Do "weak-field" and "high spin" complexes refer to the same thing? I will keep it brief; I know that there are weak field, "moderate" field, and strong field ligands and that these might contribute to a complex being high or low spin overall. My question is, does a complex having high spin indicate a weak field complex, regardless of ligand "strength" (or vice versa)?
For example, $\ce{[Fe(H2O)6]^3+}$ has a moderate strength ligand ($\ce{H2O}$) but is overall a high spin complex, so would $\ce{[Fe(H2O)6]^3+}$ be considered a weak field complex, despite the ligand strength not being weak? If not, is it only appropriate to call it a high spin complex? Thank you.

The spectrochemical series only gives info on the relative strength of the ligand field. Whether or not a metal complex is high or low spin (assuming there is an appropriate number of d electrons that both are possible) is a function of both the ligand and the metal center.
Thus, it is possible for "weak field" ligand to be in a low spin complex if the metal center favors low spin. For example, almost all complexes of second and third series transition metals are low spin, independent of ligand identities.


Detection limit for log-log calibration curves The limit of detection ($LOD$) for an analytical method can be derived from its calibration curve using the following equation:
$$ LOD = {\dfrac{(3.3 \cdot {\sigma_
\mathrm{resids}})} {m}} \tag{1} $$ [see reference at bottom]
where:

$ \sigma_\mathrm{resids} $ is the standard deviation of the residuals of the regression fit
$m$ is the slope of the regression curve.

Would the above equation still apply if I were to apply it to a calibration curve generated through linear regression of $\ln(y)$ on $\ln(x)$?
Here's an example data set that I generated (artifically - hence apparent rounding errors) to reflect what would be seen in some electrospray ionisation-mass spectrometry data (non-linear and heteroscedastic):
$$\begin{array}{|c|c|c|c|}
\hline
\text{Calibrant} & x \ (\pu{mg/L}) & y \text{ (peak area, AU)} & \ln(x) & \ln(y) \\ \hline
\text{S1} & 1 & 550 & 0 & 6.309918 \\ \hline
\text{S2} & 1 & 500 & 0 & 6.214608 \\ \hline
\text{S3} & 1 & 450 & 0 & 6.109248 \\ \hline
\text{S4} & 2 & 1100 & 0.693147 & 7.003065 \\ \hline
\text{S5} & 2 & 1200 & 0.693147 & 7.090077 \\ \hline
\text{S6} & 2 & 1300 & 0.693147 & 7.17012 \\ \hline
\text{S7} & 10 & 6500 & 2.302585 & 8.779557 \\ \hline
\text{S8} & 10 & 7000 & 2.302585 & 8.853665 \\ \hline
\text{S9} & 10 & 7500 & 2.302585 & 8.922658 \\ \hline
\text{S10} & 50 & 40000 & 3.912023 & 10.59663 \\ \hline
\text{S11} & 50 & 45000 & 3.912023 & 10.71442 \\ \hline
\text{S12} & 50 & 50000 & 3.912023 & 10.81978 \\ \hline
\text{S13} & 125 & 140000 & 4.828314 & 11.8494 \\ \hline
\text{S14} & 125 & 150000 & 4.828314 & 11.91839 \\ \hline
\text{S15} & 125 & 160000 & 4.828314 & 11.98293 \\ \hline
\text{S16} & 250 & 400000 & 5.521461 & 12.89922 \\ \hline
\text{S17} & 250 & 420000 & 5.521461 & 12.94801 \\ \hline
\text{S18} & 250 & 440000 & 5.521461 & 12.99453 \\ \hline
\text{S19} & 500 & 1000000 & 6.214608 & 13.81551 \\ \hline
\text{S20} & 500 & 1200000 & 6.214608 & 13.99783 \\ \hline
\text{S21} & 500 & 1400000 & 6.214608 & 14.15198 \\ \hline
\end{array}$$

Applying equation (1) and exponentiating the result, the $LOD$ would come out as:
$$ LOD = \exp {\left(\dfrac{3.3 \times 0.177}{1.2265} \right) } = \pu{1.61 mg/L} $$
The strange thing for me is that if I multiply the original concentrations ($x$) by 1000, say to convert to from $\pu{mg/L}$ to $\pu{\mu g/L}$,and then generate the calibration curve of $\ln(y)$ on $\ln(x)$, the $LOD$ comes out at exactly the same value, i.e. 1.61. This is because the slope and the standard deviation of the regression fit are unchanged by multiplying the concentration by 1000 - only the intercept value changes. Note, given that concentration axis was in $\pu{\mu g/L}$ before $\ln$-transformation, this would presumably mean an $LOD$ of $\pu{1.61 \mu g/L}$?
My suspicion is that I've missed a crucial step in calculating $LOD$ from the $\ln$-transformed calibration curve model.
Thanks for any pointers you can provide!
Reference: https://www.ema.europa.eu/en/documents/scientific-guideline/ich-q-2-r1-validation-analytical-procedures-text-methodology-step-5_en.pdf

I can add some points for the OP which are too long to serve as comments. I have talked to a couple of leading persons who work with a lot a of LODs. The first lesson, which is important, that there is no "the limit of a detection" of an analytical procedure. Second, some analytical chemists also state that it should be abolished because of the same reason- there is no the limit of detection. Limit of detection varies from method to method of calculation. You will find a couple of common things in legally accepted methods for LODs as the German DIN-32645, and ISO-BS 11843-2:2000 (Incorporating corrigendum October 2007). They all deal with linear calibration curves and no one want to mess around with quadratic or log transformed calibration curves. In the literature, in my opinion, LOD is the most incorrectly reported number and the experts agree with this opinion in personal conversations.
Now quadratic calibration curves are completely kosher. They are being used in atomic absorption spectrophotometers for ages (~ 40 years). Similarly, another HPLC detector based on light scattering does a small manipulation that makes the calibration curve non-linear. These detectors are used in pharmaceutical industry all the time. So, if your sole purpose is to do quantification, within 1 to 500 mg/L, this calibration curve should serve you well but do not calculate the LODs using the formula 3.3xsigma/slope. The problem is the determination of sigma, and the multiplying factor. When Kaiser developed these formulae (mostly in German papers), he never bothered to say anything about nonlinearity because nobody wanted to work with non-linear calibration curves before the computers.
The log transformation as suggested in the comments changes the noise behavior. If you really wish to determine a number called LOD for this instrument you can do the following (for the satisfaction of a novice reviewer of your paper or another higher up who wants a LOD number from you).
P.S. Do not round off concentrations and areas, the instrument readings look simulated to me because they are only changing in units of 50 and 100. Round off is a bad idea, it should only be done at the end.
a) Check the linearity range of your method. I plotted the data up to 50 mg/L. It is linear.
b) Obtain the readings in the linear range in triplicate or quadruplicate. You already have this data.
c) Look at the standard deviation of peak area vs. concentration curve. As you can see it is not a flat line (slope of 0). Hence the Kaiser's formula fails right here. You cannot apply 3.3*sigma/slope formula. The noise is heteroscedastic. Kaiser's formula is valid for homoscedastic noise.
Suggestions to obtain an LOD of your method.
d) Make about 7-8 standards in the range 0.1 mg/L to 20 mg/L. Repeat the process a to c. What we need is a standard near the estimated LOD.
e) If you still see a non-zero slope for the peak area standard deviation vs. concentration, apply a weighted least squares method. ISO-BS 11843-2:2000 has all the solved examples.


Use of geometry index for the determination of coordination environment Geometry index $\tau$ is supposed to resolve proper geometry for coordination numbers (C.N.) 4 and 5 based on its extreme values ($0$ or $1$). There is also a web app Geom which handles both cases for a structure in XYZ format.

I'd like to summarize the questionable topics regarding proper and efficient usage of this method:

Q1. I'm not sure how to address the intermediate values. Say, for $\tau_5 = 0.33$: is it a square pyramidal geometry with a character of trigonal bipyramid? Or one can just name this coordination environment square pyramid and call it a day?

Q2. Are there similar algorithms developed for the higher C.N.:

capped trigonal prism vs pentagonal bipyramid (C.N. 7);
cube vs square antiprism (C.N. 8)?

For your first question, the original paper (J. Chem. Soc., Dalton Trans. 1984, 1349–1356) that described the geometry index $\tau_5$ defined it as an "index of trigonality". For example, they write for a compound with $\tau_5=0.48$

  By this criterion, the irregular co-ordination geometry of $\ce{[Cu(bmdhp)(OH2)]2+}$ in the solid state is described as being $48\%$ along the pathway of distortion from square pyramidal toward trigonal bipyramidal.

For your second question, while I haven't been able to find a $\tau_7$ or $\tau_8$ used in the literature, it seems possible to define such parameters under the right conditions. To devise a $\tau_8$, we can see that for a regular cube $\ce{MX_8}$, there can only be bond angles of $70.5^\circ$ (between adjacent $\ce{X}$ in the same square) and $109.5^\circ$ (between opposite corner $\ce{X}$ of the same square or between corner $\ce{X}$ of different squares). However, an antiprism instead has an angle of $99.6^\circ$ separating the $\ce{X}$ of different squares.

(Image obtained from Inorganic Chemistry by Miessler and Tarr)

This suggests using a formula reminiscent of $\tau_5$ to define $\tau_8$ as the antiprismatic distortion index. One possibility is $$\tau_8=\frac{\beta-\alpha}{9.9^\circ}$$ where $\beta > \alpha$ are the two largest valence angles and $9.9^\circ$ is a normalization factor to make it between $0$ and $1$. So when $\alpha=\beta=109.5^\circ \to \tau_8=0 \to$ cubic geometry and $\alpha=109.5^\circ$ $\beta=99.6^\circ \to \tau_8=1 \to$ antiprismatic geometry.

This will only work if the structure is a regular antiprism (i.e an anticube). The same is true for defining $\tau_7$ between a pentagonal bipyramid and a monocapped trigonal prismatic. This is because the angles for these will vary if all the attached groups are not the same and so a consistent scheme based on the angles would not suffice.

I also imagine that $\tau_7$ would be harder to define in this way because I don't think there is pair of angles that on its own could describe the distortion between the two geometries.


How to temper a non-ferrous metal Can non-ferrous metals/alloys (copper, brass, bronze) be tempered (made harder) by temperature control alone, like ferrous metals? I understand it's simple to anneal copper, brass, etc., just by making them hot enough.

I'm Wondering if it's possible to temper non-ferrous metals without having to either work-harden it or media-blast it (essentially work hardening).

Most non-ferrous alloys are heated in order to form a solution.
Techniques:
Annealing:
Most non-ferrous alloys that are heat-treatable are also annealed to relieve the hardness of cold working. These may be slowly cooled to allow full precipitation of the constituents and produce a refined microstructure.
Different annealing techniques are subjected with non-ferrous alloys like  recrystallization annealing, partial annealing, full annealing, and final annealing.

Quenching
Quenching is a process of cooling a metal at a rapid rate. This is most often done to produce a martensite transformation. In ferrous alloys, this will often produce a harder metal, while non-ferrous alloys will usually become softer than normal.

Reference: Wikipedia


Is HDPE affected by long term exposure to methanol? For at least several years I'be been storing various chloride solutions with methanol as a solute in Nalgene containers, which I believe are made from high density polyethylene (HDP).

But recently two of these containers had failed. On these containers the plastic seems to have hardened and cracked (see photographs below).

The other containers still seem to be pliable, in good shape. I thought that HDPE would be resilient to methanol and provide indefinite storage. Was I wrong? Or is it perhaps the methanol in combination with the particular salt that led to failure of the container? The failed containers had strontium and potassium chloride solution. Others like lithium, sodium and copper solutions are OK.

HDPE should be fairly stable to MeOH, but it crazes in the presence of UV, when subject to oxidization or to environmental stress.

Since only some of the containers failed, were those subjected to more UV (e.g. placed by windows or at the front of a shelf in a well-lit location), placed near oxidizers, such as bleaches or were they subject to higher temperature, which would have caused the methanol liquid and vapor to expand, increasing pressure and stressing the container?

My experience has also been that apparently identical polyethylene containers have different failure rates, perhaps due to subtle differences during synthesis or manufacture.

Given polyethylene's tendency to craze, it might be best to use another type of container for long-term storage.

Note, though, that ~50% MeOH/water solution used for windshield antifreeze is usually shipped in polyethylene containers. Though the water lowers the vapor pressure, perhaps the UV susceptibility is addressed with additives.


5-Chloro-2-Methylphenol vs 2-Chloro-5-Methylbenzene? Why is the compound called 5-Chloro-2-Methylphenol rather than 2-Chloro-5-methylphenol? The occurence of the highest number 5 is exactly the same, so are they both correct?

The two compound names that you are asking about refer to different compounds. Phenol is numbered as follows:

The structures of 2-chloro-5-methylphenol and 5-chloro-2-methyl phenol are below.

Perhaps you were confused about the 2,6 or 3,5 isomers? In this case, the chloro- substituent should get the lower locant number because it is alphabetically first.


Is Fe(OH)3 a salt or an alkali? I know that a neutralization reaction (i.e. a reaction between an acid and a base) results in the formation of a salt and water, but when I came to know that $\ce{Fe(OH)3}$ is salt, I began to have doubts:

If bases are ionic compounds that break apart to form a negatively charged hydroxide ion ($\ce{OH-}$) in water, then why is $\ce{Fe(OH)3}$ is salt and not a base?

Compounds do not class themselves as either one thing or another. Looking at a specific compound, there will always be classes that it is not, but save very few very simple compounds all will fit at least two classes.

$\ce{Fe(OH)3}$ is both a base and a salt. And a transition metal compound and a hydroxide and …

Let’s take an analogy. To some, I am a friend. To others I am a student. To others again, I am a teacher. I can be all of those, depending on who looks at me.


What is the correct way to notate conformation preferences of polysaccharides? In some webpages (1 or 2) are found the way to notate the conformational preferences of monosaccharides such as furanose and pyranose.

But, it lacks to give a brief description on how to do it with polysaccharides.

For a given example:

Would it be: $^4C_1\ \phi\ \psi\ \omega\ ^4C_1$ ?

Or

Would it be mixed with Newman projections (3, 4 or 5): $^4C_1\ [a,s,t,g],\ [a,s,t,g],\ [a,s,t,g]\ ^4C_1$ ?

Or

What is the proper way to notate?

If I understood the question correctly, your first notation looks correct and less ambiguous.

Common polysaccharides (glycogen, starch, and cellulose are composed of pyranoside subunits (having six-membered rings)). Such 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 $\ce{C–O}$ bonds
  linking the residues , but 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, as can be shown on a map of energy as a function of  ϕ and Ψ:

(a)

(a) When the relative energy (Ʃ) is plotted for each value of  ϕ and Ψ with isoenergy (“same energy”) contours drawn at intervals of 1 kcal/mol above the minimum energy state, the result is a map of preferred conformations.

A map of favored conformations for oligosaccharides and polysaccharides.

The torsion angles  ϕ and Ψ which define the spatial relationship between adjacent rings, can in principle have any value from 0 to 360 degrees.
In fact, some of the torsion angles would give conformations that are sterically hindered, whereas others give conformations that maximize hydrogen bonding.

(b)

(b) Two energetic extremes for the disaccharide Gal(β-3)Gal; these values fall on the energy diagram (a) as shown by the red and blue dots. The red dot indicates the least favored conformation, the blue dot the most favored conformation.

Conformation at the glycosidic bonds of cellulose, amylose, and dextran.

The polymers are depicted as rigid pyranose rings joined by glycosidic bonds, with free rotation about these bonds. Note that in dextran there is also free rotation about the bond between C-5 and C-6 (torsion angle ѡ (omega)).

So the notation considering the glycosidic linkage will be $\ce{Glu(ϕ)-(β1-4)-Glu(Ψ)}$ or another detailed way of illustrating this is $\ce{C1(ϕ)-O4(Ψ)-C4}$ for cellulose.

There are other ways of illustrating polyasacharides notations, and their side groups e.g hydroxyl group (in newman projection)

For the precise specification of the orientation of a polyatomic ring substituent, it is necessary to specify the torsion angle about the exocyclic bond.

  The reference atom in the ring is the carbon atom with the number one
  lower than that of the substituted carbon, unless substitution is on
  the anomeric carbon, when the ring oxygen is the reference atom. The
  reference atom in the exocyclic –CH2X group is X. The exocyclic
  torsion angle is denoted by χ(chi), followed by the atoms to which it
  refers e.g χ(C1-C2-O2-H).

References

Lehninger Principles of Biochemistry 5th ed: Polysaccharides
Symbols for Specifying the Conformation of Polysaccharide Chains


IUPAC nomenclature of molecules containing alkoxide and carboxylate groups I am studying for my A levels and recently came across the above-mentioned reaction.

I tried to think up a name like disodium 4-oxidebutanoate, but this is not based on any IUPAC rule. Internet searches have revealed no name for this, at least to me.

This compound is both an alkoxide ion and an alkanoate ion. How are such compounds named?

The principal characteristic group is expressed as a suffix. The name of the carboxylate anion alone (without the alcoholate function) would be ‘butanoate’. The name of the anion derived from the hydroxy compound (without the carboxylate function) would be ‘butan-1-olate’.

For the compound that is given in the question, one anion is chosen as the parent anion and the other one expressed as an anionic substitutent group. Since the carboxylate anion is the senior characteristic group, the parent structure is butanoate.

According to Subsection P-72.6.2 of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), the prefix for the anionic group $\ce{-O^-}$ is ‘oxido’. (The prefixes is derived from the name oxide by changing the final letter ‘e’ to ‘o’.) Therefore, the name of the dianion that is shown in the question is 4-oxidobutanoate. The complete name of the salt is disodium 4-oxidobutanoate.


Is it possible to have Zero order Elementary Reaction? I was going through a question, which got me confused. Is decomposition of NH3, an elementary reaction?

I have studied that  for elementary reactions, Order = Molecularity,
if so, then why they are different in this reaction.
If this is not an elementary reaction, Then molecularity has no significance. Each step should have its own molecularity. isn't it?
So, Is it possible to have zero order elementary reaction ?

While zero order reactions are certainly possible and have considerable technological use, for example, in breathalysers used to determine alcohol content in ones breath, the question is what do you consider an 'elementary' reaction to be ?

In a pure sense the only 'elementary' reactions that occur are unimolecular and bimolecular ones (possibly also termolecular but exceptionally rare). A unimolecular reaction could be a bond dissociation or cis-trans isomerisation which can occur in an isolated molecule. Bimolecular reactions such as atom/molecule & molecule/molecule reactions are very common and strictly involve only two species.

So in this sense zero-order reactions have to have a more complex nature, such as involving an enzyme as in alcohol decomposition.

(Note also that the order of a reaction is measured experimentally as the sum of the powers to which reactants are raised in the rate expression, $rate= kA^aB^bC^c$ the order is $a+b+c$. Often it is the case that an apparently simple reaction such as $\ce{H2 + I2 }$ is only fully understood by a complex series of reactions, in this case the rate expression has fractional and reciprocal orders. The molecularity m refers to the coefficients in an elementary reaction, for example $m = a + b$ if the reaction actually occurs in a step $aA+bB \rightarrow$ so that m is a small positive number usually $1$ or $2$, for example $ m=1;\, A\rightarrow product$ or $ m=2;\, A+B\rightarrow product$)


Why is a bridgehead atom bearing a negative or a positive charge unstable? Why is a bridgehead atom bearing a negative or a positive charge unstable? Is it something with the structure of bridge bonds? I am unable to understand why.

The hybridisation of a bridgehead carbon when its not charged is sp3. Now, if you remove a hydrogen atom, its hybridisation becomes sp2.

Normally, in sp2 hybridisation, the 3 hybridised orbitals lie in a single plane for minimised repulsions. But the bridgehead carbon with positive charge would not have hybridised orbitals in a single plane. This increases repulsion and makes the compound unstable.

Here is a link - Bredt's Rule - that explains the same thing, but instead of a positively charged bridgehead, they discuss a double bond on the bridgehead atom.


How does polarity of a solvent increase conductance of electrolyte? My textbook says that electrolyte ionize more in polar solvents and greater is the polarity, larger is the conductance. But if polar solvents (lets take water) causes the formation of shells around electrolyte (say NaCl) then shouldn't effective size increase and hence conductance should decrease? What factor is more dominant? Size/Solvation or Polarity?

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

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


Why does Nitrogen gas appear in the equation for making producer gas? My book gives the equation for synthesis of producer gas as follows:

$$ \ce{2C{(s)} + O2{(g)} + N2{(g)} -> 2CO{(g)} + 4N2{(g)}} $$

And it says that the stuff on the right-hand side is called producer gas.

My question is why does $\ce{N2}$ even appear in the equation? It isn't reacting is it? It's just a mixture.

As the comments allude, the coefficient is 4 to represent the approximate 4:1 ratio of oxygen to nitrogen in air (more precisely 21:78 or 7:26).  This added nitrogen must be heated along with the oxygen and carbon.  From a stoichiometric perspective, nitrogen is just a spectator and could be excluded, but from a thermal/total system perspective as nitrogen increases the specific heat of the system requiring more heat to be inputted for heating the mixture which subsequently requires more heat dumping to cool the system, reducing the efficiency of the system compared to a system of pure reactants.


Organic chemistry - alkanes step up reaction Well I was studying organic chemistry, and I came across this reaction for increasing the length of the carbon chain. Upon asking my teacher about the following reaction, she said she was unsure about it. Is this a valid reaction? And if it is, can you please provide the name of the reagent and the actual mechanism?

I think the reagent you're talking about is ketene (R=H in the figure below).  It is a linear molecule (the central carbon is $\ce{sp}$ hybridized) and very reactive.

When mixed with an alkane and heated it dimerizes to form diketene rather than react with the alkane.

However, when irradiated with light, ketene will produce the reactive carbene "methlylene" ($\ce{CH2:}$).  When this photolysis is performed in the presence of an alkane, the methylene will insert itself into the various $\ce{C-H}$ bonds like you've drawn.  However the reaction is not synthetically useful for a number of reasons.  For example, the $\ce{C-H}$ insertion reaction is very indiscriminate and a variety of products will be formed if different $\ce{C-H}$ bonds exist in the molecule.


Are fossil fuels and petroleum the same thing? I've been reading articles about renewable energy, and in comments people talk about fossil fuels and petroleum as examples of non-renewable types of fuel. Sometimes they are used interchangeably, or even mentioned as different things. I couldn't find explicit explanation for which is which, and if one is a subset of the other. Can someone clarify this for me?

Fossil fuels are those that originate from ‘fossilised’, anaerobically partially degraded prehistoric living species (mostly plants and animals) and that take a few million years to form hence are not considered renewable.

Depending on how old the deposit is and which conditions it encountered across Earth’s history, the deposit will either turn into something gaseous, liquid or solid — and these categories are typically labelled ‘natural gas’, ‘crude oil’/‘petroleum’ and ‘coal’. Thus, petroleum is only one type of fossil fuel: the one that presently is in the liquid aggregate state.

However, equating petroleum and fossil fuels isn’t entirely unwarranted, since petroleum is both the most important fuel and the scarcest today.


Why is Mo(VI) more stable than Cr (VI)? Acids of $\ce{Mo}^\mathrm{VI}$ are more stable than those of $\ce{Cr}^\mathrm{VI}$ in terms of redox reactivity.

In p-block elements, as we go down the groups, the stability of lower oxidation state keeps on increasing. I understand how this can be explained with the inert pair effect.

On the other hand, as we go down the group from $\ce{Cr}$ to $\ce{Mo}$, why does the opposite happen and there isn't any inert pair effect? Shouldn't the lower oxidation state be more stable down the groups in d-block elements? Wouldn't there be less shielding in $\ce{Mo}$ making is less electropositive than $\ce{Cr}$?

The inert pair effect plays no particular role in the stability or instability of transition metals’ oxidation states. It is due to the large and rising energy difference between a shell’s s- and p-subshells which make the ionisation of the lower electrons more difficult. However, it is equally wrong to say that lower oxidation states be stabilised in lower periods: the most stable halide with the lowest oxidation state a halogen can have is fluoride ($\mathrm{-I}$).

All 3d-metals before iron can be oxidised up to their highest oxidation state while being strong oxidants in that state. This is not so true for the 4d-metals; an indication that the 3d electrons are better attracted to the nucleus. Indeed, the 3d orbitals do not contain radial nodes and their electrons are thus closer to the nucleus. The radial nodes in 4d and higher d-orbitals ‘add distance’ between the nucleus and the corresponding d-electrons (do not copy that wording: it is bad). It is this proximity to the nucleus and the consequential lesser extension of the orbitals into space that draws electrons into them, destabilising the higher oxidation states and stabilising slightly lower ones.

The same effect is true for f-elements; see Nicolau’s answer to a related question on f-element oxidation states’ stabilities.


How does base size affect the ratio of kinetic/thermodynamic enolate? One of the factors that favor kinetic products is bulky bases. So, I'd expect that, for a certain substrate, the use of a more bulky base to favor even more the kinetic product.

However, in the table below we see that as the base becomes even more sterically hindered, the thermodynamic product is formed. It's never the main product but it does get produced a little bit more (1% to 10%).

Why is that? Why in the case of Ph3CLi we get more B than in the case of LDA?

One problem with comparing product ratios using $\ce{LiN($i$Pr)2}$ vs $\ce{Ph3CLi}$ is that it's not really an apples-to-apples comparison - one is a lithium amide base, whereas the other is a alkyllithium that is highly stabilised by conjugation.

I think this illustrates two important points:

Things are more complicated than they first appear, ie. are the species we draw really representative of the species undergoing reaction?
Analysis of substrate and reagent structures will only get you so far, eg. for kinetically controlled processes, analysis of competing transition states is necessary.

Regarding the first point, there have been dozens of papers looking into the structure and reactivity of lithium amide and alkyllithium bases, particularly by the research group of David Collum. LDA is probably the most heavily studied. It turns out that these species are much more complex than what is implied by the structural formula. For instance, LDA in a coordinating solvent, such as THF, exists as a disolvated dimer:

How does the steric bulk of this species compare to the LDA structure we normally draw? Likewise, what is the actual structure of $\ce{Ph3CLi}$ in solution?

Even with a particular solvent and conditions, transition states have been put forward for the deprotonation event that vary markedly in structural features between substrates. For example, in a review of LDA (Angew. Chem. Int. Ed. 2007, 46, 3002-3017), Collum and coworkers show the following two different transition state structures for deprotonation of an ester using LDA in $\ce{tBuOMe}$ vs THF:

So, if we can get a good idea of what the transition state for LDA vs $\ce{Ph3CLi}$ deprotonation looks like, the complexity is such that high level calculations, eg. DFT, would be needed to explore the difference in selectivity. Even if we could qualitatively say which transition state involves the more bulky species, there may be features of the transition state, eg. conformational, stereoelectronic, that only become apparent through the modelling. Sometimes, it's not just which reagent is bigger, but details of how the two reactants come together at the transition state that determines the favoured outcome.


Carbometalation of alkenes Alkynes are capable of undergoing carbometalation, e.g. with organocopper reagents:

Why don't the corresponding alkenes undergo analogous carbometalation reactions? Is β-elimination the issue?

Alkenes are very capable of a similar carbometallation reaction as you can see in the scheme below (ref: R. M. Sultanov et al. / Tetrahedron Letters 54 (2013) 6619–6623):

I think the reasons why such reaction with alkenes is not so common is:

It is useless (correct me if i am wrong). Hence the reason why the work in the scheme above was published in a low impact journal. Why would anyone want to make alkanes in such a complicated way instead of just getting them from crude oil for example?
With alkenes, the reaction prefers to keep incorporating alkene to form really useful dimers or polymers. If one takes a look in the step wise mechanism of olefin polymerization (e.g. Ziegler-Natta) and dimerization the transmetallation/insertion step is equivalent to the one in the question posted:


Products formed on cleavage of epoxides using HI There seems to be multiple forms of $\ce{HI}$ used to cleave epoxides:
conc. $\ce{HI}$, anhydrous $\ce{HI}$, cold $\ce{HI}$, hot $\ce{HI}$, cold conc. $\ce{HI}$, hot conc. $\ce{HI}$, etc.
So, does concentration of $\ce{HI}$ and temperature have any significance in determining the product?

Cleavage of epoxides can happen in 2 ways, resulting from the cleavage with the 2 carbons to which the oxygen atom is bonded.
If the medium is acidic, it will follow the SN1 pathway, proceeding via the most stable carbocation. This makes sense, since in acidic medium, the alcohol can get protonated and break away from one of the carbon atoms.
If the medium is basic, it will proceed via the SN2 pathway.
In other words, the type of cleavage doesn’t depend on the concentration of $\ce{HI}$, the only thing that matters is the acidity or basicity of the medium.


Reduction of oxides by carbon When reducing an oxide with carbon what would be the product? I think it would produce $\ce{CO2}$ and then the $\ce{CO2}$ would react with $\ce{C}$ to give some amount of $\ce{CO}$.  For example:

$$\ce{2CuO + C -> 2Cu + CO2}$$

If the conditions of the reaction (depending on temperature, pressure and any other gases present) are such that you are producing $\ce{CO2}$, then $\ce{CuO}$ should readily oxidize any $\ce{CO}$ that gets produced all the way to $\ce{CO2}$.

The main reaction should be as you said:

$$\ce{2CuO + C -> 2Cu + CO2}$$

and if you have any formation of $\ce{CO}$, it will be oxidized as:

$$\ce{CuO + CO -> Cu + CO2}$$


Equation of water at triple point I was thinking about what exactly happens at the triple point of water, and what could be the effect of adding solutes/solvents. I feel we can represent the triple point using the following equilibrium reaction:
$$\ce{H2O (s) <=> H2O (l) <=> H2O (g)}$$

Is this correct? How would this change if I add solutes?

Note: I know that adding solutes would increase the concentration of the solution, but how exactly will it shift the equilibrium?

At the triple point the water system is assumed to be adiabatic. The system has 3 phases, ice from pure water, liquid water, and pure water vapor. The various equilibria  are shown below.

Note that the ice floats on the water. So the ice is in contact with both the liquid phase and the gas phase. Furthermore The ice can't cover the whole surface like a frozen over pond. There must be some liquid water in contact with the gas phase too.
A subtlety here... You can't make a solid phase of ice that is homogeneously say 1 molar NaCl. If you take a 1 molar NaCl and try to freeze it you get relatively pure water first, then saltier and saltier water freezing as the solid phase gets colder and colder. Since the impure ice has doesn't have a homogeneous composition, it doesn't have a specific melting point.
So for the system you proposed there is pure water ice,  a 1 molar NaCl solution, and a gas phase of water vapor. Unfortunately humans can't make a perfectly adiabatic container, so as the ice melts, the 1 molar NaCl solution becomes more dilute. Thus there is no specific triple point.
In other words... if you just throw some ice into the 1 molar NaCl solution, the final temperature of the theoretical system will depend on the relative amounts of the three phases - How much 1 molar NaCl solution, how much ice, and the volume of the gas phase.


How was the diatomic nature of many common gaseous elements originally determined? How did scientists find out that  $\ce{Cl2, H2, O2}$ atoms have a two-atomic molecular structure ?

Here I'll only provide the reasoning for $\ce{O2}$ and $\ce{H2}$ diatomic structure, I think similar reasoning was to deduce that of $\ce{Cl2}$.
The divalency of oxygen, shown by Avogadro's hypothesis.
It is known that two volumes of hydrogen gas and one volume of oxygen gas combine to form water. From Avogadro's hypothesis (equal volumes of gas contain equal numbers of molecules) it can be concluded that water must have two atoms of hydrogen and one atom of oxygen (or conceivably some multiple). Hence, oxygen is divalent (valence of 2) while hydrogen has a valence of 1:
$$\ce{2H + O \to H2O}$$
This would indicate that $\frac12$ gram of hydrogen atoms reacted with 8 grams of oxygen atoms. Since the
atomic weight of hydrogen was defined as $H = 1$, then the respective atomic weights of hydrogen and oxygen would be $H = 1$ and $O = 16$ (the values accepted today).
The diatomic nature of hydrogen and oxygen, as shown by Avogadro's hypothesis.
It is observed that when two volumes of hydrogen gas and one volume of oxygen gas combine, two volumes of water gas are formed. If hydrogen and oxygen were monatomic (as shown in the equation above), then only one volume of water would be formed. The only explanation is that hydrogen and oxygen are diatomic:
$$\ce{2H2 + O2 \to 2H2O}.$$
(Copied almost verbatim from this page)


How does hyperconjugation lead to the directing properties of alkyl group? I have explained how hyperconjugation leads to the directing properties of alkyl groups below.
Carbon joining substituent now need only single bond. So if we remove one bond between $\ce{H-C}$ along with electron from hydrogen we have a free $\ce{H+}$ ion. The bond which was removed formed a bond with carbon joining the substituent(second picture in second row). From the those two electron (on the left of the second picture in second row) we see one come from electron joining substituent(from hydrogen) and other from the carbon atom at ortho position.
How did one bond between Hydrogen and Carbon move to $\ce{C-C}$ along with one electron from hydrogen?

Now we see the electron from hydrogen jumping from ortho to para position to methyl substituent. How does this jump actually happen?
Is it that the electron( from hydrogen of $\ce{CH3}$) on Carbon at ortho jump(Tell me in detail how this jump occur) to meta position due to formation of double bond between the carbon at ortho and meta position. Due to the double bond, electron( from hydrogen of $\ce{CH3}$) on meta came to para position and so on.
How does the double bond between the carbon at ortho and meta position send electron to para position?
The explanation given above is what I think. Is it right?
In my book the following was written:

Hyperconjugation involves delocalization of $\sigma$-electron through the overlap of $\pi$ orbital of double bond with $\sigma$ orbital of the adjacent single bond.

Please explain the above extract from book using the above example of hyperconjugation of toluene.

Why methyl group is 2,4-directing? This question describes in detail why the methyl group is o,p directing.

However, I think the major point of your misunderstanding is in the interpretation of resonance structures. Resonance structures are not stable structures in their own right.
A set of resonance structures can be used to represent bonding that cannot be represented using classic Lewis structures. The structures do not exist on their own but the true structure can be viewed as an 'average' of all the contributing resonance structures.

The electrons do not jump between different atoms but are instead delocalized over all of the atoms with a slightly higher chance of being found at the ortho and para positions in the case of toluene and a slightly lower chance of being found at on the methyl carbon. This gives rise to the increased electron density of the ortho and para positions and therefore the increased negative charge and similarly the decreased electron density on the methyl carbon and the positive charge.


How does one measure the change in enthalpy of hydration? I am learning about lattice enthalpies, and my textbook talks about determining the lattice enthalpy from the standard change of enthalpy of solution and the change of enthalpy of hydration of the ions. It says that the changes in enthalpy of solution and hydration can be determined experimentally but that the lattice enthalpy cannot be.

However, I am struggling to think of how one can measure the enthalpy change when one mole of gaseous ions is dissolved in water under standard conditions. I thought one would encounter the same problem as with the lattice enthalpy, and this involves one mole of an ionic lattice being formed from its gaseous ions under standard condition.

I thought that the problem with this was getting the exact amounts of gaseous ions under standard conditions.

Usually, one would measure these changes in enthalpy calorimetricly — or at least it is a nice student excersize in physical chemistry lab courses. I was once tasked to determine the solvation enthalpy of copper(II) sulphate pentahydrate.

The calorimeter consisted of an isolated thermos in which we filled a defined amount of water and a concealed inner container in which we filled a sample of the salt; the apparatus contained a stirrer inside the isolated area. We then let the water temperature equilibrate itself, opened in the inner container from the outside (mechanically) and measured the change in temperature. From $\Delta T$ we could calculate $\Delta q$ and from that $\Delta H_\mathrm{solv}$. All corresponded to the equation

$$\ce{CuSO4 . 5 H2O + H2O -> Cu^2+ (aq) + SO4^2- (aq)} ~~~~~~ \Delta H_\mathrm{solv}(\ce{CuSO4 . 5 H2O})$$

One could then perform a second experiment using anhydrous copper(II) sulphate. According to Hess’ law, the following is true:

$$\ce{CuSO4 + H2O -> Cu^2+ (aq) + SO4^2- (aq)} ~~~~~~ \Delta H_\mathrm{solv}(\ce{CuSO4})$$
$$\ce{CuSO4 + 5 H2O -> CuSO4 . 5 H2O} ~~~~~~ \Delta H_\mathrm{hyd}(\ce{CuSO4})$$
$$\Delta H_\mathrm{solv}(\ce{CuSO4}) = \Delta H_\mathrm{hyd}(\ce{CuSO4}) + \Delta H_\mathrm{solv}(\ce{CuSO4 . 5 H2O})$$

The two $\Delta H_\mathrm{solv}$ can be determined experimentally, thus the $\Delta H_\mathrm{hyd}$ can be calculated from these two values.


Why is it that the bigger the side-chain is, its hydrogen atoms tend to be more equatorial? The most stable cyclohexane form is the chair conformation but on the other hand, the bigger the side-chain of the cyclohexane is, its hydrogen atoms become more equatorial rather than axial, which brings them closer to the cyclohexane's hydrogens.
Shouldn't they arrange in a less energetic, but a more stable conformation?

It's not exactly clear from your question what distances you consider, but I have taken snapshots of a methylcyclohexane with the methyl group in the equatorial and axial positions:

      (equatorial)

    (axial)

The distance of the methyl $\ce{H}$ atoms to the nearest cyclohexane $\ce{H}$ atoms are displayed, and you'll see that indeed the axial position has a shorted $\ce{H–H}$ distance of ~ 2 Å, while equatorial position has longer $\ce{H–H}$ distances.

Also, as others have remarked in the comments: only the groups (and $\ce{H}$ atoms) directly on the cyclohexane ring are said to be in axial or equatorial position. Atoms of side chains do not follow this terminology, in particular for side chains that can rotate freely as it means nothing there.


Why did my vinegar not create the traditional volcano? At the weekend, my son and I tried the volcano experiment. You know the one: vinegar introduced to sodium bicarb (and a little red dye) within a conical "volcano".

Not actually ours; photo borrowed from here.
So, after assiduously funnelling some (slightly clumpy) bicarb into the volcano and meticulously counting drops of red food colouring... nothing happened? You can imagine our disappointment.
As luck would have it, we have two types of vinegar in the house. So I went about swapping the "non-brewed condiment" (more on this below) that we had used previously for some fancy balsamic vinegar. Lo and behold, this time we were treated to an eruption to rival Vesuvius!
I'm left to conclude that the first vinegar we tried must have gone bad somehow. It still looked, smelled and tasted like malt vinegar, but why else would it have failed us when we needed it most?
So, my question: Why would our vinegar have failed to react with sodium bicarb?

Note 1: In the UK, "non-brewed condiment" is the glamorous name for the cheap type of vinegar that is very common in fish & chip takeaway shops. It's basically identical to malt vinegar. (In fact, I only realised mine wasn't malt vinegar when preparing this question!) It's a brown, semi-transparent liquid with a sharp taste and smell. The principle ingredients are water, acetic acid, and caramel for colouring. Since acetic acid is the same acid found in real (i.e. brewed) vinegar, surely it shouldn't matter whether it's real vinegar or not?
Note 2: For completeness, it's worth mentioning that traditional (in the UK at least) vinegar bottles aren't really very air-tight. It has occurred to me that the acetic acid has been subject to some sort of chemical/physical reaction with the air. (Perhaps oxidation? Or even just evaporation?)

Acetic acid can be decomposed by methane-producing bacteria. Note that the citation is for soil bacteria, but these methanogens can be found almost anywhere.  Indeed, there is an entire microbial community in vinegar, including nematodes (worms). The more "natural" the product, the more organic material to nourish the microbes and to speed decomposition.
That said, check the labels for acetic acid concentration when purchased. The nominal concentration for distilled vinegar is between 5% and 10%, but some types of vinegar may be only 4% acetic acid.


What is kekulization (in RDKit)? It's been a while since I've studied chemistry. Now, I am reading the documentation of RDKit. At a certain point, the term "kekulization" is mentioned. What is kekulization (in RDKit, if this is not a standard chemical term)?

This is a method to generate alternate Lewis structures of the same molecule. Here are two examples from the cumulative dissertation by Sascha Urbaczek[1]:

If you were to search for the left molecule in panel (1) using an image search or a SMILES string, you might miss the right molecule in that panel.

According to the RDkit document cited in the question, the software routinely generates the alternate position of double bonds, and then (in a second step they call "aromatization") labels the ring as aromatic. In panel (2), there are three possible Lewis structures contributing to the actual structure (i.e. there is resonance), so the software would have to generate all three to be able to search for identical structures.

Urbaczek, Sascha. A consistent cheminformatics framework for automated virtual screening. Ph.D. Thesis, Universität Hamburg, August 2014. URL: http://ediss.sub.uni-hamburg.de/volltexte/2015/7349/; URN: urn:nbn:de:gbv:18-73491; PDF via Semantic Scholar


How have I to understand in what those quantum numbers refer for ortho and para water? I need to make an English presentation at my School (Graduate school) about something people may don't know. I chose to speak about the separation of ortho and para water. I first chat here with @Orthocresol about questions I have but this is really tricky and I'm not understanding that so much. So I chose to ask the question here, it might be helpful for other people and I hope at least for me.

At first my question is to understand what the notations $\lvert0_{00}0 \rangle$ and $\lvert1_{01}M \rangle$ in the article mention before about different states of water refers too?

What are those four numbers if the ket notation, from where they come from and what does they mean? I know it makes a lot of questions but maybe are they linked?

I read this article on wikipedia and this one too, and I also found that but I don't understood a lot of things. Maybe that $J$ is a quantum number which quantify the rotation energy of a molecule. But I have a lot of troubles with the projection and I don't know what $M$ is.

I would be glad if the answer was in a part like popularisation because I guess I'll have some troubles to understand it. But I have good skills in algebra, I had a twenty hours introduction in quantum chemistry, the hardest thing I did was a presentation of the spin-orbital coupling if it can help you to jugde my level on the subject. Thank you in advance for your answer.

It's explained in the first link you gave in the previous paragraph:

"The rotational quantum states of the asymmetric rotor, i. e.,
water, can be classified by $J_{K_{a}K_{c}}M$, with the total angular
momentum quantum number $J$, the projection labels $K_{a}$
and $K_{c}$ onto the molecule-fixed a and c axes as defined in
Figure 1 respectively, and the projection quantum number
$M$ onto the space-fixed $Z$ axis."

Do you know how angular momentum in quantum mechanics works? You can look at the projection of the angular momentum onto a specific axis and the total angular momentum. Turns out the Heisenberg Uncertainty Principle prevents you from knowing the angular momentum projections along more than one axis at once...


How to bleach polyester I have a challenge. I recently bought a cotton/polyester coat with the intention of bleaching it to near-white. To my surprise, normal bleach with sodium hypochlorite had very little effect. It changed the color from light brown to mustard-yellow-ish, even after two soakings in a 30% mixture using water and 5% sodium hypochlorite for about 10 hours each.

My second attempt was better. I put a test piece of fabric in a solution and boiled it in the microwave. It made it even lighter, but still far away from my desired near-white (white-yellow or offwhite is what I hope to achieve).

I also tried to soak it in white spirit, with a small added extra bleaching effect.

Now, using common household chemicals, how can I achieve the desired bleaching? Can I increase the effect of the chlorine? Can I use peroxide, ammonia, or any other chemical to strip the fabric of its color? I'm willing to risk to have the fabric destroyed in the process.

The yellowish mustard color I would have to say was caused by your efforts to de-color this polyester, polyester is a decently strong polymer and thus is resistant to many forms of chemical tamper, and in all effects the bleach is damaging the fabric and leaving behind that yellow color, I would suggest using Sulfur Dioxide if you can get your hands on it as that can bleach polyester, you should be left with a white translucent fabric if the bleach hasn't already damaged the fabric too terribly.

Sulfur Dioxide part is iffy, you may choose to disregard that.


The open circuit voltage of a battery: Does it decrease with increase in pH of electrolyte while the voltage across a load increase? Why does the open circuit voltage of a battery decrease with increase in $\mathrm{pH}$ of electrolyte, but the voltage across a load increase?

I tested a cell which had aluminum sulfate as its anode electrolyte ($\pu{12 g}$ aluminum in $\pu{50 mL}$ water and a $\mathrm{pH}$ of 3.2), and potassium hydroxide as the cathode electrolyte ($\pu{5 g}$ of $\ce{KOH}$ in $\pu{200 g}$ of water, $\mathrm{pH}$ of 12.3). The anode electrode was aluminum foil and the cathode was platinum. The open circuit voltage (OCV) of the cell was $\pu{0.9 V}$, and when connected to a $\pu{1 k\Omega}$ resistor, the voltage was $\pu{0.324 V}$.
When I increased the concentration of $\ce{KOH}$ from $\pu{5 g}$ to $\pu{10 g}$, and all other material concentrations were kept constant, the OCV decreased from $\pu{0.9 V}$ to $\pu{0.7 V}$, but when I attached the $\pu{1 k\Omega}$ resistor, the voltage increased to $\pu{0.42V}$.
I do not understand why increasing $\mathrm{pH}$ is causing the OCV to decrease, but increase the voltage under a load to increase.
I am guessing for the voltage under the load that maybe the higher concentration of $\ce{KOH}$ leads to more ions in solution thus increasing the current flow. But if this assumption/hypothesis were true, shouldn't the OCV also increase?
Can someone please help?

Your aluminum half cell has a standard OCV (i.e., at 1 M $H^+$) of 1.676 volts. Ref. 1
You don't have standard conditions, but that doesn't matter; your conditions do not change for this half cell.
The other half cell is a reduction of water, in alkaline solution, with the electrons from the aluminum to produce hydroxide ions.

The CRC Handbook confirms -0.8277 volts for this reaction. I had intended to use a picture from Ref 1 showing the OCV, but it was incorrect by a factor of 10! Just goes to show you that you can't trust it just because it's in print, or online!
The point is, this reaction wants to consume hydroxide ions and generate electrons, but you are running it in reverse by pumping aluminum electrons into it.  When you increase the pH of the half cell, you increase its voltage by giving the half cell more reactant - but then you are fighting it because it is yielding to the aluminum voltage, so your total OCV out decreases. That's why your cell output is so low (that, and the non-standard conditions). But the good thing is that the extra ionic concentration decreases the cell resistance, so it can supply more current to your 1k resistor even with the lower OCV.
Ref 1. https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Electrochemistry/Voltaic_Cells/The_Cell_Potential


R or S configuration of propane-1,2-diol What is the specific configuration (R or S) of this compound?

My answer is R, because $\ce{H}$ is not in the back so I flipped the compound. Is my answer right?

My teacher is telling me it is S, though I'm pretty sure that it is R. That made me very confused.

What I did is that I got S at the beginning, but because $\ce{H}$ is not in the back, I flipped the compound and got the answer R. My teacher flipped the compound in a way that the answer was still S.

$\ce{OH}$ (1), $\ce{CH2OH}$ (2), $\ce{CH3}$ (3), $\ce{H}$ (4) from 1 to 3, anticlockwise which means it is S configuration. But $\ce{H}$ is not in the back, Therefore rotate the compound and the answer should be the opposite to S which means it is R.

Your comment is correct where you assign the priority.

Rather than "rotating" or "flipping" the molecule, consider "looking" at it from the angle which makes the H atom be at the back.  In the graphic it is drawn to the left side, so place yourself on the right side and look down the bond.

The priority groups 1-2-3 run anticlockwise, hence it has the S configuration.


Conductivity of molten or aqueous salts If there is a 1M aqueous solution of 1 liter NaCl (aq), and the molar conductivities of the solution and the chloride ions were found to be $x$ and $y$ respectively, is it accurate to say that the molar conductivity of sodium ions is $x-y$? If not, how else can I calculate the molar conductivity of sodium ions from the given data? (I'm just using NaCl as an example)

Is there any way to compute the molar conductivity of an ion without any experiments?

You cannot determine individual ion's molar conductivity at 1 M. The ions are not independent enough. Kohlrausch's law of independent migration of ions does not apply, it only applies at a fictitious concentration of 0 M NaCl! This value is found mathematically by extrapolation to zero concentration

Is there any way to compute the molar conductivity of an ion without
any experiments?

Yes, if you look up the tables of molar conductivities at infinite dilution, then you can determine molar conductivities of individual ions.


How does the process of protein denaturatation work? I know it's the proteins that somehow change their properties at some temparature.

But what kind of change, in terms of molecule-level structure?

Basically, the egg-white (just another protein structure) is made of long chains of amino acids. They're held in shape by weak bonds. When you heat the structure, the energy you put in, is enough to break those bonds, thereby destroy the structure. This process has a name for it.

Once the temperature is high enough, new covalent bonds can form between the amino acids, and the amino acids in some other protein and they become entangled into a much more complicated structure, which is completely different from the real one. As you cool, you get this finished structure as cooked egg...


Balancing disproportionation reactions How to balance the following disproportionation reaction
 $$\ce {XeF2 + H2O → Xe + XeO3 + HF + O2} $$
 My attempt: I tried by making total number of reductions equal to total number of oxidation but I couldn't do because Xe is both oxidised and reduced also the solvent is water so what should I do to balance  protons and oxygen on both sides?.  Please help by solving it on the basis of same procedure that is making oxidation equal to reduction

It can't be balanced unambiguously, for this is not a single reaction, but a mechanical sum of two reactions, and these can be formally mixed in arbitrary proportion.

The first reaction is indeed a disproportionation. To balance it, use the common trick: consider these two xenons different elements for a while.
$$\ce{\underbrace{XeF2}_{for\;reduction} + \underbrace{XeF2}_{for\;oxidation} +H2O->Xe + XeO3 + HF}$$
The second is a typical redox:
$$\ce{XeF2 + H2O -> Xe + O2 + HF}$$


Does anyone know what this piece of glassware is? I recently found this piece of lab glass with a jacket and spout. It says eubilicmeter but I cannot find any information about it online and I do not know what it is supposed to do.  It is hollow in the middle, sort of bell shaped. What is its name and what is it for?

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

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


Why are these molecular orbitals invalid for hexatriene? Here are the MOs (more accurately, the breakdown of the MOs) for conjugated hexatriene:

Now, when first asked to draw the MOs myself, I drew this one for $\psi_3$:

It still has two nodes, and is still symmetric. I see nothing wrong with it, except that it doesn't match the "correct" MOs.

Similarly, for $\psi^*_4$, I can draw

Now, as I understand it, MOs are formed from various combinations of atomic orbitals, subject to a few rules regarding symmetry. From what I know, the two MOs I've given above should be valid MOs for hexatriene, or at least components of an MO of hexatriene.

Why aren't these two MOs valid? Also, is there a methodical way to obtain the sequence of MOs in the general case(of a conjugated polyene)?

I read up a bit about Huckel theory. There's a pdf that explains this particularly well.

At one point, the following formula is derived at one point:

$$c_s\propto\sin\left(\frac{\pi k s}{n+1}\right)$$

which gives the coefficient (sans a positive normalisation constant) of the atomic orbital of the $s$th atom in the $k$th MO of a conjugated linear or monocyclic system on $n$ atoms.

A quick and dirty method has also been provided.

Using this, one can get a general idea of how the MO should look, and then one can get the exact +/- sequence by considering the number of nodes. Note that this is really the same thing as applying the $c_s$ formula given above, just in a graphical manner.


What is the anodic reaction of aluminum in water Standard electrode potentials are reported for materials that are not usable as actual electrodes in water because they react spontaneously and rapidly. https://www.engineeringtoolbox.com/electrode-potential-d_482.html
Some metals, like zinc, are useful as cell electrodes if pure, but if not so pure, exhibit cathodic areas that cause the metal to corrode and produce hydrogen without delivering useful external current. One way to counteract that was to coat the surface with a tiny bit of mercury, which apparently coated the cathodic sites but permitted the zinc to migrate thru as if the mercury were not there.
Other metals, like aluminum, develop a film which inhibits corrosion by developing a resistance to atom/ion movement. It has been suggested to me that the actual electrode reaction on aluminum in water is production of hydrogen, so that the half-cell using an aluminum electrode in water is the hydrogen electrode, because bare aluminum reacts immediately with water. This would be analogous to the zinc anode with surface impurities (cathodic sites), except that we do not consider zinc to be producing a hydrogen electrode, just corroding.
it seems that there is considerable usage of metals like zinc, aluminum and magnesium for cathodic protection under conditions where no hydrogen is expected, so it is reasonable to visualize the protection to be developed by reaction of the metal, possibly thru a partially protective film which is equivalent to a cell resistance rather than changing the chemistry to a hydrogen electrode. https://performancemetals.com/pages/sacrificial-anodes-faqs
Is there any evidence that metals like aluminum (or zinc or any other metals) could support a hydrogen electrode like platinum or palladium do? The standard half-cell with Pt or Pd is in 1 M HCl, which would not be appropriate for Al, Zn or Mg. Would any other conditions be better?

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.


On a periodic table it shows that Zn and Cd and Hg all have 0 electron affinity. Why is that? Question

On a periodic table it shows that Zn and Cd and Hg all have 0 electron affinity. Why is that?

What I've noticed

I've noticed that these three elements are the end of the D orbital and they are completely full, i think that it has to do something with that But i am not entirely sure

Part two of question

Why is electron affinity expressed in negative numbers?
Ex: Cs=-45.5kj/mol

They all have 2 electrons in their highest occupied s sublevel and 10 electrons in their highest occupied d sublevel (both sublevels are full).  As a result, they can not complete any orbitals by adding an electron and they can not put an electron in a sublevel that is occupied.  All of the other transition metals will either add an electron to complete the highest s sublevel, or into an already occupied d sublevel, or (in the case of palladium) add one to an s sublevel that is empty.

Cd, Hg and Zn thus are unique in that they will all have to put an unpaired electron alone into a p sublevel.


What do the prefixes meta, ortho, pyro mean in inorganic chemistry? In inorganic chemistry, when is the prefix meta used? (as in metaborate and metasulphite)
What about the terms pyro and ortho (as in orthophosphorous acid)?

For example, I recently came to know about the prefixes per and hypo which are used to refer to the highest and lowest oxidation states of an element respectively (like perchlorate and hypochlorite).

Are there any terms and conventions I should be aware of?

In the older literature, the prefixes “hypo-”, “per-”, “ortho-”, “meta-”, and “pyro-” were used in some cases for the distinction between different acids with the same characteristic element.
The prefix “hypo-” was used to denote a lower oxidation state, and the  prefix “per-” was used to designate a higher oxidation state. (The prefix “per-” is not be confused with the prefix “peroxo-”.)

$\ce{HClO}$ hypochlorous acid
$\ce{HClO2}$ chlorous acid
$\ce{HClO3}$ chloric acid
$\ce{HClO4}$ perchloric acid

The prefixes “ortho-” and “meta-” were used to distinguish acids differing in the formal content of water.

$\ce{H3BO3}$ orthoboric acid
$\ce{(HBO2)$_n$}$ metaboric acid
$\ce{H4SiO4}$ orthosilicic acid
$\ce{(H2SiO3)$_n$}$ metasilicic acid

The prefix “pyro-” was used to designate an acid that is formally formed by removing one molecule of water from two molecules of an ortho-acid.

$\ce{H4P2O7}$ pyrophosphoric acid $(\ce{2 H3PO4 -> H4P2O7 + H2O})$.

Various traditional names are retained for use in IUPAC nomenclature, though the number of retained names has been reduced with each succeeding edition of the IUPAC recommendations. The Second Edition (1970 Rules) of the IUPAC Nomenclature of Inorganic Chemistry (published in 1971) still included all of the above-mentioned prefixes; however, the approved use was limited to only some selected retained names. The further development and the current situation of the nomenclature are well explained in @Jan’s thorough answer.


Heating CrCl3(bpy)3 — isomerism and ligand exchange $\ce{CrCl3(bpy)3}$ (A) when heated forms (B) which has the formula $\ce{CrCl3(bpy)2}$ and has one free $\ce{Cl-}$ per mole of (B). B can react with ethylene diamine ($\ce{en}$) to form (C) which has the formula $\ce{CrCl3(bpy)2(en)}$.
(B) can also be converted into (D) which also has the formula $\ce{CrCl3(bpy)2}$ but cannot react with $\ce{en}$.
Identify compounds A–D and draw their structures

I know that B is cis-$\ce{[CrCl2(bpy)2]Cl}$
C is $\ce{[Cr(bpy)2(en)]Cl3}$. The $\ce{en\bond{->}metal}$ bonds are where the $\ce{Cl-}$ ions in B are.
D is the trans isomer of B.
What I dont understand is how you're meant to know what A is. I assume it's $\ce{[Cr(bpy)3]Cl3}$ as a later part of the question asks about relative molar conductivity of A and B.
Why would heating that cause a $\ce{bpy}$ to leave the complex and two chlorides to replace it? Or is it $\ce{[CrCl2(bpy)2]Cl(bpy)}$ in which case, why does heating it not form the more stable complex I've stated above?

If you are given only the options $\ce{Cr(bpy)3Cl3}$, then it is mandatory for the bipyridine ligand to actually be attached to the chromium centre. If it were not, as in your suggestion $\ce{[Cr(bpy)2Cl2]Cl . bpy}$, this additional bipyridine can dissociate away as there is nothing keeping it close to the complex. On the other hand, isolateable substances are always neutral, thus if your complex cation is $\ce{[Cr(bpy)3]^3+}$, then the three chloride anions will remain in the same crystal structure and you can isolate a complex salt of the type $\ce{[Cr(bpy)3]Cl3}$. This molecular structure can be proven by being able to precipitate three moles of chloride per mole of complex, e.g. with $\ce{AgNO3}$.

  Why would heating that cause a $\mathrm{bpy}$ to leave the complex and two chlorides to replace it? Or is it $\ce{[CrCl2(bpy)2]Cl(bpy)}$ in which case, why does heating it not form the more stable complex I've stated above?

This somehow assumes that $\ce{[Cr(bpy)3]Cl3}$ is the most stable of these complexes. However, the chemical literature thinks different. Burstall and Nyholm first reported the synthesis of $\ce{[Cr(bpy)2Cl2]Cl}$ from anhydrous chromium(III) chloride using three (!) equivalents of $\ce{bpy}$ and heating to reflux.[1]

On the other hand, traditional methods to generate $\ce{[Cr(bpy)3]^3+}$ salts — most commonly perchlorates $\ce{[Cr(bpy)3](ClO4)3}$ — typically started from a corresponding $\ce{[Cr(bpy)3]^2+}$ and employed oxidation.[1] (This was apparantly first published in 1932 by Barbieri and Tettamanzi in an obscure journal that I can’t find.)[2] Relatively recently, a new method to synthesise the $\ce{[Cr(bpy)3](ClO4)3}$ complexes from $\ce{CrCl3 . 6 H2O}$ was published — which, however, builds on completely removing any chloride from the solution by using silver(I) salts as in equation (1).[3]

$$\begin{multline}\ce{CrCl3 . 6 H2O + 3 [Ag(bpy)2](ClO4) ->\\ [Cr(bpy)3](ClO4)3 + 3 AgCl v + 3 bpy}\tag{1}\end{multline}$$

Seemingly, having chloride ions in solution around the $\ce{[Cr(bpy)3]^3+}$ ion can’t be trusted.

Partially due to SciFinder being down for maintenance as of now, I was unable to find any references for the synthesis or properties of $\ce{[Cr(bpy)3]Cl3}$, but given the Burstall-Nyholm method I would not assume the cation to be more stable than $\ce{[Cr(bpy)2Cl2]+}$.

A likely explanation for the lower stability could be the charge balance. The dichloridobis(bipyridine)chromium(III) has an overall charge of $1-$, which is generally okay. In tris(bipyridine)chromium(III), there is a very small $\ce{Cr^3+}$ ion in the centre of a complex surrounded by otherwise neutral ligands. Any method to relieve the charge could well be favourable; amoung them reduction of the metal or ligand exchange with anionic ligands. Remember that charge balance is a thing: $\ce{[CoCl4]-}$ forms when cobalt(III) reacts with chloride ions mainly due to charge balance reasons.

References:

[1]: F. H. Burstall, R. S. Nyholm, J. Chem. Soc. 1952, 3570. DOI: 10.1039/JR9520003570.

[2]: Barbieri, Tettamanzi, Atti R. Accad. Lincei. 1932, 15 877. Reference as cited in [1].

[3]: W. Kharmawphlang, S. Choudhury, A. K. Deb, S. Goswami, Inorg. Chem. 1995, 34, 3826. DOI: 10.1021/ic00118a035.


Why cotton and wool clothes shrink after washing them? I want to take measures to avoid my clothes shrinking in the washing machine and in manual washing. I read here if the water is hot that contributes more the shrinking.

The Wikipedia article is kind of as they say in Wikipedia, a "stub" (too short) and doesn't explain the reasons very scientifically.

Wool, in particular, has very "scaly" fibers that mat together forming a dense felt. Natural oils prevent this (no shrunken sheep), but cleaned wool, agitated or shaken, shrinks due to the fibers gripping each other.

"Cotton shrinks because of the tension that is applied to its yarn and fabrics during the construction of the clothing. The tension is released by heat from the washer, dryer, steam and even sunlight, which causes the fabric to be reduced to its natural size." However, mercerization, formaldehyde and other reagents can "pre-shrink" cotton and prevent subsequent shrinkage in washing.


Can tin(IV) sulfide be synthesized from tin(II) chloride via oxidation with sulfite in acidic media? Will the following reaction occur?

$$\ce{2Na2SO3 + 6SnCl2 + 12HCl → 4NaCl + 5SnCl4 + SnS2 + 6H2O}$$

I'm going to synthesizing $\ce{SnS}$ and $\ce{SnS2}$ NP's, but our supply is runned out and deadline of assignment is coming to soon, so will this reaction  occur? If it is possible, what is are the conditions of this reaction?

Yes, the reaction

$$\ce{2 Na2SO3 + 6 SnCl2 + 12 HCl → 4 NaCl + 5 SnCl4 + SnS2 + 6 H2O}$$

is described in chapter 35.6.4 Reactions of chalcogen oxo-compounds of the Russian translation [1, p. 526] of Blumenthal's Anorganikum 10th Ed. (in German).
Unfortunately, I don't posses original German edition, so the following is my English translation of the procedure:

  To the 0.5 M solution of $\ce{SnCl2}$ in 12 M $\ce{HCl}$ in three test tubes 0.2 M solutions containing $\ce{SO3^2-}$, $\ce{SeO3^2-}$ and $\ce{TeO3^2-}$ are added. In all three tubes precipitations of $\ce{SnS2}$, $\ce{Se}$, and $\ce{Te}$, respectively, are formed.

Original Russian text:

  К 0.5 M раствору $\ce{SnCl2}$ в 12 M $\ce{HCl}$, находящемуся в трех пробирках,
  приливают 0.2 M растворы, содержащие $\ce{SO3^2-}$, $\ce{SeO3^2-}$ и $\ce{TeO3^2-}$. Во всех трех пробирках наблюдается выпадение осадков, которые представляют собой $\ce{SnS2}$, $\ce{Se}$ и $\ce{Te}$ соответственно.

References

G. Blumenthal, S. Engels, I. Fitz, W. Haberditzl, K.-H. Heckner, G. Henrion, R. Landsberg, W. Schmidt, G. Scholz, P. Starke, I. Wilke, K.-Th. Wilke Anorganikum: Lehr- und Praktikumsbuch der anorganischen Chemie mit einer Einführung in die physikalische Chemie; Ed. of Russian translation: L. Kolditz; Mir: Moscow, 1984; Vol. 1.


After boiling water in an electric kettle, should I empty it? (I'm reposting my question from Physics SE where it got closed for being off-topic, but chat tells me that it might be on topic here.)

In the office I'm working in, there's an electric kettle for boiling tea water.  Every day, I boil water in it.

At one point, my boss asked me to empty the kettle after making the tea, that is, pour the unused water remaining in the kettle into the sink so that the kettle is as dry as I can easily get it.  His rationale was that this will stop the kettle from getting chalk residue from the hard water.

I obey my boss, both because he owns the office and so the kettle is his, and because he's an engineer.  But privately, I have doubts about this request.

I know from practice at home that it is very easy to clean chalk residue from an electric kettle: you just boil water with citric acid in it, then rinse the kettle properly, and this removes all the residue.  (Vinegar also works instead of citric acid, I prefer citric acid because it doesn't have the smell and is also cheap.)

On the other hand, I'm worried that if I don't leave any water in the kettle after boiling water for tea, the heating element may still be hot and its surface could become oxidized.  If that happens, that's much more diffictult to fix than the chalk.  Because of this, at my home, I always make sure that there is water in the kettle after boiling, including adding cold water from the tap if I used up all the hot water.   I have very hard water at my home, which causes a lot of difficulties cleaning the bathroom tiles.  (I do not know how hard the water is at the office.)

So what's the best practice at home?  Should I leave the kettle dry or wet after boiling water?

Your boss is somewhat right. Probably not wrong enough to argue with him.

There are several types of scaling that can be caused by mineral content in a water supply. One of the most common is called "temporary hardness" or "carbonate hardness." At high or neutral pH, calcium and magnesium carbonates have quite low solubility in water: 139 ppm for magnesium carbonate, and a mere 6.7 ppm for calcium carbonate, both at 25°C. (NOTE: Until starting this reply, I hadn't realised how little agreement there is on the solubility of calcium carbonate. Every source that I checked agrees that it is quite low; but no two sources agreed to even one sig. fig.!)

However ordinary, clean rain water often has a pH as low as 5.6, through absorption of atmospheric carbon dioxide. (Actual acid rain is even lower!) When such water runs through deposits of these very common minerals, some of them dissolve through the reaction:

$\ce{CaCO_3 + H_2O +CO_2 <-> Ca^{2+} + 2HCO_3^{-} }$

and the similar one for magnesium. Calcium bicarbonate has an aqueous solubility of 166 g/L at 20°C -- some 13,000 times more soluble than the carbonate. Of course, the dissolution process from hard rock is slow, and most runoff will not actually be saturated.

The problem is that our reaction, above, is an equilibrium; anything which tends to drive it back to the left will cause precipitation of the insoluble carbonate. One of those things is just atmospheric outgassing; this is how stalactites and stalagmites form. But a much faster method is boiling the water. That decomposes the bicarbonate ion, and rapidly drives off the freed carbon dioxide gas. The resulting precipitate of calcium carbonate is called "temporary hardness" because, as you have observed, it is easy to remove with any mild acid. Some other forms of scaling are much less forgiving.

Now, if your water hardness problem is severe, a lot of carbonate will start to precipitate as soon as the kettle is heating up. This is why another commentator suggested that it is too late to do anything after the kettle has boiled: that the scale deposits have already grown. However, there are two complications. Firstly, the supersaturated calcium carbonate solution doesn't necessarily all precipitate at once; the scale crystals may continue to grow for some time. I can't say just how long, but a process whereby large, hard crystals tend to cannibalise small, soft ones can continue for weeks.

Secondly, you will not see hardness that bad from mnay common water sources. Well water, maybe, but most municipal supplies will already be softened to the greatest degree that it is economical. If the water hardness problem is mild, some of the calcium carbonate will precipitate immediately, but some (more than 6.7 ppm!) will stay dissolved in the very hot water, until the solubility falls with falling temperature and more of it precipitates.

Long story short, your boss's concerns are not unreasonable. On the other hand, as you note, temporary hardness is easily cleaned off, and only temporary hardness is related to boiling. On the third hand, it's his kettle and he's the boss, so do as you're told!


Unit consistency in rate equations I suppose that my problem is not one of great profundity, but it is an annoying one. The problem is related to the measurement units involved in rate equations of different order. Not being a chemist myself, I have in my work encountered equations of the type: $$\frac{-\mathrm{d}[A]}{\mathrm{d}t} = k [B]^b [A]^a$$ representing the decay of some species $A$. The brackets denote concentration and $k$ is the rate constant. $B$ is some other chemical species (cooking chemical) and then there are the exponents $a$, $b$. For a first order reaction with $b = 0$ and $a = 1$ everything is still under some kind of control; if time is measured in minutes the unit of the rate constant is $\mathrm{min^{-1}}$.

However, if we have a pseudo first order reaction where the species $B$ is present albeit constant during an experimental run, problems arise. In my line of work the species $B$ is typically hydroxide ions and their concentration is traditionally given in molar, $\mathrm{mol/L}$. Even if the concentration of $B$ does not change, the initial concentration plays a role, hence the power of $[B]$ usually appears as a separate factor in the rate equation. Unfortunately, this complicates the unit balance in the equation as frequently the exponent $b$ is a non-integer number. One workaround would be to rescale the concentration to a dimensionless number, but the choice of scaling parameter would be rather arbitrary. Another, perhaps more elegant solution would be to work with molar fractions, but this would be discordant with most of the literature in the field, where the question of unit consistency mostly is disregarded.

A second order rate equation ($b = 0$, $a = 2$, or $b = 1$, $a = 1$) leads to the unit $\mathrm{L\,mol^{-1}\,min^{-1}}$ for the rate constant. I suppose this is still somehow acceptable, even if I would prefer to work with scaled concentrations (divided by the initial concentrations) with unit $\mathrm{min^{-1}}$ for the rate constant, but for fractional order rate equations there is the same problem as in the earlier described pseudo first order case.

Is there any prescribed remedy to this dilemma?

It is true that rate equations typically use molar concentrations for each species, and that for theoretical rate equations you can make sure the units work out by choosing the appropriate units for the rate constant.

There are two things that are important to understand about rate equations, though.

The first is that the "real" rate equations use activity, not concentration. Concentration is usually a "good enough" approximation that this doesn't matter in practice, but when you start thinking about the units of the reaction constant for things like fractional exponents in empirical rate laws, understanding that the concentrations should really be dimensionless activities makes the problem a lot easier.

The second thing to understand is that real reactions don't always follow theoretical rate equations. As a result, what you are often looking at in real life is an empirical rate law. With empirical rate laws, we don't usually worry about the physical significance of fractional exponents, and in turn, how they affect the units of the rate constant. The units of the rate constant become whatever they need to be to make the equation right. In other words, if you have

$\frac{-\mathrm{d}[A]}{\mathrm{d}t} = k [B]^{0.6}[A]^{1.3}$

as an empirically derived rate law that gives $\rm{mol_A}\rm{L}^{-1}\rm{min}^{-1}$, then $k$ would need to have units of $\rm{mol_A}^{-0.3}\rm{mol_B}^{-0.6}\rm{L}^{-1}\rm{min}^{-1}$. These physically nonsensical units are ok because the equation is entirely empirical - we are just describing what we observe on a macroscopic level, without attempting to give it physical significance beyond that.


Making Lyophilized Cake Lookalike using household ingredients I'm working on a machine learning model to identify flaws in vaccines in lyophilized cake form. To train the model, I need a number of samples that look something like this:

I have vials, but I'm having trouble making a suitable cake – I need something that will stick to itself when dried...

What I've tried so far:

Salt dissolved in water/isopropyl alcohol
Baking soda dissolved in water/isopropyl alcohol

Both of these turned back into powder (instead of caking) when dry.

Next, I'm considering using powdered detergent, adding water, then letting it dry...

How would you recommend making this using common household ingredients?

You may want to consider whey. Looks like Karen Smith, Dairy Processing Technologist at the Wisconsin Center for Dairy Research, already did some of the work for you. The result depends on the specific type of whey (a high score of 4 or 5 on the caking test means the material cakes readily, forming a gummy crust):

  Whey  – (Scored 2-5) –  whey exhibited a wide range of caking scores. How the whey is processed has a very large effect on the tendency of the resulting powder to cake as evident in this result. Clearly, two of the samples had large amounts of amorphous lactose and without the presence of significant amounts of protein the samples readily caked.

Conclusion: choose a whey with less protein and more amorphous sugar (such as lactose). In general you are looking for a sticky substance. The amorphous sugar is sticky (it does not form a crystalline powder) and glues the material together. Bodybuilding whey may contain too much protein. Alternately milk powder (which contains plenty of lactose) may do the trick. I'm going to the market now to get supplies and run a test.

Update: I have run a home kitchen caking test on dried skim milk. The dry solids consist of 56% sugar, 40% protein, and ~1 g fat. The milk contains 11% water.

The test was similar to that described in the linked document and consists of placing ~20 g of milk powder in a small (~3 inch diameter) aluminum baking dish and allowing it to float on hot water (~$\pu{50^\circ C}$) in a closed container for about an hour.

The caking test is of course the opposite in a sense of lyophilization, for the solid is exposed to a humid atmosphere to encourage coalescence. The following image illustrates the result:

This confirms the observation in the linked document that NFDM (lower left corner of the image array) "cakes upon storage due to presence of large amounts of amorphous lactose. The protein present limits the hardness of the cake of powder."

This verifies that milk powder cakes. However I do not have a lyophilizer at home so unfortunately cannot verify that the dried milk will remain caked once freeze dried. I might yet put it in the oven to see how this affects the structure.


Why is Fluoride anion less stable than iodide anion? Down the halogen group , electropositive character increases , then why is it that Fluoride ion is the least stable ?
The counter arguement is that fluoride ion has four lone pairs and a small size which makes it least stable.Down the group , since atomic size increases the extra electron is stabilized.
In all such stability trends there are multiple counteracting factors , how to decide which one to give more weightage ?

Actually, the answer is a bit more nuanced. If we observe the pKb values (i.e. tendency to form HX from X-, where X is a B-L base) it is pretty clear that F- would form HF very quickly.

The explanation is that I- would have a much larger area to balance out the negative charge compared to F-, and the anion would be more stable as a result.


Clarification required on whether to use $C_v$ or $C_p$ in this question on finding entropy change $1 \text{ mol}$ of an ideal monoatomic gas at $300 \text{ K}$ is subjected to a reversible isentropic compression process until final temperature reaches $600 \text{ K}$. If initial pressure is $1 \text{ atm}$. Then find $\ln (\text{P}_2)$.

My Attempt

$$\begin{aligned}\Delta S &=nC\ln\left(\dfrac{T_2}{T_1}\right)+nR\ln\left(\dfrac{P_1}{P_2}\right)=0 \\ & \boxed{\ln(P_2)= \ln(P_1)+C\ln\left(\dfrac{T_2}{T_1}\right)}\end{aligned}$$

Now if I take $C=C_v$, I get the answer $\ln(P_2)=1.5\ln(2)$, but the answer is given to be $2.5\ln(2)$ which will be obtained when we take $C=C_p$. So how to decide which one to take?

The correct form can be identified as follows. Start from the combined 1st and 2nd laws:

$$\begin{aligned} d S & = \frac{dU}{T} +  \frac{p}{T}dV \end{aligned}$$

For an ideal gas

$$\begin{aligned} d U &= nC_V dT \\ \frac{p}{T} &= \frac{nR}{V} \end{aligned}$$

such that

$$\begin{aligned} d S & = nC_V \frac{dT}{T} +  nR \frac{dV}{V} \end{aligned}$$

Integrating

$$\begin{aligned} \Delta S & = nC_V \ln \left( \frac{T_2}{T_1} \right) +  nR \ln \left( \frac{V_2}{V_1} \right) \end{aligned}$$

Substitute in the ideal gas law in the right-hand side (note $n$ is constant)

$$\begin{aligned} \Delta S & = nC_V \ln \left( \frac{T_2}{T_1} \right) +  nR \ln \left( \frac{P_1 T_2}{P_2 T_1} \right) \\ & = n(C_V +R) \ln \left( \frac{T_2}{T_1} \right) +  nR \ln \left( \frac{P_1 }{P_2 } \right) \\ & = nC_p \ln \left( \frac{T_2}{T_1} \right) +  nR \ln \left( \frac{P_1 }{P_2 } \right) \end{aligned}$$


Correlation between resonance and coordinate covalent bond So I was listening to an online chemistry lecture the other day, and the theme was "bond".

In the 'coordinate covalent bond' part, the teacher said "Examples of the coordinate covalent bond can be $\ce{SO_2}$, $\ce{HNO_3}$, etc." and I could figure out that both molecules have resonance structures: between $\ce{N}$ and $\ce{O}$ atoms in $\ce{HNO_3}$, between $\ce{O}$ and $\ce{S}$ atoms in $\ce{SO_2}$. Also I am aware of the resonance structures in the $\ce{C_6H_6}$(benzene) molecule, so I looked up the Wikipedia page and could find this sentence:

"..., while the VB description involves a superposition of resonance structures."

So I became curious about this: Is there any close correlation between resonance and coordinate covalent bond? In other words, is it safe to view a molecule as having coordinate covalent bond when it has resonance structure in it?

Based on the examples you have given, that does appear to be the case but if you look for more examples you'll find that this is not particularly true. Two well known compounds can be used to demonstrate:

(1) Carbon Monoxide: Coordinate Covalent Bond from O -> C, but no resonance structure

(2) Benzene: Resonance Structure but no Coordinate Covalent Bonds


Mechanism of Birnbaum–Simonini reaction When silver acetate reacts with iodine and $\ce{CCl4}$, it gives esters and 2 moles of $\ce{AgI}.$ But when it reacts with iodine  and lead acetate, why does it give alkyl halide, $\ce{AgI}$ and $\ce{CO2}?$
$$
\begin{align}
\ce{2CH3COOAg + I2 &->[CCl4][\phantom{(CH3COO)2Pb}] CH3COOCH3 + 2AgI + CO2}\tag{R1}\\
\ce{CH3COOAg + I2 &->[(CH3COO)2Pb] CH3I + AgI + CO2}\tag{R2}
\end{align}
$$
Why doesn't the alkyl-iodide bond break similar to when $\ce{CCl4}$ was used as solvent? What exactly lead acetate do such that alkyl-iodide bond don't break and we get alkyl iodide as product?
Also, my lecturer told that the alkyl iodide bond wont break since lead will form the complex with acetate which is not so clear. I tried to get the answer on Google but no where it shows the explanation for it.

Compiling my comments(now deleted) into an answer.

TL;DR: the products are different due to different stoichiometric ratios of the reactants and is not dependent on the solvent. You can use different solvents so that you can increase the yields.
Hunsdiecker reaction
The Hunsdiecker reaction is a reaction whereby silver salts of carboxylic acids react with a halogen to produce an organic halide.

The reaction was first demonstrated by Alexander Borodin in 1861 by preparing methyl bromide ($\ce{CH3Br}$) from silver acetate ($\ce{CH3CO2Ag}$). Shortly after, the approach was applied to the degradation of fatty acids in the laboratory of Adolf Lieben. However, it is named for Cläre Hunsdiecker and her husband Heinz Hunsdiecker, whose work in the 1930s developed it into a general method and thus the reaction was renamed to Hunsdiecker reaction. The mechanism is given below:

Birnbaum-Simonini reaction
During the discovery of this reaction by Borodin, Angelo Simonini who was working as a student of Adolf Lieben at the University of Vienna, investigating the reactions of silver carboxylates with iodine. They found that the products formed are determined by the stoichiometry within the reaction mixture. Using a carboxylate-to-iodine ratio of 1:1 leads to an alkyl iodide product, in line with Borodin's findings and the modern understanding of the Hunsdiecker reaction. However, a 2:1 ratio favours the formation of an ester product that arises from decarboxylation of one carboxylate and coupling the resulting alkyl chain with the other. It was named Birnbaum-Simonini reaction.

The mechanism is given below:

Simonini reaction
Using a 3:2 ratio of reactants leads to the formation of a 1:1 mixture of both products. This reaction is called Simonini reaction.
$$\ce{3RCO2Ag  +  2I2 -> RI + RCO2R  + 2 CO2  + 3 AgI}$$

What's the matter with different solvents?

The reaction was initially done using carbon tetrachloride. But it was seen that when a different solvent was used, the yield varied slightly. It was seen that the yield of the latter reaction improved when lead tetraacetate was used. So, this solvent was used for this reaction. (See here)


Does fumaric acid form stronger hydrogen bonds than maleic acid? My book claims:

  Fumaric acid can form intermolecular hydrogen bonds (strong) while maleic acid forms intramolecular hydrogen bonds (weak), hence fumaric acid has higher boiling point than maleic acid."

The intermolcular hydrogen bonds in fumaric acid are said to be stronger than the intramolecular hydrogen bonds in maleic acid, while I thought that intramolecular hydrogen bonds should be stronger than intermolecular hydrogen bonds. So, which one is correct?

Intramolecular hydrogen bonds do not have to be weak, but the strength of a hydrogen bond depends a lot on the $\ce{O-H\bond{....}O}$ angle. With intramolecular hydrogen bonds, it is often hard to achieve the perfect $180^\circ$ angle while the same is not a real exercise for intermolecular hydrogen bonds which have full degrees of freedom of movement on both sides.

In fact, I want to argue that maleic acid only forms intramolecular hydrogen bonds if there is no solvent (or crystal water) that can otherwise supply hydrogen bonds. If there is, all hydrogen-bonding sites — both donating and accepting — will be fully saturated with external hydrogen bonds instead; they are slightly stronger due to the better angle.

The question remains why fumaric acid — which is definitely unable to build up intramolecular hydrogen bonds — has a higher boiling point, because in theory maleic acid should also be able to form as many intermolecular hydrogen bonds. However, fumaric acid’s structure already dictates that it practically must form a network of coplanar molecules interacting with each other via hydrogen bond dimers from carboxylate to carboxylate. This is a very nice structure to keep. Maleic acid cannot form such an extensive network no matter how much it distorts. It will be restricted to much less favourable crystal structures meaning weaker intermolecular attraction.

In the liquid phase, fumaric acid will keep the near-sphere ordering somewhat intact, i.e. conserve a high number of hydrogen bonds to its neighbours. However maleic acid will probably stick with one fully intramolecular hydrogen bond that does not contribute to intermolecular attractions, thereby lowering its boiling point.


Reaction with sulfuric acid Which of the following will react with dilute sulfuric acid? : copper, copper (II) oxide, copper (II) hydroxide, copper (II) carbonate.
My answer:
Metals react with acids to give salt and hydrogen. So, copper will react with dilute sulfuric acid.
Metal oxides act as base. So, they neutralise acid to give salt and water. So, copper oxide will react with sulfuric acid.
Metal hydroxides are bases. So, they will neutralise acids. So, copper hydroxide will react with dilute sulfuric acid.
Carbonates react with acids to give salt, carbon dioxide and water. So, copper carbonate will react with dilute sulfuric acid.
So, my answer is all the given options will react with dilute sulfuric acid. Is this correct? The answers are not given in the book.
I am afraid what if there are exceptions to the general reactions that I have written above and I am not aware of that.

Metallic copper does not react with dilute sulfuric acid. Its redox potential is too positive to do it.
All other copper compounds do react with sulfuric acid, as you had suspected.


Is 98%+ purity Reagent Grade Ca-EDTA powder as safe to use as USP Grade in the body? Is $98\%+$ purity reagent grade $\ce{Ca}$-EDTA powder as safe to use as USP grade in the body?  Or, does the reagent grade have unsafe impurities that USP grade would not have even though they are the same purity?

While normally health-related questions such as this one are considered off-topic, this one is sufficiently cut-and-dried that it probably merits an answer.

NO!!!! In general, reagent-grade chemicals are NOT safe to use internally like USP grades.  The entire point of the USP specifications is to ensure, to the extent practical, that the relevant chemicals are prepared from suitable starting materials and using suitable methods such that toxic or hazardous materials are absent from the final product. This is a big reason why the USP grades are often more expensive than the lab grades.

It's certainly possible that a given reagent-grade chemical may actually be safe to ingest, but there is no real way to know for sure without conducting extensive analysis of the impurities.


What happens when chemical equilibrium is restored after being disturbed? I'm aware that, according to Le Chatelier's principle, a reaction will shift its equilibrium position in order to counteract the disturbance at equilibrium. However, I don't know what that means after equilibrium is re-established.

For example, take the generic reaction

$$\ce{a A + b B <=> cC}$$

At some point, equilibrium is disturbed, causing the forward reaction to be favored. Once equilibrium is re-established, how do the concentrations of the species compare to before equilibrium was disturbed?

Once equilibrium is re-established, how do the concentrations of the species compare to before equilibrium was disturbed?

Let's say it is a homogeneous equilibrium. You have to consider different cases:

The equilibrium was disturbed by changing one of the concentrations (removing or adding a substance). You can't get back to the same set of concentrations because there are too few or too many atoms.
The equilibrium was disturbed by adding solvent. Again, you can't get back to the same set of concentrations (all are lower than at equilibrium, and a net reaction in one or the other direction can't increase all of them at once).
You disturbed the equilibrium by changing the temperature. The equilibrium constant is at a different temperature will be different, so the old set of concentrations will not satisfy the new equilibrium constant.
You disturbed the equilibrium by intermittently changing the temperature (or - more elaborate - by attaching a power source to an electrochemical cell, and then replacing it by a wire shortcutting the cell). In this case, the concentrations change exclusively because of the chemical reaction (no "outside" disturbance), and they can return back to the original concentrations once the disturbance is removed.

In cases 1. and 2., however, the reaction quotient Q will go back to the original one (=K). In case 3., Q will be different.


Is the conjugate base of every polyprotic acid amphoteric? Since all compounds with an accessable free lone pair can act as a Brønsted-Lowry base, I was wondering if there was any specific example of a substance that when it loses a proton, it cannot accept a proton but can donate another proton.

Please note that I am not considering something like sulfuric acid which is a strong diprotic acid, as $\ce{HSO^{-}_{4}}$ still accepts a proton even if its to a limited extent.

The conjugate base of sulfuric acid, bisulfate ion, shows little amphoteric character in water solution.  Unless you drop the pH to about 2 or below, it is only an acid, and a moderately strong one at that.