Bismuth can have any oxidation state from +1 to +5 . As it is in the same group as nitrogen and phosphorous, its most commom oxidation states with strongly oxidizing substances such as halogens and oxygen are +3 and +5.

However with a strongly electopositive (reducing) element, bismuth can have an oxidation state of -3, just as nitrogen can. While it is true that the oxidizing strength reduces as we go down the group, the "softness" (tendency to of an element to adapt its oxidation state to the preferred oxidation states of other elements) goes up.

Without definite information I can't discount the possibility that $\ce{Mg3Bi2}$ is an alloy, but the fact that it has an even stoichiometric ratio of elements suggests that it is in fact a compound in which magnesium has an oxidation state of +2 and bismuth has an oxidation state of -3.


How to determine if two calibration curves are parallel, within 5%

Define the angular coefficients of the two resulting lines and evaluate the parallelism between them. Perform the statistical evaluation of the data.
For the parallelism test between the curves, considering the equations of the lines to be compared as: $y_1 = a_1 + b_1x + \varepsilon$ and $y_2 = a_2 + b_2x + \varepsilon$, the hypothesis for comparison of the lines slopes is given by:
$$H_0 : b_1 = b_2$$
$$H_1 : b_1 \neq b_2$$
When the null hypothesis ($H_0$) is true, the regression lines become: $$y_1 = a_1 + b_3x + \varepsilon$$
$$y_2 = a_2 + b_3x + \varepsilon$$
Considering a significance level of $5\ \%$, the null hypothesis will be true when the test returns a P-Value above $0.05$.

More information in the link:
http://www.real-statistics.com/regression/hypothesis-testing-significance-regression-line-slope/comparing-slopes-two-independent-samples/


A simple Google search will find you a number of informative articles.

I went and looked at one of the links and it was an abstract which I will quote below in full.  It looks like a possible method though I strongly advise against proceeding without a lot of forethought into pollution control measures.  There will be various lead compounds formed and keeping the dust and products out of the air and environment will not be easy.  As a business case this is going to be irrelevant soon as the number of CRT displays being recycled is decreasing continuously as 2nd generation LCD monitors have already reached end of life, only a small percentage of industrial control equipment still operates on CRT monitors and they will slowly be recycled over time, the big gold rush is over.

Here is a video from SWEEEP KUUSAKOSKI showing their industrial processing system, the method is not described but this started operation in 2013. The same time as the paper below was published.

Another video by Green Lyon describes a patented process to extract the lead from glass using an aqueous phase process at moderate temperatures with non-scary reagents.  Not sure if this has been reproduced elsewhere, this video was also published in 2013, a good year for CRT glass.

Extraction of Metallic Lead from Cathode Ray Tube (CRT) Funnel Glass by Thermal Reduction with Metallic Iron
Xingwen Lu†, Kaimin Shih*†, Chengshuai Liu‡, and Fei Wang†
Environ. Sci. Technol., 2013, 47 (17), pp 9972–9978
DOI: 10.1021/es401674d
Publication Date (Web): August 5, 2013

"A novel and effective process of thermal reduction treatment with the addition of metallic iron (Fe(0)) to recover lead from cathode ray tube (CRT) funnel glass is introduced. The key technological breakthrough of this process is the use of a relatively lower temperature and an inexpensive reducing agent to extract the metallic lead. The influences of temperature, the reducing agent content, and the holding time for lead reduction were examined to determine the optimal extraction efficiency. The lead extraction efficiency first increased and then decreased with increasing temperature. The maximum lead extraction efficiency occurred at 700 °C. The growth of crystalline lead first increased significantly with an increase in the Fe content, reaching maximum growth at an Fe addition of 50 wt %. The most effective treatment time was determined to be 30 min, as the vitrification of lead back to the glass matrix occurred under longer treatment times. The experimentally derived results indicate that a 58 wt % lead extraction can be achieved with the optimized operational parameters (50 wt % Fe addition, heating at 700 °C for 30 min) in a single extraction operation."


Why do we use molality on describing colligative property?

It is much easier to work with molality than molarity when measuring freezing point depression. Imagine. You weigh $10$ g, or $20$ g ice (or any other amount); you add $1$ g salt, $2$ g salt, or any other amount of any sort of solute. You drop a thermometer in the mixture, and you measure the fusion temperature. It is done quickly, and the scientist does not mind about the molarity of the obtained solution.


Let's see if a picture can help.  Newman projections can be handy in analyzing the stereochemistry of allenes.  On the left is a Newman projection of methyl allene (buta-1,2-diene).  There is a sigma plane of symmetry that is perpendicular to the screen and contains the 3 allenic carbons plus the methyl group.  Any molecule with a plane of symmetry is achiral.  On the other hand, look at the Newman projection to the right, 1,3-dimethyallene (penta-2,3-diene).  There is no plane symmetry in this case, only a $\ce{C_2}$ axis.  Draw the mirror image of the pictured allene and you will find that it is not superimposable on its mirror image. 1,3-dimethylallene is a chiral molecule.


Can we use HPLC to purify an organic reaction product?

For semiprep applications, you need larger columns, because you cannot just increase the concentration (at least not a lot) with a normal one. You can get along with an analytical column using an automated system running over the weekend, but that too fails if you want to scale up by another order of magnitude.

A HPLC pump that can run at 100x the standard flow rate is likely twenty times as expensive. The column will be much more expensive. The larger volume needs totally different safety considerations. And so on ...

As Waylander commented, you can do preparative HPLC at a pinch and save your day (or your dissertation), but to use it as a standard lab procedure is most often un..sustainable. ;)


Tried again with 365 brand (whole foods) Cream of Tartar. Worked much better.

39g Cream of Tartar
50ml distilled water
Sodium Carbonate added slowly until stirring produces clear liquid


The common confusion here is that two very different things have the same name.  $\ce O$ is a free oxygen atom and $\ce{O2}$ is two oxygen atoms chemically bound to form an oxygen molecule.

There is no common analogy for $\ce C$, but $\ce{N2}$ is called nitrogen, $\ce{H2}$ is hydrogen and $\ce{Cl2}$ is chlorine, each having the same name as that of their constituent elements.

So, to reiterate, the difference between oxygen $\left(\ce O\right)$ and oxygen $\left(\ce{O2}\right)$ is that the former is an oxygen atom while the latter consists of two $\ce O$ atoms bound together, forming a molecule also called oxygen.


Oxygen production from thermal decomposition of metal oxides

A lot of metal oxides decompose under liberation of $\ce{O2}$ when subjected to sufficiently high temperatures. For example, $\ce{HgO}$ decomposes at temperatures above $\pu{500^\circ C}$ into elemental mercury and oxygen:

$$\ce{2HgO ->~ 2Hg + O2}$$

However, disproportionation is another possible reaction pathway. An example is the thermal decomposition of iron(II) oxide to elemental iron and iron(II, III) oxide at temperatures below $\pu{575^\circ C}$ ($\ce{FeO}$ is more stable above that temperature; sources: 1, 2):

$$\ce{4FeO \rightleftharpoons Fe + Fe3O4}$$

Another example is the decompostion of $\ce{SnO}$ at temperatures between $700\mathrm -\pu{1000K}$, which proceeds via an intermediate $\ce{Sn3O4}$ mixed oxide. Tin disproportionates from the oxidation state +II in $\ce{SnO}$ to $0$ in $\ce{Sn}$ and +IV in $\ce{SnO2}$ (1, 3).

\begin{aligned}
\ce{4SnO ->~& Sn3O4 + Sn} \\
\ce{Sn3O4 ->~& 2SnO2 + Sn}
\end{aligned}


The effect of pH and solubility on coagulation in water treatment

To start, I assume you are speaking of metal salt coagulants such as aluminum sulfate (alum) and ferric chloride, as these are the most commonly used. As you have stated, their solubility is affected by pH, the simplest representation of the reaction for alum being:
$$\ce{Al^3+ + 3H2O <=> Al(OH)3,_s + 3H+}$$

Small particles in the water such as clay or bacteria are too small to settle out on their own. The purpose of coagulation is to either cause the smaller particles to aggregate and settle or to trap them during precipitation. Coagulation occurs via three main mechanisms: (1) sweep - where the metal salts precipitate (i.e., solubility) and capture particles/contaminants like a net; (2) adsorption/charge neutralization - where the positive charges on the Al and Fe cations neutralize the negatively charged e.g., clay, bacteria, etc., causing them to aggregate and settle out; (3) adsorption and bridging - where a polymer coagulant can be used to "bridge" together like charged particles, causing them to aggregate and settle out. All are affected by solubility, but sweep is the mechanism that uses precipitation of the metal salts to remove particulates.

The mechanism you will want to choose will depend on the source water and the coagulant. The mechanisms are controlled by pH and solubility. Below are the solubility diagrams for alum and ferric chloride, with solubility represented on the left y-axis, dose on the right y-axis, and pH on the x-axis. As you will see, the shaded areas are the regions for achieving optimal coagulation, depending on the mechanism you desire. For example, if your laboratory tests showed that sweep coagulation was the best for your source water and you chose alum, then you would want to be at a pH between ~7 to 8 and an alum dose of ~15 to 60 mg/L alum (the "optimal sweep" circle). This pH and dose would result in the desired solubility behavior to achieve optimum sweep conditions.

*Figure 9-11 from MWH Water Treatment: Principles and Design, 3rd Ed (Crittenden)


Yes, you have left out one very important cation from your cation/anion balance: $\ce{H+}$

The primary pathway for new particle formation in the atmosphere is the photo-oxidation, via hydroxyl radicals, of volatile reduced sulfur species to form very non-volatile sulfuric acid.  For the most part then, new particles start out as puny droplets of sulfuric acid and grow via accretion, etc. from there.

I recommend measuring the pH of the filtered aerosols, and I strongly suspect that will account for your missing cations.

If I have missed the point of the question, please feel free to ask for any clarifications in the comments.


Gas molecules in the air continuously collide with each, shoot off in different directions etc. This constant motion is like stirring a fluid. You wouldn't observe perfect density separation in a container with oil and water in it if you keep shaking/stirring it. The situation is similar here, the atmosphere is a fluid that is continuously being agitated. I believe this to be the reason, at least off the top of my head.


Dependence of the angle of rotation on the wavelength of plane polarized light

The rotation of plane polarised light by a solution of, say, sucrose depend on the ability of the oscillating magnetic filed of the light to induce an electric dipole moment in the molecule and the ability of the oscillating electric field of the light to induce a magnetic dipole moment.

For these interaction to have any magnitude it is supposed that the electrons in a molecule move in a helical path, or, alternatively that there are two linear dipoles generated by electron motion which are in planes that are at some angle to one another. Although this model is clearly artificial it emphasises the fact that movement of charge must follow somewhat crooked pathways under the influence of the radiation. The theory of this is very complicated but the result from a quantum calculation is relatively straightforwards and is that the molecular rotation M at wavelength $\lambda$ is given by

$$M_\lambda = a\sum_i \frac{\lambda_{0i}^2X}{\lambda^2-\lambda_{0i}^2}$$

where i represents all the electronic states of the molecule and the wavelength $\lambda_{0i} = c/\omega_{0i}$ where $\omega_{0i} = (E_i-E_0)/\hbar$ is the frequency of the $i^{th}$ excited state at energy $E_i$. The parameter $X=\mathrm{Im}(\mu_{m_{0i}}\cdot\mu_{e_{i0}})$ is the value imaginary part of the complex dot product of the induced magnetic and electric dipole moments and a  are a set of constants independent of wavelength.

From this formula it is seen that the polarisation rotation depends on the wavelength. (The dependence of rotation angle on wavelength has been called optical rotatory dispersion.) We can also conclude

that experiments should be performed where the molecule has little absorbance but that the rotation will be larger as an absorption band is approached. (in an absorption band elliptically polarised light can be formed)
The strength of any optical transition is not important as the dot product X depends on induced moments and the angle between them. This also means that weak optical transitions can be as important as strong one in rotation the polarisation. If the induced dipoles are perpendicular then the rotation vanishes as the dot product is zero.
That many excited states i can be involved and their effects can cancel to some extent so the value of the rotation is hard to predict.
The polarisation rotation of two mirror image molecules are equal and opposite in size and if a molecule is identical with its mirror image polarisation rotation must be zero and the term X is zero.
When the wavelength $\lambda < \lambda_{0i} $ i.e. the wavelength crosses a transition then the signal changes sign and this is called the Cotton effect.


According to H.C. Urey and G. Failla, Science  15 Mar 1935,
Vol. 81, Issue 2098, pp. 273, there's no difference in the taste of ordinary and heavy water.


Are there any other (meta)stable oxygen molecules beyond O3?

According to my German version of Holleman-Wiberg: Inorganic Chemistry there is one additional, very short lived species, the so called tetraoxygen. A short article is also available on Wikipedia.

Wiberg describes it's synthesis from dioxygen in a mass-spectrometer employing the "neutralisation-reionisation-technique" (I am translating and paraphrasing at the same time, there might be something lost in translation.) It's stable time frame is in the micro second area. The decomposition barrier of $\ce{O4}$ is about 25 kJ/mol. The structure is not known, but most likely an aggregate of two oxygen molecules, i.e. $\ce{(O2)2}$. In lower concentrations $\ce{O4}$ might be found in liquid oxygen and they may be a template for the  ε-high pressure modification of solid oxygen forming at 0.6 Mbar.

Holleman, A., Wiberg, N., Wiberg, E., et al. (2008). Lehrbuch der Anorganischen Chemie. Berlin, Boston: De Gruyter. Kapitel XIII 1.1.4 Kurzlebige Sauerstoffspezies. S. 513f.  (in german)
Translation of the 101st edition available as Egon Wiberg and Nils Wiberg (2001) Holleman, Wiberg. Inorganic Chemistry. Academic Press.


Will mixing hydrogen and oxygen without a spark result in combustion?

For hydrogen to burn or explode in air, its concentration needs to be at least 4%.

Unless you're using very high power levels, you're unlikely to reach that concentration in the room's air. You could ignite bubbles of hydrogen as soon as they pop at the surface, but if you don't, the hydrogen (with about 1/8 the density of air) will rise quickly, diffusing into the air as it goes.

If you're doing this using a battery, or a small line-operated power supply, there's very little danger of explosion.


Why does Stokes raman not relax to the ground state while anti-Stokes does?

There is no law that a transition must go to the ground state. Now, each molecule has a multitude of electronic states, and within each of them a multitude of vibrational states. Thus each transition between two pure electronic states is accompanied by a variety of related transitions between higher-to-lower and lower-to-higher vibrational states. Which of these are the Stokes lines and which are the anti-Stokes ones, you'd better figure out for yourself.


Effect of pressure on the freezing temperature of water

From the triple point (273.16K, 611.657 Pa), if pressure is increased and temperature maintained at 273.16K, all the water being liquid is the equilibrium condition.

To maintain the solid and liquid phases in equilibrium, as pressure is increased above the triple point pressure, temperature must decrease slightly (to 273.15K if the pressure has increased to atmospheric pressure, for example).


This is indeed a quite intriguing question. Thus I am afraid that my answer will be mostly incomplete and quite speculative on some aspects.

First of all,  book and litterature searches lead to find that this M-28 loss for aldehdyes is mostly attributed as a loss of ethene and not CO. This can be traced back to the 1957 article by Gilpin and McLafferty which describes the EI fragmentation of straight chain aliphatic aldehydes. In the article, they demonstrate, through $\ce{^{18}O}$ labelling of the aldehyde, that it is a neutral ethene loss and not a $\ce{CO}$ loss. I did not find a clear proposal for a mechanism leading to this ethene loss, but as it occurs on long straight chain aldehydes, I would tend to favor a mechanism occuring through multiple $\ce{H^{\bullet}}$ migration within a series of isomeric distonic ions following the scheme, taking hexanal as an example:

But of course this is not sufficient in the case of the example that is the initial question, as there is no terminal ethene loss possible. I wondered if it would apply for other aldehydes, and I was not long to find in the NIST database terminal branched aldehydes (such as for example 5-methylhexanal, for which the EI spectrum is available here) that also lead to a strong M-28 peak, which obviously cannot be obtained through the above mechanism: one would expect a M-42 peak which happens to be present, but with a similar intensity to that of the M-28 peak.

And now we enter the speculative part. Someone with access to a high resolution mass spectrometer and a collection of aldehydes could easily identify if the M-28 is loss of $\ce{CO}$ or of ethene. I will assume that in the case of the question, as well as in the case of terminally branched aldehydes it is a loss of $\ce{CO}$, but this should be experimentally checked. (If someone also finds a reference to high resolution spectra showing loss of CO, please edit this answer accordingly.)

In the case of the (2Z)-3-methylpent-2-en-4-ynal, as well as other aklene aldehydes, one could probably explain the loss of CO as a $\ce{H^{-}}$ (hydride) migration from the carbonyl carbon towards an ionized double (or triple bond), followed by a loss of $\ce{CO}$, following the mechanism:

This mechanism leads to the formation of the methyl-methylenecyclopropene ion. According to Shay et.al, the ionized methylenecyclopropene ion is the most stable structure for these types of ions, but it could also produce any of the isomers, so I chose to start with a 1,5 hydride shift, although other hydride transfers could work out fine.

Although satisfying for the initial question, this does not address the case of the branched aliphatic aldehydes, for which no hydride transfer can lead to the formation of a carbonyl group before cleaving the carbonyl - carbon bond. Considering that the final product is in this case an ionized alcane, the mechanism for the rearrangement cannot proceed through a final bond cleavage which would lead to a bond deficient (unsaturated) system. Somehow, the system has to go through a partially bonded state at one stage. The following proposal could work out:

Initially, the system proceeds though standard α bond cleavage which would usually lead to formation of the m/z 29 $\ce{HCO^{+}}$ ion (present in small abundance in the original question as well as in most aldehyde spectra). An ion-neutral intermediate complex is formed in some instances (although eventually the two steps might be not absolutely separated), which can lead to a proton transfer between the $\ce{HCO^{+}}$ ion and the alkyl radical. This transfer is thermochemically favorable and would lead to loss of $\ce{CO}$.

Point in favor of this hypothesis are:

The thermochemistry for final proton transfer is favorable. Not many radicals have thermochemical data, but one takes the propyl radical (Enthalpy of formation for the propyl radical 100 kJ/mol; Ionization energy of propane 10.94 eV; Enthalpy of formation for propane -104.7 kJ/mol; Enthalpy of formation for $\ce{H^{+}}$ 1530 kJ/mole) one can evaluate its proton affinity at 680 kJ/mol. In comparison, proton affinity of $\ce{CO}$ (at C) is of 594 kJ/mol, thus the proton transfer is exothermic by more than 80 kJ/mol. For ketones, α-cleavages lead to the formation of $\ce{RCO^{+}}$ ions, with R an alkyl group, which cannot as easily transfer a proton (proton affinity of the neutral ketenes on the order of 825 kJ/mol) which also explains why similar rearrangements would be specific of aldehydes.
Fragmentation of the ionized alcane should occur and be present in the mass spectrum. Indeed, as shown in the figure below, some ions (notably m/z 71, 43 and 27) are present in the fragmentation of ionized 2-methylpentane and also in that of 5-methylhexanal, and can harldy be explained by α-cleavage of the aldehyde or by a McLafferty rearrangement alone. Nevertheless, this is a rather weak point: loss of water is also typical of aldehyde and leads to ions which could further fragment into these ions also.

A concern with this hypothesis is that a proton transfer to a radical is not a well established process. Indeed, formation of an ionized alcane leads to an ion which cannot be directly represented using standard Lewis notation, which makes writing mechanisms involving this species a bit worrisome.

The second problem is that although ion-induced dipole interactions are not negligible, the binding between $\ce{HCO^{+}}$ and the alkyl radical is likely to be weak compared to other ion neutral complexes. Thus its lifetime might be too short to allow a proton transfer step.

Further work on this subject should, in my opinion, try to focus on, (i) experimental validation of the loss of CO, and ideally, using D-labelled compounds validate the hydrogen transfer hypothesis. MS/MS fragmentation of the M-28 ion would also allow validation that an ionized alcane is indeed formed in the process. Finally, the mechanism should be validated through some theoretical calculations, at least for the formation of an ion neutral complex and for the barrier (and ideally RRKM kinetics) for the proton transfer within the ion neutral complex.


Using cryptands to make a negatively-charged solution?

For some reason this question reminds me of Maxwell's demon paradox. You don't really separate anions and cations here; in reality the resulting metal ($\ce{M+}$) cryptand ($\ce{L}$) complex is going to be associated with anionic part ($\ce{X-}$) in organic solvent ($\ce{s}$) after it's being extracted from aqueous phase ($\ce{w}$). For the detailed process, refer to [1], for example.

There are two main processes to consider:

  $$
\begin{align}
\ce{L_\mathrm{s} + M^+_\mathrm{s} &<=> LM^+_\mathrm{s}}\label{rxn:1}\tag{1}\\
\ce{\overline{\ce{L}} + M^+_\mathrm{w} + X^-_\mathrm{w} &<=> \overline{\ce{LMX}}}\tag{2}
\end{align}$$

You would be correct if only \eqref{rxn:1} occurs, however, this is not the case.

Reference

Fyles, T. M. Can. J. Chem. 1987, 65 (4), 884–891. DOI 10.1139/v87-149 (Open Access).


When to include the mass of the dissolved substance in calorimetry problems?

There are multiple issues here, and the answer is: it depends.

How much of your substance is there relative to the solution? If the mass is, say, 1% of total and the specific heat is comparable, then your answer may be off by 1%, which is probably smaller than the systematic measurement error from heat loss in the calorimeter.

What is the specific heat of your substance? If it's very small, then it will hardly require any heat to raise the temperature of your substance.

Frequently, you have to make assumptions. Note that in your first example, it asked you to make assumptions about the specific heat of the solution. This is the specific heat for a mixture of your substance and water. That basically addresses both (1) and (2) above. Note that in this case, since it's the total solution, you should use the total mass.
In the second question, the total mass of calcium chloride is very small, so whether or not you include it is almost irrelevant.


Why don't gases escape Earth's atmosphere?

The atmosphere actually loses gases to outer space.

The average velocity $\bar v$ of gas molecules is determined by temperature $T$.
However, not all the molecules travel with the same velocity.
The probability of finding a molecule with a velocity near $v$ is described by the Maxwell distribution of speeds
$$\begin{align}
f{\left(v\right)}&=4\pi\sqrt{{\left(\frac m{2\pi kT}\right)}^3}v^2\exp\left(-\frac{m{v^2}}{2kT}\right)\\[6pt]
&=4\pi\sqrt{{\left(\frac M{2\pi RT}\right)}^3}v^2\exp\left(-\frac{M{v^2}}{2RT}\right)
\end{align}$$
where $m$ is the mass of the molecule, $k$ is the Boltzmann constant, $M$ is the molar mass of the gas, and $R$ is the molar gas constant.

Individual molecules may reach escape velocity $v_\mathrm e$ and thus be able to leave the atmosphere.

Escape velocity is the minimum velocity that is sufficient for an object to escape from the gravitational attraction of a massive body. For a planet, the escape velocity may be estimated by using the formula
$${v_\mathrm e}=\sqrt{\frac{2Gm_\text{planet}}r}$$
where $G$ is the gravitational constant, $m_\text{planet}$ is the mass of the planet, and $r$ is the distance from the centre of mass of the planet.

Therefore, atmospheric escape depends on the mass of the planet, the temperature of the atmosphere, and the molar mass of the gas.


Geometrical isomers are compounds where all atoms are connected to the same atoms but their arrangement in space differs. This does include cis-trans isomers but in the context of coordination compounds also fac-mer isomers ($\ce{[MA3B3]}$ cases).

The case of $\ce{[MA4BC]}$ is tricky. In principle, they show the same two possible arrangements that, if $\ce{B}=\ce{C}$ would be termed cis and trans. In lab practice, most chemists would use cis and trans to describe them. In IUPAC theory, that response and terminology may or may not be valid.


The "trick" is in the enviroment, not in the PCC itself.

Alcohols tend to be oxidized to carboxylic acids when they are in aqueous enviroment.

PCC was invented as a workaround to this problem: it works as a reactant in anhydrous enviroment, hence stopping the reaction at the aldheyde stage.

An alternative way to see PCC is to follow its preparation: if you add $\ce{CrO3, HCl}$ and pyridine, PCC is formed.
If you add $\ce{CrO3}$, acid and water, chromic acid is formed.

So, you see that the reagents are similar: the point is in finding at which stage the difference is.

Mechanistically speaking, the difference occurs as follows: in aqueous enviroment, water can form a geminal diol intermediate by attacking the aldheyde. This step adds the second oxygen to the aldheyde that, with further reaction with chromic acid, yelds a carboxylic acid.

When water is not present, this step doesn't occur, and the reaction stops.


Calculate the mass of Mg needed to produce 5 kg of vanadium

You answer is correct.

It is given that for every 3 moles of $\ce{Mg}$ that reacts, 2 moles of $\ce{V}$ are produced. Here's one possible way to proceed towards the answer.

First, find out how many moles of $\ce{V}$ are in a mass of $5000\,\mathrm{g}$:

$$5000\,\mathrm{g}\cdot\left({1\,\mathrm{mol}\,\ce{V}}\over 51\,\mathrm{g}\right) = 98\,\mathrm{mol}\,\ce{V}$$

Second, calculate how many moles of $\ce{Mg}$ will yield $98$ moles of $\ce{V}$ by multiplying the ratio of the stoichiometric coefficients (3:2) by the number from step 1:

$$\left({3\,\mathrm{mol}\,\ce{Mg}\over 2\,\mathrm{mol}\,\ce{V}}\right)\cdot 98\,\mathrm{mol}\,\ce{V} = 147\,\mathrm{mol}\,\ce{Mg}$$

Finally, calculate the mass of $\ce{Mg}$ by multiplying the number of moles from step 2 by the atomic mass:

$$147\,\mathrm{mol}\,\ce{Mg}\cdot\left({24\,\mathrm{g}\over 1\,\mathrm{mol}\,\ce{Mg}}\right) = 3528\,\mathrm{g} \approx 3.5\,\mathrm{kg}$$


Beckmann rearrangement v/s dehydration of Oximes

Do you think that the mechanism of "Beckmann" and "dehydration" are different?
Answer is No. They're essentially the same reaction. Aldoximes undergoes Beckmann rearrangement to form nitriles, or you can say "they dehydrate" (as hinted by Ben Norris in comments)
The mechanism for Beckmann rearrangement is as follows,

(Source: Wikipedia, Beckmann rearrangement)
Note: Just replace $\ce{R1}$ with $\ce{H}$ and then, in the intermediate (upper one) after second step, the arrow will be from $\ce{N-R1}$ bond to $\ce{N}$. That's dehydration!
Also, if you replace $\ce{P2O5}$ by $\ce{PCl5}$ or $\ce{H2SO4}$, the answer would be same for all.


Activators for non-sulphide compounds in froth floatation

A previous comment noted that sodium compounds are generally soluble and would not be strongly adsorbed on malachite and anglesite.

However, the other compound formed in the reaction between Na2S and a copper or lead salt would be CuS or PbS, both of which are very insoluble. Three cases can be visualized: 1) a sodium ion approaches the Pb or Cu salt and this aquated sodium ion forces off a lattice-bound dication, or 2) a sulfide ion approaches the Pb or Cu salt and either deposits on the crystal surface, forcing a nearby OH-, CO3-- or SO4-- to leave in order to maintain electrostatic neutrality, or 3) the low (but not zero) solubility of the Cu and Pb salts allows precipitation of PbS or CuS as tiny crystals which more frequently than not become bound to the original salt by epitaxial considerations.

In the case of CuSO4 applied to ZnS, ZnS is about 20 times more soluble than CuS, so an exchange of cations will occur, with the copper going into the insoluble phase.

Case 1 seems far-fetched. Cases 2 and 3 are similar, and may occur simultaneously. The key is that the insoluble product is what sticks to the material that you want to float. And the surfactant that forms the bubbles sticks to the activator (CuS or PbS).

It's interesting that ZnS is not easily flotable. ZnS in water may acquire an aquated layer, that is, it is activated by water, which is not selected by the surfactant, so it sinks. Both the copper carbonate and lead sulfate would seem to have surfaces more compatible with water than their sulfides, too, so converting the surface is required.


Why do we need acid when making acetone peroxide?

The role of the acid in this reaction (really a polymerization) is two-fold.  First, it protonates the acetone carbonyl oxygen making the acetone carbonyl carbon more electrophilic.  This speeds up the rate of attack by the nucleophilic hydrogen peroxide (or other hydroperoxides formed later in the reaction) on this carbon.  Second, the acid also protonates hydroxyl groups in intermediates formed later in the reaction.  This makes them better leaving groups and allows nucleophiles like hydrogen peroxide or other hydroperoxides formed in the reaction to displace them.

The acid is a catalyst, the more acid the faster the reaction, but the reaction will proceed, albeit more slowly, without the acid.  Any proton acid can be used, but as expected, the stronger the acid, the faster the reaction.

Here are links to two articles on the subject (reference 1, reference 2) along with a diagram of the proposed reaction mechanism.


Can I use litmus paper to test unknown liquid that contains a concentrated strong acid?

Yes, concentrated strong acid can destroy litmus paper.

Paper is mostly cellulose, which is a carbohydrate polymer. Concentrated sulfuric acid will pull the hydrogen and oxygen out of it as water, leaving carbon behind. The indicator dye that's supposed to change color probably won't fare too well, either.

You can dilute the unknown by adding a drop or two of it to a small amount of distilled water, then use the litmus paper to check that.

A search for "concentrated sulfuric acid on paper" will yield many images and videos showing what the acid does to paper.


How to find the strongest base using stability arguments

A vs. B

I will label your first reaction scheme, using the letters A-E for the deprotonated (i.e. as shown in the problem) form and the symbol $\ce{AH+}$ etc for the protonated form (after acting as a base and accepting a proton):

$$\ce{A (stronger base) + BH+ (stronger acid) <=> AH+ (weaker acid) + B (weaker base)}$$

Your argument about aromaticity is great; imidazolium ($\ce{BH+}$) is the stronger acid compared to cyclohexanol ($\ce{AH+}$), so the alkoxide ($\ce{A}$) is the stronger base compared to imidazole.

A vs. D

The protonated tetrahydrofuran (D) is highly reactive ("unstable") in water because it would want to get rid of a proton. That does not make it a stronger base. In your reaction scheme, you add another proton to a species that already is a strong acid (not sure if it even exists, see Why doesn't H₄O²⁺ exist?).

$$\ce{A (stronger base) + DH+ (stronger acid) <=> AH+ (weaker acid) + D (weaker base)}$$

The cyclohexoxide ($\ce{A}$) is the stronger base than protonated tetrahydrofuran ($\ce{B}$) because we know phenol ($\ce{AH+}$) exists, but we don't think twice-protonated tetrahydrofuran ($\ce{BH+}$) exists.

  the O- one is more stable (on the left side), thus it is the weaker base, meaning the OH+ one (on the right side) is the stronger base

You have to specify stable in terms of what reaction. Protonated tetrahydrofuran is very stable against protonation, so it is not a base.

  However, when I compare the two acids on each side. I see that the OH one (right side) is more stable, thus it must be the weaker acid, so its conjugate base which is the O- one must be stronger.

Now you are comparing the two in terms of either one losing a proton, which is the right thing to consider when comparing two acids.

C and E

You are right about E, the nitrogen already makes four bonds, so it can't accept another proton. I'm not sure where a proton would go in C. In any case, it is not a good base.


Materials used in cooktop heating element, and reason for black appearance

The heating element ( on the inside where you cannot see it ) is Nichrome or similar alloy ( 10 to 20 % Cr, bal. Ni ) . The surface sheath is typically Inconel 601 ( 15 Cr, 10 Fe, 0.2 Al, bal. Ni). They were Inconel 600 ( and maybe some still are 600 ) until the addition of a small amount of Al was found to increase oxidation resistance. There are several similar alloys and there are different names for the same alloy. So the black surface is oxides of Ni. Cr with some Fe. Some compositions and temperature exposure will produce a green tint in the black ( in high temperature furnace tubes of the same family of alloys). Diffraction analyses of the oxides ( from furnace tubes) often do not give specific compositions as Cr, Ni and Fe can substitute for each other.


How does volume contraction in solvent mixing work?

Yes this happens fairly regularly. It is a deviation from ideal behavior associated with the property of partial molar volume.  For an ideal solution, the total volume always equals the summed volumes of the pure components, but ideal behavior neglects several important facts about real molecules, to wit, that they may have different shapes, and may also experience stabilizing or destabilizing interactions with each other.

It's fairly easy to see how molecular shapes and sizes could affect this property. Macroscopic objects experience similar effects.  Suppose I have a bucket "full" of golf balls and a bucket full of sand, if I work carefully, I can pour quite a lot of sand into the bucket of golf balls because the smaller size of the sand grains allows them to fit into the gaps between the golf balls.  Another example: Suppose you have a 10 liter bucket of golf balls and a 10 liter bucket of tiny cubes packed tightly together. If I then try to pour those two buckets into a 20 liter container simultaneously, it almost certainly will not be able to fit all of the balls and cubes, because they will not pack as efficiently in the 20 L container, costing some space.  These are extreme examples, but illustrate one way that different shapes can play a role.

Differences in intermolecular interactions can play an even bigger role.  If you are dissolving salt in water, the individual ions making up the salt crystal can be solvated very efficiently by the polar water molecules, which experience a strong attractive interaction with the ions.  (Recall that the O and H-atoms in water have partial negative and positive charges, respectively.)  This results in the solution having a significantly smaller volume than expected based on the summed volumes of the pure salt and water used to make the solution.

Partial molar volume is not too hard to measure experimentally. For something like aqueous sodium chloride it can be determined in a fairly straightforward way from precise measurements of the densities of a series of solutions with increasing concentrations.  However it is worth noting that the volume and mass measurements used for the calculation must be very precise, yielding measured densities with 5 or more significant digits.  This requires careful use of volumetric glassware and precision balances, and the temperature of the solutions must be kept constant throughout.

For purposes of demonstrating the qualitative effect, you could just use water and ethanol (or isopropyl alcohol), which also have a significant negative excess volume. (see the first graph on the wikipedia page for partial molar volume for details).

For comparison you could use mixtures of benzene and toluene, which behave almost ideally with respect to volume effects.


The problem isn't going through gyrations just to drive you crazy. It is really the way that water analysis works. The "overall" hardness is expressed in terms of the "equivalent ppm of $\ce{CaCO3}$".

The two biggest contributors to water hardness are typically $\ce{CaCO3}$ and $\ce{MgCO3}$. Now let's say that you want to buy a water softener to replace cations $\ce{Ca^{2+}}$ and $\ce{Mg^{2+}}$ with $\ce{Na^+}$ and the $\ce{CO3^{2-}}$ with $\ce{Cl^-}$. Figuring how many gallons of water you'd use in a week and knowing your water's hardness as "equivalent ppm of $\ce{CaCO3}$" you could figure out how much ion exchange resin you'd need. So there is a functional reason that "equivalent ppm of $\ce{CaCO3}$" is helpful.


Do modern dispersion-corrected DFT methods give more accurate molecular geometries?

I am afraid I do not know of a paper specifically discussing the effects of DFT dispersion corrections on intramolecular properties. However, it should be noted that for short-range descriptors (such as bond lengths and bending angles) the dispersion corrections will not influence those as they correspond to distances below the typical distance at which the corrections take place. The various schemes differ in how they work, but typically the corrections (in Grimme's schemes) are damped below a distance of ~6 Å. It is also my experience that such local structural characteristics are not affected by dispersion corrections.

However, you are right that dispersion corrections will influence intramolecular structure at larger scale, and in particular for “soft” or “flexible” long molecules with many conformations. There, it is worth noting that the so-called “dispersion” corrections actually correspond to the lack of accounting in local DFT functionals of middle-to-long range correlation effects. Thus, it is actually expected that the corrections — or nonlocal functionals including dispersion terms — actually perform better for those too.

Unless, like Grimme's 2006 two-body dispersion correction scheme (often called “D2”), the dispersion corrections happen to be often too strong and make intramolecular distance somewhat too small (instead of much too large)!

Edit: Now that I come to think of it, one extreme example of intramolecular system that has been studied in detail w.r.t. dispersion corrections is that of helicenes. It was shown (for example there and there) that dispersion corrections improve the accuracy of computational structures and energies compared to experimental (or high-level computational) data.


Reduction of terephthalic acid to an alcohol (or phenol?)

I want to make sure that you understand the different between two commonly confused molecules (their names don't help).

Phenol is an $\ce{-OH}$ group directly attached to an aromatic ring. Because of this connectivity, phenols exhibit different chemical behavior than other alcohols. In particular they are more acidic. Phenol can be abbreviated $\ce{PhOH}$, where the $\ce{Ph}$ represents the $\ce{C6H5}$ group of the aromatic ring. As a class of molecules, phenols can be represented generically as $\ce{ArOH}$, where $\ce{Ar}$ is used not for argon (there are no compounds of argon), but for generic arenes with any substitution.

Benzyl alcohol is a different molecule. In benzyl alcohol, there is a $\ce{CH2}$ group between the benzene ring and the $\ce{OH}$. Benzyl alcohol behaves like other alcohols. Benzyl alcohol is abbreviated $\ce{PhCH2OH}$.

Conversion of benzoic acid to phenol requires loss of a carbon atom.

$$\ce{Ph{\bf C}O2H -> PhOH + {\bf CO}}$$

Lithium aluminum hydride converts carboxylic acids to alcohols, but it does not cause the loss of the carbon atom. Lithium aluminum hydride will reduce benzoic acid to benzyl alcohol. If you want the benzyl alcohol derivative, stop here. If you want the phenol, read on.

$$\ce{PhCO2H} \overset{1. \ \ce{LiAlH4}}{\underset{2. \ \ce{ H2O}}{\ce{->}}}\ce {PhCH2OH}$$

To convert benzoic acid to phenol, we need a way to loose that carbon atom. One way is to oxidize benzyl alcohol to benzaldehyde using perhaps a Swern oxidation. Then benzaldehyde can be converted to phenyl formate using a Baeyer-Villiger Oxidation. The formate could be hydrolyzed in aqueous base.

$$\ce{PhCH2OH \overset{Swern}{->} PhCHO}$$
$$\ce{PhCHO \overset{Baeyer-Villiger}{->} PhOCHO}$$
$$\ce{PhOCHO}\overset{1.\ \ce{NaOH, \  H2O}}{\underset{2. \ \ce{H3O+}}{\ce{->}}} \ce{PhOH}$$

In practice, it would be easier to buy the phenol. In your case, you want hydroquinone.


How to identify the number of pi electrons in a conjugated system to calculate the HOMO-LUMO gap with the particle in the box approach?

In a very basic first order approximation you can treat dyes with an extended conjugated π-system with the one-dimensional particle in the box approach. I simplified your system and calculated it at the DF-BP86/def2-SVP level of theory. The length of the box is about $L=1.20~\mathrm{nm}$, the following graphic shows it in Ångström.

With that there are a number of problems associated. Counting the number of electrons in the π-system should not be among them. Just assume that all non-hydrogen atoms are sp² hybridised. Then count the electrons that occupy π/p-type orbitals. Here is an image of the lowest occupied π-type orbital of that molecule; you can see that the box stretches over the whole molecule. The thumbnail next to it gives you all occupied π-type orbitals and the lowest unoccupied molecular orbital.

   In this example each phenyl ring contributes six π electrons (12), two sulfurs contribute two each (4), the nitrogen double bond (2), the carbon-carbon double bond (2) and the last nitrogen (2). That makes a total of 22 π electrons.


Is the Hartree-Fock procedure the same thing as the self-consistent-field (SCF) theory?

No.

The Hartree-Fock method encompasses a set of assumptions used to formulate and solve a series of equations embodying a mean-field, time-independent Schrödinger equation.

The self-consistent field method is a computational technique used in myriad applications where the equations to be solved are implicit—that is, where it is impossible to isolate an analytically solvable expression for a variable and its derivatives.  To the best of my knowledge, any solution method involving a "guess-and-check" approach can be considered a 'self-consistent solution method'.

The Hartree-Fock method is an example of a use-case for a self-consistent field procedure, but they are not synonymous.


This process is typically done by extracting the gold in a mixture of something like 3 parts concentrated hydrochloric acid to one part 3% hydrogen peroxide. If this is roughly the procedure you are using, which is the common metehod posted all over the internet for recovering gold from used electronics, then the answer is an unambiguous none of the above!

Unless you have experience and training for working with these chemicals in a laboratory setting, then you should not be doing this at all; it is simply not worth it.

Given that, the best answer your question is to use a fume hood, eye protection, gloves, lab coat, etc. The biggest danger you face, other than splashing the solution into your eyes, is inhaling the fumes. As this process produces toxic gases (chlorine as well as the fumes from the reagents themselves) you don't want to use a sealed container or a vacuum system that would just pump the gases out of your pump. You need an appropriate, certified fume hood for working with these materials. Relying on diffusion or wind in the "open air" is just incredibly dangerous and you could permanently damage your lungs in one unlucky breath.

Again, the answer is: none of the above, use a fume hood!


Is Bond Formation "Strictly" Exothermic?

Bond formation is alway strictly exothermic in the sense of the change of enthalpy.

  exothermic reaction A reaction for which the overall standard enthalpy change $\Delta H^\circ$ is negative.

A bond can only exist, if it needs energy to break it, i.e. the bond dissociation energy is always positive.

  bond-dissociation energy, $D$ The enthalpy (per mole) required to break a given bond of some specific molecular entity by homolysis, e.g. for $\ce{CH4 -> .CH3 + H.}$, symbolized as $D(\ce{CH3−H})$ (cf. heterolytic bond dissociation energy).

This has absolutely nothing to do with a reaction being exothermic/endothermic or exergonic/endergonic, because this is defined by the rearrangements of bonds.

Regarding noble gas diatomics, it is quite clear from MO-Theory, that there is no bond. However, even these non bonded elements have a non-zero dissociation energy. Please refer to "Why are noble gases stable" and to answers and comments within.


Identifying Donation of an Electron Pair

D donates 2 electrons, not an "electron pair." There is a subtle but important difference here because D is a reaction that takes place via the transfer of single electrons (though in the end, 2 are transferred in total).

So, A, B, and C are examples of Lewis acid-base neutralizations, but D is not. D is classical redox reaction.


High concentration of ferric ion catalyst decreases rate of decomposition of hydrogen peroxide in the Fenton reaction

In this paper, a similar reduction in rate is observed at concentrations of H2O2 above 0.03 M final H2O2 concentration (which is about 0.1 wt %). They attribute it to $\ce{HO.}$ reacting with H2O2 (to yield $\ce{HOO.}$) more quickly than it reacts with MB. The result is that $\ce{HOO.}$ is much more abundant than $\ce{HO.}$ and its lower reactivity means a slower rate of reaction with MB.

Based on their data, I would recommend repeating the experiment with much lower H2O2 concentrations.


Is +6 really the most stable form of sulfur?

I think it's a matter of activation energy and reaction kinetics. Although the product is much more thermodynamically stable than the reagent(s), the reaction path could feature a high activation energy which requires those harsh conditions and even a catalyst to make it happen. As shown in the graph below, despite having a favourable difference of free energy ($\Delta G < 0$), the activation energy $E_a$ required to go from reactants to products could be not negligible and hard to overcome.

The catalysts allows the reaction to happen more easily by lowering the free energy of the transition state (by stabilising it). The activation energy is also reduced by the high temperature, according to the Arrhenius equation:

$$k = A e^{{-E_a}/{RT}}$$


What is the anodic reaction of aluminum in water

Aluminum cannot be used easily as support for a hydrogen electrode, because it is always covered by a thin, continuous and insulating layer of aluminum oxide. As this layer is an electrical insulator, its resistance is high, and ions like $\ce{H+}$ have some difficulty in crossing it. So the measured potential of hydrogen measured on such a protected electrode is not a good measurement of the redox hydrogen potential. It corresponds to an important overpotential, which is to be avoided if you want to measure or determine a redox potential.


Why doesn't nitric acid have a positive charge?

One side of the $\ce{HNO3}$ molecule does have a positive charge, but this is neutralized by the negative charge resulting from the Oxygen atom with only one bond to the nitrogen. These charges cancel leaving an overall neutral molecule.


How can I safely handle a concentrated (fuming) nitric acid spill?

Of course, you’ll want to prevent a spillage as well as possible (making sure the bottle doesn’t fall over etc.) but sometimes you just run into bad luck. Let’s hope that your work bench is at least capable of withstanding the acid for a minute or two — if you have a tiled working surface that would be preferred since they will almost certainly withstand.

If the spill is small — a few drops or so — I would take care of it by quickly adding large amounts of water (more than ten times the volume of spilt acid). I would suggest a wash bottle or something similar. Once you have added water to dilute the acid, you can wipe it up with standard kitchen wipes.

Yes, people are always told to slowly add acid to water and not the other way around. However, if it’s only a small amount of acid and using a large amount of water (and if the acid is not sulfuric acid) you can ‘get away’ with doing it the wrong way; the water will absorb the heat quicker than it can boil.

In case you spilt larger amounts, I would also prepare some baking soda to have around. Carefully (really carefully: it will bubble vigorously) sprinkle some of the baking soda onto the spill until the gas evolution is markedly reduced. Then continue as above, i.e. add water and wipe up with kitchen wipes. This intermediate baking soda step would mainly be to reduce the acid concentration by evolving $\ce{CO2}$ — it would be an overkill to use it on small spills, too.

Have all the required material ready, before you unscrew the cap of the acid bottle.

Finally, I would not use acetone wipes. Acetone can react with nitric acid in may unpredictable ways that you really don’t want to run into. Use water and simple, dry kitchen wipes.

By the way: Having eye protection (safety goggles) is greatly recommended.


Why does nitric acid have a different molecular structure from phosphoric and arsenic acids?

Phosphorus and arsenic are much bigger atoms with lower electronegativity so they can fit more atoms around them and oxygen can bond more easily.

Under extreme pressure it is possible to make the orthonitrate ion from nitrate and oxide. It rapidly decomposes in contact with water or carbon dioxide.


If α-glucose is less stable than β-glucose, why is glucose almost always α in its compounds?

I think it has to do with configurations and conformations of sugars and biological needs.

A sugar with a six-membered ring is known as a pyranose, the simplest compound containing such a ring. Similarly, sugars with five-membered rings are designated furanoses.

The cyclization of a monosaccharide renders the former carbonyl carbon asymmetric. The resulting pair of diastereomers are known as anomers. In the α anomer, the OH substituent to the anomeric carbon is on the opposite side of the sugar ring from the CH2OH group at the chiral center that designates the D or L configuration (C5 in hexoses).

  The other anomer is known as the β form. Sugars accurately described
  as cyclic polyhydroxy hemiacetals or hemiketals, because the linear
  form exists in very minute quantities (although β forms comprises
  larger percentages in mutarotation mixtures and under physiological
  conditions)

Hexoses and pentoses may each assume pyranose or furanose forms. The equilibrium composition of a particular monosaccharide depends somewhat on conditions but mostly on the identity of the monosaccharide.

The pyranose ring, like the cyclohexane ring, may assume a boat or a chair conformation.

The relative stabilities of these various conformations depend on the stereochemical interactions between the substituents on the ring. The boat conformer crowds the substituents on its “bow” and “stern” and eclipses those along its sides, so that in cyclohexane it is ~25 kJ /mol less stable than the chair conformer. The ring substituents on the chair conformer fall into two geometrical classes: the rather close-fitting axial groups that extend parallel to the ring’s threefold rotational axis and the staggered, and therefore minimally encumbered, equatorial groups.

Since the axial and equatorial groups on a cyclohexane ring are conformationally interconvertible, a given ring has two alternative chair forms; the one that predominates usually has the lesser crowding among its axial substituents.

  The conformational situation of a group directly affects its chemical
  reactivity. For example, equatorial OH groups on pyranoses esterify
  more readily than do axial OH groups. Note that β -D-glucose is the
  only D-aldohexose that can simultaneously have all five non-H
  substituents in the equatorial position. Perhaps this is why glucose
  is the most abundant naturally occurring monosaccharide.

Glycogen and starch ingested in the diet are hydrolysed by α -amylases and glycosidases, enzymes in saliva and the intestine that break (α1n4) glycosidic bonds between glucose units. Most animals cannot utilise cellulose as a fuel source, because they lack an enzyme to hydrolyse the (β1n4) linkages.

In cellulose the glucose residues have the β configuration, whereas in amylose the glucose is in the α configuration. The glucose residues in cellulose are linked by (β1n4) glycosidic bonds, in contrast to the (α1n4) bonds of amylose. This difference gives cellulose and amylose very different structures and physical properties.

The most stable three-dimensional structure for the (α 1n4)-linked chains of starch and glycogen is a tightly coiled helix, stabilized by interchain hydrogen. For cellulose, the most stable conformation is that in which each chair is turned 180° relative to its neighbours, yielding a straight, extended chain. (In contrast, the average plane of each residue along the amylose chain forms a 60° angle with the average plane of the preceding residue, so the helical structure has six residues per turn). All ⎯OH groups are available for hydrogen bonding with neighbouring chains. With several chains lying side by side, a stabilizing network of interchain and intrachain hydrogen bonds produces straight, stable supramolecular fibers of great tensile strength.

Steric factors and hydrogen Bonding influence homopolysaccharide folding

Polysaccharide molecules can be represented as a series of rigid pyranose rings connected by an oxygen atom bridging two carbon atoms (the glycosidic bond).

  There is, in principle, free rotation about both C-O bonds linking the
  residues however, rotation about each bond is limited by steric
  hindrance by substituents. The three dimensional structures of these
  molecules can be described in terms of the dihedral angles, ϕ  and Ψ,
  about the glycosidic bond:

The bulkiness of the pyranose ring and its substituents, and electronic effects at the anomeric carbon, place constraints on the angles ϕ and Ψ; thus certain conformations are much more stable than others.

Summarising this, the abundance of the forms of sugars has a lot do with factors affecting conformations such as boat and chair – actually this the reason why , equatorial OH groups on pyranoses readily react compared to axial ones, (sterics, rotation about the C-O bond – this is limited in cellulose as it forms almost linear chains and a strong network of H bonds between adjacent molecules limiting flexibility), and other biological factors such as availability of enzymes to break glycosidic bonds connecting the molecules (as noted animals don’t have the ability to break (β1,4) glycosidic bonds).

Hope this helps

References

Lehninger Principles of Biochemistry

Biochemistry Voet and Voet


What type of reaction is the reaction between an acid and a carbonate?

The first step is a proton transfer (acid-base reaction), from HCl to carbonate, to produce hydrogen carbonate ion. The second step is also a proton transfer, from HCl to hydrogen carbonate to produce carbonic acid. The third step (or maybe better described as a two-step acid-catalyzed process) is the dehydration of carbonic acid to carbon dioxide and water.


Computational Chemistry: calculation of sterical effects

I think that you're searching for non-covalent interactions; if so, here's the usual Wikipedia link, https://en.wikipedia.org/wiki/Non-covalent_interactions

But more importantly here's the paper describing the software, http://pubs.acs.org/doi/abs/10.1021/ja100936w and here's Prof Rzepa's blog with 'non-covalent interaction' searched, http://www.ch.imperial.ac.uk/rzepa/blog/?s=non-covalent+interaction

That's the start, post again when you want to know how to implement it.


Application of limiting reagents and acid-base stoichiometry

I understand the procedure but not the questions. So there are two reactions:

$$\ce{2HCl(aq) + CaCO3(s) -> H2CO3(aq) + 2 Cl-(aq) + Ca2+(aq)}$$
$$\ce{HCl(aq) + NaOH(aq) -> H2O(l) + Cl-(aq) + Na+(aq)}$$

The first reaction has reactants in non-stoichiometric amounts. The second reaction, a titration, is done with stoichiometric amounts, yielding a neutral reaction mixture.

The strange thing is that the problem is over-determined: I can calculate the amount of HCl not neutralized by calcium carbonate either by looking at the amounts of the two (assuming the reaction goes to completion) or by the titration (assuming that neutralizing means to add sufficient NaOH to be at stoichiometry ratio with the remaining HCl).

There is also a technical detail: it would be better to boil off the carbonic acid made by protonating the carbonate. Otherwise, the analyte will contain strong and weak acid, and buffer will form during the titration (with the endpoint of the titration ill-defined: is it when the HCl is depleted, or when all acid is depleted?).

However, it seems like the question wants us to make an argument about amounts starting with the titration. We will assume that the titration and the reaction between $\ce{CaCO3 and HCl}$ go to completion, and chalk up (get it?) any inconsistencies to impure $\ce{CaCO3}$.

Amount of NaOH

$$c_{\mathrm{\ce{NaOH}}} = 0.100\  \mathrm{M}$$

$$n_{\mathrm{\ce{NaOH}}} = c_{\mathrm{\ce{NaOH}}} \cdot V_{\mathrm{\ce{NaOH}}}$$

$$\ \ \ =0.100\  \mathrm{M} \cdot 8.50\  \mathrm{mL}$$

$$\ \ \ =8.50\times 10^{-4}\  \mathrm{mol}$$

Amount of HCl

The amount of HCl in the 20 mL sample was $\pu{8.50e-4 mol}$ (from the titration). So the amount in the entire 200 mL sample after adding carbonate was $\pu{8.50e-3 mol}$. The amount used up by reacting with carbonate is the difference between the original amount, $\pu{0.021 mol}$ or $\pu{21e-3 mol}$, and what is left after the first reaction.

$$n_\mathrm{reacted} = \pu{21e-3 mol}  - \pu{8.50e-3 mol} = \pu{12.5e-3 mol}$$

Amount of $\ce{CaCO3}$

As the OP writes, there is a 2:1 stoichiometry (HCl vs $\ce{CaCO3}$), so

$$n_\ce{CaCO3} = \pu{12.5e-3 mol} / 2 = \pu{6.2e-3 mol} $$

On the other hand, if the sample were 100% and everything has 100% yield and we had boiled off the carbonic acid, we would expect a higher amount in 1.00 g of calcium carbonate, i.e. $\pu{9.99e-3 mol}$ as correctly pointed out by the OP.

Why the discrepancy?

Without more information, there is no way of knowing.


Easy-to-do experiments that clearly show outstanding properties of graphene

You may want to post this in the Physics forum, as many of the outstanding properties of graphene are expected to be physical.  See this Periodic Video: Graphene

Given this, two things to try.  First, graphene should conduct electricity, like graphite but not like carbon.  Second, graphene may dissolve in nonpolar solvents like carbon tetrachloride, similar to buckyballs, to produce colorful solutions.  See this Periodic Video: BuckyBall

Both these experiments should be checked in advance: I'm extrapolating from other experiments, but have not done these myself.


For the case of a single bond, consider a simple case like ethane, which can take an eclipsed conformation, a staggered conformation, or anything in between depending on the torsion angle of the C-C bond.  Each carbon has a region of especially high electron density where it's bonded to a hydrogen.  When the torsion angle is changed from staggered to eclipsed, these regions approach eachother and repel, and thus the eclipsed conformation is higher in energy than the staggered.

In the case of a double bond, as in ethylene, the energy associated with torsional strain is much higher because the P orbitals involved in the pi bond have to be oriented the same way.  Thus, applying torsional strain involves breaking this pi bond, and has a high energy.


Nomenclature for stereoisomer example possible wrong?

Unless I'm very much mistaken, this molecule should not be labelled with any stereochemical descriptors, whether they be relative (cis/trans) or absolute (E/Z). The double bond has three identical ethyl substituents, therefore there is no stereoisomerism possible. By way of explanation, imagine if the ethyl group and the hydrogen on carbon #4 were interchanged: the resulting molecule would be completely identical to the one pictured. This is always the case when an alkene bears two identical substituents on the same carbon atom.

To correctly use cis/trans or E/Z nomenclature, it is necessary that there not be two identical groups on the same carbon of the alkene; if there are, the labels are meaningless, since interchanging the two groups results in an identical molecule.


Program that simulates basic reactions in organic chemistry

As you mentioned, there've been quite some attempts on CAOS (Computer-Assisted Organic Synthesis), such as

WODCA from the Gasteiger group
CHIRON from the Hanessian group
LHASA from the Corey group

I might be awfully wrong, but I think they were/are all closed source.

But you might want to have a look at OSET, an open source Java project on which Ivan Tubert-Brohman worked some ten years ago.


Does chromium oxide form on Stainless-steel in 'deoxygenated' water?

From a practical standpoint, once the stainless steel item is made the passive layer immediately begins to form. Passivation with acids is a process used to accelerate this process and have parts ready to use in a shorter period of time. So if you have any high quality SS parts, they will come to you with the passive layer already developed.

Theoretically speaking, if a completely naked section of stainless was put into this "pure water", there would still be some oxide formation because the specification of <50 ppb still allows for some oxygen to be present. However, this should not have any practical implications. I suspect the name "deoxygenated water" is used to indicate there was a conscious process used to reach such a low amount of free oxygen in the water but not to claim o% free oxygen content.


What is the difference between tempering and hardening metals and their end products?

As the names imply, hardening makes the metal more rigid but more brittle, and tempering (from "temperate", moderate), forgoes some hardness for increased toughness. Iron alloys are hardened by rapid quenching and toughened by annealing, but some copper alloys are made more ductile by quenching and are hardened by working.

In brief, annealing allows atoms to migrate to a less-strained position, whereas quenching lock atoms in place; this is used to make tempered glass, as well as to harden metals.

Extremely rapid quenching can produce a metallic glass, where precipitation of constituents is prevented.

Edged-tools may need the body to be ductile, so that it does not shatter, but require for the edge to be hardened so that it retains sharpness. See http://www.smt.sandvik.com/en/products/strip-steel/strip-products/knife-steel/hardening-guide/purpose-of-hardening-and-tempering/ or http://www.stormthecastle.com/blacksmithing/blacksmithing-a-knife/stock-removal-method-of-knifemaking-part-4.htm.


Why is a cationic trialkylammonium functional group inductively withdrawing in nature?

Nitrogen is a very electronegative element and the fact that nitrogen bears a positive charge in the ammonium ion makes it even more electrophilic(as you pointed out) . Alkyl groups are only weak electron donating groups and can't help much about nitrogen's positive charge.Hence the ammonium cation is inductively withdrawing in nature.


This is difficult to explain, even face to face, but I'll give it a try as I know how frustrating this can be to grasp.  Looking at the wedge and dash model, pick one of the four atoms/groups that ISN'T the lowest priority and ISN'T in the dashed (going away from you) position.  Doesn't matter which as long as it meets the two aforementioned rules.  Personally, I prefer to pick the wedged (going towards you) group.  Now, "freeze" whichever atom/group you chose, and by this I mean this group/atom will remain in the exact same position it's already in.  In fact, go ahead and draw the central carbon atom with the dashed line, wedged line, and both normal lines, but omit the groups/atoms for now (and draw it the same way it is pictured above on the left, it makes it much easier to visualize).  So you should have a C with all 4 bonds drawn out from it.  Now, go ahead and write in the atom/group you decided to freeze in the SAME position as it was before.  Now, the objective is to get the H atom (or lowest priority group) to be at the dashed position.  And remember, the frozen atom/group cannot be moved.  There are only two other positions the H atom could assume, as it cannot occupy the frozen atom/groups location, and you don't want it in the position it's already in.  So in the case of the first molecule, it can either occupy the position of CHO or OH (if you chose the wedged group/atom to be stationary).  Well, you want the H atom to be in the dashed position, which is currently occupied by OH.  So on your new molecule, go ahead and draw in the H at the dashed position where the OH was in the original drawing.  Now, youve effectively rotated the H into the position that the OH was.  Now the final rule you must follow, is that no group EXCEPT the one you chose to freeze can remain in the same position as it was in the original drawing.  This leaves only one location for the OH to be moved to, and one location for the CHO to be moved to.  The hydrogen was already moved to a new location, so now you must move the OH group and the CHO group to new positions, and it cant be either of the already occupied positions (the positions occupied by hydrogen and the group you chose to freeze, which in this example is the wedged position).  The OH must occupy the old position of CHO, and the CHO group must occupy the old position of H (why must they occupy these exact positions?  Because there are only two unfilled positions, and remember the newest rule, all groups/atoms that arent the frozen group MUST occupy NEW positions).  So go ahead and draw them in.  Now your molecule should be ready to evaluate for R/S configuration.

Be consistent when redrawing the molecule with the original drawing.  Don't draw the wedged bond on the top of the molecule if it was on bottom in the original drawing. Hopefully this helped, this is a confusing subject though and like I said earlier it's difficult to explain even in person.  I would definitely recommend getting a set of ball and stick models as it makes visualization much easier.  Practice enough and you'll get to where you dont need them anymore.

By the way to clear up any confusion, this step-by-step is essentially rotating the molecule (you aren't changing the structure or stereochemistry).  I simply laid out detailed directions so that if you were having difficulties visualizing rotating the molecule in your head, you wouldn't have to think about it.  As long as you follow the directions and rules I have laid out, it should result in a new drawing of the original molecule from a different angle, the angle in which the lowest priority group is in the dashed (going away from you) position


What is a good resource for learning about the intermolecular bonding in water beyond high school?

Google Scholar is very helpful when you need to read advanced level of work. The key word you can search is Superheating of liquids (water is one of the common examples). Anyone who has been burnt by microwaved water knows the pain of superheated water. There are some excellent reviews on the phenomenon of superheating.

This is pretty good overview:
http://www.lassp.cornell.edu/sethna/Nucleation/

Starting from the original work in 17th Century to recent ones...

An experience to prove, that water, when agitated by fire, is infinitely more elastic than air in the same circumstances
Phil. Trans., 41 (1) (1739), pp. 162-166

Experimental superheating of water and aqueous solutions
https://www.sciencedirect.com/science/article/pii/S0016703709000994

Maximum Superheating of Water as a Measure of Negative Pressure
https://aip.scitation.org/doi/abs/10.1063/1.1722122


Simple every-day-item hydrometer for distinguishing regular soda and diet sodas

Better than a hydrometer, use an Abbe refractometer to measure the index of refraction, which varies with sugar concentration. It requires just a drop of liquid, and that small sample is unlikely to have bubbles, reducing density. You can make a refractometer to measure the sugar concentration, or you can buy one for ~$US20.

That said, diet sodas have been linked to a number of health issues, such as diabetes and diabetic retinopathy. You might be better off with unsweetened products such as seltzer, club soda, tonic water or two cents plain.

On the other hand, if you want to explore brewing or candy making, that refractometer could be useful...

À votre santé!


Fehling's solution turning yellow and then black?

In the reaction with aldehydes, Fehling's solution produces $\ce{Cu2O}$ as an orange or yellow precipitate. I don't know why the color is not always the same. The oxidation of an aldehyde produces an acid. And usually an acid cannot be more oxidized. There is one exception : methanal $\ce{HCHO}$. Because it can be oxidized in formic acid $\ce{HCOOH}$, and then into $\ce{CO2}$ . That is why Fehling's reagent produce metallic copper $\ce{Cu}$, $\ce{CO2}$ and no acid when reacting with $\ce{HCHO}$.


Reaction of propylene oxide with ammonia or hydrazoic acid

In fact, these are at most chemoselective reactions, so there will probably be a mixture of all possible products.

Ignoring stereochemistry and assuming that we want to know the 'most probable' products, as far as I can tell the main difference is that $\ce{NH3}$ is a rather nucleophilic basic compound, whereas $\ce{HN3}$ is an acid and not very nucleophilic. We are not given the solvent, so we cannot know what side reactions can happen, or what solvation effects we have. So we may assume an 'inert' solvent is used.

Within these hypotheses, I'd say that $\ce{NH3}$ will react by $\mathrm{S_N}$2, attacking mostly on the least substituted side ($\ce{CH2}$), whereas $\ce{HN3}$ will protonate the oxygen first, and the opening by the azide anion will happen by $\mathrm{S_N}$1, i.e. attacking the most stable carbocation (the one with the positive charge on the most substituted carbon. So for me, the 'correct' answer would be 2.


Software platforms with integration of multi vendor mass spectrometer device data?

Some of the MS vendors made their software work with other vendors as well. E.g. ThermoFisher's Chromeleon, Waters' Empower seems to have Data Converters for this. If you already have software bought from your vendors it may make sense to research their cross-vendor capabilities.

There are also pure software companies that build vendor-agnostic tools:

As you mentioned - ACD/Spectrus which has a wide range of vendors supported.
Virscidian's Analytical Studio.
Elsci's Peaksel (disclaimer: I'm one of the developers).

The last two have good support for handling libraries (24, 96, 384, 1536 or more samples per experiment).

There are also open-source packages and libraries:

OpenChrom, if you don't like something you can contribute or just change it to your own needs.
MZmine2. There's also a repo with MZmine3, but I don't know what's the status there.


Reaction of formic acid with bridged bicyclic compound

I leave the arrows filling and details of last step to you since it is homework and online Marvin JS did not let me do arrows the way I liked to.


The Maxwell–Boltzmann distribution is based on the principle that the distribution of kinetic energies of gas molecules is the same. So in a mixture of $\ce{N2}$, $\ce{N2}$, and $\ce{He}$, the molecules of each gas would all have the same KE distribution. However since He is much lighter, its molecules would have higher velocities.

Normally we'd think of the gas mixture as being homogeneous in a liter bottle in the laboratory. However if we consider the atmosphere then gravity has an effect. So helium will drift upwards in the atmosphere since it is less dense. In fact at the edges of the outer atmosphere a helium atom can get enough velocity to escape earth's gravity. So all helium that leaks into earth's atmosphere is ultimately lost into space.


Best reagent for reducing a ketone in the presence of an alkene

The Clemmensen reduction is really only effective at reducing ketones to alkenes when one group is aromatic. Additionally, the Clemmensen reduction uses concentrated HCl, which might cause problems. Another functional group in your molecule, the tertiary alcohol, will be removed by HCl via an elimination reaction.

Choice (b) is the Wolff–Kishner reaction, which also converts ketones into alkenes. It requires strong base (not acid) and works on almost all ketones.

Choice (c) will react with the alkene also, and so cannot be correct.


I hope I am not breaking any rules of this forum by directly quoting, but here you go:

"The cold polymer is harmless on ingestion, in contact
with the skin, or implanted in living tissues. When
heated above 250° C. it begins slowly to give rise to
small amounts of decomposition products, some of
which are toxic.
The accidental inhalation of fumes from the overheated
polymer by workmen produces an influenza-like
illness which has been described elsewhere. The
illness follows a latent interval of a few hours and
resolves within a day or two with no subsequent illeffects.
This may happen, for example, when the
polymer is heated to comparatively high temperatures
(350-400 C.) in an oven, an extruder, or some other
equipment used to fabricate it, but it is easily prevented
by the intelligent use of exhaust ventilation."

In: Harris, Kenwin. “Toxicity of Polytetrafluoroethylene.” British Medical Journal 1.5285 (1962): 1146.

https://www.ncbi.nlm.nih.gov/pmc/articles/PMC1958415/pdf/brmedj02864-0082a.pdf


As explained in Emission Spectra of Bound Helium Hydride Physical Review Letters vol. 77 pages 2941-2944:

  The ground state [of neutral HeH] is repulsive, as shown by
  numerous experiments and calculations.

but excited states of HeH that are bound exist (as first experimentally observed by this paper, but previously calculated).

Given that only excited states are bound, it don't think it proper to say HeH stably exists under any conditions natural or otherwise.

The paper does state:

  Because of the abundance of He and $\ce{H_2}$ in stellar atmospheres
  and interstellar space, HeH is expected to
  be formed there. Therefore our data may be useful to
  find the HeH lines there.

However, that it exists does not mean it stably exists.


Which chemical reactions occur between my tongue and the food I eat?

1) I would guess that all of the different interactions are possible except for ionic and covalent because those are much stronger than the rest and it would be difficult for the food molecules to disassociate from the taste receptors to be used again. Enzymes in saliva break down complex food molecules to smaller and simpler ones.

2) Started explaining in 1, molecules are not usually permanently bound to the taste receptor in order for the taste receptor to be used again (another molecule being able to bind to it). Some toxins form covalent bonds to receptors, such as curare. I'm not sure what kinds of receptors taste ones are but toxins target different specific receptors. Curare binds to nicotinic acetylcholine receptors, found in muscles.

3) We don't taste things unless a molecule binds to a taste receptor.


Finding mathematically the ground state density in DFT

What we have to do here is a functional derivative.
Lets consider the following method of finding functional derivartives:

$$F[\rho+\delta \rho]-F[\rho] = \int_\Omega \delta\rho \frac{\delta F[\rho]}{\delta\rho}\mathrm{d}\mathbf{r}+\mathscr{O}(\delta\rho^2)\tag{1}\label{FuncDer}$$

Lets now consider the Thomas-Fermi kinetic energy functional using the above equation.
The Thomas-Fermi functional is given as:

$$T_\mathrm{TF}[\rho] = C_\mathrm{TF}\int_\Omega\rho^{5/3}\mathrm{d}\mathbf{r}\tag{2}$$

Lets now try to evaluate $T_\mathrm{TF}[\rho+\delta\rho]$:

$$T_\mathrm{TF}[\rho+\delta\rho] = C_\mathrm{TF}\int_\Omega\left(\rho+\delta\rho\right)^{5/3}\mathrm{d}\mathbf{r}\tag{3}$$

Now lets do Taylor expansion of the integrand:

$$T_\mathrm{TF}[\rho+\delta\rho] = C_\mathrm{TF}\int_\Omega\rho^{5/3}+\frac{5}{3}\rho^{2/3}\delta\rho+\mathscr{O}(\delta \rho^2)\mathrm{d}\mathbf{r}\tag{4.1}$$
$$T_\mathrm{TF}[\rho+\delta\rho]= C_\mathrm{TF}\int_\Omega\rho^{5/3}\mathrm{d}\mathbf{r}+\int_\Omega C_\mathrm{TF}\frac{5}{3}\rho^{2/3}\delta\rho\mathrm{d}\mathbf{r}+\mathscr{O}(\delta \rho^2)\tag{4.2}$$

Now subtracting $T_\mathrm{TF}[\rho]$:

$$T_\mathrm{TF}[\rho+\delta\rho] - T_\mathrm{TF}[\rho]=\int_\Omega C_\mathrm{TF}\frac{5}{3}\rho^{2/3}\delta\rho\mathrm{d}\mathbf{r}+\mathscr{O}(\delta \rho^2)\tag{5}$$

Thus by comparing with Eq. (\ref{FuncDer}) it can be seen that:

$$\frac{\delta T_\mathrm{TF}[\rho]}{\delta\rho}=C_\mathrm{TF}\frac{5}{3}\rho^{2/3}\tag{6}$$

Now for the second part of the kinetic energy functional in your question, i.e. the Weizsacker kinetic energy functional:

$$T_\mathrm{W}[\rho]=C\int_\Omega\frac{\nabla\rho\cdot\nabla\rho}{\rho}\mathrm{d}\mathbf{r}\tag{7}$$

Again lets evaluate $T_\mathrm{W}[\rho+\delta\rho]$:

$$T_\mathrm{W}[\rho+\delta\rho]=C\int_\Omega\frac{\nabla(\rho+\delta\rho)\cdot\nabla(\rho+\delta\rho)}{\rho+\delta\rho}\mathrm{d}\mathbf{r}\tag{8}$$

Now doing a Taylor expansion again:

$$T_\mathrm{W}[\rho+\delta\rho] = C\int_\Omega \frac{\nabla\rho\cdot\nabla\rho}{\rho} + \frac{-2\nabla^2\rho+\frac{\nabla\rho\cdot\nabla\rho}{\rho}}{\rho}\delta\rho+\mathscr{O}(\delta\rho^2)\mathrm{d}\mathbf{r}\tag{9.1}$$

$$T_\mathrm{W}[\rho+\delta\rho] = C\int_\Omega \frac{\nabla\rho\cdot\nabla\rho}{\rho}\mathrm{d}\mathbf{r} + \int_\Omega C\left(-\frac{2\nabla^2\rho}{\rho}+\frac{\nabla\rho\cdot\nabla\rho}{\rho^2}\right)\delta\rho\mathrm{d}\mathbf{r}+\mathscr{O}(\delta\rho^2)\tag{9.2}$$

Now subtracting $T_\mathrm{W}[\rho]$:

$$T_\mathrm{W}[\rho+\delta\rho] - T_\mathrm{W}[\rho] = \int_\Omega C\left(-\frac{2\nabla^2\rho}{\rho}+\frac{\nabla\rho\cdot\nabla\rho}{\rho^2}\right)\delta\rho\mathrm{d}\mathbf{r}+\mathscr{O}(\delta\rho^2)\tag{10}$$

Thus by comparing with Eq. (\ref{FuncDer}) it can be seen that:

$$\frac{\delta T_\mathrm{W}[\rho]}{\delta\rho} = C\left(-\frac{2\nabla^2\rho}{\rho}+\frac{\nabla\rho\cdot\nabla\rho}{\rho^2}\right)\tag{11}$$

Now for the second part of your question.
For molecular calculations involving kinetic energy functionals you have to look into orbital-free density functional theory (OF-DFT).
The kinetic energy functionals are not used in Kohn-Sham density functional theory (KS-DFT).
In KS-DFT the kinetic energy are calculated from the Kohn-Sham orbitals instead of the density.

Often when working with OF-DFT the equations can be solved numerically using finte element method (FEM) for a nice introduction to this take a look at Joel Davidsson master thesis [1].
Keep in mind that doing OF-DFT calculations is not a trivial task!

[1] : https://www.diva-portal.org/smash/get/diva2:864857/FULLTEXT01.pdf


Is nucleus at the centre of the atom? If so why?

Why is the nucleus the centre of the atom?

Definition. Whatever the electron does, chemists define the position of the nucleus to be the centre of the atom meaning that electrons can be polarised away.
Quantum mechanics. When studying an isolated atom, the only thing that influences electron behaviour is the nucleus. You can calculate the ‘path’ of an atom‡ relative to the position of the nucleus with quantum mechanics and your result will be that quite a few electrons will occupy a perfect sphere whose centre is the nucleus.

The second point may require elaboration. But it really is quite simple. You have a very lightweight, very fast particle with a charge of $-1$ (your electron) and you have a very heavy, very slow particle with a charge of $+1$ (your proton). By the time the proton has moved an inch, the electron has moved a mile so we can approximate that only the electron really moves. (This has the fancy name Born-Oppenheimer approximation.) Now remember that both are attracted to each other. The proton doesn’t move, but the electron feels the ‘desire’ to ‘move towards’ the proton. (Inverted commas because this is a macroscopic view that only gets us so far when dealing with quantum effects.) If you average this over time, you will find that the electron will have spent the same amount of time on one side of the nucleus as on the other and the mean electron position will be in the nucleus. Spherically outwards from the proton, the chance of locating the atom decreases monotonously with distance; the angle does not matter. We have a perfect sphere with the proton being the centre.


how to define proper ppm concentrations for water solvent vs alcohol solvents?

I think this is why it's generally not advised to use relative units such as $\%$, $\mathrm{ppm}$ and $\mathrm{ppb}$. If they're going to be used, then it's best to indicate what's the ratio being considered, which is frequently done using $\mathrm{(v/v)}$, $\mathrm{(m/m)}$ or $\mathrm{(m/v)}$, $\mathrm{m}$ being short for mass and $\mathrm{v}$ for volume. For example:

$$\frac{\mathrm{1\ \mu g_{\ solute}}}{\mathrm{1\ mL_{\ methanol}}} = \mathrm{1\ ppm\ (m/v)} = \frac{\mathrm{1\ \mu g_{\ solute}}}{\mathrm{0.7918\ mg_{\ methanol}}} = \mathrm{1.263\ ppm\ (m/m)}$$

What's common practice may be heavily dependent on whom we're talking about. Since we usually prepare solutions in the lab by completing the volume up to a certain amount, one could say it's best to ignore the solvent density and use $\mathrm{(/v)}$ as a rule of thumb. The solute unity will depend mostly whether it's a solid being weighed or a liquid being measured.

As pointed out by @PLD, however, there might be cases where it's important to weigh the solvent, probably when very accurate measurements are necessary and specially if the solvent's volume changes considerably given changes in temperature.

One problem of using the $\mathrm{(m/v)}$ notation is that it doesn't use SI units, and might also be confusing. A more verbose way of indicating this is saying mass to volume. Doing some research for this I even found the "units" $\mathrm{ppm_v}$ and $\mathrm{ppm_m}$, which indicates the solvent's unity, but not the solute's, but definetly aren't as common and are also unclear.

The best solution here is simply to use the full unity, so instead of using $\mathrm{ppm}$ (which actually isn't accepted in many cases, for example most scientific journals), use $\mathrm{mg/mL}$ (or $\mathrm{mg\ mL^{-1}}$) or whatever was really measured. This doesn't leave room for double interpretations and indicates how the solution was prepared.


How do molecules with muons and electrons bond?

You are correct; electrons and muons are fermions with different quantum numbers (specifically, they differ in the electron number and the muon number), so the Pauli exclusion principle does not apply between them (though it of course applies among electrons and muons separately). A somewhat similar case happens with protons and neutrons (also fermions) in the nuclear shell model, which attempts to describe nuclei as containing proton and neutron shells, analogous to electron shells. The proton and neutron shells are filled independently.

Because (negative) muons are the second generation Standard Model equivalent of the electron, whatever electrons do, muons can copy - since there are "electronic" orbitals, there are also "muonic" orbitals. However, there are two main differences.

First, for electron orbitals, it's a good approximation to assume the nucleus is stationary with respect to the electrons (the Born-Oppenheimer approximation), due to the great difference in their masses (the lightest nucleus, a proton, has approximately 1836 times the mass of an electron). Because the muon is approximately 207 times heavier than an electron, now a muon has an appreciable mass relative to a proton (approximately one-ninth), and therefore the BO approximation is considerably worse. This is a scenario in-between a regular atom and positronium, where an electron "orbits" a positron "nucleus", which has the same mass. See Phys. Rep. 1982, 86 (4), 169-216 for a quantitative analysis of BO approximation errors in a simple muonic molecule.

The second and more striking difference is that, again due to the muons being 207 times heavier, muonic orbitals are accordingly around 207 times smaller meaning a muonic orbital has a typical radius of around 0.5-1 pm compared to 100-200 pm for an electronic orbital. The energies are also 207 times larger in magnitude - the 1s electronic orbital in hydrogen has an energy of -13.6 eV, whereas for "muonic hydrogen", the 1s muonic orbital has an energy of -2815 eV. These facts can be determined simply by solving the Schrödinger equation, except inputting a mass 207 times greater for the negatively-charged particle.

As you can see, this leads to a severe mismatch between the realms of electronics and muonics. There is no kind of shared electron-muon bond - at best, there would be a separate electron bond and a muon bond. However, because the energetics involved in the muon bond are so much higher, the system in the ground state is essentially equivalent to just having the muon bond, plus a small correction due to a grossly warped electron bond. The muon bond is formed normally, and the electron bond has to deal with whatever geometry is forced by the muons, however crazy.

As an example, imagine a neutral atom of regular hydrogen and a neutral atom of muonic hydrogen interacting. The muonic hydrogen atom basically pierces into the depth of the electron cloud of the regular hydrogen atom (the electron can hardly repel the muon efficiently, since it is so tightly bound to its nucleus), until both nuclei get quite close. Then the muon latches onto the other proton. The system is stabilised when the two hydrogen nuclei are approximately 0.5 pm apart, and the lone muon forms half of a muonic sigma bond.

From the "point of view" of the muon, the system looks like a singly-ionised muonic dihydrogen molecule ($\ce{\mu-H_2^+}$), with slight corrections due to the electron buzzing around, most of the time far away.

However, from the "point of view" of the electron, it sees a bizarrely elongated nucleus (from the typical ~1 fm sphere to a ~500 fm spindle) with a total charge of +1e (since the muon almost perfectly screens out a full positive charge). This spindly nucleus is still quite small relative to the electron cloud, and so the electron likely behaves similarly to a normal isolated hydrogen atom, with some corrections due to the non-spherical distribution of charge at its "nucleus" (the two separate protons bound by a muon). The electron still provides a slight amount of bonding between the protons, but it is much less than normal due to the odd geometry forced by the muon.

Muonic chemistry in its full glory would be a fascinating (and extremely dangerous!) copy of the electronic chemistry we know, but the two would operate almost completely independently. The muon is actually one of the most stable subatomic particles, with a lifetime of 2.2 µs. That sounds like almost nothing, but it's many orders of magnitude more than what is necessary to observe "chemistry". Unfortunately, it's just too difficult to produce for how ephemeral it is...


It's an ethylenediamine ligand.

Edit - elaborating as per your request: Unfortunately, you don't provide any information as to how much you already know about complex chemistry, so I'm going back to the roots here. If you already know what a complex is, skip this paragraph. As you probably know, atoms and molecules may bind to each other, forming new compounds. We all know covalent bonds in which two atoms move close together and mutually share their electrons, and ionic bonds wherein one of the partaking partners is more egocentric in terms of electrons, "stealing" it from the other but staying in the vicinity nontheless. What these types have in common is that each atom or molecule provides an electron for the formation of the bond. However, it may also happen that both electrons come fron one partner alone; an electron-rich molecule or ion approaches another atom or ion, "donating" his electrons without really giving them up. This is the situation in coordination chemistry: A "central atom" or ion -- usually, but not necessarily a transition metal -- is surrounded by a varying number of "ligands" which donate electron pairs to the central atom, all while these electron pairs remain "located" at their respective ligands (very dangerous expression, I know ...). The compound that is formed is called a complex.

So ligands are atoms or molecules which can provide lone pairs for coordination bonds, i.e. electrons in non-bonding, doubly filled orbitals. In the above example of ethylenediamine, the binding centers of the molecule are the nitrogen atoms. The en ligand is an example of a multidentate (or chelate) ligand which may form more than one coordinating bond per molecule, while the $\ce{NO2^-}$ and $\ce{CN-}$ are monodentate ligands. The chloride anions in your first compound does not bind to the copper ion, but rather stay outside its "coordination sphere" to balance the positive charge of the cobalt complex (which you might then interpret as an ionic bonding between the complex cation and the chlorine anion).

Note that there are, of course, many quirks to this. Complexes with multiple central atoms, different theories about bonding, bridging or radical ligands, charge-transfer complexes, etc. This is a very wide and important field of chemistry, after all; nothing you could exhaustively explain in one or two paragraphs.


Are Cubic Equation of states applicable to Liquids?

Cubic equations of state are summarized here.  They can, indeed, describe both the liquid and gaseous state.

Under subcritical conditions, a given temperature and pressure will give three roots.  The largest is the vapor volume, the smallest is the liquid volume and the middle one is an unstable state that must exist along with the vapor and liquid roots.  By using appropriate integrated forms we can compare the free energies associated with these roots and thus identify which is more stable.  Or, we can identify when the free energies match and thus predict vapor-liquid equilibrium.  The catch?  The equations are of only limited accuracy, especially for liquids.  Some are better than others; for more details the reader can start with the Wikipedia reference given above (https://en.m.wikipedia.org/wiki/Equation_of_state).


A lot of confusion in 1st law of thermodynamics

I'm glad that you have got good concepts but the problem is you're unable to connect the discontinuous dots. See, in order to answer your first two questions I would like to bring attention towards the difference of microscopic energy (energy possesed by the molecules present inside the system) and macroscopic energy (energy you get to see in the system). As a example, a cup of coffee is placed over ground all will say it has zero K.E and 0 P.E. But the molecules present inside the system posses both K.E and P.E. You are right with the fact that change in internal energy for ideal gas means change in K.E but the change in K.E means, difference between the total K.E(summation of K.E of every molecules present in the system) of the sytem before and after the action. In case of non ideal gas, here both P.E and K.E both will contribute.
In case of sublimation of Na, here solid is changing to gas you can't neglect volume change also you have to keep it in mind that change in enthalpy is equal to change in internal energy if the PV-work is zero and the process is isobaric.


How can the chlorophyll concentration of plants be measured?

To measure the total chlorophyll from leaves, probably not, since you need access to methyl isobutyl ether or pyridine.

But you can separate the different pigments in the leaf. From The Naked Scientist: Cut a rectangle 1 cm wide and long enough to stand up in a small jar from a coffee filter. Place your leaf over the top of the strip and roll the coin across the strip about 2 cm from the bottom. Now add a little nail varnish remover to the bottom of your jar, half a cm is plenty. Suspend your strip of paper so the bottom end is sticking a few mm into the acetone. you can hold it up by folding the top end of the strip over forming a hook which will rest over the lip of the jar. Wait a few minutes, with any luck you should see an interesting effect. Try the experiment with a strongly colored leaf like a purple one.


What reactions will their reaction rates increase with time?

Reactions that are autocatalytic can have their rate increase as a function of conversion.

One example is the decomposition of cellulose acetate by hydrolysis, also known as vinegar syndrome. The reaction is acid catalyzed, and it produces acid. In theory, the reaction gets faster. In practice, diffusion and limiting reagents many prevent a rate acceleration. Acetylsalicylic acid decomposes similarly. In the following equation, $\ce{R}$, could be the cellulose backbone or the salicylyl group. The acetyl group $\ce{-COCH3}$ is represented by $\ce{Ac}$. Thus acetic acid is $\ce{HOAc}$.

$$\ce{R-OAc + H2O ->[\ce{H+ catalyst}] R-OH + HOAc}$$

Another example is the alpha halogenation of ketones or aldehydes, which is acid catalyzed and acid-producing.

$$\ce{CH3-CO-CH3 + Br2->[\ce{H+ catalyst}] CH3-CO-CH2Br + HBr}$$

A historically interesting example is the allotropic decomposition of tin, or "tin pest", in which tin metal transforms to a brittle form in cold weather. The book Napoleon's Buttons, which is about the influence of chemistry (wittingly or otherwise) on history, opens with the conjecture that tin pest is what really stopped Napoleon's invasion of Russia.

In addition, as this answer reminds us, the rate of the reverse reaction increases to a maximum during the establishment of dynamic equilibrium.


Which part of the reaction of potassium with water releases more energy?

$$\begin{array}{c}
\ce{2K(s) +2H2O (l)->2KOH(aq) +H2(g)}\\
\ce{2H2(g) + O2(g) -> 2H2O(g)}\\
\end{array}$$

We have the two equations above, and we want to know which releases more energy. The answer is relatively simple to find: it's the standard enthalpy of formation of the products minus the standard enthalpy of formation of the reactants. This works because enthalpy is a state function and a systems enthalpy change independent of the path taken from reactants to products. So the way we do this is by imagining that we take our products, break them into their constituent elements (which have a $\Delta H_\mathrm{f}=0$ by definition), and convert those into the products. So:

$$(-482.37\ \mathrm{kJ \ mol^{-1}}) \cdot 2\ \mathrm{\ mol}-(-285.83\ \mathrm{kJ \ mol^{-1}})\cdot 2\ \mathrm{\ mol}=\frac{-393.08\ \mathrm{kJ}}{2\ \mathrm{mol}\ \ce{K}}=-196.54\ \mathrm{kJ \ mol^{-1}}$$

$$(-241.82\ \mathrm{kJ \ mol^{-1}}) \cdot 2\ \mathrm{\ mol}=\frac{−483.64\ \mathrm{kJ}}{2\ \mathrm{mol}\ \ce{H2}}=-241.82\ \mathrm{kJ \ mol^{-1}}$$

$$\begin{array}{c}
\ce{2K(s) +2H2O (l)->2KOH(aq) +H2(g)} & \Delta H=-196.54\ \mathrm{kJ \ mol^{-1}}\\
\ce{2H2(g) + O2(g) -> 2H2O(g)} & \Delta H=-241.82\ \mathrm{kJ \ mol^{-1}}^{[1]}\\
\end{array}$$

So the combustion of hydrogen gas produces more energy per mole than the reaction of potassium with water.

$^{[1]}$ Formation Enthalpies


Oxidation of alcohols by Ceric Ammonium Nitrate

In this reaction, the ion $\ce{Ce^{4+}}$ is a powerful oxidant. It needs one electron to be transformed into the more stable ion $\ce{Ce^{3+}}$. This electron may come from the central function $\ce{-CHOH-}$ of any secondary alcohol $\ce{R-CHOH-R'}$. The bond between $\ce{O}$ and $\ce{H}$ is able to loose $1$ electron and $1$ $\ce{H+}$ ion. Same thing happens with the bond $\ce{C-H}$ belonging to the same group $\ce{-CHOH-}$. As a consequence, two electrons are emitted with two $\ce{H+}$ ions, and the remaining organic molecule becomes a ketone $\ce{R-CO-R'}$. The two electrons react with two ions $\ce{Ce^{4+}}$. The two half-equations are : $$\ce{Ce^{4+} + e^- -> Ce^{3+}}$$  $$\ce{R-CHOH-R'->  R-CO-R' + 2 H+ + 2 e-}$$ so that the final equation is : $$\ce{2 Ce^{4+} + R-CHOH-R' -> 2 Ce^{3+} + R-CO-R' + 2 H^+}$$