Is the XRF spectroscopy equally sensitive for every element?

In a very very broad reasoning, XRF is much more sensitive to the "mid-range" elements.

Elements below Na are generally not detectable.

The x-ray tubes must have a beryllium window to seal the tubs which absorbs the low energy x-rays.
The lower the atomic number the less the fluorescent yield.
Low energy x-rays don't penetrate from the sample efficiently.
Low energy x-rays tend to be absorbed in the detector windows.

For the very heavy elements the detectability is poor too. Say anything about lanthanum.

The K x-rays are so energetic that a high voltage on the x-ray tube is needed to excite them.
The lanthanides and actinides tend to behave similarly chemically so geological samples tend to contain a mixture which creates peak overlaps.
Higher energy detection of x-rays is poor. EDS because the x-rays penetrate the detector. WDS because the available crystals don't diffract high energies well.
The L x-rays are not quite as useful because the greater number of x-rays leads to more peak overlap.


Which cyclohexane conformation is more stable?

By symmetry, the cyclohexane on the right requires each carbon atom to have a $\ce{C-C-C}$ angle of 120 degrees. From VSEPR theory, this is not optimal for tetrahedral carbon. A regular tetrahedron has angles between vertices of approximately 109.5 degrees and the equilibrium bond angles of a secondary carbon won't be too discrepant from this.


Which one of the compounds listed would we use in order to get the highest concentration of sulphate ion in a saturated solution?

The $K_{sp}$ values are the way to go .. the other information you mentioned are just some convenient rules of thumb for predicting relative solubilities of compounds when you don't know the $K_{sp}$ values.  Also, the 4th choice in your list .. radium .. probably also belongs on that list, since it's the next alkali earth element after barium, but it's included because it's a radioactive element, and people don't commonly use or encounter it.


Naming phosphonium cations in ionic liquids

Skepticism is an integral part of a good scientist. Remember the motto "Paper never refused ink" when reading papers and its Supporting Information. You will find incorrect references (very frequent), occasionally wrong calculations, wrong formulae, or wrong names as you read more and more papers just because scientists, like normal human beings, make mistakes.
The strategy in such cases is not to stop at two papers, because the odds are 50:50, but find at least 6-7 papers to confirm the truth.
Your notation is correct for $\ce{P2225}$.

Reference: Another paper from RSC Two phosphonium ionic liquids with high Li+ transport number, Phys. Chem. Chem. Phys., 2015,17, 23041-23051
The odds are in your favor now.


Is cyclohexane‐1,3‐dicarboxylic acid a correct IUPAC name?

The answer to this question is actually pretty interesting, I think.  @user55119 's comment that the principal chain is the cyclohexane ring is correct -- but it seems like you understand that that's what the teacher is saying, and you're asking, "why?"  The best short answer, too, is imho in your comments: @MaxW 's "naming organic compounds is just screwy" is the bottom-line answer, because there are so many crazy special cases to deal with that it sometimes seems like sorcery to come up with them.  But if one is willing to dig into it a little, 99% of them do actually have consistent reasoning behind them -- even if coming up with that reasoning can feel pretty byzantine right up to the point where you get it.  This can be particularly true for cyclic compounds, which can in some cases even have more than one valid IUPAC name.
In this particular case, the reason your answer is incorrect is because what you defined with your numbering isn't a chain.
Per IUPAC, a chain is a sequence of linked units that is bounded by precisely two "boundary units" -- either a branch point or an "end unit", which is itself defined by IUPAC as a unit* of the macromolecule that is bound to only one other unit.
Now, this clearly causes problems with cyclic compounds, so the IUPAC has some caveats included within the definition of a chain -- one being that a cyclic macromolecule with no end groups is also a chain.  So, a chain is either linear and has two boundary groups terminating it, or it's strictly cyclical and includes no end units.  That means a cycloalkane -- including the cyclohexane in your example -- is a valid chain.
Your numbering, however, contains one boundary group.  The unit at position 1 is a valid boundary group, as it is connected to only one other group.  The unit at position 7, however, is not a boundary group; it is connected to a total of two other units (#6 and #2), and worse, it's a non-boundary that you can't continue counting from either since you've already counted the other molecule it attaches to!
"Maybe it solves the problem if I start from the unit I called #3, and leave the hanging methyl group to be the last part of the chain instead of starting from it -- so the chain would start at current #3 and end at current #1, 3-4-5-6-7-2-1."  (You may want to be looking at your illustration while you read this; it'll probably make the counting I am referring to clearer than just imagining it.)  No, that also doesn't work; it obviously that just puts the problem at the other end -- it still has only one boundary unit, because while the unit at the end of that new "chain" attempt, #1, is only connected to one other, the first unit in the "chain" (#3) is connected to two units, not one, so is not a boundary unit.  Still stuck!
And of course, you can try counting from your #2 instead of your #1 or #3, but then you go all the way around and never have a way to include that methyl group hanging off of #2 -- you'd have to "step" onto #2 again, effectively counting it twice, to then "step" onto the tail CH.  (I wish I could make a video of this "pointing" at each unit in the count; it would come out much clearer, I think!)
So in effect, really the only way you have to count the components of the macromolecule and get a coherent chain out of it is to start on #2 and end on the connection between #7 and #2 -- meaning you end up with the most practical solution being exactly the caveat that IUPAC settled on in specifying that a cyclic alkene can itself also be counted as a chain.
From there, I imagine you can figure out the rest: the cyclohexane as a valid chain is also the longest chain,  you start numbering from the functional group, etc.  Hope that helped explain the teacher's answer!

Edit/addendum: Some in the comment have asked for authoritative sourcing of the above.  Good point, and thank you for bringing to my attention that I only stated what I used for sourcing but neither completely cited nor offered links to said sourcing.  First post here in the chemistry StackExchange; please forgive my error.
IUPAC's authoritative reference work for nomenclature is their so-called Gold Book -- the Compendium of Chemical Terminology.

IUPAC. Compendium of Chemical Terminology, 2nd ed. (the "Gold Book"). Compiled by A. D. McNaught and A. Wilkinson. Blackwell Scientific
Publications, Oxford (1997). Online version (2019-) created by S. J.
Chalk. ISBN 0-9678550-9-8. https://doi.org/10.1351/goldbook.

I cite various definitions from within the Gold Book, the following two being of particular importance:
IUPAC - chain (Note 2). https://goldbook.iupac.org/terms/view/C00946
IUPAC - end-group. https://goldbook.iupac.org/terms/view/E02092
To address one of the comments in particular: it should be noted that nowhere in the Gold Book is there an authoritative definition of the term "ring."  That would make claim that "according to IUPAC a ring cannot be a chain," already dubious based on the definition I cited above, even less likely to be the case -- they certainly aren't going to "prohibit" something they don't even define.
The complaint about the definition of "chain" referring to "polymers" as if it were a notation of exclusivity and implying that the Gold Book's definition somehow contradicts or does not apply to the Blue Book should be pretty handily refuted with a bit more reading in both the Blue and Gold books. The Blue Book states it doesn't deal with preferred naming of polymers, but does still define the systematic naming of organic structures whether polymeric or not.  (Blue Book P-11, p.3)  The Gold Book defines "polymer," very simply and with no caveats, as "a substance composed of macromolecules."  Since "chain" and "polymer" thus both use the term "macromolecules," one would naturally be led to read that definition as well, where along with the (hopefully) uncontroversial definition of being a high-molecular mass chain of repeated low-molecular mass components, one might stumble onto Note 2:

"If a part or the whole of the molecule has a high relative molecular mass and essentially comprises the multiple repetition of units derived, actually or conceptually, from molecules of low relative molecular mass, it may be described as either macromolecular or polymeric, or by polymer used adjectivally."

No clearer acknowledgement of the lack of some "hard line" apparently perceived by some between polymers and "regular" organic macromolecules should be necessary.
When it comes to systematic naming, especially, polymers aren't a "special class" of substance which are somehow privy to different definitions and procedures -- indeed, the Blue Book tells us that's precisely the opposite of the case.  So, for those strictly systematic purposes (which is the entire point of the question, and should be in most introductory academic settings because the point isn't to have students memorize some esoteric rules but instead to teach them something about the structure of molecules and their function within a larger compound) - the definition of "chain" given applies just fine. In fact, the definition of "polymer" is expressly defined in the Gold Book, again IUPAC's authoritative text, such that it could apply to any simple alkane with a side branch for the purposes of that text.  Rather than second-guess clearly written prose, I'd rather assume the working group knew what it was doing there by leaving it vague and allowing common usage to define terms that are inherently ambiguous.  (Also see: What is the difference between Alkanes and Polymers?)  They seem more aware that strict legalism doesn't serve their purposes than most of the practitioners of their work do.
And for those of you who skipped this addendum entirely: good for you.  As interesting as some of the contentions below are, it seems generally agreed by all that none actually change the correct answer to the original question asked anyway; whatever your take on the semantic issues around the Blue and Gold (and Purple!) books that took up half the comments, they're just that -- semantics.  One can think that an answer is based on a definition that is "irrelevant" due to its source, which I find an odd and not particularly useful criticism (but hey, that's just me); but thinking then that said definition is "incorrect" is false logic.  If someone who has never seen a math book in their life believes 2+2=4 because some hot dog vendor on a street corner told them so, it doesn't make 2+2 equal anything other than 4.  Fortunately for them, just because their source has "no relevance" doesn't mean that they can't have a grasp of basic math.


Why should epoxides be more reactive at a position where positive charge is stabilised?

To the question mark in (2): this implication is simple:  if the charge were on a neighboring carbon you'd have a state of higher energy.  "Can better stabilize" is a synonym for "gives lower overall energy".

To the objection in (4): the stability ranking refers to various electronic distributions in the possible transition states. In the transition state of an SN2 reaction, and when there is a positively charged leaving group, the carbon undergoing nucleophilic substitution will bear a partial positive charge in the transition state. So, the degree to which this positive charge is stabilised settles the matter of which carbon will be attacked preferentially.

The reactivity ranking refers to the fitness of each atom in the intermediate as a site of attack by the nucleophile. The most likely target is the one which leads to the lower-energy transition state, i.e. has a lower activation energy barrier.

So there is no contradiction.  The carbon most able to stabilize a positive charge is also the one most likely to be the target of a nucleophilic attack (other things being equal; of course there could be factors, like sterics, changing the ranking).


Reaction mechanism (SN2 and SN1 reactions)

SN1 is characterized by the possibility of formation of a stable carbocation during the reaction, whereas in the case of SN2 there cannot be formed a stable carbocation, so the most stable "intermediate" structure would be a transitional state. Stable carbocations are formed via groups with strong inductive electronic effects (+I those that repell electrons) or they can be stabilized by conjugation (the case for this one is when you have a double, triple bond or an aromatic ring directly bound to the carbon atom where the substitution takes place)
so, for example, (CH3)3C-Cl would make for an SN1 when reactiong with alcohols. Also compounds like CH2=CH-CH2-Cl.
The remaining cases, for example CH3-Cl or CH3-CH2-Cl would work via SN2, because the carbocation cannot be stabilized.
Hope that's gonna help you :)


Confusion about the oxidation state of copper in copper cyanide re-dissolved with excess cyanide

but we know that $\ce{Cu^2+}$ is more stable hence anwser should be a).

Ah, but do we? I hope you don’t ‘know’ that, because that ‘knowledge’ would defy years of chemical research.

The stability of an oxidation state, especially that of a transition metal, cannot be deduced easily or by a set of memorised rules. Many a time, a species will only be stable in one environment and unstable in another one. Take copper: In aquaeous solutions, copper(I) is usually unstable concerning oxidation to copper(II). However, copper(II) on the other hand is unstable in the presence of iodide and the two will react:

$$\ce{2Cu^2+ + 4I- -> I2 + 2CuI v}$$

So copper(I) is more stable in the presence of iodide while copper(II) is more stable in the presence of water. Notice a pattern? I don’t.

In your question specifically, the first step is a redox reaction between copper(II) and cyanide according to the following equation:

$$\ce{2 Cu^2+ + 4 CN- -> NCCN ^ + 2 CuCN v}$$

At least I notice a pattern here: Cyanide behaves much like a halide or the iodide introduced above.

By this reaction, we have established that copper(I) is more stable in the presence of cyanide than copper(II). So by adding more cyanide there is no way that we can reverse the reaction — not only because there is nothing that copper could reduce with its extra electron. So the redissolution and complex formation cannot be a redox reaction, it must be a simple complexation reaction which only allows for the correct solution to be:

$$\ce{CuCN (s) + 3 KCN (aq) <=> K3[Cu(CN)4] (aq)}$$


Irreducible representations and system states connection

The symmetry of a state is the MO (molecular orbital) product of its occupied molecular orbitals. Let's take the example of water molecule. We can see the symmetry label of each of the MO in the diagram. Each MO is assigned an irreducible symmetry label (associated with the point group of molecule) which can be deduced from the symmetry of MOs in space (or from the LCAOs). The wavefunction associated with the MOs also have the same symmetry species. Now, this arrangement of electrons represents the ground electronic state of water, so the symmetry label of the ground state of water can be obtained as

$(2a_1)^2(1b_2)^2(3a_1)^2(1b_1)^2=(A_1)^2\bigotimes (B_2)^2\bigotimes (A_1)^2\bigotimes (B_1)^2 =A_1$

Follow the same procedure to obtain ground state of $N_2^+$. Let me know if you have any questions.


Ratio of uncharged to charged side chains

I think you are simply misunderstanding the "uncharged to charged" part of the question. Imidazole is a base, so it gains a proton to become charged when the pH < pKa. That's why you got the right answer when you used $\ce{HA+}$ as the acid, as you put it. In this case, unlike with most general chemistry problems of this nature, the acid is charged while the conjugate base is uncharged.

Below pH 6, imidazole exists largely as a protonated imidazolium ($\ce{HA+}$) species. Above pH 6, it is primarily in the neutral ($\ce{A}$) imidazole form. When imidazole is protonated (i.e., in acid or $\ce{HA+}$ form) it is charged. When it is deprotonated (i.e., in base or $\ce{A}$) form, it is uncharged.

So the proper expression is $$7 = 6 + \log\frac{[\ce{A}]}{[\ce{HA+}]}$$ which should lead you to the correct answer:

$$1 = \log\frac{[\ce{A}]}{[\ce{HA+}]}$$

$$10 = \frac{[\ce{A}]}{[\ce{HA+}]}$$

So, the ratio of uncharged:charged species is $\frac{[\ce{A}]}{[\ce{HA+}]}$, or 10:1.


Classical Least Squares in UV-Vis Absorbance Experiment

That depends on the nature of the analyte and matrix. Ideally (Beer's Law) A should be linearly proportional to concentration but that may not be the case. As an example of this in the water industry barium chloride is often added to samples to determine how much sulfate ion they contain. Spectrophotometers are used to determine how much barium sulfate is in suspension. The relationship between A and sulfate concentration is not linear in this case.

What you should do is make up a series of standards spanning the range of concentrations you are interested in, measure the absorptions and plot A vs. c. Now start fitting curves, starting with linear and progressing to higher order until doing so does not make the residuals look any noisier. That gives you the order of the response curve you need and the number of standards required is one more than the order.

Ideally, the matrix wont interfere and Beer's law applies. In that case when c = 0, A = 0 and you only need one standard.


Yes we can have hydroxyferrocene.  See here for a brief summary description of possible synthesis and a nucleophile substitution reaction forming alkoxyferrocenes.


Are the names for chemical elements the same in both modern Greek and classical Greek?

For "old"  elements (i.e., known / discovered before the advent of a body like IUPAC), there was no uniform and universal rule.  You see this heritage by comparison of the name of an element assigned in different languages; languages in neighboring countries may but need not to share their approach.  E.g., $\ce{_7N}$: nitrogen (English), nitrógeno (Spanish), nitrogen (Nowegian Norsk bokmal); azoto (Portuguese), azoto (Italian), azote (French), azot (Polish), azotas (Lithuanian), aзо́т (Russian), άζωτο (Greek); stikstof (Dutch), Stickstoff (German); dusík (Czech), etc. for a few examples you may encounter in Europe.
For newly discovered / newly synthesized elements beyond atomic number 100, IUPAC published (back in 1979) a recommendation how to name them systematically.  For this you have the roots

0 = nil, 1 = un, 2 = bi, 3 = tri, 4 = quad, 5 = pent, 6 = hex, 7 = sept, 8 = oct, 9 = enn

(Pure & Appl. Chem. 51, 1979, 381-384, open access)
Thus it was until specific agreement that unnilunium ($\ce{_{101}Unu}$) eventually was named mendelevium ($\ce{_{101}Md}$), and IUPAC continues to publish how to name the new elements (example).
Despite this report, however, the roots are neither pure Latin, nor Greek, but convention.  Because there no tenners (like decem, vīgintī), or hundreds (like centum, ducentī), etc.  Even spelling the numeri only by position, you would expect unus, duo, tres, quattuor, quinque, sex, septem, octo, novem for $1\dots9$ for the former.
For the names eventually adopted, IUPAC set the rules that these

"In keeping with tradition, elements are named after

a mythological concept or character (including an astronomical object);
a mineral, or similar substance;
a place or geographical region;
a property of the element; or
a scientist.

[...] The names of all new elements should have an ending that reflects and maintains historical and chemical consistency. This would be in general “-ium” for elements belonging to groups 1–16, “-ine” for elements of group 17 and “-on” for elements of group 18. N.B. The present recommendation is here more specific than that
written in the 2002 document."
(Pure & Appl. Chem., 88, 2016, 401-505, open access)


Resonance in carbocation derived from allenes

The answer to your question might need a few clarifying points. My answer will mostly focus on the protonation (and deprotonation) of allene based on gas-phase experiments and theoretical calculations. A few references that point to elements that will be used concerning the allyl and 2-propenyl cations are as follow : G. van der Rest et al. Eur. Mass Spectrom. 3 323 (1997) and K. Raghavashari et al. J. Am. Chem. Soc. 103 5649 (1981).

Relative stability of the allyl and 2-propenyl cations

Both articles make it clear that the allyl cation (I) is the most stable structure for the $\ce{C_{3}H_{5}^{+}}$ carbocations. It is around $\pu{21 kJ mol-1}$ more stable than its 2-propenyl (II) isomer. The other possible isomers are all higher in energy.

Protonation of allene

According to the first reference above, the protonation of allene on an extremal carbon to lead to the allyl cation requires to pass a high energy barrier, whereas no high barrier is present to lead to the 2-propenyl cation. Figure 4 gives the structure of the interaction complex between $\ce{H_{3}O^{+}}$ and allene, showing clearly that the incoming proton is initially in interaction with the $\pi$ orbital on one side of the allene. Their is obviously no conjugation contribution between this protonation structure and the other $\pi$ orbital, since the proton arrival plane is perpendicular to the other $\pi$ orbital.

From this position, formation of the 2-propenyl cation (II) requires only a minimal energy barrier, requiring only a bending of the $\ce{C-C-C}$ bond and bending of the $\ce{CH_{2}}$ group. Once this is done, the $\ce{CH_{3}}$ group can rotate freely, but one should notice that it does not involve the $\ce{sp^{2}}$ vacant orbital located on the central carbon. This orbital will always remain orthogonal to the remaining $\pi$ orbital.

From this position, one could also consider direct protonation of the central carbon, followed by a rotation of the $\ce{CH_{2}}$ group since no conjugation is present at this point. But here, one should consider that stabilization by conjugation can only proceed after formation of a (non stabilized) primary carbocation. These species are much less stable than secondary carbocations (on the order of $\pu{75 kJ mol-1}$), thus leading to the presence of a large barrier for the direct formation of an allyl cation.

A last pathway that one should consider is a 1,2 hydride transfer between (II) and (I). Although hydride transfers are generally easy between carbocations, this transfer proceeds through a three-center system which can be considered a proton attached to a $\pi$ orbital. Thus, this leads us back to the initial problem, which is that no conjugation can serve as a driving force leading to the allyl cation. And only once the unstable primary carbocation is formed can a rotation of the $\ce{CH_{2}}$ group lead to formation of the allyl cation.


Copper plating with electrolyte not working (Cu(OH)2 vs CuSO4)

If you are going to do anything differently than is normally done, you should understand what you are going to do and why it should work.(*)
The essential (but often not sufficient) conditions for successful electroplating by any metal are:

The used metal salts/compounds are soluble, so ions can migrate within the solution in applied electrostatic gradient. That is not the case of copper in copper hydroxide nor alkalic copper carbonates, which are insoluble.
The metal salt has sufficient concentration, so electrolytic current reacts primarily with this salt, at least on the cathode.
The current density on the plated electrode is low enough ( typically $\pu{10 mA/cm^2}$).

The solution composition for successful creation of shiny metal surface can be tricky, based on intensive empirical reasearch, involving solution modifiers like sugar or soluble polymers. The goal is mostly decreasing ion mobility and/or the rate of electrode reaction to support even metal deposition.

(*) It does not fully apply in the basic research, which uses often "what if.." approach, but this is not the case.


Rock salt structure: chloride lattice or sodium lattice?

Unit cells are just a way of describing a given crystal. You can certainly draw $\ce{NaCl}$ the other way, with $\ce{Na+}$ in the corners and it's still correct. There are a few ways of thinking about this structure:

as two interpenetrating FCC lattices
as a $\ce{Cl-}$ lattice with $\ce{Na+}$ in the octahedral holes
as a $\ce{Na+}$ lattice with $\ce{Cl-}$ in the octahedral holes

They're all equivalent. The figure above actually shows both types of octahedral holes. Some might prefer to visualize the structure as the second case since, as you say, the chloride ions are bigger, but going the other way isn't wrong, nor does it describe anything different.


How to chose wavelength in spectrophotometry? How to use Beer-Lambert Law?

Working wavelength is chosen by analysing the spectrogram $A(\lambda)$. Such diagram is obtained by measuring absorbance $A$ in function of wavelength $\lambda$. Practically, you test a moderate concentration of your analyte and sweep the complete wavelength bandwidth of your analyser. Then you select wavelength where a maximum of absorbance occurs. If multiple maximum are present, you may prefer the highest one (best sensitivity) or a smaller one with less interferences (best selectivity). Both can happens simultaneously. The main criterion is:

$$ \underset{\lambda}{\max}\left\{A(\lambda)\right\} $$

For information: the wavelength is linked to photon energy $E$ by the following relation:

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

Where $\nu$ is the frequency of the EM wave, $c$ the speed of light and $h$ the Plank constant.

Absorbance is a property of matter, when an atom/ion/molecule absorbs a photon (visible or nearly visible IR/UV), it can promote electron to a highest energy level (UV, highest energy) or increase vibrational energy (IR, lowest energy) of the compound and the photon is destroyed.

A spectrophotometer assess the transmittance $T$ (ratio of transmitted $\phi$ and incident flux $\phi_0$ expressed as powers) of light beam through a sample cell.

$$ T = \frac{\phi}{\phi_0} $$

And transmittance (dimensionless) is linked to absorbance (dimensionless) in the following way (functional scaling to achieve linearisation between signal and physical quantity):

$$ A = -\log_{10}(T)$$

Beer-Lambert Law is a limit law conditioned by several hypothesis and it is only valid for low concentrations. When Beer-Lambert applies you can linearly link concentration $x$ with absorbance.

$$ A(x) = \epsilon(\lambda)lx $$

Where $\epsilon$ is a proportional coefficient (with suitable units) called extinction coefficient which describes how matter absorbs light at wavelength $\lambda$. The quantity $l$ stands for the optical path length through the sample cell.

Numerous substances validate Beer-Lambert Law but some does not. Therefore you must assess linearity before quantifying unknown concentrations. This is usually done when performing calibration curve. Then you know your dynamic range and sensitivity of your analytical method and you can regress concentration from absorbance.

I suggest you to read more about those concepts in a reference book such as Skoog & West, Fundamentals of Analytical Chemistry or Vogels, Textbook of Quantitative Chemical Analysis. There is also a free electronic book available on internet called Analytical Chemistry 2.0.


The link you gave in the question, actually shows the equation of how this sensor works. You don't need to produce any electrons, you just need to measure the voltage.

Try to write the Nernst equation for the above overall reaction and that will show you what is going on.


Do Na or Li need heat in order to react with Cl₂?

From personal experience, Na(s) and Cl2(g) react at room temperature without added energy. Li(s) and Cl2(g) should do so also.

Consider the following heat of formation data harvested from the NIST Chemistry Webbook:

For NaCl:
$$\ce{Na(s) + } \frac 1 2\ce{Cl2(g)->NaCl(s)}\quad \Delta H_f^o=-411.12 \; kJ/mol $$

For LiCl:
$$\ce{Li(s) + } \frac 1 2\ce{Cl2(g)->LiCl(s)}\quad \Delta H_f^o=-408.27 \; kJ/mol $$

These data suggest that both reactions should occur without added heat. However, the rate of reaction is not determinable from these data. The reaction between Li(s) and Cl2(g) will likely be slower.


How does partial reduction of esters to aldehydes occur?

DIBAL exists as a bridged dimer, and it becomes a reducing agent only after it has formed a Lewis acid–base complex, so it reduces electron-rich carbonyl groups most rapidly.

DIBAL will reduce esters even at –70 °C, and at this temperature the tetrahedral intermediate that forms during the reaction may be stable. Only in the aqueous acidic work-up does it collapse to the aldehyde (by expelling EtOH in the example reaction below) when excess DIBAL has been destroyed so that no further reduction to the alcohol is possible.

A stable tetrahedral intermediate is more likely in the reduction of lactones, and DIBAL is most reliable in the reduction of lactones to lactols.

Edit:

As requested by the OP here are some more details about the collapse of the tetrahedral intermediate: During the aqueous acidic work-up the alcohol group is turned into a good leaving group via protonation. Thus, it is easily expelled from the molecule, when a water molecule removes the $\ce{Al^{$i$}Bu_2}$ group and the $\ce{C=O}$ double bond is reformed.

A different route, where the former carbonyl-oxygen is protonated twice - thus being turned into a good leaving group - and afterwards expelled from the molecule to give an oxonium ion is possible too. But such an oxonium ion is rather unstable and highly reactive, so that the reverse reaction is very fast and the oxonium ion is not produced in any appreciable amount.


Points needed to come up with a calibration curve

There is not really a simple answer to this—it depends on what you're doing and what level of error your application can tolerate. If you've characterized the response of your instrument and you know that it's linear within a certain range but the response has a variable offset, even a single-point calibration may be sufficient. On the other hand, if you just built an instrument and have no idea of its response characteristics but need very accurate results, you're best to take measurements at many different points (and over a length of time to see how stable the response is) to accommodate the possibility of non-linear response in the range of interest.

Occasionally, if you need to follow a measurement standard (e.g. the ITS-90 standard for temperature), you may be constrained to doing it a certain way so your data can be directly compared with that of other labs but in general, do enough to get the results you need, but not so much that you're just wasting your time. This might mean taking a few well-separated calibration points to get a ball-park value for an unknown sample, then bracketing the value with more calibration points near it to get a better measurement.


Look at it in this way. Consider the energy to break a bond to be arbitrarily assumed to be at $\pu{0 J/mol}$. Now, there are two bonds, $\ce{A-B}$ and $\ce{C-D}$. Let's say they have energies of $\pu{-242 kJ/mol}$ and $\pu{-378 kJ/mol}$. Clearly, the $\ce{C-D}$ bond has the lower energy.

Now, you can say that a stronger bond needs more energy to break it. So, this means that you need to supply $\pu{242 kJ/mol}$ of energy to break $\ce{A-B}$ and $\pu{378 kJ/mol}$ of energy to break $\ce{C-D}$. You can definitely see that $\ce{C-D}$ needs more energy to break it, and to bring the energy to $\pu{ 0 J/mol}$.

Hope this answer helps you.


Observations when mixing potassium Triiodide and Sodium thiosulfate

The keyword you should search is iodometric titration. Triiodide solutions are dark brown, just like the tincture of iodine. If you were to add thiosulfate ion (say from a buret) to  the triiodide solution would become colorless. The change is not that sharp. Therefore this is not the way end point is detected. A small drop of starch is added. Starch makes a deep bluish purple complex with free iodine. The moment all the iodine (in the form of triiodie $\ce{I2.I-}$ ) is consumed by thiosulfate, there is a very sharp transition from bluish purple to a colorless solution.


Why is water not part of the equilibrium constant?

There is no IUPAC definition of $K_a$, however, many reputable texts like Levine's Physical Chemistry define $K_a$ as:

$$\frac {a(\ce{A-})a(\ce{H3O+})}{ a(\ce{HA})a(\ce{H2O})}$$

where "a(X)" is the activity of species "X".

Then various approximations can be made, such as approximating $a(\ce{H2O}) = 1$ and approximating the activity of the solutes as the concentration of the solutes.

These approximations are only reasonable in dilute solutions.


Is it possible to find fire outside the Earth?

Many non-organic substances burn. For example:

Reactive metals such as potassium burn when exposed to oxygen. (Thanks to @Jon Custer in his comment)
Non metal elements such as carbon, sulphur and phosphorous also burn in oxygen when heated.

So yes fire is definitely possible outside earth.

Additionally simple organic compounds such as methane are known to exist in large quantities on other bodies such as Titan, a moon of Saturn, which has a whole 'hydrological' cycle based around methane, and to a lesser extent ethane and heavier hydrocarbons, with a methane atmosphere and lakes of methane on the surface. (Thanks to @ron in his comment)


The second one, $(2)$, $\ce{<->}$, indicates resonance structures.

The difference between the second two is explained in another post -

What are the correct equilibrium arrows?

Loong basically says that $\ce{<-->}$, $(4)$, indicates forward and reverse elementary steps happening in equilibrium, that is that those are the only two things actually reacting with each other. On the other hand, $\ce{<=>}$, $(3)$, indicates a net reaction, that is that there may be multiple steps in between, leading to this net equation. He does a more in depth explanation of this in his post.

Case $(5)$ indicates a net equilibrium where products are favored, and $(6)$ indicates a net equilibrium where reactants are favored.


Calculating charge on amino acid from pKa

For this, you can use the Henderson-Hasselbalch equation. Using the degree of dissociation, $\alpha$, this can be written as
$$\mathrm{pH} = \mathrm{p}K_\mathrm{a} + \log\frac{\alpha}{1-  \alpha}$$
Rewriting to solve for $\alpha$:
$$\alpha = \frac{1}{10^{\mathrm{p}K_\mathrm{a} - \mathrm{pH}} + 1}$$
As stated above, $\alpha$ is the degree of dissociation, meaning the degree at which $\ce{H+}$ is dissociated from the ionizable group. At $\mathrm{pH} = \mathrm{p}K_\mathrm{a}$, $\alpha = 0.5$, meaning that $50\%$ of the ionizable groups in question are deprotonated. If $\mathrm{pH} = \mathrm{p}K_\mathrm{a} + 1$, about $90\%$ of all groups are deprotonated and if $\mathrm{pH} = \mathrm{p}K_\mathrm{a} + 2$, about $99\%$ of all groups are deprotonated. (If you want to, take a look at the shape of the $\alpha$ distribution for $\ce{-SH}$ in cysteine at Wolfram Alpha.)

From a textbook I found the following $\mathrm{p}K_\mathrm{a}$ values for cysteine:
\begin{align}
 \mathrm{p}K_\mathrm{a}(\ce{-COOH})   &= 1.9\\
 \mathrm{p}K_\mathrm{a}(\ce{-NH_3+}) &= 8.35\\
 \mathrm{p}K_\mathrm{a}(\ce{-SH})     &= 10.5
\end{align}

From these values, $\alpha$ can be calculated for each ionizable group at the desired pH and this will give you the net charge of the amino acid. Upon deprotonation, the following changes in charge occur for the ionizable groups:
\begin{array}{lclcr}
 \ce{-COOH} &:& 0 &\rightarrow &-\\
 \ce{-NH_3+}&:& + &\rightarrow &0\\
 \ce{-SH}   &:& 0 &\rightarrow &-\\
\end{array}

As an example, let's calculate the charge of cysteine at pH 10. Using the Henderson-Hasselbalch in a spreadsheet yields the following $\alpha$ values:
\begin{array}{lcr}
 \ce{-COOH}  &:& 0.9999\\
 \ce{-NH_3+} &:& 0.9781\\
 \ce{-SH}    &:& 0.2403\\
\end{array}

This gives the following total charge for cysteine at pH 10:
$$(-1)\cdot0.9999+(+1)\cdot(1-0.9781)+(-1)\cdot0.2403=-1.218$$


Standard Heats and Free Energies of Formation and Absolute Entropies of 2-Chloro-2-methylpropane

$\Delta H_\mathrm{f}^\circ = -211.3~\mathrm{kJmol^{-1}}$

Neither source provided entropy or free energy data though but this is a start.

http://courses.chem.indiana.edu/c360/documents/thermodynamicdata.pdf
    http://webbook.nist.gov/cgi/cbook.cgi?ID=C507200&Units=SI&Mask=2#Thermo-Condensed


Why is cyclopentadiene anion is aromatic but cycloheptatrienyl anion is not?

That happens because of Huckel's rule of aromaticity. The cycloheptatrienyl anion has 8 electrons which translates as 4n electrons, not the 4n+2 as implied by Huckel. If you are about to construct the MOs for both cycles, you'll find that putting 4n electrons will rise in a diradical molecule, not stable as it already sounds. A quick trick for doing that is using Frost's circle. All you need to do is inscribe the correspondent polygon in a circle and make sure that you have a vertex touching the circle as low as possible. The vertices will give you the qualitative energy of the MOs.


Steric effect of t-butyl group on SN1 on a cyclohexane ring

I would argue that the reasoning makes sense.
There are some important things to discuss here.
The cis isomer ($C$) and the trans isomer ($T$) will both react via the same carbocation intermediate. This means the energies of the intermediate are the same. Chloride has an A-value of 0.43, so $C$ is higher in energy than $T$. Since the formation of cation is endothermic, we may invoke the Hammond Postulate: the first step has a late transition state that structurally looks like the intermediate and is similar to it in energy. Finally, we conclude that the transition state for the reaction from $C$ has a lower barrier and thus a faster reaction.

Now, there is no way to "prove" that this or any other rationale is correct. There is no way to prove any model is correct. That's not how science works. You can only have consistent models that are good enough for now. They may even be "right," but you will never know. On the other hand, you could a reject the model based on inconsistency with experimental evidence, but there's no contradictory experimental evidence here.


I don't have specific product recommendations (and I think they're discouraged in general on SE), but these are the general principles:

You'll only get condensation on goggles (or anything else) if they're cooler than the dewpoint of the air contacting them. If your goggles are cold, you'll get condensation.

If you're sweaty, it'll be worse.

If the goggles are glass or another high-thermal-mass material, they'll stay cold longer, making the problem worse.

If the goggles are ventilated, humid air can escape from the space between your goggles and your face, making the problem better.

Does the problem only arise after you've been working in the lab for a while, or does it happen as soon as you put the goggles on? If the latter, you might be able to fix it by just putting the goggles in a warm interior pocket for a few minutes before you start work.

There are also spray-on anti-fog products for goggles. If you can't find one of those, try rinsing the goggle interior surfaces with soapy water, then not rinsing the soap off. The soapy film left on the goggles won't prevent condensation, but it will slow the formation of droplets, which will slow the fogging effect.


Short answer: The two curves are the distribution of kinetic energy of molecules at two different temperatures. A catalyst won't change the distribution; only temperature change will change the energy distribution. The line marked Ea is the activation energy; a molecule has to have more than this to react. A catalyst could change the activation energy; since there is only one Ea, we can eliminate catalytic activity as an issue. All the molecules with less energy than this just bounce around, not reacting.

The curve with the higher peak represents the lower temperature. The one with the lower peak has a longer and higher tail out to higher energies, so the shaded area represents the greater number of high energy (reactive) molecules at the higher temperature. Answer: A.


The property of "moldability" can be based on several properties of plastics
There are broadly two types of polymer commonly used. Polymers that set and thermoplastics that can be molded when hot. Both of these types can have the desirable property of being liquids that can fill a complex mold before becoming rigid (though by far the commonest is the thermoplastic that merely requires heat to make it flow).
Thermoplastics can be supplied to manufacturers as beads of the already polymerised compound that can be melted in the feed to the moulding machine. Examples include polystyrene, polypropylene, polyethylene and many others. These form flowable liquids when hot enough so that a mold can be filled easily during manufacturing and the polymer will set into a rigid shape when cooled. This makes for very easy manufacturing.
This is how a very wide variety of complex shaped can be made. PET bottles, Lego bricks, toy soldiers, plastic shells for computers and printers, car bumpers and many, many other complex components are made this way.
But other complex components can be made from plastics that set from chemical reactions and cannot be thermally molded. These work as long as the setting reaction is slow enough and the components are liquid. There are a variety of such materials which usually involve mixing a (fairly) liquid monomer or oligomer (examples include methyl methacrylate, some silicones, epoxies and some polyesters. Usually the process involves mixing a controlled amount of the monomer (eg methyl methacrylate which was often the basis for home modelling kits) with another ingredient that triggers polymerisation (often a radical initiator). With the right mix of the two ingredients, the polymerisation speed can be controlled so the speed of polymerisation is slow enough that the liquid can fill a complex mold before starting to set. This process is more often used in small batches as it isn't always well adapted to speed because the curing process can take minutes to hours.
The thing that matters in both thermoplastics and thermosets is that the components are liquids that can fill complex molds before setting rigid. In some cases the setting mechanism is based on cooling; in others it is based on chemical reactions.


Carbon monoxide detection without a detector

Surf a civilian aircraft supply.  They sell a little absorbant white tab (chemical spot detector) containing $\ce{PdCl2}$.  When exposed to CO it goes black.

http://www.aircraftspruce.com/menus/ps/carbonmonoxidedetectors.html

Canary.  Goldfish won't work - they can survive on anaerobic respiration,

http://aquaticcommons.org/3663/1/14P089.pdf
   who needs powdered alcohol?


Oxide is a term reserved for the oxide anion $\ce{O^2-}$. This anion is typically characterised as rather stable since the very electronegative element oxygen is in its maximum negative oxidation state. Dioxide means, that there are two oxides present in whichever structure we are looking at.[1] A typical example for a dioxide would be manganese dioxide, also called manganese(IV) oxide $\ce{MnO2}$. As shown in the image below, it crystallises in the rutile structure, like titanium dioxide $\ce{TiO2}$.

Figure 1: Crystal structure of rutile, $\ce{TiO2}$ which is equivalent to β-$\ce{MnO2}$. Red spheres represent oxides, grey ones the metal in question. Image taken from Wikipedia, where a full list of authors is available.

Hydrogen peroxide, on the other hand, as is evident from the structure in figure 2 below, contains not two $\ce{O^2-}$ units but a single $\ce{O2^2-}$ unit, composed of two oxygens bound to each other. This means that oxygen’s oxidation state is $\mathrm{-I}$ rather than $\mathrm{-II}$ and that hydrogen peroxide is readily reduced. (It can also easily be oxidised back into $\ce{O2}$. The functional group $\ce{O-O}$ is labelled peroxide and it would not make sense calling it dioxide since it does not feature oxides.

Figure 2: Structure of hydrogen peroxide. Red spheres are oxygen, white ones hydrogen. Image taken from Wikipedia, where a full list of authors is available.

Aside from that, IUPAC also allows for nomenclature to entirely ignore the molecular structure and name entirely by molecular composition. In that compositional nomenclature, dihydrogen dioxide is an acceptable name like dihydrogen monooxide would be for water. Note, however, that this only tells us the number and type of atoms involved while not giving us any clue to how they are connected. It is thus typically better to refer to a peroxide as a peroxide to emphasise the actual functional group present.


Selective nitro reduction of poly nitro compounds

Is there any way to predict which nitro group will be reduced

After having gone through three pages of Google Scholar and reading several research papers looking for an answer, there is definitely no way to predict the product here for every general case. The most detailed paper on the topic of Zinin reduction by far is Porter, H. K. Organic Reactions 2011, 455–481 (link). Even this paper went as far as to say that (p458):

  It is evident that much remains to be learned about the mechanism and that studies using modern techniques and instruments would be desirable.

However, it does give reasonably detailed examples of of selective reduction of substituted dinitrobenzene which are worth mentioning. To begin with:

  The positions of the substituents play an important role in the progress of the reaction. ... In polynitro compounds, the relative positions of the groups determine whether reduction gives monoamino or polyamino derivatives and determine the isomer distribution of the monoamino derivatives.

For example: (p459):

  In substituted dinitro- and trinitro-benzenes, the least hindered nitro group is preferentially reduced.

  In dinitro- and trinitro-phenols and their ethers, a nitro group ortho to hydroxy or alkoxy is preferentially reduced.

This is the most you need to know about this reduction. The other details are more technical, and are attached below are interested readers.

Further detailing:

In the end of the paper, Porter has provided huge tables detailing several (substituted) mono-, di-, tri-nitrobenzene derivatives and their major products on Zinin reduction. So, I studied nearly a hundred different reductions through those tables, and tried to devise other reasonable ways to predict the major product than the two already mentioned above, especially in asymmetrical polynitrobenzene derivatives. I've not listed below mononitrobenzenes since they have only a single nitro group to be reduced. Further, symmetrical polynitrobenzenes (like 2,4,6-trinitromesitylene) have all their nitro groups equivalent, so the product obtained on reduction of either group is same. I've ignored biphenyls and napthalenes in my calculations to keep results simple. Further, exceptional products are not listed here, as they have already been detailed in the paper (p460-p464 under "Side reactions").

The following findings may be used to predict the major product in simple polynitrobenzene derivatives. Keep in mind that exceptional side-reactions do occur, so the prediction may not always be accurate. Yet, a substantially large list of predictable products is obtained as shown below. Examples of compounds are enclosed in parentheses.

The less sterically hindered $\ce{-NO2}$ is reduced in alkyl derivatives of polytnitrobenzene (tables 2 and 3).
Example: 5-t-butyl-2,4,6-trinitro-m-xylene yields 4-t-butyl-3,5-dinitro-2,6-xylidene i.e. a fancy way to say that the $\ce{-NO2}$ group para to the tert-butyl group got reduced.
When a hydroxy or alkoxy group is present, then the ortho nitro group is reduced, even in presence of other halogens on the ring.
(table 4: 2,4-dinitrophenol, 2-chloro-4,6-dinitrophenol, 4-bromo-2,6-dinitrophenol)
If both an alkyl substituent and a hydroxy/alkoxy group are present, and: (table 4)

a single nitro group is ortho to the hydroxy/alkoxy, then that ortho nitro group is reduced
(2,4-dinitroanisole, 2,4-dinitrophenetole, 2,3-dimethoxy-5,6-dinitrotoluene)
no nitro group is ortho/both nitro groups are ortho to the hydroxy/alkoxy group, then the least sterically nitro group hindered is reduced
(2,3-dimethoxy-5,6-dinitrotoluene; 1,2,4-trimethoxy-5,6-dinitrobenzene yields 3,4,6-trimethoxy-2-nitroaniline)

Table 5, 6, 7, 8, 10, 11 do not have any analyseable compounds.
Table 9 lists nitro derivatives of aniline.

preference in reduction is given to the nitro group ortho to the amino group.
(2,4-dinitroaniline, 2,6-dinitroaniline, 3,5-dinitroaniline)
even in presence of other alkoxy or alkyl substituents, the nitro group ortho to the amino group is only reduced.
(4,6-dinitro-o-toluidine, 4,6-dinitro-o-anisidine, 4,6-dinitro-2,5-xylidene)
the ortho nitro group is reduced even in case of N-alkyl anilines, though they are much more bulky.
(N-ethyl-2,4-dinitroaniline, N-methyl-2,4-dinitroaniline)

There are a few cases which I was unable to resolve. The author has not noted them as an exception either. Suggestions are welcome:

Table 4 states that 2,4-dinitroanisole yields 5-nitro-o-anisidine. It is not entirely clear to me why the expected 4-nitro-o-anisidine was not formed. The same yield is observed in 2,4-dinitrophenetole yieldings 5-nitro-o-phenetidine.
Table 7 gives the reduction of 2,4-dinitro-$\alpha$-toluenesulfonic acid as 2-amino-4-nitro-$\alpha$-toluenesulfonic acid, which is not clear why, given the steric constraints at the ortho position.
Table 9 states reduction of 3,5-dinitro-2,4-xylidene gives 2,4-dimethyl-5-nitro-m-phenylenediamine. It is not clear to me why the more sterically hindered nitro group is reduced. It isn't even ortho to the aniline group.

I believe this is a reasonably detailed analysis of the selective reduction of the polynitro compounds.


What is the Lewis structure of Ozonide ion?

From a cursory e-book search:

  Ozonide ion is paramagnetic and is shown to have magnetic
  susceptibility. Bond length is 135 Å and bond angle is 108°. The
  structure is:


Many compounds are good hydride donors, if it results in carbonyl bond formation or extended conjugation. Sometimes, a compound can donate H- ion if it results in stable pi bond formation.
(e.g. Tertiary butyl carbanion)


Difference of Ascorbic acid dissolved or precipitated in Glycerol?

To acknowledge OP's request, I am putting this as my answer:

What OP's initially referring to (US20110008445A1) was a patent application. Since it was a 2011 application, I searched for the patent and found it (Ref.1). It's abstract states that:

  A Suspension of ascorbic acid in glycerol or in glycerol comprising diglycerol, in which the content of ascorbic acid is 13% by mass or greater, and further in which 8 to 12% by mass of the ascorbic acid is dissolved in glycerol or in glycerol comprising diglycerol, and the rest of ascorbic acid is precipitated in the form of microcrystals having a particle diameter of $\pu{25 \mu m}$ or smaller and is uniformly dispersed in the suspension. The suspension of ascorbic acid in glycerol is useful as a base material for cosmetics containing ascorbic acid which exhibits excellent feel in the use (spreadability and Smooth feel on application to the skin).

The patent described the preparation of ascorbic acid suspentions in alcohol or glycerol. Based on my knowledge in ascorbic acid, all forms of suspensions contain ascorbic acid in  in its usable form. However, one should be aware that ascorbic acid tends to have browning effect due to slow formation of related furan derivative on long standing (Ref.2).

References:

Mikiharu Okumura, Takanori Ohta, “Suspension of Ascorbic Acid in Glycerin and Process for Production Thereof,” United States Patent 2012, 8,211,472 B2 (https://patentimages.storage.googleapis.com/62/29/bf/f95fde91d869cd/US8211472.pdf).
T. Lalikainen, M. A. Joslyn, C. O. Chichester, “Browning Reactions, Mechanism of Browning of Ascorbic Acid-Citric Acid-Glycine Systems,” J. Agric. Food Chem. 1958, 6(2), 135–139 (DOI: 10.1021/jf60084a011).


What makes oceanic desalination so expensive?

Reason is salt dissolve very easily with water, forming very strong bond. Not only salt there are also other minerals. To break this bond we need technology as well as energy and both of them are very expensive(today).

There are two methods by which we can break this bond:

Thermal distillation:
Thermal distillation involves heat: Boiling water turns it into vapor—leaving the salt behind—that is collected and condensed back into water by cooling it down
Membrane distillation: The most common type of membrane separation is called reverse osmosis. Seawater is forced through a semipermeable membrane that separates salt from water. Because the technology typically requires less energy than thermal distillation, most new plants, now use reverse osmosis.

Today about 10 to 13 billion gallon  water are desalinated worldwide per day. That's only about 0.2% of global water consumption.

Membrane distillation is bit more cheaper than thermal distillation. Even though  large amounts of energy are needed to generate the high pressure that forces the water through the filter. Current methods require about 14 kilowatt-hours of energy to produce 1,000 gallons of desalinated seawater.

So we can conclude that cost of distillation is high because we need notable large amount of electricity to heat water in thermal plant and generate high pressure. Also other technology(like say burner, pumps, etc) are also expensive.

Source: ScientificAmerica and LiveScience

Also note we are trying to decrease the cost of distillation and day by day number of seawater distillation  are increasing.


What is the mechanism of AChE inhibition by Onchidal?

Q.1: How and why does Onchidal act as an AChE inhibitor?
Onchidal is an antipredatory, feeding deterrent product secreted by marine organisms such as Onchidella binneyi and Elysia cripata, not having the protection of an external shell and thus relying on the production of a defensive secretion. [Gavignin et al., 2000, 1996, 1997; Ireland et al. 1978] To mammals, it is a neurotoxin that acts as an irreversible inhibitor of AChE, with a Kd of 300 μM, where it forms a covalent species with the enzyme.[Abramson, 1989]
Q.2: What does the adduct of Onchidal with AChE look like?
This molecule was isolated from the mucus obtained by squeezing the mollusc O.binneyi. [Ireland et al. 1978] It is a sesquiterpene acetate containing an α/β-unsaturated aldehyde and an acetate ester as shown above. It has also been speculated that onchidal’s active form is believed to be ancistrodial.

Abramson (1989) proposed multiple modes of inhibition:
(i) Schiff base formation with a lysine;
(ii) attack on the beta-unsaturated carbon in conjugation with the aldehydes by a Cys, Lys, His, or Tyr; and
(iii) the formation of 1,4-dialdehyde, which reacts with a Lys to form a pyrrole covalent adduct.
A look at the AChE co-crystal structure with an inhibitor (paraoxon in this case to locate the active site) as deposited with RCSB as 5HF5.pdb (DOI: 10.2210/pdb5HF5/pdb) , indicates that the closest reactive ε-amino group of a Lys residue is ~18  Å away. This distance is presumed too much for a covalent attachment of onchidal to AChE active site and thereby affecting AChE’s catalytic function. Likewise, there is no free sulhydryl available in the active site either, so a different residue may be involved as the point of attachment. The following figure below shows AChE with paraoxon shown with speres; Lys in green; and Cys in orange. The distance is in Å on the following figure I prepared:

Stoddard (2014) proposed three docking poses for onchidal. Here is the figure S2 from their paper:

Then, Stoddard (2014) further noted:

Only one His, the active site His-440, is located in the gorge. Several Tyr are located in the gorge; Tyr 442, 334, 130, 121, and 70. Based on the crystal structure and Abramson’s findings, the most probable mechanism is attack of the beta-unsaturated carbon in conjugation with the aldehyde by a Tyr; however a candidate Tyr for this mechanism was not identified from our docking analysis.

I’m wondering whether non-specific cellular esterases may convert Onchidal to an alcohol? If true, my speculation may change the docking analyses and ensuing inference. In any case, I’m not sure we have the full picture of Onchidal molecular mode of action, just yet.
Q.3: Why is the inhibition of AChE by Onchidal irreversible?
Onchidal contains an aldehyde and vinyls. Such reactive functionalities have literature precedents describing irreversible adduct formation with specific protein residues.

References
Abramson S.N., Radic Z., Manker D., Faulkner D.J., Taylor P. Onchidal (1989) “A naturally occurring irreversible inhibitor of acetylcholinesterase with a novel mechanism of action.” Mol. Pharmacol. 36:349–354.
Gavagnin M., Mollo E., Castelluccio F., Montanaro D., Ortea J., Cimino G. (1997) “A novel dietary sesquiterpene from the marine sacoglossan Tridachia crispate. Nat. Prod. Lett. 10:151–156. doi: https://10.1080/10575639708043731
Gavagnin M., Mollo E., Cimino G., Ortea J. (1996) “A new γ-dihydropyrone-propionate from the Caribbean sascoglossan Tridachia crispate” Tetrahedron Lett. 37:4259–4262. doi: 10.1016/0040-4039(96)00811-8
Gavagnin M., Mollo E., Montanaro D., Ortea J., Cimino G. (2000) “Chemical studies of Caribbean sacoglossans: Dietary relationships with green algae and ecological implications.” J. Chem. Ecol. 26:1563–1578. doi: 10.1023/A:1005526526884
Ireland C., Faulkner D.J. (1978) “The defensive secretion of the opisthobranch mollusc Onchidella binneyi. Bioorg. Chem. 7:125–131. doi: 10.1016/0045-2068(78)90043-3
Stoddard, S.V., Hamann, M.T., and Wadkins, R.M. (2014) “Insights and Ideas Garnered from Marine Metabolites for Development of Dual-Function Acetylcholinesterase and Amyloid-β Aggregation Inhibitors.”Mar Drugs 12:2114-2131. doi: 10.3390/md12042114


Halogenation of an alpha-halogen ketone in acidic conditions

The same mechanism discussed in my answer to your earlier question on acid catalyzed halogenation alpha to a carbonyl still applies here.  For reference, here is the same link supplied earlier that shows the mechanism for this acid catalyzed halogenation.

In that earlier answer I said that the enol that "produces the more stable double bond (the one that is more highly substituted)" is the enol that will form and react nucleophilically with the halogen to produce the alpha-halo carbonyl product.  In light of your new question, the statement that more highly substituted double bonds are more stable needs some qualification.

The closer in electronegativity a substituent is to carbon, the more it will stabilize a double bond.  Consequently, the more alkyl groups attached to a double bond, the more stable the double bond (barring large steric effects).  On the other hand, fluorine destabilizes a double bond.  It is so electronegative that it prefers to bond with orbitals high in p character.  When forced to bond to a double bond $\ce{sp^2}$ orbital (that is enriched in s character) destabilization results (see this earlier fluorobullvalene example for more on why fluorine "hates" to be on a double bond).

Based on its electronegativity, chlorine should be somewhere between fluorine and an alkyl group in terms of its ability to stabilize a double bond.  If we compare the heats of formation of 1-butene (-0.1 kJ/mol) and trans-2-butene (-11.2 kJ/mol) we see that placing the methyl group on the double bond stabilizes the molecule by 11.1 kJ/mol.  If we compare the heats of formation of allyl chloride (-5.6 kJ/mol) and trans-1-chloropropene (-12 kJ/mol) we see that the placing the chlorine group on the double bond stabilizes the molecule less, only 6.4 kJ/mol.

These differences mean that we will have much more of the enol without the chlorine on the enol double bond, then we would have of the enol without the methyl group on the enol double bond.  So we have more of the unsubstituted enol in the chlorine case and the enol with the chlorine on the double bond is much less nucleophilic (less reactive) due to the electron withdrawing inductive effect of the chlorine substituent.  These two factors combine to make the second alpha-chlorination take place on the unsubstituted side of the molecule.

As to steric effects, they probably don't play a significant role.  One metric commonly used to compare "steric size" is the cyclohexane A value.  The A value for chlorine is 0.43, much smaller than the A value for methyl (1.7), suggesting that a chlorine substituent is smaller than methyl substituent.


Taste same but why not LiCl a substitute for NaCl (common salt)?

The reason why anything is ever considered a substitute for normal salt (NaCl) is that some people take too much salt with their diet and this is widely thought to be bad for your blood pressure. This, however, is disputed by some mainly on the grounds that only a small proportion of people are sensitive to hight sodium in the diet and the rest of us can cope with widely differing amounts with no discernible harm.

However, this doesn't mean that anything that tastes like salt is necessarily a good substitute. Not having enough salt is also a problem as the Na/K balance in cells is important and a serious lack of sodium can be an acute danger. For most people with an adequate diet, using a salt substitute (for table salt) can reduce their sodium intake but, since there is already a lot of sodium in the diet, won't cause a problem. But not for all people.

The most common table salt substitute is KCl, which, again, won't be harmful to people with a normal diet. The body is good at processing potassium and sodium and can regulate their levels within reasonable bounds. So KCl is safe as the body is used to coping with it.

This isn't true for lithium. There is no major biological role for lithium in the body, and, while lithium is regarded as an essential trace element, nobody really knows why. Whatever the normal role, the body is not designed to cope with significant amounts of it (unlike K and Na which are major components of many biological systems). So this might suggest that using large amounts a salt substitute has an unknown risk for harm far greater than the use of, say, potassium. Moreover, lithium does has significant known effects on the nervous system, though the mechanisms are poorly understood. This is illustrated by the use of controlled amounts of lithium salts as a significant treatment for bipolar disorder. This fact alone (regardless of any other toxic effects in high doses) would suggest it isn't a safe alternative when used as a substitute for table salt in uncontrolled doses.

The use of lithium as a treatment for bipolar disorder is relatively recent (started in the early 1950s) so it might have looked like a harmless alternative to sodium chloride in earlier times.


No, as the ice is melting, the water is cooling down. Ice melts in water because the water molecules in solution collide with those in the ice, giving them energy. In exchange, the water molecules in solution slow down a little during the collision. While the Ice does not change temperature as it is melting, the energy from the water is going into breaking the crystal structure of the ice, turning it into water. This process consumes energy, and the energy to fuel this process comes from the water in solution. Another factor to consider is conservation of energy - if energy is being put into breaking the ice, where else could that energy come from but from the water?


But.....6 member lactones are super stable so I wouldn't be surprised if only 1 ring opened/ some recyclized


What does this representation of organic aromatic compounds mean?

This is the cyclooctatetraene dianion. One way to get to it is to start with cyclooctatriene and remove one hydrogen ion each from the two $\ce{- CH2 -}$ groups. This is related to the cyclopentadienyl anion, the more common aromatic anion.

Image source: https://www.masterorganicchemistry.com/2017/05/17/frost-circles/


Equilibrium constant Kp for vaporization of water?

Recall the origin of the equilibrium constant.  It comes from the fact that the chemical potential of an ideal gas can be written in the form u(T,p0) + RTln(p/p0), with p0 some reference pressure.  What this lets you do is measure the equilibrium composition at a reference pressure, then calculate the equilibrium composition at a slightly different pressure with a simple formula, without having to know the general dependence of chemical potential on pressure and temperature.  Very handy!  But the existence of the reference pressure is essential.  It goes into the numerical value of the equilibrium constant.  For example,  for just two reactants with stoichiometric coefficients c1 and -c2, i.e. the equilibrium is c1 R1 = c2 R2, you have:

dG = 0 = c1 (u1 + RTln(p1/p0)) - c2 (u2 + RTln(p2/p0))

-(c1 u1 - c2 u2)/RT = ln((p1/p0)^c1 / (p2/p0)^c2)

exp(-(c1 u1 - c2 u2)/RT) = (p1/p0)^c1 / (p2/p0)^c2

The thing on the left is just a function of T and the reference pressure p0, and we set it equal to the measured equilibrium constant K(T,p0), which of course we get by measuring the equilibrium composition at T and p0.  The thing on the right you'll then recognize as the usual expression for Kp.

K(T,p0) = (p1/p0)^c1 / (p2/p0)^c2

With this equation in hand (and the value of K) we can calculate the equilibrium composition {p1,p2} at some temperature and pressure not too
far away from the reference T and p0, using the usual freshman chemistry tricks.

Now you can certainly choose a different reference pressure if you like, it doesn't have to be 1atm, that's just convention.  If you chose a different pressure, K would be different, but the expression for Kp would be the same, only you'd be dividing by the new reference pressure.  But you have to have some reference pressure for this method to work.


What are some books, websites, etc. where I can find the detailed protocol for standardization of sodium thiosulfate with dichromate?

My go-to for this is typically Kolthoff & Sandell's Textbook of Quantitative Inorganic Analysis.  They unfortunately do not report a detailed protocol for dichromate standardization of thiosulfate solutions therein, but instead cite Volumetric Analysis, Vol III by Kolthoff and Stenger (Amazon and Google Books).  I don't have access to this volume; WorldCat shows numerous copies in university libraries in my area of the U.S.; more than likely there's a library where you are that has a copy.


Why does CO2 have higher boiling point than CO?

CO2 has more electrons than CO. This means that it has a much larger electron cloud as compared to CO, so its more easily polarised and thus, the ease of forming instataneous dipole-induced dipole bonds increases. Even though CO is a polar molecule and it forms permanent dipole-permanent dipole bonds, in this case the id-id bonds are stronger.

P.s. Just a 17 year old A Level Chem student here, I might be wrong


These devices have been reported to work in some situations, but not in others.

The underlying theory involves altering the growth of calcium carbonate scale with an ordinary magnetic field. The magnetic field changes which form of calcium carbonate crystal (calcite, aragonite, and vaterite) is favored. Certain forms of crystal make tough scale while others can pass through a water system without scaling.

The growth and precipitation of carbonate crystals are affected by many other things such as calcium concentration, carbon dioxide concentration, temperature, pH, and the presence of other metallic ions. Exposure time to the magnetic field, magnetic field strength, and magnet orientation also have an effect.

The different forms of carbonate crystals can be tested to have specific properties under controlled laboratory conditions.  Under uncontrolled conditions, the scientific literature is full of mixed results and mixed explanations.  Because so many things affect the formation of scale, it is impossible to say if a magnet or device generating a magnetic field will work to thwart scaling in any specific situation.  It may work wonderfully in some situations but not at all in others.

The best way to tell if such a device is worthwhile would be to see if several people using your same water supply report it as having worked lately.  If water management practices change (e. g. factors affecting the formation of scale are controlled) then such a device may cease to work.  If no one using your same supply can attest to its operation then this would be a risky purchase.


As Poutnik said, the melting point is well defined for pure substances. In the case of egg yolk, it is not a pure substance and also when you heat it, some molecules decompose. So we can't measure or even define a melting point for it. Also, the temperature is not constant, when the egg is melting.
A substances' phase doesn't get influenced only by temperature. Maybe it is strange that heating the egg yolk in a pan, makes it to behave like a solid. Actually, some complex changes in intermolecular forces are getting done.
If you want to cause an egg yolk to melt, I think that the only work to do is to heat it under the normal pressure.


Multiple solutions in the balancing of redox reactions

The first thing that you should realise is that under specific conditions (here: acidic ones) only one product is formed in this reaction. Permanganate is not reduced to both manganese(IV) oxide and manganese(II) cations, it will either cleanly give manganese(IV) oxide (if the acid concentration is very low) or cleanly give manganese(II) cations (if the acid concentration is sufficient). The only way to generate both is to have sufficient acid at first which is subsequently used up by the reaction and drops below a threshold. Thus, if you were titrating and notice the dark precipitate of $\ce{MnO2}$, immediately stop, throw away the analysis and start again as it has become worthless.

Therefore, there is only one simplest solution to the problem, depending on which half-reaction you prefer:

$$\begin{align}\ce{MnO4- + 8 H+ + 5 e- &-> Mn^2+ + 4 H2O}\tag{Red1}\\[0.6em]
\ce{H2O2 &-> O2 + 2 H+ + 2e-}\tag{Ox}\\[0.6em]
\ce{2 MnO4- + 6 H+ + 5 H2O2 &-> 2 Mn^2+ + 8 H2O + 5 O2}\tag{Redox1}\\[1.6em]
\ce{MnO4- + 3 e- + 4 H+ &-> MnO2 + 2 H2O}\tag{Red2}\\[0.6em]
\ce{H2O2 &-> O2 + 2 H+ + 2 e-}\tag{Ox}\\[0.6em]
\ce{2 MnO4- + 2 H+ + 3 H2O2 &-> 2 MnO2 + 4 H2O}\tag{Redox2}\end{align}$$


In addition to the answer and the comment given, the use of mercury in thermometers had another advantage frequently used in the lab:

Mercury, as a metal, has a high electrical conductivity. Imagine two thin wires inside the capillary of your thermometer. One of them is fixed.
The other wire is can move along a threaded rod. Typically, this was realized via magnetical coupling to a freely rotating knob on top of the (hermetically sealed) thermometer. Upon heating, the mercury expands and finally reaches the upper wire previously adjusted to a particular working temperature. At this point, an electrical current may flow through the thermometer.

Source: http://upload.wikimedia.org/wikipedia/commons/4/48/Kontth.jpg

These contact thermometers, nowadays replaced by mercury-free electronic thermoswitches, once were the working horses in the lab when it came to controlled heating of oil baths or monitoring cooling circuits for uv lamps.


Best chair configuration for trimethylcyclohexane matching a specific Haworth projection?

While you are correct that B should be the most stable conformer of trimethylcylcohexane, the question asked which one best represents the one shown in the Haworth projection. Answer choice B has R stereochemistry on carbon B (see below), while carbon B has S stereochemistry in the original projection. As such, the initially drawn compound is different from the answer choice B. On the other hand, answer choice A has the correct stereochemistry, and also has the maximum number of possible methyl groups in equatorial positions, causing it to be the best depiction of the Haworth projection.

Note: Instead of finding the absolute stereochemistry by using R/S notation, you could also notice that the Haworth projection and answer choice B have different relative stereo chemistry. In the original molecule, carbons B and C are cis, while in answer choice B, these carbons are trans.


Azobenzenes are one of the most widely used moietes in photochemically  switchable systems. Nevertheless, the pathways for the E-Z isomerisation of azobenzene in the $S_1$ and the $S_2$ state have been in debate for decades and I'm not quite sure whether this is finally solved.

  However, in the $n \rightarrow \pi^*$ excited state ($S_1$), the rotation is no longer hindered and it can isomerize.

Yes and no.

It is true that the $S_1$ state arises from a $n \rightarrow \pi^*$ transition in a weak, broad band with $\lambda_{\mathrm{max}} = 432\,\mathrm{nm}$ and $\epsilon \approx 400\,\mathrm{L\,mol^{-1}\,cm^{-1}}$ in hexane.

By the way, E-azobenzene exhibits another, much stronger absorption at $\lambda_{\mathrm{max}} = 319\,\mathrm{nm}$ with $\epsilon \approx 22800\,\mathrm{L\,mol^{-1}\,cm^{-1}}$ in the same solvent. This $\pi\rightarrow\pi^*$ transition to the $S_2$ state would indeed allow for an isomerisation through an one-bond-flip rotational mechanism.

(For the UV data, see 1 and 2).

But - you saw that coming after this lengthy introduction - the situation is much more complicated!

You have neglected the fact that E-Z isomerisations can proceed by other mechanisms than just the simple one-bond-flip that is usually postulated for alkenes!

Forgive me for not getting into all details on this, but for polyenes, to give an example, the so-called Bicycle Pedal (BP) and Hula twist (HT) pathways have been proposed and confirmed as space-conserving alternatives to the one-bon-flip mechanism. The latter can be visualized as "let's just flip the other side of the former double bond around like a rotor".

Back to the azobenzenes. Even from more recent works (2, 3, 4, 5), it seems that the E-Z isomerisation is still not fully explained, although a plain rotation around the former double bond can be excluded.

The two main fractions in the debate support mechanisms in which a contribution of inversion through a change of the $\ce{C-N-N}$ angle in the excited state is assumed. Whether this is the only effect and the reaction proceeds through a transition state with a linear $\ce{C-N-N}$ arrangement or whether an inversion-assisted torsional pathway provided the beter explanation is still unclear.

References

1 George Zimmerman, Lue-Yung Chow, and Un-Jin Paik, J. Am. Chem. Soc., 1958, 80, 3528-3531.

2  Alessandro Cembran, Fernando Bernardi, Marco Garavelli, Laura Gagliardi, and Giorgio Orlandi, J. Am. Chem. Soc., 2004, 126, 3234-3243.

3 Eric Wei-Guang Diau, J. Phys. Chem. A, 2004, 108, 950-956.

4 Eric M. M. Tan, Saeed Amirjalayer, Szymon Smolarek, Alexander Vdovin,    Francesco Zerbetto, and Wybren Jan Buma, Nature Communications, 2015, 6, Article number: 5860.

5 Junfeng Shao, Yibo Lei, Zhenyi Wen, Yusheng Dou, and Zhisong Wang,
J. Chem. Phys., 2008, 129, 164111


Do alkali metals not form diatomic molecules?

Diatomic molecules of alkali metals are detected in the gas phase. However, it's so happens that the bond in them is very weak and at the temperature the alkali metal vapors develop only a few percents of the metal in the vapors exists as diatomic molecules. It so happens, that metal bonding allows to achieve an overall more energetically favored state.

Outer orbitals of the alkali metal atoms are very diffuse, so their bonds are weak. Only Lithium, the smallest of the alkali metals, has practically meaningful covalent chemistry to speak of.


Field Effect and it's effect on basicity on amines

The basicity of amines has to do not only with the degree of substitution with electron-releasing alkyl groups but also factors such as steric hindrance and degree of solvation.

The electronegative nitrogen is able to isolate electron density from the alkyl groups. So having more electron-releasing alkyl groups to stabilize the positive formal charge created by the abstraction of a proton from the protic solvent might seem like a good thing.

On the other hand steric hindrance due to bulky alkyl groups reduces basicity.

Also, having more alkyl groups reduces solvation due to hydrogen "bonding." Consider ammonia, $\ce{NH_3}$. This molecule can form four hydrogen bonds. On the other hand, a tertiary amine that has been protonated can form only "good" hydrogen bond; one might argue that the hydrogens on the alkyl groups can form hydrogen "bonds", but these are obviously going to be weaker "bonds" compared to the ones among ammonia molecules.


Order of filling electrons in Jahn-Teller energy diagram

This was just an illustrative example. The tendency of a complex to favor low-spin or high-spin configurations is dictated by the relative magnitudes of the pairing energy, P, and the splitting energy, $\Delta$, between $e_g$ and $t_{2g}$.*

If $\Delta$ is greater than the energy required to pair two electrons, then electrons pair in $t_{2g}$ before occupying $e_g$. In the example you've shown of compression, this would be the $d_{xy}$ orbital.

*For octahedral complexes. Jahn-Teller can also happen for tetrahedral complexes.


How to calculate the amount charge transfer from a natural population analysis?

I will evade answering the main question because I am not feeling expert enough, but I want to point out that Natural population analysis (NPA) is very different (and better) from the Loewdin one. Loewdin is computed from orbitals of a special basis set (although it comes from the density, it is still free-atom-derived basis set which is used to obtain the density matrix). NPA (part of NBO) is from density matrix which is adjusted to the particular molecular environment, hence better representing the "real" atomic population. NPA is available in the commercial NBO program, old version of which is also included with Gaussian package (maybe also in other packages, I don't know). You can also try AIM charges which are absolutely basis-independent (if Your density is good enough).

Personally I would think the analysis You proposed is fine with NPA or AIM charges, but I state again that I am not an expert in the topic.

See https://www.q-chem.com/qchem-website/manual/qchem43_manual/sect0061.html and also http://www.huntresearchgroup.org.uk/teaching/teaching_comp_chem_year4/L7_bonding.pdf .


Lysol Multipurpose Cleaner with hydrogen peroxide contains only 0.5% hydrogen peroxide. This site (using-hydrogen-peroxide.com) suggests using up to 35% hydrogen peroxide for cleaning toilets. This hydrogen peroxide website (and who knew it had one) may be biased, but that's quite a margin between the product you want to use and what the website says is safe. I would feel pretty comfortable using this Lysol product for cleaning toilets (except that it may actually not be strong enough!).


Find the amount of atoms for a given mass

For the elements A and B, the relation between the number of atoms n, the mass m, and the atomic mass M is given by

  $$n_A = \frac{m_A}{M_A}\quad,\quad n_B = \frac{m_B}{M_B}$$

Your assignment states:
$$m_B = 2\cdot m_A $$
$$M_B = 2\cdot M_A $$

Solve for $n_B$ ;-)

I seriously suggest to practise these calculations a lot. Ask your tutor for more worksheets or find a workbook (with solutions). Solve it alone, or work in groups. No matter how, you have to practise.

If you are in doubt, where to divide and where to multiply, remember:

Units are your friends!


I this, you are totally confuse with meaning hexagonal unit cell and hexagonal crystal system. First of all, I make you to clear about these two terms.

Hexagonal unit cell is single entity of hexagonal crystal system. It means there are a no. of repetitive hexagonal unit cells inside a hexagonal crystal system.

Yes, the diagram given in the question is of hcp i.e. hexagonal unit cell.

But, your next statement : In the unit cell of hcp there are 17 atoms whereas in unit cell of hexagonal crystal system there are 8.Is this true?

is incorrect. One thing I want to clear you that as per saying there are 17 atoms in unit cell of hcp is wrong. We can make the statement after correction is : To make a unit cell of hcp, we need 17 atoms. It means that there are 17 atoms present in unit cell hcp but to make a unit cell of hcp we require 17 atoms.

Now, we calculate the no. of atoms present inside the unit cell of hcp or we can say this as no. of effective atoms of unit cell of hcp.

Calculation of no. of effective atoms of unit cell of hcp :

No. of effective atoms of unit cell of hcp = 3 (there are a total of three atoms present inside the unit cell of hcp) * 1 (as all three atoms are in main unit cell of hcp, no part of atom is outside the unit cell of hcp) + 2 (there are two atoms present a centre of each base of hexagonal unit cell) * 1/2 ( half part of each atom is inside the main hexagonal unit cell) + 12 (there are a total of 12 atoms present at 12 corners of each base of hexagonal unit cell) * 1/6 ( one-sixth part of each atom is inside the main hexagonal unit cell) = 6 (not 8)


Soaps come in different colors, but why is soap lather always white?

Though @DHMO's answer is quite interesting, you shouldn't take it without a pinch of salt. I've to disagree with it in some aspects.

Soap colorants range from various kinds of Mica to dyes found naturally in plants. Now typically when you dye anything the quantity of dye used in comparison to the quantity of the substance to be dyed is very small. Too much dye, and it could possibly stain your hands/clothes, so soap manufacturers use as little dye as possible, but just enough to give the soap a good color.

Now for the rest of this answer, I will assume (reasonably enough) that by 'colored' soap you are referring to soap that goes by colors other than white.

Now when you lather using soap, only a really teeny-tiny bit actually goes into the water. As @DHMO correctly points out, the dye/pigment used to color the soap is greatly diluted. Why don't you carve off a piece of colored soap (having more or less the same size as as an almond) and dissolve it a mug of water by gently stirring it. I emphasize on "gently" so that you don't stir up a lather.

Have a look at the water. You'll see that the soap, apart from making it more cloudy, has not visibly imparted any particular color to the water. This is not unexpected, as I've already mentioned, the quantity of dye used is very small.

Now if you go ahead and agitate that soap solution, it will give rise to white lather, and this shouldn't be surprising anymore.

Now @DHMO's goes on to mention that Total Internal Reflection (TIR) is what imparts the white color to the foam. But this is grossly incorrect. While TIR can result 'white light', it is not the dominant phenomenon acting here (I'm not saying it's completely absent here either).

What is largely giving foam it's white appearance is another phenomenon called Scattering.

Now you might ask, 'Then why isn't soap water white?'. Well, since the foam is made up of lots of teeny tiny bubbles, light passing through it will have to encounter several surfaces, and it's these surfaces that scatter the light in so many directions.

So to say it straight DHMO's answer is incorrect. (No offence DHMO)

Remember I said you can't see any visible coloration in the water because of the dye is present in really small quantities? Well here's a way to validate that claim: Simply combine a teaspoon of red food-color with a bit of hand-wash (now that's really concentrated). Now you can lather the soap and lo behold! You have red foam.

[Credit to @ACuriousMind from Physics.SE, for that bit on why soap water isn't white. Also he confirmed that the white color of foam is due to scattering]

Edit- More kudos to @ACuriousMind for providing this amazing link on scattering.


How get a dilution at 150g/l of H2SO4 at 96%?

First of all: Never add water to sulphuric acid. The heat of the reaction may lead to an explosion. Use safety glasses, safety gloves, and wear protective clothing.

The densities at 20°C are

Water: $\rho_{water} = 0.9982~\mathrm{g~cm^{-3}}$
15% $\ce{H2SO4}$: $\rho_1 = 1.10~\mathrm{g~cm^{-3}}$
96% $\ce{H2SO4}$: $\rho_2 = 1.84~\mathrm{g~cm^{-3}}$

To get $5000~\mathrm{cm^3}$ 15% $\ce{H2SO4}$ with a mass of

$m = \rho_1 \cdot V = 1.10~\mathrm{g~cm^{-3}} \cdot 5000~\mathrm{cm^3} = 5500~\mathrm{g}$

you need

$m = 0.15 \cdot 5500~\mathrm{g} = 825~\mathrm{g}$ pure $\ce{H2SO4}$

These are contained in

$m = \frac{825~\mathrm{g}}{0.96} = 859.4~\mathrm{g}$ of 96% $\ce{H2SO4}$.

i.e.

$V = \frac{859.4~\mathrm{g}}{1.84~\mathrm{g~cm^{-3}}} = 467~\mathrm{cm^3}$ of 96% $\ce{H2SO4}$.

Finally, the amount of water you need to pour the acid in is

$m = 5500~\mathrm{g} - 859.4~\mathrm{g} = 4640.6~\mathrm{g}$ water, i.e. $4649.0~\mathrm{cm^3}$

In short: gently pour, while stirring, $467~\mathrm{cm^3}$ of 96% $\ce{H2SO4}$ into $4649~\mathrm{cm^3}$ distilled water to get $5000~\mathrm{cm^3}$ 15% $\ce{H2SO4}$. Break the procedure if the temperature gets too high.

Disclaimer: I do not take any responsibility for the correctness of the above calculation.


Why is phosphorous pentoxide a dehydrating agent?

First of all, your reaction is wrong. $\ce{HPO3}$ does not exist as a single entity. It is actually a cylic molecule of formula $\ce{(HPO3)_n}$. The simplest such compound is $\ce{(HPO3)3}$ which is called trimetaphosphoric acid. See this answer for more details.

So, what happens is that phosphorus pentoxide is very powerful dehydrating agent and can rip off water molecules from compounds. Phosphorus pentoxide can dehydrate nitric acid to form its anhydride, dinitrogen pentoxide.
$$\ce{2HNO3 ->[P2O5][- H2O] N2O5}$$

Why dinitrogen pentoxide is not used as a dehydrating agent?

My take on this is that hydrolysis of dinitrogen pentoxide is not that exothermic as compared to hydroysis of phosphorus pentoxide. Hydrolysis of phosphorus pentoxide is very exothermic, releasing so much energy to dehydrate compounds to form corresponding anhydrides and phosphoric acid.
$$\ce{P4O10 + 6 H2O → 4 H3PO4   (–177 kJ)}$$
Phosphoric acid is hygroscopic which further sucks remaining water from compound. If we compare all the dehydrating agents, phosphorus pentoxide will be on a different league. Dinitrogen pentoxide is nothing.
Nitric acid also hydrolyses to form nitric acid but the reaction is not that exothermic. Also, nitric acid is not that hygroscopic as compared to phosphoric acid for which it is not that great of a dehydrating agent. But hydrolysis of dinitrogen pentoxide has its importance. A solution of dinitrogen pentoxide and nitric acid is a good nitrating agent and hence $\ce{N2O5}$ is known as a nitrating agent than a dehydrating agent. Also, this hydrolysis reaction is a basis of forming atmospheric aerosols ($\ce{HNO3}$ is a good $\ce{NO_x}$ reservoir).


What are the crystals that have formed around the top of a conc. nitric acid bottle?

Chemicals cupboard often produce quite extraordinary reactions, much like hospitals are home to some unusual and severe infections.

Nitric acid does not normally exhibit such a bizarre behavior. Yet in your case it did. How so? My bet is on some unknown additional component which came from the same cupboard, but from a different bottle. For instance, it could be some lower aliphatic amine (Upd. or simply ammonia, as was brought to my attention by other answers). They are volatile, and they produce crystalline salts when they meet with strong acids.

A similar thing is known to happen with $\ce{NH3}$ and $\ce{HCl}$:  if they are stored nearby, especially in poorly corked bottles, they both (as well as everything around) get covered in a white residue of $\ce{NH4Cl}$. It doesn't form nice crystals, though; it looks more like a layer of hoarfrost. But that doesn't mean much. Related compounds do not necessarily have related crystal structures, let alone related crystal shapes, so it looks perfectly plausible to me that (substituted or not) ammonia nitrate would grow into fine needles.

The prevalence of amine/ammonia in the air within the cupboard (which could happen if the other bottle was corked worse than this one) would naturally lead to $\ce{HNO3}$ molecules fighting this battle at their own doorstep, that is, at the very bottleneck of this bottle and nowhere else. This would explain the accumulation of residue at one spot.


Your name is correct, while the given one is not. When the set of locants for subsituents would be the same no matter what, you would assign the lowest locant to the substituent that appears first in the name, which, by alphabetical order, it would be the ethyl group in this case. Hence 4-ethyl-5-methyloctane is the right name.


Why hybridization occurs in diatomic molecule

I thought having the lone pair in a 2s orbital would be favored over
  2sp

That's true, placing the lone pair in a 2s orbital would be put the lone pair electrons at lower energy then if they were in a 2sp orbital.  However you're only looking at a part of the puzzle.  As you note, in this case the sigma bonding electrons would be in a pure p orbital and they would then be higher in energy then if they were in an sp orbital. You need to optimize the total energy of the molecule, so you need to consider all electrons and find the hybridization that gives the lowest possible energy to both the bonding and non-bonding electrons.

Which hybridization scheme a molecule adopts is the result of a delicate balance between electron stabilization and other factors such as bond strengths, steric and electrostatic effects.  The balance of these factors can shift and that is why some molecules hybridize one way and other molecules hybridize differently.


How does Ammonia and water mixture works in an absorption refrigeration?

The difference between the absorption refrigerator and the more common refrigerator is that there is no compressor in the absorption refrigerator.  The Absorption refrigerator still needs a way to convert the evaporated refrigerant back to a liquid state.  Instead of compressing it in one step, the evaporated refrigerant is absorbed by a liquid.  The liquid is heated and the refrigerant evaporates from the liquid.  The refrigerant is cooled in a heat exchanger and condensed.  It is now a liquid and ready to go back through the cycle.

The equation you have written is a reaction between ammonia and water.  This reaction will occur in an absorption refrigerator that uses water as the absorber and ammonia as the refrigerant.  The equilibrium constant of the reaction is low so it is not accounted for in a discussion of how an absorption refrigerator works.


Relation Between Degree of Hydration and Electropositivity

The degree of hydration in solution does not directly depend on electropositivity. It depends on the ratio of charge to volume. The higher charged and the smaller a fragment is, the more water molecules hydrate it.

Note that almost every salt in solution can be considered as separated ionic pairs. For those that do not seperate, i.e. have a significantly large covalent bonding proportion, you would be looking at the solvation of the non-dissociated ion pair meaning that you can’t properly compare that hydration to simple salts’ hydration.

Metals get more electropositive the closer they are to the bottom left of the periodic table with cesium being the (non-radioactive) corner. Electropositivity tends to increase with an atom’s size. But the larger the size, the less the degree of hydration is, because the charge per volume of the metal ion will be smaller!

Also, the most electropositive metals are alkaline metals, meaning the only create $\ce{M+}$ ions. However, the stronger the charge, the smaller an ion gets: $\ce{Ga^3+}$ has an ionic radius of $76\,\mathrm{pm}$ — smaller than $\ce{Li+}$ whose ionic radius is $90\,\mathrm{pm}$ although they are two periods apart! So $\ce{Ga^3+}$ will be a lot stronger hydrated than $\ce{Li+}$ because it contains much more charge per volume.


Why does heating an atom make it emit certain frequencies?

Absorbing photons is just one of ways atoms and their electrons, or molecules, can gain energy. Other ways are e.g mechanical collisions and chemical reactions.

Note that atoms do not reflect light they do not absorb, but let it pass through like if they were not there. If they had reflected it,  air would have been an opaque fog. A very small number of photons are dispersed by the Rayleigh scattering, what gives the sky it's blue colour.

A flame  contains mixture of

atoms
atomic ions
free electrons
molecules
radicals(molecules with unpaired electrons}
molecular ions ( and ion radicals )

Atoms and atomic ions do have spectral lines in their emission or absorption spectra, related to particular energies, frequencies and wavelengths ( somewhat broaden, mostly  by the Doppler effect ).

But molecules, radicals and molecular ions have rather broad, diffuse bands.

Photons in UV, VIS and near IR have energies typical for electron excitations/transitions.

Far IR and MW have energies needed to change molecular bond vibrations and bond/molecule rotations, as these are quantized as well.

High energetic collision and mainly chemics reactions frequently lead to a  molecule with excited electron. This energy gets dissipated either as mechanical energy passed to other molecules, or, it can be emitted.

Special cases are free electrons and solid matter in flames, that are not limited to particular frequencies at all.

Electrons can emit photons of any frequency (limited by electron energy ).

Solids ( smoke, large scale atomic structures ) have continuos emission spectrum.


Why are the atomic radii of iron, cobalt, nickel, and copper almost the same?

I think the point of this question is for you to realise that options 1 and 2 can't be the correct answer.

As you go down a group, new electron shells are occupied which extend further from the nucleus, increasing the atomic radius. Therefore option 1 must be wrong.

Effective nuclear charge increases across a period because the nuclear charge increases but the shielding stays roughly the same (at least until you get to transition metals). Therefore atomic radius will decrease across the group and so option 2 cannot be correct and the answer must be option 3.


When we are talking about zinc... We have to think about period trends. So in periods along moving from left to right covalent radius decreases. But in case of transition metal..

Two effects works first is $Z_{eff}$. Another effect is electronic repulsion.At the starting the no of electron in $d-$ orbital are less so $Z_{eff}$ is dominant over electronic repulsion but as we are moving alosng d-orbital electronic effect become dominant factor over $Z_{eff}$, So in the middle redius become almost constant and at the end it increases.

When we are talking about gallium and Aluminium radius comparision thats mean we are talking about group trends ,So in group trend abnormal size is due to poor shielding effect and relativistic effect. Normal trend on going down the size increases from aluminium to gallium size decreases because here a poor shielding of 3-d orbital present so $Z_{eff}$ penetarate and valence shell contract so size decreases.


Two chemicals are used in lead test kits: sodium sulfide or a rhodizonate salt. Sodium sulfide should be easy to obtain (or synthesize), and is useful for light-colored paint, as it produces a gray or black color change. Potassium rhodizonate can be synthesized from inositol, but you might be better off getting sodium rhodizonate, e.g. from Sigma-Aldrich for ~US$35 for 5 grams.


Bleach turned towel bright pink before fading to white

Although Dr. Laundry says it only happens to the incoming brighteners within a detergent and not to those fixed to the fabric, I think that your new towels should have enough brighteners - deposited into and/or onto them - to react as well.

Optical brightners and bleach reaction:  https://www.clorox.com/dr-laundry/detergentbleach-reaction/

Additional note : brighteners absorb light in the near UV and emit in the blu tail of the Vis spectrum, so making the fabric "extrawhite". Search for Optical brighteners in Google or Wikipedia.

Perahps someone here could know or propose the ongoing reaction and is transient behaviour. As the OP said in a comment, the towels were washed with a detergent containing a stilbenedisulfonic triazine derivative(s).


Buffers calculating amounts of acid & base

I'm going to address only the first part since that was in the original question.

You were given a buffer of pH 4.5 and the pKa of the acid used is 4.79. You worked out the volumes of the acid to be 9.7 mL and the volume of base to be 5.3 mL. You are now asked to get the amount of acid and the amount of base (i.e. the number of moles of each in the buffer).

$n_a$ means the number of moles of the acid.

This is equal to the concentration of the acid multiplied by the volume of the acid, namely:

$n_a = [\ce{HA}].V_a$

Since we know that the concentration of the acid, $[\ce{HA}]$ is 0.1 M and the volume of acid used is 9.7 mL, the number of moles of acid becomes:

$n_a = 0.1\ \text{mol/L}\times 9.7\ \text{mL} \times\dfrac{1\ \text{L}}{1000\ \text{mL}}$

The $\dfrac{1\ \text{L}}{1000\ \text{mL}}$ is only to convert mL to L. Do the same for the number of moles of the conjugate base.

For the second part, you need to realise that the pH you get that way will be the combined pH of the buffer and the acid. You will probably have to use the assumption that 1 drop = 1 mL, unless the question or your tutor already mentioned how to consider the volume of the added acid.