Can NaOH and NaH2PO2 exist together in aqueous solution?

Your question: Can $\ce{NaOH}$ and $\ce{NaH2PO2}$ exist together in aqueous solution?

I assume what you meant is your textbook is saying when two separate solutions of $\ce{NaOH}$ and $\ce{NaH2PO2}$ are combined, there should be no reaction. If that is what you meant "exist together," then your textbook is correct because there won't be an acid-base reaction since $\ce{NaH2PO2}$ has no "acidic hydrogen" contrast to $\ce{NaH2PO4}$ and $\ce{Na2HPO4}$ as Poutnik mentioned in his comment (more descriptive presentation would be $\ce{Na^+^-OP(=O)H2}$). The following diagram summerizes this point:

As diagram shows $\ce{NaH2PO2}$ (sodium hypophosphite) does not have any acidic protons to react with strong bases like $\ce{NaOH}$. The two hydrogens in $\ce{NaH2PO2}$  have attach ed to phosphorous atom in covalent bonds. And also showing is sodium bicarbonate, which has one acidic proton to react with $\ce{NaOH}$.


transpiration of trees and the cooling of the air

You seem to misunderstand the thermodynamics of evaporation: "Evaporation is an endothermic process, in that heat is absorbed during evaporation." It appears you are confusing evaporative cooling with a heat pump. In a heat pump, the compression half of the cycle changes the vapor back to liquid, which does make things warmer.

Consider boiling water at sea level: Water liquid at 100°C becomes water vapor at 100°C, requiring ~540 cal/g energy to change state from liquid to vapor at the same temperature. At lower temperatures, as water evaporates, thermal energy is withdrawn from the liquid, cooling it below ambient temperature. "Swamp coolers" work on that principle.


A non-life process making glucose from carbon dioxide and water using electricity or light?

Popular research these days for CO2 fixation involves the following enzymes with associated conversion rate:

RuBisCO - 10 molecules/sec
ECR(Enoxyl-CoA carboxylase/reductase) - 80
molecules/sec

The major noncarbohydrate precursors are lactate, amino acids, and glycerol.
Here are some links to current research:

TiO2-catalyzed synthesis of sugars from formaldehyde
CO2 reduction -> Hydrocarbon eg. Formaldehydes/acids (CH2O).
Then Formose/Butlerov reaction (catalyst:base+Ca/Mg) -> sugar
Volume 1 of Comprehensive Glycoscience
CO2 to Glucose Whitepaper

Some topics for research:

Gluconeogenesis
Formose reaction
Calvin cycle (absence of light)


What is the correct way to pronounce iodine?

Manshu is correct,in British English we pronounce iodine as in seen while in North American English we pronounce it as in mine.


What exactly does the term "gram atoms" mean?

$\ce{MgB2}$ is 1 gram-molecule of $\ce{MgB2}$ but 3 gram-atoms of $\ce{MgB2}$.

Why is this so? Let us consider this problem in the familiar mole:

If we were asked how many moles of the substance (molecule) were in this, we would state 1 mole, if there were $\pu{45.3 g}$ of substance. However, if we were asked how many moles of atoms were in the substance, we would accordingly state 3. Thus, in one molecule, there are 3 atoms, so if we simply consider the number of atoms, there would be the necessity of counting the number. of atoms in a molecule, as a molecule is comprised of more than a single atom

* This is purely hypothetical. One would probably not consider  $\ce{MgB2}$, as J. LS pointed out, it decomposes . However, the concept can be applied to other compounds


Can sulfuric acid itself be electrolysed?

After some searching, I found this 1956 Thesis(Reference), which describes the production of peroxydisulfuric by electrolyzing concentrated sulfuric acid. The yield varied according to the concentration of the acid, temperature, current, and the nature of the electrode. In the Thesis, the chemist noted that discharged $\ce{HSO4-}$ united together to form peroxydisulfuric acid ($\ce{H2S2O8}$). It further stated that  oxidation of sulfate ions by anodic oxygen was responsible for the peroxydisulfuric acid formation. Also, hydroxyl ion at anode played an important role in the acid formation.

The reaction mechanism was confirmed by electrolyzing sulfuric acid and ammonium sulfate at various concentration at $\pu{16\!- \! 20 ^\circ C}$ using platinum electrode and the reaction occurred in accordance to the theory. It was also confirmed that amount of peroxidic oxygen increased with increase in ammonium sulfate. The specific gravity of sulfuric acid and the current density was varied and accordingly the highest yield of 78.4% was obtained. It was also noted that increasing the current density increases the yield up to a certain point as  peroxydisulfuric acid was found to convert into peroxymonosulfuric acid. At that moment, a small trace of hydrogen peroxide was noted.

So, the chemical species formed during electrolysis are $\ce{HSO4-}$, $\ce{SO4^2-}$, peroxide ion, protons, trace of hydrogen peroxide, and other side products.

Reference:

Sin-Chou Fan, “The electrolytic production of peroxydisulfuric acid using periodically reversed direct current and alternating current superimposed on direct current,” MS Thesis, Virginia Polytechnic Institute, 1956.


Why does formaldehyde exist primarily as the gem-diol in aqueous solution?

In acetone the carbonyl form is more stable than the gem-diol, whereas with formaldehyde the situation is reversed and the gem-diol form is preferred. Overall, the equilibrium shifts by 6 powers of 10 between these two compounds.

It takes $\pu{\approx1.4 kcal/mol}$ to shift an equilibrium by 1 power of 10. Therefore in the acetone-formaldehyde comparison there is a stabilization change of $\pu{\approx8.4 kcal/mol}$.

It is difficult to see how changing the substituents in the gem-diol could have such a large effect, after all the gem-diol carbon is a saturated  $\mathrm{sp^3}$ carbon, no p-orbitals are avaliable for resonance interaction. Considering steric effects as proposed in the other answer, methyl groups and hydroxyl groups are not particularly large, nor is the gem-diol carbon especially congested. Hence it is difficult to see how steric factors could provide the necessary $\pu{8 kcal/mol}$ change in stabilization. For reference, Cis-di-t-butylethylene only has $\pu{9.3 kcal/mol}$ of strain energy.(1)

On the other hand we might expect the substituents on the carbonyl carbon (an unsaturated $\mathrm{sp^2}$ carbon with a p-orbital available for interaction) to have a significant effect on stability. Just as alkyl substituents stabilize a carbon-carbon double bond through both an electron-releasing inductive effect (2) and a hyperconjugative effect, we would expect the same effects to operate with a carbon-oxygen double bond.  In fact, we might anticipate that the alkyl stabilization effect will be even greater with the carbonyl double bond since the carbonyl carbon is more electron deficient than the carbon in a carbon-carbon double bond.

If we compare the heats of combustion for the isomers acetone ($\pu{-427.9 kcal/mol}$) and propanal ($\pu{-434.1 kcal/mol}$), we see that attaching a second alkyl group to the carbonyl carbon stabilizes the system by $\pu{6.2 kcal/mol}$.(3) Of course, adding the "first" alkyl group will also provide significant stabilization. Hence, it is clear that adding two alkyl groups to formaldehyde will provide enough electronic stabilization (inductive and hyperconjugative) to shift the equilibrium by $10^6$ from the enol to the keto form.

see here
the alkyl group has an $\mathrm{sp^3}$ carbon while the double bond carbon is $\mathrm{sp^2}$ hybridized; since the latter is more electronegative, electron density will flow inductively (through sigma bonds) from the alkyl group to the $\mathrm{sp^2}$ carbon.
data from K. Peter C. Vollhardt, Neil E. Schore, "Organic Chemistry, Fourth Edition: Structure and Function" p. 125; see here


How to calculate the osmotic pressure of a sucrose solution?

I did it in this step: First I use ΔTb=kbm And substitute following Tb=101.45 and kb=0.512 I get m=198.144 mol/kg

This should be the first clue that something went wrong.  198 mol of sucrose is about 67 kilograms of sucrose, so your implied molality means that for every kg of water, there are 67 kg of sucrose.  That can't be right.  (It would be about the same as adding a tablespoon or two of water to a 3-lb bag of sucrose and expecting the whole thing to dissovlve.)

I think you forgot to use the change in boiling point temperature in your $\Delta T_b = k_b m$ equation.  The change is only 1.45 °C, not 101.45 °C.

What happens if you plug in that value and redo your calculations?

Update: Now that you've done that, your answer is at least chemically reasonable but there is still one problem.

  Density equal to 1.036 g/ml equals 1.036 kg/L, which means 1.036 kg for 1 L and 1 kg=0.965 L.

This is true, but keep in mind the definition of molality: it is moles of solute per kg of solvent.  However the density is obviously referring to the kg of total solution.  So, it is inappropriate to find the molarity by simply dividing the molality by the density.  As I described in a past answer on chem.se, if the molarity is $C$, the molality is $\frac{1000}{f}\frac{\frac{C f}{\rho}}{1-\frac{Cf}{\rho}}$, where $f$ is the formula weight of the solute and $\rho$ is the density.  So you will have to invert this relationship to find the molarity from the molality.


Bonding and coordination of oxygen in a Ga2O3 crystal structure

Does it help if you think of this as an ionic compound? In ionic compounds, there are no directed bonds. The coordination number depends on the size of the ions, coupled with the need for electroneutrality.

For example, in calcium oxide, both calcium and oxygen (in red) have octahedral coordination, with 6 neighbors each:

Granted, the electronegativity difference in calcium oxide is 2.44, while in gallium oxide it is only 1.65, and not different from that of silicon oxide, which has a network covalent structure.

Also, consider structure like that of hexagonal ice, where each oxygen is surrounded by four hydrogen atoms, two connected by covalent bonds, and two connected by hydrogen bonds. In this case, the atomic distances are clearly different, and you would be correct to say that oxygen makes two bonds, with a tetrahedral coordination if you consider the hydrogen atoms from neighboring atoms.

To distinguish these three extremes (ionic bonds in one case, covalent bonds plus contacts with neighboring atoms, network covalent structure) and to categorize what is going on with gallium oxide, you would have to look at the gallium oxygen bond lengths, which should be available in the paper you cite.


Schrödinger equation for multi-electron atoms

That probably depends on what you consider to be the "form" of the Schrödinger equation.

As Todd succinctly said, the Schrödinger equation[s]

$$\mathrm i\hbar \frac{\mathrm d|\psi\rangle}{\mathrm dt} = \hat{H}|\psi\rangle \quad \text{(time-dependent)} \tag{1}$$

or

$$\hat{H}|\psi\rangle = E|\psi\rangle \quad\text{(time-independent)} \tag{2}$$

don't change. The Hamiltonian will, of course, be different, but the question is how different? The general form of the (non-relativistic) Hamiltonian for an atom has three components

$$\hat{H} = \color{red}{\underbrace{-\frac{\hbar^2}{2m_N}\nabla_N^2}_{\text{KE of nucleus}}} \,\,\,\, \color{blue}{\underbrace{- \sum_i \frac{\hbar^2}{2m_e}\nabla_i^2}_{\text{KE of electrons}}} \,\,\,\, \color{green}{\underbrace{-\sum_i\frac{Ze^2}{4\pi\varepsilon_0r_i}}_{\text{nucleus-electron PE}}}\,\,\,\, \color{purple}{\underbrace{+ \sum_i\sum_{j>i} \frac{e^2}{4\pi\varepsilon_0r_{ij}}}_{\text{electron-electron PE}}} \tag{3}$$

where the nucleus has mass $m_N$ and charge $Ze$; and $r_i$ denotes the distance between electron $i$ and the nucleus, and $r_{ij}$ denotes the distance between electrons $i$ and $j$.

Whether you are dealing with a hydrogen atom or a silver atom, this form remains valid. The only special thing about the hydrogen atom (or in a hydrogenic ion) is that there is only one electron, so the first two sums only have one term, and the last sum has no terms. So, the Hamiltonian for the hydrogen atom can be simplified to

$$\hat{H}_\text{H atom} = \color{red}{-\frac{\hbar^2}{2m_p}\nabla_p^2} \color{blue}{- \frac{\hbar^2}{2m_e}\nabla_e^2} \color{green}{- \frac{e^2}{4\pi\varepsilon_0r}} \tag{4}$$

but I'm of the opinion that the form of the Hamiltonian

$$\hat{H} = \color{red}{\hat{T}_n} + \color{blue}{\hat{T}_e} + \color{green}{\hat{V}_{en}} + \color{purple}{\hat{V}_{ee}} \tag{5}$$

doesn't change, it's just that there are fewer terms in each sum. You might disagree, but I don't really care, as long as you write the correct Hamiltonian out.


Meaning of "MS1 and MS2 resolution" and "MS1 and MS2: Unit" in a tandem mass-spectrometry parameter list

The term "unit" probably means the size of the isolation window used in the QQQ.  The term "unit" means the nominal isolation window is 1 Da (i.e. one mass "unit") wide.
The "resolution" is a standard term in mass spectrometry, and here they are saying it is about 1 Da.


I have separated C1 and C3 into unique structures. For C1, it is clear that the oxygen of the hydroxyl group has the top priority while the methylene group has the lowest priority. A digraph is constructed such that we delineate two paths around each of the unsaturated rings. Thus, non-duplicate atoms C3 and C5 terminate with their respective duplicate atoms C(3) and C(5). The cyclopentene ring has locants in sphere 3 of {(7),7,H} which is a tie with {(4),4,H} in sphere 3 of the cyclopropene ring. But the locants {8,H,H} of the cyclopentene lose to {(3),3,H} of the cyclopropene ring. The priorities are O>cyclopropene>cyclopentene>methylene. C1 has the (R)-configuration. continued...

A different digraph for C3 is constructed by dissecting the three rings into a total of six pathways, two for each ring. The rings are designated with red, blue and black dots. The red and blue rings take precedent over the black ring because each one has locants {O,O,H} at the branchs while the black ring has {O,C,H}. To distinguish between blue and red, we move to the next atoms on the chain. Blue has two oxygens while red has two carbons. The priorities are blue>red>black>methylene. C3 has the (S)-configuration.

Related issues:

What are the CIP rules for cyclic substituents?
How to check for geometrical isomerism in cyclic compounds?
More on Rings and Digraphs


There are two enantiomeric forms of glucose, D- and L, with D-glucose being the naturally occurring form. Dextrose is just another common name for D-glucose, so if a microorganism can metabolise D-glucose then it can by definition metabolise dextrose as they are just two words for the same thing.

As for the question of whether the enzyme would recognise both optical isomers (enantiomers), this would probably not be the case. Enzymes are 'chiral', as are the two enantiomers of glucose. The active sites of enzymes/proteins are often specific for a given isomer, hence why many drugs are only biologically active in one enantiomeric form, with the other enantiomer being either inactive, less active, or in certain cases toxic.

In particular, the first step in glucose metabolism is phosphorylation of the C-6 hydroxyl group by the enzyme hexokinase. However, hexokinase is incapable of phosphorylating L-glucose.


Is the heat for constant pressure the same as for constant volume in the formation of nitric oxide?

Kevin is right. In that reaction, the number of moles does not change. So carrying it at constant pressure or at constant volume does not make any difference. And $q_v$ = $q_p$


Electrons always fill in the lowest energy configuration possible.

Cr and Cu, as well as Cu and Ag, are exceptions in the "typical" filling order.  In the case of Cr and Cu, they are stabilized by having 2 half filled orbitals, which maximizes exchange energy and minimizes electron repulsion.  In their case, the energy to promote an s electron to the d orbitals is compensated for by the effects of exchange energy and no repulsion.  This effect is called "the stability of half filled subshells", or something to that effect, in most textbooks.

The energy cost to promote 2 electrons from the s subshell, as would be the case for Vanadium, is too much to be compensated for by the exchange energy gained and the repulsion lost, so the lower energy configuration is to have two electrons in the s subshell and 3 in the d.

Here's a nice website that shows the electron configuration for all elements: with http://www.ptable.com/#Orbital


Can I disable "Display visual cues" by default in Avogadro?

Sure. There's a big checkbox in the Navigate Tool settings:

When you quit, the settings should be remembered for next launch.

If not, please e-mail the Avogadro mailing list and we can figure out what's going on.


Calculating the electrons an atom wants to gain/lose to reach a noble gas

The book's explanation about a noble gas configuration is somewhat accurate, but fairly incomplete.

The elements on the right and on the left of the periodic table (the alkali (earth) metals, the halogens, the chalcogens (the group that starts with Oxygen) and the pnictogens (Nitrogen group)) have electron configurations that make it somewhat easier to lose/gain electrons to "look" like noble gasses (this is why alkali metals and halogens are so reactive: the destabilization from the charge is very minimal compared to the stability of the electron configuration).

However, as you have observed, the book goes to some effort to avoid talking about transition metals. There is a reason for that.

Noble gas configurations are a subset of the stable configurations of electrons. In reality, what's actually being aimed for is an element with no incomplete electron shells.

What is an electron shell? The actual quantum mechanical definition may be a bit more complicated than you need, but for your purposes, it suffices to say that it's how electrons will tend to be distributed around a nucleus. Note that this is extremely distinct from saying that the electron "orbits" the nucleus.

If you look at a periodic table, there are four general regions to look at:

The labels "s", "p", "d", and "f" all refer to the various electron orbitals, and the blocks all indicate what the outermost (or "valence") shell is.

So while elements in the s and p blocks can lose a small number of electrons get complete shells (which will resemble the noble gases), transition metals are a bit more complicated. And by a bit more complicated, I mean that there are very rough rules in place that have almost more exceptions than they have actual examples of following the rule.

But in very general terms, a transition metal will lose electrons. If they are in the first half of the transition metals in their row, they can lose a number of electrons less than or equal to the number of d electrons plus the number of s electrons. This maximum number of electrons lost will make them resemble the noble gasses in configuration, but it's only appreciably stable when the transition metal is in the first half of transition metals in its group ($\ce{Mn^{7+}}$ is well documented in ionic compounds, but $\ce{Fe^{8+}}$ doesn't exist a lot because it has too many electrons to try to lose to comfortably fall into a noble gas configuration). But transition metals can lose electrons up to this point. The issue is, most transition metals can adopt multiple charge configurations for various reasons, so generally, when writing an ionic compound with a transition metal in words, you need to specify the charge of the transition metal (actually, its oxidation state, but in an ionic compound that's mostly splitting hairs). So you can have "Iron (II) Oxide", which is a black powder and "Iron (III) Oxide" which is rust. These compounds will have different chemistries because of the charge on the transition metal.

Suffice to say, your textbook is avoiding transition metals because transition metals are weird.


Do the dimensions of the cathode and anode matter in electrochemistry?

The dimensions of your electrodes should not effect the open circuit voltage, since this is determined solely by the identity and concentrations of the redox species.

Once the cell is running, the dimensions may affect how much the voltage drops off as you draw current, known as "overpotential". If your cell is limited by its ability to deliver ions to the surface of the electrode, or to transfer charge into the electrode, then the higher surface area of a large electrode is likely to result in a more efficient battery.


How to calculate the amount of substances product produces from the masses and amounts of starting materials?

From the equation,
$$\ce{C + O2 -> CO2},$$
we can see that one mole of $\ce{C}$ reacts with one mole of $\ce{O2}$.

Therefore, let us calculate the amount of each reactant present.

The amount of $\ce{C}$ is already given: $n(\ce{C}) = \pu{1 mol}$.

The molar mass of $\ce{O2}$ is $M(\ce{O2}) = \pu{32 g//mol}$, so $\pu{16 g}$ of $\ce{O2}$ are $\pu{0.5 mol}$.

Since one mole of $\ce{C}$ only reacts with one mole of $\ce{O2}$, there remain $\pu{0.5 mol}$ of unreacted $\ce{C}$, while $\pu{0.5 mol}$ of $\ce{C}$ react with $\pu{0.5 mol}$ of $\ce{O2}$.

In formal terms, oxygen is called the limiting reactant as it limits the amount of product formed.

From the equation, one mole of $\ce{CO2}$ is produced per each mole of $\ce{C}$. Therefore, $\pu{0.5 mol}$ of $\ce{CO2}$ is produced.


How the plate height is assumed to be the standard deviation?

I feel that you misunderstood what the authors were trying to say in that textbook. Forget about everything for the time being.
Imagine if you injected a single analyte band into an HPLC column of length L at a constant flow rate. You waited for some time, and now the band is exiting the column. For simplicity, assume that the band is rectangular.
Ask yourself, how much average distance did the analyte band travel? Indeed, it must have traveled a distance L if it has to exit the column. Most modern HPLC columns are 10 cm long; assume analyte traveled an average of  L=10 cm.
Now a band cannot be infinitely thin inside a chromatographic column. The analyte spreads in space, i.e., its width increases as it passes through the column due to diffusion and other processes.
So let us say the band is now a 0.05 cm wide rectangle when it reached the end of the column. Half of it is inside the column, and half of it has exited the column. What width of the band is inside the column, and what width is outside the column? You would say 0.05/2 cm of the band is inside the column, and 0.05/2 cm of the band is outside the column.

Total distance traveled by the portion of the band which has exited = 10 + 0.05/2 cm
Total distance traveled by the portion of the band which is inside = 10 - 0.05/2 cm

Now in chromatography, as you have seen, peaks are not rectangles, but they are Gaussians. Furthermore, Gaussians are characterized by their "standard deviation," which is a measure of their spread. If you apply the sample logic

The mean (=center of the Gaussian) has traveled a distance of 10 cm or L
The portion of the Gaussian which is still inside traveled a distance L-1*standard deviation
The portion of the Gaussian which exited the column traveled a distance L+1*standard deviation.

Note that conventionally a full Gaussian band at the base is considered to have a width of 4xstd. dev.


Is there a connection between UV-degradation and the photoelectric effect

Yes and no.

In a metal, the photoelectric effect is when a UV photon has enough energy to push an electron from it's current energy level to at least the $n=\infty$ level. Extra energy goes into electron kinetic energy.

For UV degradation, we are promoting elections to higher energy orbitals in an analogous way. This produces an energized state that is highly reactive and may undergo radical reactions that ultimately lead to the formation of different species.

The difference is that we assume the electron is essentially removed from the atom in the photoelectric effect. Elections are weakly enough bound in a metal for a UV photon to achieve this. But in general, for UV degradation, the UV photon can't actually remove the electron from the molecule. But the photon can energize electrons that lead to degradation reactions.


What is the difference between grafting and bonding in reference to polymers?

Grafting a polymer B on a polymer A is attaching  polymer B to the linear backbone of polymer A (co-polymer).
If A and B are identical, we obtain a branched polymer (Homopolymer).


Why can Cu have an oxidation number of +2?

Atomic copper has the electron configuration $\ce{[Ar] 3d^{10} 4s^1}$.  By removing one electron and producing $\ce{Cu^{+1}}$, an inert gas configuration $\ce{[Ar] 3d^{10} 4s^0}$ is produced.  While it does take a lot more energy to remove the second electron from copper (first IP=745 kJ/mol, second IP=1,958 kJ/mol), if this energy can be offset by the energy gained through bond formation and lattice energy (or hydration energy) then $\ce{Cu^{+2}}$ compounds will form.  In fact $\ce{Cu^{+2}}$ is the most common oxidation state of copper so the energetics must generally work out that the energy gained by forming more than one bond to copper and gaining additional lattice (or hydration) stabilization, more than offsets the energy cost of removing that second electron.  Removal of additional electrons can also occur with copper to form $\ce{Cu^{+3}}$ and $\ce{Cu^{+4}}$ compounds, for example, $\ce{K3CuF6}$ and $\ce{Cs2CuF6}$.


To answer your question, we can use the idea of radioactive decay;

According to the IB Physics Book from Oxford University, radioactive decay is "a naturally occurring process in which the nucleus of an unstable atom will spontaneously change into a different nuclear configuration by the emission of combinations of alpha particles, beta particles, and gamma radiation. Alpha particles are simply a Helium atom while beta particles consist of either electrons or positrons (the antiparticle of the electron).

Radioactive decay is a random process and therefore it is impossible to predict which exact nucleus may decay next. However, with a large sample of nuclei, say of your igneous rock, it is highly probable that in a given amount of time, a predictable amount of nuclei will decay, even if we cannot predict which ones exactly. Hence we have the idea of half-life, which is the amount of time it takes for the half of the initial nuclei of a sample to decay.

  approximately what percentage of the atoms (or protons, etc) in the rock might be the same ones that were in it 10,000 years ago?

It depends on what the igneous rock consists of; if it is made out of ununoctium-294, an infinitesimal amount of it will remain, as the half-life of it is 5 milliseconds. On the other hand, if the rock consists of Uranium-238, after 10 thousand years the composition of the atomic composition of the rock will have barely changed, as Uranium-238 has a half-life of 4500000000 years. So to answer your question, you would need to look at the half-life of your igneous rock. For example, if your igneous rock has a half life of 10 thousand years, then only 50% of the nuclei would be what they were 10 thousand years ago. However, it would also depend on how much of the rock is radioactive, as MaxW mentioned.

  Is it reasonable to even think of a "particular", individual atom?

As already mentioned, it is difficult to predict which particular nucleus will decay next, therefore, looking at the substance as whole facilitates predicting how much will decay in a certain period of time.

  Is it the "formation" of particles that persists over time, or is it the actual "matter"?

I do not quite understand this question. Nonetheless, I would say that it is the matter that persists over time, as the matter consists of these particles. In the photo provided below, you can see that the nuclei are decaying, but they are not decaying row for row, yet more in a random manner. Hence, I would not consider it to be the formation of particles that consists over time, but the matter as a whole.


Chair conformation. Wedged lines vs no wedged lines?

Cyclohexane is often drawn as a regular hexagon even though its prevailing chair conformation is not planar.

Perspective diagrams are preferred when discussing conformation (since there is no other reasonable way to depict boat and chair forms) and acceptable when it is just desired to emphasize configuration. (For the depiction of carbohydrates and structurally related molecules, Haworth projections can be also preferred.)

The use of wedged and bold bonds in such perspective diagrams is discussed in Graphical Representation of Stereochemical Configuration (IUPAC Recommendations 2006):

  ST-3.3 Use of bold bonds to emphasize perspective

  Since perspective diagrams imply a single absolute configuration, it is not necessary to use solid wedged or bold bonds to emphasize that configuration further. In some cases, such bonds may help an inexperienced viewer to understand the perspective used. (…)

However, note that

  Only solid wedged and bold bonds should be used for the purpose of emphasizing perspective. Hashed wedged bonds and hashed unwedged bonds should not be used, either in isolation or in combination with bold bonds.


Carbocation rearrangement with expansion of five-membered ring?

Your proposed mechanism involves a ring expansion from cyclopentane to cyclohexane. This gives very little driving force in terms of relieving strain in the system. Cyclopentane, in a twisted or "envelope" conformation, has some angle strain ($102-106^o$ vs $109.5^o$ $^{ref~1}$) and it does also has some torsional strain down some of its bonds.

This is mostly alleviated by the move to cyclohexane, if it is in a full chair conformation, but the stabilization of the system is minimal ($25-29kJmol^{-1} $ $^{ref~2}$). Here is a link to a question that deals with ring strain and the calculation of it from $\Delta H$ of combustion.

Along with this, your mechanism proposes this ring opening involving an alkyl shift that moves the carbocation from tertiary to secondary. This involves a destabilizing of the cation due to the reduced hyperconjugation (and inductive effect). Studies have shown that a secondary carbocation is around $67-75kJmol^{-1} $less stable than a tertiary, for simple substituents (and in the gas phase$^{ref~3}$).

In total, the ring opening from a five to six-membered ring in your mechanism does not result in an overall increase in stability. Hence why the second mechanism is suggested to happen instead.


Why is brine solution more effective than granular salt in melting ice on a surface?

This is because the brine solution is already in solution, whereas granular salt has to dissolve first. If ice is present instead of water, this is difficult, as there is no liquid water for the salt to dissolve in.
Secondly, salt dissolution is an endothermic process, so it will further reduce the temperature of the surroundings. For example, if you have an ice and water solution and add salt crystals you can bring the temperature of the ice/water combination to a much lower temperature (even below $\pu{-10^\circ C}$!).


If the process we're discussing is an equilibrium, and if the forward reaction is "unfavorable" (I take that to mean that the products are of higher energy then the reactants), then there are still ways to shift the equilibrium in whichever direction you want.  For example

A <-> B + C

using LeChatelier's principle, if we increase the pressure in our vessel, we will push the equilibrium back towards A, the opposite will occur if we decrease the pressure.  Temperature, catalysts, reagent concentrations, etc. can all affect the position of an equilibrium.


In order to be a gas does the pressure exerted by the gas has to be more than the atmospheric pressure?

Your teacher is right. For example, at room temperature, the pressure of water vapor over a pond or lake might be around 25 torr; that's less than atmospheric pressure, but the water vapor is undeniably a gas.

Maybe your confusion stems from a belief that gases must be produced by boiling liquids (since a liquid must have a vapor pressure equal to the external pressure before it begins to boil). But a liquid can also become a gas through evaporation, no matter what the external pressure is.

In evaporation, molecules escape from the liquid's surface and become a gas.
Molecules in the liquid are in constant motion, and they frequently collide with each other. Every now and then one will get knocked off the surface of the liquid by the others. This can happen at any temperature or pressure the liquid exists at.

In boiling, bubbles of gas form throughout the liquid. This can only happen when the temperature is high enough to make the vapor pressure equal to the external pressure.


Obtaining activity coefficients of conjugate acids of some common carboxylic acid molecules

If no specific activity coefficient data are available, I would use the general formula for ionic compounds of small enough ion cincentations:

$$
\begin{align}
I &= \frac{1}{2}\sum_{{\rm i}=1}^{n} c_{\rm i}z_{\rm i}^{2} \\
\log{\gamma} &= -0.509 . ( z_+^m . |z_-|^n )^{\frac{1}{m+n}}. \sqrt{I} \quad (*)\\
\end{align}
$$

$c_i$ is the ion concentration.
$z_i$ is the ion charge.
$I$ is the Ionic strength.
$\gamma$ is the activity ciefficient by the Debye-Hückel equation.
(*) - See the link for more info and more complex formulas for the medium $I$.


Polyethylene terephthalate reaction with tap water

It is probably not a chemical reaction that's the problem, it's that the container should be cleaned before it's refilled, and an unclean container might or will have bacteria.  While you're thinking 'refilled' might be in a matter if minutes, someone labeling the bottle is thinking about it sitting on a shelf with the cap off for months before reuse.  At that point there could be any number of unhealthy contaminants.


Yes, they are all valid resonance structures.  In the first series, there is a third resonance structure you could draw with the negative charge alpha to the carbonyl carbon. In the second series, the phenoxide anion, you can draw two more structures.  One with the negative charge in the other ortho position and one with the negative charge in the para position.  This significant resonance stabilization of the phenoxide anion explains why phenol is such a strong acid.  In your third series there is also a third structure with the positive charge situated alpha to the nitrogen.


Understanding Equivalent Wyckoff positions

It seems to me that care has to be taken with the words 'equivalent'  and 'same' in the definitions above. Equivalent positions do not mean the same position. In the case, say, of a 4 fold axis any point say in the first quadrant can be replicated to the other three quadrants and these are equivalent points but not the same point, clearly. In some cases such as with reflections in a plane, two sets of identical (or equivalent) points may be produced and so and the number of equivalent points is reduced. The number of equivalent sets is called the multiplicity. These have been called special positions when they lie on a symmetry element

In the Wycoff notation equivalent positions are labelled consecutively $a, b, c\cdots$ starting with $a$ for the highest symmetry. A position 2f, for example, indicates two equivalent atoms at position given by symmetry f. For each space group a table of general and special positions can be made.  This includes point symmetries, positions multiplicities and Wycoff symbols. The table below shows examples.

Table copied from  page 75 of 'Introduction to Crystallography' by  D. Sands, publ. Dover


A more appropriate name for the ionic factor is the van't Hoff factor of colligative properties. In the simplest terms, it is the number of particles created after being dissolved in the solution. Since glucose is not an ionic compound, it does not dissociate, therefore $i=1$ (meaning one glucose molecule stays one glucose molecule).


How exactly are ions in a Mass Spectrometer accelerated?

The acceleration of the freshly generated ions is stepwise (similar to a linear particle accelerator), i.e. the beam of ions is guided an array of electrodes with a small opening in their centre, as schematically depicted below:

(image source)

Hence when ions approach the first accelerator electrode from the left hand side, their interaction with the subsequent accelerating electrode already phases in.  Ions passing the slit of the first accelerating electrode will experience an even stronger accelerating force by the subsequent accelerating electrode.  Eventually, their momentum will increase further; and on the same time, the ions preferentially will keep their direction of propagation towards the mass dispersive element (here: the magnet), regardless the attractive force between them and the last accelerating electrode.


Electronic configuration of excited states of iron

The electronic configurations of the excited states of neutral iron are listed on this NIST Database page:

3d64s2 — a 5D — ground state
3d7(4F)4s1 — a 5F — 6928.268 cm–1
3d7(4F)4s1 — a 3F — 11976.238 cm–1
3d7(4P)4s1 — a 5P — 17550.180 cm–1
3d64s2 — a 3P — 18378.185 cm–1
3d6(5D)4s14p1(3P) — z 7D0 — 19350.892 cm–1

The reference give for this data is: G. Nave, S. Johansson, R. C. M. Learner, A. P. Thorne, and J. W. Brault, Astrophys. J. Suppl. Ser. 94, 221 (1994).

So, while in general predicting electronic configurations for excited states is a tricky business, you are right: the first excited electronic configuration is to “promote” one 4s electron into the 3d orbitals. And the second excited state configuration (regardless of the different spectroscopic terms and sublevels) is to promote the 4s electron into 4p.


Why doesn't hydroxide concentration equal concentration of carbonic acid and bicarbonate in a sodium bicarbonate solution?

Your misunderstanding appears to be some confusion between stoichiometry and equilibrium concentrations. The chemical equation describes the stoichiometry or the ratio of the species in the reaction. In other words, each time the reaction happens, the equation describes how many of each reactant are consumed and how many are produced. In your example:

$$\ce{HCO3- + H2O <=> H2CO3 + OH-}$$

Each time this reaction occurs in the forward direction, one bicarbonate anion and one water molecule react to form one carbonic acid molecule and one hydroxide anion. In the reverse direction, one carbonic acid molecule and one hydroxide anion react to form one bicarbonate anion and one water molecule.

Since this reaction is reversible, the equilibrium constant tells you something about the extent of the reaction. In other words, what portion of the overall system is sitting on the reactant side and what portion sis sitting on the product side. Put another way, the system is at equilibrium when the rates of the forward and reverse reactions are equal. These rates are dependent on the concentrations of the species on the reactant and product sides of the equation and on rate constants.

$$\mathrm{rate_{forward}=rate_{reverse}}\\
k_\mathrm{f}[\ce{HCO3-}][\ce{H2O}]=k_\mathrm{r}[\ce{H2CO3}][\ce{OH-}]$$

The equilibrium constant is the ratio of the rate constants:

$$K=\frac{k_\mathrm{r}}{k_\mathrm{f}}=\frac{[\ce{H2CO3}][\ce{OH-}]}{[\ce{HCO3-}][\ce{H2O}]}$$

The equilibrium constant is thus a law of mass action expression for the situation where the forward and reverse rates are equal. At this state, both forward and reverse reactions are occurring following the stoichiometric ratios, but the concentrations of each species has reached a steady state.


What exactly is hydrolysis reaction in organic chemistry?What mechanism does it follow?

Your textbook is almost correct. Here is the corrected version of its statement:

  Hydrolysis is a special type of nucleophilic substitution where water acts as both nucleophile and a solvent molecule.

Note me omitting the $\mathrm{S_N1}$ part. This is because a hydrolysis mechanism can be any nucleophilic substitution, whether $\mathrm{S_N1, S_N2, S_N}$ for heavy atoms or $\mathrm{S_NAr}$.

That also explains the given order of reactivity. tert-Butyl bromide is very fast at $\mathrm{S_N1}$ reactions. Bromomethane very rapidly undergoes $\mathrm{S_N2}$ reactions. The other two are in-between with bromoethane being just a tad better at $\mathrm{S_N2}$ reactions than 2-bromopropane is at $\mathrm{S_N1}$ reactions.

($\mathrm{S_N1}$ is preferred in 2-bromopropane as the solvent is polar protic and it can stabilize the intermediate carbocation well)


What is the precise definition of Ramachandran angles?

Definition

Protein backbone dihedral angles are called φ (phi), involving the backbone atoms COn-1 - Nn - Cαn - COn , ψ (psi) involving the backbone atoms Nn - Cαn - COn - Nn+1) and ω (omega) involving the backbone atoms Cαn - COn - Nn+1 - Cαn+1). Thus, φ controls the COn-1 - COn distance, ψ controls the Nn - Nn+1 distance and ω controls the Cαn - Cαn+1 distance. Where n is the current atom for which you want to calculate the dihedral angles. Dihedral angle varies between -180 to +180. It is positive if rotation between atoms is clockwise and negative if rotation is anticlockwise.

Source: http://mebinfotalk.blogspot.com/2013/06/how-to-calculate-dihedral-angle-using.html
And here is an illustration from the same source:

To calculate, use the cross product to get vectors normal to the (v2, v1) and the (v2, v3) planes, and calculate angles between them (it is easy to get the direction wrong, so I would test with known cases).
In the question, the OP states:

They rely on the fact that two bonds on one side are in the same plane as the rotating bond.

No, the plane is between the rotating bond and the adjacent bond. Two bonds of one atom always form a plane. Here is an illustration: Dihedral angle of gaseous and crystalline HOOH
3D demonstration
A nice interactive tutorial can be found at Proteopedia: https://proteopedia.org/wiki/index.php/Tutorial:Ramachandran_principle_and_phi_psi_angles
Omega
Omega is the torsion angle of the peptide plane. It is most often near 180 (trans peptide) but sometimes near zero (cis peptide).
In a PNAS paper from 2012 (https://doi.org/10.1073/pnas.1107115108), the authors state in the abstract:

Analyses as a function of the φ,ψ-backbone dihedral angles show that the expected value deviates by ± 8° from planar as a systematic function of conformation, but that the large majority of variation in planarity depends on tertiary effects.

In protein structure determinations, the omega angle is often restrained, especially in low resolution structures. Looking at structures deposited in the protein data bank, it might therefore seem that the distribution of the omega torsion around 180 is more tight (less varied) than it is in reality.
First and last dihedral
The first amino acid has no phi angle and the last has no psi angle. Exceptions are when the N-terminus or C-terminus are modified.
Bond angle tau
Different from the torsion angles phi and psi, the angle tau is a bond angle (depends on three atom position, not four). It is the angle at the C-alpha carbon between the nitrogen and the carbonyl carbon. As an $sp^3$ carbon, the C-alpha carbon is expected to have an angle around $109^\circ$. The larger the angle, the more (phi, psi) combinations are possible without steric clashes, i.e. the larger the "allowed" regions in the Ramachandran plot.


Change in enthalpy and entropy when sugar dissolves in water

The entropy change for ideal mixing is $\Delta S=-R(x_A\ln{x_A}+x_B\ln{x_B})$


How to solve for concentration of two equilibrium reactions with finite equilibrium constants?

Completely accurate answer :

The main reactions which take place are
$$
\begin{array}{ll}
\ce{ H2O <=> H+ + OH-} &\quad\left(K_\mathrm{w}= 1\times 10^{-14}\right) \\
\ce{HNO2 <=> H+ + NO2-} &\quad\left(K_\mathrm{a}= 4\times10^{-4}\right) \\
\end{array}
$$
$\ce{HNO2}$ acid is a weak acid, which does not dissociate completely in water. This means it does not give all its hydrogen ions into the water, the resulting solution contains four species ${\ce{H+}, \ce{OH−},\ce{NO2−}}$, and undissociated acid $\ce{HNO2}$. In order to specify the concentrations of the four species
present in an aqueous solution of $\ce{HNO2}$, we need four independent equations
 between them. These are :

1- Equilibria. We have two equilibrium equations:

The dissociation equilibrium of water
$$\ce{ K_\mathrm{w}=[H+][OH−]}\tag{1}$$
The dissociation equilibrium of acid
$$\mathrm{ K_\mathrm{a}}=\frac{[H^+]_e[NO_2^−]_e}{ [HNO_2]_e}\tag{2}$$
2- Mass balance. The mass balance equation that relates the concentrations of the various dissociation products of the substance to its initial concentration, which we designate here as $\ce[HNO_2]_0$. For the solution of $\ce{HNO_2}$, this equation would be
$$\ce{[HNO2]_0  = [HNO2]_e + [NO2−]_e}\tag{3}$$
3-Charge balance. In any ionic solution, the sum of the positive and negative electric charges must be zero; in other words, all solutions are electrically neutral. This is known as the electroneutrality principle.
$$\ce{[H+]_e = [NO2−]_e + OH−}_e\tag{4}$$
We can use ${\text{Eq}\mathrm{3}}$ to get an expression for $\ce{[HNO2]_e}$, and this can be solved for$\ce{ [NO2−]_e}$; these are substituted into $\text{ Eq}\mathrm{2}$  to yield
$$ K_\mathrm{a}=\frac{\ce{ [H+]_e} {([H^+]_e - [OH^-]_e)} }{[HNO_2]_0 - [H^+]_e + [OH^-]_e}  \tag{5}$$
This equation is cubic in $\ce{[H+]_e}$ when $\ce{[OH−]}$ is expressed as $\frac{K_\mathrm{w}}{\ce{[H+]_e}}$; it becomes as

$$ K_\mathrm{a}=\frac{\ce{[H+]_e} {([H^+]_e – K_\mathrm{w}/{[H^+]_e})} }{[HNO_2]_0 - [H^+]_e + K_\mathrm {w}/{[H^+]_e}}  \tag{6a}$$
Or
$$\
 [\ce{H+}]_e^3
  + K_\mathrm{a} [\ce{H+}]_e^2
  -(K_\mathrm{w} + [\ce{HNO2}]_0K_\mathrm{a})[\ce{H+}]_e
   - K_\mathrm{w} K_\mathrm{a}
  \tag{6b}
$$
 For most practical applications, we can make approximations that eliminate the need to solve a cubic equation.

a- Unless the acid is extremely weak or the solution is very dilute, the concentration of $\ce{OH−}$ can be neglected in comparison to that of $\ce {[H+]_e}$. If we assume that $\ce{[OH−]_e<<[H+]_e}$, then $\text{Eq}\mathrm{5}$ can be simplified to
$$K_\mathrm{a }=\frac{[H^+]_e^2}{[HNO_2]_0 –[H^+]_e}\tag{7}$$
which is a quadratic equation:
$$\ce{[H^+]_e^2 + K_\mathrm{a}[H+]_e – K_\mathrm{a}[HNO_2]_0 = 0}\tag{8}$$
b-If the acid is fairly concentrated (usually more than $\mathrm{10^{−3}}\text{M}$), a further simplification can frequently be achieved by making the assumption that$\ce{ [H^+]_e << [HNO_2]_0}$. This is justified when most of the acid remains in its protonated form$\ce{ [HNO_2]_e}$, so that relatively little $\ce{H+}$ is produced. In this event,$\text{Eq}\mathrm{7}$ reduces to
$$K_\mathrm{a}=\frac{[H^+]_e^2}{[HNO_2]_0}\tag{9}$$
Or
$$\ce{[H^+]}  \approx  \sqrt{K_\mathrm{a}\ce{[HNO_2]_0}}\tag{10}$$
-   Comparing of the$\ce{ [H+]_e }$values  resulted from solution the three
 derived equations when the initial acid concentration equal $\mathrm{0.1}\text{M}$

1-  The completely accurate answer solution using cubic equation $\mathrm{(6b)}$by wolframalpha

$\ce{[H+]_e} = 0.00612772\text{M}$

2- The approximate  answer solution using quadratic equation $\mathrm{(8)}$

$\ce{[H+]_e} = 0.00612771680\text{M}$

3- The further approximate  answer solution using the reduced equation $\mathrm{(10)}$
$\ce{[H+]_e} = 0.00632455532\text{M}$


Substitute for alkali-iodide-azide reagent for BOD test of water samples

There are two types of oxygen demand of water: easily oxidated substances and  oxidated with difficulties. Thus potassium permanganate, $\ce{KMnO4}$ (permanganate index) are used for first group or potassium bichromate $\ce{K2Cr2O7}$ (Chemical oxygen demand)  for both groups. Both $\ce{KMnO4}$ and $\ce{K2Cr2O7}$ are used in acidic environment.

Alternatives are iodates and cerium(IV) compounds.

I recommend using $\ce{KMnO4}$, but it is prohibited as precursor in some countries. also it is not primary standard, thus standardization by titration of solution of primary standard like oxalic acid needed.

$\ce{K2Cr2O7}$ is primary standard, but it is highly toxic, carcinogen. Thus be very careful with that chemical.

You can try iodates for determination but they are not very common at school. Some bleaching agents contain iodates, but I can not guarantee their usability for analysis.


Comparing the basicity of aniline and ortho anisidine

Why you considering the methyl group on $\ce{-OCH3}$ to be "locked" in position, facing away from the $\ce{-NH2}$ group? It is free to rotate in all directions, and it will come at a position next to the $\ce{-NH2}$ group, where the ortho effect will apply.

For a visualization, here is a similar image:

(taken from Martin's excellent post)

PS: I believe that Steric Inhibition of Resonance should not be invoked here, because both sides of the $\ce{-NH2}$ group are not blocked by the $\ce{-OCH3}$ group, and $\ce{-OCH3}$ is certainly not as bulky a group as t-butyl.


How is it determined how many electrons are transferred in redox reactions?

In short, the species you mention ($\ce{Sn^{5+}}$ and $\ce{SO_4^{5-}}$) do not exist in any chemical system at anywhere near ordinary conditions. The number of electrons shuffled in the reaction is not chosen arbitrarily, but is based on the initial and final oxidation numbers of the elements in the reaction, after the equations are balanced. The only stable compound with formula $\ce{SnSO_4}$ is made of $\ce{Sn^{2+}}$ and $\ce{SO_4^{2-}}$ ions. Hence in your reaction, $\ce{Sn}$ atoms start with oxidation number +2 and end with 0 (therefore gaining two electrons) and $\ce{Fe}$ atoms start with oxidation number 0 and end with +2 (therefore losing two electrons).

As another example of electron counting, here's the reaction between $\ce{Ba}$ and $\ce{Al^3+}$:

Oxidation: $\ce{Ba^0 -> Ba^{2+} + 2e^{-}}$

Reduction: $\ce{Al^{3+} + 3e^{-} -> Al^0}$

To balance the equations, the oxidation half-reaction must be multiplied by 3 and the reduction half-reaction must be multiplied by two. Therefore:

Oxidation: $\ce{3Ba^0 -> 3Ba^{2+} + 6e^{-}}$

Reduction: $\ce{2Al^{3+} + 6e^{-} -> 2Al^0}$

Global: $\ce{3Ba^0 + 2Al^{3+} -> 3Ba^{2+} + 2Al^0}$

In the global reaction, six electrons are involved.


How are half lifes calculated for radioactive substances that decay over many years

Suppose you have a sample of $\ce{N}$ atoms of a radioactive material. As it is radioactive, you may state that a small fraction k is decomposed by second. This fraction does not depend on the total number of atoms. If this sample is placed in a Geiger counter, each decomposition of its atoms produces a ray that crosses the inner volume of the Geiger counter, and it will be detected and counted. In a given time, the number of rays is equal to the number of disintegrations. If dN/dt is the number of atoms disintegrated per second (or per minute), the factor k can be calculated by dividing the number of rays detected every second in the counter, by the total number of atoms N of the radioactive material. k = $\ce{\frac {dN}{dt}·\frac{1}{N}}$
Once k is known, the half-life T can be derived from the integrated law, if $\ce{N_o}$ is the original number to atoms at t = $\ce{0}$  $$\ce{N/N_o = e^{-kt}}$$ The half-life is the time T after which $\ce{N = N_o/2}$, which gives $$\ce{1/2 = e^{-kT}}$$ with the following result : $$\ce{T = \frac{ln 2}{k}} = 0.693/k $$
So the period may be obtained from the knowledge of k. It is not necessary to wait up to the radioactive material has decreased to one half.
For example, if $0.0238$ g, or $\ce{6.02·10^{19}}$ atoms U-$238$ is inserted in a Geiger counter, $292$ rays will be detected per second. From this measurements, the constant k can be calculated : $$k\ce{ = \frac{292 s^{-1}}{6.02·10^{19}} = 4.87·10^{-18} s^{-1}}$$ and the half-life is $$\ce{T = \frac{0.693}{4.87·10^{-18}} = 1.423·10^{17} s = 4.51· 10^{9} years}$$


Is it possible, chemically, to change a dextrorotatory compound to its levorotatory counterpart?

Yes,this process is called stereoconversion. Both SN1 and SN2 reactions can invert the chirality of a carbon center. In the case of the SN1 mechanism, departure of a leaving group generates a prochiral intermediate which is then open to nucleophilic attack on either face. If both faces are equally acessible to the nucleophile this will result in racemisation of the molecule. In the case of the SN2 reaction the nucleophile will enter a trigonal bipyramidal intermediate with the molecule. Assuming the nucleophile has the same stereochemical precedence as the leaving group, the stereochemistry will be inverted.


Measure the concentration of lactose in dairy products

If you are looking for economical methods then you have the option of titration and spectrophotometry. See this paper by Folin THE DETERMINATION OF LACTOSE J. Biol. Chem. 1918, 33, 521-524. The titration is based on the reducing ability of lactose.

Another reliable titrimetric method can be seen in this paper by Hites, Determination of Lactose in Milk Products J. Anal. Chem. 1949, 21(8), 993-995.


Producing carbon nanotubes from graphite

Graphite does not contain CNTs, a simple extraction therefore is not possible.
Graphite seemingly can be converted to CNTs by a "simple" oxidative method published by DW Lee and JW Seo on arxiv.org in 2010.
The authors claim to have used commercially available graphite powder (45 µm, Aldrich) of high purity (>99.995%), but did not provide any information on the yields.
Pencils typically contain a mixture of graphite powder and various amounts of clay as a binder.


Why does dipole must lie parallel to a rotation symmetry axis?

No, it would not

A dipole means that there is a difference (in charge) from one "side" to another "side".

If we have determined that something is symmetric, it means the "sides" are equal to each other — and that as you rotate $\left(\frac{1}{n}\right)$ around the symmetry axis, everything stays the same. If there is anything that shifts out of position, that symmetry is not there (it is broken).

Now, your image is a bit hard for me to read, is it supposed to be linear or is it on purpose not? I am going to assume it is not linear, but that the light grey are symmetrically placed… So in your image, there is only a $C_2$ rotational axis, in the approximate "up/down" direction. By placing a fluorine in one end  (and something else at the other end) you would destroy the $C_2$ symmetry. If you rotate it $180$ degrees around (if you could see it through the backside of your monitor…) the fluorine would be on the other side, clearly not identical to $E$, the original. No symmetry.

If it was linear, the light grey and dark gray arranged as a line, then it would still retain a $C_\infty$ rotational axis along the dark grey atoms. Rotating around this in any fraction or direction would still keep everything looking the same.


Heating a hydrocarbon with $\ce{CuO}$ generally evolves $\ce{CO2}$ and $\ce{H2O}$. Urea is no exception for this rule.
The $\ce{CO2}$ evolved will turn lime water milky(this is a test for $\ce{CO2}$ gas).


This is not a database, and I don't know about how small molecules you are looking for, I found that article on Jacs Kinetics and thermodynamics of keto-enol tautomerism of simple carbonyl compounds: an approach based on a kinetic study of halogenation at low halogen concentrations

If you can access the file it will be better for you, if not I let you some matrix of values. I hope it can help ! I give you them in the same the order of the publication.

I added some other publication at the end.

Links

Tautomerization of 2-Acetylcyclohexanone. 1. Characterization of Keto−Enol/Enolate Equilibria and Reaction Rates in Water

Kinetic studies of keto-enol and other tautomeric
equilibria by flash photolysis (a lot of constants in this one)


Hair styling gels and 'liquid gels' - effect of formulation / composition on viscosity and performance

Question 1: Carbomer
Question 2: Not sure about the question, but anyway. Sure you can create a spray (theoretically), but the process is really hard (please, do not make it at your house) because you need a propellant. By the way, such a high concentration of Parfum.
Question 3:
Part1:
According to your observations, in my own opinion, the concentration of these components maybe are too high: Octylacrylamide/Acrylates/Butylaminoethyl Methacrylate Copolymer, Niacinamide, Hydroxyisohexyl 3-Cyclohexene Carboxaldehyde.
Part2:
Maybe because of the oxidation of hexyl cinnamaldehyde to Cinnamic acid you can feel a very unpleasant smell. But it is hard to tell because everything you describe is not in a detail (for example, the scent, what reminds and etc..)
Part3:
VP/VA Copolymer, PVP makes your hair shiny and not sticky.
Panthenol makes your hair not dry.
JUST A REMAINDER:
It does not mean, that those components which I have mentioned are very good or very toxic. It all depends on concentration and experimental data. We use poisons' as drugs, but lethal and effective concentrations differ.
I answered what could possibly be in my own subjective, but a trained opinion based on your poor objective/experimental data. Try to say more about what you smell, what you feel, and there many databases of components like (found some reliable databases for you):
Official Cosmetics ingredients review database: https://www.cir-safety.org/ingredients
Cosmetics and their function (easy to use database): https://www.cir-safety.org/ingredients https://www.ewg.org/skindeep/
Chemical compounds database: https://pubchem.ncbi.nlm.nih.gov/


Distinguishing pentanol isomers in 13C DEPT NMR spectroscopy

In an ideal non-moving world at $0\,\mathrm{K}$, every atom would give exactly one peak in NMR spectra. So the actual question is why some of these spectra have less signals. Partly this is due to local symmetry. If, e.g., two methyl groups are attached to the same carbon atom, these two are said to be chemically equivalent, i.e. one can rotate bonds to overlap one over the other. A different example of chemical equivalence is pentan-3-ol which you can rotate around the $\ce{C-OH}$ axis to give an almost identical molecule, putting the $\ce{CH2}$ groups on top of each other and the methyl groups, too.

With that, you did the circling right, and we should expect:

3 spectra with five signals;
3 spectra with four signals; and
2 spectra with three signals.

However we see one spectrum with two signals and are lacking one with four signals. (And technically, if you look at spectrum 2 closely, that could be interpreted as having five signals, not four.)

In a standard $\ce{^{13}C}$-DEPT spectrum (i.e. the ones you showed), carbons from $\ce{CH}$ and $\ce{CH3}$ groups point upward and those from $\ce{CH2}$ groups downward — and there is your answer. There are carbons in the structures that do not have hydrogens bound to them and which thus do not show up in the DEPT spectra.

In case you’re interested, I will give you the appropriate IUPAC names of the pentanol isomers that were measured in the following spoiler tag.

   2-methylbutan-2-ol. Carbon 2 cannot be seen.

   3-methylbutan-2-ol. The two methyl groups at the end are not completely identical and thus give two signals.

   pentan-3-ol

   pentan-1-ol

   2,2-dimethylpropan-1-ol. Carbon 2 cannot be seen.

   2-methylbutan-1-ol

   pentan-2-ol

   3-methylbutan-1-ol


Collisonally induced intersystem crossing

There are two factors that I recall that can enhance intersystem crossing (ISC, the conversion of an excited singlet state to an excited triplet state, or vice-versa; $\ce{S_1 <=> T_1}$).  One is the presence of a "heavy atom" such as bromine or iodine in the colliding molecule (or the excited molecule itself).  The other is the presence of a paramagnetic species.

Both types of molecules enhance ISC by relaxing the quantum rules which forbid  singlet-triplet conversions. When a heavy atom or paramagnetic molecule collides with a molecule in its excited state it can facilitate the electron spin flip (e.g. converting paired electrons to unpaired electrons, or vice versa) through spin-orbit coupling.  It is this spin-orbit coupling that relaxes the rules forbidding direct conversion between $\ce{S1}$ and $\ce{T1}$ states.  In essence spin-orbit coupling leads to a mixing of the singlet and triplet wavefunctions, thereby allowing singlet-triplet conversion.


How to distinguish diastereomers of unsaturated ketones by NMR?

According to Noack and Jones, unsaturated ketones with an α substituent — even if it is only methyl — generally adopt an s-trans conformation due to steric reasons. Only if the α substituent is hydrogen can an equilibrium between s-cis and s-trans be observed.[1] Therefore, we should better draw the structure this way:
$\hspace{30ex}$
$\hspace{28ex}$s-trans configuration of α-methylenones
I already explicitly added the unsaturated hydrogen atom to the image. As we can see, it is either rather close to the keto group or pointing away from it. If the alkene’s configuration is (E) and the hydrogen is thus pointing ‘inward’ it experiences a larger deshielding due to the close carbonyl group. Therefore, its signal in 1H-NMR is shifted to lower field when compared to the (Z)-configured double bond.
You can also compare experimental shifts. Unfortunately I didn’t find a pair of identical molecules differing only in the double bond orientation, and not even the solvent is the same, but I think the data still shows a clear trend, namely that for (E) enones the chemical shift is in the high alkene range while for the (Z) enones it is in the low range:
$\hspace{33ex}$
(E)-3-methylhex-3-en-2-one. 1H-NMR ($500~\mathrm{MHz}$, $\ce{CDCl3}$) $\delta =$ $\mathbf{6.59}$ (tq, $J = 7.2,1.3~\mathrm{Hz}$, 1 H), $2.28$ (s, 3 H), $2.24$ (app. qnt, $J = 7.4~\mathrm{Hz}$, 2 H), $1.73$ (s, 3 H), $1.05$ (t, $J = 7.6~\mathrm{Hz}$, 3 H).[2]
$\hspace{33ex}$
(Z)-3-methylnona-3,8-dien-2-one. 1H-NMR ($300~\mathrm{MHz}$, $\ce{C6D6}$) $\delta =$ $5.71$ (ddt, $J = 17.1, 10.5, 6.6~\mathrm{Hz}$, 1H), $\mathbf{5.38}$ (tt, $J = 7.5,  1.5~\mathrm{Hz}$, 1 H), $4.92$–$5.02$ (m, 2 H), $2.29$ (q, $J = 7.5~\mathrm{Hz}$, 2 H), $1.92$ (q, $J = 7.2~\mathrm{Hz}$, 2 H), $1.85$ (s, 3 H), $1.60$ (t, $J = 2.7~\mathrm{Hz}$, 3 H), $1.35$ (qnt, $J = 7.5~\mathrm{Hz}$, 2 H).[3]

References:
[1]: K. Noack, R. N. Jones, Can. J. Chem. 1961, 39, 2225. DOI: 10.1139/v61-294.
[2]: W. F. Austin, Y. Zhang, R. L. Danheiser, Org. Lett. 2005, 7, 3905. DOI: 10.1021/ol051307b.
[3]: L. Jiao, C. Yuan, Z.-X. Yu, J. Am. Chem. Soc. 2008, 130, 4421. DOI: 10.1021/ja7100449.


Helicity in Molecular motors and topological isomers

Any pretty non-old organic stereochemistry book should work.

M/P notation

You present some organic compounds, so let's see definition from the Nomenclature of Organic Chemistry: IUPAC Recommendations and Preferred Names 2013:

  P-92.1.1.2 The helicity rule: stereodescriptors ‘M’ and ‘P’

  Helicity is the chiral sense of a helical, propeller, or screw-shaped molecular entity (see ref. 37). The ‘helicity’ rule is expressed by Prelog and Helmchen (see subsection 5.1, ref. 36) as: “Depending on Whether the identified helix is left- or right-handed it is designated ‘minus’ and marked ‘M’ or ‘plus’ and marked ‘P’ ”.

  The application of this system to the description of conformations and configurations considers the torsion angle between two specified (fiducial) groups are attached to the atoms linked by that bond. The sign of the smaller torsion angle between the fiducial groups defines the sense of chirality sense of the helix (see torsion angle, P-94).

  P-92.1.1.2.1 The chirality of hexahelicenes is denoted by the stereodescriptors ‘P’ and ‘M’.

  (P)-hexahelicene (PIN)                  (M)-hexahelicene (PIN)

  (…)

The description follows further, the rule can be applied also for biphenyls, allenes, or similar structures, I'd say where along the axis some two approx. perpendicular groups are in clockwise or anticlockwise rotated positions.

The arrows show the helicity orientation in the direction from the observer; but may be confusing(opposite), when one applies common hand and thumb methods for determining a screw thread handedness, but results in the same correct helicity orientation/descriptor.

Δ/Λ notation

In the question you're mentioning molecular motors. These can be also propeller like structures. If they were from inorganic chemistry world, they could use different notation Δ/Λ; from IUPAC Red Book 2005:

  IR-9.3.4.11 The skew-lines convention

  Tris(bidentate) complexes constitute a general family for which a useful, unambiguous convention has been developed based on the orientation of skew lines which define a helix.
  Examples 1 and 2 represent the delta (Δ) and lambda (Λ) forms of a complex such as $\ce{[Co(NH2CH2CH2NH2)3]^{3+}}$.   (…)

            2.
  delta (Δ)                      lambda (Λ)

This one is the Δ isomer:

Δ/Λ likely corresponds to organic's P/M respectively (however, in some structures, there might be doubts which parts of the structure should be interpreted as the propeller blades.. see e.g. Q. Molecular knots absolute configuration)

The structures in your ask, 3,3′-dimethyl-2,2′,3,3′-tetrahydro-1H,1′H-4,4′-biphenanthrylidene, do have a combination of a geometrical isomerism at the connector double bond, and two chirality axes (perpendicular to the double bond connector).


What software was used to create these organic reaction scheme figures?

$\mathrm\LaTeX$ to the rescue
I reproduced these reaction schemes with $\mathrm\LaTeX$ using the following particular packages:

chemfig for drawing 2D chemical structures (note that I enabled sans-serif font for the labels like the original did);
chemmacros for the typesetting of chemical formulas and chemistry-related notations;
siunitx for handling physical quantities (here, temperatures);
stix for loading STIX fonts, which in my opinion are the best choice for technical writing due to canonical and coherent appearance of the glyphs and great coverage for special symbols.

All of this is free for any type of use, works on all major operating systems used in production and the source code is available.
Result

Source code
% !TEX program = pdflatex

\documentclass{report}
\usepackage[T1]{fontenc}
\usepackage{stix}
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{mathtools}
\usepackage{siunitx}
\usepackage{chemmacros}
\chemsetup{
    modules = all,
}

\newcommand{\bndlen}{1.7em}
\newcommand{\bndlenshort}{0.5}
\newcommand{\bndlenshorter}{0.75}
\newcommand{\bndlenlonger}{1.5}
\newcommand{\bndlenlong}{1.75}
\newcommand{\lnwidth}{0.8pt}
\newcommand{\cramwid}{0.3em}

\usepackage{chemfig}
\setchemfig{
    angle increment = 15,
    atom sep = \bndlen,
    bond offset = 1pt,
    double bond sep = 3pt,
    cram width = \cramwid,
    compound sep = 5.0em,
    arrow coeff = 1,
    arrow offset = 12pt,
    scheme debug = false,
    bond join = true,
    chemfig style={line width=\lnwidth},
    arrow style={line width=\lnwidth},
}
\renewcommand*\printatom[1]{\ensuremath{\mathsf{#1}}}

\begin{document}

\schemestart
    \chemfig{MeO-[::30]*6(-=-(-~--[::-60]CHO)=-=)}
    \arrow{->[\parbox{5cm}{\centering
        i) \ch{Hg^2+}, dil.~\ch{H2SO4}\\
        ii) \ch{AgNO3}, \ch{NH4OH}
    }][
        iii) \ch{Zn}-\ch{Hg}, conc.~\ch{HCl}]}[,2.5]
    \textbf{Q}
    \arrow{->[\parbox{5cm}{\centering
        i)~\ch{SOCl2}\\
        \hphantom{i)~}pyridine
    }][
        ii) \ch{AlCl3}]}[,1.5]
    \textbf{R}
    \arrow{->[\parbox{5cm}{\centering
        \ch{Zn}-\ch{Hg}\\
        conc.~\ch{HCl}
    }]}[,1.5]
    \textbf{S}
\schemestop

\bigskip

\schemestart
\chemfig{(-[::120]*5(---=-))-~-(-[::-60]*6(---=--))}
\arrow{->[
        i) \ch{H2}, \ch{Pd}-\ch{BaSO4}, quinoline
    ][
    ii) dil.~\ch{KMnO4}~(excess), \SI{273}{\kelvin}]}[,3.5]
\textbf{P}
\schemestop

\bigskip

\schemestart
    \chemfig{H_3C-[::-30](-[::-60]Br)-[::60]-[::-60]Br}
    \arrow{->[\parbox{5cm}{\centering
        i) alc.~\ch{KOH}\\
        ii) \ch{NaNH2}
    }][
        iii) red hot iron tube, \SI{873}{\kelvin}]}[,3]
\schemestop

\end{document}


Which one, Mulliken charge distribution and NBO, is more reliable?

As I do not want to drag this out in the comments, I decided to give a brief answer. Let me first cite the very helpful comments to keep the for (relative) eternity and add my thoughts at the end.

There is no proper, strict and agreed definition of atom border, that's the real problem. Gaussian includes option to fit charges on atoms (assumed for the sake of the procedure as points) so they produced electrostatic field as accurately fitting once calculated from the electronic density distribution as possible, and maybe (I haven't used Gaussian for a long time, I'm not sure) dipoles too. –  permeakra

NBO is more accurate than Mulliken in very many ways. However, NBO is not always accurate. Generally speaking, you would never want to use Mulliken charges for any sort of production level work unless its been calibrated. –  LordStryker

Mulliken is of course the cheapest and fastest way to compute charges. However, this method tends to give qualitative results at best. The reason for this is obvious. It will divide the canonical orbitals equally amongst the participating atoms. There is no polarisation whatsoever. Another problem is, that they are very basis set dependent. And that being said, the worst part is, that the description becomes worse by increasing the basis set.
NBO charges are much more reliable, since the operate on the electron density instead. Localised natural atomic orbitals can be used to describe the computed electron density. Polarisation of bonds is therefore considered. This method obviously only works very well for systems, that can actually be (quite well) described by a Lewis structure. However, they usually give a nice qualitative picture and are quite robust when it comes to increasing the basis set.
There are a couple of more population analysis tools, that can be used and should be considered when it is necessary to derive some conclusions from charges.
Quite common is also the Hirshfeld charges approach, which divides the electron density into spherical basins around the nucleus. Maybe interesting to read is "Are the Hirshfeld and Mulliken population analysis schemes consistent with chemical intuition?" (I can't tell, I do not have access.)
There are also several schemes to obtain partial charges from the electrostatic potential.
Probably the most rigorous approach to atomic partial charges is the Quantum Theory of Atoms in Molecules  (QTAIM). Here the Laplacian of the electron density is used to describe the partitioning in electronic basins. The best feature of this analysis is, that it can be also used for experimentally obtained electron densities. However, as everything, this has also major drawbacks. One of the most important is the computational effort. Unfortunately the integration scheme is not always completely robust when it comes to some particular molecules. However, it is almost completely basis set independent and results could be cross referenced with experimental data.
Since permeakra's initial statement holds true, especially for the agreed part, the best way to deal with partial charges is to be careful and maybe check with different methods. In gaussian there are several schemes implemented, most commonly, Mulliken, Hirshfeld and NBO (3.1) and some others, see the population keyword for more detail.


The energy on the y-axis is the total energy of the system. It uses a reference of E = 0 for when 2 H atoms are infinitely far apart.

As to why the plot looks the way it does...

When the H atoms are infinitely far apart, they are not interacting at all. When brought together, you get a covalent bond which forms. This can be treated at various levels - In the simple classical picture, the 2 electrons are "shared" between the two atoms, leading to lots of electron density in between the atoms, which the 2 protons are then attracted to, and this is what holds the 2 H atoms together. Now we have a slight problem - why do they stop? You could say due to the nuclear repulsions, but if you have a simple situation, using  $\ce {H2+}$:

$\ce{H+}$---e---$\ce{H+}$

With the electron just sat halfway between 2 nuclei. Now let's analyse this situation with just classical mechanics, all units set = 1 for convenience. q = Charge on electron and nucleus. d = Distance from nuclei to nuclei:

Force on each $\ce{H+}: $
$\ F = q^2/d^2 - q^2/(d/2)^2 =  -3q^2/d^2  $

We have $\ce{q^2/d^2}$ term for the repulsion between the 2 nuclei and the negative $\ce{-q^2/(d/2)^2}$ term is due to attraction between nucleus and electron. In this situation the covalent bond results in an attraction of both nuclei together for all values of d - IE eventually we get a singularity. This is why the covalent bond can only really be understood in a quantum mechanical framework.

Taking again the situation of $\ce{H2}$ but this time in a quantum mechanical way.

We start off with 2 hydrogen atoms infinitely far apart. They each have an electron in H 1s, with OPPOSITE spins. When they are brought closer together the wavefunctions change. Now what happens is this. The electron in H 1s of one atom is able to tunnel into the potential well of the other hydrogen atom. By doing this the total volume in which the electron resides is larger - Obviously the electron can go anywhere in the universe, as per usual QM, but the H atom is localised less. This leads to a decrease in the electron's kinetic energy - It is spread out more, and so uncertainty in position is larger, so uncertainty in momentum can be less, and so leads to a lower KE. The potential energy which the electron experiences won't change a huge amount - Yes there are 2 potential wells but it doesn't get closer to them. This is the story at first glance.

However there is the virial theorem which states:
$2KE+PE+R(dE/dR)=0$

So the decrease in KE by bringing 2 H atoms closer together must be balanced out somehow - Either an increase in KE by another means, or an increase in potential energy (PE). If we shrink the 1s orbitals contributing to the bond, this in turns increases KE and decreases PE - PE is negative, and so becomes more so. Thus we find the orbitals shrink in bonding. This all leads to a picture like this:

As 2H atoms draw closer together, electron on one can tunnel onto the other. This leads to a decrease in KE, and as at the equilibrium bond distance $2KE+PE = 0$ this decrease in KE must be balanced out by a contraction in the H 1s orbitals contributing to the bond, leading to both a larger KE term and a more negative PE term. In other words a covalent bond forms because it reduces the kinetic energy of the contributing electrons and increases the potential they feel, due to the contraction of orbitals via the Virial Theorem pathway.

Now back to your plot. At far distances, E=0, because no interaction, and this is taken as 0 point - IE referenced relative to infinite separation. As 2 H atoms brought together, get interaction described as above. At around 2 bond lengths, the KE term reaches a minimum. It then rises - the 2H atoms are so close now that more overlap - IE more tunneling of electron from well to the other well - has no more effect. Instead the wavefunction of the electron is found to contract. This raises the KE term - electron now confined to a smaller space = Larger KE term - but the PE term becomes more negative by a larger amount. At the equilibrium bond length, the virial equation applies as written above:

$2KE+PE=0$

now applies. Thus the KE term is positive with value x, and the PE term is negative, with value 2x. Thus if
$E=PE+KE$
we have a negative binding energy, as expected.

Interestingly the involvement of KE in bonding leads to a new class of bonds, called charge shift bonds - the prime example of which is difluorine, $\ce{F2}$. This molecule has an overall bond order of 1 according to MO theory, but has a large antibonding effect due to filling of antibonding pi orbitals. Calculations using VB theory indicate that the covalent contribution to the structure never actually has a minimum - IE antibonding at all distances. In this case adding in ionic structures is entirely responsible for the bond observed.

This stackexchange answer explains it well, if you're interested, and there's a number of interesting papers listed there as well.

What is charge shift bonding?


Effect of anisotropy on electrical conductivity of crystalline solids

There are indeed compounds with very different conductivity in different directions
Depending on the structure of the crystal, it is possible that a compound has very different properties in different directions. Some crystal structures cannot show this as they are essentially the same in all spatial directions, but others can. Metals don't usually have anisotropic conductivity due to the nature of the conduction bands but other conductors can show it.
The archetypal compound that shows this property is graphite. Graphite consists of multiple flat layers of covalently bonded C-6 hexagons in an extended plane stacked on top of each other by much weaker forces. The delocalised electrons in the plane can conduct electricity well but there is little scope for electrons to move freely perpendicular to the plane. So the conductivity is about 1,000 times higher along the plane than it is perpendicular to it (see Electrical conductivity of graphite).
So it is certainly possible that electrical conductivity can be different in different crystal directions. But this requires particular structural features. Other forms of anisotropy are far more common (like birefringence).


Equilibrium constant in terms of mole fractions

The initial mole fraction of $\ce{CO}$ present must be $y=0.5$.  Then the initial mole fraction of $\ce{H2O}$ present is also $y=0.5$, and the initial moles of both $\ce{H2}$ and $\ce{CO2}$ are 0.

After equilibration, both $\ce{H2}$ and $\ce{CO2}$ are present in the amount $x$ and so by stoichiometry, $\ce{H2O}$ and $\ce{CO}$ must be present in the amount $y-x=0.5-x=\frac{1}{2}-x$.

Since the reaction doesn't change the number of moles of gas, it doesn't matter much if we use $K_p$ or $K_c$, so $K_p=\frac{x^2}{(\frac{1}{2}-x)^2}=\frac{x^2}{\frac{1}{4}-x+x^2}=~...$

I'm sure you can figure out the rest.


Which is more stable in gas phase, free radicals or ions?

It is difficult to comment on such a broad statement because often it depends on the specifics. So instead, I will talk about a single example: Will the HCl molecule dissociate into radicals (atoms, in this case) or ions?

$$\tag{1}\ce{H-Cl -> H+ + Cl-}$$

$$\tag{2}\ce{H-Cl -> H. + Cl.}$$

Under many circumstances, neither (1) nor (2) happens, H-Cl is the major species.

In water, where the dipole of water solvates chloride ions and the hydrogen ion "reacts" with water to form hydronium ions and large structures, HCl ionizes completely (strong acid). In other solvents, it ionizes less but would also act as acid if there is an appropriate base.

$$\ce{HCl(aq) + H2O(l) -> H3O+(aq) + Cl-(aq)}$$

In the gas phase, ions of opposite charge have a strong long-range attraction, so reaction (1) is unlikely. Reaction (2) is endothermic (bond dissociation energy is defined as homolytic cleavage into radicals, and it has a positive value). If you raise the temperature, you favor the products in reaction (2), so at higher temperatures, the concentration of radicals would increase.

To directly compare ions and radicals in the gas phase, you would consider the following reaction:

$$\tag{3}\ce{H. (g) + Cl. (g) <=> H+(g) + Cl-(g)}$$

Taking away an electron from hydrogen corresponds to the ionization energy of hydrogen atoms (1313. kJ/mol, i.e. it takes energy to remove the electron from the electrostatic field of the nucleus). Adding an electron to the chlorine atom corresponds to the electron affinity of chlorine atoms (-359 kJ/mol, i.e. energy is released when an additional electron encounters a neutral chlorine). Overall, the process is endothermic. Different from reactions (1) or (2), however, the number of reactant particles is equal to the number of product particles, so in this case, it is not clear if the change in entropy is positive or negative, but it won't dominate the reaction at modest temperature. So at room temperature, this reaction will not happen, the radicals are favored over the ions.

  What is stabler in gas phase, free radicals or ions?

For the example I chose, free radicals are more stable than ions.

  Is there a different trend in liquid or solid phase?

For the example I chose, it is the opposite in aqueous solution.


Why do liquids vaporise spontaneously at temperatures below boiling point despite ∆G being positive?

Start with a closed container with a quantity of liquid water, and no vapor. Maintain the container at room temperature. What will happen? If the temperature does not change, the overall Gibbs free energy will not change in equilibrium. Now, $G = H - TS$, and at constant temperature $\Delta G = \Delta H - T\Delta S$.

Converting liquid (take as the reference state) to vapor costs enthalpy, the $\Delta H$. But, by allowing vapor to occupy the empty space in the container you have gained a great deal of entropy, $\Delta S$. The first water molecule to go in to the vapor gains a huge amount of entropy - it has the entire volume to roam around in - overwhelming any enthalpy required. The reaction of liquid -> vapor is, in fact, highly spontaneous because of this! But, the next little bit of vapor gets a smaller entropy gain - there is already vapor occupying the head space in the closed vessel. And the next little bit gets even less entropy. Yet, it keeps costing the same amount of enthalpy to convert the liquid to vapor.

As more and more liquid goes in to the vapor phase, the additional entropy gain decreases, while the enthalpy cost remains the same. Eventually the reaction reaches an equilibrium point when $\Delta G = 0 = \Delta H - T \Delta S$ for the overall reaction. The $\Delta H$ is the total enthalpy required to make the equilibrium vapor concentration, and the $\Delta S$ is the total entropy gained by doing so. You have converted $x$ moles of liquid in to vapor, and it is now in equilibrium with the remaining $y$ moles of liquid.

Perhaps that gives you a better perspective.


Could neutralisation of safe chemicals produce dangerous ones?

While not an acid-base reaction, an example of mixing two relatively "safe" chemicals to form dangerous ones is bleach (sodium hypochlorite) plus ammonia - both common household chemicals. This reaction yields chloramine, dichloramine, and nitrogen trichloride, all more toxic and hazardous than bleach or ammonia.


Is ammonium hydroxide safe to be used for sanitizing food?

In pure form, aqueous solutions of ammonia ($\ce{NH3}$) cannot be consumed. Solutions containing 5 % ammonia or more are considered “corrosive” (causing severe skin burns and eye damage).

However, in reasonable quantities or in dilute form, aqueous solutions of ammonia can be generally used safely in food. They are sometimes named “ammonium hydroxide”, “ammonia aqueous”, or “aqua ammonia”. The food additive code (the E number, which is commonly found on food labels) of aqueous solutions of ammonia is E 527.

This food additive is approved in the European Union as well as in the United States.

For example, aqueous solutions of ammonia are used in food processing as pH control agent (acidity regulator), leavening agent, surface-finishing agent, or additive to boiling water.


Can a benzyl ether be removed in the presence of an ester?

In Wutz's book on protecting groups,[1] I found an example of almost instantaneous reductive debenzylation which leaves an ester untouched:[2]

The use of BCl3 is also possible:[3]

References:

Wuts, P. G. M. Greene's Protective Groups in Organic Synthesis, 5th ed.; Wiley: Hoboken, NJ, 2014, p 131.
Sulikowski, M. M.; Davies, G. E. R. E.; Smith, A. B., III. Avermectin–milbemycin synthetic studies. Part 7. An approach to the southern hemisphere of milbemycin α1. J. Chem. Soc., Perkin Trans. 1 1992, 979–989. DOI: 10.1039/P19920000979.
Williams, D. R.; Brown, D. L.; Benbow, J. W. Studies of Stemona alkaloids. Total synthesis of (+)-croomine. J. Am. Chem. Soc. 1989, 111 (5), 1923–1925. DOI: 10.1021/ja00187a081.


Suggestions for simulating transport through membranes with Molecular Dynamics (MD) and Monte Carlo (MC)?

I always find AMBER to be the most streamlined for conducting MD, however it (as well as CHARMM) are proprietary.  Note that both are Langevin dynamics based, whereas LAMMPS is brownian dynamics.

That being said, an MD simulation is only as good as the parameters for the force field; the engine is of no matter (unless we're discussing speed).  Basically, I would use a QM program (Gaussian is my goto, but nwchem or psi4 are excellent open source options) to optimize the geometry of the membrane.  Because this is rather heavy, you'd most likely be best to go with a semi-empircal method as ab-initio could take weeks (depending on your computational power).  Once optimized, use the QM software to derive force constants as well as ESP charges (partial charges) for each atom.  Take this data, and parameterize a force field with it (I'm partial to using the Amber FF's...google GAFF or FF99SB... Theres a paper out there from Cornell, Merz, et. al. that should give you insight).

I hope this helps give you an idea of what you're looking at for a respectable MD simulation.


How to prepare iron(II) carbonate from iron(III) chloride?


Why can't we use 30 mL of water and 70 mL of alcohol to make a 70% (v/v) solution of Isopropyl Alcohol?

Assuming you are talking about either isopropyl alcohol or ethyl alcohol (ethanol) and not both. They are different beasts, they are both used as disinfectants and solvents, but isopropyl is way more toxic to humans (who more or less regularily drink etanol solutions).
70ml ethanol and 30ml water, whem mixed, do not make 100ml mixture, but somewhat less.
That's what "non additive" volume means.
I don't have the numbers at hand, but the "contraction" is pretty much measurable and in particular important in the mass production of desinfectants and drinks.


Finding concentration and moles given final and initial pH

Both the cases are similar, so I'll show you the calculation for the fist case only:

The $[\ce{H+}]$ ion concentration changes from $\pu{10^-2~M}$ to $\pu{10^-11~M}$. Assuming water at $\pu{25^\circ C}$, this implies the change in $[\ce{OH-}]$ ion  concentration from $\pu{10^-12~M}$ to $\pu{10^-3~M}$. Since the volume of the solution is fixed at $\pu{500L}$, the amount of $\ce{OH-}$ ions has changed from $500\times\pu{10^-12 moles}$ to $500\times\pu{10^-3 moles}$. This change is exactly the number of moles of $\ce{NaOH}$ that need to be added to this solution, since $\ce{NaOH}$ is a strong base and dissociates with $100\%$ yield.


Which reaction between Lugol and Hydrogen peroxide?

I'll try to figure out the color of every compound separately, but there is a strong interaction between them and the different parameters so this could be quite tricky:

$\ce{I+ , IO3-}$ don't absorb in the visible region
$\ce{I3-}$  I think should be red
$\ce{I2}$ is blue if solid but has different colors in solution depending from the type of solvent and $\ce{H+}$ concentration see Hildebrand (1909). In polar solvent is brown, however I don't exclude this is due to the interaction with $\ce{I3-}$.

Because there is an equilibrium:

$$ \ce{I_2 + I^- <=> I_3^-}$$

So there is always a little bit of iodine in the solution and this could cause the brown color (however keep in mind that is $I$ is not so soluble in water).
Adding $H_2O_2$ should oxidize the iodide in this way:
$$ \ce{2I^- + 2H^+ + H_2O_2 -> I_2 + 2H_2O}$$
It occurs the reverse reaction too but should be negligible. In this case is probably a cinetic matter,  you can make the solution transparent reducing the colored compounds. If you have a look to the iodine clock reactions iodine can be reduced by bisulfite:
$$ \ce{I_2 + HSO_3^- + H2O -> 2I^- + HSO4^- +2H^+}$$

But I exclude that a reduction could occur in an oxidizing ambient like your. So my hypothesis are two:

FIRST: You are oxidizing the colored species. Adding $\ce{H2SO4}$ and $\ce{H2O2}$ you could obtain one of the most oxidant agent, Peroxymonosulfuric acid ($\ce{H2SO5}$),  see Piranha solution (I think this could be very dangerous please take all the precautions!). But now I don't think that the equation that if I wrote before (see J. Am. Chem. Soc., 1931, 53 (1), pp 38–44 and
J. Am. Chem. Soc., 1978, 100 (1), pp 87–91):

$$ \ce{I_2 + 5H2O2 -> 2H^+ + 2IO3^- + 4H2O}$$

could occur. I know iodic acid $\ce{HIO3}$can be produced oxidating $\ce{I2}$ but I never found a reaction involving $\ce{H2SO4}$ and hydrogen peroxide. It could be a more complex reaction maybe involving the others ions. I was thinking about potassium hydrogen iodate ($\ce{KH(IO3)2}$). I try to figure out a stechiometric reaction but is quite an hell!

SECOND: simply you are  modifying the interaction between the solvent and $\ce{I2}$ and $\ce{I3-}$ avoiding the  light absorption to take place.

I'm sorry this is now quite not an answer, but I hope it helps!

Iodine color reaction are wonderful (but very complex)!


You can use any triglyceride to make soap.  The process is called saponification.  In this process fatty acid molecules are cleaved from glycerol (glycerine).  In order for the reaction to proceed it is heated.  Regardless of the fatty acid source, the soap will probably not taste very good.


How would you add saccharide residue information to PDB files?

Short answer: your workflow doesn't include residue perception, so you would need to do this manually.

I don't use CORINA, but I suspect the answer is the same as if you generated coordinates using Open Babel.

You're using ChemDraw, which as far as I know, doesn't include any sort of residue information in the file, then creating SMILES and/or SD files, which definitely don't include residue information.

Then you create 3D coordinates with CORINA.

But nowhere along that chain is there a step to create / perceive / retain residue information. So I'm not surprised that you get one large unknown residue.

Open Babel does have code that does amino acid and nucleic acid residue perception when translating to PDB or Mol2 or other format that allows storing residue information. It does not have support for perceiving saccharides. Indeed, I'm not even sure what algorithms exist for automatically perceiving sugars.

My suggestion would be to either:

Number / label the saccharides by hand (yuck)
Write some sort of script in Python to label the residues - since presumably the atoms come in a particular order
Use a software builder that creates the residue information with the 3D coordinates.

For the latter case, a quick search suggested a few possibilities, although I haven't tried any of them:

Glycam
Yasara


Chemical Spill in Charleston, West Virginia

The name MCHM or "4-methycycloheaxe methanol" (as newspapers reported it) is not entirely a systematic name and is a little ambiguous. The actual compound is better called  4-methylcyclohexylmethanol and the structure is shown below:

Specific details about its toxicity are not easy to find, though it has been patented as an air freshener ingredient. It has also been around for a long time and is used in the coal industry as a frothing agent in significant quantities so if it had some particularly nasty property we would probably know by now. Some safety information is available (pdf) on the related 1,4-cyclohexanedimethanol and that isn't particularly toxic. So there doesn't seem to be any strong reason to worry about it, though it will taint the flavour of water. There are no obvious paths for it to known nasties in low concentrations in water, so the biggest worry is probably the taint.

Having said all this, the company should not have let it get into the water supply.


Brown particles in electrolysis of brine

If your tap water is groundwater, it will contain a lot of metals that will start rusting if reacting with O2 and Cl (your salt). The reaction causes the brown particles (rust/ Ferric Oxide). The only way to prevent this is to filter the water continuously. This reaction is also common with salt-electrolyzed pools. Here is a forum thread that has exactly the same issue you have.

This was the answer:

  That looks like rust, Ferric Oxide, exactly!
  However, the cell should be Titanium, it contains no iron, but that doesn't mean it can't oxidize. Or "Rust", all metals do, Ti's oxidation would be white and powdery.

  So either there is a non Ti component inside that cell that contains iron, such as steel that is breaking down, or there's something upstream of it, a Ferrous metal that's rusting, thus producing Ferric Oxide and dribbling down the pipe and into the cell. Is there any on the upstream side of the cell?
  If not, then it has to be localized to inside the cell.

  Another though is that you have a very high iron content in your water, mixed with the electrolysis process is causing it to oxidize on the anode, where it combines /w oxygen molecules in the water to form rust, Fe2O3.
  This would keep it localized to the cell, and not be upstream, only at the cell, since that's the catalyst, or source and slightly downstream as the flow of generated Ferric Oxide tapers off.
  This shouldn't effect the Ti plates though, just simply clean them with MA and you're golden!!
  Ti is virtually indestructible, so they should be fine.

  EDIT: Hayward says to submerge it just to the top of the cell, so the wiring harness, the electronics on the cell are not touching the acid, coiling the harness may be helpful prior to immersing. 4:1 dilute, 1 gallon of water to 1 quart of MA. And remember, always, always add the acid to the water, never, ever the other way around.
  The manuf. says to soak for a few, then rinse with a high pressure hose, then repeat if needed, until the plates are clean.
  BTW, Acid attacks Oxides, including Ferric Oxide.

Full thread here

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Salt product in water electrolysis


Helmholtz Free Energy Minimization Proof

Who says that $dS_{bath}=\frac{dU_{bath}}{T}=-\frac{dU_{system}}{T}$?  The equation should correctly read $\Delta S_{bath}=\frac{\Delta U_{bath}}{T_{bath}}=\frac{Q_{bath}}{T_{bath}}=-\frac{Q_{syst}}{T_{bath}}$, where the Q's are for the irreversible path.  The Q of the system for the reversible path is different from that for the irreversible path because of both temperature gradients and viscous dissipation of mechanical energy to internal energy during the irreversible path (both of which generate entropy).  For the reversible path of the system, these entropy generation effects are not present.


Naming Ionic Compounds with Transition Metals

As stated in the Wikipedia article for iron:

Like other group 8 elements, iron exists in a wide range of oxidation states, −2 to +6, although +2 and +3 are the most common.

Thus, assigning $\ce{FeO2}$ as an iron(IV) compound is not something that is impossible, but it is unlikely. There are only a few compounds with that are iron(IV).
A more likely scenario is that $\ce{FeO2}$ may contain other anions of oxygen:

Oxide $\ce{O^2-}$
Superoxide $\ce{O2^-}$
Peroxide $\ce{O2^2-}$
Ozonide $\ce{O3^-}$

While oxide is the most common anion, there is another on the list above that is reasonably common. Which is it? If you have that anion, what is your oxidation number for iron?


You are surely right in that the labeled oxygen will stay with the alcohol and not with the acid. Now is the time to recall what do you know about the comparative acidity of alcohol and... wait, what does the word "aqueous" stand for, really?


Why perborate is greener bleaching agent?

According to Wikipedia, "Chlorination... can oxidize organic contaminants, producing chloroform and other trihalomethanes, which are carcinogenic, and many hundreds of [other]... by-products.

[W]hen household bleach and wastewater were mixed, 1–2% of the available chlorine was observed to form [halogenated] organic compounds."

Perborates, percarbonates and similar bleaching agents, while strong oxidizers, will not create these chlorocarbon chemicals (unless the chlorine is also supplied).

However, boron and perborates are not entirely "green". Boron compounds are somewhat toxic. If the prices were the same, I'd prefer percarbonate bleach to perborate.


Stereoselectivity of Roush crotylboration

This figure from this reference (p. 170). might help you visualize things.

As you can see, the stereochemistry is given by electronic repulsion between the carbonyl group and the incoming aldehyde.