Why should space whisky taste different to Earth whisky?

From the press release from Suntory, the Japanese company which sent 6 vials of their whisky into space, which details their study:

  Tokyo, Japan, 30 July -- Suntory Global Innovation Center is about to embark upon space experiments on the “development of mellowness in alcoholic beverage through the use of a microgravity environment.” This research will be conducted in the International Space Station’s Japanese Experiment Module (nicknamed “Kibo”), with the cooperation of the Japan Aerospace Exploration Agency (JAXA).
  H-II Transfer Vehicle No. 5, commonly known as “Kounotori5” or HTV5, is scheduled to be launched from JAXA’s Tanegashima Space Center on August 16 (Sunday) carrying alcohol beverages produced by Suntory to the Japanese Experiment Module aboard the International Space Station, where experiments on the “development of mellowness” will be conducted for a period of about one year in Group 1 and for two or more years (undecided) in Group 2.

  With the exception of some items like beer, alcoholic beverages are widely known to develop a mellow flavor when aged for a long time. Although researchers have taken a variety of scientific approaches to elucidating the underlying mechanism, we still do not have a full picture of how this occurs.

  Our company has hypothesized that “the formation of high-dimensional molecular structure consisting of water, ethanol, and other ingredients in alcoholic beverages contributes to the development of mellowness,” and we have been conducting collaborative researches on this topic with research groups of Professor Shigenao Maruyama of the Institute of Fluid Science, Tohoku University and Professor Mitsuhiro Shibayama of the Institute for Solid State Physics, the University of Tokyo, the Japan Synchrotron Radiation Research Institute and Suntory Foundation for Life Sciences. The results of these collaborative researches have suggested the probability that mellowness develops by promoted formation of the high-dimensional molecular structure in the alcoholic beverage in environments where liquid convection is suppressed.

  On the basis of these results, the space experiments will be conducted to verify the effect of the convection-free state created by a microgravity environment to the mellowing of alcoholic beverage.

  The Suntory Group aims to use these experiments to help find a scientific explanation for the “mechanism that makes alcohol mellow.”

It appears that the working hypothesis is that the lack of gravity in space precludes liquid convection and thus alters the characteristics of liquids aged in that environment ${\it vs.}$ that on Earth. As you indicate, it's allegedly one way that promotes the formation of compounds which are otherwise not formed when gravity and convection within the ageing liquid are present.

From the same press release, details about the study:

  Title of study
          Elucidating the Mechanism Mellowing Alcoholic Beverage

  Experimental period
          Group 1: August 16, 2015 (Sunday) (planned) to September 2016 (planned)
  Group 2: August 16, 2015 (Sunday) (planned) to undecided date two or more years later

  Content of experiment
          One set of samples consisting of various alcoholic beverages will be stored in a convection-free state in Japanese Experiment Module “Kibo” on the International Space Station, and another set of the identical samples will be stored in Japan for the same period of time. The following methods will be then used to analyze and compare the two sets of samples.

  Measurement of substance diffusion coefficient with the use of a phase shifting interferometer$^{1}$ in cooperation with the Institute of Fluid Science, Tohoku University.
  Detection of high-dimensional structure by small angle X-ray scattering$^{2}$ using SPring-8, in cooperation with the Japan Synchrotron Radiation Research Institute and the Institute for Solid State Physics, the University of Tokyo.
  Measurement of substance diffusion with the use of the NMR method$^{3}$ in cooperation with Suntory Foundation for Life Sciences.

  [1] Phase shifting interferometer: Uses the properties of light (visible light) waves to visualize temperature and concentration distributions, etc. that cannot be seen by the eyes, making it possible to detect slight changes in the ingredients of alcoholic beverages and determine how they are distributed.

  [2] Small angle X-ray scattering: Can be used to obtain nanoscale (1 millionth of a meter) structure by irradiating a substance with X-rays and observing the scattered X-rays in an angular region of a few degrees or less.

  [3] NMR method: Uses the phenomenon of nuclear magnetic resonance (NMR) to extract information from the organic compounds that are the object of measurement. Can be used to obtain information on the molecular structure of a compound and its dynamic properties.

  Samples
          Five types of distilled spirits differing aging periods and 40% ethanol: Total of six samples

Suntory is not alone. Ardbeg scotch made a similar trip into space, and you can read about the experiment and results here. They used GC, HPLC, GC-MS, and organoleptic assessment to analyze their samples.

As stated in their summary:

  The key aims of the experiment were as follows:-

  (i) To determine if conditions of micro-gravity would have an impact on the range of naturally occurring terpene compounds and on the extraction
  of key flavour-active compounds in oak wood by Scotch malt distillate.

  (ii) To assess the impact of maturation in conditions of micro-gravity on the flavour profile of oak matured distillate.

  (iii) To investigate the presence of novel compounds and flavours in model maturation systems for Scotch malt distillate, as a result of conditions
  of micro-gravity.

  The key findings of the above three aims would then be considered in conjunction with actual maturation conditions on Earth to investigate
  techniques for developing novel flavours in Ardbeg Islay Single Malt Scotch Whisky.

In the Results and Discussion section (I do encourage interested parties to jump to the link, and there are several tables and too much information to post here), they note:

  GC analysis of major volatile congeners demonstrated no significant differences between the International Space Station samples and the control
  samples on Earth. Since this group of compounds comprising primarily alcohols, aldehydes, ketones and esters, are produced during fermentation
  and concentrated up, or in some cases, catalysed during distillation, their presence and quantities should have been identical in each set of samples
  to start off with. The conditions of micro-gravity on the ISS have not impacted on their levels, as demonstrated by the similar levels in the Earth
  based samples (Table 1.)

  A more detailed investigation of these compounds, along with phenolic compounds, was carried out on our behalf by external whisky experts and
  scientists, using GC-MS. In the resulting chromatograms, the areas of 75 separate peaks were integrated, most of which were identified as alcohols,
  esters, acetals and phenols. Again, no major differences were found between the ISS and Earth samples. From an Ardbeg perspective, this was
  particularly interesting as the conditions on the ISS have not impacted on the absolute quantities of the various phenolic compounds. These are
  the compounds which contribute the powerful, smoky, medicinal flavours to Ardbeg spirit.

  The most significant variable between the ISS and Earth samples was discovered when the results of the HPLC analysis of key maturation related
  congeners (wood extractives) were considered. The absolute concentration of these compounds was far higher than would normally be expected
  in standard, barrel-matured spirit, almost certainly as a result of the much higher surface area of woody material that the spirit was exposed to in
  the MixStix. However, when the two samples, ISS and Earth, were compared to each other, the lignin breakdown products - which are the major
  flavour active derivatives from oak wood - were present in higher concentrations in the control samples than in the space samples. This indicates
  that conditions of micro-gravity have inhibited their extraction from oak. On the other hand, ellagic acid and gallic acid, which are tannin breakdown
  products, both appeared to be higher in the ISS samples. The levels of these compounds are detailed in Table 2.

  By drilling down further into these results, and subjecting them to some statistical analysis, the relative abundance of the main, easily extractable
  lignin breakdown products (vanillin, coniferaldehyde and sinapaldehyde) expressed as a percentage of the total wood extractives, is broadly
  similar in each sample. However, other extractives, such as the syringyl compounds, are very different in their total abundance, with much lower
  percentages in the ISS samples. This result means that while the overall extraction of wood-derived compounds is negatively impacted by micro-gravity,
  the impact is not equal; the compounds can be subdivided into ‘readily extractable’ and ‘less readily extractable’.

  While it is clear that micro-gravity is inhibiting the release of wood extractives, not all compounds are affected equally, as would be expected
  in normal conditions. This has resulted in an unusual ratio between the readily extractable compounds and the less readily extractable ones in
  the ISS samples. This offers the intriguing possibility of using the ratios of such compounds as a potential marker to identify ‘unusual’ maturation
  characteristics, and thus a means of identifying for example, spurious age claims in whiskies. This possibility may be further investigated by the Scotch
  Whisky Research Institute.

  For studies such as this, where we are examining the impact of differing conditions on either model systems, or on the full scale in actual barrels,
  the key question is always ‘What does this actually mean for the flavour?’, so the organoleptic analysis was of great interest to us. The results from
  the key volatile congeners, wood extractives and phenols analysis did not lead us to believe that from an organoleptic perspective, there would
  be large differences between the ISS and control samples. However, when we finally nosed and tasted the samples in the sensory laboratory,
  the results were unexpected.

  Triangle tests gave a high number of correct responses, with almost all participants picking out a difference between the ISS and control samples.
  When we examined the actual aroma and flavour profiles, the differences between the two were as remarkable as they were unexpected; all tasters
  described the Earth control samples as being of the general Ardbeg whisky style, while the ISS samples displayed a quite different set of flavours.
  These have been summarised in the detailed descriptions below.

The above-mentioned MixStix are described in the write-up:

  MixStix™ technology, which provides a convenient platform for small scale sample experimentation, is a trade-marked piece of technology used
  by NanoRacks. Essentially, they are small glass vials which can hold a range of materials; in our case these materials were 6ml of Ardbeg new spirit
  distillate, along with a small quantity of charred oak wood shavings (from a once used American oak Bourbon barrel), c. 1mm x 1mm x 6cm in
  size. The two materials are separated from each other by a glass dividing partition, which is broken by bending the Teflon outside covering of the
  MixStix™ at time of experiment activation, thus allowing the two separate components to mix together.


But what are anti-bonding pi-orbital? In search for an intuitive explanation

First, remember that the + and - on the lobes of an electronic wavefunction do not ever represent or suggest electronic charge. An electron is always negative; a proton (or nucleus) is always positive. In order to imbed this firmly in your mind without + or - signs, imagine one lobe of a p-orbital flashing red while the other flashes blue. In this picture, an s-orbital would flash either red OR blue.
Plus and minus signs are just the way mathematicians describe the orbitals: they don't use red and blue pens, they use + and -. Now take the orbitals and move them about. Wavefunctions have a wave nature; waves have a frequency; so two orbitals of the same kind on similar atoms will have the same frequency. Two hydrogen s-orbitals flashing red at the same frequency look like one object, that is, they seem to be bonded into a hydrogen molecule (if they are close enough). Their waveforms are constructive.
But if one flashes red while the other flashes blue, the picture is not so attractive - the individual atoms are not in sync, not bonded. If I now switch the metaphor to a musical (audio) frequency rather than a visual wavelength, the idea of wave addition or neutralization images better (see picture).

To clarify, the red and blue orbitals don't totally collapse each other and disappear - just where they might "touch". So each orbital, red and blue, still contains one (negative) electron, but there is no continuity of electron travel from and around one nucleus to the other, because there is a gap, a node, a location where there is no electron density, where they might come in contact. This represents two  atoms that have been dissociated from their bonded state as a molecule. They could have been bonded together using two red-flashing orbitals, but got excited by some energy input that brought one electron up to a blue energy which is now out of sync with the red orbital, and finds it repulsive. A $pi$ bond between two carbon atoms would show the two upper lobes of each atomic orbital to have the same color and the two lower lobes of each atomic orbital to have the other color: both upper and lower regions would be bonding. A $pi$ antibond would have a red and a blue in the upper region and a blue and a red in the lower region: repulsive!
So we usually show the direction of bonding or repulsion in bonding or antibonding molecular orbital by using + and - signs on the atomic (and molecular) orbitals, saving red and blue colors for PowerPoint presentations.


Why can't iodide be oxidised to iodate in acidic medium?

Permanganate oxidizes iodide to iodine $\ce{I2}$ at all pH values. In acidic conditions, the reaction is finished here. But in basic solution, the iodine $\ce{I2}$ is transformed into iodide and iodate according to $$\ce{3I2 + 6 OH^- -> 5 I- + IO3^- + 3 H2O}$$The mixture  iodide + iodate reacts the other way as soon as the solution becomes acidic : $$\ce{5I^- + IO3^- + 6 H+ -> 3 I2 + 3 H2O}$$


Interplanar distance given Miller indice of the planes

Your adjacent plane isn't.

Worth a thousand words, they say.


Why does my metal phone ring stand becomes iridescent, and turns back to normal after a while?

This is an interference phenomenon, more specifically interference of light between two thin layers with an air gap. Recall the beautiful soap bubble as an example. Whatever coating your have on this object gets very very slightly separated from the surface upon mild heating. This generates an air gap between two layers.

Iridescence is a different phenomenon, which is angle dependent in the sense that you see different colors at different angles.

If you have over heated an empty stainless steel frying pan by mistake, that also forms a interference pattern (purple tinge). This is because of an oxide coating on the surface but it is not reversible. You have to scrub it away.


Interaction between metalic nanoparticles and aggregation

Metallic nanoparticles, considered as colloids, are attracted to each other by van der Waals forces. To prevent their aggregation, we normally use dispersant using electrostatic repulsion or steric hindrance to stabilize the colloids.


Calculating Enthalpy Changes at Different Temperatures

@Chet already gave the qualitative answer but the actual equation it is based on is:

$$
\Delta H^\circ(T)=\Delta H{^\circ}_{T_1} + \int_{T_1}^{T} \Delta C_p{^\circ}(T^\prime ) \,\mathrm{d}T^\prime$$

If you know the enthalpy at a reference temperature and the heat capacity as a function of temperature you can calculate the enthalpy change at any other temperature.

Actually d) and e) together are not wrong. We all know these equations:
$$\Delta G= - RT\ln (K) \qquad \Delta G= \Delta H - T\Delta S $$

With d) you can calculate $\Delta G$ and with e) you can calculate $\Delta H$.

Lets look deeper into: $\Delta G^\circ$, $\Delta G$, $K$ and $Q$.
To answer your second question completely I would recommend to look again into a textbook of your choice. Especially the definitions of activity and chemical potential.

If you know the chemical potential $\mu^\circ$ of a pure substance you can calculate the chemical potential of mixtures if you know the activities:
$$\mu= \mu^\circ + RT \ln(a)$$

Now we make two assumptions:

We have constant temperature and pressure. ($\mu = \Delta G$)
We can calculate the activities simply over the partial pressures or concentrations.

This leads then to your equation:
$$
\Delta G= \Delta G^\circ + RT \ln(Q)
$$
where $Q$ is the reaction quotient.
Since we assume equilibrium,  we know that $\Delta G=0$ and $Q = K$. This leads to:
$$ 0= \Delta G^\circ + RT \ln(K) \Rightarrow \Delta G^\circ = -RT \ln(K) $$

The last equation is the handy one you use for calculating equilibrium constants from energies and vice versa.
So to be really precise one has to write:
$$ \Delta G^\circ= \Delta H^\circ - T\Delta S^\circ $$

But one usually skips the superscript $^\circ$.
At equilibrium $Q$ and $K$ are essentially the same.


Oxidation of 1,4,5,8-tetrahydronaphthalene by meta-chloroperoxybenzoic acid

As the following figure indicates, the terminal oxygen in a peracid (the oxygen involved in the epoxidation) has a partial positive charge ; therefore this oxygen is electrophilic.

(image source)

The central double bond in your compound is tetrasubsituted, the lateral bonds are only disubstituted with alkyl substituents (remember that alkyl groups are electron donating).  Consequently the central double bond is the most electron rich and will be the preferred site of attack for the electrophilic oxygen.


PE expression for vibrational spectra of molecules using Newton's second law

It's probably simplest to start with the familiar example of the 1D harmonic oscillator. Here
$$F=-\frac{dV}{dx}=-\frac{d\left(\frac{1}{2}kx^2\right)}{dx}=-kx\tag{1}$$
which is often called Hooke's law, $F=-kx$, where x is the displacement from the equilibrium position, and the potential is assumed quadratic (parabolic) in the position coordinate. The derivation above requires the substitution $V(x) = \frac12 kx^2$, which can be obtained by trimming a Taylor expansion of the potential about the energy minimum (assumed = 0) at the quadratic term, and the force constant is defined as
$$k=\left(\frac{d^2V}{dx^2} \right)_e\tag{2}$$
Also,
$$F=ma=m\frac{d^2x}{dt^2}\tag{3}$$
If we transform to mass-weighted coordinates (or equivalently set m=1), we therefore can write that
$$\frac{d^2x}{dt^2} = -\frac{dV}{dx}=-\left(\frac{d^2V}{dx^2} \right)_e x\tag{4}$$
The derivation in your book does essentially the same thing. Note that since the potential is assumed quadratic and the other listed conditions are imposed, the potential can be written as
$$V = \frac12 \sum_{i = 1}^{3N}\sum_{j = 1}^{3N} \left( \dfrac{\partial^2{V}}{\partial{q_i}\partial{q_j}} \right)_e q_i q_j \tag{5}$$
The double summation can be collapsed into a single one when you take the derivative with respect to a spatial coordinate, say $q_k$:
$$\begin{align} \dfrac{d^2 q_k}{d t^2}  &=-\left( \dfrac{\partial{V}}{\partial{q_k}} \right)  \\&=-\left(\dfrac{\partial{}}{\partial{q_k}} \left[  \frac12 q_k^2  \left( \dfrac{\partial^2{V}}{\partial{q_k^2}} \right)_e   + \frac12 q_k\sum_{j \neq k }^{3N} \left( \dfrac{\partial^2{V}}{\partial{q_k}\partial{q_j}} \right)_e  q_j  + \frac12 q_k \sum_{i \neq k }^{3N} \left( \dfrac{\partial^2{V}}{\partial{q_i}\partial{q_k}} \right)_e q_i  + \frac12 \sum_{i \neq k }^{3N}\sum_{j \neq k}^{3N} \left( \dfrac{\partial^2{V}}{\partial{q_i}\partial{q_j}} \right)_e q_i q_j  \right] \right)  \\ &= -\left( \left[ \left( \dfrac{\partial^2{V}}{\partial{q_k^2}} \right)_e  q_k  + \frac12 \sum_{j \neq k}^{3N} \left( \dfrac{\partial^2{V}}{\partial{q_k}\partial{q_j}} \right)_e  q_j  + \frac12 \sum_{i \neq k }^{3N} \left( \dfrac{\partial^2{V}}{\partial{q_i}\partial{q_k}} \right)_e q_i    \right] \right)    \\ &= - \sum_{j = 1}^{3N} \left( \dfrac{\partial^2{V}}{\partial{q_k} \partial{q_j}} \right)_e q_j \end{align}$$


Is water acidic enough for an redox reaction involving H+?

Yes the reaction would be spontaneous at standard conditions.  However standard conditions for electrochemistry are at pH 0, i.e. an $[\ce{H+}]$ concentration of 1 molar.

Solutions of lead nitrate by themselves are not going to give you pH 0.  Apparently very very concentrated lead nitrate solutions will reach pH 3 or 4, due to the weakly acidic nature of hydrated lead ions.

The electrochemical potential of the reaction you are interested in at pH 3 can be calculated from the standard (i.e. pH 0) potential via the Nernst equation.

$$E_\mathrm{cathode} = E_0\;– \frac{RT}{nF}\ln{Q}$$

Here, $Q$ is the "half"-reaction quotient, given by
$$ Q = \frac{\mathrm{p}_{\ce{N2O4}}}{[\ce{NO3-}]^2[\ce{H+}]^4}$$ Now assume that nitrate concentration and the partial pressure of dinitrogen tetroxide are in their standard states.

If you go through the math (see examples here or here), then you will see that there is a 118 millivolt-per-decade decrease in the potential, i.e. every increase in the pH by 1 unit decreases the potential by 118 mV.  So already at pH 3 the reaction will no longer be favorable.

Copper is unstable to concentrated nitric acid, but the acid needs to be concentrated.  Dilute acids cannot oxidize copper so easily.


The benzene ring in phenol is highly activated toward electrophilic substitution and hence attempts to directly nitrate it result in charring and copious evolution of oxides of nitrogen. The reaction is highly exothermic and difficult to control.
To reduce the reactivity, the phenol is first mono-sulfonated ( some of the product which is substituted may also be used). The products are ortho- and para-isomers. The para-isomer is separated and then nitrated. The nitration is comparatively far smoother (easier to handle).
Ipso substitution of $\ce{SO3}$ groups occur.

Links: Wikipedia, Research paper
Link: (For ipso substitution): PDF, Wikipedia (see subsection: Ipso substitution)


Can we have an adiabatic process in a sealed metal container?

Adiabatic means that energy is transferred between the system and the surroundings only as work and not as heat.

Let's look at your process. You have a non-isothermal process, so $\Delta T_{sys} \ne 0$ for the process. An adiabatic process does not need to be isothermal, so this condition is okay. What makes your process non-adiabatic is the condition that constant temperature is somehow maintained (likely through a constant temperature bath which serves as the surroundings). This requires heat exchange between the system inside the metal container and the surroundings outside the container.

If you remove the constraint that the overall temperature is held constant, then you can have an adiabatic process.


Is probability of finding electron changing here?

I swear I read the question many times, but I still not quite sure what it is about.  So I will try to clarify some things hoping that they are relevant.

First off, any complex multiple of a wave function indeed describes exactly the same state of a system. This directly follows from the postulate of quantum mechanics which states that the expectation value of an observable $A$ represented by a self-adjoint operator $\hat{A}$ is given as follows,
$$
\langle A \rangle
=
\frac{\langle \Psi | \hat{A} | \Psi \rangle}{\langle \Psi | \Psi \rangle} \, ,
$$
where $\Psi$ is the wave function. It is indeed easy to show that given $\Psi$ we can multiply it by any complex number $c$ and this would not change the expectation value of any observable,
$$
\langle A \rangle
=
\frac{\langle c \Psi | \hat{A} | c \Psi \rangle}{\langle c \Psi | c \Psi \rangle}
=
\frac{c^{*} c \langle \Psi | \hat{A} | \Psi \rangle}{c^{*} c \langle \Psi | \Psi \rangle}
=
\frac{\langle \Psi | \hat{A} | \Psi \rangle}{\langle \Psi | \Psi \rangle} \, ,
$$
since $c^{*} c$ in numerator and denominator perfectly cancels each other. Also note that the following definition of the expectation value
$$
\langle A \rangle = \langle \Psi | \hat{A} | \Psi \rangle \, ,
$$
holds true only for a normalized wave function $\Psi$ and trivially follows from the above mentioned more general definition.

Secondly, for the simplest case of one particle in one spatial dimension, if one would like to interpret the wave function $\Psi(x,t)$ as a probability amplitude that the particle is at $x$, and consequently, it square modulus, $ \left|\Psi(x, t)\right|^2 = {\Psi(x, t)}^{*}\Psi(x, t)$ as the probability density that the particle is at $x$, he/she have to normalize it as follows,
$$
\int\limits_{-\infty}^\infty d x \, |\Psi(x,t)|^2 = 1 \, .
$$
This is required because one of the axioms of the probability theory requires that

  the probability that some elementary event in the entire sample space
  will occur is 1.

The sample space is just a set of all possible outcomes in an experiment, which in our case is the set of events of finding a particle at any possible position, i.e anywhere between $-\infty$ and $+\infty$. Thus, in our case the probability of finding the particle anywhere between $-\infty$ and $+\infty$ have to be 1, and consequently, the wave function has to be normalized. Otherwise its interpretation as a probability amplitude would be inconsistent with the probability theory.

Hope it helps.


How do I solve oxidation numbers for unknown elements?

You shouldn’t really need to use the shortcut rules (fluorine having $\mathrm{-I}$, hydrogen having $\mathrm{+I}$ and oxygen having $\mathrm{+II}$ ‘except in peroxides’) with the exception of really simple compounds where it all falls into place immediately. Instead:

If the compound is molecular or contains a molecular ion, draw the Lewis structure.
Assign each bonding electron pair to the more electronegative bond partner.
Count the electrons each atom ends up with, substract from the number of electrons it should have according to its position in the periodic table, invert that number’s sign.
If at step one you are only left with ions, just take the ion’s charge as oxidation state.
You did it right if the overall sum of oxidation states matches the molecular ion’s charge or zero for the overall structure.

(Note: I am simplifying here by assuming that compounds are either molecular or ionic. That, of course, is most certainly not the case but the very crude generalisation gets us very far in this case.)

To work through one of your examples:
$\ce{KSCN}$ is an ionic compound consisting of potassium cations and thiocyanate anions. According to step 4, we can assign potassium the oxidation state of $\mathrm{+I}$.

The structure of thiocyanate is $$\ce{S=C=\overset{-}{N} <-> \overset{-}{S}-C#N}$$
In both cases, the terminal atoms have enough lone pairs to complete their octet.
The electronegativites are $\ce{N}>\ce{S}>\ce{C}$. Thus, we assign the bonds so that:

sulphur and nitrogen each have eight valence electrons
carbon has zero

This is independent of the resonance structure we used.
Calculating:

Sulphur:
eight electrons, should have six.
$\displaystyle-(8-6) = \mathrm{-II}$
Carbon:
zero electrons, should have four.
$\displaystyle-(0-4) = \mathrm{+IV}$
Nitrogen:
eight electrons, should have five.
$\displaystyle-(8-5) = \mathrm{-III}$

Has been done before.
To check:

$\displaystyle \mathrm{+I + -II + +IV + -III = \pm0}$ (for $\ce{KSCN}$)

$\displaystyle \mathrm{-II + +IV - III = -I}$ (for $\ce{SCN-}$)


Steric inhibition of resonance vs. hydrogen bonding

I think the main effect in play here is straight electron withdrawing potential of the oxygens. In my opinion this is a classical BSc question that tries to avoid the grey zones of pka eye balling.

I would invite you to review the concept of steric inhibition of resonance since I'm not entirely sure I understand your argument there.

Here's a good starting point if you do not have a text book covering the subject
https://www.quora.com/What-is-steric-inhibition-of-resonance


There are no wavefunctions in DFT.

I don't like that "wave function" is used here in the plural form and I feel like it reflects OP's misconception. The right way to say this is

  There is no wave function in DFT.

See, there exist only one wave function, the function that describes the state of a system in question and can be used to calculate its properties. When it comes to a many-electron system, the wave function is a function that describes the state of all the electrons present in the system interacting with each other. This is the wave function people have in their minds when they say the it is absent in DFT. And that is true: the whole point of DFT is to get rid of this function as a description of state since it is too complex to work with and difficult to visualise.

But that is not the whole story. In practise the Kohn-Sham DFT is used which introduces a fictious system of non-interacting electrons. Each electron in the fictious system is described by a Kohn-Sham orbital, a single-particle wave function, while the whole fictious system is described by the Kohn–Sham wave function, a Slater determinant constructed from a set of Kohn-Sham orbitals. Still there is no wave function in KS-DFT in the above mentioned sense:

KS orbitals don't qualify as the wave function just because they do not describe the state of a whole system;
the Kohn–Sham wave function doesn't qualify because it describes a different system (fictious non-interacting rather than real interacting).

So, the KS wave function is not the wave function, since it doesn't describe the (real) system. You can't use it to calculate the properties of the (real) system, you have to use the electron density to do so. In this sence, the statement that there is no wave function in DFT holds true for KS-DFT as well.

Can the KS wave functions (both single- and many-particle one) be somehow useful in any other way is a different story. See, for instance, the discussion of the usefulness of KS-orbitals here.


How to know if a reaction forms H3O+ ions or OH- ions?

H3O+ is just the combination of a a $\ce{H+}$ ion, which we know to be released from the dissociation of an acid in an aqueous solution, with a water molecule. It happens because the $\ce{H+}$ is so, so positive (and therefore so reactive) that the water molecules, with its lone pairs (which are locally negatively charged), are willing to form a dative covalent bond with the $\ce{H+}$ ion.
So like Ivan said, whenever you see an acid in an aqueous solution you can consider $\ce{H+ \equiv H3O+}$. The key there though, is that it must be in aqueous solution so that it can dissociate (and the $\ce{H3O+}$ form in the first place). That is the condition that is needed to form it. This is the same for $\ce{OH-}$, which requires an aqueous solution to form and thus release the ions in say NaOH, forming $\ce{Na+}$ and $\ce{OH-}$.


Have you taken into account that there is a $\ce{-CH3}$ group attached to the top N which has +I effect i.e., electron donating effect?
$\ce{-CH3}$ group will destabilize the negative charge on N formed after the release of proton.
Electron releasing group decreases acidic strength of a compound. So in this question proton of primary amine is more acidic and that of secondary amine is less acidic.


Synthesis via Sn2/Sn1, E2, E1: Choice of solvent

Your thinking is not wrong. For (e) DMF is a better choice of solvent - this doesn't mean that MeOH won't work just that DMF is better and is typically used in these reactions. For even better results you can use DMSO. If you dare, as DMSO/cyanide is a very toxic cocktail.
(j) I don't think it is a practical synthetic reaction since it can possibly give you elimination and/or $S_N1$ (ie no inversion of stereochemistry) products.


You should find it in any analytic chemistry textbook, considering volumetric methods.
We did so in the high school.

In short, a solution of known molar concentration of well defined acidic substance is made, that get titrated by the $\ce{NaOH}$ solution.

Typically oxalic acid $\ce{(COOH)_2 . 2 H2O}$, weighted at good scales, preparing solution $\pu{0.05 mol / l}$.

When you Google "acidimetric titration sodium hydroxide oxalic acid" , you get tons of references, from illustrating videos to PDF documents.


$\ce{NC-CN <=> 2CN}$

Which side is favored at low pressure?


I know that ibuprofen is sold as racemic mixture, it's a drug for fever treatment. It's very cheap in Brazil, one dollar each box (400mg per tablet).


Freezing Point Depression and Temperature Decrease

Phase change is not at constant temperature, phase equilibrium is

  [OP] it's a phase transition so the temperature of the ice should remain constant

Here is a counter example: If you add ice cubes to hot water, the ice will melt, cooling down the hot water. In this system, there is no thermal equilibrium, so not all of the components are at the melting point of water. If you wait a while, you might see the following: All the ice is gone, or some ice and some water remains, or (if the ice was really cold) all the water is frozen. Only if there is some ice and some water left do you know the equilibrium temperature - equal to the melting point.

A system can gain thermal energy from other types of energy

If you stir a viscous solution, it will get measurably warmer. (This is a classic experiment of thermodynamics.) There is no heat transfer - nothing gets colder. For melting ice, it is a conversion of thermal energy into chemical energy (breaking hydrogen bonds).

  How can the ice be cooling as it melts?

It can't, at least not for pure water (see below for adding water). The temperature of the ice is below the melting point (or at it), and the temperature of the water is above the melting point (or at it, or a bit lower if it is super-cooled). The thermal energy will come from the liquid water, cooling it. If everything is at the melting point, melting and freezing will happen at the same rate, and nothing changes temperature. In a real system where surrounding is warmer, thermal energy from the surrounding flows into the system, melting the ice while the system stays at the melting point (there will be small fluctuations so that the thermal energy reaches the ice).

Scenario of adding salt

Starting with an ice/water equilibrium (at melting temperature), we can add salt (at the same temperature, and let's say there is no enthalpy of dissolution). After adding the salt, the temperature is still the same, but the melting temperature is lower. With the temperature higher than the melting temperature, water molecules break away from the ice, which requires energy. This energy comes from thermal energy in the liquid and in the remaining ice. As a result, the entire system cools down.

Will the temperature change?

  In your example where ice cubes were added to hot water, wouldn't the pure water ice cube remain at 32 F until it fully melted? How is this any different than having an ice cube on a hot plate and melting it? The temperature during the heating of the cube doesn't increase because the absorbed heat is being used to break the chemical bonds of the solid ice.

If the system is not at thermal equilibrium (different temperatures within the system), some parts of the system have to change temperature before equilibrium is reached. The hot water will certainly decrease in temperature. If the ice cubes are colder than the melting point at the start, they will warm up until they have reached the melting point. (Ice can have any temperature below the melting point. For example, ice from a freezer is at about $\pu{-20^\circ C}$.)

If the system is at thermal equilibrium from the start but not at phase equilibrium (or not a chemical equilibrium for a reaction that is endothermic or exothermic), the temperature will also change. In my example of adding salt to a water: ice mixture, the process of melting is endothermic, so the temperature at equilibrium will be colder (and equal to the freezing point of the salt solution). As ice melts, the temperature decreases and the freezing point increases (because the pure melt-water dilutes the salt solution). Once the two temperatures match, the system is at equilibrium (thermal equilibrium and phase equilibrium).


Weight of sodium acetate required to get to certain pH

It's not really a matter of remembering a certain formula, this is not a conceptual thinking and should be avoided. When approaching these types of questions, always start with a balanced chemical equation. In this case we deal with a salt formed by a weak acid and a strong base, which hydrolyses according to the following net ionic equation:

$$\ce{CH3COO- + H2O <=>[$K_\mathrm{h}$] CH3COOH + OH-}$$

$$
\begin{align}
K_\mathrm{h} &= \left.\frac{[\ce{CH3COOH}][\ce{OH-}]}{[\ce{CH3COO-}]}\qquad\right|\cdot\frac{[\ce{H+}]}{[\ce{H+}]}\\
&= \frac{[\ce{CH3COOH}][\ce{OH-}][\ce{H+}]}{[\ce{CH3COO-}][\ce{H+}]}\\
&= \frac{K_\mathrm{w}}{K_\mathrm{a}}\label{eq:1}\tag{1}
\end{align}
$$

At the same time disregarding autoprotolysis of water, and taking into account that $[\ce{CH3COOH}]\approx [\ce{OH-}]$ and $[\ce{CH3COO-}]\approx c$ ($c$ is the concentration of sodium acetate), hydrolysis constant is

$$K_\mathrm{h} = \frac{[\ce{OH-}]^2}{c}\label{eq:2}\tag{2}$$

Knowing that

$$[\ce{OH-}] = \frac{K_\mathrm{w}}{[\ce{H+}]} = \frac{K_\mathrm{w}}{10^{-\mathrm{pH}}}\label{eq:3}\tag{3}$$

one can equate \eqref{eq:1} and \eqref{eq:2}, substitute $[\ce{OH-}]$ with \eqref{eq:3}, and express concentration as

$$
\require{cancel}
c = \frac{K_\mathrm{a}[\ce{OH-}]^2}{K_\mathrm{w}} = \frac{K_\mathrm{a}K_\mathrm{w}^{\cancel{2}1}}{10^{-2\mathrm{pH}}\cancel{K_\mathrm{w}}} = \frac{K_\mathrm{a}K_\mathrm{w}}{10^{-2\mathrm{pH}}}\tag{4}
$$

Finally, mass $m$ of sodium acetate can be found from its molar mass $M$ and volume of the solution $V$:

$$m = cVM = \frac{VMK_\mathrm{a}K_\mathrm{w}}{10^{-2\mathrm{pH}}} = \frac{\pu{2.50 L}\cdot\pu{82.03 g mol-1}\cdot 1.75\cdot 10^{-5} \cdot 10^{-14}}{\pu{10^{-2\cdot 9} mol-1 L}} = \pu{35.89 g}\tag{5}$$


Changes in pH when adding substances other than acids or bases

You could use an alcohol or acetone but they are slightly acidic (though much less so than acetic acid; also alcohols react to form esters). Now effects from acidic nature can't completely avoided as even a molecule as non-acidic as methane still has a pka of 50. I think an aliphatic ether such as tetrahydrofuran (THF) or diethylether would be the best substance to add as the ethers are only very slightly basic and are miscible with water.

More simply:

  what substance could I add to change the pH considerably?

Why does it have to be a substance? The $K_\mathrm w = 10^{14}$ at $\pu{25^\circ C}$. If you add heat the $K_\mathrm w$ will increase and thus the pH will increase.


How do scientists count the exact number of electrons, protons, neutrons in the atom?

For charge neutrality, the number of electrons has to equal the number of protons.  So, the question really is how to know how many protons and how many electrons. This is actually a very interesting question dating back to the early days (and even before) of nuclear physics and chemistry.

Now, one can do it fairly easily with an ion accelerator and a magnet. With the right set up, one can get a beam of mono-energetic ions out of the accelerator, send them through a stripper to remove electrons, and then send them through a magnet to separate the components out by velocity (which depends on the ion mass through the energy). From the magnet deflections one can get a charge-to-mass ratios, and get back to number of protons (charge on a fully stripped ion) and number of neutrons (from number of protons and the mass). This is the heart of mass spectrometry (although one can use time-of-flight techniques rather than, or in addition to, magnetic separation).


If the final volume of the solution is .5L then what is important here is how many moles/grams of the substance would be used to make the solution. In general when making a solution, the solution is raised to the final volume after the mixture. So whether .5L of $\ce{H2O}$ is used, or .4998L of $\ce{H2O}$ is used doesn't matter.


Why does ionization increase from Li to Be?

The s-orbitals have a much greater degree of penetration than the p-orbitals, so they experience the nuclear charge more strongly.  This electrostatic attraction is sufficiently greater than the electrostatic repulsion between the two electrons that the ionisation energy increases.

Because the p-orbitals do not penetrate through the s-orbitals to the nucleus as much there is less electrostatic attraction.  By placing a second electron in a p-orbital the repulsion is now more significant compared to the attractive force, and the ionisation energy is less than expected.

However, each new element also increases the nuclear charge, which in turn increases the attraction that each electron has to the nucleus.  When the next p-orbital becomes doubly occupied this stronger attraction once again dominates, increasing the ionisation energy.


Hazards of trivalent chromium and hexavalent chromium

Hexavalent chromium is a genotoxic carcinogen. Due to the similarity of the chromate(VI) and sulfate ions, it is transported around the body via the sulfate channels. Due to this it is a major health hazard. Trivalent chromium on the other hand is not generally regarded as a health hazard due to its poor solubility in water. With this information you should report both separately.


Electrolysis of calcium hydroxide in glycerol

Calcium hydroxide is probably slightly soluble in glycerine due to tight chelation of Ca++ ion by the three hydroxyls of (probably 2) glycerine(s). But the hydroxyls will be present too, and on hydrolysis, will yield O2 plus H2O.

But beyond that, imagine you produce some calcium metal: it will be in glycerine, which has hydroxyl groups that will react with the Ca to produce H2. It could be a competition where some conditions allow production of measurable amounts of Ca metal, but it doesn't look promising. PEG 400 (polyethylene glycol with a mw of 400) has been used as a solubility enhancer for Ca(OH)2 (similar to glycerine), but has many fewer hydroxyls, and might do the job better. https://www.mdpi.com › pdf yields a pdf with more information.

Glycerine and PEG seem to be used to increase solubility of Ca(OH)2 in water rather than in the other solvent alone. I suspect increased viscosity is an issue.

Another way to get some calcium would be to use mercury as the cathode; then calcium will deposit and amalgamate (I'm pretty sure. I've done it with sodium and ammonium, but not with calcium.).


Understanding the difference between physical and chemical changes

Yes, (B) is a physical change. Stretching the balloon simply rearranges the polymer strands in the balloon material, making them straighten out instead of being kinked up. When you let the balloon go, they just pull back to their former kinked shape.

Your reasoning on (C) and (D) is solid.

The answer is indeed (A). See Wikipedia's page on hydrolysis for more information.


Short answer: The reverse reaction can occur using an electrolytic cell.

Long answer: There are two types of cell-electrochemical cell and electrolytic cell.

Electrochemical cell uses a spontaneous reaction to convert chemical energy to electrical energy.

$\ce{CuSO4 + Zn -> ZnSO4 + Cu}$

This reaction is spontaneous and the cell setup for this reaction is an electrochemical cell.

Electrolytic cell is just the opposite of electrochemical cell. Non spontaneous reactions can be made to happen using an electrolytic cell.

$\ce{Cu + ZnSO4 -> Zn + CuSO4}$

This reaction is non spontaneous. Energy from an external source will be used to make this reaction happen. If you want to know how these cells work, refer to this site.


Alternative explanation of ethene's pi bond

Here is the meaning of the words ""vulnerable attack" and "negative region of space".
The nucleus of the six atoms (2 C and 4 H) are situated in the same plane, which is usually defined as horizontal. The electrons of the 4 H atoms and the three first electrons from each Carbon atom are all together included in sigma bonds, and their maximum probability of presence is situated in this plan. These bonds are made of negative charges, that should be able to attract positive charges. But these bonds can hardly attract any arriving positively charged ion, because this attraction would be entirely compensated by the repulsion due to the neighboring nuclei.
The pi electrons are very different. Their maximum probability of presence is not situated in the plan. They make a sort of double long cloud high above and much under the plan of the molecule. They produce a sort of negative cloud (or a negative region of space) surrounding the molecule above and under the plan.
Electrons are negatively charged. So they may attract positive charges. If a positive ion like $\ce{H+}$ or $\ce{Br+}$ approaches an ethene molecule, it will be attracted by this negative $\pi$ cloud. If this approach leads to the formation of a new bond between the ethene molecule and the coming ion, it makes sense to state that the coming ion attacks the ethene. One may also state that the $\pi$ cloud attacks the coming ion. In other words, it also makes sense to say that the $\pi$  bond of ethene is vulnerable to attack, at least more vulnerable than the sigma bonds in the plane.


Reaction of Metallic Tin with NH4Cl and treatment with Sulfur

Note: this is an incomplete answer because question is unclear till now

  Metallic $\ce{Sn}$ is reacted with $\ce{NH4Cl}$..

$$\ce{2NH4Cl + Sn → SnCl2 (white ppt.) + 2NH3 (both aq and g.) + H2 ^}$$

Ammonium chloride react with tin to produce tin(II) chloride, ammonia
and hydrogen. (source)

Some tin(II) chloride will react with aqueous ammonia to form tin (II) hydroxide and amm. chloride.(source)

$$\ce{SnCl2 + 2(NH3•H2O) or NH3(aq) → Sn(OH)2 or SnO.H2O(white ppt.)  + 2NH4Cl (white ppt.)}$$

  ..and the resulting solution is treated with Sulfur.

I am suspecting that is hydrogen sulfide and not sulfur.(Are you sure?)

$$\ce{SnCl2 + H2S → SnS(dark brown ppt.) + HCl}$$

Tin(II) chloride react with hydrogen sulfide to produce tin(II) sulfide and hydrogen chloride. Tin(II) chloride - concentrated solution. Hydrogen sulfide - saturated solution. (source)

So, answer to your questions:

What is the gaseous product?

  Hydrogen, Some ammonia gas.

What is the color of final precipitate?

  Stannous sulfide ($\ce{SnS}$)- dark brown ppt.

The above resulting solution is added to $\ce{AuCl3}$, resulting in a purple coloured solution. What is the composition of the solution?

  $$\ce{2[AuCl4]- + 4Sn^2+ + 2H2O -> [2Au + Sn(OH)2)](purple ppt.) +
3Sn^4+ + 2H+ + 8Cl-}$$

  (Vogel)

  Note: $\ce{AuCl3}$ exist as $\ce{AuCl4-}$ ions in solution.


The instrument used is called a polarimeter. You basically shine monochromatic linear polarized light through a solution of known concentration and you measure the rotation of the light on the other end of a tube with defined length. You then normalize the rotation according to concentration and length of your tube to get the specific rotation which is then independent from concentration and length and only depends on temperature, solvent and wavelength of the light.

You can determine if it's rotated left or right by changing the concentration since the observed rotation is directly proportional to concentration. You also need to do this to determine if your rotation is more than a full 360°, since a measurement of 10° could also be 370° or 730° or 1090° or also −350°, −710°, and so on.

Let's look at this using a simple example and let us assume our tube has a length of 10 cm (which is the length used for the specific rotation, so we don't need to worry about this). Specific rotation is also normalized to a concentration of 1 g/100 mL.

To determine our specific rotation we could do 3 measurements at 1 g/100 mL, 2 g/100 mL and 0.5 g/100 mL.

1 g/100 mL : +160°
2 g/100 mL : −40°
0.5 g/100 mL : −100°

Now what does this tell us? If the specific rotation would be +160° then double the concentration should be 320° and half should be +80°, so that's clearly not the case. If it would be −200° instead of +160°, then double the concentration would be −400° which would show up as −40° and half the concentration would be −100°. So our specific rotation in this case is −200°.


Do all gases occupy same volume at equal temperature and pressure conditions?

Why do all gases occupy 22.4 L [per mol] at STP?

The question is based on a false premise. Only ideal gases are guaranteed to occupy 22.4 L/mol at STP.  There are many gases that are not ideal.

  So going by this, all gases should occupy same $x$ L at some other temperature-pressure conditions. Is this true?

No, again this would be true only for ideal gases, not all gases.

At STP, many commonly encountered gases on Earth behave nearly ideally, but certainly not all.  For example, vapor of acetic acid and other simple carboxylic acids show strong deviations from ideality.

Wikipedia is a good place to read about the assumptions of the ideal gas law and the kinetic theory of gases.  The reason that ideal gases share a molar volume of 22.4 L at STP is not related to diffusion, but rather to the assumption of: (a) non-interacting gas particles that are (b) effectively zero-sized relative to the mean distance between them.

You can read about equations of state for non-ideal gases on Wikipedia as well.  Popular equations to describe non-ideal gases include the van der Waals equation, the Redlich-Kwong equation, and a whole slew of others listed at Wikipedia's real gas page.


Organic Nomenclature: But-1-en-3-yne or But-3-yne-1-en

The IUCN Blue Book says that:

P-31.1.1.1 The presence of one or more double or triple bonds in an otherwise saturated parent hydride [...] is denoted by changing the ending ‘ane’ of the name of a saturated parent hydride to ‘ene’ or ‘yne’. Locants as low as possible are given to multiple bonds as a set, even though this may at times give ‘yne’ endings lower locants than ‘ene’ endings. If a choice remains, preference for low locants is given to the double bonds. In names, the ending ‘ene’ always precedes ‘yne’, with elision of the final letter ‘e’ in ‘ene’. Only the lower locant for a multiple bond is cited, except when the numerical difference between the two locants is greater than one, in which case the higher locant is enclosed in parentheses.

Therefore the double bond gets priority whatever be its locants (numbering) i.e., '-ene' comes before '-yne'. And in this case it's "but-1-en-3-yne".


Calculating the molarity of DNA in a cell

There is one particle of dsDNA in the cell. Divide by $N_\mathrm{A}$ to get the amount of substance, and divide by the volume to get the concentration matching the concentration given in the answer.
\begin{align}
  V_{\mathrm{cell}}
    &= 4 \cdot {(\pu{0.5 μm})}^{3}\\[0.5ex]
    &= \pu{0.5 μm3}\\[0.5ex]
    &= \pu{5E-16 L}\\[3.5ex]
  N_{\mathrm{A}}
    &= \pu{6E23 1//mol}\\[3.5ex]
  n_{\mathrm{DNA}}
    &= \frac{1}{N_{\mathrm{A}}}\\[0.5ex]
    &= \pu{1.7E-24 mol}\\[3.5ex]
  c_{\mathrm{DNA}}
    &= \frac{n_{\mathrm{DNA}}}{V_{\mathrm{cell}}}\\[1.5ex]
    &= \frac{\pu{1.7E-24 mol}}{\pu{5E-16 L}}\\[1.5ex]
    &= \pu{3E-9 mol//L}
\end{align}


Is it safe to mix calcium carbonate and sodium chloride?

Safe in context of chemistry, physiology or long term health care prevention ?

With zero knowledge, it is safer to stay with commercial products, until you know, what you are doing.

For mouthwash, calcium carbonate is quite ineffective, being insoluble solid. Better would be solution of sodium bicarbonate ( baking soda ), present in some toothpastes as well.

Salt is not great desinfectant and must be used in really high concentration, that can be very disturbing if swallowed.


Balancing by oxidation numbers method multiple atoms

In some (many?) cases you get a more parsimonious description by considering an entire reactant as an oxidant or reductant.  Separating out individual atoms makes you miss the forest because you get confused by the trees.

Take the ammonium perchlorate-aluminum reaction. Since aluminum is being oxidized let us render the ammonium perchlorate as the oxidizing agent, ignoring the fact that only some of its atoms are oxidizing.  You then have a reaction:

$\ce{NH_4ClO_4}\rightarrow \ce{N^0}+4\ce{H^I}+\ce{Cl^{-I}}+4\ce{O^{-II}}$

where the Roman numerals indicate oxidation states in the products.  These add up to $-5$ for all the product whereas the ammonium perchlorate, as a neutral compound, started with the states adding up to zero.  So five electrons must be added to the reactants:

$\ce{NH_4ClO_4}+5 e^-\rightarrow \ce{N^0}+4\ce{H^I}+\ce{Cl^{-I}}+4\ce{O^{-II}}$

Thus although some atoms might be oxidized, others reduced, and still others are just along for the ride, we identify a net oxidizing effect of the ammonium perchlorate as a whole, numerically five electrons per formula unit.

Now we know that the redox stoichiometry will be $3\ce{NH_4ClO_4}+5\ce{Al}$ and we can balance the reaction accordingly.  (We will need to double the coefficients to eliminate a fraction, that's just algebra.)


What is the meaning of the J value in term symbols

For multi-electron atoms where the spin-orbit coupling is weak, it can be presumed that the orbital angular momenta of the individual electrons add to form a resultant orbital angular momentum $\bf{L}$.

Likewise, the individual spin angular momenta are presumed to couple to produce a resultant spin angular momentum $\bf{S}$ . Then $\bf{L}$ and $\bf{S}$ combine to form the total angular momentum $\bf{J} = \bf{L}+ \bf{S}$.


Why is there a doublet instead of a singlet in this Wittig reaction salt configuration NMR?

Phosphorous exists entirely as $\ce{^31P}$, and $\ce{^31P}$ has a magnetic moment of 1/2 (just like a proton). Therefore, a single phosphorous will split any protons coupled to it into a doublet - as you observed. If you looked at the corresponding $\ce{^31P}$ NMR spectrum you would see that the phosphorous signal would be split into a triplet by the two protons.


I think the importance of Bohr model comes from the fact that it's the first atomic model that uses the hypothesis of quantization. More precisely Energy Quantization. It introduces the first quantum number $n$. It allows to interpret the atomic emission spectrum of hydrogen.

So, it is an interesting pedagogic tool to teach students how the atomic models had evolved to arrive to the modern atomic theory (based on quantum physics).

It's also a beautiful example to teach students how empirical science, like chemistry, evolved by making observation (emission spectrum of hydrogen), formulate hypotheses (atomic model based on the model of solar system but with electrostatic forces), testing the model (it works for interpreting the discrete emission spectrum), but it fails to interpret the behaviour of the atom under magnetic field (so, scientists thought about rejecting the model and used another one, the modern model of atoms).


How do you Synthesise Glycosides of flavonoids and stilbenoids?

I suggest to forget about the enzymes.

Pepare for a probably lengthy synthesis - you will need patience and a clever protection/deprotection scheme.

The following is definitely not applicable to all starting materials but might give some directions.

If the flavonoid contains a catechol moiety, just like quercetin (1), the vicinal phenolic $\ce{OH}$ groups can be protected as an acetal by heating the starting material with $\alpha,\alpha$-dichlordiphenylmethane in diphenylether (DOI). The protecting groups is easily removed by hydrogenation over $\ce{Pd/C}$ at 1 atm.

Note that the PMB (4-methoxybenzyl) group is another protecting group for (phenolic) $\ce{OH}$ that gets off by hydrogenation.

As far as selectivities for the protection of the other $\ce{OH}$ groups in 1 are concerned, have a look at the 1962 paper by Leonard Jurd, who described that the methylation of quercetine pentaacetate in dry acetone predominantly furnishes 7-O-methylquercetin. There are quite some more recent articles citing this!


Why are lactate and lactic acid used synonymously in biochemistry?

If you were to produce any lactic acid in the body, it would pretty much immediately get deprotonated, so by right it's always going to be lactate. The $\mathrm pK_\mathrm{a}$ of a typical carboxylic acid is at most 5, and at physiological pH levels (~ 7, give or take) the ratio of $\ce{HA}:\ce{A-}$ is on the order of $1:100$. (As getafix mentioned, lactic acid is somewhat more acidic than usual, because of the electron-withdrawing hydroxyl group, so the ratio would be expected to be even larger. But these are just meant to be very rough numbers, for illustration.)

You could say that it's sloppiness on the part of the biologists, but past a certain point you don't really care whether it's $\ce{HA}$ or $\ce{A-}$, and that's why the distinction isn't really enforced. From a biochemical perspective the more important thing is the oxidation state as well as other byproducts (ATP, carbon dioxide, etc.) Likewise the citric acid cycle is called that despite citrate being the actual compound in the cycle.

And perhaps the most egregious case of sloppiness: DNA stands for deoxyribonucleic acid, but in the body the phosphate groups are deprotonated, so chemically speaking it really should be called the conjugate base of DNA: "deoxyribonucleate". But again, nobody really cares - as long as we know that it has a negative charge, we can call it by a strictly incorrect name and it's fine.

There are probably tons of physicists out there who hate us chemists for using non-rigorous terminology, too!


Whats the difference between reversible process in thermodynamics and reversible chemical reactions?

There is no misunderstanding at all.

The second law of thermodynamics can be roughly expressed as: in a given process the entropy of the universe will always increase (defining universe as system + surroundings).

Reversible processes

For reversible processes the entropy of the universe remains constant. Most natural occurring processes are irreversibles.

A reversible process is a process where you can change its direction by inducing infinitesimal changes to the system. Throughout the entire process the system is at equilibrium, so from one state A to a state B you will not observe macroscopic changes.

The system goes from state A to state B in the following way: A + dA = B where B is infinitesimally different from A. These changes are applied over properties of the system like temperature and pressure by modifying its surroundings.

Reversible chemical reaction

In the case of a reversible chemical reaction it means that if you combine the products of that reaction you’ll obtain at least some of the reactants again. Usually these reactions require the same amount of energy that was consumed or released to obtain the products (This means that you must take into account the enthalpy of the forward reaction).

A typical example are lead batteries where lead reacts with sulfuric acid to form lead sulfate and $\ce{H+}$ ions while the battery is discharging and then you apply a direct current to revert the reaction and obtain lead, sulfuric acid and lead oxide again.

Even when a reaction is thermodynamically possible you must take into account the kinetics parameters. For example: it is thermodynamically possible to turn graphite into diamond but kinetically at standard conditions (1 atm, 25$^\circ$C) it will happen in millions of years.


When pH increases, does the effect of increased [OH-] or reduced [H+] control precipitation of CaC2O4

In the body we'll assume that:

There is some low level of $\ce{Ca^{2+}}$ present in the blood or urine.
The pH isn't nearly high enough to precipitate out $\ce{Ca(OH)2}$.
Under "normal" bodily conditions calcium oxlate doesn't precipitate out either so the overall oxalate concentration is low.

Let's let $x$ be the concentration of all the oxalate species. By knowing $x$, the pH and the two pka's we can calculate the exact concentrations of all three oxalate species.

$x = \ce{[HOOC-COOH] + [HOOC-COO^-] + [^-OOC-COO^-] }$

Now the pKa values for oxalic acid are far enough apart that at most two of the three oxlate species will have an appreciable concentration.
For normal bodily functions only $\ce{[HOOC-COO^-]}$ and $\ce{[^-OOC-COO^-]}$ will be the only significant species. Since only a couple of significant figures will be of interest, we can use the Henderson–Hasselbalch equation to calculate the relative concentration of the two species to reasonable accuracy. (The essence is that the third species has a negligibly small concentration.)

pH = $\text{pK}_{a2} + \text{log}(\dfrac{\ce{[^-OOC-COO^-]}}{\ce{[HOOC-COO^-]}})$

Thus as the pH increases relatively more $\ce{[^-OOC-COO^-]}$ is formed and less $\ce{Ca^{2+}}$ will be soluble.

But normal blood is alkaline already. Thus there would be a negligibly small amount of $\ce{[HOOC-COO^-]}$ in the blood to start with since the $\text{pK}_{a2} = 4.2$. So calcium oxalate will be unlikley to ppt in the blood.

However the kidneys make urine which has a pH from 4.5 to 8. So if the pH in the urine changes as much as from 4.5 to 8 then there could be a greater tendency to ppt calcium oxalate in the urine.

ANSWER

(4) is out since if body was that alkaline your tissue would dissolve.

(3) is out since you can see from the HH equation specified that the presence of another buffer would effect the absolute pH but not the relative oxalate species at that pH.

(2) is true, but the answer doesn't note any effect of pH. So the answer doesn't really address the problem.

Thus the answer must be (1).


Hydrochloric acid is an aqueous solution of hydrogen chloride.


Energy levels transitions of an electron?

If I remember correctly, you can use the equation:

$$E = \frac{hc}{\lambda}$$

and

$$\nu = \frac{c}{\lambda}$$

to solve for the frequency, wavelength, or energy of the photon. Note that $E$ is energy, $h$ is Planck's constant, $c$ is the speed of light, $\lambda$ is the wavelength, and $\nu$ is frequency.

The energy, (actually 11.97eV not 11.2eV)converted into joules (2.719 $*10^{-18}$ J) and then put into the first equation:

1.918 $×10^{-18}$ J = $\frac{6.626× 10^{-34} × 2.998 × 10^7 m×J}{\lambda}$

then lambda can be solved for:

$\lambda$ = $\frac{6.626× 10^{-34} × 2.998 × 10^7 m×J}{1.918×10^{-18} J}$

$\lambda$ = $1.036×10^{-7} m $

= 103.6 nanometers.

This solves for your wavelength like you asked for.


Don't overthink it.

The oxygens have a partial negative charge in nitrate from resonance.  The protons in water have a partial positive charge from dipoles.  Positives attract negatives.

Molecule polarity becomes an issue when discussing neutral molecules interacting with neutral molecules.  Molecular polarity is one thing, and bond polarity is another.  Transformations / arrangements / alignments of molecules such that the dipoles (the things that make a molecule or bond 'polar') align to maximize interaction are certainly important for dissolution, but when a species is a bona fide free ion, water will tend to dissolve it.


Can all species that undergo aldol reactions also undergo Cannizzaro reactions?

No. The Cannizzaro reaction only works for aldehydes with no $\alpha$-protons, benzaldehyde being a classic example.

For a start, ketones and any other species that can form stable enolates, such as esters or carboxylic acids, can undergo the aldol reaction or similar reactions like the Claisen condensation. However, the Cannizzaro reaction is limited to aldehydes because only they can disproportionate to form alcohols and carboxylic acids.

Secondly, the Cannizzaro reaction involves strongly basic conditions, typically using sodium or potassium hydroxide. Under these conditions an aldehyde with $\alpha$-protons will undergo deprotonation to form an enolate much more readily than nucleophilic attack on the the acyl carbon by a hydroxide ion.


The differences in acceleration due to gravity is not the main factor in comparing how accurate the approximation is for each planet.

The main factor is the mass of gas each planet's atmosphere contains.

Mercury has almost no atmosphere.  The total mass of all gas in Mercury's atmosphere is only 10000 kg! The pressure is less than $10^{-14}$ bar. The approximation will be very accurate.

The mass of Venus's atmosphere is $4.8 \times 10^{24}$kg and pressure is 92 bar, so PV = nRT will be much worse of an approximation than on earth.

Mars has a small atmosphere, less than Earth more than Mercury.

Jupiter, Saturn, Uranus and Neptune are gas giants with no distinct surface.


Ok.  You have five equations, two are differential equations and three are algebraic.  And you have the explicit assumption that the equilibria are fast enough so that the two equilibrium equations are always satisfied.

You also have five unknowns, A, B, C, AB, and BC.  One method of solution is to use the algebraic equations to eliminate three of the variables.  This should leave you with two differential equations with two unknowns.  If you are very lucky (I've not worked out the algebra) you will get two equations that can be integrated independently.  More likely you will have two simultaneous differential equations.

In either case the result is apt to be a mess.  So I'd cheat.  If this is a textbook problem or homework problem, there is probably a way to combine the equations to get simple enough expressions.  Otherwise, you may have serious troubles solving the set.


Why might the O—H stretch in the IR of MY triphenylmethanol product be less intense than expected?

A possible explanation would be the concentration of sample you used. Of course, if you are confident that you had utilized sufficient concentration then there may be another reason for the decreased intensity. However, some of the other peaks are also somewhat decreased in intensity so it is a possibility.

This point is noted here (http://chemwiki.ucdavis.edu/Physical_Chemistry/Spectroscopy/Vibrational_Spectroscopy/Infrared_Spectroscopy/Infrared%3A_Interpretation#The_Origin_of_Peak_Intensities)


Taiwan propylene explosions one week ago

Propene gas is flammable and has a higher density ($1.81 \ \mathrm{kg/m^3}$) than air. It therefore easily forms explosive mixtures with air when it accumulates near ground level or, like in this case, inside an underground channel without much ventilation. While propene itself is color- and odourless, it is described in this article that "eyewitnesses reported a smell of gas (...) coming out of manholes". This smell comes from volatile compounds which are added to the gas in small amounts to make the detection of leaking gas easier. However, by the time that the smell was noticed, enough propene had already accumulated underground to exceed explosion limits.

Detection of underground gas leaks by odour alone is therefore not a reliable means to prevent possible explosions, namely when the gas is heavier than air. It is also not possible to exactly localize the leak when the gas escapes from multiple manholes which are spread over a large area. Gas sensors in the underground pipeline network could provide earlier warnings about leaks, and gas accumulation could be prevented in this case by shutting down the affected pipeline. However, prevention of leaks in the first place (by regular inspections) should still have higher priority.

The MSDS data sheet for propene states that shutting down gas supply and letting the fire burn out while cooling the surroundings with water is the recommended way of extinguishing burning propene. So using water to cool down the roads with the aim to reduce ground temperature was not completely off, but trying to fight the fire directly with water is unadvisable.


Reasons for dry conditions in substituition reactions with benzene

Electrophiles
are reactive cations, they will react with water to form
eg an alcohol (or some weak acid etc)  and protons.
No reaction with benzene.


Is the hydrogenation of carbon dioxide to methane a solution or a problem?

The problem is that the carbon dioxide which is released from burning is often mixed with other things which include air. If we were to want to convert carbon dioxide into methane then the oxygen of the air would make it harder.

While one option might be chemical looping combustion to obtain a purer carbon dioxide stream we still have the problem of where would we get hydrogen from. The normal method in industry to get the hydrogen gas needed for things like hydrogen peroxide production is to use steam reforming of methane.

$\ce{H2O + CH4 -> CO + 3H2}$

Using the water gas shift reaction the carbon monoxide is converted into carbon dioxide and hydrogen (using more steam). My worry about an industrial process for the conversion of unwanted carbon dioxide into methane is that unless we use hydropower, solar power or nuclear power to obtain electricity for the production of hydrogen from water then we will never be able to make a net reduction in the amount of carbon dioxide as the process will form as much carbon dioxide as it consumes.

There is some electrochemistry in which carbon dioxide can be converted into other things by reduction at the cathode but I think that this is currently only at a very small scale.


Is there an easy way to remember charges on ions?

I don't believe there is any simple way to determine what they are except by just plain memorization. I have given some tricks below that have helped me.

When I write down my poly-atomic ions, I generally put them in similar groups in a certain order. From least amount of oxygen to most amount oxygen.

hypo (name of ion) ite -> least oxygen
(name of ion) ite -> one more oxygen than above
(name of ion) ate -> one more oxygen than 'ite'
per (name of ion) ate -> one more oxygen than 'ate'

Understanding the above helps most with 'ite' and 'ate'. 'Ate' generally being 1 more than 'ite' for oxygen.

Then I list them from least charge to most charge. I order them this way:

Nitr ite $\ce{NO2-}$
Nitr ate $\ce{NO3-}$
Sulf ite $\ce{SO3^2-}$
Sulf ate $\ce{SO4^2-}$

You can see that 'ate' has one more oxygen, and that I have listed them in charge order. Sometimes looking at the trends in oxygen and charge will help you memorize. I have had to remember them by writing them down several times and now I am comfortable.

Here is another example: Hypochlorite is
$\ce{ClO^-}$

So, chlorite should be?

   $\ce{ClO2-}$, because it has one more oxygen, and

Chlorate should be?

   $\ce{ClO3-}$, because it has one more oxygen than the above, and

$\ce{ClO4-}$, will be called?

   Per chlor ate, because it has the maximum amount of oxygen.

The above is not a hard and fast rule. One exception I can think of is manganate and permanganate.

Manganate is $\ce{MnO4^2-}$ but permanganate is $\ce{MnO4^-}$; same amount of oxygen, just a different charge.

Trying to figure out any of the above from a periodic table didn't make a difference to me because my professor tests us without it. Yikes!


This compound (formally called "hexadecahydronaphtho[1,8-de]naphthalene", but in common parlance also "[6.6.6.6]fenestrane" - thanks Mithoron!) has never been actually synthesised before, but some of its properties have been studied computationally.

It does not particularly matter, but it is worth pointing out that this compound possesses stereochemistry which was not indicated in your original drawing. Hence your question is essentially asking about a collection of stereoisomers.

Gund and Gund1 investigated the strain energies of a few $\ce{C17H28}$ isomers, shown above, using empirical force field methods. The compound you are interested in is 4b in the scheme (the various 4a are constitutional - as opposed to configurational isomers - of 4b). The numbers next to the names are the strain energies in kcal/mol. The higher the energy, the more strained; hence it seems that 4b, if synthesised, would be generally thermodynamically unstable with respect to 4a (kinetic stability is of course entirely separate). The letters 'c' (chair), 't' (twist boat), and 'b' (boat) refer to the conformations of each six-membered ring in the structure. The labels cis and trans refer to the relationship between each neighbouring pair of fused six-membered rings.

The potential for systems containing fused six-membered rings to undergo "ring flips" was also investigated computationally (using DFT) in a recent paper by Alder et al.2 In particular, the authors were interested in the all-cis isomer of your compound, i.e. ccct/4b6 in the screenshot above - which incidentally has a rare $S_4$ symmetry. They found that inversion has a whopping barrier of +88.9 kJ/mol (B3LYP-gCP-D3/6-31G*//B3LYP/6-
31G*) - almost double that of cyclohexane (~45 kJ/mol). The authors conclude that this isomer of [6.6.6.6]fenestrane "probably has the highest inversion barrier for any nonpolymeric multi-ring structure built from cyclohexane rings".

Although the compound has never been made, the authors propose that at high temperature the inversion process (which interconverts equatorial and axial hydrogens) may be studied by NMR spectroscopy. However, due to the significantly slower rate of inversion relative to cyclohexane, temperatures of 100–200 °C would be needed (as compared to cyclohexane which requires low temperatures for NMR studies).

References

Gund, P.; Gund, T. M. How many rings can share a quaternary atom? J. Am. Chem. Soc. 1981, 103 (15), 4458–4465. DOI: 10.1021/ja00405a028.
Alder, R. W.; Butts, C. P.; Sessions, R. B. Perhydrohelicenes and other diamond-lattice based hydrocarbons: the choreography of inversion. Chem. Sci. 2017, 8 (9), 6389–6399. DOI: 10.1039/C7SC01759F.


How to determine whether paper was dipped in acetone?

If significant amounts of acetone are left (according to my old lab manual, the analysis works with a 50 mg sample), you could use the reaction with 2-nitrobenzaldehyde in the presence of sodium hydroxide, which generates indigo dye.

See also: Baeyer–Drewson indigo synthesis


How are samples prepared for analysis with ICP-OES? Can it analyse gases?

About the measurement of solids

Injections tubes are indeed often made out of fused quartz. Dealing with aggressive solvents is actually not that much of an issue because of the extremely low amounts of material that can be detected with ICP-OES. Therefore, you can simply dilute your aggressive solvent before use. You can look for example at this research note where they dissolve a sample in $\ce{HCl}$ and $\ce{HNO3}$ and then dilute it before use.

About the measurement of gases

Yes, it is possible to measure gases directly, but it is not recommended and some adaptations to the normal procedure are needed. In this research paper they investigated aerosol analysis by ICP. The main issue was keeping sampling lines at elevated temperatures to allow proper calibration. A secondary problem arose because their target analyte, metallic compound in flue gases, which are volatile at (or close to) room temperature, are typically highly unstable or explosive. Therefore, they recommend dissolving areosols in a liquid solution of water and some organic solvent (1-methoxy-2-propanol in their case) and use this as feed.

For the measurement of 'normal' gases it seems that the problems they encounter will not occur, because you don't need elevated temperatures in your calibration. So in principle that means that you can indeed measure air. It should be noted that in some machines a shear flow of dried air or nitrogen is used to cut the plasma (see wikipedia) which will disallow measurement of air, as this Perkin Elmer instrument (overview table in slide 7 of this presentation). Some other instruments, like the one in the research paper I referenced, do not use such a shear flow and will thus be able to detect air.

Gas samples are not at all 'too thin', because the normal nebulized feed that goes into the plasma is similarly dilute.


What chemicals are used in receipt paper?

Have a look at wikipedia article on Thermal Paper. The whole trick is in the three components, which are mixed as solid substances and applied as coating to the paper. When you heat the paper, the solid substances melt and the coloring reaction occurs.

For this you need fine pulverized chemicals, stirred in a form of suspension in some volatile solvent and apply them on the canvas. It should not be a big problem, see e.g. patent on Thermal Paper.

If all this effort is for an artistic purpose, I would suggest as a first try to approach the companies manufacturing the thermal paper (there should be lot of them) and ask them for a bottle of the prepared suspension, telling them your plans. The success rate would not be high, but hopefully non-zero and it saves you 99% of work.

Additionally, the painting itself could be quite fun, I would expect the components to be UV-active, so the black light could help you.


Try Open Babel, the windows GUI has a lot of features that may be of help depending on what are those specifics you need in your Z-matrix.


On the use of hexane in the organic reduction by DIBAL-H

Unlike other reducing metal hydrides (e.g., $\ce{NaBH4}$ and $\ce{LiAlH4}$), diisobutylaluminum hydride (DIBAL-H) is a liquid at room temperature and dissolve in many hydrocarbons such as toluene and hexanes, which also have very low freezing points. For example, hydrocarbons toluene and hexanes both have freezing points around $\pu{-95 ^\circ C}$. Thus, they are good solvents for reaction involving reduction by DIBAL-H at low temperatures such as the example given in the question ($\pu{-70 ^\circ C}$). As shown in the equation, the low reaction temperatures help stop the reduction at aldehyde level. At room temperature with excess DIBAL-H, the reduction may continue further to give corresponding alcohols or even corresponding alkanes.
DIBAL-H reduces alkyl or aryl nitriles to their corresponding imines at low temperatures and resultant imines converted to corresponding aldehydes upon acid work-up. The same principle mentioned above apply here as well. Here, at high temperatures, imines reduced further to give corresponding amines.
Note: To see the mechanism of the reduction, see here.


Open resource for spectroscopic raw data

As a quick fix: To access the IR spectral data of $\ce{HCl}$, I opted for a seach by (Hill) formula  on NIST's web page, which yielded three entries here.  For $\ce{HCl}$ in particular (here) I opted for the first one here, indeed offering the spectrum as image file or as JCAMP-DX.

As in the comment by @EdV, this JCAMP-DX an ASCII file indeed.  Note JDXview by Norbert Haider (Department of Drug Synthesis, University of Vienna, Austria) is a very handy program (only 560 kb, installation free executable) for a quick display of this file by NIST. Via File -> Save, the export of the data in a two-column, tabulator separated .txt file is just a click away.  The program works perfectly well in Windows, and with wine libraries equally in Linux, too.  (The author equally distributes the underlying Delphi / Pascal code.)

Note 1: Because of the tags physical chemistry, and spectroscopy, you may be interested in a python module like jcamp to access either IR, or NMR spectra in the JCAMP-DX formats (there are multiple).

Note 2: Sometimes, all you have indeed is an image (a scanned) spectrum.  If you pay attention and proceed in sections (e.g., $4000 - 2000$, then $2000 - 800\,\pu{cm^{-1}}$) to account for the break on the abscissa in IR spectroscopy, a digitizer like WebPlotDigtizer may be worth to use.  If you want, the program works via a web browser, installation free, with «normal image files» like .png, and since some time with images embedded in a .pdf (think about publications), too.


Nitrogen is equally electronegative as chlorine but also more compact.  Whether you consider hydrogen bonding as purely electrostatic or having a covalent character (The lone pair on atom X overlaps the molecular orbital in the H-Y bond of an adjacent molecule, thus partially delocalizing the bond), the more compact structure of nitrogen offers an advantage.


There are multiple reasons:

To improve our experimental prowess: There's no single catch-all process to create higher elements. While the basic process is to smash stuff together, once can't get much more specific than that. For each new element, the techniques used differ. And, as with any experiment, the complications which arise differ. Solving these lead to a deeper understanding of how to conduct experiments. Note that there have been gaps of years between the synthesis of various transuranium elements. This is not because scientists find it a useless task; it is because designing an experiment to synthesize these elements is hard.
To test stability: We have various nuclear models that can predict the stability of the elements. Creating new atoms helps validate these theories.
Islands of stability: Some higher elements are considered islands of stability in that they are more stable than their neighbors. If we can improve our experimental techniques to the extent that these can be generated, we may find some use for them. Of course, these may or may not be useful — these still are short-lived, just not extremely short-lived
Because we can!


Are vitamins in tablet are more efficient than the vitamins in their natural source?

For some vitamins, such as Vitamin C, which are always found in the form of a unique chemical substance, there is no chemical difference between the vitamin in a tablet--synthetic or otherwise--and the same vitamin as found in a foodstuff. Still, how well you absorb a vitamin might be affected by what else you consume at the time - for example a given amount of Vitamin C might be absorbed more or less efficiently as an ingredient in orange juice than as a pill, even though it is the same substance in both cases. But these differences might not necessarily be significant.

Other vitamins, such as Vitamin A or Vitamin D, are found in multiple closely related chemical forms, any of which the body can use to fulfill its need for the vitamin. In that case, the body may absorb or utilize the vitamin more efficiently when it is in one form relative to another. Fats in the diet can further influence your absorption of a fatty vitamin like Vitamin D.

Minerals, such as iron, are notorious for being made easier or harder to absorb depending on their chemical form and how they are bound to other components of the food. So, for example, iron in greens is known to be slower/more inefficient for the body to extract than iron in liver.


Are there safe inks that allow one to eat a fish after painting it?

I Googled "edible ink" and found this item, which is a set of edible ink-containing pens.
The keyword is "edible." I'd stay away from anything not specifically labeled "edible."
Finally, it is safe to eat fish adorned with ink specifically certified to be safe to eat.
Disclaimer: There might still be some issues with the fish, so keep your eye/nose on that.


How many electrons can have the quantum number set n=5, l=3, ml=-1?

This post answers the question of the of what the other quantum numbers mean:

What do the quantum numbers actually signify?

As for how many electrons can have $n=5,\ l=3,\ m_l=-1$...

The Pauli Exclusion Principle states that two identical fermions (an eletron is a type of fermion) cannot occupy the same quantum state.

  In other words, two electrons (in the same atom) cannot have identical quantum numbers.

For each $n$, the values of $l=0,1,2,...,n$
For each $l$, the values of $m_l=-l,...,-1,0,+1,...,+l$

For each $m_l$ there are only two allowed values of $m_s$, namely $m_s = + \frac{1}{2}$ and $m_s = -\frac{1}{2}$. Thus, each set of $\{n,l,m_l\}$ can only describe two electrons. In your case, we have two electrons with $n=5,\ l=3,\ m_l=-1$:

$\{n,l,m_l,m_s\}=\{5,3,-1,+\frac{1}{2}\}$
$\{n,l,m_l,m_s\}=\{5,3,-1,-\frac{1}{2}\}$


How does fluorouracil inhibit thymidylate synthase?

The inhibition comes about because:

Uracil and $5$-fluorouracil are sufficiently similar that both moieties have a strong binding affinity for the active site of thymidylate synthase.
The presence of the fluorine prevents $5$-methylation of the uracil core by a folate cofactor, which otherwise would turn it into a thymine moiety.
This inability of the enzyme to convert the $5$-fluorouracil moiety into a thymine moiety (which has a much lower affinity for the active site) means that the enzyme has no way to reject the $5$-fluorouracil or the methylating folate species, once bound.

The particulars of this interaction are nicely spelled out in Mishanina et al.$^\dagger$, conveniently available in preprint:

  [5-fluorodeoxyuridylate] reacts [as normal] with the active site cysteine and undergoes condensation with [N,N-methylene-tetrahydrofolate] but prevents proton abstraction and elimination of the tetrahydrofolate [product], as a consequence stalling the reaction at the covalent ternary complex [of enzyme, inhibitor, and folate species].

Scheme 4A from the preprint linked above illustrates well the specifics of the chemistry.  (I refrain from reproducing it here due to uncertainty regarding its copyright/license status.)

As well, Figure 2 from the Mechanism Description section of the Wikipedia page on thymidylate synthase shows very nicely the enzyme interactions with the normal deoxyuridylate substrate (click image for larger version):

$\hspace{20pt}$
Image by CJerc (Own work) [CC BY-SA 3.0 (http://creativecommons.org/licenses/by-sa/3.0 )], via Wikimedia Commons

$^\dagger$ Mishanina et al. Bioorganic Chemistry 43: 37 (2012). doi:10.1016/j.bioorg.2011.11.005


Why Benzene is a good solvent even while it is not polar

Benzene is a good solvent for many organic compounds, the term good solvent to me does not mean a nice solvent with a good all round profile.

To me the perfect solvent would be non toxic, not flammable but combustable, very resistant to radiation, inert to nitric acid, be easy to distill, be cheap and easy to obtain. It would also have to be able to dissolve a useful range of different things.

Benzene is more able to dissolve many organic substances than a saturated hydrocarbon such as hexane, this is becuase the pi cloud of the benzene ring is able to interact with the pi clouds of various solutes. For example polystyrene will be solvated better by benzene than by hexane for this reason.

This good ability to dissolve things is why I regard benzene as a good solvent. It is a good solvent to dissolve things but often a bad thing to work with in the lab. Benzene has one special use, as a NMR solvent. One of my favourite tricks is if my student can not understand a proton NMR spectrum due to overlapping peaks. I tell them to repeat the measurement in d6 benzene. Often a lot of the chemical shifts for protons change, with a little luck the resulting spectrum is more easy to understand. Sometimes you need to look at both the chloroform and benzene solution NMR spectra to be able to understand enough of the spectrum to understand what your structure is.

While benzene is an unpopular solvent due to the health issues and flammability, there are plenty of aromatic solvents which have better health and fire profiles. For example Solvesso 150ND is an aromatic kerosene which is marketed by Exxon, this has different solvent properties to the aliphatic saturated kerosene marketed by Statoil under the name of "solvent 70".

This can have some advantages, if I want to dilute the ionic liquid aliquat 336 to 30 % in a diluent and then use this solution to extract metals such as copper, cobalt and iron from aqueous chloride media. Then with solvent 70 as the diluent (diluting agent) then the organic phase is very prone to separate into a layer of ionic liquid with metal complexes as the anions and a diluent rich layer. But when an aromatic solvent such as ethyl benzene or solvesso 150 is used then the organic phase can hold a lot more metal before this unwanted phase splitting occurs.

The aromatic solvent (diluents) tend to be more toxic than the saturated aliphatic (alkane) diluents. If you look at "A comparison of two methods of recovering cobalt from a deep eutectic solvent: Implications for battery recycling" by F.J. Albler et. al., Journal of Cleaner Production, 2017, volume 167, pages 806-814 it contains an assessment of leaving a puddle on the floor of an unventillated room is made for both solvent 70 and ethyl benzene / toluene is made. The solvent 70 level in the air is likely to pose a far smaller threat to health than the aromatics. But aromatics like solvesso 150 are likely to pose a far smaller carcinogenic threat than benzene.

It is important to not regard all alkanes or aromatics as being the same, hexane is a popular solvent in chemistry but it is very toxic. In the body it is converted to hexane2.5-dione which is very bad for you. But the other longer alkanes do not have the same harmful effect. You also need to consider the exposure route. For example a puddle of diesel fuel on the floor of a garage is far less harmful than exposure of a person's hand to a high pressure spray of diesel which injects it into the person's body.


How to use salt spray data in real environments?

I have to agree with Klaus.  ASTM B117 is, at best, a QC procedure.  In my experience, most corrosion and coatings scientist who use B117 do so because it is very well established, but it does not correlate to a real-world measure of performance.  ASTM D5894 is a better cabinet test and there is also an SAE procedure developed for automobiles.


Water electrolysis - what is happening to an iron anode?

On an anode made of iron, some of the removed electrons may be used oxidizing iron metal to ferrous iron(II) ions, or oxidizing iron(II) ions to iron(III), rather than bubbling off oxygen gas.  Pushing more current through the cell is likely to result in gas evolution alongside the iron oxidation, since all three of these processes will run simultaneously.

$$\ce{Fe(metal) -> Fe^{2+}(aq) + 2 e^- }$$
$$\ce{Fe^{2+}(aq) -> Fe^{3+}(aq) + e^- }$$
$$\ce{2H_2O(liquid) -> O_2 (gas) + 4H^+ (aq) + 4e^- }$$

Ferrous iron(II) ion solutions are a pale green, but rapidly oxidize in air to vivid red-brown ferric iron(III).  Iron oxides and hydroxides form precipitates and slimes in a variety of colors including black (magnetite), red (iron(III) oxide), and yellow (iron(III) hydroxide).


I assume you are aware of the common clock-reactions, such as the iodine clock, etc. By varying concentration of reagents and pH of the solution, you could run several repetitions to generate a trend-line, and from that extrapolate the estimated pH and reagent concentration to provide you with the time delay you are looking for. The iodine clock is a two-step reaction as outlined below:

In the first, slow reaction, iodine is produced:

$$\ce{H2O2 + 2I- + 2H+ -> I2 + 2H2O}$$

In the second, fast reaction, iodine is reconverted to 2 iodide ions by the thiosulfate:

$$\ce{2S2O3^2- + I2 -> S4O6^2- + 2I-}$$

By decreasing pH in reaction 1 and increasing the thiosulfate level in reaction 2, you should be able to lengthen the ‘clock time’. Let me know how things turn out!


Reading number of outer shell electrons and other properties from periodic table?

Start with a periodic table that shows the electron configurations. You might try the Dynamic Periodic Table; as you mouse over an element, its electron configuration is presented.

Then go to Electron Configurations in the Periodic Table, Electronic Structure of Atoms or Block (periodic table) to see how shells and orbitals are added. It's pretty straightforward until the transition metals, and things get much more complicated for the lanthanides and actinides, where a new shell may start before an inner shell is complete.

The general rule is: an orbital is more "stable" when filled, or half-filled, so fluorine, for example, can readily accept one more electron to finish its outer 2s2 2p5 shell and it will then have the configuration of neon, 2s2 2p6. Many periodic tables don't present the inner shells, using instead the shorthand of [Ne] 3s2 3p5 for chlorine, rather than the full 1s2 2s2 2p6 3s2 3p5.

In fact, you're not alone in trying to understand the electron structure; see the current debate on the position of Lw in the table: New data on synthetic element trigger rethink of periodic table.

BTW, note there are different ways a shell can be completed. Carbon will share its four outer electrons with four hydrogen atoms; this is called a covalent bond. Chlorine outright "steals" an electron from sodium; this is an ionic bond. See Covalent Lewis Dot Structures and Ionic Bonding and Electron Transfer for more info.


Why is a solution of PCl3 more acidic than MgCl2?

There is no such thing as a water solution of $\ce{PCl3}$. It reacts with water, and quite vigorously. Once the reaction is over, there is no more $\ce{PCl3}$ in the solution. Arguably, there is no $\ce{MgCl2}$ in the other solution either, but it would form (as a hexahydrate) if you evaporate the solution; $\ce{PCl3}$ would not.

The reason for this is that $\ce{Mg}$ is a metal and $\ce{P}$ is a nonmetal. They are different elements, about as different as two elements can be. They have nothing in common, except for a few sparse random traits, like being able to burn brightly in air when ignited. No wonder they are different in this regard as well. $\ce{MgCl2}$ hydrolyzes only very slightly (say, one part in a million). $\ce{PCl3}$ hydrolyzes completely.


Recommendations for electrostatic potential surfaces

You have probably found out from playing around that there are many options. For coloring there is no standard, blue and red are common because they are the easiest to see since they contrast well. As far as the density cutoff, there are some papers that recommend values but the main idea is to choose a value that gets close to the Van Der Waals radius for the atoms. This is after all what is typically is of interest in a chemical sense. The electron density at the Van Der Walls surface is what others atoms/molecules "see". Typically I will just plot both the Van Der Waals and electrostatic surface and modify the electrostatic cutoff to make them close. Then you can set the limits for positive and negative colors.


Is propane-1,2,3-tricarbaldehyde the correct IUPAC name of CHO-CH2-CH(CHO)-CH2-CHO?

From the Nomenclature of Organic Chemistry: IUPAC Recommendations and Preferred Names 2013, let's cite one of the rules for usage of the -carbaldehyde suffix, the relevant one:

  P-66.6.1.1.2 The suffix ‘carbaldehyde’ is used when more than two $\ce{-CHO}$ groups are attached to an alkane.

  Example:

  $\ce{OHC-\overset{1}{C}H2-\overset{2}{C}H(CHO)-\overset{3}{C}H2-\overset{4}{C}H2-CHO}$
  butane-1,2,4-tricarbaldehyde (PIN)

Thus the correct name is propane-1,2,3-tricarbaldehyde.


how to separate gamma-butyrolactone from N-methyl-2-pyrrolidone

Cud [sic] a person wanting to separate these just emerse [sic] them into liquid nitro [sic] or any [sic] to get the desired chemical to freeze then filter off the unfrozen butyrolacton [sic]?

With liquid nitrogen this would not be possible because liquid nitrogen condenses/boils at -195 °C.  It would freeze both the pyrrolidone and the lactone.


How to find the volume of oxygen release by the decomposition if potassium chlorate?

1 kg for 1 liter is, in fact, liquid water at room temperature.

You need to use the ideal gas law $\ce {PV=nRT}$.


How to calculate the concentration of lead(II) liberated from lead(IV) oxide under acidic conditions?

I think that the problem here is one of notation. I'm pretty sure that the logarithm intended to be used is the natural logarithm, $\log_e(K)$ also represented as $\ln(K)$, and not the base 10 one. Otherwise your method is good.

If you use this, you get a much more reasonable answer (work it out). It also makes sense that the question was worded in this way, as the equations in which one tends to use the equilibrium constant $K$ actually use its natural logarithm, for example the Gibbs's free energy and the Nernst equation.

By the way, this reaction is also part of the reduction process at the cathode of a lead-acid battery.


What exactly is peptone (tryptone?) and why does it not get precipitated?

Sigma-Aldrich has listed long list of their Basic Ingredients of Microbiology, which are called Protein Sources (Peptones):

Peptones and extracts are excellent natural sources of amino acids, peptides, and proteins in growth media. They are most often obtained by enzymatic digestion or acid hydrolysis of natural products such as animal tissues, milk, and plants or microbial cultures. The number of available peptones and extract is enormous, and can promote and sustain the growth of most organisms.

Some protones are animal-based and some are plant-based. For instance, peptone water is a microbial growth medium composed of peptic digest of animal tissue and sodium chloride. The $\mathrm{pH}$ of the medium is $7.2\pm 0.2$ at $\pu{25 ^\circ C}$ and is rich in tryptophan. Peptone water is also a non-selective broth medium which can be used as a primary enrichment medium for the growth of bacteria (Ref.1).
References:

J.E.L. Corry, G.D.W. Curtis, R.M. Baird, Eds, Culture Media for Food Microbiology, Volume 34; 1st Edition, Elsevier Science: London, UK, 1995 (ISBN: 9780444814982).