the case of two substituents at the same carbon, cannot be made.


Given the concentration and density of a sample, calculate moles of bleach?

The key issue to address this kind of problems is to determine the mass of the pure compound in the given volume of the solution. So, we'll use first the density to determine the mass of the solution. Then, the mass concentration to determine the mass of the pure compound. After that, we can readily compute the number of moles of the pure compound using the molar mass.

The mass of $1\,\mathrm{mL}$ of bleach solution is $1.07\,\mathrm{g}$.
So, the mass of $4~\mathrm{mL}$ of bleach solution is $4 \cdot 1.07\,\mathrm{g}$

Or, each $100\,\mathrm{g}$ of bleach solution has $6\,\mathrm{g}$ of pure bleach.
So, the mass of  $4 \times 1.07\mathrm{g}$ of bleach solution has $\frac{4\cdot 6 \cdot 1.07}{100}\,\mathrm{g}$ of pure bleach.

The number of moles in  $4\mathrm{mL}$ of bleach solution is: $$\frac{4\cdot 6 \cdot 1.07}{100\cdot 74.44}= 0.0034\,\mathrm{mol}$$


To clear up what are becoming confusing comments.

An atom is a nucleus of protons and neutrons surrounded by electrons.

An atom can decay by fission to make two or more atoms with a smaller number of protons and neutrons in each new atom.  The electrons get distributed between them.  Now in physics we'd normally describe this as a nuclear decay, rather than thinking about electrons around the nucleus which, to a certain extent, are irrelevant to the underlying process.

Now depending on how you want to think about this you're either increasing the number of atoms by splitting one into two, or you're creating two (or more) smaller sized atoms which you might loosely describe as halves of the original.

Fission products (the results of fission) do not generally contain exactly half the number of protons and neutrons and electrons in the original atom, even when those numbers are even, so we can't say they're generally exact halves.

Fission can also release neutrons - so a nucleus looses e.g. one neutron but still has the same number of protons left.  Even when the split is closer to 50-50, the process can still release some neutrons, and in fact that's exactly what is used in nuclear weapons and reactors.

Another typical decay type is by alpha particle emission.  An alpha particle is, in fact, a Helium nucleus, so again that's a split to two different nuclei.


How to determine the composition of a gas mixture after a combustion?

This is clearly a homework question. Yet, since OP has given good effort to solve the question, I'd like to give OP some insight to solve these kinds of questions, taking current question as an example.

Suppose the mixture contains $x\;\pu {g}$ of $\ce{CH4}$ and $y\;\pu {g}$ of $\ce{C2H4}$. The combustion reactions of $\ce{CH4}$ and $\ce{C2H4}$ can be given as:

$$\ce{CH4 + 2 O2 -> CO2 + 2H2O} \tag{1}$$
$$\ce{C2H4 + 3 O2 -> 2 CO2 + 2 H2O} \tag{2}$$

Thus, the amount of $\ce{CO2}$ produced from the reaction $(1)$ is:

$$x\,\pu{g}\,\ce{CH4} \times\frac{\pu{1 mol}\,\ce{CH4}}{\pu{16 g}\,\ce{CH4}} \times \frac{\pu{1 mol}\,\ce{CO2}}{\pu{1 mol}\,\ce{CH4}}\times \frac{\pu{44 g} \,\ce{CO2}}{\pu{1 mol}\,\ce{CO2}}=\frac{11}{4}x\,\pu{g} \,\ce{CO2}$$

And, the amount of $\ce{H2O}$ produced from the reaction $(1)$ is:

$$x\,\pu{g}\,\ce{CH4} \times\frac{\pu{1 mol}\,\ce{CH4}}{\pu{16 g}\,\ce{CH4}} \times \frac{\pu{2 mol}\,\ce{H2O}}{\pu{1 mol}\,\ce{CH4}}\times \frac{\pu{18 g} \,\ce{H2O}}{\pu{1 mol}\,\ce{H2O}}=\frac{9}{4}x\,\pu{g} \,\ce{H2O}$$

Similarly, the amount of $\ce{CO2}$ produced from the reaction $(2)$ is:

$$y\,\pu{g}\,\ce{C2H4} \times\frac{\pu{1 mol}\,\ce{C2H4}}{\pu{28 g}\,\ce{C2H4}} \times \frac{\pu{2 mol}\,\ce{CO2}}{\pu{1 mol}\,\ce{C2H4}}\times \frac{\pu{44 g} \,\ce{CO2}}{\pu{1 mol}\,\ce{CO2}}=\frac{22}{7}y\,\pu{g} \,\ce{CO2}$$

And, the amount of $\ce{H2O}$ produced from the reaction $(2)$ is:

$$y\,\pu{g}\,\ce{C2H4} \times\frac{\pu{1 mol}\,\ce{C2H4}}{\pu{28 g}\,\ce{C2H4}} \times \frac{\pu{2 mol}\,\ce{H2O}}{\pu{1 mol}\,\ce{C2H4}}\times \frac{\pu{18 g} \,\ce{H2O}}{\pu{1 mol}\,\ce{H2O}}=\frac{9}{7}y\,\pu{g} \,\ce{H2O}$$

Since the mixture produced $\pu{1.185 g}$ of $\ce{CO2}$ and $\pu{0.6 g}$ of $\ce{H2O}$, we can derive two equations with two variables, $x$ and $y$:

$$\frac{11}{4}x+\frac{22}{7}y=1.185$$
and,
$$\frac{9}{4}x+\frac{9}{7}y=0.6$$

Up to this point OP has taken correct path. Now, let's consider these two questions. These equations can be simplify as:

$$77 x + 88y=1.185\times 28$$
or
$$ 7x + 8y=\frac{1.185\times 28}{11}=3.016 \tag {3}$$
and,
$$ 7x + 4y=\frac{0.6\times 28}{9} = 1.867 \tag {4}$$

Thus, $(3)-(4)$ gives you:

$$4y = 3.016 - 1.867 = 1.149 $$
$$\therefore \; y = \frac{1.149}{4} = \pu{0.2873 g}\,\ce{C2H4}$$

And $(4)\times 2-(3)$ gives you:

$$7x = 1.867 \times 2 - 3.016 = \pu{0.718 g}\,\ce{CH4}$$
$$\therefore \; x = \frac{0.718}{7} = \pu{0.1026 g}\,\ce{CH4}$$

Thus, it is clear that you were still in the correct path on solving the same two equations for $x$ and $y$.

According to my solution, mass percentage of $\ce{CH4}$ is:

$$\frac{0.1026}{0.1026 + 0.2873} \times 100 = 26.31\% \; (w/w)$$

And mass percentage of $\ce{C2H4}$ is:

$$\frac{0.2873}{0.1026 + 0.2873} \times 100 = 73.69\% \; (w/w)$$

Since this is not the one of the given answers, let's check for mol% like what you have done:

$$n_{\ce{CH4}}=\frac{0.1026}{16}= \pu{0.0064 mol}$$

$$n_{\ce{C2H4}}=\frac{0.2873}{28}= \pu{0.0103 mol}$$

Thus, the percentage of composition for $\ce{CH4}$ would be:

$$\frac{0.0064}{0.0064+0.0103}\times 100 = 38.32\%\: \pu{mol}$$

Similarly, the percentage of composition for $\ce{C2H4}$ would be:

$$\frac{0.0103}{0.0064+0.0103}\times 100 = 61.68\%\: \pu{mol}$$

Accordingly, the most closest answer is (5), although both given values are $\approx 2\%$ off from calculated values. It is noteworthy that OP's calculations are correct to the end.

Note: This question is not well written. That's why I have to go through two calculations to see which $\%$ is they are talking about. Also, the sum of compositions given in answer choice (4) is not $100\%$.


How to properly relate quantum mechanical energy to the classical world

*So we cannot stop here as quantum theory and classical mechanics are virtually indistinguishable (with respect to raw energies) at this point correct?

This is wrong. Just because you can't immediately, instinctively comprehend the difference between the $10^{-18}\  \mathrm{J}$ and $10^{-71}\  \mathrm{J}$ and therefore think they are both "indistinguishably small" doesn't mean it's true. The difference is massive. Our brains just haven't evolved to deal with numbers whose logarithms are very far from 0, because for almost all of our existence as a species, it never had to. Perhaps this site may give some very qualitative though fun context on how brains perceive the magnitude of a number.

In order to cope with this limitation of ours, we have developed some strategies. One can be understood as a form of chunking, where we simply modify numbers in a way so they become closer to something we understand. You probably won't be able to immediately give an example of an object that weighs 200,000 ounces or $3\times 10^{-27}$ solar masses, though you may find it quite a bit easier if I say 6 tonnes instead. This is because even if we inherently know the concept of an ounce or a solar mass, we can't really comprehend the magnitudes of 200,000 or $3\times 10^{-27}$ (the latter of course being much worse), and in order to get our heads around it more easily, we instead invent a new unit to measure weights in which we can express the same information using numbers closer to 1. The whole idea of scientific notation for numbers and SI prefixes for units can be understood as a consequence of this.

Another way is the one you use; instead of reframing the units and numbers, reframe the situation. You may feel there is no difference between $10^{-18}\  \mathrm{J}$ and $10^{-71}\  \mathrm{J}$, but the 53 orders of magnitude in difference is about the same as the number of atoms in a human body and the number of atoms in the entire observable Universe! Notice how I simply took the ratio into a slightly more familiar context (you know the Earth is just a small part of the observable Universe, and a human being is puny in comparison to the Earth).

So all of your calculations are correct, but if Homo sapiens were smarter, they would be entirely unnecessary!


Unitary Transformation of the HF equations

You are going to want to check out Szabo and Ostlund page 143, they explain all the nitty-gritty for this matrix. The math itself is on page 22 and 23. But if you don't have a copy...

The $\mathbf{S}^{-1/2}$ comes from the need to orthogonalize your basis so that you get your equation (5) instead of the $\mathbf{FC}=\epsilon\mathbf{SC}$. It's one method of several to achieve this orthogonalization, and it's called symmetric orthogonalization.

So how do you compute this function of a Hermitian matrix? Well, functions on diagonal matrices are super easy; you just apply the function to each diagonal element. For non-diagonal matrices, this is not true. More to your question, you can't just take one over the square root of each element of $\mathbf{S}$. But you can if it's diagonal. And you can make any Hermitian matrix diagonal via a unitary transformation.

The trick, then, is to diagonalize $\mathbf{S}$ first, apply the inverse square root to each eigenvalue, then transform back using the eigenvectors. So you're going to want to

Solve $\mathbf{SU} = \mathbf{sU}$. Where $\mathbf{s}$ are the eigenvalues of $\mathbf{S}$, and $\mathbf{U}$ are the eigenvectors.
For each $s$ in $\mathbf{s}$ do $s^{-1/2}$.
Transform back to original basis by $\mathbf{S}^{-1/2} = \mathbf{Us}^{-1/2}\mathbf{U}^{\dagger}$


Why does [Co(NO2)6]4- ion have 3 unpaired electron as opposed to 1?

According to Elliot, Hathaway & Slade (1966) cited by this paper, Co(II) with nitrite ligands is low spin.

So maybe it is just a typo. For d7 low spin, you would expect one unpaired electron.


The problem with looking for information about stereochemistry on the internet is that many sources use outdated terminology, imprecise definitions or have downright erroneous information!

Here's a flow chart that I put together that I believe to be accurate, yet hopefully straightforward:

When using the flow chart, it is really important to remember that each decision on the flow chart is related to comparing two molecules, ie. it doesn't make sense to ask "what is compound A", but it does make sense to ask "how are compounds A and B related"? A molecule may be the enantiomer of another molecule, but a diastereomer of another.

Implicit in the flow chart is that stereoisomers have different 3-dimensional arrangements of their atoms, while having the same connectivity of atoms to each other.

The term "geometrical isomer" is now deprecated and its use is "strongly discouraged". It is an older term for cis-trans isomers, which are just a subset of diastereomers.

Back to your original post - the part in bold about diastereomers is problematic because the definition of diastereomers does not relate directly to stereogenic centers. It emerges that compounds with two or more chirality centers (the correct term for what are often referred to as stereogenic centers or stereocenters) can often exist as diastereomers, but chiral centers are not a pre-requisite. For example, cis- and trans-2-butene are diastereomers according to the definition.

A meso-compound is an achiral compound (the mirror image is super-posable), which is a member of a set of diastereomers, where at least one of those diastereomers is chiral. A meso-compound most commonly arises when the compound has two or more chiral centers, yet is not chiral due to an internal plane of symmetry. Meso-compounds don't fit neatly on the flow chart, which is based on comparing two molecules, but are a particular type of diastereomer.


Thermometer fluid is separated, defective?

No, it is not defective, just slightly damaged, possibly from shipping. You should call customer service at Cole-Palmer and see if they will replace (or repair) the thermometer.

If not, there are 2 ways to fix it. I haven't tried either one, so I don't know how well they work.

If there is a bulb at the top and the thermometer is low range (under 150°C), then you would heat the thermometer in an upright position until all the bubbles are pushed up into the bulb.

The other method is to cool the thermometer down until all the liquid is in the bottom bulb. The second method may not work if the liquid freezes before all the liquid has moved into the bulb.


The odour and aroma of wine consists of about 600–800 components. Many of these compounds have been described with a range of odour/flavour sensations, even in more than one category, and often have differing flavour impressions, dependent upon their concentrations in aqueous solution or in the air-space above.

The odours and flavours in wine tasting are typically divided into groups, for example: Floral, Woody, Rustic/Vegetal, Balsamic, Fruity, Animal, Empyreumatic, Chemical, Spicy, and Etherish. [Bakker, J.; Clarke, R. J. Wine flavour chemistry; 2nd ed.; Wiley-Blackwell 2012]

What you describe as “strong taste of soot” probably belongs to the Chemical group, which is associated with synthetic chemical manufacture.

Typical compounds in this group that might be responsible for the soot aroma are volatile phenols:

3-methylphenol (tarry/leathery odour)
4-methylphenol (tarry/smoky odour)
2-ethylphenol (smoky odour, phenolic flavour)
2-methoxyphenol (smoky/woody odour)

A special substance that is believed to be responsible for a strong, petrolly kerosene-like aroma developed during aging is 1,1,6-trimethyl-1,2-dihydronaphthalene, which is formed from carotene breakdown.


Is this the lowest chair conformer of Beta D-galactose

The chair conformation that you have drawn (4C1) is likely to be the most stable one, as it minimizes the number of heavy axial groups.

To determine the chair conformation of a hexose, it is generally easiest to draw it and compare it with β-D-glucose, where all heavy groups are equatorial and the conformation is 4C1. If the number of heavy axial groups becomes smaller when the conformation is changed to 1C4 (all equatorial groups in 4C1 become axial and vice versa), then it is likely that the conformation is 1C4.


How is the reaction rate constant affected by concentration?

...I also understand that the value of k, the rate constant, does not depend/is not correlated with the concentration of the reactant(s).

It would be easier to represent this in calculus, but I assume you have not gotten there yet.

Anyway, you are correct, $k$ is a constant.

$$\mathrm{rate} = k [A]^x \implies \mathrm{rate} \propto k$$

thus as $k$ increases, the rate will increase for a given concentration of $\ce A$. The question did provide that $x = 1$ but this is not needed to solve the problem. Think of rate as:
$$\mathrm{rate} = -\frac{\Delta [A]}{\Delta t}$$
(the reaction is removing $\ce A$) which is in the form of a slope. thus the larger the rate value, the lower the slope of the curve. Rate is dependend on concentration, but for the same concentration, a reaction with a higher rate constant will proceed faster, thus remove $\ce A$ fast and have a more negative slope.

Lets compare for the two $k$ values, $k_{upper}$ and $k_{lower}$

$$\mathrm{rate}_{upper}  = k_{upper}[\ce A]= -\frac{\Delta [A]}{\Delta t} \\
\text{and }\\
\mathrm{rate}_{lower}  = k_{lower}[\ce A]= -\frac{\Delta [A]}{\Delta t} $$

Lets say arbitrarily that two reaction rates are: $k_{1} = \pu{2 s-1}$ and $k_{2} = \pu{4 s-1}$ and the instantaneous concentration is $[\ce A] = \pu{0.6M}$ then the reaction rates would be as follows:

$$\mathrm{rate}_{1}  = \pu{2 s-1}\times \pu{0.6M} = \pu{1.2 M s-1} = -\frac{\Delta [A]}{\Delta t} \implies \frac{\Delta [A]}{\Delta t} = -\pu{1.2 M s-1}\\
\text{and }\\
\mathrm{rate}_{2}  = \pu{4 s-1}\times \pu{0.6M} = \pu{2.4 M s-1} = -\frac{\Delta [A]}{\Delta t} \implies \frac{\Delta [A]}{\Delta t} = -\pu{2.4 M s-1}$$

Since a reaction of $\mathrm{rate}_2$ has a more negative slope, it must correspond to a curve  that decreases faster than a reaction of $\mathrm{rate}_1$ and the concentration would be lower sooner, thus $\mathrm{rate}_2$ is the lower curve and $\mathrm{rate}_1$ is the upper curve.


What would Kp be if there is no gas reactant

Kp by definition is the product of the partial pressures of the gaseous products divided by the gaseous reactants raised to the power of their respective stoichiometric coefficient in the balanced chemical equation.

To answer your question, this definition by itself isn't enough. You need to go a little deeper. Since your reactants are not gaseous they got to be either liquid or solids. Compared to the gaseous products, the active mass of the liquid or solid does not change much. We assume the concentration of such substances to be constant value hence we shift them to the Kp to get a new Kp (actually we consider this Kp to be the correct constant). Or in other words, we can safely substitute 1 in place of all non-gaseous reactants while writing the Kp formula.

Therefore, Kp for your question is

= (P.Pressure of gaseous productA)$^a$ (P.Pressure of gaseous productB)$^b...$


As a home tool, but not necessarily a definitive solution, I would suggest the dissolution of your copper metal/alloy in a galvanic cell where the Cu is the anode and as a noble cathode, use a source of graphite.

This approach for the dissolution has been favorably argued in commercial patents in a 1999 U.S. Patent 6,692,632. To quote in parts:

  The method is preferably used in synthesis, but can also be useful in the destruction of pollutants etc. In certain embodiments of this nature, the use of external current applied to the system is not necessary, and a simple galvanic cell can be set up...

On copper particularly, for the home chemist, I have demonstrated a so-called Bleach Battery approach as an electrosynthesis path discussed in 'Subject: Experimental Variants of Bleach Battery to Obtain Metal Oxides (like CuO). However, I would replace the cathode, as in the current case as Nickel could be present as an impurity also in your copper metal sample, with very noble graphite in its path to CuO and other metal oxides that may become evident (or be chemically identified).

As the copper can be alloyed with metals like Zinc, Tin, Aluminum, Nickel, Silicon,..., with the first three creating respective oxides from the dissolving a small piece of the copper metal/alloy, potentially revealing distinct metal oxides as, for example, CuO is black, ZnO is white, SnO is blue-black (as produced by heating the tin(II) oxide hydrate, SnO·xH2O
where x<1), and Al2O3 varies from white to a clear transparent gel. The metals Nickel and Silicon are expected in their metallic state.

I say, convert a bit of it into metal oxides and proceed from there!


Orbital size of Carbon and Oxygen in Methanal

The orbital size in the picture shows the coefficient in the linear combination of atomic orbitals, i.e the pi bonding MO has a larger contribution from the oxygen atomic orbital, and the pi anti-bonding MO has a larger contribution from the carbon atomic orbital.

Here is another picture for a different molecule where they use the same convention, but also show the coefficients explicitly (source):

  Is the diagram for π bonding drawn correctly in terms of orbital size?

An orbital is not a "thing" so it does not have a size. It is a 3-dimensional function. The picture just shows the symmetry of that function and the node (where it is zero). For a slightly more realistic depiction, you can check out molcalc.


There are two other widely used acid/base theories that apply here: Brønsted–Lowry theory and Lewis theory.

A Brønsted base is defined as a proton acceptor. Technically, $\ce{CuO}$ is capable of reacting with $\ce{H+}$ to form hydroxide ions and eventually $\ce{H2O}$. Although $\ce{CuO}$ is insoluble in water (as you said), this reaction can occur in other solvents or in a biphasic system (where the proton is in solution and the $\ce{CuO}$ is present as an undissolved solid).

A Lewis base is defined as an electron donor. Generally, this means it has a lone pair of electrons that is available for reactions. The oxygen atom in $\ce{CuO}$ has lone pairs of electrons available.

$\ce{CuO}$ is considered a base because it fits these two descriptions.


Both reasons given are valid and contribut to alcohols having higher boiling points than amines, with one caveat. It is not stated whether the amine is primary, secondary or tertiary, and a tertiary amine has no $\ce{N-H}$ bond.

Although the $\ce{O-H}$ bond is more polar than the $\ce{N-H}$ bond and can therefore form somewhat stronger hydrogen bonds, the fact that an alcohol has two lone pairs is the main reason for the higher boiling point.

Even though a primary amine has two hydrogens that can act as hydrogen bond donors, their already weaker $\delta +$ is spread among two hydrogens, making those hydrogen bonds very weak, effectively limiting them to act as a single hydrogen bond donor.

So, in general, an amine can only participate as a single hydrogen bond acceptor and donor (except for tertiary amines), while an alcohol's two lone pairs and single hydrogen allow for a much stronger hydrogen bonding network.


Firstly, we need to clearly define "atomic volume" and "molar volume". Typically, one would understand "atomic volume" to be the volume of an atom of the element and it is directly related to the radius of the atom. Molar volume is a macroscopic property that is defined by Wikipedia as the ratio of the molar mass to the mass density of the substance at a certain temperature and pressure. As such, molar volume would not only be dependent on atomic radius but would also depend on the structure and bonding of the atoms.
With this understanding, it is easier to explain the trend that you have provided. The molar masses of the group 1 metals are considerably large but yet they have a relatively low mass density. As such, they are seen as spikes in the graph, with the highest molar volumes in their respective periods. Mass density generally increases from group 1 to group 13 resulting in the fall in molar volume across the period initially while for the transition metals, they stay roughly constant and take relatively large values. However, once we reach the non-metals, the trend seems to reverse. The density values for the non-metals seem to display some variability so I am not sure how the their molar volume trend can be explained. Obviously, their molar volume would be higher than that of the transition metals that come before them as those are significantly more dense.
Density data for the elements can be found here. More detailed molar volume data for the elements can also be found on another page at the same website.


It is possible and is not unheard to perform an extraction assisted by microwaves (even "dry ones" e.g.).  Either you seal the reaction vessel with a special cap to clamp, and monitor temperature and pressure evolution in function of time and power applied by the dedicated microwave oven.  This has the potential advantage of both rapid heating, and "pressure cooking".  (Depending on the material, using the reaction tube sealed a twice may be discouraged, though.)

Or you opt-in for a system where the the reaction flask is in the microwave oven, and the reflux condenser passes across the walls of the oven.  While in principle ovens designed for use in a kitchen may be used, e.g.,

(source)

it is both safer and more reproducible to use ovens dedicated to this (especially: focussing on the spot where your reactor is situated) and providing proper shielding and monitoring of the process parameters.

(source)


How is the unit of measure "equivalent" abbreviated?

The first place to check would be standards that are used frequently globally, such as CAS. According to the American Chemical Society page CAS Standard Abbreviations & Acronyms, the abbreviation for the unit of measurement 'equivalent' is equiv, while that of the adjective 'equivalent' is equiv. (with a dot).

Also, the ACS page Standard Abbreviations and Acronyms, shows equation abbreviated to eq and equivalent as equiv.

However, some journals, such as the Journal of Biological Chemistry page Table I: Abbreviations of units of measurement and of physical and chemical quantities state that the abbreviation of equivalent is as you used, eq, so if you are intending to publishing, just be aware of what the journal expects in terms of abbreviations.


The overall reaction is really better given as: $$\ce{2H2SO4 + CaC2O4 -> H2C2O4 + Ca^{2+} + 2HSO4^{-}}$$ since at that strong of a pH the predominate sulfuric acid species is $\ce{HSO4^{-}}$.

Better yet the reaction could be written as  $$\ce{2H^+ + CaC2O4 ->C[{0.2 N \ H2SO4, 70 °C}]\ H2C2O4 + Ca^{2+}}$$

As was already pointed out in the comments, calcium sulfate has a fair solubility 0.2 g per 100 ml of water at at 20 °C, and the $K_\text{sp} = 4\times10^{-5}$.

The solubility varies with temperature. The solubility of calcium sulfate dihydrate increases from 0.223 g/100 ml at 0 C to about 0.265 g/100 ml at 40 C, then decreases to 0.205 g/100 ml at 100 C.


A spontaneous reaction, as contrasted with a non-spontaneous reaction, is a reaction quantified by a negative free energy change. A reaction is said to be spontaneous if it occurs without being driven by an outside force. So, we can say:  In any spontaenous process, the path between the reactants and products is irreversible


First, it looks like you supercooled water. This at least explains the dips in temperature below zero degrees Celsius. It is not difficult to supercool water up to a few degrees below the normal freezing point. It remains metastable provided you avoid perturbing it, say by shaking. See for instance: http://www1.lsbu.ac.uk/water/supercooled_water.html. And yes, availability of nucleation points (small crystals) accelerates freezing.

The plateau observed around $\pu{2^\circ C}$ may be due to density differentials across the sample. At about $\pu{4^\circ C}$ water reaches its maximum density. Above this temperature, the colder water at the surface sinks, creating convection currents and accelerating the cooling process. As you dip below about $\pu{4^\circ C}$ the convective currents cease (around 3000 s), since the density gradient is stable: colder, less dense water is at the top. The water at the top of the sample continues to cool and decrease in density without sinking. If the thermometer is placed away from the air interface, it should at this point measure a slowly changing temperature as the temperature gradient toward the interface builds up. The temperature behavior will be a response to heat diffusion instead of heat convection.

You may gain additional information if you add a little dye near the thermometer and attempt to follow its movement, either before or after the convective behavior is presumed to have stopped. Convection should result in mass flow, mixing the dye more rapidly. In the diffusion regime dye should disperse more slowly and locally.

This discussion makes various assumptions of course with regard to the purity of the water and the geometry and insulation of the container it's in. It also assumes that the water has not frozen at all before about 4800 s. It also makes assumptions about the location of the thermometer.


Why doesn't diamond conduct electricity but silicon does?

I'm not a solid-state physicist. Yet, this is how I understand this problem, so bare with me.

According to this article:

  Every solid has its own characteristic energy-band structure. This variation in band structure is responsible for the wide range of electrical characteristics observed in various materials. In semiconductors and insulators, electrons are confined to a number of bands of energy, and forbidden from other regions. The term "band gap" refers to the energy difference between the top of the valence band and the bottom of the conduction band. Electrons are able to jump from one band to another. However, in order for an electron to jump from a valence band to a conduction band, it requires a specific minimum amount of energy for the transition. The required energy differs with different materials. Electrons can gain enough energy to jump to the conduction band by absorbing either a phonon (heat) or a photon (light).

This forbidden gap plays a major role in determining the electrical conductivity of material. The materials can be classified into three types based on the forbidden gap. they are:

Insulators: The materials, which does not allow the flow of electric current through them are called insulators. The band gap between the valence band and conduction band is very large in insulators, which is approximately equal to $\pu{15 eV}$. Normally, in insulators the valence band is fully occupied with electrons due to sharing of outer most orbit electrons with the neighboring atoms whereas its conduction band is empty, i.e., no electrons are present in the conduction band.
Semiconductors: The material, which has shown electrical conductivity between that of a conductor and an insulator is called a semiconductor. The band gap of semiconductors is is very small, and it falls between that of valence band and conduction band. The value of this forbidden gap is about $\pu{1-3 eV}$. As a rule of thumb, if the material has $\Delta E_\mathrm{gap} \le \pu{3.0 eV}$, it is considered as a semiconductor (Semiconductors- Band Gaps, Colors, Conductivity and Doping).
Conductors: In a conductor, the valence and conduction bands are overlapped each other. Therefore, there is no forbidden gap in a conductor (the value is generally negative). Thus, the valence band electrons to move into conduction band needs only a little amount of applied external energy.

Experimental values of $\Delta E_\mathrm{gap}$ for diamond and silicon are $\pu{5.48 eV}$ and $\pu{1.17 eV}$, respectively (Ref.1). Keep in mind that band gap get narrowed when going down on a group in periodic table. For example, $\Delta E_\mathrm{gap}$ of germanium is $\pu{0.74 eV}$ (Ref.1).

Therefore, silicon and germanium behave like semiconductors, while diamonds fall into insulator category in general (yet, please refer to  Ed V's comment above for better definition).

References:

Fabien Tran, Peter Blaha, "Accurate Band Gaps of Semiconductors and Insulators with a Semilocal Exchange-Correlation Potential," Phys. Rev. Lett. 2009, 102, 226401, 4 pages (DOI: 10.1103/PhysRevLett.102.226401).


There is very well-written chapter "The chemistry of monovalent copper in aqueous solutions" in Advances in inorganic chemistry, Volume 64 [1, pp. 220–223, DOI: 10.1016/B978-0-12-396462-5.00007-6] which extensively covers as to why copper(I) is unlikely to exist in aqueous solution and why nitrate is a poor ligand for the purpose of preserving monovalent copper:

  The redox potentials of copper in acidic aqueous solutions are as follows:

  $$
\begin{align}
\ce{Cu+(aq) + e- &-> Cu^0(s)}, &\quad E^\circ &= \pu{0.521 V} \tag{1}\\
\ce{Cu^2+(aq) + e- &-> Cu^+(s)}, &\quad E^\circ &= \pu{0.153 V} \tag{2}\\
\ce{Cu^2+(aq) + 2 e- &-> Cu^0(s)}, &\quad E^\circ &= \pu{0.342 V} \tag{3}\\
\end{align}
$$

  From these values, it is clear that $\ce{Cu+(aq)}$ is unstable and
  disproportionates in aqueous solutions:

  $$\ce{2Cu+(aq) <=> Cu^2+ + Cu^0(s)}, \quad K_{1.4} = \pu{(1 - 1.7) × 10^{6} M-1} \label{rxn:4}\tag{4}$$

  However, if $\ce{Cu+(aq)}$ is formed in a clean aqueous medium, the disproportionation reaction has often a relatively long induction period as reaction \eqref{rxn:4a} is endothermic due to the $\ce{Cu^0(s)}$ lattice energy.

  $$\ce{2Cu+(aq) -> Cu^2+(aq) + Cu^0(aq)}\label{rxn:4a}\tag{4a}$$

  Furthermore, the concentration of $\ce{Cu^0(aq)}$ in nonacidic aqueous solutions is limited due to the low solubility of $\ce{CuOH}/\ce{Cu2O}:$

  $$\ce{Cu+(aq) + OH- <=> CuOH(s)/Cu2O(s)} \quad K_\mathrm{SP} = \pu{1E-14 M-2} \label{rxn:5}\tag{5}$$

  Naturally, the concentration of Cu(I) species in aqueous solutions
  can be increased by adding appropriate ligands.

  […]

  …the instability of $\ce{Cu+(aq)}$ stems from the difference between the solvation energies of the two cations and the second ionization potential of copper.
  This is correct for all transition-metal ions, but as the ionization energies increase along the series faster than the solvation energies, the lower oxidation states become relatively more stable.
  In order to shift reaction \eqref{rxn:4} to the left ligands which
  preferentially bind to $\ce{Cu(I)},$ must be added to the solution.

  […]

  $\ce{Cu(I)L_n}$ complexes can be stabilized in aqueous solutions via
  employing ligands with one or more of the following properties:

  Ligands that have a lower coordination number with $\ce{Cu(I)}$
  complexes than with $\ce{Cu(II)}$ complexes; these $\ce{Cu(I)}$ complexes are stabilized due the entropy gain in reaction \eqref{rxn:7}:

  $$\ce{Cu(II)L_m + e- -> Cu(I)L_n + $(m - n)$ L} \label{rxn:7}\tag{7}$$

  Ligands that are π acids and therefore prefer $\ce{Cu(I)}$ over $\ce{Cu(II)}$ and shift the redox potential of the $\ce{Cu(II)/Cu(I)}$ couples anodically.
  Hydrophobic ligands, which often also increase the radius of
  the complex, decrease the stabilization of $\ce{Cu(II)}$ by decreasing
  its solvation energy.
  Ligands which enforce a tetrahedral coordination sphere, or
  one which approaches this geometry, on the central cation.
  Ligands that are soft bases, for example, thiols. Thus, for example, the biological copper-trafficking proteins, the metallochaperone Atx1, and the copper-transporting P-type ATPases have a highly conserved CXXC metal binding motif which binds a single $\ce{Cu(I)}$ ion via a two-coordinate site consisting of two cysteines.

  All these types of ligands stabilize $\ce{Cu(I)}$ complexes by shifting the redox potentials of the $\ce{Cu(II)/Cu^0(s)}$ and $\ce{Cu(I)/Cu^0(s)}$ couples and thus shifting the equilibrium constant of reaction \eqref{rxn:4} to the
  left.
  However, ligands might stabilize $\ce{Cu(I)}$ complexes also
  kinetically by inhibiting the disproportionation reaction by
  inhibiting the formation of $\ce{Cu^0(s)}$ via blocking the approach of two copper atoms, this is, for example, the mechanism by which
  some $\ce{Cu(I)}$ enzymes are stabilized.

  Naturally, all these ligands also increase the solubility of $\ce{Cu(I)}$ by competing with reaction \eqref{rxn:5}.

References

Inorganic/Bioinorganic Reaction Mechanisms, 1st ed.; Eldik, R. van, Ivanović-Burmazović, I., Eds.; Advances in inorganic chemistry; Academic Press: London; Waltham, MA, 2012; Vol. 64. ISBN 978-0-12-396462-5.


How does simulation software work at a fundamental level?

Well, that's a rather diffuse question. From what I have seen, the software packages you mention do Molecular Mechanics (MM), which is a treatment of molecules with classical physics. For each atom in a given molecule, the forces acting upon it due to chemical bonding with other atoms is calculated using a set of empirically determined parameters (a force field), and then an estimate of the potential energy is provided. But given that this question is tagged with "Computational Chemistry", I also want to mention quantum chemistry, i.e. the application of quantum mechanics to atoms and molecules. This field is both far more interesting and far more difficult than the classical MM treatment.

A first shot would be the Wikipedia article about Computational Chemistry which already points you to other relevant topics, i.e. the article about Molecular Mechanics, which is quite concise about the topic (although I don't know how exhaustive it is in this regard, since I don't deal with MM in my research). For quantum chemistry, the necessary knowledge is more spread-out; relevant Wikipedia articles would be Eigenproblems, the Schrödinger equation and Ab initio methods. I don't know any literature about MM, but if you want to look deeper into quantum chemistry, I would recommend the following two books (and lots and lots of time and patience for reading them):

Attila Szabo, Neil Ostlund: "Modern Quantum Chemistry". Provides a good mathematical introduction and thorough treatments of the Hartree-Fock and Moller-Plesset methods. It's cheap, also.
Ira Levine: "Quantum Chemistry". Covers a wider range of methods, but is in my opinion a bit more difficult to consume than Szabo/Ostlund.


What is the role of NaHSO3 in syn diol formation from alkenes with OsO4?

After you add the $\ce{OsO_4}$ to make the reaction, you'll need to reduce the remaining $\ce{OsO_4}$ maintaining the $\ce{pH}$ of the solution constant.

In addition:

  The hydrolysis of the osmate ester proceeds by simple substitution
  reactions. The $\ce{NaHSO3}$, a mild reducing agent, facilitates ligand
  subsitution by reducing the $\ce{Os}$ further.

The Art of Writing Reasonable Organic Reaction Mechanisms, Robert B. Grossman, p 293

And $\ce{NaHSO_3}$ can do both and so you will use it. You also can add $\ce{Na_2SO_3}$.

I found also something interesting on JACS, it's mentioned at least four times in the document that $\ce{NaHCO3}$ is used as buffer. Also there is a lot of information in the references of this article.

  While the reaction is normally run under basic conditions
  ($\ce{K_2CO_3}$, $\ce{pH} = 12.2$, aqueous layer), it is possible to
  buffer the system with $3$ equiv of $\ce{NaHCO3}$ ($\ce{pH} = 10.3$,
  aqueous layer).$^{38, 39}$ Buffering of the reaction mixture does not
  affect the ee, but it can have a beneficial effect on the yield when
  base-sensitive substrates are used or base-sensitive products are
  formed. Thus, the AD of allyl or cinnamyl halides (cf. eqs $1$ and
  $3$) should be performed under buffered conditions to minimize epoxide
  formation (see eq $1$, the yield with allyl bromide as substrate is
  between $61$ and $74%$ under buffered conditions; in the absence of
  $\ce{NaHCO3}$ only $40-50$ % of diol is obtained38). Unfortunately,
  the AD reaction does not turn over if $\ce{K_2CO_3}$ is replaced
  entirely by $\ce{NaHCO3}$.

Hartmuth C. Kolb, Michael S. VanNieuwenhze, K. Barry Sharpless
Chem. Rev., 1994, 94 (8), p 2494


There are a few things that influence the properties of polymers:

Molecular weight of the chains, i.e. how many monomers are joined in each chain, can drastically alter the polymer's properties. For example, for polyethylene, UHMWPE has extremely long chains and produces a very strong material suitable for bullet-resistant vests, whereas LLDPE has much shorter chains that produce a weaker material that is much more ductile (think about how easy it is to stretch a plastic shopping bag or the like).
Another factor that affects a polymer's properties is branching. Branching is when a polymer is not made up of nice straight chains, but chains that occasionally fork off into several different chains. The difference between LLDPE and LDPE is that LDPE has a lot more branching. This keeps the chains from packing as nicely so LDPE isn't as strong.
The other big factor (other than adding another monomer to make a copolymer) is the introduction of crosslinking. Crosslinking is where different chains are connected to one another, often by addition of a molecule that can bond to a chain at both ends, e.g. rubber vulcanization crosslinks the different chains with linear sulfur chains. Natural rubber is very different than what most people think; it's a soft sticky material that doesn't hold its shape. The polyisoprene chains it's composed of aren't crosslinked and can easily slide past one another, but once crosslinked, the chains are connected and aren't free to move as much. Depending on the degree of crosslinking, you can go from something somewhat soft, like a bouncy ball, to something very rigid, like a car tire.

Hydrogels are not simple linear polymers. While they're mostly made from a single monomer, small additions of related molecules are used to control how they polymerize. N.B. These apply generally to silicones, other polymers may have other branching/termination processes. If the monomer only has two sites at which it can connect to other monomers, naturally the only structure that can be formed is a linear chain. One can control the length of the chains by adding a small amount of a terminator—a monomer that can only make one connection. If a terminator reacts with a growing chain, it stops the growth as it can't connect to anything else. The more terminator added, the shorter chains are, on average. Similarly, if you add a monomer that has more than two reactive groups, branching and crosslinking will occur. Often, a non-crosslinked polymer is supplied as a liquid, then a crosslinking agent is added to harden the polymer. Dow Corning has a great primer on the actual chemistry involved.

What makes a hydrogel special is that it's a polymer network that encapsulates water, i.e. it's an extensively crosslinked polymer that is hydrophilic. If the polymer was just a series of individual chains, it would just dissolve in water since nothing is holding the different chains together. When crosslinked, water can get in, but the mesh of polymer is linked together enough to hold its shape, though it may swell to accommodate the water. Depending on the degree of crosslinking, more or less water can fit within the structure. A less linked polymer can fit more water since there are fewer crosslinks resisting, but is less rigid when hydrated for the same reason (think Jell-O). A more crosslinked polymer won't absorb as much water, but will be much more rigid (like a contact lens). Contact lenses have to balance holding their shape, oxygen permeability, surface hydrophilicity, etc. so there are many reasons why, despite being based on the same primary monomer, two different brands may have a different blend of the other stuff that lets one control the hydrogel's properties.


How to calculate the abundances of certain gases from the peak areas, GC-MS

If you only have a single calibration standard concentration for each compound as your question implies, then you will simply need to calculate a calibration factor for each compound then multiply your measured peak area by the calibration factor.

This is also know as a single point calibration and it is important that the peak area of the standard is greater than that of your sample or you might be out of the linear response range of the instrument.  Also, a single point calibration assumes that the instrument response is linear from zero up to the concentration of the standard, or at least that the concentration of your sample is very close to that of your standard.

In your example you would calculate the calibration factor as:

  $\mathrm{539.894\ ppt / 28335299\ m^2 = 1.905\ x\ 10^{-5}\ ppt/m^2}$

Then if you had a sample with a peak area of 18335299.12, the concentration of CFC - 12 would be:

  $\mathrm{18335299.12\ m^2\ x\  1.905\ x\ 10^{-5}\ ppt/m^2\ =349\ ppt}$

Note: The calibration factor is often calculated in the inverse manner of my example above.  If calculated in that way, then of course you would divide your peak area by the calibration factor.


Maximum Local Value for the Electron Density

Single electron systems

I believe that particularly high values of n[V] are unlikely because as you pack more and more electrons in the same region, their repulsion would start to overpower their attraction to the nuclei, meaning the electron density would likely spread out more.

The main constraint on high electron density is not the repulsion between electrons, but the Heisenberg uncertainty principle. Even a single electron has a limited average electron density, given its average momentum (see e.g. here). For single electron systems (hydrogen-like ions), you get the highest electron density by increasing the nuclear charge. According to the virial theorem, as the potential energy of the electron increases (the magnitude, so tighter binding), the kinetic energy also increases. This will go along with a higher momentum, allowing the electron to occupy a smaller volume (i.e. have a higher density).
Multi-electron systems
For multiple electrons, theory gets complicated. X-ray diffraction gives experimental data related to electron density, but you need extremely high resolution (and extremely low static and dynamic disorder) to be able to map the electron density near the nucleus. In this study on LiF and NaF, the electron density at the lithium nucleus is estimated as 50 electrons per cubic angstrom using maximum entropy methods (panel b).

Fluorine, with a higher atomic number, will have higher electron density at the nucleus, but the density is cut off in the figure so you can't see the maximum value.

I understand that such a value would vary depending on promixity to the atomic nuclei (which would shield electrons from their own mutual interactions)

The innermost electrons would not be shielded, so the maximum electron density for a uranium nucleus with one electron would not be much different from one with 92 electrons. In fact, the binding energy of the inner electrons was used to figure out atomic numbers using Moseley's law.

Is there a maximum value for n[V] above which the mutual repulsion of electrons would mean the density was unreal?

Electron repulsion is not the main constraint on high electron density. To get high electron density, you need a high nuclear charge. At some point, relativistic effects become important, and for super-heavy nuclei, electrons get captured.
There was a comment from Martin about the use of the word "unreal". I agree with Martin that if you observe a certain high electron density, that is as real as it gets. You might run into a calculation claiming a ridiculously high electron density, and then might conclude that results are not plausible and warrant further research.
So where can I find high electron density?
That would be near the nucleus of the heaviest elements as long as they are stable at least for a while. Using the arguments made above using the uncertainty principle, the binding energy of inner electrons, and the associated momentum of the electron (or looking at the Bohr model), the radius $a_0$ should be proportional to the reciprocal of the atomic number, meaning the electron density increases with the cube of the atomic number.


What is the result of this elimination reaction?

I am not an expert but I think you are correct. The alkylhalide is secondary, the nucleophile is a strong base and the solvent is polar protic. That leads to the conclusion that the most likely reaction to happen is E2. According to Zaitsev rule the major product should be the most substituted alkene because its transition state is lower in energy. That means that the main product would be 2-octene. In reference to the cis and trans structure, the trans one would be more stable. Taking all of the above under consideration, the main product of the reaction would be trans-2-octene and the minor products: cis-2-octene and 1-octene.


Magnesium cannot give one or two electrons to the chloride ion. Because Chloride ions have the maximum number of 8 electrons in their external shells. There is no possibility for a new electron to stay on a Chloride ion. And the Chloride ion will be reluctant to accept or to loose one of its electrons, because it has the ideal conformation (8 outer electrons). So there are no "multiple ways" of reacting, as you imagine. The only thing that may happen is for magnesium to loose two electrons per atom, because at the end of this departure, it will have 8 outer electrons, and this seems to be to ultimate goal for all atoms. These electrons will be caught by $H^+$ ions, which are transformed into $H$ atoms and then into $H_2$ molecules.


It is not implausible that bamboo should have antibacterial properties, after all plants generate lots of protective chemicals. However, the use of this particular word (kun or kunh) by a number of commercial textile websites seems like a promotional gimmick without a strong scientific basis.

Claims of an intrinsic antimicrobial property of processed bamboo fibers are likely exaggerated, as the excerpt below (1) suggests:

  3.3. Antibacterial Properties

  The antibacterial properties of the control sample (OB) and the
  composites (ZB, GB, ZGB) were qualitatively investigated by the
  bacteriostatic circle method against one Gram-negative (E. coli) and
  one Gram-positive (B. subtilis), and the results were shown in Figure
  6. Figure 6A displayed the antibacterial activities of bamboo samples against E. coli. It can be observed that OB had little antibacterial
  activity without an inhibition zone.

This is not to say that bamboo lacks antimicrobial properties. Plants employ a myriad of chemical substances as sources of physical and chemical protection. Extracts of bamboo (2) apparently contain bactericidal compounds, although the concentration in processed fibers appears insufficient to grant these significant antibacterial properties.

It's also worth adding that the above refer to specific species of bamboo and there may be significant variety in antibacterial properties across species.

In conclusion it is difficult if not impossible to answer the OPs question as it is unclear what "kun" or "kunh" refers to, it may just be a catchphrase to describe unknown substances granting antibacterial properties to raw bamboo.

References

Junyi Zhang, Bo Zhang, Xiufang Chen, Bingbing Mi, Penglian Wei, Benhua Fei, and Xindong Mu. Antimicrobial Bamboo Materials Functionalized with ZnO and Graphene Oxide Nanocomposites. Materials (Basel). 2017 Mar; 10(3): 239.
doi: 10.3390/ma10030239
The origin of the antibacterial property of bamboo. T. Afrin , T. Tsuzuki , R.K. Kanwar & X. Wang.  The Journal of The Textile Institute. 2012; 103(8): 844-8491. doi: 10.1080/00405000.2011.614742


Does the ignition point of flammable substances changes when provided a pure oxygen atmosphere?

No, but because of a technical point.

From Wikipedia: The autoignition temperature or kindling point of a substance is the lowest temperature at which it will spontaneously ignite in normal atmosphere without an external source of ignition, such as a flame or spark.

So the definition of the autoignition temperature is based on air not pure oxygen.

For the ignition temperature in general the Wikipedia reference goes on: The temperature at which a chemical will ignite decreases as the pressure or oxygen concentration increases.


How can I lower the flash point of stearic acid (Palm Based Fatty Acids)

could you have a small amount of a more volatile liquid with the hexadodecanoic acid - such as white spirit or kerosene? This would support a flame and raise the temperature to the point at which the hexadodecanoic acid acid will burn.


Colour due to F-centre is same as that produced in a flame test, why?

Why are the colours the same as the metal. Basically, the anion doesn't affect the colour, but the cation does. And since the cation is just the metal missing an electron or several, of course, they exhibit similar behaviour!

Colours we see are determined by the transitions made by electrons "falling" from a high energy state to a lower energy state, and in the process releasing one or some photons with a frequency in what we call the visible spectrum.

When an F-centre is formed in the described conditions, we have effectively extended the lattice with additional cations the same as those already present. But rather than additional anions that match as well, free electrons take the place in the lattice that would be occupied by one.

Now the anion already has its orbitals filled - there is nowhere for an associated electron to move into when excited, nor somewhere for other electrons to enter when relaxed, and thus there is no way for them to release energy as photons that we sense as light. So the anion doesn't affect the colour.

However the cation has empty orbitals when neutral, let alone when positively charged (i.e. has an electron(s) removed). This means any excitation or relaxation will involve these orbitals.

The result is that the electron movements in the  F-centre lattice are the same as those that would occur in the ionisation of the metal which  the cation is derived from, and hence the colours are also the same.

Why those specific colours for the respective compounds? That would be a completely different question, and probably found in a quantum mechanics discussion rather than a chemistry area.


How can I find whether there is oxygen present inside a well of diameter 1 m at 50 m's deep?

The ideal solution would be a single gas personal detector, such as this one (it's only the first example I found, I don't know this brand, but there are many equivalent apparatuses from other vendors).

While digital personal detectors are expensive ($200+ range), there are simpler systems which also provide some basic safety (better than nothing). The Davy lamp is perhaps the best known example of these, and it was a huge breakthrough in mine safety when it was popularized after 1815.

(Source: http://lindal-in-furness.co.uk/MinersLamps/MinersLamp.jpg)


Reactivity comparison of phenol and toluene towards electrophilic substitution

You have to make a comparison between phenol and toluene. You have identified what toluene does, but you haven't described what phenol does.

The hydroxyl group in phenol increases electron density in the ring by resonance (+M effect). Note that the methyl group in toluene increases electron density only by hyperconjugation and inductive effect.

Resonance effects are generally far more superior to hyperconjugation. Hence, without doubt, phenol is more activated towards EAS as compared to toluene.


Is the reactivity of Benzene with an -N(R)₃⁺ group greater than its reactivity with an -NH₃⁺ group?

$\ce{Ph-NR_3^+}$  is likely weaker than $\ce{Ph-NH_3^+}$ because the mechanism of deactivation is based on the amount of positive charge localized on the Nitrogen:  Alkylgroups will stabilized the positive charge through sigma donation.

That being said, the difference will be small.  Additionally, there are deprotonation side-reactions to consider for anilium ($\ce{Ph-NH_3^+}$) ions, which may compromise any discussion of deactivation.


How does polarity of a solvent increase conductance of electrolyte?

The important parameter is the dielectric constant (relative permitivitty) $\epsilon$ of the solvent because this controls the interaction energy $E$ between two ions at a fixed distance $r$ apart, i.e. $\displaystyle E\sim \frac{1}{\epsilon r}$. If the dielectric constant is large, for water it is 78 but for benzene 2.3 you can see that the interaction energy for benzene will be far larger than for water. This means that any pair of ions formed in benzene will be attracted to one another from a far greater distance (with the same energy) than the same ions in water would be. The chance that the ions separate is given relative to the thermal energy of the solvent which is just $RT$, so if the interaction energy is far larger than $RT$ the chance of separating is small and the ion does not dissolve and vice versa. The thermal energy of the solvent tries to jostle the ions apart.

Notes. The interaction energy is  $\displaystyle E(r,\epsilon)= \frac{Z_1Z_2}{4\pi\epsilon_0}\frac{e^2}{\epsilon r}$ where $Z$ are the charges, $e$ the charge on the electron and $\epsilon_0$ the permitivitty of free space.  The thermal energy per molecule at temperature $T$ is $k_BT$ where $k_B$ is the Boltzmann constant and is $RT$ per mole. The probability of being at separation $r$ with dielectric constant $\epsilon$ is $\sim\exp(-E(r,\epsilon)/k_BT)$.


What do these labels for molecular electronic states mean?

As stated in this answer, these are irrep (irreducible representation) labels for molecular symmetry point groups. In the context of chemistry, point groups are usually introduced when learning about structural symmetry (atoms and bonds). This is a broad topic with many technical points, too many for a single answer, so I won't cover the basics but hopefully the few things needed to make the connection between what you read in a book and what's present in this paper.

The key is recognizing that the structural symmetry of molecules leads to symmetry in their molecular orbitals. Take the water molecule as an example. Water's molecular structure can be described at best by the $C_{2v}$ point group, so we say it has $C_{2v}$ symmetry. Composing its molecular orbitals from atomic orbitals is actually an exercise in symmetry, as only atomic orbitals of identical symmetry may combine. Here is the LCAO-MO diagram for water, with the oxygen 1s orbital left out due to its extremely low energy; it won't mix with the hydrogen AOs.

Here is the character table for the $C_{2v}$ point group.

AOs of identical symmetry (they transform identically for any given symmetry operation) are shown in the same color. A few things are worth pointing out:

s orbitals are perfectly symmetric (a sphere), so they will always transform as the totally symmetric irrep of a molecular point group. In $C_{2v}$, that's $A_1$. Once you learn what the operations $E$, $C_{2}$, $\sigma_{v}(xz)$, and $\sigma_{v}(yz)$ do, you can convince yourself that the $\sigma$ combination of $\ce{H2}$ AOs is also completely symmetric under those operations, so it also belongs to the $A_1$ irrep. The p orbitals are figured out from the 3rd column, d from the 4th.
The 2s and $\sigma$ AOs combine to give the $3a_1$ MO. The "3" is because it's the 3rd such MO with $a_1$ symmetry. The symmetry label is in lowercase because it for a 1-electron state. Symmetry labels are in uppercase for many-electron states, such as the total wavefunction, as you see in the linked paper. This is a matter of convention, nothing more.

Neutral water is a closed-shell molecule; there aren't any unpaired electrons, so it has singlet spin multiplicity. How do you determine the symmetry of the total wavefunction? A Slater determinant can help, or even just the Hartree product part of it. Take a fictitious system with 4 electrons in 2 orbitals. The Hartree product will look something like

$$
\Psi_{\text{Hartree}} = [\phi_{1}(1)\alpha(1)][\phi_{1}(2)\beta(2)][\phi_{2}(3)\alpha(3)][\phi_{2}(4)\beta(4)]
$$

This is a product wavefunction that might be made from an electron configuration such as $1s^{2}2s^{2}$, $1s^{2}2p_{x}^{2}$, $4p_{x}^{2}4p_{z}^{2}$. Each of these MOs has a certain symmetry; the total symmetry is given by the product of the individual symmetries. For this step one needs the point group product table (again for $C_{2v}$):

$$
\begin{array}{|l|l|l|l|l|}
\hline
            & \mathbf{A_1} & \mathbf{A_2} & \mathbf{B_1} & \mathbf{B_2} \\ \hline
\mathbf{A_1} & A_1          & A_2          & B_1          & B_2          \\ \hline
\mathbf{A_2} & A_2          & A_1          & B_2          & A_1          \\ \hline
\mathbf{B_1} & B_1          & B_2          & A_1          & A_2          \\ \hline
\mathbf{B_2} & B_2          & B_1          & A_2          & A_1          \\ \hline
\end{array}
$$

If the electron configuration of the system is indeed $4p_{x}^{2}4p_{z}^{2}$, this results in a total wavefunction symmetry of

$$
B_{1} \times B_{1} \times A_{1} \times A_{1} = A_{1}
$$

The spin components don't contribute, only the spatial components. Note that in any product table, the product of any irrep with itself is always the totally symmetric representation in that group, hence the diagonal being all $A_{1}$. Product tables are also symmetric, since irreps are commutative under multiplication. Since this is a singlet, the ground-state wavefunction can be specified as $^{1}\!A_{1}$. For a closed-shell system (RHF), due to the pairs of electrons always leading to $A_{1}$ symmetry, it will always have a total state symmetry of $^{1}\!A_{1}$. If you have an open-shell system of any kind in a ground or excited state, with either an even or odd number of electrons, the total wavefunction symmetry will be determined by electrons in those orbitals alone, since the product between fully-occupied orbitals will always be of $A_{1}$. In the simplest case of one unpaired electron, the total symmetry of the ground state will be the symmetry of the SOMO.

Now, on to the paper. The most important part for our understanding of the state label assignments is Table 3:

All the systems considered have either $C_{2v}$ or $C_{s}$ symmetry. Note how going from a point group of higher symmetry ($C_{2v}$, 4 irreps) to one of lower symmetry ($C_{s}$, 2 irreps) leads to states collapsing into fewer irreps due to fewer symmetry operations that can distinguish them. For the plots (here the last row of Figure 6),

there is the isosurface of the molecular orbital (really a natural orbital), the orbital number and symmetry, and the natural orbital occupation number (NOON) for that NO in that state.

There are a number of good Dover publications (1, 2, 3) for learning about group theory in chemistry and spectroscopy that are highly recommended, since the primary peer-reviewed literature isn't usually a good resource for learning this. An excellent resource for character tables is WebQC.


Participation of Pi-bonds and lone pairs in resonance

You have probably misunderstood what the sentence actually means.

It means that only one, either $\pi$ bond or lone pair will participate in resonance if the atom has both.

For example, in pyridine, the nitrogen has lone pair as well as is attached with a $\pi$ bond.

While resonance, its $\pi$ bond participates in resonance because it (and not the lone pair) is in the plane perpendicular to the molecule.

Observe that the lone pair remains at the place it is meaning it doesn't participate.

Conclusion: The pair of electrons present in the orbital perpendicular to the plane of molecule participates in resonance if it is present.


Let's list what's going on here. Orthophosphoric acid dissociates stepwise:

\begin{align}
\ce{H3PO4 + H2O & <=>[$K_\mathrm{a1}$] H3O+ + H2PO4–} \tag{R1}\\
\ce{H2PO4- + H2O & <=>[$K_\mathrm{a2}$] H3O+ + HPO4^2–} \tag{R2}\\
\ce{HPO4^2- + H2O & <=>[$K_\mathrm{a3}$] H3O+ + PO4^3–} \tag{R3}
\end{align}

Mass balances for protons and phosporous, respectively:

\begin{align}
C_\ce{H} & = [\ce{H3O+}] + [\ce{HPO4^2–}] + [\ce{H2PO4–}] + [\ce{H3PO4}] = [\ce{H3O+}] + \sum_{n=1}^3 [\ce{H_$n$PO4^$(3-n)-$}] \tag{1}\\
C_\ce{P} & = [\ce{PO4^3-}] + [\ce{HPO4^2–}] + [\ce{H2PO4–}] + \ce{H3PO4} \tag{2}
\end{align}

Now we compare the equilibrium constants:

\begin{align}
K_\mathrm{a1} & = \frac{[\ce{H3O+}][\ce{H2PO4-}]}{[\ce{H3PO4}]} = 7.1 \times 10^{−3} \tag{3}\\
K_\mathrm{a2} & = \frac{[\ce{H3O+}][\ce{HPO4^2-}]}{[\ce{H2PO4-}]} = 6.3 \times 10^{−8} \tag{4}\\
K_\mathrm{a3} & = \frac{[\ce{H3O+}][\ce{PO4^3-}]}{[\ce{HPO4^2-}]} = 4.2 \times 10^{−13} \tag{5}
\end{align}

Notice that $K_\mathrm{a1} >> K_\mathrm{a2}$ (more that $10^5$), therefore by the first dissociation step phosphoric acid behaves as an acid of medium strength, producing equal amounts of $\ce{H3O+}$ and $\ce{H2PO4-}$ (considering charge balance):

$$[\ce{H3O+}] \approx [\ce{H2PO4-}] \tag{6}$$

which directly leads us to the answer. Using expression for $K_\mathrm{a2}$:

$$[\ce{HPO4^{2-}}] = K_\mathrm{a2} \times \frac{[\ce{H2PO4-}]}{[\ce{H3O+}]} = K_\mathrm{a2} \tag{7}$$

That's it, $[\ce{HPO4^{2-}}] = \pu{6.3 \times 10^{−8} (mol/L)}$

Now for the fun part, if you really want to solve quadratic equation, we can also find solution pH and $[\ce{PO4^{3-}}]$.

Based on the values of equilibrium constants, it is safe to assume that pretty much only first dissociation step defines solution's pH. We can rewrite equation for $K_\mathrm{a1}$ as follows:

$$K_\mathrm{a1} = \frac{[\ce{H3O+}]^2}{C_0 - [\ce{H3O+}]},$$

where $C_0$ is the concentration of the solution ($\pu{0.10 M}$). Using the given numbers we have

$$7.1 \times 10^{-3} = \frac{[\ce{H3O+}]^2}{0.10 - [\ce{H3O+}]}$$
$$[\ce{H3O+}]^2 + 7.1 \times 10^{-3} [\ce{H3O+}] - 7.1 \times 10^{-4} = 0 $$
$$[\ce{H3O+}] = \frac{-7.1 \times 10^{-3} + \sqrt{(7.1 \times 10^{-3})^2 + 4 \times 7.1 \times 10^{-4}}}{2} = \pu{2.33 \times 10^{-2} (mol/L)}$$

$$\mathrm{pH} = - \log{[\ce{H3O+}]} = 1.63$$

WolframAlpha thinks the same.

Now we know $[\ce{H2PO4-}] = \pu{2.33 \times 10^{-2} (mol/L)}$ and we can finally find $[\ce{PO4^3-}]$ from the expression for $K_\mathrm{a3}$:

$$[\ce{PO4^{3-}}] = K_\mathrm{a3} \times \frac{[\ce{HPO4^2-}]}{[\ce{H3O+}]} = 4.2 \times 10^{−13} \times \frac{6.3 \times 10^{−8}}{2.33 \times 10^{-2}} = \pu{1.1 \times 10^{-18} (mol/L)}$$

This demonstrates that in pretty concentrated solution of phosphoric acid anions $\ce{PO4^3-}$ are practically non-existing.


Air contains 20 % O2 by volume. How many cm³ of air will be required for oxidation of 100 cm³ of acetylene?

Let's begin with your answer:
$2\ \mathrm{mol}$ $\ce{C2H2}$ will react with $5\ \mathrm{mol}$ of oxygen.
$1\ \mathrm{mol}$ $\ce{C2H2}$ will react with $5/2\ \mathrm{mol}$ of $\ce{O2}$.
If we suppose that the gases are under standard conditions, then
$22.4\ \mathrm{L}$ $\ce{C2H2}$ will react with  $\frac{5}{2} \times 22.4\ \mathrm{L  } ~ \ce{O2}$.
This means that:
$0.1\ \mathrm{L}$ $\ce{C2H2}$ will react with $\frac{5}{2} \times 0.1\ \mathrm{L}$ $\ce{O2}$.
$0.1\ \mathrm{L}$ $\ce{C2H2}$ will react with $\frac{5}{2} \times 0.1\times 5\ \mathrm{L}$ of air.
So, the volume of air is $1.25\ \mathrm{L}= 1250\ \mathrm{cm^3}$


How do we know if this following molecule is a superimposable mirror image or non-superimposable mirror image ? [See Image attached]

Take 2 round, soft objects like potatoes or apples.
Pin into each 4 pins of 3 different kinds ( 1 + 2 + 1 ),
pointing to the most different directions.
Call the 1st pin bromine, the 2nd one hydrogen, the 3rd one ethyl.
Do the same for the other potato to be the mirror image of the first one.
Try to rotate one of them to look like the other.
If you succeed, they are superimposable.
Now, replace one of the same pins on both objects by another -
4th kind - of pins in such a way that they are still mutual mirror images.
Try to match them the same way you tried previously.
If you succeed, they are superimposable.

The summary:

While there are 2 hydrogen atoms on a carbon atoms - or more generally, there are only 3 kinds of groups on a C atom - the groups/atoms of the same kind are interchangeble and you can rotate the molecule to match the mirror image.
If there are 4 different groups on C atoms, you cannot make such a match.
Said by other way, the former case has a plane of symmetry, the latter case does not.


How to distinguish palmitic acid from lauric acid?

Consider the reaction which happens, what is produced and then what is different about the products? The key has to be in their properties.

The saponification reaction which takes place will produce a salt, in this case it will make sodium palmitate and sodium laurate. These two salts have very different solubilities and this makes a marked difference in the lathering ability. Sodium laurate is higher and thus lathers more readily.

For more details :

Solubility data

Saponification

Scientific Soapmaking: The Chemistry of the Cold Process
By Kevin M. Dunn


homogeneous mixture - Are they truly uniform in composition and proportion?

Maybe it is just a question of what you mean by "evenly dissolved". True enough, in your example, the proportion of alcohol to water is not "even" - rather the solution is indeed mostly water containing only a little bit of alcohol. But for a true solution such as this, the alcohol is dispersed evenly (uniformly) throughout the bulk of the solution, so that if you repeatedly withdrew a sample and analyzed it, you would get the same result no matter which part of the liquid you sampled.

That would not be the case for oil and water - they would exist in separate phases and which phase you sampled would give a different analysis result.


I'm inclined to suspect that your identical hotplates were less identical than you think.  They can have substantial differences in heating rate as well as the actual temperature they reach, particularly if they have been around for a while.  You could try boiling another pair of similar beakers and check the time they take to boil to confirm.

Usually lab glass breaks because of thermal shock - either hot glass with cold water or cold glass with hot water.  However, if you heated the beaker steadily for 40 minutes, it definitely wasn't shocked into breaking.  An exception might be if the hotplate is malfunctioning and get extremely hot extremely quickly, but that doesn't seem likely.  It is definitely possible that there was already a small crack in the beaker and that heating caused it to open further.

A contaminant is very unlikely to have been the source of the problem.  You should be able to boil dirty beakers all day long without them popping on you.

Unfortunately it's pretty much impossible to tell what happened for sure unless it winds up being reproducible.


Steric effects of a lone pair and piperidine

The energy difference between the two conformations in your example is extremely small.  This makes it difficult to say that "this" is the reason why things are the way they are. Nonetheless, it is a very interesting question. Perhaps, like in the "how many angels can you draw on the head of a pin" question, the discussion around the question might be as significant as the answer.

What we need is more data.  The phosphorus analogue of piperidine, phosphacyclohexane, has also been examined and the hydrogen resides in the axial position (no evidence for the equatorial isomer was observed).

There are important differences between these nitrogen and phosphorus analogues.  To start with, phosphorus-carbon and phosphorus-hydrogen bond lengths (180-85 pm and 145 pm respectively) are typically longer than the analogous nitrogen-carbon and nitrogen-hydrogen bonds lengths (150 pm and 102 pm respectively).  More important is the fact that the phosphorus is basically unhybridized, so the lone pair of electrons around phosphorous exists in a non-directional s-orbital.  This makes conformational analysis of phosphacyclohexane easier because we don't have competing (hydrogen and lone pair) substituent effects to consider and dissect.

The observation that the hydrogen in phosphacyclohexane exists in the axial position suggests that there is an attractive 1-3 synaxial H-H interaction and/or a sterically repulsive 1-2 H-H gauche interaction.  Given the long length of bonds to phosphorous, repulsive interactions seem unlikely and the attractive 1-3 synaxial H-H explanation was favored.

Another data point comes with N-methyl piperidine where the lone pair strongly prefers the axial position.  This was attributed to a repulsive 1-3 synaxial Me-H interaction.  The fact that the methyl group adopts the equatorial position also shows that the repulsive 1-2 Me-H gauche interaction must be of little consequence and any repulsion in the corresponding H-H case would be even smaller.

Piperidine itself appears to be somewhere in the middle, in between N-methyl piperidine and phosphacyclohexane.  Certainly the repulsive 1-3 synaxial Me-H interaction observed in N-methyl piperidine has been reduced, but not eliminated.

Whether just the repulsive 1-3 synaxial H-H interaction has been reduced or whether there is now also a contribution from the attractive 1-3 synaxial H-H interaction seen in phosphacyclohexane cannot be determined.


What happens when a body of water and a pice of ice are placed in a pricessly controlled zero degree environment

My understanding is, if you put ice into a >0 degree environment it
  will be 0 degrees as it melts and if you put water into a <0 degree
  environment it will 0 degrees as it freezes.

Correct. If you place ice in a container where the temperature is above  $0\ \mathrm{^\circ C}$, the ice will first heat up to $0\ \mathrm{^\circ C}$, and then undergo phase transition to a liquid. The same works the other way around, as you stated. Placing water in a sub-zero temperature will cause the water to cool down to $0\ \mathrm{^\circ C}$ and then it will freeze (form a solid). See picture below:

  But want happens if you put water and ice in an environment maintained exactly at 0 degrees.

This really depends on the other qualities of water. If you add salt in to the water, it will remain as a liquid at $0\ \mathrm{^\circ C}$. This is due to freezing-point depression, which is described reasonably well in Wikipedia.

If you are considering pure water (no salt or other solvents mixed), then you will get an equilibrium of water and ice. It might be easier to understand, if you keep in mind that although the overall temperature is $0\ \mathrm{^\circ C}$, there will be some variation between single molecules, some having a lower amount of thermal energy and some having a higher amount.


Why are sigma bonds stronger than pi bonds?

The reason behind this is the orientation of the overlapped orbitals. Sigma bonds result from head-on(co-axial) overlapping while pi bonds are outcome of lateral(para-axial) overlapping. Here is a pictorial representation of ethene(sp2 hybridized C atoms) :

The greater the extent of overlapping, the higher the probability of finding the valence  electrons in between the nuclei and hence the bond will be stronger & shorter.

In MOT, this can be explained using Overlap Integral. This is how Atkins depicts it :

In simple terms, after forming a sigma-bond (a pre-requisite for pi-bonds), the two atoms get locked along the inter-nuclear axis. As a result, the orbitals available for pi-bonding can only partially overlap, thus forming a weaker bond.


Group 15... Lewis Acid, Lewis Base or Not At All?

For anyone else with a similar question above... I have been researching this for the last day or two, I have found Group 15 should be spilt into two groups - either nitrogen or heavier G15 elements when considering basicity or acidity.

Nitrogen - Trihalides

NF3 < NCl3 < NBr3 < NI3 (with NI3 as the most basic due to increasing relative electronegativity of the nitrogen). (These compounds are generally only weak bases.)

In my research I have found the greater the electronegativity of the lone pair atom in comparison to its ligands (i.e. the halides in this instance), the more free the lone pair is to participate in a dative bond to form a Lewis adduct with a Lewis acid. The more free the lone pair is the stronger the base is, hence NI3 is the strongest base out of the trihalides.
This is why NH3 is such a great base as the lone pair on N is not inductively pulled towards the ligands because H is relatively less electronegative. Whereas in NF3 the lone pair is so drawn away from nitrogen this compound is sometimes not even classed as a base, hence it is the weakest of the trihalides.
The reason nitrogen is on its own is because it has max coordination number 4.

P, As and Sb - Trihalides or Pentahalides

I have found halides, either tri or penta, are generally classed as Lewis acids.
PX3, AsX3, SbX3 (X - F, Cl, Br or I)

First the trihalides, these act as mild Lewis acids towards trialkylamines and halides. This is even though the P, As or Sb will have a lone pair. As before,  this is partly due to electronegativity reducing the availability of the lone pair.

My textbook points out PF3 is probably a little bit unusual in this group because it has empty d-orbitals (making it a weak sigma donor yet a strong pi acceptor.)

The pentahalides of heavier group 15 elements are pretty strong acids
PX5, AsX5, SbX5 (X - F, Cl, Br or I)

The trihaldies of nitrogen are explained nicely here -
https://courses.lumenlearning.com/introchem/chapter/trihalides-boron-halogen-compounds/

Also a clever link about NCl3 -
http://www.chm.bris.ac.uk/motm/ncl3/ncl3h.htm

The heavier G15 information and pictyre came from a textbook Inorganic Chemistry (Weller, Overton, Rouke, Armstrong)


Does anyone know what this piece of glassware is?

I believe this is an ebulliometer, from the Latin ebullientem (to boil).

It's a used to measure boiling points. Clearly, there are some spelling issues with the sharpie label here.


Splitting of multiplets in ¹H NMR spectra?

Which nuclei you should consider as neighboring atoms for the multiplet splitting depends on the size of the coupling constant. The coupling constant are usually named after the number of bonds between the coupling nuclei, so a coupling between two hydrogens that are three bonds apart from each other would be a $\mathrm{^3J_{HH}}$-coupling. In an alkyl chain like CH3CH2 the coupling between the CH2 and the CH3 hydrogens would be a $\mathrm{^3J}$-coupling that is about 7 Hz large. Couplings over more than three bonds are usually not observed, because they are smaller than the linewidth of the signals in your NMR spectrum. The exception to that you're likely to encounter are couplings along C=C double bonds, where you can even see $\mathrm{^4J}$-couplings or more.

So the simplified rule would be that couplings along three bonds are visible, couplings along more than three bonds are only visible when there is at least one C=C double bond along the way.

The concepts of chemical and magnetic equivalence are essential to understanding how different multiplicities arise. Chemical equivalence means there exists a symmetry operation that exchanges those nuclei. Magnetically equivalent nuclei additionally need to have identical couplings to all other spins in the molecule. Magnetically equivalent nuclei don't couple to each other.

In propane the hydrogens of both CH3 groups are three bonds away from the hydrogens of the CH2 group, so the coupling between those is visible in the NMR spectrum. The hydrogens of each CH3 group are magnetically equivalent, due to the fast rotation along the C–C bond, and the two CH3 groups should also be magnetically equivalant. So you have 6 magnetically equivalent hydrogens that couple to your CH2 hydrogens. The result of that is a splitting into a septet (7).

The OH-group is an interesting exception, as you would expect it to lead to a visible coupling on hydrogens connected to the same carbon, but you don't observe that under most conditions. The reason is that the OH is acidic enough that the hydrogen exchanges quickly with the solvent, so the hydrogen dissasociates and associates quickly. This happens too fast for NMR, so the other nuclei only see the average OH-hydrogen. This eliminates the coupling to the OH, and it is also the reason why the OH-signal is often very broad or even completely gone in NMR spectra.


A "100% ionic" bond would be a bond whose bonding electron(s) were never in the vicinity of the cation, but rather always in unperturbed valence orbitals of the anion.  That's not far from the truth in a paradigmatic ionic compound like $\ce{NaCl}$, but no matter how electronegative the anion is, the bonding electrons will still experience some attraction toward the positive charge on the cation, and so a detailed model of their wave functions will show a small but nonzero amplitude near the cation.  That's just basic electrostatics; you have a positive charge and a negative charge not that far away from each other, they will attract.

Despite that, we can still talk about 100% ionic bonds as a theoretical limit and an acceptable approximation for modeling some compounds.  It's kind of like how we can use absolute zero as the theoretical zero point of a temperature scale even though nothing in the universe can ever be that cold.


Can any other common reagent other than than H3PO4 cause this conversion?

Eaton's reagent - $\ce{P2O5}$ in Methanesulfonic Acid - is a good alternative.
JOC 62(1997) 3552

Conc. Nitric acid risks nitrating the aromatic ring. I have never seen conc. Nitric or conc. $\ce{HCl}$ used for this


I believe that the order of stability is II > III > I. The adduct should be formed between the lone pair on nitrogen and the Lewis acid. From there, I would apply the M+, I+ effects of the methyl groups as are known from benzene derivatives (although the substitution chemistry of pyridine is different from benzene).


Finding final temperature for adiabatic expansion

The equations you are proposing to use are valid strictly for an adiabatic reversible expansion. In the present adiabatic expansion, the external force per unit area is dropped suddenly from 5 atm to 1 atm, and held at that value until the volume doubles.  This takes place rapidly and spontaneously, and is not reversible, so the equations for an adiabatic reversible expansion do not hold.  For an irreversible expansion, the ideal gas law (or other equation of state) cannot be used to describe the force per unit area acting on the piston face, because the equation of state only applies to a system in thermodynamic equilibrium (or, since a reversible expansion consists of a continuous sequence of thermodynamic equilibrium states, to a reversible expansion).  In an irreversible expansion, the force at the piston face (where work is occurring) is determined not only by the amount of volume change, but by the rate at which the volume is changing.  This is because of viscous stresses that are important in rapid expansions and compressions.  In addition, the mass (inertia) of the gas comes into play, causing the pressure and temperature to be spatially non-uniform within the gas during its deformation.  So there is no single uniform value of the pressure or the temperature that can be assigned to the gas.


Chemical Kinetics- Net Number of Particles

Actually, the rate law for the species $\ce{C}$ is written as:

$$\frac{\mathrm{d}\ce{C}}{\mathrm{d}t} = k \ce{A} \ce{B} - k' \ce{C} \ce{y}$$

If you integrate this, you would get an equation in terms of the number of particles, as you require. However, this is a very typical integral, and there would be a number of other arbitrary constants, which would make the final result not equal to the one you've shown.

What you've written is a mistake on replacing the rate of formation of $\ce{C}$ with its number of particles itself. Both are not equivalent.


Elastic scattering of x rays in 3D: the principle for XRD analysis

In 3D you will see spots not cones from a single crystal.  If you Google for "XRD spot pattern" images, you will probably find yourself enlightened about how many different shapes, and forms diffraction patterns can take.

Debye-Scherrer rings are one special case which occurs when you have a diffraction pattern from a very fine powder.  In such a powder you are summing the diffraction patterns from thousands to millions of individual crystallites.  In this case, you can imagine a large number of spot patterns rotated randomly around the beam axis so each spot projects into a cone.  Each individual crystallite, however will produce a spot pattern, and there is never anything but a spot pattern from a single crystallite.  (Sometimes spots can be smeared when there is strain in the crystal, or when the crystal is too small to produce high quality diffraction, but these are really cases of imperfect crystals.)

You can see diffractions from rays transmitted through the crystal.  However, X-rays tend to be absorbed within a few microns to millimeters in most materials so that limits the size of crystal you can use for transmission X-ray diffraction.

Finally, I noticed you said "Two parallel and in-phase x-rays..."  I just want to note that diffraction occurs with a single photon.  A single photon can scatter, and produce constructive/destructive interference with ITSELF.  You need to do this with many photons to see a full interference pattern, however, and this is usually the case in experimental situations.  If this is new to you, then take a look at Young's double slit experiment.

Hopefully this clarifies things a bit.


how are colloids and suspensions solutions?

The definition of solution is a little bit cloudy. IUPAC define a solution in this way:

  A liquid or solid phase containing more than one substance, when for
  convenience one (or more) substance, which is called the solvent, is
  treated differently from the other substances, which are called
  solutes.

And indeed this is a quite general approch that takes in account that we usaually call solutions systems (such as metallic solution) that are not compose of single molecules or ions dissolved in a solvent. Besides this approch chemists generally call solution a system with the folowing charachteristics:

It is stable.
It has only one phase.
It has to be an homogeneous mixture of two or more substances.
The solute can't be distinguished from the solvent (no scattering, can't be filtered)

And so I would not define colloids and suspension solution because they have a dispersed phase and often they are not stable.